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IN   MEMORIAM 
FLORIAN  CAJORI 


EUCLID'S 
ELEMENTS  OF  GEOMETRY 

BOOKS  I— VI. 


SonDon:   C  j.  clay  and  SONS, 
CAMBKIDGE   UNIVEKSITY  PEESS  WAREHOUSE, 
AVE   MARIA  LANE. 


CAMBRIDGE:   DEIGHTON,  BELL,  AND  CO. 

LEIPZIG:   F.   A.   BROCKHAUS. 

NEW  YORK :    MACMILLAN  AND  CO. 


SPECIMEN  COZY, 
^itt  iress  iMatftematiral  ^mt& 


EUCLID'S 
ELEMENTS    OF    GEOMETKY 

EDITED  FOR   THE  SYNDICS  OF  THE  PRESS 


H.   M.   TAYLOR,   M.A. 

FELLOW   AND   FORMERLY   TUTOR   OF  TRINITY   COLLEGE,    CAMBRIDGE. 


BOOKS   I— VI. 


CAMBRIDGE 

AT   THE   UNIVEESITY  PRESS 
1893 

[All  Rifihts  I'eserved.] 


Cambritifie : 

PBINTED   BY   C.    J.    CLAY,    M.A.    AND   SONS, 
AT    THE    UNIVERSITY   PRESS. 


\ 


r3 


NOTE. 

The  Special  Board  for  Mathematics  in  the  University 
of  Cambridge  in  a  Eeport  on  Geometrical  Teaching  dated 
May  10,  1887,  state  as  follows: 

'  The  majority  of  the  Board  are  of  opinion  that  the  rigid  adherence 
to  Euclid's  texts  is  prejudicial  to  the  interests  of  education,  and 
that  greater  freedom  in  the  method  of  teaching  Geometry  is 
desirable.  As  it  appears  that  this  greater  freedom  cannot  be 
attained  while  a  knowledge  of  Euclid's  text  is  insisted  upon  in 
the  examinations  of  the  University,  they  consider  that  such 
alterations  should  be  made  in  the  regulations  of  the  examina- 
tions as  to  admit  other  proofs  besides  those  of  Euclid,  while 
following  however  his  general  sequence  of  propositions,  so  that 
no  proof  of  any  proposition  occurring  in  Euclid  should  be 
accepted  in  which  a  subsequent  proposition  in  Euclid's  order 
is  assumed.' 

On  March  8,  1888,  Amended  Regulations  for  the 
Previous  Examination,  which  contained  the  following 
provision,  were  approved  by  the  Senate  : 

'  Euclid's  definitions  will  be  required,  and  no  axioms  or  postulates 
except  Euclid's  may  be  assumed.  The  actual  proofs  of  propo- 
sitions as  given  in  Euclid  will  not  be  required,  but  no  proof  of 
any  proposition  occurring  in  Euclid  will  be  admitted  in  which 
use  is  made  of  any  proposition  which  in  Euclid's  order  occurs 
subsequently.' 

And  in  the  Regulations  for  the  Local  Examinations 
conducted  by  the  University  of  Cambridge  it  is  provided 
that : 

•Proofs  other  than  Euclid's  will  be  admitted,  but  Euclid's  Axioms 
will  be  required,  and  no  proof  of  any  proposition  will  be 
accepted  which  assumes  anything  not  proved  in  preceding 
propositions  in  Euclid.' 


PREFACE  TO  BOOKS  I.    AND  II. 


IT  was  with  extreme  diffidence  that  I  accepted  an  invi- 
tation from  the  Syndics  of  the  Cambridge  University 
Press  to  undertake  for  them  a  new  edition  of  the  Elements 
of  Euclid.  Though  I  was  deeply  sensible  of  the  honour, 
which  the  invitation  conferred,  I  could  not  but  recognise 
the  great  responsibility,  which  the  acceptance  of  it  would 
entail. 

The  invitation  of  the  Syndics  was  in  itself,  to  my  mind, 
a  sign  of  a  widely  felt  conviction  that  the  editions  in 
common  use  were  capable  of  improvement.  Now  improve- 
ment necessitates  change,  and  every  change  made  in  a 
work,  which  has  been  a  text  book  for  centuries,  must  run 
the  gauntlet  of  severe  criticism,  for  while  some  will  view 
every  alteration  with  aversion,  others  will  consider  that 
every  change  demands  an  apology  for  the  absence  of  more 
and  greater  changes. 

I  will  here  give  a  short  account  of  the  chief  points,  in 
which  this  edition  differs  from  the  best  known  editions  of 
the  Elements  of  Euclid  at  present  in  use  in  England. 

While  the  texts  of  the  editions  of  Potts  and  Todhunter 
are  confessedly  little  more  than  reprints  of  Simson's  English 
version  of  the  Elements  published  in  175G,  the  text  of  the 
present  edition  does  not  profess  to  be  a  translation  from 
the  Greek.  I  began  by  retranslating  the  First  Book :  but 
there  proved  to  be  so  many  points,  in  which  I  thought  it 

M3061G5  * 


vi  PREFACE. 

desirable  to  depart  from  the  original,  tiiat  it  seemed  best 
to  give  up  all  idea  of  simple  translation  and  to  retain 
merely  the  substance  of  the  work,  following  closely  Euclid's 
sequence  of  Propositions  in  Books  I.  and  II.  at  all  events. 

Some  of  the  definitions  of  Euclid,  for  instance  trapezium^ 
rhomboid,  gnomon  are  omitted  altogether  as  unnecessary. 
The  word  trapezium  is  defined  in  the  Greek  to  mean  "  ant/ 
four  sided  figure  other  than  those  already  defined,^'  but  in 
many  modern  works  it  is  defined  to  be  "«  quadrilateral, 
which  has  one  pair  of  parallel  sides."  The  first  of  these 
definitions  is  obsolete,  the  second  is  not  universally  ac- 
cepted. On  the  other  hand  definitions  are  added  of  several 
words  in  general  use,  such  as  ^^eWme^er,  parallelogram, 
diagonal,  which  do  not  occur  in  Euclid's  list. 

The  chief  alteration  in  the  definitions  is  in  that  of  the 
word  figure,  which  is  in  the  Greek  text  defined  to  be  "  that 
ivhich  is  enclosed  hy  one  or  more  boundaries."  I  have 
preferred  to  define  a  figure  as  "a  combination  of  points, 
lines  and  surfaces."  That  Euclid's  definition  leads  to  diffi- 
culty is  seen  from  the  fact  that,  though  Euclid  defines  a 
circle  as  "a  figure  contained  by  one  line...",  he  demands  in 
his  postulate  that  "a  circle  may  be  described...".  Now  it  is 
the  circumference  of  a  circle  which  is  described  and  not 
the  surface.  Again,  when  two  circles  intersect,  it  is  the 
circumferences  which  intersect  and  not  the  surfaces. 

I  have  rejected  the  ordinarily  received  definition  of  a 
square  as  "  a  quadrilateral,  whose  sides  are  equal,  and  whose 
angles  are  right  angles."  There  is  no  doubt  that,  when  we 
define  any  geometrical  figure,  we  postulate  the  possibility 
of  the  figure ;  but  it  is  useless  to  embrace  in  the  definition 
more  properties  than  are  requisite  to  determine  the  figure. 

The  word  axiom  is  used  in  many  modern  works  as 
applicable  both  to  simple  geometrical  propositions,  such  as 
"  two  straight  lines  cannot  enclose  a  space,"  and  to  proposi- 


PREFACE.  vii 

tions,  other  than  geometrical,  accepted  without  demonstra- 
tion and  true  universally,  such  as  ^Hhe  whole  of  a  thing  is 
greater  than  a  part  "  These  two  classes  of  propositions  are 
often  distinguished  by  the  terms  "geometrical  axioms"  and 
"general  axioms."  I  prefer  to  use  the  word  axiom  as  appli- 
cable to  the  latter  class  only,  that  is,  to  simple  propositions, 
true  of  magnitudes  of  all  kinds  (for  instance  'Hhings  which 
are  equal  to  the  same  thing  are  equal  to  one  anothefi'"'^),  and 
to  use  the  term  postulate  for  a  simple  geometrical  proposi- 
tion, whose  truth  we  assume. 

When  a  child  is  told  that  A  weighs  exactly  as  much 
as  B,  and  B  weighs  exactly  as  much  as  C,  he  without 
hesitation  arrives  at  the  conclusion  that  A  weighs  exactly 
as  much  as  C.  His  conviction  of  the  validity  of  his 
conclusion  would  not  be  strengthened,  and  possibly  his 
confidence  in  his  conclusion  might  be  impaired,  by  his 
being  directed  to  appeal  to  the  authority  of  the  general 
proposition  "  things  which  are  equal  to  the  same  thing  are 
equal  to  one  another."  I  have  therefore,  as  a  rule,  omitted 
in  the  text  all  reference  to  the  general  statements  of 
axioms,  and  have  only  introduced  such  a  statement  occa- 
sionally, where  its  introduction  seemed  to  me  the  shortest 
way  of  explaining  the  nature  of  the  next  step  in  the 
demonstration. 

If  it  be  objected  that  all  axioms  used  should  be  clearly 
stated,  and  that  their  number  should  not  be  unnecessarily 
extended,  my  reply  is  that  neither  the  Greek  text  nor  any 
edition  of  it,  with  Avhich  I  am  acquainted,  has  attempted 
to  make  its  list  of  axioms  perfect  in  either  of  these  respects. 
The  lists  err  in  excess,  inasmuch  as  some  of  the  axioms 
therein  can  be  deduced  from  others  :  they  err  in  defect, 
inasmuch  as  in  the  demonstrations  of  Propositions  conclu- 
sions are  often  drawn,  to  support  the  validity  of  which  no 
appeal  can  be  made  to  any  axiom  in  the  lists. 


viii  PREFACE. 

Under  the  term  postulate  I  have  included  not  only  what 
may  be  called  the  postulates  of  geometrical  operation,  such 
as  ^Ht  is  assumed  that  a  straight  line  may  he  drawn  from 
any  point  to  any  other  point^^  but  also  geometrical  theorems, 
the  truth  of  which  we  assume,  such  as  ''Hwo  straight  lines 
cannot  have  a  common  part.^^ 

The  postulates  of  this  edition  are  nine  in  number. 

Postulates  3,  4,  6  are  the  postulates  of  geometrical 
operation,  which  are  common  to  all  editions  of  the  Elements 
of  Euclid.  Postulates  1,  5,  9  are  the  Axioms  10,  11,  12  of 
modern  editions.  Postulates  2,  7,  8  do  not  appear  under 
the  head  either  of  axioms  or  of  postulates  in  Euclid's  text, 
but  the  substance  of  them  is  assumed  in  the  demonstrations 
of  his  propositions. 

Postulate  9  has  been  postponed  until  page  51,  as  it 
seemed  undesirable  to  trouble  the  student  with  an  attempt 
to  unravel  its  meaning,  until  he  was  prepared  to  accept  it 
as  the  converse  of  a  theorem,  with  the  proof  of  which  he 
had  already  been  made  acquainted. 

It  may  be  mentioned  that  a  proof  of  Postulate  5,  ^^all 
right  angles  are  equal"  is  given  in  the  text  (Proposition  lOB), 
and  that  therefore  the  number  of  the  Postulates  might 
have  been  diminished  by  one :  it  was  however  thought 
necessary  to  retain  this  Postulate  in  the  list,  so  that  it 
might  be  used  as  a  postulate  by  any  person  who  might 
prefer  to  adhere  closely  to  the  original  text  of  Euclid. 

One  important  feature  in  the  present  edition  is  the 
greater  freedom  in  the  direct  use  of  "the  method  of  super- 
position "  in  the  proofs  of  the  Propositions.  The  method  is 
used  directly  by  Euclid  in  his  proof  of  Proposition  4  of 
Book  I.,  and  indirectly  in  his  proofs  of  Proposition  5  and 
of  every  other  Proposition,  in  which  the  theorem  of 
Proposition  4  is  quoted.  It  seems  therefore  but  a  slight 
alteration  to  adopt  the  direct  use  of  this  method  in  the 


PREFACE.  ix 

proofs  of  any  theorems,  in  the  proofs  of  which,  in  Euclid's 
text,  the  theorem  of  Proposition  4  is  quoted. 

It  may  of  course  be  fairly  objected  that  it  would  be 
more  logical  for  a  writer,  who  uses  with  freedom  the 
method  of  superposition,  to  omit  the  first  three  Propo- 
sitions of  Book  I.  To  this  objection  my  reply  must  be 
that  it  is  considered  undesirable  to  alter  the  numbering 
of  the  Propositions  in  Books  I.  and  II.  at  all  events.  No 
doubt  a  work  written  merely  for  the  teaching  of  geometry, 
without  immediate  reference  to  the  requirements  of  candi- 
dates preparing  for  examination,  might  well  omit  the  first 
three  Propositions  and  assume  as  a  postulate  that  "a  circle 
may  he  described  with  any  point  as  centre^  and  with  a 
length  equal  to  any  given  straight  line  as  radius^"  instead 
of  the  postulate  of  Euclid's  text  (Postulate  6  of  the  pre- 
sent edition),  "  a  circle  may  he  described  with  any  point  as 
centre  and  with  any  straight  line  drawn  from  that  j)oi7it  as 
radius." 

The  use  of  the  words  "each  to  each"  has  been  aban- 
doned. The  statement  that  two  things  are  equal  to  two 
other  things  each  to  each,  seems  to  imply,  according  to  the 
natural  meaning  of  the  words,  that  all  four  things  are 
equal  to  each  other.  Where  we  wish  to  state  briefly  that 
A  has  a  certain  relation  to  a,  B  has  the  same  relation  to 
6,  and  C  has  the  same  relation  to  c,  we  prefer  to  say  that 
A,  B,  C  have  this  relation  to  «,  h,  c  respectively. 

The  enunciations  of  the  Propositions  in  Books  I.  and 
II.  have  been,  with  some  few  slight  exceptions,  retained 
throughout,  and  the  order  of  the  Propositions  remains 
unaltered,  but  different  methods  of  proof  have  been  adopted 
in  many  cases.  The  chief  instances  of  alteration  are  to  be 
found  in  Propositions  5  and  6  of  Book  I.,  and  in  Book  II. 

The  use  of  what  may  be  called  impossible  figures,  such 
as  occurred  in  Euclid's  text  in  the  proofs  of  Propositions 


X  PREFACE. 

6  and  7  of  Book  I.  has  been  avoided.  It  seems  better  to 
prove  that  a  line  cannot  be  drawn  satisfying  a  certain  condi- 
tion without  making  a  pretence  of  doing  what  is  impossible. 

Two  Propositions  (10  A  and  10  B),  have  been  introduced 
to  shew  that,  if  the  method  of  superposition  be  used,  we 
need  not  take  as  a  postulate  ^^  all  rigid  angles  are  equal  to 
one  another,'^  but  that  we  may  deduce  this  theorem  from 
other  postulates  which  have  been  already  assumed. 

Another  new  Proposition  introduced  into  the  text  is 
Proposition  26  A,  '[if  two  triangles  have  two  sides  equal  to 
two  sides,  and  the  angles  opposite  to  one  ^;azV  of  equal  sides 
equal,  the  angles  opposite  to  the  other  pair  are  either  equal  or 
supplementary/,"  which  may  be  described,  with  reference  to 
Euclid's  text,  as  the  missing  case  of  the  equality  of  two 
triangles.  It  is  intimately  connected  with  what  is  called 
in  Trigonometry  "  the  ambiguous  case  "  in  the  solution  of 
triangles. 

Another  new  Proposition  (41  A)  is  the  solution  of  the 
problem  "  to  construct  a  triangle  equal  to  a  given  rectilineal 
figure."  It  appears  to  be  a  more  practical  method  of  solving 
the  general  problem  of  Proposition  45  "^o  construct  aj)aral- 
lelogram  equal  to  a  given  rectilineal  figure,  having  a  side 
equal  to  a  given  straight  line,  and  having  an  angle  equal  to 
a  given  angle,"  to  begin  with  the  construction  of  a  triangle 
equal  to  the  given  figure  rather  than  to  follow  the  exact 
sequence  of  Euclid's  propositions. 

In  the  notes  a  few  ''Additional  Propositions"  have 
been  introduced  containing  important  theorems,  which  did 
not  occur  in  Euclid's  text,  but  with  which  it  is  desirable 
that  the  student  should  become  familiar  as  early  as 
possible.  Also  outlines  have  been  given  of  some  of  tho 
many  different  proofs  which  have  been  discovered  of 
■  Pythagoras's  Theorem.  They  may  be  found  interesting  and 
useful  as  exercises  for  the  student. 


PREFACE.  xi 

Euclid's  proofs  of  many  of  the  Propositions  of  Book  II. 
are  unnecessarily  long.  His  use  of  the  diagonal  of  the 
square  in  his  constructions  in  Propositions  4  to  8  can 
scarcely  be  considered  elegant. 

It  is  curious  to  notice  that  Euclid  after  giving  a 
demonstration  of  Proposition  1  makes  no  use  whatever 
of  the  theorem.  It  seems  more  logical  to  deduce  from 
Proposition  1  those  of  the  subsequent  Propositions  whicli 
can  be  readily  so  deduced. 

In  Book  II.  outlines  of  alternative  proofs  of  several  of 
the  Propositions  have  been  given,  which  may  be  developed 
more  fully  and  used  in  examinations,  in  place  of  the 
proofs  given  in  the  text.  Some  of  these  proofs  are  not, 
so  far  as  I  know,  to  be  found  in  English  text  books. 
The  most  interesting  ones  are  those  of  Propositions  12 
and  13.  Some,  which  I  thought  at  first  were  new,  I  have 
since  found  in  foreign  text  books. 

The  Propositions  in  the  text  have  not  been  distin- 
guished by  the  words  "Theorem"  and  "Problem."  The 
student  may  be  informed  once  for  all  that  the  word 
theorem  is  used  of  a  geometrical  truth  which  is  to  be 
demonstrated,  and  that  the  word  problem  is  used  of  a 
geometrical  construction  which  is  to  be  performed. 

Although  Euclid  always  sums  up  the  result  of  a  Propo- 
sition by  the  words  oinp  ISct  Sct^ai  or  ottc/j  t3et  Trotryo-ai, 
there  seems  to  be  no  utility  in  putting  the  letters  q.e.d. 
or  Q.E.F.  at  the  end  of  a  Proposition  in  an  English  text- 
book. The  words  "  Quod  erat  demonstrandum  "  or  "  Quod 
erat  faciendum  "  in  a  Latin  text  were  not  out  of  place. 

"When  the  book  is  opened,  the  reader  will  see  as  a  rule 
on  the  left  hand  page  a  Pi-oposition,  and  on  the  opposite 
page  notes  or  exercises.  The  notes  are  either  appropriate 
to  the  Proposition  they  face  or  introductory  to  the  one 
next  succeeding.     The  exercises  on  the  right  hand  page  are. 


xii  PREFACE, 

it  is  hoped,  in  all  cases  capable  of  being  solved  by  means  of 
the  Proposition  on  the  adjoining  page  and  of  preceding 
Propositions.  They  have  been  chosen  with  care  and  with 
the  special  view  of  inducing  the  student  from  the  com- 
mencement of  his  reading  to  attempt  for  himself  the 
solution  of  exercises. 

For  many  Propositions  it  has  been  difficult  to  find 
suitable  exercises :  consequently  many  of  the  exercises  have 
been  specially  manufactured  for  the  Propositions  to  which 
they  are  attached.  Great  pains  have  been  taken  to  verify 
the  exercises,  but  notwithstanding  it  can  scarcely  be  hoped 
that  all  trace  of  error  has  been  eliminated. 

It  is  with  pleasure  that  I  record  here  my  deep  sense 
of  obligation  to  many  friends,  who  have  aided  me  by 
valuable  hints  and  suggestions,  and  more  especially  to 
A.  R.  Forsyth,  M.A.,  Fellow  and  Assistant  Tutor  of 
Trinity  College,  Charles  Smith,  M.A.,  Fellow  and  Tutor  of 
Sidney  Sussex  College,  R.  T.  Wright,  M.A.,  formerly 
Fellow  and  Tutor  of  Christ's  College,  my  brother-in-law 
the  Reverend  T.  J.  Sanderson,  M.A.,  formerly  Fellow  of 
Clare  College,  and  my  brother-  W.  W.  Taylor,  M.A., 
formerly  Scholar  of  Queen's  College,  Oxford,  and  after- 
wards Scholar  of  Trinity  College,  Cambridge.  The  time 
and  trouble  ungrudgingly  spent  by  these  gentlemen  on 
this  edition  have  saved  it  from  many  blemishes,  which 
would  otherwise  have  disfigured  its  pages. 

I  shall  be  grateful  for  any  corrections  or  criticisms, 
which  may  be  forwarded  to  me  in  connection  either  with 
the  exercises  or  with  any  other  part  of  the  work. 


H.   M.   TAYLOR. 


Trinity  College,  Cambridge, 
October  1,  1889. 


PREFACE  TO  BOOKS  III  AND  IV. 


IN  Book  III.  the  chief  deviation  from  Euclid's  text  will 
be  found  in  the  first  twelve  Propositions,  where  a  good 
deal  of  rearrangement  has  been  thought  desirable.  This 
rearrangement  has  led  to  some  changes  in  the  sequence 
of  Propositions  as  well  as  in  the  Propositions  themselves ; 
but,  even  with  these  changes,  the  first  twelve  Propositions 
will  be  found  to  include  the  substance  of  the  whole  of  the 
first  twelve  of  Euclid's  text. 

The  Propositions  from  13  to  37  are,  except  in  unim- 
portant details,  unchanged  in  substance  and  in  order. 

The  enunciation  of  the  theorem  of  Proposition  36  has 
been  altered  to  make  it  more  closely  resemble  that  of  the 
complementary  theorem  of  Proposition  35. 

An  additional  Proposition  has  been  introduced  on  page 
186  involving  the  principle  of  the  rotation  of  a  plane 
figure  about  a  point  in  its  plane.  It  is  a  principle  of 
which  extensive  use  might  with  advantage  be  made  in  the 
proof  of  some  of  the  simpler'  properties  of  the  circle. 
It  has  not  however  been  thought  desirable  to  do  more 
in  this  edition  than  to  introduce  the  student  to  this 
method  and  by  a  selection  of  exercises,  which  can  readily 
be  solved  by  its  means,  to  indicate  the  importance  of  the 
method. 


PREFACE   TO   BOOKS    V.  AND    VI. 

of  the  Greek  text  is  printed  in  brackets  at  the  lieacl  of  each 
Proposition. 

In  Book  VI  a  slight  departure  from  Euclid's  text  is 
made  in  the  treatment  of  similar  figures.  The  definition  of 
similar  polygons  which  is  adopted  in  this  work  brings  into 
prominence  the  important  property  of  the  fixed  ratio  of 
their  corresponding  sides.  Its  use  has  the  great  merit  of 
tending  at  once  to  simplicity  and  brevity  in  the  proofs  of 
many  theorems. 

The  numbering  of  the  Propositions  in  Book  VI  remains 
unchanged :  Propositions  27,  28,  29  are  omitted  as  in  many 
of  the  recent  English  editions  of  Euclid,  and  iti  several  cases 
a  Proposition  which  consists  of  a  theorem  and  its  converse 
is  divided  into  two  Parts.  Proposition  32  of  Euclid's  text, 
which  is  a  very  special  case  of  no  great  interest,  has  been 
replaced  by  a  simple  but  important  theorem  in  the  theory 
of  similar  and  similarly  situate  figures. 

The  chief  difficulty  with  respect  to  the  additions  which 
have  been  made  to  Book  VI  was  the  immense  number  of 
known  theorems  from  which  a  selection  had  to  be  made. 

I  have  attempted  by  means  of  two  or  three  series  of 
Propositions  arranged  in  something  like  logical  sequence  to 
introduce  the  student  to  important  general  methods  or 
well-known  interesting  results. 

One  series  gives  a  sketch  of  the  theory  of  transversals, 
and  the  properties  of  harmonic  and  anharmonic  ranges  and 
pencils,  and  leads  up  to  Pascal's  Theorem.  Another  series 
deals  with  similar  and  similarly  situate  figures  and  leads  up 


PREFACE   TO   BOOKS   V.  AND    VI. 

to  Gergoime's  elegant  solution  of  the  problem  to  describe  a 
circle  to  touch  three  given  circles.  These  are  followed  by 
an  introduction  to  the  method  of  Inversion,  an  account  of 
Casey's  extension  of  Ptolemy's  Theorem,  some  of  the  im- 
portant properties  of  coaxial  circles,  and  Poncelet's  Theorems 
relating  to  the  porisms  connected  with  a  series  of  coaxial 
circles. 

No  attempt  has  been  made  to  represent  the  very  large 
and  still  increasing  collection  of  theorems  connected  with 
the  "Modern  Geometry  of  the  Triangle." 

I  hereby  acknowledge  the  great  help  I  have  received  in 
this  portion  of  my  work  from  friends,  and  especially  from 
Dr  Forsyth  and  from  my  brother  Mr  J.  H.  Taylor.  To  the 
latter  I  am  indebted  for  the  Index  to  Books  I — VI,  which 
I  hope  may  prove  of  some  assistance  to  persons  using  this 
edition. 

H.  M.  T. 

TuiNiTY  College,  Cambridge, 
March  16,  1893. 


THE  ELEMENTS  OF  GEOMETRY. 


T.  E. 


BOOK  I. 


Definition  1.  That  'which  has  ^wsition  but  not  magni- 
tude is  called  a  point. 

The  word  point  is  used  in  many  different  senses.  We  speak  in 
ordinary  language  of  the  point  of  a  pin,  of  a  pen  or  of  a  pencil. 
Any  mark  made  with  such  a  point  on  paper  is  of  some  definite  size 
and  is  in  some  definite  position.  A  small  mark  is  often  called  a  spot 
or  a  dot.  Suppose  such  a  spot  to  become  smaller  and  smaller ;  the 
smaller  it  becomes  the  more  nearly  it  resembles  a  geometrical  point : 
but  it  is  only  when  the  spot  has  become  so  small  that  it  is  on  the 
point  of  vanishing  altogether,  i.e.  when  in  fact  the  spot  still  has 
position  but  has  no  magnitude,  that  it  answers  to  the  geometrical 
definition  of  a  point. 

A  point  is  generally  denoted  by  a  single  letter  of  the  alphabet: 
for  instance  we  speak  of  the  point  A. 

Definition  2.  That  which  has  position  and  length  but 
neither  breadth  nor  thickness  is  called  a  line. 


The  extremities  of  a  line  are  points. 
The  intersections  of  lines  are  points 


DEFINITIONS.  3 

The  word  line  also  is  used  in  many  different  senses  in  ordinary 
language,  and  in  most  of  these  senses  the  main  idea  suggested  is  that 
of  length.  For  instance  we  speak  of  a  line  of  railway  as  connecting 
two  distant  towns,  or  of  a  sounding  line  as  reaching  from  the  bottom 
of  the  sea  to  the  surface,  and  in  so  speaking  we  seldom  think  of  the 
breadth  of  the  railway  or  of  the  thickness  of  the  sounding  line. 

When  we  speak  of  a  geometrical  line,  we  regard  merely  the  length: 
we  exclude  the  idea  of  breadth  and  thickness  altogether :  in  fact  we 
consider  that  the  cross-section  of  the  line  is  of  no  size,  or  in  other 
words  that  the  cross-section  is  a  geometrical  point. 

If  a  point  move  with  a  continuous  motion  from  one  position  to 
another,  the  path  which  it  describes  during  the  motion  is  a  line. 


Definition  3.  That  which  has  position,  length  and 
breadth  hut  not  thickness  is  called  a  surface. 

The  boundaries  of  a  surface  are  lines. 

The  intersections  of  surfaces  are  lines. 

The  word  surface  in  ordinary  language  conveys  the  idea  of  ex- 
tension in  two  directions :  for  instance  we  speak  of  the  surface  of  the 
Earth,  the  surface  of  the  sea,  the  surface  of  a  sheet  of  paper. 
Although  in  some  cases  the  idea  of  the  thickness  or  the  depth  of  the 
thing  spoken  of  may  be  present  in  the  speaker's  mind,  yet  as  a  rule 
no  stress  is  laid  on  depth  or  thickness.  When  we  speak  of  a  geome- 
trical surface  we  put  aside  the  idea  of  depth  and  thickness  altogether. 
We  are  told  that  it  takes  more  than  300,000  sheets  of  gold  leaf  to 
make  an  inch  of  thickness ;  but  although  the  gold  leaf  is  so  thin,  it 
must  not  be  regarded  as  a  geometrical  surface.  In  fact  each  leaf 
however  thin  has  always  two  bounding  surfaces.  The  geometrical 
surface  is  to  be  regarded  as  absolutely  devoid  of  thickness,  and  no 
number  of  surfaces  put  together  would  make  any  thickness  whatever. 


Definition  4.     That  tohich  has  jjosition,  length,  breadth 
and  thickness  is  called  a  solid. 

The  boundaries  of  solids  are  surfaces. 

1—2 


4  BOOK  I. 

Definition  5.  Any  combination  of  points,  lines,  and 
surfaces  is  called  a  figure. 

Definition  6.  A  line  which  lies  evenly  between  points 
on  it  is  called  a  straight  line. 

This  is  Euclid's  definition  of  a  straight  line.  It  cannot  be  turned 
to  practical  use  by  itself.  We  supplement  the  definition,  as  Euclid 
did,  by  making  some  assumptions  the  nature  of  which  wUl  be  seen 
hereafter. 

Postulates.  There  are  a  few  geometrical  propositions 
so  obvious  that  we  take  the  truth  of  them  for  granted,  and 
a  few  geometrical  operations  so  simple  that  we  assume  we 
may  perform  them  when  we  please  without  giving  any 
explanation  of  the  process.  The  claim  we  make  to  use  any 
one  of  these  propositions,  or  to  perform  any  one  of  these 
operations,  is  called  a  postulate. 

Postulate  1.     Two  straight  lines  cannot  enclose  a  space. 

This  postulate  is  equivalent  to 

Two  straight  lines  cannot  intersect  in  more  than  one  point. 

Postulate  2.  Two  straight  lines  cannot  have  a  common 
pa/rt. 

If  two  straight  lines  have  two  points  ^ ,  jB  in  common,  they  must 
coincide  between  A  and  5,  since,  if  they  did  not,  the  two  straight 
lines  would  enclose  a  space.  Again,  they  must  coincide  beyond  A 
and  B,  since,  if  they  did  not,  the  two  straight  lines  would  have  a 
common  part.    Hence  we  conclude  that 

Tioo  straight  lines,  which  have  two  poi^its  in  common,  are 
coincident  throughout  their  length. 

Thus  two  points  on  a  straight  line  completely  fix  the  position  of  the 
line.  Hence  we  generally  denote  a  straight  line  by  mentioning  two 
points  on  it,  and  when  the  straight  line  is  of  finite  length,  we  generally 
denote  it  by  mentioning  the  points  which  are  its  two  extremities. 


DEFINITIONS.  5 

For  instance,  if  P  and  Q  be  two  points  on  a  straight  line,  the  Hne 
is  called  the  straight  line  PQ  or  the  straight  line  QP,  or  sometimes 
more  shortly  PQ  or  QP:  and  the  straight  line  which  is  terminated 
by  two  points  P  and  Q  is  called  in  the  same  way  PQ  or  QP. 

It  may  be  remarked  that,  when  merely  the  actual  length  of  the 
straight  line  is  under  discussion,  we  use  PQ  or  QP  indifferently: 
but  that,  when  we  wish  to  consider  the  direction  of  the  line,  we 
must  carefully  distinguish  between  PQ  and  QP. 


Postulate  3.     A  straight  line  may  he  drawn  from  any 
point  to  any  other  point. 


Postulate  4.  A  finite  straight  line  may  be  produced 
at  either  extremity  to  any  length. 

The  demands  made  in  Postulates  3  and  4  are  in  practical  geometry 
equivalent  to  saying  that  a  '  straight  edge '  may  be  used  for  drawing 
a  straight  line  from  one  point  to  another  and  for  producing  a  straight 
line  to  any  length. 

We  assume,  as  Euclid  did,  that  it  is  possible  to 
shift  any  geometrical  figure  from  its  initial  position 
unchanged  in  shape  and  size  into  another  position. 

Test  of  Equality  of  Geometrical  Figures.  The  criterion  of  the 
equality  of  two  geometrical  figures,  which  we  shall  use  in  most  cases, 
is  the  possibility  of  shifting  one  of  the  figures,  unchanged  in  shape 
and  size,  so  that  it  exactly  fits  the  place  which  the  other  of  the  figures 
occupies.     (See  Def.  21.) 

This  method  of  testing  the  equality  of  geometrical 
figures  is  generally  known  as  the  method  of  superposition. 

Test  of  equality  of  straight  lines.  Two  straight  lines  AB,  CD 
are  said  to  be  equal,  when  it  is  possible  to  shift  either  of  them,  say 
AB,  so  that  it  coincides  with  the  other  CD,  the  end  A  on  0  and  the 
end  B  on  I),  or  the  end  A  on  1)  and  the  end  B  on  C. 


6  BOOK  L 

Addition  of  Lines.     Having  defined  the  equality  of  straight  lines, 
we  proceed  to  explain  what  is  meant  by  the  addition  of  straight  lines. 


D 


If  in  a  straight  line  we  take  points  ^,  B,  C,  D  in  order,  we  say 
that  the  straight  line  ^C  is  the  sum  of  the  two  straight  lines  AB^  BG 
(or  of  any  two  straight  lines  equal  to  them), 

and  that  the  straight  line  AB  is  the  difference  of  the  two  straight 
lines  AC,  BC  (or  of  any  two  straight  lines  equal  to  them). 
In  the  same  way  we  say  that  the  straight  line  AD  is  the  sum  of  the 
three  straight  lines  AB,  BG,  GD. 

Again,  if  AB  be  equal  to  BG,  we  say  th&t  AG  is  double  of  AB  or 
ofBG. 

Definition  7.  A  surface  which  lies  evenly  between 
straight  lines  on  if  is  called  a  plane. 

This  is  Euclid's  definition  of  a  plane  :  there  is  the  same  difficulty 
in  making  use  of  it  that  there  is  in  making  use  of  his  definition  of 
a  straight  line. 

Consequently  this  definition  has  by  many  modern  editors  been 
replaced  by  the  following,  which  perhaps  merely  expresses  Euclid's 
meaning  in  other  words  : 

A  surface  such  that  the  straight  li7ie  joining  any  two 
j)oints  in  the  surface  lies  wholly  in  the  surface  is  called 
a  plane. 

Definition  8.  A  figure,  which  lies  xoholly  in  one  plane, 
is  called  a  plane  figure. 

All  the  geometrical  propositions  in  the  first  six  books 
of  the  Elements  of  Euclid  relate  to  figures  in  one  plane. 
This  part  of  Geometry  is  called  Plane  Geometry. 


DEFINITIONS.  7 

Definition  9.  Two  straight  lines  in  the  same  plane, 
which  do  not  meet  however  far  tliey  may  he  jn-oducecl  both 
ways,  are  said  to  he  parallel*  to  one  another. 


D 


Definition  10.  A  plane  angle  is  the  inclination  to  one 
another  of  tico  straight  lines  which  meet  hut  are  not  in  tlie 
same  straight  line. 

The  idea  of  an  angle  is  one  which  it  is  very  difficult  to  convey  by 
the  words  of  a  definition.  We  will  content  ourselves  by  explaining 
some  few  things  connected  with  angles. 

If  two  straight  lines  AB,  AG  meet 
at  A,  the  amount  of  their  divergence 
from  one  another  or  their  inclination 
to  one  another  is  called  the  angle 
which  the  lines  make  loith  one  another 
or  the  angle  hetiueen  the  lines,  or  the  angle  contained  by  the  lines. 

The  angle  formed  by  the  straight  lines  AB,  AG  ia  generally  de- 
nominated BAG,  or  GAD,  the  middle  letter  always  denoting  the  point 
where  the  lines  meet,  and  the  letters  B  and  G  denoting  any  two  points 
in  the  straight  lines  AB,  AC.  It  must  be  carefully  noted  that  the 
magnitude  of  the  angle  is  not  affected  by  the  length  of  the  straight 
hnes^B,  AG. 

The  point  A,  where  the  two  straight  lines  AB  and  AC,  which 
form  the  angle  BAG,  meet,  is  called  the  vertex  of  the  angle  BAG. 

If  there  be  only  two  straight  lines  meeting  at  a  point  A,  the  angle 
formed  by  the  lines  is  sometimes  denoted  by  the  single  letter  A. 


*  Derived  from  Trapd  "by  the  side  of"  undaWriXas  "one  another' 
irapdWtjXoL  ypafxfxai  "lines  side  by  side". 


8  BOOK  T. 

Test  of  Equality  of  Angles.  Two  angles  are  said  to  be  equal, 
when  it  is  possible  to  shift  the  straight  lines  forming  one  of  the 
angles,  unchanged  in  position  relative  to  each  other,  so  as  to  exactly 
coincide  in  direction  with  the  straight  lines  forming  the  other  angle. 

D 
A 


/B  G 

For  instance,  the  angles  ABC,  DEF  will  be  equal,  if  it  be  pos- 
sible to  shift  AB,  BG  unchanged  in  position  relative  to  each  other, 
so  that  B  coincides  with  E,  and  so  that  also  either  BA  coincides  in 
direction  with  ED  and  BG  with  EF,  or  BA  coincides  in  direction 
with  EF  and  BG  with  ED. 

If  a  straight  line  move  in  a  plane,  while  one  point  in  the  line 
remains  fixed,  the  line  is  said  to  turn  or  revolve  about  the  fixed  point. 
If  the  revolving  line  move  from  any  one  position  to  any  other 
position,  it  generates  an  angle,  and  the  amount  of  turning  from  one 
position  to  the  other  is  the  measure  of  the  magnitude  of  the  angle 
between  the  two  positions  of  the  line. 

For  instance  each  hand  of  a  watch,  as  long  as  the  watch  is  going, 
is  turning  uniformly  round  its  fixed  extremity,  and  is  generating  an 
angle  uniformly. 

This  mode  of  regarding  angles  enables  us  to  realize  that  angles 
are  capable  of  growing  to  any  size  and  need  not  be  limited  (as  in 
most  of  the  propositions  in  Euclid's  Elements  they  are  supposed  to 
be)  to  magnitudes  less  than  two  right  angles.     (See  Def.  11.) 

Addition  of  Angles.  If  three  straight 
lines  AB,  AG,  AD  meet  at  the  same  point, 
we  say  that  the  angle  BAD  is  the  sum  of 
the  two  angles  BAG,  GAD  (or  of  any  two 
angles  equal  to  them). 

In  the  same  way  we  say  that  the  angle 
BAG  is  the  difference  of  the  two  angles  "^  "" 

BAD,  GAD  (or  of  any  two  angles  equal  to  them). 

Two  angles  such  as  BAG,  GAD,  which  have  a  common  vertex  and 
one  common  bounding  line,  are  called  adjacent  angles. 


DEFINITIONS.  9 

Definition  11.  If  two  adjacent  angles  made  hy  two. 
straight  lines  at  the  pohit  where 
they  meet  be  equal,  each  of  these 
angles  is  called  a  right  angle,  aiid 
the  straight  lines  are  said  to  be  at 
right  angles  to  each  other. 


Either  of  two  straight  lines  which  Ci  I> 

are  at  right  angles  to  each  other  is 
said  to  be  perpendicular  to  the  other.  b 

If  a  straight  line  AE  be  drawn  from  a  point  A  at  right  angles 
to  a  given  straight  line  CD,  the  part  AE  intercepted  between  the 
point  and  the  straight  line  is  commonly  called  tlie  perpendicular 
from  the  point  A  on  the  straight  line  CD. 

Euclid  uses  as  a  postulate. 

Postulate  5.    All  right  angles  are  equal  to  one  another. 
It  is  not  necessary  to  assume  this  proposition,  since  it  can  be  proved 
by  the  method  of  superposition.     A  proof  will  be  found  on  a  sub- 
sequent page.     (p.  37) 

Definition  12.  An  angle  less  than  a  right  angle  is  called 
an  acute  angle. 

A71  angle  greater  tha^i  a  right  angle  and  less  than  two 
right  angles  is  called  an  obtuse  angle. 

Definition  13.  A  line,  ivhich  is  such  that  it  can  be 
described  by  a  moving  point  starti7ig  from  any  j)oint  of  the 
line  and  returning  to  it  again,  is  called  a  closed  line. 

A  figure  composed  wholly  of  straight  lines  is  called 
a  rectilineal  figure. 

llie  straight  lines,  which  form  a  closed  rectilineal  figure, 
are  called  the  sides  of  the  figure. 

The  sum  of  the  lengths  of  the  sides  of  any  figure  is 
called  the  perimeter  of  the  figure. 

The  poifit,  where  two  adjacent  sides  meet,  is  called  a 
vertex  or  an  angular  point  of  the  figure. 

The  angle  formed  by  two  adjacent  sides  is  called  an  angle 
of  the  figure. 


10  BOOK  I. 

A  straight  line  joining  any  two  vertices  of  a  closed  recti- 
lineal figure^  which  are  not  extremities  of  the  same  side,  is 
called  a  diagonal*. 

The  surface  contained  within  a  closed  figure  is  called 
the  area  of  the  figure. 

A  closed  rectilineal  figure,  which  is  such  that  the  whole 
figure  lies  07i  one  side  of  each  of  the  sides  of  the  figure,  is 
called  a  convex  figure. 

A  closed  rectilineal  figure  is  in 
general  denoted  by  naming  the  letters, 
which  denote  its  vertices,  in  order: 
for  instance  the  five-sided  figure  in 
the  diagram  is  denoted  by  the  letters 
A,  B,  G,  D,  E,  in  order:  i.e.  it  might 
be  called  the  figure  ABODE,  or  the 
figure  CBAED. 

A,  B,  G,  D,  E  are  its  vertices. 

AB,  BG,  GD,  BE,  EA  are  its  sides. 

ABG,  BGD,  GDE,  BE  A,  EAB  are  its  angles. 

AG,  AD  are  two  of  its  diagonals. 

It  will  be  observed  that  a  closed  figure  has  the  same  number  of 
angles  as  it  has  sides. 

If  a  closed  figure  have  an  even  number  of  sides,  we  speak  of 
a  pair  of  sides  as  being  opposite,  and  of  a  pair  of  angles  as  being 
opposite. 

If  a  closed  figure  have  an  odd  number  of  sides,  we  speak  of  an 
angle  as  being  opposite  to  a  side  and  vice  versa. 

For  instance  in  the  quadrilateral  ABGD  the  side  AD  is  said  to  be 
opposite  to  the  side  BG,  and  the  angle  BAD  opposite  to  the  angle 
BGD,  but  in  the  five-sided  figure  ABGDE  the  side  GD  is  said  to  be 
opposite  to  the  angle  BAE,  and  the  angle  AED  opposite  to  the 
side  BG. 


Derived  from  5td  "through",  and  yosvla  "an  angle". 


DEFINITIONS. 


11 


Definition  14.  A  figure,  all  the  sides  of  which  are  equals 
is  called  equilateral. 

A  figure,  all  the  angles  of  which  are  equal,  is  called 
equiangular. 

A  figure,  which  is  both  equilateral  and  equiangular,  is 
called  regidar. 

Definition  15.  A  closed  rectilineal  figure,  which  has 
three  sides*,  is  called  a  triangle. 

A  closed  rectilineal  figure,  which  has  four  sides,  is  called 
a  quadrilateraL 

A  closed  rectilineal  figure,  which  has  more  than  four 
sides,  is  called  a  polygon  f- 

Definition  16.  A  triangle,  which  has  two  sides  equal, 
is  called  isosceles  J. 


A  triangle,  which  has  a  right  angle,  is  called  right-angled. 
The  side  opposite  to  the  right  angle  is  called  the  hypo- 


tenuse §• 


*  A  figure,  which  has  three  sides,  must  also  have  three  angles. 
It  is  for  this  reason  called  a  triangle. 

t  Derived  from  ttoXi^s  "much"  and  -^wvla  "an  angle". 

X  Derived  from  taos  "equal"  and  ctk^Xos  "a  leg". 

§  Derived  from  vir6  "under"  and  TeLveiv  "to  stretch",  r/  viro- 
relvovcra  ypa/x/j.'/i  "the  line  subtending"  or  "stretching  across"  (the 
right  angle). 


12 


BOOK  I. 


A  triangle^  which  has  an  obtuse  angle^  is  called  obtuse- 
angled. 


A  tria7igle,  which  has  three  acute  angles,  is  called  acute- 
angled. 


Definition  17.  A  quadrilateral^ 
which  has  four  sides  equal,  is  called 
a  rhombus. 


Definition  18.  A  quadri- 
lateral, whose  opposite  sides  are 
parallel,  is  called  a  parallelo- 
gram. 


.  Definition  19.  ^  parallelogra7n, 
one  of  whose  angles  is  a  right  angle, 
is  called  a  rectangle. 


It  will  be  proved  later  that  each  angle  of  a  rectangle  is  a  right 
angle. 


Definition  20.  A  rectangle,  which 
has  two  adjacent  sides  equal,  is  called 
a  square. 

It  will  be  proved  later  that  all  the  sides  of  a 
square  are  equal. 


DEFINITIONS. 


13 


Definition  21.  Two  figures  are  said  to  be  ec[ual  in  all 
respects,  when  it  is  possible  to  shift  one  unchanged  in  shajye 
and  size  so  as  to  coincide  with  the  other. 


The  figures  ABODE,  FGHKL  are  equal  in  all  respects,  if  it  be 
possible  to  shift  ABODE  so  that  the  vertices  A,  B,  O,  D,  E  may 
coincide  with  the  vertices  J^,  G,  H,  K,  L  respectively :  in  which  case 
the  sides  of  the  two  figures  must  be  equal,  AB,  BO,  OD,  DE,  EA  to 
EG,  GH,  HK,  KL,  LF  respectively,  and  the  angles  must  be  equal, 
ABO,  BOD,  ODE,  DEA,  EAB  to  FGH,  GHK,  HKL,  RLE,  LEG 
respectively. 

Definition  22.     A  plane  closed  lioie,  which  is  such  that 
all  straight  lines  drawn  to  it  from  a 
fixed  point  are  equal,  is  called  a  circle. 


This  point  is  called  the  centre 
the  circle. 


oj 


It  will  be  proved  hereafter  that  a  circle 
has  only  one  centre. 


A  straight  line  drawn  from  the  centre  of  the  circle  to  the 
circle  is  called  a  radius. 

A  straight  line  drawn  through  the  centre  and  terminated 
both  ways  by  the  circle  is  called  a,  diameter. 

It  will  be  proved  hereafter  that  three  points  on  a  circle  completely 
fix  the  position  and  magnitude  of  the  circle:  hence  we  generally 
denote  a  circle  by  mentioning  three  points  on  it;  for  instance  the 
circle  in  the  diagram  might  be  called  the  circle  BDE,  or  the  circle  DBO. 

The  one  assumption  which  we  make  with  reference  to  describing  a 
circle  is  contained  in  the  following  postulate  : 


u 


BOOK  I. 


Postulate  6.  A  circle  inay  he  described  with  any  point 
as  centre  and  with  any  straight  line  drawn  from  that  point 
as  radius. 

Postulate  7.  Any  straight  line  drawn  through  a  point 
within  a  closed  Jigure  Tnust,  if  produced  far  enough,  intersect 
the  figure  in  two  points  at  least. 


In  the  diagram  we  have  three  specimens  of  closed  figures  each 
with  a  point  A  inside  the  figure. 

It  is  easily  seen  that  any  straight  line  through  A  must  intersect 
the  figure  in  two  points  at  least:  in  the  case  of  two  of  the  figures 
a  straight  line  cannot  intersect  the  figure  in  more  than  two  points : 
but  in  the  third  case,  a  straight  line  can  be  drawn  to  intersect  the 
figure  in  four  points. 

Postulate  8.  Any  line  joining  two  points  one  within 
and  the  other  without  a  closed  figure  must  intersect  the 
figure  in  one  point  at  least. 

It  follows  that 

Any  closed  line  drawn  through  two  points  one  within  and  the  other 
witJiout  a  closed  Jigure  must  intersect  the  Jigure  in  ttco  points  at  least. 


--^^..^B 


In  the  diagram  we  have  three  specimens  of  closed  figures  with  two 
points  A^B,  one  inside  and  the  other  outside  the  figure. 


DEFINITIONS.  15 

It  is  easily  seen  that  any  line  joining  A  and  B  must  intersect  the' 
figure  in  one  point  at  least,  and  that  any  closed  line  drawn  through  A 
and  B  must  intersect  the  figure  in  two  points  at  least :  in  two  of  the 
cases  in  the  diagram  either  of  the  paths  represented  by  part  of  the 
dotted  line  joining  A  and  B  intersects  the  figure  in  one  point  only  and 
the  closed  line  drawn  intersects  the  figure  in  two  points  only :  but  in 
the  third  case  one  of  the  paths  from  ^  to  i?  represented  by  part  of 
the  dotted  line  intersects  the  figure  in  one  point  only,  while  the  path 
represented  by  the  other  part  of  the  dotted  line  intersects  the  figure  in 
three  points,  and  the  closed  line  drawn  through  A  and  B  intersects 
the  figure  in  four  points. 

Axioms.  There  are  a  number  of  simple  propositions 
generally  admitted  to  be  true  universally,  i.e.  with  reference 
to  magnitudes  of  all  kinds. 

Such  propositions  were  called  by  Euclid  koivoX  evvoiat, 
"common  notions":  they  are  now  usually  denominated 
axioms'*',  a^najxara,  as  being  propositions  claimed  without 
demonstration. 

The  following  are  examples  of  such  axioms : 

Things  which  are  equal  to  tlie  same  thing  are  equal  to  one 
another. 

If  equals  he  added  to  equals,  the  wholes  are  equal. 

If  equals  he  taken  froin  equals,  the  remaiiiders  are  equal. 

Doubles  of  equals  are  equal. 

Halves  of  equals  are  equal. 

The  whole  of  a  thing  is  greater  than  a  part. 

If  one  thing  he  greater  than  a  second  and  the  second 
greater  than  a  third,  the  first  is  greater  than  the  third. 

Such  propositions  as  the  above  we  shall  use  freely  in 
the  following  pages  without  further  remark. 

*  Dr  Johnson  in  his  English  Dictionary  defined  an  axiom  as  "a 
proposition  evident  at  first  sight,  that  cannot  be  made  plainer  by 
demonstration." 


16  BOOK  I. 


PROPOSITION  1. 


On  a  given  finite  straight  liite  to  construct  an  equilateral 
triangle. 

Let  ABhe  the  given  finite  straight  line: 
it  is  required  to  construct  an  equilateral  triangle  on  AB. 

Construction.     With  A  as  centre  and  AB  as  radius, 

describe  the  circle  BCD.  (Post.  6.) 

With  B  as  centre  and  BA  as  radius,  describe  the  circle  ACE. 

These  circles  must  intersect :  (Post.-  8.) 

let  them  intersect  in  C.  VH^^  \  ^ 

Draw  the  straight  lines  CA,  CB  :        (Post.  3.) 

then  ABC  is  a  triangle  constructed  as  required. 


Proof.     Because  A  is  the  centre  of  the  circle  BCD^ 

AC  \&  equal  to  AB.  (Def.  22.) 

And  because  B  is  the  centre  of  the  circle  ACE, 
BC  is  equal  to  BA. 
Therefore  CA,  AB,  BC  are  all  equal. 
Wherefore,   the  triangle  ABC  is  equilateral,  and  it  has 
been  constructed  on  the  given  finite  straight  line  AB. 


I'ROPOSITION-  1.  17 

It  is  assumed  in  this  proposition  that  the  two  circles  intersect.  It 
is  easily  seen  that  they  must  intersect  in  two  points.  "We  can  take 
either  of  these  points  as  the  third  angular  point  of  an  equilateral 
triangle  on  the  given  straight  line ;  there  are  thus  two  triangles  which 
can  be  constructed  satisfying  the  requirements  of  the  proposition. 
We  say  therefore  that  the  problem  put  before  us  in  this  proposition 
admits  of  two  solutions. 

We  shall  often  have  occasion  to  notice  that  a  geometrical  problem 
admits  of  more  than  one  solution,  and  it  is  a  very  useful  exercise  to 
consider  the  number  of  possible  solutions  of  a  particular  problem. 

For  the  future  we  shall  generally  use  the  abbreviated  expression 
"draw  AB^'  instead  of  "draw  the  straight  line  AB'''  or  "draw  a 
straight  line  from  the  point  A  to  the  point  £." 


EXEKCISES. 

1.  Produce  a  straight  line  so  as  to  be  (a)  twice,  {b)  three  times, 
(c)  five  times,  its  original  length. 

2.  Construct  on  a  given  straight  line  an  isosceles  triangle,  such 
that  each  of  its  equal  sides  shall  be  (a)  twice,  {h)  three  times,  (c)  six 
times,  the  length  of  the  given  line. 

3.  Prove  that,  if  two  circles,  whose  centres  are  A,  B,  and  whose 
radii  are  equal,  intersect  in  C,  D,  the  figure  ABCD  is  a  rhombus. 


T,  B. 


18 


BOOK  L 


PROPOSITION  2. 

From  a  given  point  to  draw  a  straight  line  equal  to  a  given 
straight  line. 

Let  A  be  the  given  point,  and  BC  the  given  straight  line : 
it  is  required  to  draw  from  A  a  straight  line  equal  to  BG. 

Construction.     Draw  AB,  the  straight  line  from  A  to 

one  of  the  extremities  of  BG  ',  (Post.  3.) 

on  it  construct  an  equilateral  triangle  ABB.    (Prop.  1.) 

With  B  as  centre  and  BG  as  radius,  describe  the  circle  GEF^ 

(Post.  6.) 
meeting  BB  (produced  if  necessary)  at  E.    (Post.  7.) 
With  B  as  centre  and  BE  as  radius,  describe  the  circle  EGH^ 
meeting  DA  (produced  if  necessary)  at  G  : 
then  AG  i^  Si,  straight  line  drawn  as  required. 


Proof.     Because  B  is  the  centre  of  the  circle  GEF^ 

BG  is  equal  to  BE.  (Def.  22.) 

Again,  because  D  is  the  centre  of  the  circle  EGH^ 
ZXy  is  equal  to  BE; 
and  because  ABB  is  an  equilateral  triangle, 

BA  is  equal  to  BB ;  (Def.  14.) 

therefore  AG  \^  equal  to  BE. 
And  it  has  been  proved  that  BG  is  equal  to  BE; 
therefore  AG  \^  equal  to  BG. 
Whorefore, /rom  the  given  point  A  a  straight  line  AG 
has  been  drawn  equal  to  the  given  straight  line  BG. 


PROPOSITION  2.  19 

It  is  assumed  in  this  proposition  that  the  straight  line  BB  inter- 
sects the  circle  GEF.  It  is  easily  seen  that  it  must  intersect  it  in  tioo 
points. 

It  will  be  noticed  that  in  the  construction  of  this  proposition 
there  are  several  steps  at  which  a  choice  of  two  alternatives  is 
afforded:  (1)  we  can  draw  either  AB  ov  AC  as  the  straight  line  on 
which  to  construct  an  equilateral  triangle :  (2)  we  can  construct  an 
equilateral  triangle  on  either  side  of  AB:  (3)  if  DB  cut  the  circle  in  E 
and  J,  we  can  choose  either  DE  or  DI  as  the  radius  of  the  circle 
which  we  describe  with  D  as  centre. 

There  are  therefore  three  steps  in  the  construction,  at  each  of 
which  there  is  a  choice  of  two  alternatives:  the  total  number  of 
solutions  of  the  problem  is  therefore  2  x  2  x  2  or  eight. 

On  the  opposite  page  two  diagrams  are  drawn,  to  represent  two 
out  of  these  eight  possible  solutions.  It  will  be  a  useful  exercise 
for  the  student  to  draw  diagrams  corresponding  to  some  of  the 
remaining  six. 


EXEKCISES. 

1.  Draw  a  diagram  for  the  case  in  which  the  given  point  is  the 
middle  point  of  the  given  straight  line. 

2.  Draw  a  diagram  for  the  case  in  which  the  given  point  is  in 
the  given  straight  line  produced. 

3.  Draw  from  a  given  point  a  straight  line  (a)  twice,  (&)  three 
times  the  length  of  a  given  straight  line. 

4.  Draw  from  D  in  any  one  of  the  diagrams  of  Proposition  2 
a  straight  line,  so  that  the  part  of  it  intercepted  between  the  two 
circles  may  be  equal  to  the  given  straight  line.  Is  a  solution  always 
possible  ? 


20  BOOK  I. 


PROPOSITION  3. 

From  the  greater  of  two  given  straight  lines  to  cut  off 
a  i^art  equal  to  the  less. 

Let  AB  and  CD  be  the  two  given  straight  lines,  of  whicli 
AB  i^  the  greater  : 
it  is  required  to  cut  off  from  AB  d^  part  equal  to  CD. 

Construction.      From    A    draw    a    straight    line    AE 
equal  to  CD;  (Prop.  2.) 

with  A  as  centre  and  ^jS'  as  radius, 

describe  the  circle  EFG.  (Post.  6.) 

The  circle  must  intersect  AB  between  A  and  B, 

for  AB  \^  greater  than  AE. 

Let  F  be  the  point  of  intersection  : 

then  AF  is  the  part  required. 


Proof.     Because  A  is  the  centre  of  the  circle  EFG^ 

AE  is  equal  to  AF.  (Del  22.) 

But  AE  was  made  equal  to  CD]   (Construction.) 
therefore  AF  \&  equal  to  CD. 
Wherefore,  from  AB  the  greater  of  two  given  straight 
lines  a  part  A  F  has  heeii  cut  off  equal  to  CD  the  less. 


PROPOSITION  3.  21 


The  demand  made  in  Postulate  6,  that  "a  circle  maybe  described 
with  any  point  as  centre  and  with  any  straight  line  drawn  from  that 
point  as  radius,"  is  equivalent,  in  practical  geometry,  to  saying  that 
a  pair  of  compasses  may  be  used  in  the  following  manner:  the  ex- 
tremity of  one  leg  of  a  pair  of  compasses  may  be  put  down  on  any 
point  A,  the  compasses  may  then  be  opened  so  that  the  extremity  of 
the  other  leg  comes  to  any  other  point  and  then  a  circle  may  be 
swept  out  by  the  extremity  of  the  second  leg  of  the  compasses,  the 
extremity  of  the  first  leg  remaining  throughout  the  motion  on  the 
point  A. 

Compasses  are  also  used  practically  for  carrying  a  given  length 
from  any  one  position  to  any  other :  for  instance,  they  would  gene- 
rally be  used  to  solve  the  problem  of  Proposition  3  by  opening  the 
compasses  out  till  the  extremities  of  the  legs  came  to  the  points 
C,  D :  they  would  then  be  shifted,  without  any  change  in  the  opening 
of  the  legs,  until  the  extremity  of  one  leg  was  on  A  and  the  extremity 
of  the  other  in  the  straight  line  AB. 

Euclid  restricted  himself  much  in  the  same  way  as  a  draughtsman 
would,  if  he  allowed  himself  only  the  first  mentioned  use  of  the  com- 
passes :  the  first  three  propositions  shew  how  Euclid  with  this  self- 
imposed  restriction  solved  the  problem,  which  without  such  a  restric- 
tion could  have  been  solved  more  readily. 

After  the  problems  in  the  first  three  propositions  have  been  solved, 
we  may  assume  that  we  can  draw  a  circle,  as  a  practical  draughts- 
man would,  with  any  point  as  centre  and  with  a  length  equal  to  any 
given  straight  line  as  radius. 


EXEECISES. 

1.  On  a  given  straight  line  describe  an  isosceles  triangle  having 
each  of  the  equal  sides  equal  to  a  second  given  straight  line. 

2.  Construct  upon   a  given  straight  line  an  isosceles  triangle 
having  each  of  the  equal  sides  double  of  a  second  given  straight  line. 

3.  Construct  a  rhombus  having  a  given  angle  for  one  of  its 
angles,  and  having  its  sides  each  equal  to  a  given  straight  line. 


22  BOOK  L 


PROPOSITION  4. 

If  two  triangles  have  two  sides  of  iJie  one  equal  to  two 
sides  of  the  other,  and  also  the  angles  contained  hy  those  sides 
equal,  the  two  triangles  are  equal  m  all  resjjects. 

(See  Def.  21.) 

Let  ABC,  DEF  be  two  triangles,  in  which  AB  is  equal 
to  DE,  and  AC  to  DF,  and  the  angle  BAG  is  equal  to  the 
angle  ^i>i^: 

it  is  required  to  prove  that  the  triangles  ABC,  DEF  are 
equal  in  all  respects. 


Proof.     Because  the  angles  BAG,  EDF  are  equal, 
it  is  possible  to  shift  the  triangle  ABC 

so  that  A  coincides  with  D, 
and  AB  coincides  in  direction  with  DE, 

and  AC  with  DF.      (Test  of  Equality,  page  8.) 
If  this  be  done, 

because  ^^  is  equal  to  DE, 

B  must  coincide  with  E  \ 

and  because  AC  is  equal  to  DF, 

C  must  coincide  with  F. 

Again  because  B  coincides  with  E  and  G  with  F, 

BG  coincides  with  EF -,  (Post.  2.) 

therefore  the  triangle  ABC  coincides  with  the  triangle  DEF, 
and  is  equal  to  it  in  all  respects. 

Wherefore,  if  two  triangles  &c. 


PROPOSITION  4.  23 

The  proof  of  this  proposition  holds  good  not  only  for  a  pair  of 
triangles  such  as  ABC,  DEF  in  the  diagram:  it  holds  good  equally 
for  a  pair  such  as  ABC,  D'E'F',  one  of  which  must  be  reversed  or 
turned  over  before  the  triangles  can  be  made  to  coincide  or  fit  exactly. 

In  this  proposition  Euclid  assumed  Postulate  2,  that  two  straight 
lines  cannot  have  a  common  part.  When  the  triangle  ABC  is 
shifted,  so  that  ^  is  on  D  and  AB  is  on  DE,  there  would  be  no 
justification  for  the  conclusion  that  B  must  coincide  with  E,  because 
^^  is  equal  to  DE,  if  it  were  possible  for  two  straight  lines  to  have  a 
common  part.  In  fact,  two  curved  lines  might  be  drawn  from  the 
point  D  starting  in  the  same  direction  DE  but  leading  to  two  totally 
distinct  points  E  and  F  although  the  lines  were  of  the  same  length. 
It  is  tacitly  assumed,  that  if  the  lines  be  straight  lines,  this  is  im- 
possible. 


EXERCISES. 

1.  If  the  straight  line  joining  the  middle  points  of  two  opposite 
sides  of  a  quadrilateral  be  at  right  angles  to  each  of  these  sides,  the 
other  two  sides  are  equal. 

2.  If  in  a  quadrilateral  ABCD  the  sides  AB,  CD  be  equal  and 
the  angles  ABC,  BCD  be  equal,  the  diagonals  AC,  BD  are  equal. 

3.  If  in  a  quadrilateral  two  opposite  sides  be  equal,  and  the 
angles  which  a  third  side  makes  with  the  equal  sides  be  equal,  the 
other  angles  are  equal. 

4.  Prove  by  the  method  of  superposition  that,  if  in  two  quadri- 
laterals ABCD,  A'B'C'D',  the  sides  AB,  BC,  CD  be  equal  to  the  sides 
A'B',  B'C,  CD'  respectively,  and  the  angles  ABC,  BCD  equal  to  the 
angles  A'B'C,  B'C'D'  respectively,  the  quadrilaterals  are  equal  in  all 
respects. 


24 


BOOK  I. 


PROPOSITION  5. 

If  two  sides  of  a  triangle  he  equal,  the  angles  opjjosite  to 
these  sides  are  equal,  and  the  angles  made  hy  producing 
these  sides  beyond  the  third  side  are  equal. 

Let  ABC  be  a  triangle,  in  which  AB  is  equal  to  AC, 
and  AB,  AC  are  produced  to  D,  E  : 

it  is  required   to   prove    that   the  angle  ACB  is  equal  to 
the  angle  ABC,  and  the  angle  BCE  to  the  angle  CBJ). 

Construction.  Let  the  6gure  ABODE  be  turned  over 
and  shifted  unchanged  in  shape  and  size  to  the  position 
ahcde,  A  to  a,  B  to  b,  C  to  c,  D  to  d  and  E  to  e. 

A 

A 


Proof.     Because  the  angles  DAE,  ead  are  ecjual,  it  is 
possible  to  shift  the  figure  abode 

so  that  a  coincides  with  A, 
and  ae  coincides  in  direction  with  AD, 

and  ad  with  AE.     (Test  of  Equality,  page  8.) 
If  this  be  done, 

because  ac  is  equal  to  AB, 

c  must  coincide  with  B  ; 

and  because  ab  is  equal  to  AC, 

b  must  coincide  with  C ; 

hence  cb  coincides  with  BC.  (Post.  2.) 

Now  because  ace  coincides  in  direction  with  A  BD, 

and  cb  with  BC, 

the  angle  acb  coincides  with  the  angle  ABC, 

and  the  angle  bee  with  the  angle  CBD ; 

therefore  the  angle  acb  is  equal  to  the  angle  ABC, 

and  the  angle  bee  to  the  angle  CBD. 


PROPOSITION  5.  25 

But  the  angle  ach  is  equal  to  the  angle  AGB^ 

and  the  angle  hce  to  the  angle  BCE ; 

therefore  the  angle  ACB  is  equal  to  the  angle  ABC^ 

and  the  angle  BCE  to  the  angle  GBD. 
Wherefore,  if  two  sides  &c. 

Corollary  1.  An  equilateral  triangle  is  also  equi- 
angular. 

Corollary  2.  If  two  angles  of  a  triangle  be  unequal.^ 
the  sides  opposite  to  these  angles  are  unequal. 


EXEKCISES. 

1.  The  opposite  angles  of  a  rhombus  are  equal. 

2.  If  a  quadrilateral  have  two  pairs  of  equal  adjacent  sides,  it 
has  one  pair  of  opposite  angles  equal. 

3.  If  in  a  quadrilateral  ABGD,  AB  be  equal  to  AD  and  EC  to 
DC,  the  diagonal  AG  bisects  each  of  the  angles  BAD,  BCD. 

4.  If  in  a  quadrilateral  ABCD,  AB  be  equal  to  ^D  and  jBC  to 
DC,  the  diagonal  BD  is  bisected  at  right  angles  by  the  diagonal  AC. 

5.  Prove  that  the  triangle,  whose  vertices  are  the  middle  points 
of  the  sides  of  an  equilateral  triangle,  is  equilateral. 

6.  Prove  that  the  triangle,  formed  by  joiuing  the  middle  points 
of  the  sides  of  an  isosceles  triangle,  is  isosceles. 

7.  Prove  by  the  method  of  superposition  that,  if  in  a  convex 
quadrilateral  ABCD,  AB  be  equal  to  CD  and  the  angle  ABC  to  the 
angle  BCD,  AD  is  parallel  to  BC 


26  BOOK  I. 

PROPOSITION  6. 

If  two  angles  of  a  triangle  he  equal,  tlie  sides  opposite  to 
these  angles  are  equal. 

Let  ABC  be  a  triangle,  in  which  tlie  angle  ABC  is 
equal  to  the  angle  ACB  : 

it  is  required  to  prove  that  AC  is  equal  to  AB. 

Construction.  Let  the  triangle  ABC  be  turned  over 
and. shifted  unchanged  in  shape  and  size  to  the  position 
ahcpA  to  a,  B  to  6,  and  C  to  c. 


Pkoof.     Because  the  sides  BC,  cb  are  equal, 
it  is  possible  to  shift  the  triangle  acb 
so  that  cb  coincides  with  BC,  c  with  B,  and  b  with  C, 

(Test  of  Equality,  page  5.) 
and  so  that  the  triangles  acb,  ABC  are  on  the  same  side 

of  BC. 
If  this  be  done, 

because  the  angles  ABC,  acb  are  equal, 
ca  must  coincide  in  direction  with  BA  ; 
and  because  the  angles  ACB,  abc  are  equal, 
ba  must  coincide  in  direction  with  CA. 
And  because  two  straight  lines  cannot  intersect  in  more 
than  one  point,  (Post.  1.) 

the  point  a,  which  is  the  intersection  of  ca  and  ba,  must 
coincide  with  the  point  A,  which  is  the  intersection  of 
BA  and  CA. 

Now  because  a  coincides  with  A  and  c  with  B, 
ac  coincides  with  AB  and  is  equal  to  it. 
But  ac  is  equal  to  AC ; 
therefore  ^0  is  equal  to  AB. 
Wherefore  if  two  angles  &c. 
Corollary.    An  equiangular  triangle  is  also  equilateral. 


PROPOSITION  6.  27 

When  in  two  propositions  the  hypothesis  of  each  is  the 
conclusion  of  the  other,  each  proposition  is  said  to  be  the 
converse  of  the  other. 

The  theorems  in  Propositions  5  and  6  are  the  converses  of  each 
other. 

It  must  not  be  assumed  that  the  converse  of  a  proposition  is 
necessarily  true. 


EXEECISES. 

1.  Shew  that,  if  the  angles  ABC  and  ACB  at  the  base  of  an 
isosceles  triangle  be  bisected  by  the  straight  lines  JSD  and  CD,  DEC 
will  be  an  isosceles  triangle. 

2.  BAG  is  a  triangle  having  the  angle  B  double  of  the  angle  A. 
If  BB  bisect  the  angle  B  and  meet  ^C  at  D,  BB  is  equal  to  AB. 

3.  Prove  by  the  method  of  superposition  that,  if  in  two  triangles 
ABC,  A'B'C  the  angles  ABC,  BCA  be  equal  to  the  angles  A'B'C, 
B'C'A'  respectively  and  the  sides  BC,B'C'  he  equal,  the  triangles  are 
equal  in  all  respects. 

4.  Prove  by  the  method  of  superposition  that,  if  in  two  quadri- 
laterals ABCB,  A'B'C'B'  the  angles  BAB,  ABC,BCB  be  equal  to  the 
angles  B'A'B',  A'B'C,  B'C'B'  respectively,  and  the  sides  AB,  BC  be 
equal  to  the  sides  A'B',  B'C  respectively,  the  quadrilaterals  are  equal 
in  all  respects. 

5.  If  in  a  quadrilateral  ABCB,  AB  be  equal  to  AB  and  the  angle 
ABC  to  the  angle  ABC,  then  BC  is  equal  to  BC,  and  the  diagonal 
AC  bisects  the  quadrilateral  and  two  of  its  angles. 


28 


BOOK  I. 


PROPOSITION  7. 

If  two  points  on  the  saine  side  of  a  straight  line  he  equi- 
distant from  one  point  in  the  line,  they  cannot  he  equidistant 
from  any  other  point  in  the  liiie. 

Let  AB  he>  ^  given  straight  line,  and  C,  D  be  two 
points  on  the  same  side  of  it  equidistant  from  the  point  A : 
it  is  required  to  prove  that  C,  D  cannot  be  equidistant 
from  any  other  point  in  the  line. 

Construction.     Take  any  other  point  B  in  the  line,  and 
draw  BC,  BD. 


Proof.  Because  C  and  D  are  two  different  points,  either 

(1)  the  vertex  of  each  of  the  triangles  ABC,  ABD  must  be 
outside  the  other  triangle, 

(2)  the  vertex  of  one  triangle  must  be  inside  the  other, 

or  (3)  the  vertex  of  one  triangle  must  be  on  a  side  of  the 
other. 

(1)  First  let  the  vertex  of  each  triangle  be  without  the 
other. 

Because  AD  is  equal  to  AC, 

the  angle  ACD  is  equal  to  the  angle  ADC.     (Prop.  5.) 

But  the  an^le  ACD  is  greater  than  the  angle  BCD, 

and  the  angle  BDC  is  greater  than  the  angle  ADC : 

therefore  the  angle  BDC  is  greater  than  the  angle  BCD ; 

therefore  BC,  BD  are  unequal.     (Prop.  5,  Coroll.  2.) 

(2)  Next  let  the  vertex  D  of  one  triangle  ABD  be  within 
the  other  triangle  ABC  : 

produce  AC,  AD  to  B,  F.  (Post.  4.) 


PROPOSITION  7.  29 

Then  because  in  the  triangle  ACD,  AC  is.  equal  to  AD, 
the  angles  ECD,  FDC  made  by  producing  the  sides  AC,  AD 
are  equal.  (Prop.  5.) 

But  the  angle  ECD  is  greater  than  the  angle  BCD, 
and  the  angle  BDC  is  greater  than  the  angle  FDC ; 
therefore  the  angle  BDC  is  greater  than  the  angle  BCD; 
therefore  BC,  BD  are  unequal,    (Prop.  5,  Coroll.  2.) 
(3)     Next  let  the  vertex  D  of  one  triangle  lie  on  one  of  the 
sides  BC  of  the  other  : 

then  BC,  BD  are  unequal. 

Wherefore,  if  two  points  07i  the  same  side  &c. 


30  BOOK  I. 


PROPOSITION  8. 

If  two  triangles  have  three  sides  of  the  one  equal  to  three 
sides  of  the  other,  the  triangles  are  equal  in  all  respects. 

Let  ABC,  DEF  be  two  triangles,  in  which  AB  i^  equal 
to  DE,  AG  to  DF,  and  BC  to  EF\ 

it  is  required  to  prove  that  the  triangles  ABC,  DEF  are 
equal  in  all  respects. 


Proof.     Because  the  sides  BC,  EF  are  equal, 
it  is  possible  to  shift  the  triangle  ABC, 
so  that  BC  coincides  with  EF,  B  with  E  and  C  with  F, 

(Test  of  Equality,  page  5.) 
and  so  that  the  triangles  ABC,  DEF  are  on  the  same  side 
of  EF. 
If  this  1)6  done, 

A  must  coincide  with  D : 
for  there  cannot  be  two  points  on  the  same  side  of  the 
straight  line  EF  equidistant  from  E 

and  also  equidistant  from  F,  (Prop.  7.) 

Now  because  A  coincides  with  i>, 

and  B  coincides  with  E,  (Constr.) 

AB  coincides  with  BE.  (Post.  2.) 

Similarly  it  can  be  proved  that 
AC  coincides  with  DF. 
Therefore  the  triangle  ABC  coincides  with  the  triangle  DEF, 
and  is  equal  to  it  in  all  respects. 
Wherefore,  if  two  tricmgles  <fec. 


PROPOSITION  8.  31 


EXEECISES. 

1.  If  a  quadrilateral  have  two  pairs  of  equal  sides,  it  must  have 
one  pair  and  may  have  two  pairs  of  equal  angles. 

2.  ABC,  BBC  are  two  isosceles  triangles  on  the  same  base  BC, 
and  on  the  same  side  of  it :  shew  that  AD  bisects  the  vertical  angles 
of  the  triangles. 

3.  If  the  opposite  sides  of  a  quadrilateral  be  equal,  the  opposite 
angles  are  equal. 

4.  If  in  a  quadrilateral  two  opposite  sides  be  equal,  and  the 
diagonals  be  equal,  the  quadrilateral  has  two  pairs  of  equal  angles. 

5.  If  in  a  quadrilateral  the  sides  AB,  CD  be  equal  and  the 
angles  ABC,  BCD  be  equal,  the  angles  CD  A,  DAB  are  equal. 

6.  The  sides  AB,  AD  of  a  quadrilateral  ABCD  are  equal,  and 
the  diagonal  AC  bisects  the  angle  BAD ;  shew  that  the  sides  CB,  CD 
are  equal,  and  that  the  diagonal  AC  bisects  the  angle  BCD. 

7.  ACB,  ADB  are  two  triangles  on  the  same  side  of  AB,  such 
that  AC  i^  equal  to  BD,  and  AD  is  equal  to  BC,  and  AD  and  BC 
intersect  at  0  :  shew  that  the  triangle  AOB  is  isosceles. 

8.  A  diagonal  of  a  rhombus  bisects  each  of  the  angles  through 
which  it  passes. 


32  BOOK  I. 

PROPOSITION  9. 
To  bisect  a  given  angle. 

Let  BAG  be  the  given  angle: 
it  is  required  to  bisect  it. 
Construction.      Take  any  point  D  in  AB, 

and  from  AC  cut  off  AE  equal  to  AD.    (Prop.  3.) 
Draw  DE, 
and  on  DE,  on  the  side  away  from  A^  construct  the  equi- 
lateral triangle  DEF.  (Prop.  1.) 
Draw  AF: 
then  AF \^  the  required  bisector  of  the  angle  BAG. 


Proof.     Because  in  the  triangles  DAF,  EAF, 
DA  is  equal  to  EA, 
AF  to  AF, 
and  FD  to  FE, 
the  triangles  are  equal  in  all  respects  ;     (Prop.  8. ) 
therefore  the  angle  DAF  is  equal  to  the  angle  EAF. 
Wherefore,  the  given  angle  BAG  is  bisected  by  the  straight 
AF. 


PROPOSITION  9.  33 

In  the  construction  for  this  proposition  it  is  said  that  the  equi- 
lateral triangle  DEF  is  to  be  constructed  on  the  side  of  BE  away 
from  A.  This  restriction  is  introduced  in  order  to  prevent  the  possi- 
bility of  the  point  F  coinciding  with  the  point  A. 

In  practical  geometry  it  is  always  desirable  to  obtain  two  points, 
which  determine  a  straight  line,  as  far  apart  as  possible,  as  then  an 
error  in  the  position  of  one  of  the  points  causes  less  error  in  the 
position  of  the  straight  line. 


EXERCISES. 

1.  A  straight  line,  bisecting  the  angle  contained  by  two  equal 
sides  of  a  triangle,  bisects  the  third  side. 

2.  The  bisectors  of  the  angles  ABC,  AGB  oi&  triangle  ABC  meet 
in  D:  prove  that,  if  BB,  BC  be  equal,  AB,  AC  are  equal. 

3.  Prove  that  there  is  only  one  bisector  of  a  given  angle. 

4.  Prove  that  the  bisectors  of  the  angles  of  an  equilateral  triangle 
meet  in  a  point. 

5.  ^  Prove  that  the  bisectors  of  the  angles  of  an  isosceles  triangle 
meet  in  a  point. 

6.  BAC  is  a  given  angle ;  cut  off  AB,  AG  equal  to  one  another  : 
with  centres  B,  C  describe  circles  having  equal  radii :  if  the  circles 
intersect  at  D,  AB  bisects  the  angle  BAC. 

7.  Prove  by  the  method  of  superposition  that,  if  one  diagonal 
of  a  quadrilateral  bisect  each  of  the  angles  through  which  it  passes, 
the  two  diagonals  are  at  right  angles  to  each  other. 


T.  E. 


34 


BOOK  L 


PROPOSITION  10. 
To  bisect  a  yive7i  finite  straight  line. 

Let  AB  he  the  given  finite  straight  line: 

it  is  required  to  bisect  it. 

Construction.    OnAB  construct  an  equilateral  triangle 

ABC,  (Prop.  1.) 

and  bisect  the  angle  ACB  by  the  straight  line  CD,  meeting 

ABsitD:  (Prop.  9.) 

then  AB  is  bisected  as  required  at  D. 


Proof.     Because  in  the  triangles  ACD,  BCD, 
AG  is  equal  to  BC\ 
and  CD  to  CD, 
and  the  angle  ACD  is  equal  to  the  angle  BCD, 

the  triangles  are  equal  in  all  respects;     (Prop.  4.) 
therefore  -4i>  is  equal  to  BD. 
Wherefore,  the  given  finite  straight  line  AB  is  bisected 
at  the  point  D. 


PROPOSITION  10.  35 


EXEBCISES. 

1.  Prove  that  there  is  only  one  point  of  bisection  of  a  given 
finite  straight  line. 

2.  If  two  circles  intersect,  then  the  straight  line  joining  their 
centres  bisects  at  right  angles  the  straight  line  joining  their  points  of 
intersection. 

3.  Draw  from  the  vertex  of  a  triangle  to  the  opposite  side  a 
straight  line,  which  shall  exceed  the  smaller  of  the  other  sides  as 
much  as  it  is  exceeded  by  the  greater. 


3—2 


u 


BOOK  I. 


PEOPOSITION    10  A. 

From  the  same  point  in  a  given  straight  line  and  on  the 
same  side  of  it,  only  one  straight  line  can  be  drawn  at  right 
angles  to  the  given  straight  line. 

From  the   point    C   in   the    straight   line  AB  let  the 
straight  line  CD  be  drawn  at  right  angles  to  AB : 
it  is  required  to  prove  that  no  other  straight  line  can  be 
drawn  from  G  at  right  angles  to  AB  on  the  same  side  of  it. 

Construction.  Draw  from  C  within  the  angle  ACB 
any  straight  line  CE, 


D 


Proof.     Because  the  angles  DOB,  DC  A  are  equal, 
and  the  angle  ECB  is  greater  than  the  angle  DCB, 
and  the  angle  DC  A  is  greater  than  the  angle  EC  A  ; 
therefore  the  angle  ECB  is  greater  than  the  angle  EC  A  ; 
therefore  CE  is  not  at  right  angles  to  AB.    (Del  11.) 
Similarly   it  can  be  proved   that  no  straight  line   drawn 
from   C  within  the  angle  DCB  can  be  at  right  angles 
to^^. 
Therefore  no  straight  line  other  than  CD  drawn  from  C 
can  be  at  right  angles  to  ^^  on  the  same  side  of  it. 
Wherefore,  from  the  same  point  kc. 


PROPOSITION  ion.  37 


PROPOSITION    10  B. 

All  right  angles  are  equal  to  one  another. 

Let  the  straight  lines  AB^   CD  meet  at  E  and  make 
the  angles  CEA,  CEB  right  angles;  (Def.  11.) 

and  let  the  straight  lines  FG,  HK  meet  at  L  and  make 

the  angles  HLF^  II LG  right  angles  : 
it  is  required  to  prove  that  the  angle  CEA  is  equal  to  the 

aniifle  II LF. 


Proof.     If  the  figure  ABODE  be  shifted,  so  that  E 
coincides  with  L,  and  the  line  ^^  in  direction  with  FG^  and 
so  that  ECy  LII  are  on  the  same  side  of  EG  : 
then  EC  must  coincide  with  LH,  for  at  the  same  point  L 
in  FG  on  the  same  side  of  it  there  cannot  be  two  straight 
lines  at  right  angles  to  EG;  (Prop.  10  A.) 

therefore  the  angle  A  EC  coincides  with  the  angle  FLH, 
and  is  equal  to  it. 
Wherefore,  all  right  angles  &c. 


38  BOOK  I. 


PROPOSITION  11. 

To  draw  a  straight  line  at  right  angles  to  a  given  straight 
line  from  a  given  point  in  it. 

Let  ABh^  the  given  straight  line,  and  G  the  given  point 
in  it : 

it  is  required  to  draw  from  C  a  straight  line  at  right  angles 
to^^. 

Construction.     Take  any  point  D  in.  AG ^ 

and  from  GB  cut  off  GE  equal  to  GD.  (Prop.  3.) 
On  DE  construct  an  equilateral  triangle  DFE^    (Prop.  1.) 
and  draw  GF : 
then  GF  is  a  straight  line  drawn  as  required. 


Z> 


Proof.     Because  in  the  triangles  DGF,  EGF, 
DG  is  equal  to  EG, 
GF  to  GF, 
and  FD  to  FE,  (Constr.) 

the  triangles  are  equal  in  all  respects ;  (Prop.  8.) 

therefore  the  angle  DGF  is  equal  to  the  angle  EOF, 
and  they  are  adjacent  angles; 
therefore  each  of  these  angles  is  a  right  angle ;  and  the 
straight  lines  are  at  right  angles  to  each  other.  (Def.  11.) 
Wherefore,    GF  has  been  drawn  at  right  angles  to  the 
given  straight  line  AB,fro7n  the  given  point  G  in  it. 


PROPOSITION  11.  39 

In  the  definition  of  a  circle  (Def.  22)  we  meet  with  the  idea  of 
a  point,  which  moves  subject  to  a  given  condition,  the  condition 
being  that  the  point  is  always  to  be  at  a  given  distance  from  a 
given  point,  i.e.  from  the  centre  of  the  circle.  The  'path  of  such  a 
moving  point,  or  the  place  {locus},  at  some  position  on  which  the 
point  must  always  be  and  at  any  position  on  which  the  point  may 
be,  is  called  the  locus  of  the  point.  Hence  we  say  in  the  case  just 
mentioned  that  the  locus  of  a  point  which  is  at  a  given  distance  from 
a  given  point  is  a  circle. 

As  a  further  illustration  of  this  idea,  let  us  consider  the  locus  of  a 
point,  which  moves  subject  to  the  condition  that  it  is  always  to  be 
equidistant  from  two  given  points. 

Let  A,  B  he  two  given  points,  and  let 
P  be  a  point  equidistant  from  A  and  B, 
i.e.  let  PA  be  equal  to  PB, 

Draw  AB  and  take  G  the  middle  point 
otAB. 

Draw  PC. 

Then  because  in  the  triangles  PGA, 
PGB, 

PA  is  equal  to  PB,  PC  to  PC,  and  ^C  to  BG, 
the  two  triangles  are  equal  in  all  respects:         (Prop.  8.) 
therefore  the  angle  PGA  is  equal  to  the  angle  PCB, 
and  since  they  are  adjacent  angles  each  is  a  right  angle. 

It  follows  that,  if  P  be  equidistant  from  A  and  B,  it  must  lie  on 
the  straight  line  GP  which  bisects  AB  a.t  right  angles.  Every  point 
on  GP  satisfies  this  condition. 

We  may  state  the  result  of  this  proposition  thus :  The  locus  of  a 
point  equidistant  from  tico  given  points  is  the  straight  line  which 
bisects  at  right  angles  the  straight  line  joining  the  given  points. 

EXEECISES. 

1.  The  diagonals  of  a  rhombus  bisect  each  other  at  right  angles. 

2.  Find  in  a  given  straight  line  a  point  equidistant  from  two 
given  points.      Is  a  solution  always  possible? 

3.  Find  a  point  equidistant  from  three  given  points. 

4.  In  the  base  JBC  of  a  triangle  ABG  any  point  D  is  taken. 
Draw  a  straight  line  such  that,  if  the  triangle  ABG  be  folded  along 
this  straight  line,  the  point  A  shall  fall  upon  the  point  D, 


40 


BOOK   T. 


PROPOSITION  12. 

To  draw  a  straight  line  at  right  angles  to  a  given  straight 
line  from  a  given  point  without  it. 

Let  AB  be  the  given  straight  line,  and   C  the  given 
point  without  it : 

it  is  required  to  draw  from  C  a  straight  line  at  right  angles 
to^^. 

Construction.     Take  any  point  D  on  the  side  of  AB 
away  from  (7,  draw  C7i>, 

and  with  G  as  centre  and  CD  as  radius,  describe  the  circle 

J^EF,    _  (Post.  6.) 

meeting  AB  (produced  if  necessary)  at  E  and  F. 

Draw  CE,  OF, 

and  bisect  the  angle  EOF  by  the  straight  line  CG  meeting 

AB  s^tG'.  (Prop.  9.) 

then  (76^  is  a  straight  line  drawn  as  required. 


Proof.     Because  in  the  triangles  EGG,  FGG, 
EC  is  equal  to  EG, 
and  CG  to  CG, 
and  the  angle  EGG  is  equal  to  the  angle  FGG, 

the  triangles  are  equal  in  all  respects ;     (Prop.  4.) 
therefore  the  angle  CGE  is  equal  to  the  angle  CGF, 
and  they  are  adjacent  angles. 
Therefore  the   straight  lines  CG,  AB  are  at  right  angles 
to  each  other.  (Def.  11.) 

Wherefore,  CG  has  been  drawn  at  right   angles    to   the 
given  straight  line  AB  from  the  given  point  C  without  it. 


PROPOSITION   12.  41 

In  the  construction  for  this  proposition  it  is  said  that  the  point  D 
is  to  be  taken  on  the  side  of  AB  mcay  from  G.  This  restriction  is 
introduced  as  a  means  of  ensuring  the  intersection  of  the  circle  DEF 
with  the  straight  line  AB. 


EXEKCISES. 

1.  Through  two  given  points  on  opposite  sides  of  a  given  straight 
line  draw  two  straight  lines,  which  shall  meet  in  the  given  line  and 
include  an  angle  bisected  by  that  line.  In  what  case  can  there  be 
more  than  one  solution  ? 

2.  From  two  given  points  on  the  same  side  of  a  given  straight 
line,  draw  two  straight  lines,  which  shall  meet  at  a  point  in  the  given 
line  and  make  equal  angles  with  it. 

3.  Prove  by  the  method  of  superposition  that,  if  the  perpendiculars 
on  a  given  straight  line  from  two  points  on  the  same  side  of  it  be 
equal,  the  straight  line  joining  the  points  is  parallel  to  the  given  line. 


42  BOOK  I. 

PROPOSITION  13. 

The  sum  of  the  angles,  which  one  straight  line  makes 
with  another  straight  line  on  one  side  of  it,  is  equal  to  two 
right  angles. 

Let  the  straight  line  AB  make  with  the  straight  line 
CD,  on  one  side  of  it,  the  angles  ABC,  ABD  : 
it  is  required  to  prove  that  the  sum  of  these  angles  is  equal 
to  two  right  angles. 

If  the  angle  ABC  be  equal  to  the  angle  ABD, 

each  of  them  is  a  right  angle, 

and  their  sum  is  equal  to  two  right  angles. 

Construction.     If  the  angles  ABC,  ABD  be  not  equal, 

from  the  point  B  draw  BE  at  right  angles  to  CD ;  (Prop.  11.) 

BE  cannot  coincide  with  BA ; 

let  it  lie  within  the  angle  ABD. 


D 


Proof.     Now  the  angle  CBE  is  the  sum  of  the  angles 
CBA,  ABE; 

to  each  of  these  equals  add  the  angle  EBD ; 
then  the  sum  of  the  angles  CBE,  EBD  is  equal  to  the  sum 

of  the  angles  CBA,  ABE,  EBD. 
Again,  the  angle  DBA  is  equal  to  the  sum  of  the  angles 
DBE,  EBA ; 

to  each  of  these  equals  add  the  angle  ABC ; 
then  the  sum  of  the  angles  DBA,  ABC  is  equal  to  the  sum 

of  the  angles  DBE,  EBA,  ABC. 
And  the  sum  of  the  angles  CBE,  EBD  has  been  proved  to 
be  equal  to  the  sum  of  the  same  three  angles. 


PROPOSITION  13.  43 

Therefore  the  sum  of  the  angles  CBE,  EBD  is  equal  to 
the  sum  of  the  angles  DBA,  ABC. 

But  CBE,  EBD  are  two  right  angles ;  (Constr.) 
therefore  the  sum  of  the  angles  DBA,  ABC  is  equal  to  two 
right  angles. 

"Wherefore,  the  sum  of  the  angles  &c. 

Corollary.  The  sum  of  the  four  angles,  which  two 
intersecting  straight  lines  make  ivith  one  another,  is  equal  to 
four  right  angles. 


EXEECISES. 

1.  Prove  in  the  manner  of  Proposition  13  that,  if  A,  B,  C,  D  be 
four  points  in  order  on  a  straight  line,  the  sum  of  AB,  BD  is  equal  to 
the  sum  of  ^C,  CD. 

2.  If  one  of  the  four  angles,  which  two  intersecting  straight 
lines  make  with  one  another,  be  a  right  angle,  all  the  others  are 
right  angles. 

3.  Prove  by  the  method  of  superposition  that  only  one  perpen- 
dicular can  be  drawn  to  a  given  straight  line  from  a  given  point 

without  it. 

4.  Prove  by  the  method  of  superposition  that,  if  two  right- 
angled  triangles  have  their  hypotenuses  equal  and  two  other  angles 
equal,  the  triangles  are  equal  in  aU  respects. 

5.  A  given  angle  BAC  is  bisected;  if  CA  be  produced  to  G  and 
the  angle  BA  G  bisected,  the  two  bisecting  lines  are  at  right  angles. 


44  BOOK  I. 


PROPOSITION   14. 

If,  at  a  point  in  a  straight  line,  two  other  straight 
lines,  on  opposite  sides  of  it,  make  the  adjacent  angles  together 
equal  to  two  right  angles,  these  two  straight  lines  are  in  the 
sa7ne  straight  line. 

At  the  point  B  in  the  straight  line  AB,  let  the  two 
straight  lines  BC,  BD,  on  opposite  sides  of  AB,  make  the 
adjacent  angles  ABC,  ABD  together  equal  to  two  right 
angles : 

it  is  required  to  prove  that  BB  is  in  the  same  straight  line 
with  CB. 

Construction.     Produce  GB  to  B. 


Proof.     Because  the  straight  line  AB  makes  with  the 
straight  line  GBE,  on  one  side  of  it,  the  angles  ABC,  ABE, 
the  sum  of  these  angles  is  equal  to  two  right  angles. 

(Prop.  13.) 
But  the  sum  of  the  angles  ABC,  ABD  is  equal  to  two  right 

angles. 
Therefore  the  sum  of  the  angles  ABC,  ABB  is  equal  to  the 
sum  of  the  angles  ABC,  ABE. 
From  each  of  these  equals  take  away  the  angle  ABC ; 

then  the  angle  ABD  is  equal  to  the  angle  ABE ; 
therefore  the  line  BD  coincides  in  direction  witli  BE, 

and  is  in  the  same  straight  line  with  CB. 
Wherefore,  if  at  a  poirit  ikc. 


PROPOSITION  14.  45 

Definition.  Two  angles,  which  are  together  equal  to 
two  right  angles,  are  called  supplementary  angles,  and  each 
angle  is  said  to  be  the  supplement  of  the  other. 

Two  angles,  which  are  together  equal  to  one  right  angle, 
are  called  complementary  angles,  a7id  each  angle  is  said  to 
he  the  complement  of  the  other. 


EXERCISES. 

1.  If  E  be  the  middle  point  of  the  diagonal  AC  of  a  quadri- 
lateral ABCD,  whose  opposite  sides  are  equal,  B,  E,  D  lie  on  a  straight 
line. 

2.  If  OA,  OB,  OG,  OD  be  four  straight  lines  drawn  in  order 
from  O,  such  that  the  angles  BOG,  BOA  are  equal  and  also  the  angles 
AOB,  GOD,  then  the  lines  OA,  OG  are  in  the  same  straight  line  and 
also  the  Hues  OB,  OD. 

3.  If  it  be  possible  within  a  quadrilateral  ABGD,  whose  opposite 
sides  are  equal,  to  find  a  point  E  such  that  EA,  EC  are  equal,  and 
EB,  ED  are  equal,  then  AEG,  BED  are  straight  lines. 

4.  If  it  be  possible  within  a  quadrilateral  ABGD,  whose  opposite 
sides  are  equal,  to  find  a  point  E,  such  that  EA,  EB,  EC,  ED  are 
equal,  then  the  quadrilateral  is  equiangular. 


46  BOOK  I. 


PKOPOSITION   15. 

If  two  straight  lines  intersect,  vertically  opposite  angles  are 
equal. 

Let  the  two  straight  lines  AB,  CD  intersect  at  E : 
it  is  required  to  prove  that  the  angle  AEG  is  equal  to  the 
angle  DEB^  and  the  angle  CEB  to  the  angle  AED. 


Proof.  The  sum  of  the  angles  CEA,  AED,  which  AE 
makes  with  CD  on  one  side  of  it,  is  equal  to  two  right 
angles.  (Prop.  13.) 

Again,  the  sum  of  the  angles  AED,  DEB,   which  DE 
makes  with  ^i^  on  one  side  of  it,  is  equal  to  two  right 
angles. 
Therefore  the  sum  of  the  angles  CEA,  AED  is  equal  to  the 

sum  of  the  angles  AED,  DEB. 
From  each  of  these  equals  take  away  the  common  angle 
AED; 

then  the  angle  CEA  is  equal  to  the  angle  DEB. 
Similarly  it  may  be  proved  that  the  angle  CEB  is  equal 
to  the  angle  AED. 

Wherefore,  if  two  straight  lines  &c. 


PROPOSITION  15.  47 


EXEECISES. 

1.  If  tlie  diagonals  of  a  quadrilateral  bisect  one  another,  oppo- 
site sides  are  equal. 

2.  In  a  given  straight  line  find  a  point  such  that  the  straight 
lines,  joining  it  to  each  of  two  given  points  on  the  same  side  of  the 
line,  make  equal  angles  with  it. 

3.  A,  B  are  two  given  points;  CD,  DE  two  given  straight 
lines :  find  points  P,  Q  in  CD,  DE,  such  that  AP,  PQ  are  equally 
inclined  to  CD,  and  PQ,  QB  equally  inclined  to  DE. 

4.  A  straight  line  is  drawn  terminated  by  one  of  the  sides  of  an 
isosceles  triangle,  and  by  the  other  side  produced,  and  bisected  by  the 
base:  prove  that  the  straight  lines  thus  intercepted  between  the 
vertex  of  the  isosceles  triangle,  and  this  straight  line,  are  together 
equal  to  the  two  equal  sides  of  the  triangle. 


48  BOOK  I. 


PROPOSITION  16. 

An  exterior  miyle  of  a  triangle  is  greater  than  either  of 
the  interior  opposite  angles. 

Let  ABC  be  a  triangle,  and  let  ACD  be  the  exterior 
angle  made  by  producing  the  side  BG  to  D  : 
it  is  required  to  prove  that  the  angle  ACB  is  greater  than 
either  of  the  interior  opposite  angles  CBA^  BAG. 

Construction.       Bisect  AG  -^.t  E.  (Prop.  10.) 

Draw  BE  and  produce  it  to  F^ 

making  EF  equal  to  EB^  (Prop.  3.) 

and  draw  FG. 


Proof.     Because  in  the  triangles  AEB,  GEF^ 
AE  'w,  equal  to  GE^ 
and  EB  to  EF, 
and  the  angle  AEB  is  equal  to  the  angle  GEF, 

the  triangles  are  equal  in  all  respects ;   (Prop.  4.) 
therefore  the  angle  BAE  (or  BAG)  is  equal  to  the  angle  FGE. 
Now  the  angle  EGD  (or  AGD)  is  greater  than  the  angle 
EGF. 
Therefore  the  angle  ACD  is  greater  than  the  angle  i^^C. 
Similarly  it  can  be  proved  that  the  angle  BGG^  which  is 
made  by  producing  AG  and  is  equal  to  the  angle  AGD,  is 
greater  than  the  angle  ABG 

Wherefore,  an  exterior  angle  <fec. 


PROPOSITION  16.  49 


EXEECISES. 

1.  Only  one  perpendicular  can  be  drawn  to  a  given  straight  line 
from  a  given  point  without  it. 

2.  Shew  by  joining  the  angular  point  ^  of  a  triangle  to  any  point 
in  the  opposite  side  BC  between  B  and  G  that  the  angles  ABG^  BCA 
are  together  less  than  two  right  angles. 

3.  Not  more  than  two  equal  straight  lines  can  be  drawn  from  a 
given  point  to  a  given  straight  line. 

4.  Prove  by  the  method  of  superposition  that,  if  a  quadrilateral  be 
equiangular,  its  opposite  sides  are  equal. 

.5.  Prove  by  the  method  of  superposition  that  two  right-angled 
triangles,  which  have  their  hypotenuses  equal  and  one  side  equal  to 
one  side,  are  equal  in  all  respects. 


T.  E. 


50  •  BOOK  L 


PROPOSITION   17. 


The  su7n  of  any  two  angles  of  a  triangle  is  less  than  two 
right  angles. 

Let  ABC  be  a  triangle  : 
it  is  required  to  prove  that  the  sum  of  any  two  of  its  angles 
is  less  than  two  right  angles. 

Construction.     Produce  any  side  BC  to  1). 


Proof.     Because   ACD   is    an    exterior   angle   of    the 
triangle  ABC, 

it  is  greater  than  the  interior  opposite  angle  ABC. 

(Prop.  16.) 
To  each  of  these  unequals  add  the  angle  ACB\ 
then  the  sum  of  the  angles  ACD,  ACB  is  greater  than  the 

sum  of  the  angles  ABC,  ACB. 
But  the  sum  of  the  angles  A  CD,  A  CB  is  equal  to  two  right 
angles.  (Prop.  13.) 

Therefore  the  sum  of  the  angles  ABC,  ACB  is  less  than  two 

right  angles. 
Similarly  it  can    be  proved   that    the   sum    of    the  angles 
BAC,  ACB  is  less  than  two  right  angles; 

and  also  the  sum  of  the  angles  CAB,  ABC. 

Wherefore,  the  sum  of  any  two  angles  &c. 


PROPOSITION  17.  51 

The  theorem  established  in  this  proposition  may  be  stated  thus : 
If  from  two  points  B,  G  in  the  straight  line  BG  two  straight  lines  be 
drawn  which  meet  at  any  point  A,  then  the  sum  of  the  angles  ABG, 
AGB  is  less  than  two  right  angles. 


B  O 

We  shall  assume  as  a  postulate  the  converse  of  this  theorem, 
which  may  be  stated  thus 

If  from  two  points  B,G  va.  the  straight  line  BG  two  straight  lines 
BP,  CQ  be  drawn  making  the  sum  of  the  angles  PBG,  QGB  on  the 
same  side  of  BG  less  than  two  right  angles,  the  two  lines  BP,  GQ  will 
meet  if  produced  far  enough. 

It  may  be  observed  that  the  theorem  established  in  Proposition  17 
proves  that  the  lines  PB,  QG  cannot  meet  when  produced  beyond  B 
and  G :  if  therefore  the  postulate  just  stated  be  allowed,  it  follows 
that  the  lines  JSP,  GQ  must  meet  when  produced  beyond  P  and  Q. 

The  postulate  which  we  here  assume  may  be  stated  in  general 
terms  as  follows 

Postulate  9.  If  the  sum  of  the  two  interior  angles, 
which  two  straight  liiies  make  with  a  given  straight  line  on 
the  same  side  of  it,  be  not  equal  to  two  right  angles,  the  two 
straight  lines  are  not  parallel. 


EXEECISES. 

1.  A  triangle  must  have  at  least  two  acute  angles. 

2.  Assuming  Postulate  9,  prove  that  any  two  straight  lines 
drawn  at  right  angles  to  two  given  intersecting  straight  lines  must 
intersect. 

3.  Prove  that  a  straight  line  drawn  at  right  angles  to  a  given 
straight  line  must  intersect  all  straight  lines  which  are  not  at  right 
angles  to  the  given  straight  line. 


4—2 


52  BOOK  I, 


PROPOSITION   18. 

When  two  sides  of  a  triangle  are  unequal,  the  greater 
side  has  the  greater  angle  opposite  to  it. 

Let  ABC  be  a  triangle,  of  which  the  side  AC  i^  greater 
than  the  side  AB  : 

it  is  required  to  prove  that  the  angle  ABC  is  greater  than 
the  angle  ACB. 

Construction.     From  AC  the  greater  of  tlie  two  sides 
cut  oiY  AD  equal  to  AB  the  less.  (Prop.  3.) 

Draw  BD. 


Proof.     Because    ABB   is   an    exterior   angle   of   the 
triangle  BDC, 

it  is  greater  than  the  interior  opposite  angle  DCB. 

(Prop.  16.) 

And  because  AB  is  equal  to  AD, 

the  angle  y(/)7>  is  equal  to  the  angle  ADD.    (Prop.  5.) 

Therefore  the  angle  ABD  is  greater  than  the  angle  ACB. 

But   the    angle    ABC   is   greater    than   the   angle   ABD; 

therefore  the  angle  ABC  is  greater  than  the  angle  ACB. 

Wherefore,  when  two  sides  &c. 


PROPOSITION  18.  53 

ADDITIONAL  PEOPOSITION. 

The  straight  lines  draicn  at  right  angles  to  the  sides  of  a  triangle 
at  their  middle  points  meet  in  a  point. 

Let  ABC  be  a  triangle  and  D,  E,  F  the  middle  points  of  the 
sides  jBC,  CA,  AB. 

Draw  EO,  FO*  at  right  angles  to  CA,  AB. 
Draw  OA,  OB,  OC,  OD. 


Because  in  the  triangles  AEO,  GEO, 
AE  is  equal  to  CE,  EO  common,  and  the  angle  AEO  is  equal  to  the 
angle  CEO, 

the  triangles  are  equal  in  all  respects;  (Proi?.  4.) 

therefore  ^0  is  equal  to  CO. 
Similarly  it  can  be  proved  that  ylO  is  equal  to  BO; 

therefore  jBO  is  equal  to  GO. 
Next  because  in  the  triangles  BOD,  COD, 

BO  is  equal  to  CO  and  BD  to  CD  and  OD  is  common, 

the  triangles  are  equal  in  all  respects ;  (Prop.  8.) 

therefore  the  angle  BDO  is  equal  to  the  angle  CDO, 
and  01)  is  at  right  angles  to  BC. 

Wherefore  the  straight  line  drawn  at  right  angles  to  BC  at  its 
middle  point  D  passes  through  0,  the  intersection  of  the  straight  lines 
drawn  at  right  angles  to  the  other  two  sides  at  their  middle  points. 

EXEKCISES. 

1.  ABC  is  a  triangle  and  the  angle  A  i^  bisected  by  a  straight  line 
which  meets  BC  at  D;  shew  that  BA  is  greater  than  BD,  and  CA 
greater  than  CD. 

2.  Prove  that,  if  D  be  any  point  in  the  base  BC  between  B  and 
C  of  an  isosceles  triangle  ABC,  AD  is  less  than  AB. 

3.  Prove  that,  if  AB,  AG,  AD  be  equal  straight  lines,  and  AG 
fall  within  the  angle  BAD,  BD  is  greater  than  either  BC  or  CD. 

4.  ABCD  is  a  quadrilateral  of  which  AD  is  the  longest  side  and 
BC  the  shortest ;  shew  that  the  angle  ABC  is  greater  than  the  angle 
ADC,  and  that  the  angle  BCD  is  greater  than  the  angle  BAD. 

5.  If  the  angle  C  of  a  triangle  be  equal  to  the  sum  of  the  angles  A 
and  B,  the  side  AB  is  equal  to  twice  the  straight  line  joining  G  to  the 
middle  point  of  AB. 

*  "We  assume  that  the  straight  lines  drawn  at  right  angles  to  CA, 
AB  at  E  and  F  meet.     (See  Exercise  2,  page  51.) 


54  BOOK  L 


PROPOSITION  19. 

When  two  angles  of  a  triangle  are  unequal,  the  greater 
angle  has  the  greater  side  opposite  to  it. 

Let  ABC  be  a  triangle,    of    which   the  angle  ABC  is 
greater  than  the  angle  ACB  : 

it  is  required  to  prove  that  the  side  ^C  is  greater  than  the 
side  AB. 


Proof.     AG  must   be   either  less    than,    equal  to,    or 
greater  than  AB. 

li  AC  were  less  than  AB, 

the  angle  ABC  would  be  less  than  the  angle  -4C7i;  (Prop.  18.) 

but  it  is  not ; 

therefore  ^C  is  not  less  than  AB. 

li  AC  were  equal  to  AB, 

the  angle  ABC  would  be  equal  to  the  angle  ACB  ;  (Prop.  5.) 

but  it  is  not ; 

therefore  ^C  is  not  equal  to  AB. 

Therefore  AC  must  be  greater  than  AB. 

Wherefore,  when  two  angles  &c. 


PBOPOSITIOJV  19.  55 

We  leave  it  to  the  student  to  prove  that,  while  a  point  P  is 
moving  along  a  straight  line  XY,  the  distance  OP  of  the  point  P 
from  a  fixed  point  O  outside  the  line  is  decreasing  when  P  is  moving 
towards  H  the  foot  of  the  perpendicular  from  0  on  the  line,  and  that 
OP  is  increasing  when  P  is  moving  away  from  H.     Assuming  the 


^J"' 


truth  of  this  proposition,  it  follows  that  OH  is  less  than  each  of 
the  two  straight  lines  OP^,  OP^  where  Pj,  P.j  are  two  positions  of  the 
point  P  close  to  H  on  either  side  of  it.  For"  this  reason  we  say  that 
OH  is  a  mlTilTmim  value  of  OP. 

In  the  same  way,  if  a  geometrical  quantity  vary  continuously,  its 
magnitude  in  a  position,  where  it  is  greater  than  in  the  positions 
close  to  it  on  either  side,  is  called  a  maximum  value. 

It  will  be  seen  that,  if  a  quantity  vary  continuously,  there  must 
be  between  any  two  equal  values  of  the  quantity  at  least  one  maximum 
or  minimum  value. 


EXEKCISES. 

1.  Prove  that  the  hypotenuse  of  a  right-angled  triangle  is  greater 
than  either  of  the  other  sides. 

2.  The  base  of  a  triangle  is  divided  into  two  parts  by  the  perpen- 
dicular from  the  opposite  vertex :  prove  that  each  part  of  the  base  is 
less  than  the  adjacent  side  of  the  triangle. 

3.  A  straight  line  drawn  from  the  vertex  of  an  isosceles  triangle 
to  any  point  in  the  base  produced  is  greater  than  either  of  the  equal 
sides. 

4.  If  D  be  any  point  in  the  side  BG  of  a  triangle  ABC,  then  the 
greater  of  the  sides  AB,  AG  ia  greater  than  AD. 

5.  The  perpendicular  is  the  shortest  straight  line  which  can  be 
drawn  from  a  given  point  to  a  given  straight  line ;  and,  of  any  two 
others,  that  which  makes  the  smaDer  angle  with  the  perpendicular 
is  the  shorter. 

6.  The  base  of  a  triangle  whose  sides  are  unequal  is  divided  into 
two  parts  by  the  straight  line  bisecting  the  vertical  angle :  prove  that 
the  greater  part  is  adjacent  to  the  greater  side. 


56  BOOK  I. 


PROPOSITION   20. 

The  sum  of  any  two  sides  of  a  tria7igle  is  greater'  than 
tJie  third  side. 

Let  ABC  be  a  triangle  : 
it  is  required  to  prove  that  the  sum  of  any  two  sides  of  it 
is  greater  than  the  third  side; 

namely,  the  sum  of  CA,  AB  greater  than  BG ', 
the  sum  of  AB,  BC  greater  than  CA  ; 
the  sum  of  BC,  CA  greater  than  AB. 
Construction.     Produce  any  side  BA  to  D, 

making  AD  equal  to  AC.  (Prop,  3.) 

Draw  DC. 


Proof.     Because  AC  i^  equal  to  AD, 

the  angle  ADC  is  equal  to  the  angle  A  CD.      (Prop.  5.) 

But  the  angle  BCD  is  greater  than  the  angle  ACD. 

Therefore  the  angle  BCD  is  greater  than  the  angle  BDC. 

And  because  in  the  triangle  BCD, 

the  angle  BCD  is  greater  than  the  angle  BDC ; 

BD  is  greater  than  BC.  (Prop.  19.) 

Now  because  DA  is  equal  to  A  C, 
BD,  which  is  the  sum  of  BA,  AD,  is  equal  to  the  sum  of 
CA,  AB. 

Therefore  the  sum  of  CA,  AB  is  greater  than  BC. 
Similarly  it  can  be  proved  that  the  sum  of  AB,  BC 
is  greater  than  CA ;  and  that  the  sum  of  BC,  CA  is  greater 
than  AB. 

Wherefore,  the  sura  of  any  two  sides  &c. 


PROPOSITION  20.  57 

The  result  of  this  proposition  enables  us  to  solve  a  great  number 
of  problems,  of  which  the  following  is  a  specimen — To  find  in  a  given 
straight  line  XY  a  jioint  P  such  that  the  sum  of  its  distances  PA,  PB 
from  two  given poi7its  A,  B  is  a  minimum. 


If  the  points  ^,  7i  be  on  opposite  sides  of  AT,  the  straight  line 
AB  intersects  AT  in  the  point  required. 

If  ^,  5  be  on  the  same  side  of  AT,  draw  AH  perpendicular  to 
XY;   produce  AH  to  C,  so  that  HC  is  equal  to  HA. 
Take  any  point  P  in  AT. 

Draw  BDC,  DA,  PA,  PB,  PC.     Then  it  is  easily  proved  that  AP 
is  equal  to  CP,  and  AD  to  CD. 

Therefore  the  sum  of  AP,  PB  is  equal  to  the  sum  of  CP,  PB, 
and  this  is  a  minimum  when  P  coincides  with  D.     (Prop.  20.) 
Therefore  D  is  the  point  required. 
From  the  diagram  it  is  seen  that  the  angle  BDY  is  equal  to  the 
angle  CDX,  which  is  equal  to  the  angle  ADX. 
It  appears  therefore  that  when  the  sum  of  PA,  PB  is  a  minimum, 
the  lines  PA,  PB  make  equal  angles  with  XY. 


EXERCISES. 

1.  Prove  that  any  three  sides  of  any  quadrilateral  are  greater  than 
the  fourth  side. 

2.  If  D  be  any  point  within  a  triangle  ABC,  the  sum  of  DA,  DB, 
DC  is  greater  than  half  the  perimeter  of  the  triangle. 

3.  The  sum  of  the  four  sides  of  any  quadrilateral  is  greater  than 
the  sum  of  its  two  diagonals, 

4.  In  a  convex  quadrilateral  the  sum  of  the  diagonals  is  greater 
than  the  sum  of  either  pair  of  opposite  sides. 

5.  D  is  the  rdiddle  point  oi  BC  the  base  of  an  isosceles  triangle 
ABC,  and  E  any  point  in  AC.  Prove  that  the  difference  of  BD,  DE 
is  less  than  the  difference  of  AB,  AE. 

6.  The  two  sides  of  a  triangle  are  together  greater  than  twice  the 
straight  line  drawn  from  the  vertex  to  the  middle  point  of  the  base. 

7.  Find  in  a  given  straight  line  a  point  such  that  the  difference 
of  its  distances  from  two  fixed  points  is  a  maximum. 


58  BOOK  I. 


PROPOSITION  21. 


If  from  the  ends  of  the  side  of  a  triangle  there  he  drawn 
two  straight  lines  to  a  point  within  the  triangle,  the  sum 
of  these  lines  is  less  titan  the  sum  of  the  other  two  sides  of 
the  triangle,  but  they  contain  a  greater  angle. 

Let  ABC  be  a  triangle;  and  from  B,  C,  the  ends  of  the 
side  BC,  let  the  two  straight  lines  BJ),  CD  be  drawn  to 
a  point  D  within  the  triangle  : 
it  is  required  to  prove  that  the  sum  of  BD,  DC  is  less  than 

the  sum  of  BA,  AC,  but  the  angle  BDC  is  greater  than 

the  angle  BAC. 

Construction.     Produce  BD  to  meet  AC  n.t  E. 


Proof.     The  sum  of  the  two  sides  BA,  AE  oi  the  tri- 
angle BAE  is  greater  than  the  third  side  BE.        (Prop.  20.) 

To  each  of  these  unequals  add  EC ; 
then  the  sum  of  BA,  ^C  is  greater  than  the  sum  of  BE,  EC. 
Again,  the  sum  of  the  two  sides  CE,  ED  of  the  triangle 
CED  is  greater  than  the  third  side  CD. 

To  each  of  these  unequals  add  DB; 
then  the  sum  of  CE,  EB  is  greater  than  the  sum  of  CD,  DB. 
And  it  has  been  proved  that  the  sum  of  BA,  AC  is  greater 

than  the  sum  of  BE,  EC ; 
therefore  the  sum  of  BA,  AC  is  greater  than  the  sum  of 

BD,  DC. 
Again,   the  exterior  angle  BDC  of   the  triangle  CDE  is 
greater  than  the  interior  opposite  angle  CED.  (Prop.  16.) 
And  the  exterior  angle  CEB  of  the  triangle  ABE  is  greater 
than  the  interior  opposite  angle  BAE; 
therefore  the  angle  BDC  is  greater  than  the  angle  BAC. 
Wherefore,  if  from  the  ends  &c. 


P BO  POSITION  21.  59 


EXEKCISES. 


1.  If  D  be  any  point  within  a  triangle  ABC,  the  sum  of  I) A,  DB^ 
DC  is  less  than  the  perimeter  of  the  triangle  and  greater  than  half 
the  perimeter. 

2.  Prove  that  the  perimeter  of  a  triangle  is  less  than  the  peri- 
meter of  any  triangle  which  is  drawn  completely  surrounding  it. 

3.  If  two  triangles  have  a  common  base  and  equal  vertical  angles, 
the  vertex  of  each  triangle  lies  outside  the  other  triangle. 

4.  If  from  the  angles  of  a  triangle  ABC,  straight  lines  AOD, 
BOE,  COF  be  drawn  through  a  point  0  within  the  triangle  to  meet 
the  opposite  sides,  the  perimeter  of  the  triangle  ABC  is  greater  than 
two-thirds  of  the  sum  of  AD,  BE,  CF. 

5.  ABD,  ACD  are  two  triangles  on  the  same  side  of  AD  in  which 
^C  is  greater  than  AB.  Prove  that,  if  the  angles  ABD,  ACD  be  both 
right  angles  or  be  equal  obtuse  angles,  then  BD  is  greater  than  DC. 


60 


BOOK  I. 


PROPOSITION  22. 

'fo   coixstruct  a   triangle  having  its  sides  equal  to  three 
given  straight  Ihies. 

Let  AB,  CD,  EF  be  the  three  given  lines  : 
it  is  required  to  construct  a  triangle  whose  sides  are  equal 
to  AB,  CD,  EF, 

Construction.    Produce  one  of  the  given  lines  CD  both 
ways, 

and  cut  off  CG  equal  to  AB,  (Prop.  3.) 

and  DR  to  EF. 
With  C  as  centre  and  CG  as  radius  describe  the  circle  GKL, 
and  with  D  as  centre  and  Dll  as  radius  describe  the  circle 
HKM. 

Let  these  circles  intersect  in  K : 

Drawee,  DK: 

then  CKD  is  a  triangle  drawn  as  required. 


Proof.     Because  G  is  the  centre  of  the  circle  GKL, 
CK  is  equal  to  CG ; 
and  CG  is  equal  to  AB.  (Constr.) 

Therefore  CK  is  equal  to  AB. 
Again,  because  D  is  the  centre  of  the  circle  HKM, 
D  K  is  equalto  D  H ; 
and  DH  is  equal  to  EF.  (Constr.) 

Therefore  DK  is  equal  to  EF. 
Therefore  the  three  lines  KC,  CD,  DK  are  equal  to  the 
,  three  AB,  CD,  EF  respectively. 

Wherefore,  the  triangle  KCD  has  been  constructed  having 
its  sides  equal  to  the  three  given  straight  lines  AB,  CD,  EF. 


PROPOSITION  22.  61 

It  may  l?e  observed  that  it  is  not  possible  to  construct  a  triangle 
which  shall  have  its  sides  equal  to  any  three  given  straight  lines.  In 
Proposition  20  it  has  been  proved  that  any  two  sides  of  a  triangle  are 
together  greater  than  the  third  side.  It  follows  therefore  that  it  is 
impossible  to  construct  a  triangle  having  its  sides  equal  to  three 
given  straight  lines,  except  when  the  given  straight  lines  are  such 
that  amj  two  of  them  are  greater  than  the  third  or  the  greatest 
line  is  less  tJian  the  sum  of  the  other  tioo. 

We  see  therefore  that  in  this  proposition  we  have  to  solve  a  pro- 
blem, which  admits  of  solution  only  when  the  given  lines  satisfy  a 
certain  condition. 

We  shall  meet  with  many  other  problems  in  which  the  geometrical 
quantities  given  in  the  i)roblem  (for  that  reason  generally  called  the 
data),  must  satisfy  some  condition  in  order  that  the  problem  may 
admit  of  solution.  It  will  be  a  useful  exercise  for  the  student  to 
investigate  such  conditions  when  they  exist. 


EXEECISES. 

1.  Prove  that  the  two  circles  drawn  in  the  construction  of  Pro- 
position 22  will  always  intersect,  provided  that  the  sum  of  any  two 
of  the  given  straight  lines  is  greater  than  the  third. 

2.  How  many  different  shaped  triangles  could  be  made  of  8  dif- 
ferent lines  whose  lengths  are  respectively  2,  2,  2,  3,  3,  4,  4,  5  inches? 

3.  Construct  a  right-angled  triangle,  having  given  the  hypotenuse 
and  one  side. 

4.  Construct  a  quadrilateral  equal  in  all  respects  to  a  given 
quadrilateral. 


62  BOOK  I. 

PROPOSITION  23. 

From  a  given  point  in  a  given  straight  line  to  draw  a 
straight  line  making  with  the  given  straight  line  an  angle 
equal  to  a  given  angle. 

Let  ABC  be  the  given  straight  line,  B  the  given  point 
in  it,  and  DBF  the  given  angle  : 

it  is  required  to  draw  from  B  a  straight  line  making  with 
ABC  an  angle  equal  to  the  angle  DEF. 

Construction.     In  ED^  EF  take  any  points  G^  H^ 

and  draw  GH. 

From  BC  cut  oiff  BK  equal  to  EH,       (Prop.  3.) 

and  construct  the  triangle  LBK, 

having  the  side  BK  equal  to  EH, 

BL  equal  to  EG, 

and  KL  equal  to  HG  :  (Prop.  22.) 

then  BL  is  a  straight  line  drawn  as  required. 


Proof.     Because  in  the  triangles  BLK,  EGH, 
KB  is  equal  to  HE, 
BL  to  EG, 
and  LK  to  GH, 
the  triangles  are  equal  in  all  respects ;   (Prop.  8. ) 
therefore  the  angle   KBL  (or  CBL^  is  equal  to  the  angle 
HEG  (or  FED). 
"Wherefore,  from  the  given  point  B  in  the  given  straight 
line  ABC  a  straight  line  BL  has  been  drawn  7naking  with 
the  straight  line  ABC   an  angle  KBL  equal  to  the  given 
angle  FED. 


PROPOSITION  23.  63 


EXEKCISES. 

1.  Construct  a  triangle,  having  given  the  base  and  each  of  the 
angles  at  the  base. 

2.  Make  an  angle  double  of  a  given  angle. 

3.  If  one  angle  of  a  triangle  be  equal  to  the  sum  of  the  other 
two,  the  triangle  can  be  divided  into  two  isosceles  triangles. 

4.  Construct  a  triangle,  having  given  the  base,  one  of  the  angles 
at  the  base,  and  the  sum  of  the  sides. 


64 


BOOK  I. 


PROPOSITION  24. 

If  two  sides  of  one  triangle  he  equal  to  two  sides  of 
anotJier  and  the  angle  contained' hy  the  two  sides  of  the  one 
he  greater  than  the  angle  contained  hy  the  two  sides  of  the 
other,  the  third  side  of  the  one  is  greater  than  the  third  side 
of  the  other. 

Let  ABC,  DEF  be  two  triangles,  in  which  AB  is  equal 
to  DE  and  AC  to  DF,  and  the  angle  BAC  is  greater  than 
the  angle  EDF : 

it  is  required  to  prove  that  the  third  side  BC  is  greater 
than  the  third  side  EF. 
Construction.     Of  the  two  sides  DE,  DF  let  DF  be 
one  which  is  not  less  than  the  other.    From  the  point  D  in 
the  straight  line  DE,  draw  DC  making  with  BE 

the  angle  EDG  equal  to  the  angle  BAC,    (Prop.  23.) 
and  make  DG  equal  to  DF.  (Prop.  3.) 

Draw  EG  meeting  DF  in  //. 


Proof.     Because  DF  is.  not  less  than  DE, 

and  DG  is  equal  to  DF, 

DG  is  not  less  than  DE. 

And  because  in  the  triangle  DEG, 

DG  is  not  less  than  DE, 

the  angle  DEG  is  not  less  than  the  angle  DGE. 

(Props.  5  and  18.) 
Next,  because  DHG  is  the  exterior  angle  of  the  triangle  i>-E'^, 
it  is  greater  than  the  interior  opposite  angle  DEG. 

(Prop.  16.) 
Therefore  the  angle  DUG  is  greater  than  the  angle  DGH. 
And  because  in  the  triangle  DHG, 
the  angle  DHG  is  greater  than  the  angle  DGH, 

DG  is  greater  than  DH.  (Prop.  19.) 


PROPOSITION  24.  65 

But  DG  is  equal  to  DF. 

Therefore  DF  is  greater  than  DH, 

or  the  point  F  lies  outside  the  triangle  DEG. 

Next  because  the  sum  of  DH,  HG,  two  sides  of  the  triangle 

DHG,  is  greater  than  the  third  side  DG, 

and  the  sum  of  FII,  HE,  two  sides  of  the  triangle  ElIF^ 

is  greater  than  the  third  side  EF ; 

the  sum  of  DH,  HG,  FH,  HE  is  greater  than  the  sum 

of  DG,  EF; 

i.e.  the  sum  of  DF,  EG  is  greater  than  the  sum  of  DG,  EF. 

Take  away  the  equals  DF,  DG ; 

then  EG  is  greater  than  EF. 

Now  the  triangles  EDG,  BAG  are  equal  in  all  respects. 

(Prop.  4.) 
Therefore  JBC,  which  is  equal  to  EG,  is  greater  than  EF. 
Wherefore,  if  two  sides  (fee. 


EXEECISES. 

A  point  P  moves  along  the  circumference  of  a  circle  from  one 
extremity  A  of  a,  diameter  AB  to  the  other  extremity  B ;  prove  that 
throughout  the  motion 

(a)  AP  is  increasing  and  BP  is  decreasing  ; 

(b)  if  0  be  any  point  in  AB  nearer  A  than  i>,  OP  is  increasing  ; 

(c)  if  0  be  any  point  in  BA  produced,  OP  is  increasing. 


T.  E. 


BOOK  I 


PROPOSITION  25. 

If  two  sides  of  one  triangle  be  equal  to  two  sides  of 
another^  and  the  third  side  of  the  one  he  greater  than  the 
third  side  of  the  other,  the  angle  opposite  to  the  third  side 
of  the  one  is  greater  than  the  angle  opposite  to  the  third  side 
of  the  other. 

Let  ABC,  DEF  be  two  triangles,  in  which  AB  i's,  equal 
to  DE,  and  AG  to  DF,  and  BC  is  greater  than  EF : 
it  is  required  to  prove  that  the  angle  BA  C  is  greater  than 
the  angle  EDF. 


Proof.     The  angle  BAG  must  be  either 
greater  than,  equal  to,  or  less  than  the  angle  EDF. 
If  the  angle  BAG  were  equal  to  the  angle  EDF, 

BG  would  be  equal  to  EF',  (Prop.  4.) 

but  it  is  not ; 
therefore  the  angle  BAG  is  not  equal  to  the  angle  EDF. 
Again,  if  the  angle  BAG  were  less  than  the  angle  EDF, 
BG  would  be  less  than  EF;         (Prop.  24.) 
but  it  is  not ; 
therefore  the  angle  BAG  is  not  less  than  the  angle  EDF. 
Therefore  the  angle  BAG  is  greater  than  the  angle  EDF. 
Wherefore,  if  two  sides  &c. 


PROPOSITION-  25.  67 


EXEKCISES. 

1.  If  D  be  the  middle  point  of  the  side  BC  oi  &  triangle  ABC, 
in  which  ^O  is  greater  than  AB,  the  angle  ADC  is  an  obtuse  angle. 

2.  If  in  the  sides  AB,  AC  of  a  triangle  ABC,  in  which  AC  is 
greater  than  AB,  points  D,  E  be  taken  such  that  BD,  CE  are  equal, 
CD  is  greater  than  BE. 

3.  If  in  the  sides  AB,  AC  produced  of  a  triangle  ABC,  in  which 
AC  IB  greater  than  AB,  points  D,  E  be  taken  such  that  BD,  GE 
are  equal,  BE  is  greater  than  CD. 

4.  If  in  the  side  AB  and  the  side  AC  produced  of  a  triangle  ABC 
points  D  and  E  be  taken,  such  that  BD,  CE  are  equal,  BE  is  greater 
than  CD. 


5—2 


68  BOOK  I. 

PROPOSITION  26.     Part  1. 

If  two  triangles  have  two  angles  of  the  one  equal  to  tioo 
angles  of  the  other ^  and  the  side  adjacent  to  the  angles  in  the 
one  equal  to  the  side  adjacent  to  the  angles  in  the  other,  the 
triangles  are  equal  in  all  respects. 

Let  ABC,  DEF  be  two  triangles,  in  which  the  angle 
ABC  is  equal  to  the  angle  DEF,  and  the  angle  BCA  is 
equal  to  the  angle  EFD,  and  the  side  BG  adjacent  to  the 
angles  ABC,  BCA  is  equal  to  the  side  EF  adjacent  to  the 
angles  DEF,  EFD : 

it  is  required  to  prove  that  the  triangles  ABC,  DEF  are 
equal  in  all  respects. 


Proof.     Because  the  sides  BC,   EF  are   equal,  it  is 
possible  to  shift  the  triangle  ABC, 
so  that  BC  coincides  with  EF,  B  with  E  and  C  with  F, 

(Test  of  Equality,  page  5.) 
and  the  triangles  are  on  the  same  side  of  EF. 
If  this  be  done, 

because  BC  coincides  with  EF, 
and  the  angle  ABC  is  equal  to  the  angle  DEF, 
BA  must  coincide  in  direction  with  ED. 
Similarly  it  may  be  proved    that    CA    must   coincide 
in  direction  with  FD. 

Therefore  the  point  A,  which  is  the  intersection  of  BA,  CA, 

must  coincide  with  D,  which  is  the  intersection  of  ED,  FD. 

Next,  because  A  coincides  with  D,  and  B  with  E, 

AB  must  coincide  with  DE.  (Post.  2.) 

Similarly  AG  must  coincide  with  DF. 


PROPOSITION  26.     PART  1.  69 

Therefore  the  triangle  ABC  coincides  with  the  triangle  DEF, 
and  is  equal  to  it  in  all  respects. 
Wherefore,  if  two  triangles  &c. 


EXEKCISES. 

1.  If  AD  be  the  bisector  of  the  angle  BAG,  and  BDG  be  drawn 
at  right  angles  io  AB,  AB  is  equal  to  AG. 

2.  AB,  AG  are  any  two  straight  lines  meeting  at  A  :  through  any 
point  P  draw  a  straight  line  meeting  them  at  E  and  F,  such  that  AE 
may  be  equal  to  AF. 

3.  If  upon  the  same  base  AB  two  triangles  BAG,  ABB  be  con- 
structed, having  the  angle  BAG  equal  to  ABB,  and  ABG  equal  to 
BAD,  then  the  triangles  BDG,  AGD  are  equal  in  all  respects. 

4.  If  the  opposite  sides  of  a  quadrilateral  be  equal,  the  diagonals 
bisect  each  other. 

5.  If  the  straight  line  bisecting  the  vertical  angle  of  a  triangle  be 
at  right  angles  to  the  base,  the  triangle  is  isosceles. 


70  BOOK  I. 

PROPOSITION   26.     Part  2. 

If  two  triangles  have  two  angles  of  the  one  equal  to  two 
angles  of  the  other ^  and  the  sides  opposite  to  a  pair  of  equal 
angles  equal,  the  triangles  are  equal  in  all  respects. 

Let  ABC,  DEF  be  two  triangles,  in  which  the  angle 
ABC  is  equal  to  the  angle  DEF,  and  the  angle  BOA  equal 
to  the  angle  EFD,  and  BA  the  side  opposite  to  the  angle 
BCA  is  equal  to  ED  the  side  opposite  to  the  angle  EFD : 
it  is  required  to  prove  that  the  triangles  ABC,  DEF  are 
equal  in  all  respects. 


Proof.     Because  the  sides   AB,   DE  are  equal,   it  is 
possible  to  shift  the  triangle  DEF, 
so  that  DE  coincides  with  AB,  D  with  A  and  E  with  By 
and  the  triangles  are  on  the  same  side  of  AB. 
If  this  be  done, 

because  ED  coincides  with  BA, 
and  the  angle  DEF  is  equal  to  the  angle  ABC, 

EF  must  coincide  in  direction  with  BC. 
Now  F  cannot  coincide  with  any  point  G  in  BC, 
since  the  angle  AGB  the  exterior  angle  of  the  triangle  AGO 
is  greater  than  the  interior  and  opposite  angle  ACB, 

(Prop.  16.) 

which  is  equal  to  the  angle  DFE. 

Again  F  cannot  coincide  with  any  point  H  in  BC  produced, 

since  the  interior  and  opposite  angle  AHB  of  the  triangle 

ACH  is  less  than  the  exterior  angle  ACB,        (Prop.  16.) 

which  is  equal  to  the  angle  DFE. 


PROPOSITION  26.     PART  2.  71 

Therefore  F  must  coincide  with  6', 

EF  with  BC,  and  DF  with  ^(7; 

therefore  the  triangle  DEF  coincides  with  the  triangle  ABC, 

and  is  equal  to  it  in  all  respects. 

Wherefore,  if  two  triangles  kc. 

ADDITIONAL  PEOPOSITION. 
The  straight  lines,  which  bisect  the  angles  of  a  triangle,  meet  in  a 
point. 

Let  ABC  be  a  triangle. 

Bisect  the  angles  ABC,  BCA  by  the  straight  lines  BI,  CI*. 

Draw  IL,  IM,  IN  perpendicular  to  the  sides. 


Because  in  the  triangles  IBN,  IBL 

the  angle  IBN  is  equal  to  the  angle  IBL, 

and  the  angle  INB  to  the  angle  ILB, 

and  BI  is  common, 

the  triangles  are  equal  in  all  respects  :  (Prop.  26,  Part  2.) 

therefore  IN  is  equal  to  IL. 

Similarly  it  can  be  proved  that  IM  is  equal  to  IL  : 

therefore  IN  is  equal  to  131. 

Next  because  in  the  right-angled  triangles  IAN,  lAM 

the  hypotenuse  lA  is  common, 

and  IN  is  equal  to  IM, 

the  triangles  are  equal  in  all  respects :  (Exercise  5,  page  49.) 

therefore  the  angle  IAN  is  equal  to  the  angle  I  AM, 

and  I  A  is  the  bisector  of  the  angle  BAG. 

Therefore  the  bisector  of  the  angle  BA  C  passes  through  the  intersection 

of  the  bisectors  of  the  angles  ABC,  BCA. 

EXERCISES. 

1.  The  perpendiculars  let  fall  on  two  sides  of  a  triangle  from 
any  point  in  the  straight  line  bisecting  the  angle  between  them  are 
equal  to  each  other. 

2.  In  a  given  straight  line  find  a  point  such  that  the  perpendicu- 
lars drawn  from  it  to  two  given  straight  lines  which  intersect  are  equal. 

3.  Through  a  given  point  draw  a  straight  line  such  that  the  per- 
pendiculars on  it  from  two  given  points  may  be  on  opposite  sides  of 
it  and  equal  to  each  other. 

*  It  is  assumed  that  these  lines  intersect. 


72  BOOK  I. 

PROPOSITION  26  A. 

If  two  triangles  have  two  sides  equal  to  two  sides,  and 
the  angles  opposite  to  07ie  p)ciir  of  equal  sides  equals  the  angles 
opposite  to  the  other  pair  are  either  equal  or  supplementary. 

Let  ABC,  DEF  be  two  triangles,  in  which  AB  is  equal 
to  DE,  and  BC  to  EF,  and  the  angle  BAG  is  equal  to  the 
angle  EDF : 

it  is  required  to  prove   that  the  angles  ACB,   DEE  are 

either  equal  or  supplementary.  (Def.  page  45.) 

Of  the  two  sides  AG,  DE,  let  AG  h^  not  greater  than  BE. 

B 

Fig.  1. 


Proof.     Because  the  sides  AB,   DE  are  equal,   it  is 
possible  to  shift  the  triangle  ABC, 

so  that  AB  coincides  with  DE, 
A  with  D  and  B  with  E,     (Test  of  Equality,  page  5.) 
and  so  that  the  two  triangles  ABC,  DEF  are  on  the  same 
side  of  DE. 
If  this  be  done, 

because  AB  coincides  with  DE, 

and  the  angle  BAG  is  equal  to  the  angle  EDF, 

AG  must  coincide  in  direction  with  DE. 

Because  AG  is,  not  greater  than  DE, 

G  must  coincide  either  (1)  with  F  or  (2)  with  G  some  point 

iuDF. 

(Fig   1.)     If  G  coincide  with  F, 

then  BC  coincides  with  EF,  (Post.  2.) 

and  the  triangle  ABC  with  the  triangle  DEF, 
and  the  two  triangles  are  equal  in  all  respects ; 
therefore  the  angle  A  GB  is  equal  to  the  angle  DFE. 

(Fig.  2.)     Again,  if  G  coincide  with  G, 
because  BC  is  equal  to  EG,  and  EF  is  equal  to  BC, 
EG  is  equal  to  EF, 


PROPOSITION-  26  A.  73 

And  because  in  the  triangle  EFG, 

EG  is  equal  to  EF, 

the  angle  EFG  is  equal  to  the  angle  EGF.     (Prop.  5.) 

Now  the  angles  DGE,  EGF  are  together  equal  to  two  right 

angles,  i.e.  are  supplementary;  (Prop.  13.) 

therefore  the  angles  DGE^  EFG  are  supplementary ; 

and  the  angle  DGE  is  equal  to  the  angle  AGB; 
therefore  the  angles  ACB,  DFE  are  supplementary. 
Wherefore,  if  two  triangles  &c. 

Corollary.  When  two  triangles  have  two  sides  equal 
to  two  sides,  and  the  angles  opposite  to  one  pair  of  equal  sides 
equal  to  one  another,  they  are  equal  in  all  respects,  provided 
that  of  the  angles  opposite  to  the  second  pair  of  equal  sides, 

(1)  each  be  less  than  a  right  angle, 

(2)  each  be  greater  than  a  right  angle, 
or    (3)     one  of  them  be  a  right  angle. 


EXEKCISES. 

1.  If  the  straight  line  bisecting  the  vertical  angle  of  a  triangle 
also  bisect  the  base,  the  triangle  is  isosceles. 

2.  If  two  given  straight  lines  intersect,  and  a  point  be  taken 
equally  distant  from  each  of  them,  it  lies  on  one  or  other  of  the  two 
straight  lines  which  bisect  the  angles  between  the  given  straight 
lines. 

3.  Prove  that  two  right-angled  triangles  are  equal  in  all  respects, 
if  the  hypotenuse  and  a  side  of  the  one  be  respectively  equal  to  the 
hypotenuse  and  a  side  of  the  other. 

4.  If  two  exterior  angles  of  a  triangle  be  bisected,  and  from  the 
point  of  intersection  of  the  bisecting  lines  a  straight  line  be  drawn 
to  the  third  angle,  it  bisects  that  angle. 

5.  If  two  triangles  have  two  sides  equal  to  two  sides,  and  the 
angles  opposite  to  the  greater  sides  equal,  the  triangles  are  equal 
in  all  respects. 

6.  Construct  a  triangle  having  given  two  sides  and  the  angle 
opposite  to  one  of  them.    Is  this  always  possible  ? 


74  BOOK  I. 


On  Equal  Triangles. 

It  is  in  many  cases  convenient  to  denominate  the  sides 
BC,  CA,  AB  oi  a  triangle  ABC  by  the  small  letters  a,  6,  c 
respectively.  Here  a,  b,  c  stand  for  the  sides  of  the  triangle 
opposite  to  the  angles  A,  B,  C  respectively. 

Using  this  notation  we  may  sum  up  the  results  of 
Propositions  4,  8,  26  Part  1,  26  Part  2,  and  26  A  as  follows  : 

Two  triangles  ABC^  A'B'C  are  equal  in  all  respects, 

(I)  if  a  =  a\     h^h\  and  C=C",  (Prop.  4.) 

(II)  if  a  =  a',     h  =  h',  and  c  =  c\   (Prop.  8.) 

(III)  if  ^  =  ^',  B  =  B',  and  c  =  c',  (Prop.  26,  Part  1.) 
(IY)if^=^',  B  =  B\s,iid  a  =  a\  (Prop.  26,  Part  2.) 
(Y)     a  a  =  a'j    b  =  b',  and  A  =  A'j  and  if  in  addition 

(1)  -5  and  B'  be  each  less  than  a  right  angle, 
or  (2)  B  and  B'  be  each  greater  than  a  right  angle, 
or  (3)  either  B  or  B'  be  a  right  angle.         (Prop.  26  A.) 

The  six  quantities,  the  angles  A,  B,  C,  and  the  sides 
a,  b,  c,  are  often  denominated  the  parts  of  the  triangle 
ABC. 

It  will  be  observed  that  the  equality  of  three  pairs  of 
parts  is  always  required  to  ensure  the  equality  in  all  respects 
of  two  triangles,  but  that  the  equality  of  three  pairs  of 
parts  is  not  always  sufficient. 

By  the  theorem  of  Proposition  32  it  can  be  shewn  that, 
if  any  two  of  the  equations  A  =  A',  B  =  B\  C  =  C\  be  true, 
the  third  is  also  true  :  from  this  we  conclude  that  the  set 
of  equations  A=A',  B^B\  C  =-C\  is  insufficient  to  deter- 
mine the  equality  of  the  triangles,  and  that  the  two  cases 
III.  and  IV.  are  virtually  the  same. 


75 


On  the  angles  made  by  one  straight  line  with  two 

OTHERS. 

When    a    straight    line   ABGD    intersects    two   other 
straight  lines  EBF,  GO II, 


the  angles  ABE,  ABF,  DCG,  BCH  outside  the  two  lines 

EFj  Gil  are  called  exterior  angles  ; 
the  angles   GBB,  CBF,  BCG,  BCH  inside  the  two  lines 

FF,  GH  are  called  interior  angles ; 
a  pair  of  interior  angles  on  opposite  sides  of  ABGD  are 

called  alternate  angles. 
There  are  two  pairs  of  alternate  angles  in  the  diagram,  EBC,  BCH; 

CBF,  BCG. 

A  pair  of  angles,  one  at  B  and  the  other  at  C,  one  exterior 
and  the  other  interior,  on  the  same  side  of  ABGD  are 
sometimes  called  corresponding  angles. 

There  are  four  pairs  of  corresponding  angles,  ABF,  BCH;  ABE,  BCG ; 
BCH,  CBF;  and  DCG,  CBE,  the  first  angle  in  each  pair  being 
an  exterior  angle,  and  the  second  the  interior. 


76  BOOK  1. 


PROPOSITION  27. 

If  a  straight  line,  meeting  two  other  straight  lines  in  the 
same  plane,  make  two  alternate  angles  equal,  the  two  straight 
lines  are  parallel. 

Let  the  straight  line  EF,  meeting  the  two  straight 
lines  AB,  CD  in  the  same  plane,  make  the  alternate  angles 
AEi\  EFD  equal  to  one  another : 

it  is  required  to  prove  that  AB,  CD  are  parallel. 


Proof.  AB,  CD  cannot  meet  when  produced  beyond 
B  and  D ;  for  if  they  did,  the  exterior  angle  AEF  of  the 
triangle  formed  by  them  and  EF  would  be  greater  than 
the  interior  opposite  angle  EFD  (Prop.  16.); 
but  it  is  not. 
Similarly  it  can  be  proved  that  AB,  CD  cannot  meet 
when  produced  beyond  A  and  C. 

But  those  straight  lines  in  the  same  plane  which  do  not 
meet  however  far  they  may  be  produced  both  ways,  are 
parallel.  (Del  9.) 

Therefore  AB,  CD  are  parallel. 
Wherefore,  if  a  straight  lifie  &c. 


PROPOSITION  27.  77 


EXEKCISES. 

1.  No  two  straight  lines  drawn  from  two  angles  of  a  triangle 
and  terminated  by  the  opposite  sides  can  bisect  one  another. 

2.  Two  straight  lines  at  right  angles  to  the  same  straight  line 
are  parallel. 

3.  Prove  Proposition  27  by  the  method  of  superposition. 


78  BOOK  L 

PROPOSITION  28. 

If  a  straight  line  intersecting  two  other  straight  lines, 
'make  an  exterior  angle  equal  to  the  interior  and  opposite 
angle  on  the  same  side  of  the  line ;  or  if  it  make  two  interior 
angles  on  the  same  side  together  equal  to  two  right  angles, 
the  two  straight  lines  are  parallel. 

Let  the  straight  line  EF,  intersecting  the  two  straight 
lines  AB,  CD, 

(1)  make  the  exterior  angle  EGB  equal  to  the  interior 
and  opposite  angle  on  the  same  side  GHD, 
or     (2)  make  the  interior  angles  on  the  same  side  BGH, 
GHD  together  equal  to  two  right  angles  : 
it  is  required  to  prove  that  AB,  CD  are  parallel. 


Proof.     (1)  Because  the  angle  EGB  is  equal  to  the 
angle  GHD, 

and  the  angle  EGB  is  equal  to  the  angle  AGH,    (Prop.  15.) 
the  angle  AGH  is  equal  to  the  angle  GHD  ; 
and  they  are  alternate  angles ; 
therefore  AB,  CD  are  parallel.        (Prop.  27.) 
(2)  Because  the  angles  BGH,  GHD  are  together  equal 
to  two  right  angles, 

and  the  angles  AGH,  BGH  are  together  equal  to  two  right 
angles,  (Prop.  13.) 

the  angles  AGH,  BGH  are   together  equal  to  the  angles 
BGH,  GHD. 

Take  away  the  common  angle  BGH; 
then  the  angle  AGH  is  equal  to  the  angle  GHD ; 
and  they  are  alternate  angles ; 
therefore  AB,  CD  are  parallel.        (Prop.  27.) 
Wherefore,  if  a  straight  line  &c. 


PROPOSITION  28.  79 


EXEECISES. 

1.  If  a  straight  line  intersecting  two  other  straight  lines  make 
two  external  angles  on  the  same  side  of  the  line  together  equal  to 
two  right  angles,  the  two  straight  lines  are  parallel. 

2.  If  a  straight  line  intersecting  two  other  straight  lines  make  two 
external  angles  on  opposite  sides  of  the  line  equal,  the  two  straight 
lines  are  parallel. 

3.  If  a  straight  line  intersecting  two  other  straight  lines  make 
two  corresponding  angles  equal,  the  two  straight  lines  are  parallel. 


80  BOOK  I. 

PROPOSITION  29. 

If  a  straight  line  intersect  two  parallel  straight  lines,  it 
makes  alternate  angles  equal,  it  makes  each  exterior  angle 
equal  to  the  interior  and  opposite  angle  on  the  same  side  of 
the  liiie,  and  it  also  makes  interior  angles  on  the  same  side 
together  equal  to  two  right  angles. 

Let   the   straight   line  EF  intersect  the  two  parallel 
straight  lines  AB,  CD  : 
it  is  required  to  prove  that 

(1)  the  alternate  angles  AGff,  GHD  are  equal, 

(2)  the  exterior  angle  EGB  is  equal  to  the  interior 

and  opposite  angle  GHD  on  the  same  side  of  EF, 

and  (3)  the  two  interior  angles  BGH,  GHD  on  the  same 

side  of  EF  are  together  equal  to  two  right  angles. 


Proof.     (1)    Because  AGTT,  BGII  are  the  angles  which 
EF  makes  with  AB  on  one  side  of  it, 

the  sum  of  the  angles  AGH,  BGII  is  equal  to  two  right 
angles.  (Prop.  13.) 

Therefore,  if  the  angles  AG II,  GHD  were  unequal, 
the  sum  of  the  angles  BGH,  GHD  would  not  be  equal  to 

two  right  angles ; 
and  since  these  are  the  interior  angles  which  the  straight 
lines  AB,  CD  make  with  EF  on  one  side  of  it, 

AB,  CD  would  not  be  parallel.     (Post.  9,  page  51.) 
But  AB,  CD  are  parallel ; 
therefore  the  angle  A  Gil  is  equal  to  the  angle  GHD. 

(2)  But  the  angle  AGH  is  equal  to  the  angle  EGB; 

(Prop.  15.) 
therefore  the  angle  EGB  is  equal  to  the  angle  GHD. 

(3)  Add  to  each  of  the  equal  angles  EGB,  GHD  the 
angle  BGII; 


PROPOSITION  29.  81 

then   the  angles  EGB^   BGH  are   together   equal  to   the 
angles  BGH,  GHD. 

But  the  angles  EGB,  BGH  are  together  equal  to  two  right 

angles. 
Therefore  the  angles  BGH^  GHD  are  together  equal  to  two 

right  angles. 

Wherefore,  if  a  straight  line  (fee. 

Corollary.     All   the   angles   of  a   rectangle  are   right 
angles.  (See  Def.  19.) 


EXEKCISES. 

1.  Any  straight  line  parallel  to  the  base  of  an  isosceles  triangle 
makes  equal  angles  with  the  sides. 

2.  If  through  any  point  equidistant  from  two  parallel  straight 
lines,  two  straight  lines  be  drawn  cutting  the  parallel  straight  lines, 
they  will  intercept  equal  portions  of  these  parallel  straight  lines. 

3.  If  the  straight  line  bisecting  an  exterior  angle  of  a  triangle 
be  parallel  to  a  side,  the  triangle  is  isosceles. 

4.  If  DE,  DF  drawn  from  D  any  point  in  the  base  ^C  of  an 
isosceles  triangle  ABC,  to  meet  AB,  AG  in  E,  F  he  parallel  to  AG, 
AB,  the  perimeter  of  the  parallelogram  AEDF  is  constant. 


T.  E. 


82 


BOOK  I. 


PROPOSITION  30. 

Straight   lines  parallel   to    the   same    straight   line   are 
parallel  to  each  other. 

Let  each  of  the  straight  lines  AB,  CD  be  parallel  to  EF : 
it  is  required  to  prove  that  AB^  CD  are  parallel  to  one 
another. 
Construction.     Draw  a  straight  line  GHK  intersecting 
AB,  CD,  EFin  G,  H,  K  respectively. 


Proof.     Because  GHK  intersects  the  parallels  AB,  EF^ 
the  angle  GKF  is  equal  to  the  angle  AGH.  (Prop.  29.) 
Again,  because  GK  intersects  the  parallels  CD,  EF, 

the  angle  GHD  is  equal  to  the  angle  GKF.   (Prop.  29.) 
Therefore  the  angle  A  GH  is  equal  to  the  angle  GHD ; 
and  they  are  alternate  angles ; 
therefore  AB  is  parallel  to  CD.     (Prop.  27.) 
Wherefore,  straight  lines  &c. 


PROPOSITION  30.  83 


EXERCISES. 

1.  Two  intersecting  straight  lines  cannot  both  be  parallel  to 
the  same  straight  line. 

2.  Only  one  straight  line  can  be  drawn  through  a  given  point 
parallel  to  a  given  straight  line.  » 

3.  If  two  straight  lines,  each  of  which  is  parallel  to  a  third 
straight  line,  meet,  the  two  lines  are  coincident  throughout  their 
length. 

4.  If  a  straight  line  intersect  one  of  two  parallel  straight  lines, 
it  must  intersect  the  other. 


G— 2 


84  BOOK  L 


PROPOSITION  31. 

To  draw  through  a  given  point  a  straight  line  parallel 
to  a  given  straight  line. 

Let  A  be  the  given  point,  and  BC  the  given  straight 
line: 

it  is  required  to  draw  through  A  a  straight  line   parallel 

to  BO, 

Construction.     In  BO  take  any  point  D,  and  draw 

AD ;  from  the  point  A  in  the  straight  line  ^i>  on  the  side 

of  AD  remote  from  0  draw  A£J  making  the  angle  DAE 

equal  to  the  angle  ADO ;  (Prop.  23.) 

and  produce  the  straight  line  UA  to  F : 

then  EF  ig  the  straight  line  required. 


Proof.     Because  the  straight  line  AD  meets  the  two 
straight  lines  BO,  FF, 

and  makes  the  alternate  angles  FAD,  ADO  equal, 

FF  is  parallel  to  BO.  (Prop.  27.) 

Wherefore,  the  straight  line  FAF  has  been  drawn  through 
the  given  point  A,  parallel  to  the  given  straight  line  BO. 


PROPOSITION  31.  85 


EXERCISES. 

1.  Find  a  point  5  in  a  given  straight  line  CD,  such  that,  if  AB 
be  drawn  to  B  from  a  given  point  Ay  the  angle  ABC  will  be  equal 
to  a  given  angle. 

2.  Draw  through  a  given  point  between  two  intersecting  straight 
lines  a  straight  line  so  that  it  is  bisected  at  the  point. 

3.  ABCD  is  a  quadrilateral  having  BG  parallel  to  AD ;  shew  that 
its  area  is  the  same  as  that  of  the  parallelogram  which  can  be  formed 
by  drawing  through  the  middle  point  of  DC  a  straight  line  parallel 
to  AB. 

4.  AG,  BG  are  two  given  straight  lines:  it  is  required  to  draw  a 
straight  line  from  a  given  point  P  to  AC,  so  that  it  is  bisected  hyBG. 

5.  Construct  a  triangle  having  given  two  angles,  and  the  length 
of  the  perpendicular  from  the  third  angle  on  the  opposite  side. 

6.  Construct  a  right-angled  triangle,  having  given  one  side  and 
the  angle  opposite. 


86  BOOK  I. 

PROPOSITION  32. 

An  exterior  angle  of  a  triangle  is  equal  to  the  sum  of 
the  two  interior  opposite  angles ;  and  the  sum  of  the  three 
interior  angles  of  a  triangle  is  equal  to  two  right  angles. 

Let  ABC  be  a  triangle  : 

it  is  required  to  prove  that  (1)  the  exterior  angle  ACD 
made  by  producing  the  side  BG  is  equal  to  the  sum  of 
the  two  interior  opposite  angles  CAB^  ABC, 

and  (2)  the  sum  of  the  three  interior  angles  ABC^  BCA^ 
CAB  is  equal  to  two  right  angles. 

Construction.  Through  the  point  C  draw  CE  parallel 
to^^.  (Prop.  31.) 


Proof.     (1)     Because  AC  meets  the  parallels  BA,  CE, 

the  alternate  angles  BAC,  ACE  are  equal.    (Prop.  29.) 

Again,  because  BD  meets  the  parallels  BA,  CE, 

the  exterior  angle  ECD  is  equal  to  the  interior  opposite 

angle  ABC.  (Prop.  29.) 

And  the  angle  ACE  was  proved  to  be  equal  to  the  angle 

BAC] 
therefore  the  whole  angle  ACD  is  equal  to  the  sum  of  the 
two  angles  CAB,  ABC. 
(2)     To  each  of  these  equals  add  the  angle  BCA  ; 
then  the  sum  of  the  angles  ACE,  ACB  is  equal  to  the  sum 

of  the  three  angles  ABC,  BCA,  CAB. 
But  the  sum  of  the  angles  ACE,  ACB  is  equal  to  two  right 
angles;  (Prop.  13.) 

therefore  also  the  sum  of  the  angles  ABC,  BCA,  CAB  is 
equal  to  two  right  angles. 
Wherefore,  An  exterior  angle  &lq. 


FEOPOSITIO.Y  32.  87 

Corollary.  2%e  sum  of  the  interior  angles  of  any  con- 
vex rectilineal  figure  of  n  sides  is  less  hy  four  right  angles 
than  2n  right  angles. 

This  may  be  proved  in  either  of  the  following  ways  : 
In  fig.  (1),  where  straight  lines  are  drawn  from  any  point  0  within 

the  figure  to  the  vertices,  the  angles  of  the  n  triangles  so  formed 

are  equal  to  the  angles  of  the  figure  together  with  the  angles  at  0, 

which  are  equal  to  four  right  angles. 
In  fig.  (2),  where  all  the  diagonals  from  one  vertex  E  are  drawn,  the 

angles  of  the  w  -  2  triangles  so  formed  are  together  equal  to  the 

angles  of  the  figure. 

B 

(2) 


D 


EXEKCISES. 

1.  Straight  lines  AD,  BE,  CF  are  drawn  within  the  triangle 
ABC  making  the  angles  DAB,  EBC,  FGA  all  equal  to  one  another. 
If  AD,  BE,  CF  do  not  meet  in  a  point,  the  angles  of  the  triangle 
formed  by  them  are  equal  to  those  of  the  triangle  ABC. 

2.  Trisect  a  right  angle. 

3.  Trisect  a  quarter  of  a  right  angle. 

4.  If  A  be  the  vertex  of  an  isosceles  triangle  ABC,  and  BA  be 
produced  to  D,  so  that  AD  is  equal  to  BA,  and  DC  be  drawn:  then 
BCD  is  a  right  angle. 

5.  A  straight  line  drawn  at  right  angles  to  BC  the  base  of  an 
isosceles  triangle  ABC  cuts  AB  in  D  and  CA  produced  in  E:  prove 
that  AED  is  an  isosceles  triangle. 

6.  Construct  a  right-angled  triangle  having  given  the  hypotenuse 
and  the  sum  of  the  sides. 

7.  The  line  joining  the  right  angle  of  a  right-angled  triangle  to 
the  middle  point  of  the  hypotenuse  is  equal  to  half  the  hypotenuse. 

8.  The  locus  of  the  vertices  of  all  right-angled  triangles  which 
have  a  common  hypotenuse  is  a  circle. 


88  BOOK  I. 


PROPOSITION  33. 

If  two  sides  of  a  convex   quadrilateral  he   equal  and 
parallel,  the  other  sides  are  equal  and  parallel. 

Let  ABDG  be  a  quadrilateral,  in  which  the  sides  AB, 
CD  are  equal  and  parallel : 

it  is  required  to  prove  that  the  sides  AC,  BD  are  equal  and 
parallel. 

Construction.     Draw  one  of  the  diao^onals  BC. 


Proof.     Because  ^^  is  parallel  to  CD, 

and  BC  meets  them, 

the  alternate  angles  ABC,  BCD  are  equal.      (Prop.  29.) 

Because  in  the  triangles  ABC,  DCB, 

AB  is  equal  to  DC, 

and  BC  to  CB, 

and  the  angle  ABC  to  the  angle  DCB, 

the  triangles  are  equal  in  all  respects;    (Prop.  4.) 

therefore  the  angle  ACB  is  equal  to  the  angle  DBC, 

and  CA  to  BD. 
And  because  the  straight  line  BC  meets  the  two  straight 
lines  AC,  BD,  and  makes  the  alternate  angles  ACB,  CBD 
equal  to  one  another, 

ACis  parallel  to  BD.  (Prop.  27.) 

And  it  was  proved  to  be  equal  to  it. 
Wherefore,  if  two  sides  &c. 


PROPOSITION  33.  89 


EXEECISES. 

1.  Draw  a  straight  line  so  that  the  part  intercepted  between  two 
given  straight  lines  is  equal  to  one  given  straight  line  and  parallel  to 
another. 

2.  If  a  quadrilateral  have  two  of  its  opposite  sides  parallel,  and 
the  two  others  equal  but  not  parallel,  any  two  of  its  opposite  angles 
are  together  equal  to  two  right  angles. 

3.  If  a  straight  line  which  joins  the  extremities  of  two  equal 
straight  lines,  not  parallel,  make  the  angles  on  the  same  side  of  it 
equal  to  each  other,  the  straight  line  which  joins  the  other  extremities 
will  be  parallel  to  the  first. 

4.  If  from  D  any  point  in  the  base  BC  of  an  isosceles  triangle 
ABC,  BE,  DF  be  drawn  perpendicular  to  the  sides,  then  the  sum  of 
DE,  DF  is  constant. 


90  BOOK  I. 

PROPOSITION  34. 

Opposite  sides  of  a  parallelogram  are  equal,  and  opposite 
angles  are  equal ;  and  a  diagonal  of  a  parallelogram  bisects 
its  area. 

Let    ACDB   be    a    parallelogram,  of    which   BC    is    a 
diagonal : 
it  is  required  to  prove  that  (1)  opposite  sides  are  equal,  AB 

to  CD,  and  AC  to  BD ; 
(2)  opposite  angles  are  equal,  B  AC  to  BDC  and  A  BD  to  ACD ; 
aiid  (3)  the  diagonal  BC  bisects  the  area  of  the  parallelogram. 


Proof.      Because  ^^  is  parallel  to  CJJ,  and  BC  meets 
them, 

the  alternate  angles  ABC,  BCD  are  equal.     (Prop.  29.) 

And  because  ^C  is  parallel  to  BD,  and  BC  meets  them, 

the  alternate  angles  ACB,  CBD  are  equal.    (Prop.  29.) 

Now  because  in  the  two  triangles  ABC,  DCB, 

the  angle  ABC  is  equal  to  the  angle  DCB, 

and  the  angle  BCA  to  the  angle  CBD, 

and  the  side  BC  adjacent  to  the   equal  angles  in  each  is 

common  to  both, 

the  triangles  are  equal  in  all  respects.     (Prop.  26,  Part  1.) 

Therefore  ^^  is  equal  to  DC,  AC  equal  to  DB, 

and  the  angle  BAC  equal  to  the  angle  CDB. 

And  because  the  angle  ABC  is  equal  to  the  angle  DCB, 

and  the  angle  CBD  to  the  angle  BCA, 

the  whole  angle  ABD  is  equal  to  the  whole  angle  DCA, 

And  the  angle  BAC  has  been  proved  to  be  equal  to  the 

angle  CDB. 
Therefore  in  the  parallelogram  AD  (1)  opposite  sides  are 
equal  and  (2)  opposite  angles  are  equal. 


PROPOSITION  34.  91 

Again,  it  has  been  proved 
that  the  triangles  ABC,  DCB  are  equal  in  all  respects: 
therefore  (3)  the  diagonal  BC  bisects  the  area  of  the  paral- 
lelogram AD. 
Wherefore,  opposite  sides  tkc. 

Corollary  1.     All  the  sides  of  a  square  are  equal. 

Corollary  2.  The  angles  made  hy  a  pair  of  straight 
lines  are  equal  to  the  angles  made  hy  any  pair  of  straight 
lines  parallel  to  them. 

A  parallelogram  ABCD  is  often  spoken  of  as  the  parallelogram  AG, 
or  the  parallelogram  BD,  or  more  simply  as  -40  or  BD,  when  there  is 
no  danger  of  confusion  with  the  diagonal  ^C  or  with  the  diagonal  BD. 


EXERCISES. 

1.  Prove  that,  if  the  diagonals  of  a  quadrilateral  bisect  one 
another,  the  quadrilateral  is  a  parallelogram.  Prove  also  the  converse. 

2.  If  two  sides  of  a  quadrilateral  be  parallel  and  the  other  two 
equal  but  not  parallel,  the  diagonals  are  equal. 

3.  If  in  a  quadrilateral  the  diagonals  be  equal  and  two  sides 
be  parallel,  the  other  sides  are  equal. 

4.  Find  in  a  side  of  a  triangle  the  point  from  which  straight  lines 
drawn  parallel  to  the  other  sides  of  the  triangle  and  terminated  by 
them  are  equal. 

5.  Prove  that  every  straight  line  which  bisects  the  area  of  a 
parallelogram  must  pass  through  the  intersection  of  its  diagonals. 

6.  Construct  a  triangle  whose  angles  shall  be  equal  to  those  of  a 
given  triangle,  and  whose  area  shall  be  four  times  the  area  of  the 
given  triangle. 

7.  ABCD  is  a  parallelogram  having  the  side  ^D  double  oiAB: 
the  side  ^jB  is  produced  both  ways  to  E  and  F  till  each  produced  part 
equals  AB,  and  straight  lines  are  drawn  from  G  and  D  to  E  and  F  so 
as  to  cross  within  the  figure :  shew  that  they  will  meet  at  right 
angles. 

8.  If  0  be  any  point  within  a  parallelogram  ABGD,  the  sum  of 
the  triangles  OAB,  OCD  is  half  the  parallelogram. 

9.  Divide  a  given  straight  line  into  n  equal  parts,  where  n  is 
a  whole  number. 


92  BOOK  I. 


PROPOSITIOISr  35. 

Two  parallelograms^  which  have  one  side  common  and 
the  sides  opposite  to  the  common  side  in  a  straight  line,  are 
equal  in  area. 

Let  A  BCD,  EBGF  be  two  parallelograins,  which  have 
a  common  side  BC,  and  the  sides  AD,  EF  in  a  straight' 
line : 

it  is  required  to  prove  that  ABGD,  EBGF  are  equal  in  area. 
A     B      D      F   A  D      :E  fa  DE  F 


Peoof.     Because  ABGD  is  a  parallelogram, 

AB  is  equal  to  DG,  (Prop.  34.) 

and  because  EBGF  is  a  parallelogram, 

BE  is  equal  to  GF ; 

and  because  AB  is  parallel  to  DG, 

and  BE  to  GF, 

the  angle  ABE  is  equal  to  the  angle  DGF. 

(Prop.  34,  Coroll.  2.) 
And  because  in  the  triangles  ABE,  DGF, 
AB\^  equal  to  DG, 
and  BE  to  GF, 
and  the  angle  ABE  to  the  angle  DGF, 
the  triangles  are  equal  in  all  respects.  (Prop.  4.) 
Take  from  the  area  ABGF,  the  equal  areas  FDG,  EAB\ 
then  the  remainders  are  equal, 
that  is,  the  parallelograms  ABGD,  EBGF  are  equal  in  area. 
Wherefore,  two  parallelograms  &c. 


PROPOSITION  35.  93 

The  propositions  in  the  remaining  part  of  the  First 
Book  of  Euclid  and  those  in  the  Second  Book  relate  chiefly 
to  cases  of  equality  of  the  areas  of  two  figures. 

The  test  of  equality  to  which  we  have  hitherto  always  appealed 
has  been  that  of  the  possibility  of  shifting  one  figure  so  that  it 
exactly  coincides  with  the  other.  In  this  case  the  figures  are  equal 
in  all  respects,  but  we  say  that  two  figures  are  equal  in  area  also, 
when  it  is  possible  to  shift  all  the  parts  of  the  area  of  one  figure, 
so  that  they  together  exactly  fit  the  area  of  the  second  figure. 

It  will  be  observed  that  this  is  the  test  made  use  of  in  Pro- 
position 35. 

For  the  future  we  shall  often,  when  there  is  no  danger 
of  ambiguity,  speak  of  the  equality  of  two  figures  when  we 
mean  only  equality  of  area,  and  we  shall  often  speak  of 
a  figure  when  we  mean  only  the  area  of  the  figure. 


EXEKCISES. 

1.  Construct  a  rectangle  equal  to  a  given  parallelogram. 

2.  Construct  a  rhombus  equal  to  a  given  parallelogram. 

3.  Construct  a  parallelogram  to  be  equal  to  a  given  parallelogram 
in  area  and  to  have  its  sides  equal  to  two  given  straight  lines.  Is 
this  always  possible  ? 


94  BOOK  I. 


PROPOSITION  36. 

Two  j^cLt'cdl^logramSj  which  have  two  sides  equal  and  in 
a  straight  lirie  and  also  have  the  sides  opposite  to  the  equal 
sides  in  a  straight  line,  are  equal. 

Let  A  BCD,  EFGH  be  two  parallelograms,  which  have 
their  sides  BC,  FG  equal  and  in  a  straight  line,  and  also 
their  sides  AD,  EH  in  a  straight  line : 
it  is  required  to  prove  that  ABCD,  EFGH  are  equal. 

Construction.     Draw  BE,  GH. 


Proof.     Because  BC  is  equal  to  FG, 

and  FG  to  EH,  (Prop.  34.) 

BC  is  equal  to  EH; 
and  they  are  parallel. 
Because  the  two  sides  BC,  EH  of  the  convex  quadrilateral 
EBCH  are  equal  and  parallel, 

the  other  sides  BE,  CH  are  equal  and  parallel ; 

(Prop.  33.) 
therefore  EBCH  is  a  parallelogram. 
Now   because   EBCH  and   ABCD   have  the  side  BC 
common,  and  the  sides  AD,  EH  in  a  straight  line, 

EBCH  is  equal  to  ABCD.  (Prop.  35.) 

Similarly  it  can  be  proved  that  EBCH  is  equal  to  EFGH, 
Therefore  the  parallelograms  ABCD,  EFGH  are  equal. 
Wherefore,  two  parallelograms  &c. 


PROPOSITION  36.  95 

ADDITIONAL  PROPOSITION. 

The  straight  lines,  drawn  from  the  vertices  of  a  triangle  per- 
pendicular to  the  opposite  sides,  meet  in  a  point*. 

Let  ABC  be  a  triangle,  and  AL,  BM,  GN  be  drawn  perpendicular 
to  BG,  GA,  AB  respectively. 

Draw  the  straight  lines  FAE,  DBF,  BCD  parallel  to BC,  GA,  AB 
respectively. 


Because  BE  is  a  parallelogram, 

AEis  equal  to  BC ;  (Prop.  34.) 

and  because  GF  is  a  parallelogram, 

FA  is  equal  to  JSC; 

therefore  FA  is  equal  to  AE. 

Again,  because  AL  meets  the  parallels  FAE,  BLG, 

the  angle  FAL  is  equal  to  the  angle  ALG.      (Prop.  29.) 

But  the  angle  ^L(7  is  a  right  angle; 
therefore  the  angle  FAL  is  a  right  angle. 
Therefore  AL  is  the  straight  line  drawn  at  right  angles  to  FE  at  its 
middle  point. 
Similarly  it  can  be  proved  that  BM,  GN  are  the  straight  lines 
drawn  at  right  angles  to  FD,  BE  at  their  middle  points. 
Now  AL,  BM,  GN  the  straight  lines  drawn  at  right  angles  to  the 
sides  of  the  triangle  DEF  at  their  middle  points  meet  in  a  point. 

(Add.  Prop.,  page  53.) 
Therefore  AL,  BM,  GN  the  straight  lines  drawn  from  the  vertices 
of  the  triangle  ABG  perpendicular  to  the  opposite  sides  meet  in  a 
point. 

EXERCISES. 

1.  Construct  a  parallelogram  to  be  equal  to  a  given  parallelogram 
and  to  have  one  of  its  sides  in  a  given  straight  line. 

2.  Construct  a  parallelogram  to  be  equal  to  a  given  parallelogram 
and  to  have  two  of  its  sides  in  two  given  straight  lines. 

*  This  point  is  often  called  the  ortliocentre  of  the  triangle. 


96  BOOK  I. 

PROPOSITION  37. 

Two  triangles,  wliich  have  one  side  common  and  the 
angular  points  opposite  to  the  common  side  on  a  straight 
line  parallel  to  it,  are  equal. 

Let  ABC,  DBC  be  two  triangles,  which  have  a  common 
side  BC,  and  their  angular  points  A,  D  on  a,  straight  line 
AD  parallel  to  BG  : 

it  is  required  to  prove  that  the  triangles  ABC,  DBC  are 
equal. 

Construction.     Through  B  draw  BE  parallel  to  GA, 
and  through  C  draw  Ci^  parallel  to  BD,    (Prop.  31.) 
meeting  AD  (produced  if  necessary)  in  B  and  F. 


Proof.  Because  the  parallelograms  EBGA,  DBGF 
have  a  common  side  BG  and  the  sides  EA,  DF  in  a 
straight  line, 

EBGA  is  equal  to  DBGF.         (Prop.  35.) 
And  because  the   diagonal  AB  bisects  the  parallelogram 
EBGA, 

the  triangle  ABC  is  half  of  EBGA  ;     (Prop.  34.) 
and   because   the  diagonal  DC   bisects  the  parallelogram 
DBGF, 

the  triangle  DBG  is  half  of  DBGF. 
Now  the  halves  of  equals  are  equal. 
Therefore  the  triangles  ABC,  DBG  are  equal. 
Wherefore,  two  triangles  &c. 


PROPOSITION  37.  97 


EXEKCISES. 

1.  If  P  be  a  point  within  a  parallelogram  ABCD,  the  difference 
of  the  triangles  PAB,  PAD  is  equal  to  the  triangle  PAC. 

2.  If  P  be  a  point  outside  a  parallelogram  ABCD,  the  sum  of 
the  triangles  PAB,  PAD  is  equal  to  the  triangle  PAC. 

3.  AB  and  ECD  are  two  parallel  straight  lines:  BF,  BF  are 
drawn  parallel  to  AD,  AE  respectively:  prove  that  the  triangles ^BC, 
DBF  are  equal  to  one  another. 

4.  ABC  is  a  given  triangle:  construct  a  triangle  of  equal  area, 
having  AB  for  base  and  its  vertex  in  a  given  straight  line. 

5.  Points  A,  B,  G  are  taken,  one  on  each  of  three  parallel 
straight  lines:  BG,  GA,  AB  meet  the  lines  through  A,  B,  G  respec- 
tively in  a,  b,  c:  prove  that  each  of  the  triangles  ABG,  Abe,  BcUy 
Gab,  is  equal  to  half  the  triangle  abc. 


T.  E. 


98  BOOK  I. 


PROPOSITION  38. 

Two  triangles^  which  have  two  sides  equal  and  in  a 
straight  line  and  also  have  the  angular  points  opposite  to 
the  equal  sides  on  a  straight  line  parallel  to  it,  are  equal. 

Let  ABC,  DEF  be  two   triangles,    which  have   their 
sides   BC,  EF  equal    and   in  a    straight   line,    and   their 
angular  points  A,  D,  on  a  straight  line  AD  parallel  to  BF\ 
it  is  required  to  prove  that  the  triangles  ABC,  DEF  are 
equal. 
Construction.     Through  B  draw  BG  parallel  to  CA, 
and  through  F  draw  FII  parallel  to  ED, 
meeting  AD  (produced  if  necessary)  in  G  and  H. 

G  A  T)  H 


Proof.     Because    the    parallelograms    GBCA,    DEFH 
have  their  sides  BG,  EF  equal  and  in  a  straight  line, 
and  also  their  sides  GA,  DH  in  a  straight  line, 

they  are  equal  to  one  another.        (Prop.  36.) 

Because  the  diagonal  AB  bisects  the  parallelogram  GBCA, 

the  triangle  ABC  is  half  of  GBCA  ]     (Prop.  34.) 

and  because  the  diagonal  DF  bisects   the   parallelogram 

DEFH, 

the  triangle  DEF  is  half  of  DEFH. 
Now  the  halves  of  equals  are  equal ; 
therefore  the  triangles  ABC,  DEF  are  equal. 
Wherefore,  two  triangles  &c. 

Corollary.  Two  triangles,  which  have  two  sides  equal 
and  in  a  straight  line  and  also  have  the  angular  points 
opposite  to  the  equal  sides  coincident,  are  equal. 


PROPOSITION  38.  99 


EXEECISES. 

1.  ABCD  is  a  parallelogram;  from  any  point  P  in  the  diagonal 
BD  the  straight  lines  PA,  PC  are  drawn.  Shew  that  the  triangles 
PAB  and  PGB  are  equal  in  area. 

2.  The  three  sides  of  a  triangle  are  bisected,  and  the  points 
of  bisection  are  joined;  prove  that  the  triangle  is  divided  into  four 
triangles,  which  are  all  equal  to  one  another. 

3.  If  the  sides  BG,  CA,  AB  of  a  triangle  ABC  be  produced  to 
A',  B',  C  respectively,  so  that  CA'  =  BC,  AB'  =  CA,  AB  =  BC',  prove 
that  the  area  of  the  triangle  A'B'C  is  seven  times  that  of  the  triangle 
ABC. 

4.  Make  a  triangle  such  as  to  be  equal  to  a  given  parallelogram, 
and  to  have  one  of  its  angles  equal  to  a  given  angle. 

5.  If  the  sides  AB,  BC,  CA  of  a  triangle  ABC  be  respectively 
bisected  in  c,  a,  h,  and  Aa,  Cc  intersect  in  P:  then  BPb  is  a  straight 
line. 

6.  The  sides  AB,  AC  oi  a  triangle  are  bisected  in  D,  E:  CD, 
BE  intersect  in  F.  Prove  that  the  triangle  BFC  is  equal  to  the 
quadrilateral  ADFE. 

7.  If  AB,   PQRS,    CD   be    three   parallel    straight    lines    and^' 
P,  Q,  R,  S  be  situate  on  AC,  AD,  BC,  BD  respectively,  then  PQ 
is  equal  to  RS,  and  PR  to  QS. 

8.  A',  B',  C  are  the  middle  points  of  the  sides  of  the  triangle 
ABC,  and  through  A,  B,  C  are  drawn  three  parallel  straight  lines 
meeting  B'C,  CA',  A'B'  va.  a,  h,  c  respectively;  prove  that  the  tri- 
angle ahc  is  half  the  triangle  ABC  and  that  he  passes  through  A,  ca 
through  B,  ah  through  G. 


7—2 


100  BOOK  I. 


PROPOSITION  39. 


If  two  equal  triangles  have  a  common  side  and  lie  on 
the  same  side  of  it,  the  angular  points  opposite  to  the  common 
side  lie  on  a  straight  line  parallel  to  it. 

Let  ABC,  BBC  be  two  equal  triangles,  which  have  a 
common  side  BG,  and  lie  on  the  same  side  of  BC : 
it  is  required  to  prove  that  the  angular  points  A,  D  oppo- 
site to  the  side  BC  lie  on  a  straight  line  parallel  to  BC. 

Construction.  Draw  AD,  and  in  BD  or  BD  produced 
take  any  point  E  other  than  D,  and  draw  AE,  EC. 


Proof.    Because  the  triangle  DBC  is  not  equal  to  the 
triangle  EBC, 

and  the  triangle  ABC  is,  equal  to  the  triangle  DBC, 

the  triangle  ABC  is  not  equal  to  the  triangle  EBC. 

If  AE  were  parallel  to  BC, 

the  triangle  A  BC  would  be  equal  to  the  triangle  EBC ; 

(Prop.  37.) 
but  they  are  not  equal ; 
therefore  ^^  is  not  parallel  to  BC. 
But  it  is  possible  to  draw  a  straight  line  through  A  parallel 
to  BC;  (Prop.  31.) 

therefore  AD  is  parallel  to  BC. 
Wherefore,  if  two  equal  triangles  &c. 


PROPOSITION  39.  101 

Additional  Proposition. 

Each  side  of  a  triangle  is  double  of  the  straight  line  joining  the 
middle  points  of  the  other  sides  and  is  parallel  to  it. 

Let  ABC  be  a  triangle,  D,  E,  F  the  middle  points  of  the  sides 
BG,  GA,  AB. 

Draw  BE,  GF,  EF,  FD,  DE. 


Because  the  two  triangles  BFG,  AFG  have  their  sides  BF,  AF 
equal  and  in  a  straight  line,  and  the  point  G  common, 

the  triangles  are  equal ;  (Prop.  38,  Coroll.) 

therefore  the  triangle  BFG  is  half  of  the  triangle  ABG, 
Similarly  it  can  be  proved  that  the  triangle  BEG  is  half  of  the 
triangle  ABG. 

Therefore  the  triangle  BFG  is  equal  to  the  triangle  BEG. 

Next  because  the   triangles  BFG,  BEG  are  equal  and  have  a 

common   side  BG,   the  straight  line  FE  joining  their  vertices  is 

parallel  to  BG.  (Prop.  39.) 

Similarly  it  can  be  proved  that  DF  is  parallel  to  GA  and  ED 

to  AB. 

Again  because  BFED  is  a  parallelogram, 

BD  is  equal  to  FE.  (Prop.  34.) 

And  because  DFEG  is  a  parallelogram, 

DC  is  equal  to  FE: 
therefore  BG  is  double  of  FE. 

EXERCISES. 

1.  The  middle  points  of  the  sides  of  any  quadrilateral  are  the 
angular  points  of  a  parallelogram. 

2.  Of  equal  triangles  on  the  same  base,  the  isosceles  triangle  has 
the  least  perimeter. 

3.  Two  triangles  of  equal  area  stand  on  the  same  base  and  on 
opposite  sides:  shew  that  the  straight  line  joining  their  vertices  is 
bisected  by  the  base  or  the  base  produced. 

4.  The  triangle  ABG  is  double  of  the  triangle  EBG:  shew  that, 
if  AE,  BG  produced  if  necessary  meet  in  D,  then  AE  is  equal  to  ED. 

5.  If  the  straight  lines  joining  the  middle  points  of  two  of  the 
sides  of  a  triangle  to  the  opposite  vertices  be  equal,  the  triangle  is 
isosceles. 


102  BOOK  L 


PROPOSITION  40. 


If  two  equal  triangles  have  two  sides  equal  and  in  a 
straight  line,  and  if  the  triangles  lie  on  the  same  side  of 
this  line,  the  angular  points  opposite  to  the  equal  sides  lie 
on  a  straight  line  parallel  to  the  first  straight  line. 

Let  ABC,  DEF  be  two  equal  triangles,  which  have 
equal  sides  BC,  EF  in  a  straight  line  and  lie  on  the  same 
side  of /ii^: 

it  is  required  to  prove  that  the  angular  points  A,  D  oppo- 
site to  BC,  EF  lie  on  a  straight  line  parallel  to  BF. 

Construction.  Draw  AD,  and  in  ED  or  ED  produced 
take  any  point  G  other  than  D,  and  draw  AG,  GF. 


Proof.     Because  the  triangle  DEF  is  not  equal  to  the 
triangle  GEF, 

and  the  triangle  ABC  is  equal  to  the  triangle  DEF, 

the  triangle  ABC  is  not  equal  to  the  triangle  GEF. 

Ji  AG  were  parallel  to  BF, 

the  triangle  ABC  would  be  equal  to  the  triangle  GEF ; 

(Prop.  38.) 
but  they  are  not  equal ; 
therefore  AG  is  not  parallel  to  BF. 
But  it  is  possible  to  draw  a  straight  line  through  A  parallel 
to  BF;  (Prop.  31.) 

therefore  AD  is  parallel  to  BF. 
Wherefore,  if  two  equal  triangles  &c. 


PROPOSITION  40.  103 


Additional  Proposition. 

The  straight  lines  joining  the  vertices  of  a  triangle  to  the  middle 
-points  of  the  opposite  sides  meet  in  a  point*  which  is  for  each  line  the 
point  of  trisection  further  from  the  vertex. 

Let  ABC  be  a  triangle,  and  D,  E,  F  he  the  middle  points  of  the 
sides  BC,  CA,  AB. 

Draw  BE,  GF  and  let  them  intersect  in  G. 
Bisect  BG,  CG  in  M,  N,  and  draw  F3I,  MN,  NE. 


In  the  triangle  ABC, 

BG  is  double  of  FE  and  is  parallel  to  it.    (Add.  Prop.,  page  101.) 
In  the  triangle  GBC, 

BG  is  double  of  MN  and  is  parallel  to  it. 
Therefore  FE  is  equal  and  parallel  to  MN.      {Prop.  30.) 
Therefore  FMNE  is  a  parallelogram.  (Prop.  33.) 

Now  the  diagonals  of  a  parallelogram  bisect  each  other. 

(Exercise  1,  page  91.) 
Therefore  GE  is  equal  to  GM,  which  is  equal  to  MB. 
Therefore  BG  is  double  of  GE. 
Similarly  GG  is  double  of  GF. 
Similarly  it  can  be  proved  that  AD  passes  through  G,  and  that 
^6^  is  double  of  GD. 


EXEECISES. 

1.  A  point  P  is  taken  within  a  quadrilateral  ABCD:  prove  that, 
if  the  sum  of  the  areas  of  the  triangles  PAB,  PCD  be  independent 
of  the  position  of  P,  ABCD  is  a  parallelogram. 

2.  The  locus  of  a  point  P  such  that  the  sum  of  the  areas  of  the 
two  triangles  PAB,  PBC  is  constant,  is  a  straight  line  parallel  to  AC. 

3.  AB,  CD  are  two  given  straight  lines:  the  locus  of  a  point  P 
such  that  the  sum  of  the  two  triangles  PAB,  PCD  is  constant,  is  a 
straight  line. 

4.  Trisect  a  given  straight  line. 

*  This  point  is  often  called  the  centre  of  gravity  or  the  centroid 

of  the  triangle. 


104  BOOK  I. 


PROPOSITION  41. 

If  a  parallelogram  and  a  triangle  have  a  common  side, 
and  the  angular  point  of  the  triangle  opposite  to  the  common 
side  lie  on  the  same  straight  line  as  the  opposite  side  of  the 
parallelogram^  the  panrallelogram  is  double  of  the  triangle. 

Let  ABCD  be  a  parallelogram  and  EBG  be  a  triangle 
which  have  a  common  side  BO,  and  let  the  angular  point  E 
of  the  triangle  lie  in  the  same  straight  line  as  the  side  AD 
of  the  parallelogram  : 

it  is  required  to  prove  that  the  parallelogram  ABCD  is 
double  of  the  triangle  EBC. 
Construction.  Draw  AC. 


^  E 


Proof.      Because   the    triangles   ABC^    EBG^    have    a 
common  side  BC^  and  ^^  is  parallel  to  BC^ 

the  triangles  ABC^  EBC  are  equal.    (Prop.  37.) 
And  because  the   diagonal   AC  bisects  the  parallelogram 
ABCD, 

the  parallelogram  ABCD  is  double  of  the  triangle  ABC. 

(Prop.  34.) 
Therefore  the  parallelogram  ABCD  is  double  of  the  triangle 
EBC. 
Wherefore,  if  a  parallelogrann  &c. 


PROPOSITION  41.  105 


EXERCISES. 

1.  ABCD  is  a  parallelogram;  from  D  draw  any  straight  line  DFG 
meeting  jBC  at  F  and  AB  produced  at  G;  draw  AF  and  CG :  shew 
that  the  triangles  ABF,  CFG  are  equal. 

2.  If  P  be  a  point  in  the  side  AB,  and  Q  a  point  in  the  side  DC 
of  a  parallelogram  ABCD,  then  the  triangles  PCD,  QAB  are  equal  in 
area. 

3.  The  area  of  any  convex  quadrilateral  is  double  that  of  the 
parallelogram  whose  vertices  are  the  middle  points  of  the  sides  of 
the  quadrilateral. 

4.  The  sides  BC,  CA,  AB  of  a  triangle  ABC  are  trisected  in  the 
points  D,  d;  E,  e;  F,  f  respectively:  prove  that  the  area  of  the 
hexagon  DdEeFf  is  two-thirds  that  of  the  triangle  ABC. 


106 


BOOK  I. 


PROPOSITION  41  A. 

To  construct  a  triangle  equal  to  a  given  rectilineal 
figure. 

Let  ABGDEFG  be  the  given  rectilineal  figure : 
it  is  required  to  construct  a  triangle  equal  to  ABGDEFG, 

Construction.  Draw  one  of  the  diagonals  AG  such 
that  with  two  adjacent  sides  of  the  figure  AB^  BC  it  forms 
a  triangle  ABG. 

Through  the  vertex  B  draw  BF  parallel  to  GA^  to  meet 
GA  produced  in  P.     Draw  PC. 


Proof.  Because  the  triangles  FAG^  BAG  have  a 
common  side  AG^  and  their  angular  points  F,  B  on  a. 
straight  line  parallel  to  ^C: 

the  two  triangles  FAG,  BAG  are  equal.    (Prop.  37.) 
Add  to  each  the  figure  AGDEFG ; 
then  the  figure  FGDEFG  is  equal  to  the  figure  ABGDEFG. 

Now  the  sides  of  the  figure  FGDEFG  are  fewer  by  one 
than  the  sides  of  the  figure  ABGDEFG ;  therefore  by  con- 
tinued application  of  this  process  we  can  construct  a  series 
of  figures  all  equal  to  the  given  figure,  the  sides  of  each 
figure  being  fewer  by  one  than  the  sides  of  the  figure  last 
preceding. 

We  shall  thus  ultimately  obtain  a  triangle  equal  to  the 
given  rectilineal  figure. 


PROPOSITION  41  A.  107 

It  will  be  seen  that  by  the  method  adopted  in  Proposition  41 A 
a  triangle  can  be  constructed  equal  to  a  given  rectilineal  figure  of 
4  sides  by  using  the  process  once,  to  a  figure  of  5  sides  by  using  it 
twice,  and  to  a  figure  of  n  sides  by  using  it  n  — 3  times. 


EXEECISES. 

1.  On  one  side  of  a  given  triangle  construct  an  isosceles  triangle 
equal  to  the  given  triangle. 

2.  On  one  side  of  a  given  quadrilateral  construct  a  rectangle 
equal  to  the  quadrilateral. 

3.  Construct  a  triangle  equal  in  area  to  a  given  convex  five-sided 
figure  ABODE :  AB  is  to  be  one  side  of  the  triangle  and  AE  the 
direction  of  one  of  the  other  sides. 

4.  Bisect  a  given  (1)  parallelogram,  (2)  triangle,  (3)  quadri- 
lateral by  a  straight  line  drawn  through  a  given  point  in  one  side  of 
the  figure. 

5.  ABCD  is  a  given  quadrilateral:  construct  a  quadrilateral  of 
equal  area,  having  AB  for  one  side,  and  another  side  on  a  given 
straight  line  parallel  to  AB. 

6.  ABCD  is  a  given  quadrilateral :  construct  a  triangle,  whose 
base  shall  be  in  the  same  straight  line  as  AB,  its  vertex  at  a  given 
point  P  in  CD,  and  its  area  equal  to  that  of  the  given  quadrilateral. 


108  BOOK  I. 

PROPOSITION  42. 

To  construct  a  parallelogram  equal  to  a  given  triangle, 
and  having  an  angle  equal  to  a  given  angle. 

Let  ABC  be  the  given  triangle,  and  D  the  given  angle : 

it  is  required  to  construct  a  parallelogram  equal  to  ABC, 

and  having  an  angle  equal  to  D. 

Construction.       Bisect  BC  at  E : 

draw  AB, 

and  from  the  point  B,  in  the  straight  line  BC,  draw  BF 

making  the  angle  CBF  equal  to  the  angle  I> ;      (Prop.  23.) 

through  A  draw  AFG  parallel  to  BC  meeting  BF  in  F, 

and  through  C  draw  CG  parallel  to  BF  meeting  A  FG  in  G : 

then  FBCG  is  a  parallelogram  constructed  as  required. 


Proof.     Because  the  opposite  sides  of  the  quadrilateral 
FBCG  are  parallel, 

FBCG  is  a  parallelogram. 
Because  the  triangles  ABB,  ABC  have  the  sides  BB,  BC  equal 
and  in  a  straight  line,  and  the  angular  point  A  common, 
the  triangle  ABB  is  equal  to  the  triangle  ABC ; 

(Prop.  38,  Coroll.) 

therefore  the  triangle  ABC  is  double  of  the  triangle  ABC. 

Because  the  parallelogram  FBCG  and  the  triangle  ABC 

have  a  common  side  BC  and  the  point  A   lies  on  the 

same  straight  line  as  the  side  FG,  (Prop.  41.) 

the  parallelogram  FBCG  is  double  of  the  triangle  ABC. 

Therefore  the  parallelogram  FBCG  is  equal  to  the  triangle 

ABC,  and  it  has  an  angle  C^^  equal  to  thfe  given  angle  D. 

Wherefore  a  parallelogram  FBCG  has  been  constructed 

equal  to  the  given  triangle  ABC,  and  having  an  angle  CBF 

equal  to  the  given  angle  J). 


PROPOSITION  42.  109 


EXERCISES. 

1.  On  one  side  of  a  given  triangle  construct  a  rectangle  equal  to 
the  triangle. 

2.  On  one  side  of  a  given  triangle  construct  a  rhombus  equal  to 
the  triangle.     Is  this  always  possible  ? 

3.  On  one  side  of  a  given  triangle  as  diagonal  construct  a  rhom- 
bus equal  to  the  triangle. 


110  BOOK  I. 


PROPOSITION  43. 


Complements  of  parallelograms  about  a  diagonal  of  a 
parallelogram,  are  equal. 

Let  ABCD  be  a  parallelogram,  of  which  AC  is  a  dia- 
gonal ;  and  EH,  GF  are  parallelograms  about  A  C ;  and 
KB,  KD  the  complements  :  (See  note  on  page  111.) 

it  is  required  to  prove  that  KB  is  equal  to  KD. 


Proof.     Because  BD  is   a   parallelogram,   and   AC  ^ 
diagonal, 

the  triangle  ABC  is  equal  to  the  triangle  ADC.   (Prop.  34.) 
Now  the  triangle  ABC  is  equal  to  the  two  triangles  AEK, 

KGC  and  the  parallelogram  KB ; 
and  the  triangle  ADC  is  equal  to  the  two  triangles  AUK, 

KFC  and  the  parallelogram  KD. 
Therefore  the  two  triangles  AEK,  KGC  and  the  parallelo- 
gram KB  are  together  equal  to  the  two  triangles  AHK, 
KFC  and  the  parallelogram  KD. 
Again,  because  EH  is  a  parallelogram  and  AK  2^,  diagonal, 
the  triangle  AEK  is  equal  to  the  triangle  AHK; 

(Prop.  34.) 
and  because  GF  is  a  parallelogram,  and  KC  a  diagonal, 
the  triangle  KGC  is  equal  to  the  triangle  KFC.    (Prop.  34.) 
Therefore  taking  away  equals  from  equals,  the  remainder, 
the  complement  KB,  is  equal  to  the  remainder,  the  com- 
plement KD. 
Wherefore,  complements  of  parallelograms  &c. 


PROPOSITION   43.  Ill 

If  through  a  point  ^  on  a  diagonal  ^C  of  a  parallelogram  ABC  I), 
straight  lines  HKG,  EKF  be  drawn  parallel  to  the  sides  AB,  BG 
respectively  to  meet  the  sides  AT),  BG,  AB,  DG  in  H,  G,  E,  F  respec- 
tively ;  then  EH,  OF  are  called  parallelograms  about  the  diagonal 
AG,  and  the  parallelograms  EG,  FH  are  called  complements  of  these 
parallelograms. 


EXERCISES. 

1.  Prove  that  in  the  diagram  of  Proposition  43,  the  following 
are  pairs  of  equal  triangles:  ABK  and  ADK;  AEG  and  AHG;  AKG 
and  AKF. 

2.  The  diagonals  of  parallelograms  about  a  diagonal  of  a  paral- 
lelogram are  parallel. 

3.  Parallelograms  about  a  diagonal  of  a  square  are  squares. 

4.  If  ABGD,  AEFG  be  two  squares  so  placed  that  the  angles 
at  A  coincide,  then  A,  F,  G  lie  on  a  straight  line. 

5.  If  through  E  a  point  within  a  parallelogram  ABGD  straight 
lines  be  drawn  parallel  to  AB,  BG,  and  the  parallelograms  AE,  EG 
be  equal,  the  point  E  lies  in  the  diagonal  BD. 


112 


BOOK  I. 


PROPOSITION   44. 

To  construct  a  'parallelogram  equal  to  a  given  'parallelo- 
gram, having  an  wiigle  equal  to  an  angle  of  the  given  paral- 
lelogram, and  having  a  side  equal  to  a  given  straight  line. 

Let  ABCD  be  the  given  parallelogram,  and  EF  the 
given  straight  line  : 

it  is  required  to  construct  a  parallelogram  equal  to  ABGB, 
having  an  angle  equal  to  the  angle  BAD,  and  having  a 
side  equal  to  EF. 
Construction.     Produce  DA  to  G,  and  make  AG  equal 
to  EF.  (Prop.  3.) 

Through  G  draw  HGK  parallel  to  AB  meeting  CB  pro- 
duced in  K.  (Prop.  31.) 

Draw  KA  and  produce  it  to  meet  CD  produced  in  L, 
and  through  L  draw  LMH  parallel  to  DAG  to  meet  BA 
produced  in  M  and  HGK  in  H  : 
then  MA  GH  is  a  parallelogram  constructed  as  required. 


Proof.  Because  LGKH  is  a  parallelogram,  KL  a  dia- 
gonal, and  MG,  BD  complements  of  parallelograms  about 
the  diagonal  KLy 

MG  is  equal  to  BD.  (Prop.  43.) 

Again,  because  the  straight  lines  BAM,  DAG  intersect  at  A, 

the  angle  MAG  is  equal  to  the  vertically  opposite  angle 

BAD.  (Prop.  15.) 

And  AGi^  equal  to  EF.  (Constr.) 

Wherefore,  a  parallelogram  MAG H  has  been  constructed 

equal  to  the  given  parallelogram  ABCD,  having  an  angle 

MAG  equal  to  the  angle  BAD  and  having  a  side  AG  equal 

to  the  given  straight  line  EF. 


FliOFOSITION  i4.  113 


EXERCISES. 

1.  On  a  given  straight  line  construct  a  rectangle  equal  to  a  given 
rectangle. 

2.  On  a  given  straight  line  construct  a  rhombus  equal  to  a  given 
triangle.     Is  this  always  possible  ? 

3.  Construct  a  rectangle  equal  to  the  sum  of  two  given  rect- 
angles. 

4.  Construct  a  rectangle  equal  to  the  difference  of  two  given 
rectangles. 


T.  E. 


114  BOOK  L 

PROPOSITION   45. 

To  construct  a  'parallelogrmn  equal  to  a  given  7'ectilineal 
figure,  having  a  side  equal  to  a  given  straight  line,  and 
having  an  angle  equal  to  a  given  angle. 

Let  A  be  tlie  given  rectilineal  figure,  B  the  given  angle, 
and  C  the  given  straight  line  : 

it  is  required  to  construct  a  parallelogram   equal  to  the 

ligure  A,  having  an  angle  equal  to   the   angle  B,   and 

liaving  a  side  equal  to  C. 

Construction.     Construct  the  triangle  DEF  equal  to 

the  figure  A.  (Prop.  41  A.) 

Construct  the  parallelogram  GIIKL  equal  to  the  triangle 

DEF,  having  the  angle  GHK  equal  to  the  angle  B. 

(Prop.  42.) 
Construct  the  parallelogram  MNPQ  equal  to  the  parallelo- 
gram GIIKL,  having  an  angle  MNP  equal  to  the  angle 
GHK,  and  having  the  side  Jl/iV  equal  to  C.      (Prop.  43.) 


■E  F  H  K      N  F 

Proof.      Because  the   triangle    DEF  is  equal  to   the 
figure  A,  (Constr.) 

and  the  parallelogram  GK  is  equal  to  the  triangle  DEF, 

(Constr.) 
and  the  parallelogram  MP  is  equal  to  the  parallelogram  GK, 

(Constr.) 

the  parallelogram  MP  is  equal  to  the  figure  A. 

Because  the  angle  GHK  is  equal  to  the  angle  B,     (Constr.) 

and  the  angle  MNP  is  equal  to  the  angle  GHK,      (Constr.) 

the  angle  MNP  is  equal  to  the  angle  B. 

And  MN  is  equal  to  C  (Constr.) 

Wherefore  a  parallelogram  MNPQ  has  been  constructed 

equal  to  the  given  rectilineal  figure  A,  having  the  side  MN 

equal  to  tlie  given  straight    line  G,  and   having  the  angle 

MNP  equal  to  the  given  angle  B. 


PROPOSITION   45.  115 


EXEECISES. 

1.  Ou  a  given  straight  line  as  diagonal,  construct  a  rhombus 
equal  to  a  given  triangle. 

2.  Construct  a  right-angled  triangle,  having  given  the  hypotenuse 
and  the  perpendicular  from  the  right  angle  on  it.  (See  Exercise  8, 
page  87.) 

3.  Construct  a  rectangle  equal  to  a  given  rectangle,  and  having 
a  diagonal  equal  to  a  given  straight  line. 

4.  Construct  a  rectangle  equal  to  the  sum  of  two  given  triangles. 


8—2 


116 


BOOK  I. 


PROPOSITION  46. 

On  a  given  straight  line  to  construct  a  square. 

Let  -4^  be  the  given  straight  line : 
it  is  required  to  construct  a  square  on  AB. 
Construction.     From  the  point  A  draw  ^C  at  right 
angles  to  ^jB;  (Prop.  11.) 

and  make  AC  equal  to  AB  ]  (Prop.  3.) 

through  B  draw  BD  parallel  to  AG,     (Prop.  31.) 
and  through  C  draw  CD  parallel  to  AB  meeting  BD  in  D : 
then  ABDC  is  a  square  constructed  as  required. 


Proof.     Because  CD  is  parallel  to  AB, 
ajid  BD  to  A  Q 
the  figure  ABDC  is  a  parallelogram.     (Def.  18.) 
Again  the  angle  CAB  is  a  right  angle  j 
therefore  the  parallelogram  ABDC  is  a  rectangle.    (Def.  19.) 
Again,  the  adjacent  sides  AC,  AB  are  equal ;     (Constr.) 
therefore  the  rectangle  ABDC  is  a  square.     (Del  20.) 
Wherefore,  ABDC  is  a  square  constructed  on  the  given 
straight  line  AB. 


PROPOSITION  46.  117 


EXEECISES. 

1.  If  two  squares  be  equal  in  area,  their  sides  are  equal. 

2.  The  squares  on  two  equal  straight  lines  are  equal  in  all 
respects. 

3.  If  in  the  sides  AB,  BC,  CD,  DA  of  a  square  points  E,  F,  G, 
H  be  taken  so  that  AE,  BF,  CG,  DH  are  equal :  then  EFGH  is  a 
square. 

4.  If  the  diagonals  of  a  quadrilateral  be  equal  and  bisect  each 
other  at  right  angles,  the  quadrilateral  is  a  square. 

5.  On  the  sides  AC,  BC  of  a  triangle  ABC,  squares  ACDE, 
BCFG  are  constructed  :  shew  that  the  straight  lines  AF  and  BD  are 
equal. 

6.  Construct  a  square  so  that  one  side  shall  be  in  a  given  straight 
line  and  two  other  sides  shall  pass  through  two  given  points. 

7.  Construct  a  square  so  that  two  opposite  sides  shall  pass 
through  two  given  points,  and  its  diagonals  intersect  at  a  third  given 
point. 

8.  Prove  that  the  straight  line,  bisecting  the  right  angle  of  a 
right-angled  triangle,  passes  through  the  intersection  of  the  diagonals 
of  the  square  constructed  on  the  outer  side  of  the  hypotenuse. 


118  BOOK  I. 


PROPOSITION  47. 


In  a  right-angled  triangle,  the  square  on  the  hypotenuse 
is  equal  to  the  sum  of  the  squares  on  the  other  sides. 

Let  ABC  be  a  right-angled  triangle,  having  the  right 
angle  BA  C : 

it  is  required  to  prove  that  the  square  on  BC  is  equal  to 
the  sum  of  the  squares  on  BA^  AC. 

Construction.  On  BC  on  the  side  away  from  A  con- 
struct the  square  BDEC,  (Prop.  46.) 
and  similarly  on  BA,  AC  construct  the  squares  BAGF, 
ACKH; 

through  A  draw  A  L  parallel  to  BD  meeting  DE  in  L  ; 

(Prop.  31.) 
and  draw  AD,  EC. 


Proof.  Because  each  of  the  angles  BAC,  BAG  is  a 
right  angle,  the  two  straight  lines  AC,  AG,  on  opposite 
sides  of  AB,  make  with  it  at  A  the  adjacent  angles 
together  equal  to  two  right  angles; 

therefore  CA  is  in  the  same  straight  line  with  AG. 

(Prop.  14.) 
The  angle  DBC  is  equal  to  the  angle  EBA, 

for  each  of  them  is  a  right  angle.    (Prop.  10  A.) 


PROPOSITION  47.  119 

Add  to  each  of  these  equals  the  angle  A  EC ; 
then  the  angle  DBA  is  equal  to  the  angle  FBC. 
And  because  in  the  triangles  ABD,  FBC, 
AB  is  equal  to  FB  and  BD  to  BC  ■ 
and  the  angle  ABD  is  equal  to  the  angle  FBC ] 
the  triangles  ABD,  FBC  are  equal  in  all  respects.    (Prop.  4.) 
Because  the  parallelogram  BL  and  the  triangle  ABD  have 
a  common  side  BD  and  A  is  in  the  same  straight  line 
as  the  side  of  BL  opposite  to  BD,  (Prop.  41.) 

the  parallelogram  BL  is  double  of  the  triangle  ABD. 
And  because  the  square  GB  and  the  triangle  FBC  have  a 
common  side  FB,  and  C  is  in  the  same  straight  line  as 
the  side  of  GB  opposite  to  FB,  (Prop.  41.) 

the  square  GB  is  double  of  the  triangle  FBC. 
Now  the  doubles  of  equals  are  equal. 
Therefore  the  parallelogram  BL  is  equal  to  the  square  GB. 
Similarly  it  can  be  proved  that  the  parallelogram  CL  is 

equal  to  the  square  HC. 
Therefore  the  whole  square  BDEC,  which  is  the  sum  of  the 
rectangles  BL,  CL,  is  equal  to  the  sum  of  the  two  squares 
GB,  HC. 
And  the  square  BDEC  is  constructed  on  BC,  and  the  squares 

GB,HConBA,AC. 
Therefore  the  square  on  the  side  BC  is  equal  to  the  sum  of 
the  squares  on  the  sides  BA,  AC. 
Wherefore,  in  a  right-angled  triangle  kc. 

EXEKCISES. 

1.  Construct  a  square  equal  to  the  difference  of  two  given  squares. 

2.  The  diagonals  of  a  quadrilateral  intersect  at  right  angles. 
Prove  that  the  sum  of  the  squares  on  one  pair  of  opposite  sides  is 
equal  to  the  sum  of  the  squares  on  the  other  pair. 

3.  If  0  be  the  point  of  intersection  of  the  perpendiculars  drawn 
from  the  angles  of  a  triangle  upon  the  opposite  sides,  the  squares  on 
OA  and  BC  are  together  equal  to  the  squares  on  OB  and  GA^  and 
also  to  the  squares  on  0(7  and  AB. 

4.  Divide  a  given  straight  line  into  two  parts  so  that  the  sum 
of  the  squares  on  the  parts  may  be  equal  to  a  given  square. 

5.  Divide  a  given  straight  line  so  that  the  difference  of  the 
squares  on  the  parts  is  equal  to  a  given  square. 


120 


BOOK  L 


The  proof  of  the  theorem  "the  square  on  the  hypotenuse 
of  a  right-angled  triangle  is  equal  to  the  sum  of  the  squares 
on  the  other  sides,"  which  we  have  given  in  the  text  of  the 
47th  proposition,  is  attributed  to  Euclid. 

Tradition  however  says  that  the  first  person  to  dis- 
cover a  proof  of  the  truth  of  the  theorem  was  Pythagoras, 
a  Greek  philosopher  who  lived  between  570  and  500  B.C. 
The  theorem  is  in  consequence  often  quoted  as  the  Theorem 
of  Pythagoras.  "What  was  the  nature  of  Pythagoras'  proof 
is  not  known. 

The  theorem  is  one  of  great  importance  and  a  large 
number  of  proofs  of  its  truth  have  been  discovered.  It 
is  advisable  that  the  student  should  be  made  acquainted 
with  some  proofs  besides  the  one  given  in  the  text. 

We  have  made  a  selection  of  live  proofs  of  the  theorem  : 
in  each  case  not  attempting  to  give  the  complete  proof, 
but  merely  giving  hints  of  the  line  of  argument  to  be 
used,  and  leaving  the  student  to  develope  it  more  fully. 

Proof  I.  Take  a  right-angled  triangle  ABC,  and  on  the  side 
AB  away  from  C  construct  the  square  ABDE,  and  on  the  hypotenuse 
^C  on  the  same  side  as  B  construct  the  square  ACFG.  From  F 
draw  FH  perpendicular  to  BD,  and  FK  perpendicular  to  ED  pro- 
duced. 


It  may  be  proved  that 

(1)  CBD  is  a  straight  line, 

(2)  G  lies  in  DJ^;, 

(3)  the  triangles  ABC,  AEG,  CHF,  GKF  are  all  equal, 

(4)  HK  is  a  square  and  equal  to  the  square  on  BC. 


PROPOSITIOK  47.  121 


Proof  II.  Take  a  right-angled  triangle  ABC  and  on  the  sides 
AB,  BC,  GA  away  from  C,  A,  B  construct  the  squares  BADE,  CBFG, 
ACHK. 

Through  L  the  intersection  of  the  diagonals  AE,  BD  of  the  square 
on  the  larger  side  AB,  draw  iHLTf.  perpendicular  to  CA  and  OLP 
parallel  to  CA . 

Take  Q,  R,  S,  T  the  middle  points  of  the  sides  AC,  CH,  HK,  KA 
of  the  square  on  the  hypotenuse. 

Through  Q,  S  draw  QUV,  SWX  parallel  to  BC,  and  through  R,  T 
draw  RVW,  TXU  parallel  to  AB. 


It  may  be  proved  that 

(1)  all    the    quadrilaterals    LMEO,   LOBN,  LNAP,    LPDM, 
AQUT,  CRVQ,  HSWR,  KTXS  are  equal  to  one  another, 

(2)  the  quadrilateral  UVWX  is  a  square, 

(3)  the  squares  CF,  UW  B.re  equal. 


122 


BOOK  7. 


Proof  III.     Take  two  equal  squares  ABCD,  EFGH. 

Take  any  point  I  in  AT),  and  measure  off  BK,  CL,  DM,  EN,  EO 
each  equal  to  AI. 

Draw  IK,  KL,  LM,  MI ;  through  N  draw  NQP  parallel  to  EF, 
and  through  0  draw  OQR  parallel  to  EH.     Draw  QF,  QH. 


E    N 


H 


M     O 


Q 


F    P 


G 


It  may  be  proved  that 

(1)  the  square  ABCD  is  divided  into  one  square  IKLM  and 
four  equal  right-angled  triangles, 

(2)  the   square  EFGH  is   divided  into   two   squares  EOQN, 
QPGR  and  four  equal  triangles, 

(3)  all  the  triangles  are  equal  to  each  other, 

(4)  the  square  IL  is  equal  to  the  sum  of  the  squares  ON,  PR, 

(5)  the  three  squares  IL,  ON,  PR  are  squares  on  the  hypotenuse 
and  on  the  sides  of  one  or  other  of  the  equal  triangles. 

Proof  IV.  Take  a  right-angled  triangle  ABC  and  on  the  hypo- 
tenuse BC  on  the  same  side  as  A  construct  the  square  BCED,  and 
on  the  sides  CA,  AB  away  from  B,  C  construct  the  squares  CAHK, 
ABFG. 

Through  A  draw  ML  AN  perpendicular  to  BC,  and  produce  EG, 
KH  to  meet  MEAN. 


G    N 


PROPOSITION-  47.  123 

It  may  be  proved  that 

(1)  D  lies  in  FG, 

(2)  E  lies  in  KH  produced, 

(3)  the   rectangle  BL,  and   the   square  AF  are   each  equal 

to  the  parallelogram  AD^ 

(4)  the   rectangle    GL   and  the  square    AK  are   each   equal 

to  the  parallelogram  AE. 

Proof  V.  Take  a  right-angled  triangle  ABG :  on  the  sides  AB, 
BC,  CA  away  from  C,  A,  B  construct  the  squares  BADE,  GBFG, 
ACHK. 

On  HK  construct  a  triangle  HLK  equal  in  all  respects  to  the 
triangle  ABC  having  HL  parallel  to  AB,  and  KL  to  CB. 

Draw  FE,  GB,  BD,  BL. 


It  may  be  proved  that 

(1)  GBD  is  a  straight  line. 

(2)  the  triangles  FBE,  GBA  are  equal, 

(3)  all  the  quadrilaterals  GFEB,   GCAD,  BCHL,  LKAB  are 
equal  to  one  another. 


124  BOOK  I. 

PROPOSITION  48. 

If  the  square  on  one  side  of  a  triangle  he  equal  to  the 
sum  of  the  squares  on  the  other  sides,  the  angle  contained 
hy  these  two  sides  is  a  right  angle. 

Let  the  square  on  BC,  one  of  the  sides  of  the  triangle 
ABC,  be  equal  to  the  sum  of  the  squares  on  the  other  sides 
BA,  AC: 
it  is  required  to  prove  that  the  angle  BAC  is  a  right  angle. 

Construction.  From  the  point  A  draw  AD  at  right 
angles  to  ^  C ;  (Prop  .11.) 

and  make  AD  equal  to  BA  ;  (Prop.  3.) 

and  draw  DC. 


Proof.     Because  DA  is  equal  to  BA, 

the  square  on  DA  is  equal  to  the  square  on  BA. 
To  each  of  these  equals  add  the  square  on  AC; 
then  the  sum  of  the  squares  on  DA,  AC  \^  equal  to  the 
sum  of  the  squares  on  BA,  AC. 

Now  because  the  angle  DAC  is  a  right  angle, 
the  square  on  DC  is  equal  to  the  sum  of  the  squares  on 
DA,  AC.  (Prop.  47.) 

And  the  square  on  BC  is  equal  to  the  sum  of  the  squares 
on  BA,  AC.  (Hypothesis.) 

Therefore  the  square  on  DC  is  equal  to  the  square  on  BC, 
and  DC  is  equal  to  BC. 
And  because  in  the  triangles  DAC,  BAC, 
DA  is  equal  to  BA, 
AC  to  AC, 
and  CD  to  CB, 
the  triangles  are  equal  in  all  respects  ;     (Prop.  8.) 


PROPOSITION  48.  125 

therefore  the  angle  DAG  is  equal  to  the  angle  BAG. 

Now  DAG  is  a  right  angle ;  (Constr.) 

therefore  BAG  is  a  right  angle. 
Wherefore,  if  the  square  ttc. 


EXEKCISES. 

1.  If  the  difference  of  the  squares  on  two  sides  of  a  triangle  be 
equal  to  the  square  on  the  third  side,  the  triangle  is  right  angled. 

2.  The  locus  of  a  point,  such  that  the  difference  of  the  squares 
on  its  distances  from  two  given  points  is  equal  to  a  given  square,  is 
a  straight  line. 

3.  Prove  by  means  of  Proposition  48  that  the  straight  lines, 
drawn  from  the  vertices  of  a  triangle  perpendicular  to  the  opposite 
sides,  meet  in  a  point. 

4.  If  the  sum  of  the  squares  on  two  opposite  sides  of  a  quadri- 
lateral be  equal  to  the  sum  of  the  squares  on  the  other  two  sides,  the 
diagonals  of  the  quadrilateral  intersect  at  right  angles. 

5.  ABCD  is  a  quadrilateral  such  that,  if  any  point  P  be  joined 
to  A,  B,  C,  D,  the  sum  of  the  squares  on  PA,  PC  is  equal  to  the  sum 
of  the  squares  on  PB,  PD :  prove  that  ABCD  is  a  rectangle. 


126  BOOK  /. 

MISCELLANEOUS   EXERCISES. 

1.  How  many  diagonals  can  be  drawn  through  the  same  vertex 
in  (1)  a  quadrilateral,  (2)  a  hexagon,  (3)  a  polygon  of  n  sides? 

2.  How  many  different  diagonals  can  be  drawn  to  (1)  a  quadri- 
lateral, (2)  a  hexagon,  (3)  a  polygon  of  n  sides? 

3.  If  two  straight  lines  bisect  each  other  at  right  angles,  any 
point  in  either  of  them  is  equidistant  from  the  extremities  of  the 
other. 

4.  A  straight  line  drawn  bisecting  the  angle  contained  by  two 
equal  sides  of  a  triangle  bisects  the  third  side  at  right  angles. 

5.  Two  isosceles  triangles  GAB^  DAB  are  on  the  same  base  AB: 
shew  that  the  triangles  ACD,  BCD  are  equal  in  all  respects. 

6.  Prove  by  the  method  of  superposition  that,  if  two  isosceles 
triangles  have  the  same  vertical  angle,  their  bases  are  parallel. 

7.  It  ABC,  DBC  he  two  triangles  equal  in  all  respects  on  oppo- 
site sides  of  BC,  then  AD  is  perpendicular  to  BC  and  is  bisected 
by  it. 

8.  The  angle  BAG  of  a  triangle  ABG  is  bisected  by  a  straight 
line  which  meets  BC  in  D,  and  from  AB  on  AB  produced  AE  is  cut 
off  equal  to  ^C :  prove  that  DE  is  equal  to  DG. 

9.  If  two  circles  whose  radii  are  equal  cut  in  A,  B  and  if  the  line 
joining  their  centres  when  produced  meet  the  circumference  in  C,  D, 
prove  that  AGBD  is  a  rhombus. 

10.  Prove  by  the  method  of  superposition  that,  if  the  opposite 
angles  of  a  quadrilateral  be  equal  and  one  pair  of  opposite  sides  be 
equal,  the  other  sides  are  equal. 

11.  Prove  by  the  method  of  superposition  that,  if  a  quadrilateral 
has  two  pairs  of  adjacent  angles  equal,  it  has  one  pair  of  opposite 
sides  equal. 

12.  Two  adjacent  sides  of  a  quadrilateral  are  equal,  and  the  two 
angles  which  they  form  with  the  other  sides  are  together  equal  to  the 
angle  between  the  other  sides.  Prove  that  one  diagonal  of  the  quadri- 
lateral is  equal  to  a  side. 

13.  In  a  triangle  BAG  is  the  greatest  angle.  Prove  that,  if  a 
point  D  be  taken  in  AB,  and  a  point  E  in  AG,  DE  is  less  than  BG. 

14.  If  AD  be  drawn  perpendicular  from  the  vertex  A  to  the 
opposite  side  BG  of  a  triangle  ABG  in  which  ^C  is  greater  than  AB, 
then  DC  is  greater  than  BD,  and  the  angle  DAG  is  greater  than  the 
angle  BAD. 

15.  How  many  different  triangles  can  be  formed  by  taking  three 
lines  out  of  six  lines  whose  lengths  are  2,  3,  4,  5,  6,  7  inches  re- 
spectively ? 


MISCELLANEOUS  EXERCISES.  127 

16.  Find  the  point  the  sum  of  whose  distances  from  the  four 
angular  points  of  a  convex  quadrilateral  is  a  minimum. 

17.  AB  is  a  given  finite  straight  line.  From  C  the  middle  point 
of  AB,  CI)  is  drawn  in  any  direction  and  of  any  length.  Prove  that 
AD,  BIJ  together  are  greater  than  twice  CD. 

18.  From  a  given  point  draw  a  straight  line  making  equal  angles 
with  two  given  intersecting  straight  lines.  How  many  such  lines  can 
be  drawn? 

19.  In  a  given  straight  line  find  a  point  equally  distant  from  two 
given  straight  lines.     In  what  case  is  a  solution  impossible  ? 

20.  How  many  equalities  must  be  given  between  the  sides  and 
the  angles  of  (1)  two  quadrilaterals,  (2)  two  hexagons,  (3)  two 
polygons  of  n  sides,  before  the  conclusion  can  be  drawn  that  the 
figures  are  equal  in  all  respects  ? 

21.  In  the  triangle  ABC  the  angles  at  B  and  C  are  equal ;  m  and  I 
are  points  in  AC  produced  and  on  AB  resi^eetively,  and  Im  is  joined 
cutting  BC  in  O.  Prove  that,  if  the  sum  ofAl  and  Am  be  double  AB, 
then  jBO  is  greater  than  CO. 

22.  The  side  BC  of  a  triangle  ABC  is  produced  to  a  point  D ; 
the  angle  ACB  is  bisected  by  the  straight  line  CE  which  meets  AB  at 
E.  A  straight  line  is  drawn  through  E  parallel  to  BC,  meeting  AC 
at  F,  and  the  straight  line  bisecting  the  exterior  angle  ACD  at  G. 
Shew  that  EF  is  equal  to  FG. 

23.  A  straight  line  drawn  at  right  angles  to  BC  the  base  of 
an  isosceles  triangle  ABC  cuts  the  side  AB  at  D  and  CA  produced 
atE  :  shew  that  AED  is  an  isosceles  triangle. 

24.  If  in  the  base  of  a  triangle  ABC,  there  be  taken  any  two 
points  P  and  Q  equidistant  from  the  extremities  of  the  base,  and 
if  through  each  of  the  points  P,  Q  two  straight  lines  be  drawn  parallel 
to  AB,  AC,  so  as  to  form  two  parallelograms  having  PA,  QA  for 
diagonals ;  these  two  parallelograms  are  equal  in  area. 

25.  Find  a  point  equidistant  from  each  of  three  straight  lines  in 
a  plane  which  do  not  coincide  in  direction  with  the  sides  of  any 
triangle  that  can  be  drawn  in  the  plane.  Is  the  construction  required 
in  this  last  problem  always  possible  ? 

26.  From  a  point  P  outside  an  angle  BA  C  draw  a  straight  line 
cutting  the  straight  lines  AB,  ^C  in  points  D  and  E  such  that  PD 
may  be  equal  to  DE. 

27.  If  one  acute  angle  of  a  triangle  be  double  of  another,  the 
triangle  can  be  divided  into  two  isosceles  triangles. 

28.  If  in  a  triangle  ABC,  ACB  be  a  right  angle,  and  the  angle 
CAB  be  double  the  angle  ABC,  then  AB  is  double  AC. 


128  BOOK  I. 

29.  P  is  a  point  in  the  side  CD  of  a  square  ABGD  such  that  AP 
is  equal  to  the  sum  of  PC  and  CB  and  Q  is  the  middle  point  of  CD. 
Prove  that  the  angle  BAP  is  twice  the  angle  QAD. 

30.  AOB,  COD  are  two  indefinite  straight  lines  intersecting  each 
other  in  the  point  O,  and  P  is  a  given  point  in  the  plane  of  these 
lines.  It  is  required  to  draw  through  the  point  P  a  straight  line 
PXY  cutting  AB  in  X  and  CD  in  Y,  in  such  a  manner  that  OX  may 
be  equal  to  OY.    Can  this  problem  be  solved  in  more  than  one  way? 

31.  Construct  an  equilateral  triangle  one  of  whose  angular  points 
is  given  and  the  other  two  lie  one  on  each  of  two  given  straight  lines. 

32.  The  sides  AB,  AC  of  a,  triangle  are  bisected  in  D,  E,  and  BE, 
CD  are  produced  to  F,  G,  so  that  EF  is  equal  to  BE,  and  DG  to  CD : 
prove  that  FAG  is  a  straight  line. 

33.  In  a  plane  triangle  an  angle  is  acute,  right  or  obtuse,  according 
as  the  straight  line  joining  the  angle  to  the  middle  point  of  the  opposite 
side  is  greater  than,  equal  to,  or  less  than  half  that  side. 

34.  The  difference  of  the  angles  at  the  base  of  a  triangle  is  double 
the  angle  between  the  perpendicular  to  the  base,  and  the  bisector  of 
the  vertical  angle. 

35.  AC  is  the  longest  side  of  the  triangle  ABC.  Find  in  AC  a. 
point  D  such  that  the  angle  ADB  shall  be  equal  to  twice  the  angle 
ACB. 

36.  ABCD  is  a  quadrilateral:  the  bisectors  of  the  angles  ABD, 
ACD  meet  at  F:  prove  that  the  angle BFC  is  half  the  sum  of  the  angles 
BAG,  BDC. 

37.  Prove  that,  if  points  L,  M,  N  be  taken  in  the  sides  BC,  CA, 
AB  of  a  triangle  ABC,  such  that  the  triangles  ANM,  NBL,  MLG  are 
equiangular  to  each  other,  they  are  equiangular  also  to  the  triangles 
ABC  and  LMN. 

38.  If  one  angle  at  the  base  of  a  triangle  be  double  the  other,  the 
less  side  is  equal  to  the  sum  or  the  difference  of  the  segments  of  the 
base  made  by  the  perpendicular  from  the  vertex,  according  as  the  first 
angle  is  greater  or  less  than  a  right  angle. 

39.  Any  convex  pentagon  ABCDE  has  each  of  its  angular  points 
joined  to  the  non-contiguous  points  and  a  star-shaped  figure  ACEBD 
is  formed:  prove  that  the  sum  of  the  angles  of  the  figure  ACEBD  is 
two  right  angles. 

40.  In  the  figure  of  Proposition  1,  if  AB  produced  both  ways  meet 
the  circles  again  in  D  and  E  and  CD,  CE  be  drawn,  CDE  will  be  an 
isosceles  triangle  having  one  angle  four  times  each  of  the  other  angles. 

41.  From  the  extremities  of  the  base  of  an  isosceles  triangle 
straight  lines  are  drawn  perpendicular  to  the  sides;  shew  that  the 
angles  made  by  them  with  the  base  are  each  equal  to  half  the  vertical 
angle. 


MISCELLANEOUS  EXERCISES.  129 

42.  The  sides  AB,  AG  of  &  given  triangle  ABC  are  bisected  at 
the  points  E,  F;  &  perpendicular  is  drawn  from  A  to  the  opposite 
side,  meeting  it  at  D.  Shew  that  the  angle  FDE  is  equal  to  the 
angle  BAG. 

43.  AB,  AG  are  two  given  straight  lines,  and  P  is  a  given  point 
in  the  former :  it  is  required  to  draw  through  P  a  straight  line  to  meet 
^C  at  Q,  so  that  the  angle  APQ  may  be  three  times  the  angle  AQP. 

44.  Construct  a  right-angled  triangle  having  given  the  hypotenuse 
and  the  difference  of  the  sides. 

45.  From  a  given  point  it  is  required  to  draw  to  two  parallel 
straight  lines,  two  equal  straight  lines  at  right  angles  to  each  other. 

46.  Construct  a  triangle  of  given  perimeter,  having  its  angles 
equal  to  those  of  a  given  triangle. 

47.  Given  one  angle,  and  the  opposite  side,  and  the  sum  of  the 
other  sides,  construct  the  triangle. 

48.  If  two  triangles  on  the  same  side  of  a  common  base  have 
their  sides  terminated  in  opposite  extremities  of  the  base  equal,  the 
line  joining  the  vertices  will  be  parallel  to  the  common  base. 

49.  If  an  exterior  angle  of  a  triangle  be  bisected,  and  also  one  of 
the  interior  and  opposite  angles,  the  angle  contained  by  the  bisecting 
lines  is  equal  to  half  the  other  interior  and  opposite  angle. 

50.  If  the  perpendicular  drawn  from  one  of  the  equal  angles  of 
an  isosceles  triangle  upon  the  opposite  side  divide  the  angle  into  two 
parts,  one  of  which  is  double  of  the  other,  the  vertical  angle  of  the 
triangle  is  either  one  half  or  four-fifths  of  a  right  angle. 

51.  The  sides  AB,  AG  oi  a,  triangle  ABG  are  produced  to  E,  JF, 
and  the  angles  GBE,  BGF  are  bisected  by  the  straight  Hues  BD,  GD: 
prove  that  the  angle  BDG  is  half  the  sum  of  the  angles  ABC,  AGB. 

52.  In  a  triangle  ABG  points  D,  E  are  taken  in  ^C7  such  that 
AD  is  equal  to  AB,  and  that  GE  is  equal  to  GB,  and  a  point  F  is 
taken  in  ^jB  such  that  BF  is  equal  to  i^C:  prove  that  the  angle  EBD 
is  equal  to  the  angle  BGF. 

53.  ABG  is  a  triangle,  and  from  a  point  D  in  AB  the  straight 
line  DEF  is  drawn  meeting  BG  in  E  and  AG  produced  in  F;  shew 
that  the  angles  between  the  straight  lines  bisecting  the  angles  ABE, 
ADE  are  equal  to  the  angles  between  the  straight  lines  bisecting  the 
angles  AGE,  AFE. 

54.  Find  a  point  in  a  given  straight  line  such  that  its  distance 
from  a  given  point  is  double  of  its  distance  from  a  given  straight  line 
through  the  given  point. 

55.  Construct  an  equilateral  triangle,  having  given  its  altitude. 
T.  E.  9 


130  BOOK  I. 

56.  The  straight  line  bisecting  the  exterior  angle  at  the  vertex  A 
of  a  triangle  ABC  will  meet  the  base  BG  produced  beyond  B,  or 
beyond  0,  according  as  ^C  is  greater  or  less  than  AB. 

57.  If  from  any  point  within  an  isosceles  triangle  perpendiculars 
be  let  fall  on  the  base  and  the  sides,  the  sum  of  these  perpendiculars 
is  less  than  the  altitude  of  the  triangle,  if  the  vertical  angle  be  less 
than  the  angle  of  an  equilateral  triangle. 

58.  ABC  is  a  triangle  right-angled  at  A,  and  on  AB,  AG  are 
described  two  equilateral  triangles  ABD,  AGE  (both  on  the  outside 
or  both  on  the  inside),  and  DB  and  EG  or  those  produced  meet  in  F; 
shew  that  A  is  the  ortho-centre  of  the  triangle  DEF. 

59.  If  the  angle  between  two  adjacent  sides  of  a  parallelogram  be 
increased,  while  their  lengths  do  not  alter,  the  diagonal  through  their 
point  of  intersection  will  diminish. 

60.  If  straight  lines  be  drawn  from  the  angles  of  any  parallelo- 
gram perpendicular  to  any  straight  line  which  is  outside  the  parallelo- 
gram, the  sum  of  those  from  one  pair  of  opposite  angles  is  equal  to 
the  sum  of  those  from  the  other  pair  of  opposite  angles. 

61.  If  a  six-sided  plane  rectilineal  figure  have  its  opposite  sides 
equal  and  parallel,  the  three  straight  lines  joining  the  opposite  angles 
will  meet  at  a  point. 

62.  The  vertical  angle  GAB  of  a  triangle  ABG  is  bisected  by 
ADE,  and  BE,  GF  are  drawn  perpendicular  to  ADE :  prove  that  the 
middle  point  of  BG  is  equidistant  from  E  and  F. 

63.  Find  in  a  side  of  a  triangle  a  point  such  that  the  sum  of 
two  straight  lines  drawn  from  the  point  parallel  to  the  other  sides  and 
terminated  by  them  is  equal  to  a  given  straight  line. 

64.  If  the  angular  points  of  one  parallelogram  lie  on  the  sides  of 
another  parallelogram,  the  diagonals  of  both  parallelograms  pass 
through  the  same  point. 

65.  If  the  straight  line  joining  two  opposite  angles  of  a  parallelo- 
gram bisect  the  angles,  the  parallelogram  is  a  rhombus. 

66.  Draw  a  straight  line  through  a  given  point  such  that  the  part 
of  it  intercepted  between  two  given  parallel  straight  lines  may  be  of 
given  length. 

67.  Bisect  a  parallelogram  by  a  straight  line  drawn  through  a 
given  point  within  it. 

68.  Shew  that  the  four  triangles  into  which  a  parallelogram  is 
divided  by  its  diagonals  are  equal  in  area. 

69.  Straight  lines  bisecting  two  adjacent  angles  of  a  parallelogram 
intersect  at  right  angles. 

70.  Straight  lines  bisecting  two  opposite  angles  of  a  parallelogram 
are  either  parallel  or  coincident. 


MISCELLANEOUS  EXERCISES.  131 

71.  Find  a  point  such  that  the  perpendiculars  let  fall  from  it  on 
two  given  straight  lines  shall  be  respectively  equal  to  two  given  straight 
lines.    How  many  such  points  are  there  ? 

72.  ABCD  is  a  quadrilateral  having  BG  parallel  to  AD,  E  is  the 
middle  point  of  DC ;  shew  that  the  triangle  AEB  is  half  the  quadri- 
lateral. 

73.  If  the  sides  of  a  triangle  be  trisected  and  straight  lines  be 
drawn  through  the  points  of  section  adjacent  to  each  angle  so  as  to 
form  another  triangle,  this  is  equal  to  the  original  triangle  in  all 
respects. 

74.  If  two  opposite  sides  of  a  parallelogram  be  bisected  and  two 
straight  lines  be  drawn  from  the  points  of  bisection  to  two  opposite 
angles,  the  two  lines  trisect  the  diagonal  which  passes  through  the 
other  two  angular  points. 

75.  BAG  is  a  right-angled  triangle,  A  being  the  right  angle. 
AGDE,  BGFG  are  squares  on  ^O  and  BG.  AG  produced  meets  DF 
in  K.     Prove  that  DF  is  bisected  in  K,  and  that  AB  is  double  of  GK. 

76.  In  a  triangle  ABG  on  AG,  BG  on  the  sides  of  them  away 
from  B,  A,  squares  AGDE,  BGFG  are  constructed ;  prove  that,  if  AG 
produced  bisect  DF,  BAG  is  a  right  angle. 

77.  A  straight  line  PQ  drawn  parallel  to  the  diagonal  ^C  of  a 
parallelogram  ABGD  meets  AB  in  P  and  BG  in  Q;  shew  that  the 
other  diagonal  BD  bisects  the  quadrilateral  BPDQ. 

78.  If  the  opposite  angles  of  a  quadrilateral  be  equal,  the  oppo- 
site sides  are  equal. 

79.  If  in  two  parallelograms  ABGD,  EFGH,  AB  be  equal  to  EF, 
BG  to  FO,  and  the  angle  ABG  to  the  angle  EFG,  then  the  parallelo- 
grams are  equal  in  all  respects. 

80.  If  the  straight  line  AB  be  bisected  in  G,  and  AD  and  BE  be 
drawn  at  right  angles  to  AB,  and  AD  be  taken  equal  to  AG,  and  DE 
be  drawn  at  right  angles  to  DG,  DE  is  double  of  DG. 

81.  ABG  is  a  given  triangle :  construct  a  triangle  of  equal  area, 
having  its  vertex  at  a  given  point  in  BG  and  its  base  in  the  same 
straight  line  as  AB. 

82.  In  the  right-angled  triangle  ABG,  the  sides  AB,  AG  which 
contain  the  right  angle  are  equal.  A  second  right-angled  triangle  is 
constructed  having  the  sides  containing  the  right  angle  together 
equal  to  AB,  AG,  but  not  equal  to  one  another.  Prove  that  this 
triangle  is  less  than  the  triangle  ABG. 

83.  If  two  triangles  have  two  sides  of  the  one  equal  to  two  sides 
of  the  other,  and  the  sum  of  the  two  angles  contained  by  these  sides 
equal  to  two  right  angles,  the  triangles  are  equal  in  area. 

9-2 


132  BOOK  I. 

84.  If  a  triangle  be  described  having  two  of  its  sides  equal  to  the 
diagonals  of  any  quadrilateral,  and  the  included  angle  equal  to  either 
of  the  angles  between  these  diagonals,  then  the  area  of  the  triangle 
is  equal  to  the  area  of  the  quadrilateral. 

85.  ABC  is  a  triangle ;  find  the  locus  of  a  point  0  such  that  the 
sum  of  the  areas  OAB,  OBG,  OCA  is  constant  and  greater  than  the 
area  of  ^£(7. 

86.  Any  point  P  is  joined  to  0  the  middle  point  of  AD.  On  AP, 
DP  squares  AFQQ',  DP  RE'  are  described  on  the  sides  remote  from 
D,  A  respectively.  Prove  that  QR  is  double  of  and  perpendicular 
to  OP. 

87.  ABGD  is  a  parallelogram;  E  the  point  of  bisection  of  AB\ 
prove  that  AG,  DE  being  joined  will  each  pass  through  a  point  of 
trisection  of  the  other. 

88.  In  every  quadrilateral  the  intersection  of  the  straight  lines 
which  join  the  middle  points  of  opposite  sides  is  the  middle  point  of 
the  straight  line  which  joins  the  middle  points  of  the  diagonals. 

89.  The  line  joining  the  middle  points  of  the  diagonals  of  the 
quadrilateral  ABGD  meets  AD  and  BG  in.  E  and  F.  Shew  that  the 
triangles  EBG,  FAD  are  each  half  the  quadrilateral. 

90.  PQR  is  a  straight  line  parallel  and  equal  to  the  base  BG  of  a 
triangle  meeting  the  sides  in  P  and  Q.  Shew  that  the  triangles  BPQ, 
AQR  are  equal. 

91.  Two  straight  lines  AB  and  GD  intersect  at  E,  and  the  tri- 
angle AEG  is  equal  to  the  triangle  BED;  shew  that  BG  ia  parallel 
to^D. 

92.  Construct  the  smallest  triangle,  which  has  a  given  vertical 
angle,  and  whose  base  passes  through  a  given  point. 

93.  In  the  base  JO  of  a  triangle  take  any  point  D  ;  bisect  AD, 
DG,  AB,  BG  at  the  points  E,  F,  G,  H  respectively:  shew  that  EG  is 
equal  and  parallel  to  FH. 

94.  Given  the  middle  points  of  the  sides  of  a  triangle,  construct 
the  triangle. 

95.  ABG  ia  &  triangle.  The  side  GA  is  bisected  in  D  and  E  is 
the  point  of  trisection  of  the  side  BG  which  is  nearer  to  B.  Shew 
that  the  line  joining  A  to  E  bisects  the  line  joining  B  to  D. 

96.  Shew  how  by  means  of  Prop.  40  to  produce  a  finite  straight 
line  beyond  an  obstacle  which  cannot  be  passed  through  directly. 

97.  BAG  is  a  fixed  angle  of  a  triangle,  and  (i)  the  sum,  (ii)  the 
difference  of  the  sides  AB,  AG  is  given;  shew  that  in  either  case 
the  locus  of  the  middle  point  of  UC  is  a  straight  line. 


MISCELLANEOUS  EXERCISES.  133 

98.  Through  a  fixed  point  0  two  straight  lines  OPQ,  OP'Q'  are 
drawn  meeting  two  fixed  parallel  straight  lines.  Prove  that,  if  PQ' 
and  P'Q  meet  in  R,  the  locus  of  22  is  a  straight  line,  and  that  OR 
bisects  PP'  and  QQ'. 

99.  ABC  is  a  triangle  and  on  the  side  BG  &  parallelogram 
BDEC  is  described,  and  the  parallelogram  whose  adjacent  sides  are 
AI),  AB  is  completed  and  also  that  whose  adjacent  sides  are  AE^  AG; 
shew  that  the  sum  or  the  difference  of  the  two  latter  parallelograms 
is  equal  to  the  first. 

100.  ABG  is  a  triangle;  ADF,  GFE  are  the  perpendiculars  let 
fall  from  A  and  G,  one  on  the  internal  bisector  of  the  angle  B  and  the 
other  on  the  external  bisector:  the  area  of  the  rectangle  BDFE  is 
equal  to  that  of  the  triangle. 

101.  If  on  the  sides  AB,  BG,  GA  of  any  triangle,  squares  ABEF, 
BGGH,  GAEL  be  constructed,  and  EH,  GL,  KF  be  drawn,  then 
all  the  triangles  ABG,  BEH,  GGL,  AKF  are  equal  to  one  another. 

102.  Construct  a  square,  which  shall  have  two  adjacent  sides 
passing  through  two  given  points,  and  the  intersection  of  diagonals 
at  a  third  given  point. 

103.  ABG  is  a  triangle  right-angled  at  A  and  K  is  the  corner 
of  the  square  on  AB  opposite  to  A  and  H  the  corner  of  the  square 
on  AG  opposite  to  A.  If  AB  be  produced  to  D,  so  that  AD  is  equal 
to  GK  and  ^  C  be  produced  to  E,  so  that  AE  is  equal  to  BH,  then 
GD  is  equal  to  BE. 

104.  Upon  BG,  GA,  AB,  sides  of  the  triangle  ABG,  perpen- 
diculars are  drawn  from  a  point  D,  meeting  the  sides,  or  the  sides 
produced,  in  E,  F,  G  respectively.  Prove  that  the  sum  of  the  squares 
on  AG,  BE,  GF  is  equal  to  the  sum  of  the  squares  on  BG,  GE,  AF. 

105.  Divide  a  given  straight  line  into  two  parts  such  that  the 
square  on  one  of  them  may  be  double  the  square  on  the  other. 


BOOK    11. 


DEFINITIONS. 

1)efinition  1.  A  rectangle  is  said  to  be  contained  "by 
the  two  sides  which  contain  any  one  of  its  angles. 

The  expression,  that  a  rectangle  is  contained  hy  two  straight  lines 
AB,  BG  is  of  course  a  faulty  one,  as  the  area  of  the  rectangle  is 
contained  by  the  four  sides  of  the  figure.  The  expression  must  be 
considered  merely  as  an  abbreviated  form  of  the  statement  that  the 
rectangle  has  for  two  of  its  adjacent  sides  the  two  straight  lines 
AB,  BG. 

It  can  easily  be  proved  that,  if  two  adjacent  sides  of 
one  rectangle  be  equal  to  two  adjacent  sides  of  another 
rectangle,  the  two  rectangles  are  equal  in  all  respects. 

(See  Exercise  79,  page  131.) 

A  rectangle,  two  of  whose  adjacent  sides  are  equal  to 
two  straight  lines  AB,  CD,  is  therefore  often  said  to  be  the 
rectangle  contained  by  AB,  CD ;  such  a  rectangle  is  often 
denominated  simply  the  rectangle  AB,  CD. 

It  is  clear  that  the  rectangle  AB^  CD  is  the  same  as 
the  rectangle  CD,  AB. 

Also  it  may  be  noticed  that,  ii  AB  h^  equal  to  CD,  the 
rectangle  AB,  CD  is  equal  to  the  square  on  AB,  or  to  the 
square  on  CD. 


135 


In  Arithmetic  or  in  Algebra,  if  we  wish  to  represent  a  given 
length,  we  take  a  definite  length,  for  instance  an  inch,  as  a  unit  of 
length  and  we  express  the  given  length  by  the  number  of  units  of 
length  contained  in  it. 

In  the  same  way,  if  we  wish  to  represent  a  given  area,  we  take 
a  definite  area,  for  instance  a  square  inch,  as  a  unit  of  area,  and 
we  express  the  given  area  by  the  number  of  units  of  area  contained 
in  it. 

It  is  easily  seen  that,  if  a  rectangle  have  2  inches  in  one  side 
and  3  inches  in  an  adjacent  side,  its  area  consists  of  2  x  3  or  6  squares, 
each  square  having  one  inch  as  its  side,  and  similarly  that,  if  a 
rectangle  have  m  units  of  length  in  one  side  and  n  units  of  length 
in  an  adjacent  side,  its  area  consists  of  wm  squares,  each  square  having 
a  unit  of  length  as  its  side. 

Thus  in  Arithmetic  or  in  Algebra  the  area  of  a  rectangle  is  repre- 
sented by  the  product  of  the  numbers,  which  represent  the  lengths 
of  two  adjacent  sides. 

If  the  rectangle  be  a  square,  its  area  is  represented  by  the  square 
of  the  number,  which  represents  the  length  of  a  side. 

In  consequence  of  this  connection  between  Algebra  and  Geometry, 
there  is  a  certain  correspondence  between  the  theorems  and  problems 
of  the  Second  Book  of  Euclid,  and  theorems  and  problems  in  Algebra. 

A  short  statement  of  a  corresponding  proposition  in  Algebra  is 
given  as  a  note  to  each  Proposition,  in  which  statement  each  straight 
line  is  represented  by  a  corresponding  letter,  and  each  area  by  a 
corresponding  product. 


136 


BOOK  II. 


PROPOSITION  1. 

If  there  he  two  straight  lines,  one  of  which  is  divided 
into  any  two  parts,  the  rectangle  contained  hy  the  two  straight 
lines  is  equal  to  the  sum  of  the  rectangles  contained  hy  the 
undivided  line,  and  each  of  the  parts  of  the  divided  line*. 

Let  AB  and  CD  be  two  straight  lines ;  and  let  CD  be 
divided  into  any  two  parts  at  the  point  E : 
it  is  required  to  prove  that  the   rectangle  contained  by 

AB,  CD  is  equal  to  the  sum  of  the  rectangles  contained 

by  AB,  CE,  and  by  AB,  ED. 

Construction.     From  the  point  C  draw  CF  at  right 

angles  to  CD;  (I.  Prop.  11.) 

and  make  CG  equal  to  AB ;        (I.  Prop.  3.) 

through  G  draw  GHK  parallel  to  CD ;     (I.  Prop.  31.) 

and  through  D,  E  draw  DK,  EH  parallel  to  CG  meeting 

GHK  in  K  II. 


E 


D 


Proof.     Each  of  the  figures  CK,  CH,  EK  is  a  parallel- 
ogram, (Constr.) 
and  each  of  the  figures  has  one  angle  a  right  angle ; 

(I.  Prop.  29,   Coroll.) 
therefore  each  of  the  figures  is  a  rectangle. 
Now  the  rectangle  CK  is  equal  to  the  sum  of  the  rect- 
angles CH,  EK. 

*  The  algebraical  equivalent  of  this  theorem  is  the  equation 
a{h-\-c)=.ah-\-ac. 


PROPOSITION  1.  137 

But  CK  is  contained  by  AB,  CD^ 
for  it  is  contained  by  CG^  CD,  and  CG  is  equal  to  AB. 

And  CH  is  contained  liy  yl^,  0^, 
for  it  is  contained  by  CG,  CE,  and  CG  is  equal  to  AB. 

And  ^A'^  is  contained  by  AB,  ED, 
for  it  is  contained  by  EH,  ED,  and  EH  is  equal  to  CG, 
which  is  equal  to  AB. 
Therefore  the  rectangle  contained  by  AB,  CD  is  equal  to 
the  sum  of  the  rectangles  contained  by  AB,  CE,  and  by 
AB,  ED. 
Wherefore,  iftJiere  he  two  straight  lines  <fec. 

Corollary  1.  If  there  he  two  straight  lines,  one  of 
which  is  divided  into  any  numher  of  j^arts,  the  rectangle 
contained  hy  the  two  straight  lines  is  equal  to  the  sum  of 
the  rectangles  contained  hy  the  undivided  line  and  each  of 
the  parts  of  the  divided  line. 

If  CD  be  divided  into  three  parts  at  the  points  E,  F: 
the  rectangle  AB,  CD  is  equal  to  the  sum  of  the  rectangles 
AB,  CE,  and  AB,  ED  :  (Prop.  1.) 

and  the  rectangle  AB,  ED  is  equal  to  the  sum  of  the  rect- 
angles AB,  EF  and  AB,  FD  : 
therefore  the  rectangle  AB,  CD  is  equal  to  the  sum  of  tlie 

rectangles  AB,  CE;  AB,  EF  and  AB,  FD. 
And  so  on  for  any  number  of  points  of  division. 

Corollary  2.  If  there  he  two  straight  lines,  one  of  which 
is  divided  into  any  two  parts,  the  rectangle  contained  hy  the 
undivided  line  and  one  of  the  parts  of  the  divided  line  is 
equal  to  the  difference  of  the  rectangles  contained  hy  the  un- 
divided line  and  the  whole  of  the  divided  line  and  hy  the 
undivided  line  and  the  reinaining  part  of  the  divided  line. 

EXERCISES. 

1.  If  ^,  B,  C,  D  be  four  points  in  order  in  a  straight  line,  then 
the  sum  of  the  rectangles  AB,  CD,  and  AD,  BC  is  equal  to  the  rect- 
angle AC,  BD. 

2.  If  there  be  two  straight  lines,  each  of  which  is  divided  into 
any  number  of  parts,  the  rectangle  contained  by  the  two  straight 
lines  is  equal  to  the  sum  of  the  rectangles  contained  by  each  of  the 
parts  of  the  first  line  and  each  of  the  parts  of  the  second  line. 


138  BOOK  II, 


PROPOSITION  2. 

If  a  straight  line  he  divided  into  any  two  parts,  the 
square  on  the  whole  line  is  equal  to  the  sum  of  the  rectangles 
contained  hy  the  whole  and  each  of  the  parts*. 

Let  the  straight  line  ABhe  divided  into  any  two  parts 
at  the  point  C : 
it  is  required  to  prove  that  the  square  on  AB  is  equal  to 

the  sura  of  the  rectangles  contained  by  AB^  ^C'and  by 

AB,  CB. 

Construction.     Take  DE  a  straight  line  equal  to  AB. 


-? — -^ 


E 


Proof.     The  rectangle  DE,  AB  i^  equal  to  the   sura 
of  the  rectangles  DE,  AC  and  DE,  CB.  (Prop.  1.) 

Because  DE  is  equal  to  AB, 
the  rectangle  DE,  AB  \^  equal  to  the  square  on  AB, 

the  rectangle  DE,  AC  to  the  rectangle  AB,AC, 
and  the  rectangle  DE,  CB  to  the  rectangle  AB,  CB  : 
therefore  the  square  on  ^^  is  equal  to  the  sum  of  the  rect- 
angles AB,  AC,  and  AB,  CB. 
Wherefore,  if  a  straight  line  (fee. 

*  The  algebraical  equivalent  of  this  theorem  is  the  equation 


PROPOSITION  2. 


139 


Outline  of  Alternative  Proof. 

On  AB  construct  the  square  AEDB, 

and  draw  CF  parallel  to  AE  to  meet  ED  in  F. 


G       B 


F 


D 


It  may  be  proved  that 
(1)     ^i^is  the  rectangle  AB,  AC, 
and  (2)     CD  is  the  rectangle  AB,  CB, 

and  hence  that  the  square  on  AB  is  equal  to  the  sum  of 
the  rectangles  AB,  AC  and  AB,  CB. 


EXEECISES. 

1.  D  is  a  point  in  the  hypotenuse  BC  of  a.  right-angled  triangle 
ABC :  prove  that,  if  the  rectangle  BD,  BG  be  equal  to  the  square  on 
AC,  then  the  rectangle  BG,  DG  is  equal  to  the  square  on  AB. 

2.  A  point  D  is  taken  in  the  side  BG  of  a  triangle  ABG:  prove 
that,  if  the  rectangles  BD,  BG  and  BG,  DG  be  equal  to  the  squares 
on  AB,  AG  respectively,  the  angle  BAG  is  a  right  angle. 


140  BOOK   11. 


PROPOSITION  3. 

If  a  straight  line  he  divided  into  any  two  parts,  the  rect- 
angle contained  hy  the  whole  line  and  one  of  the  parts  is 
equal  to  the  sum  of  the  square  on  that  part  and  the  rect- 
angle contained  by  the  two  parts*. 

Let  the  straight  line  ^^  be  divided  into  any  two  parts 
at  the  point  C  : 
it  is  required  to  prove  that  the  rectangle  AB,  AC  i^  equal  to 
the  sum  of  the  square  on  AC  and  the  rectangle  AC,  CB. 
Construction.     Take  DE  a  straight  line  equal  to  AG. 

AG  B 

i 


Proof.     The  rectangle  DE,  AB  i^  equal  to  the  sum  of 
the  rectangles  DE,  AC,  and  DE,  CB.  (Prop.  1.) 

Because  DE  is  equal  to  A  C, 
the  rectangle  DE,  AB  is  equal  to  the  rectangle  AC,  AB, 
the  rectangle  DEy  AC  to  the  square  on  AC, 
and  the  rectangle  DE,  CB  to  the  rectangle  AC,  CB ; 
therefore  the  rectangle  AB,  AC  i^  equal  to  the  sum  of  the 
square  on  AC  and  the  rectangle  AC,  CB. 
Wherefore,  if  a  straight  line  &c. 

*  The  algebraical  equivalent  of  this  theorem  is  the  equation 


PROPOSITION  3. 


141 


Outline  of  Alternative  Proof. 

On  AC  construct  the  square  ADEG, 
and  draw  ^^  parallel  to  AD  to  meet  DE  produced  in  F. 


G       B 


E 

It  may  be  proved  that 
(1)     ^i^  is  the  rectangle  AB,  AC, 
and  (2)     CF  is  the  rectangle  AC,  CB, 

and  hence  that  the  rectangle  AB,  AC  i^  equal  to  the  sum 
of  the  square  on  AC  and  the  rectangle  AC,  CB. 


EXERCISES. 


1.  AD  is  drawn  perpendicular  to  the  hypotenuse  BC  of  a  right- 
angled  triangle  ABC:  prove  that  the  rectangle  BD,  DC  is  equal  to 
the  square  on  AD. 

2.  In  the  triangle  ABC,  AD  is  the  perpendicular  from  A  on  BC: 
prove  that,  if  the  rectangle  BD,  DC  he  equal  to  the  square  on  AD, 
the  angle  ^  is  a  right  angle. 

3.  In  the  triangle  ABC,  AD  is  the  perpendicular  from  A  on  BC: 
prove  that,  if  the  rectangle  BD,  BC  be  equal  to  the  square  on  AB, 
the  angle  ^  is  a  right  angle. 


U2  BOOK  11. 


PROPOSITION  4. 

If  a  straight  line  be  divided  into  any  two  parts,  the 
square  on  the  whole  line  is  equal  to  the  sum  of  the  squares 
on  the  parts  and  twice  the  rectangle  co7itained  by  the  jjarts*. 

Let  the  straight  line  AB  be  divided  into  any  two  parts 
at  the  point  G  : 
it  is  required  to  prove  that  the  square  on  ^^  is  equal  to 
the  sum  of  the  squares  on  AC  and  CB,  and  twice  the 
rectangle  contained  by  AC,  CB. 


Proof.     Because  ^^  is  divided  into  two  parts  at  C, 
the  square  on  AB  is  equal  to  the  sum  of  the 

rectangles  AB,  AC  and  AB,  CB,  (Prop.  2.) 

and  the  rectangle  AB,  AC  vs,  equal  to  the  sum  of  the  square 

on  ^C  and  the  rectangle  AC,  CB,  (Prop.  3.) 

and  the  rectangle  AB,  CB  is  equal  to  the  sum  of  the  square 

on  CB  and  the  rectangle  AC,  CB.  (Prop.  3.) 

Therefore  the  square  on  AB  is  equal  to  the  sum  of  the 

squares  on  ^C  and  CB,  and  twice  the  rectangle  AC,  CB. 

Wherefore,  if  a  straight  line  &c. 

*  The  algebraical  equivalent  of  this  theorem  is  the  equation 
(a+&)2  =  a2  +  62  +  2a6. 


PROPOSITION  4.  143 

Outline  of  Alternative  Proof. 

On  AB  construct  the  square  ADEB  : 
draw  C^ri^  parallel  to  AD :  make  DH  equal  to  AC^ 
and  draw  HGK  parallel  to  AB. 

D         F  E 


H 


Q 

It  may  be  proved  that 

(1)  DG  is  equal  to  the  square  on  AG, 

(2)  GB  is  the  square  on  GB, 

and     (3)     IIG,  FK  are  each  equal  to  the  rectangle  A  (7,  C5, 

and  hence  that  the  square  on  AB  is  equal  to  the  sum  of 

the  squares  on  AC,  GB  and  twice  tlie  rectangle  AG,  GB. 


EXEKCISES. 

1.  The  square  on  a  straight  line  is  four  times  the  square  on  half 
of  the  line. 

2.  If  the  sides  BC,  CA,  AB  of  a  right-angled  triangle  ABC  be 
bisected  in  the  points  D,  E,  F  respectively,  twice  the  squares  on  AD, 
BE  and  CF  are  together  equal  to  three  times  the  square  on  the 
hypotenuse. 

3.  If,  in  an  acute-angled  triangle  ABC,  AD  be  drawn  perpen- 
dicular to  BC,  then  the  sum  of  the  squares  on  AB,  AC  and  twice 
the  rectangle  BD,  DO  is  equal  to  the  sum  of  the  square  on  BC  and 
twice  the  square  on  AD. 

4.  If  BAC  be  an  obtuse  angle  and  BD,  CE  be  drawn  at  right 
angles  to  CA,  BA  respectively,  then  the  rectangle  BA,  AE  is  equal 
to  the  rectangle  CA,  AD. 

5.  ABCD  is  a  square  and  E,  F,  G,  H  are  points  on  the  sides 
AB,  BC,  CD,  DA  respectively:  prove  that,  if  EFGH  be  a  rectangle, 
it  is  either  double  of  the  rectangle  AE,  EB,  or  equal  to  the  sum  of 
the  squares  on  AE,  EB. 


144  BOOK  II. 


PROPOSITION  5. 

If  a  straight  line  he  divided  into  two  equal  parts  and  also 
into  two  unequal  parts,  the  sum  of  the  rectangle  contained 
hy  the  unequal  parts  and  the  square  on  the  line  between  the 
points  of  section  is  equal  to  the  square  on  half  the  line*. 

Let  the  straight  line  ^^  be  divided  into  two  equal  parts 
at  the  point  C,  and  into  two  unequal  parts  at  the  point  D  : 
it  is  required  to  prove  that  the  sum  of  the  rectangle  AD,  DB 
and  the  square  on  DC,  is  equal  to  the  square  on  AC. 

A ,D       ,g B 


Proof.     Because  DB  is  divided  into  two  parts  at  0, 
the  rectangle  AD,  DB  is  equal  to  the  sum  of  the 

rectangles  AD,  DC  and  AD,  CB,  (Prop.  1.) 

that  is  to  the  sum  of  the  rectangles  AD,  DC,  and  AD,  AC, 
since  AC  is  equal  to  CB.         (Hypothesis.) 
And  because  ^C  is  divided  into  two  parts  at  D, 
the  rectangle  AD,  ^C  is  equal  to  the  sum  of  the  square  on 
AD  and  the  rectangle  AD,  DC.  (Prop.  3.) 

Therefore  the  rectangle  AD,  DB  is  equal  to  the  sum  of  the 
square  on  AD  and  twice  the  rectangle  AD,  DC. 
Add  to  each  of  these  equals  the  square  on  DC ; 
then  the  sum  of  the  square  on  DC  and  the  rectangle  AD,  DB 
is  equal  to  the  sum  of  the  squares  on  AD,  DC  and  twice 
the  rectangle  AD,  DC,  which  sum  is  equal  to  the  square 
on  AC.  (Prop.  4.) 

Therefore  the  rectangle  AD,  DB,  together  with  the  square 
on  DC,  is  equal  to  the  square  on  AC, 
Wherefore,  if  a  straight  line  &c. 

*  The  algebraical  equivalent  of  this  theorem  is  the  equation 
{a-b){a  +  b)  +  b^  =  a^. 


PROPOSITION  5.  145 


The  theorems  of  Propositions  5  and  6  may  both  be 
included  in  one  enunciation,  thus,  The  difference  of  the 
squares  on  two  given  straiyht  lines  is  equal  to  the  rectangle 
contained  by  the  sum  and  the  difference  of  the  lines. 

The  straight  lines  in  both  propositions  are  AD,  BD :  the  only 
difference  being  that  in  Prop.  5.  BD  is  the  greater  and  in  Prop.  6. 
AD  ia  the  greater. 

For  an  outline  of  an  alternative  proof  of  Propositions  5  and  6, 
see  page  147. 


EXEKCISES. 

1.  A  straight  line  is  divided  into  two  parts;  shew  that,  if  twice 
the  rectangle  of  the  parts  be  equal  to  the  sum  of  the  squares  on  the 
parts,  the  straight  line  is  bisected. 

2.  Divide  a  given  straight  line  into  two  parts  such  that  the 
rectangle  contained  by  them  shall  be  the  greatest  possible. 

3.  Divide  a  given  straight  line  into  two  parts  such  that  the  sum 
of  the  squares  on  the  two  parts  may  be  the  least  possible. 

4.  Divide  a  given  straight  line  into  three  parts  so  that  the  sum 
of  the  squares  on  them  may  be  the  least  possible. 

5.  ABC  is  an  equilateral  triangle  and  D  is  any  point  in  the  side 
BG.  Prove  that  the  square  on  2? C  is  equal  to  the  rectangle  contained 
by  BD,  DC,  together  with  the  square  on  ^D. 

6.  A  point  D  is  taken  on  the  hypotenuse  BC  of  a  right-angled 
triangle  yl/^ 6';  prove  that,  if  the  rectangle  BD,  DC  be  equal  to  the 
square  on  AD,  D  is  either  the  middle  point  ot  BC  or  the  foot  of  the 
perpendicular  from  A  on  BC. 


10 


146  BOOK  It. 

PROPOSITION  6. 

If  a  straight  line  he  bisected,  and  produced  to  any  pointy 
the  sum  of  the  rectangle  contained  hy  the  whole  line  thus 
produced  ajid  the  part  of  it  produced,  and  the  square  on 
half  the  line  bisected,  is  equal  to  the  square  on  the  straight 
line  which  is  made  up  of  the  half  and  the  part  produced  *. 

Let  the  straight  line  AB  be  bisected  at  the  point  C, 

and  produced  to  the  point  JD  : 

it  is  required  to  prove  that  the  sum  of  the  rect- 
angle AD,  BD,  and  the  square  on  GB  is  equal  to  the 
square  on  CD. 

A ^ ,B     D 


Proof.     Because  AD  i^  divided  into  two  parts  at  B, 

the    rectangle    AD,    BD    is    equal    to    the    sum    of    the 

rectangle  AB,  BD  and  the  square  on  BD.  (Prop.  3.) 

Because  AB  i^  bisected  in  C, 

the  rectangle  AB,  BD  is  double  of  the  rectangle  CB,  BD ; 

(Prop.  1.) 
therefore  the  rectangle  AD,  BD  is  equal  to  the  sum  of  the 
square  on  BD  and  twice  the  rectangle  GB,  BD. 
Add  to  each  of  these  equals  the  square  on  CB ; 
then  the  sum  of  the  square  on  CB  and  the  rectangle  AD,  BD 
is  equal  to  the  sum  of  the  squares  on  CB,  BD  and  twice 
the  rectangle  CB,  BD. 
And  the  sum  of  the  squares  on  CB,  BD  and  twice  the 
rectangle  CB,  BD  is  equal  to  the  square  on  CD', 

(Prop.  4.) 
therefore  the  sum  of  the  rectangle  AD,  BD  and  the  square 
on  CB  is  equal  to  the  square  on  CD. 
Wherefore,  if  a  straight  line  (fee. 

*  The  algebraical  equivalent  of  this  theorem  is  the  equation 
(a+&)(&-a)+a2=b2, 
or  (2a  +  &)b  +  a2  =  (a  +  &)2. 


PROPOSITION  6. 


U7 


Outline  of  Alternative  Proof  of  Propositions  5  and  6. 

Let  A,  B,  C  he  any  three  points  in  a  straight  line. 
Draw  AFE,  BGK,  CHD  at  right  angles  to  ABC. 


K      D 


H 


Take  AF  equal  to  AB  and  FE  to  BG,  and  draw  EKD,  FGH 
parallel  to  ABG. 
It  may  be  proved  that 

(1)  FB  is  the  square  on  AB^ 

(2)  KII  is  equal  to  the  square  on  BG, 

(3)  EB  is  the  rectangle  AB,  AG, 
and     (4)     ^^  is  the  rectangle  ^(7,  ^(7, 

and  hence  that  the  difference  of  the  squares  on  AB,  BG, 
which  is  equal  to  the  difference  of  the  rectangles  EB,  EH, 
is  equal  to  the  rectangle  contained  by  the  sum  and  the 
difference  of  AB  and  BG. 


EXERCISES. 

1.  A  straight  line  is  divided  into  two  equal  and  also  into  two 
unequal  parts;  prove  that  the  difference  of  the  squares  on  the  two 
unequal  parts  is  equal  to  twice  the  rectangle  contained  by  the  whole 
line  and  the  part  between  the  points  of  section. 

2.  The  straight  line  AB  is  bisected  in  C  and  produced  to  D,  GE 
is  drawn  perpendicular  io  AB  and  equal  to  BD,  and  a  point  F  is  taken 
in  BD  so  that  EF  is  equal  to  CD ;  prove  that  the  rectangle  J)F,  DA 
together  with  the  rectangle  DF,  FB  is  equal  to  the  square  on  BD. 


10—2 


148  BOOK  II. 


PROPOSITION  7. 

If  a  straight  line  he  divided  into  any  two  parts,  the 
suin  of  the  squares  on  the  whole  line  and  on  one  of  the  2)arts 
is  equal  to  the  S2im  of  twice  the  rectangle  contained  hy  the 
whole  line  and  that  part  and  the  square  on  the  other  part^. 

Let  the  straight  line  -4^  be  divided  into  any  two  parts 
at  the  point  C  : 
it  is  required  to  prove  that  the  sum  of  the  squares  on  AB, 

GB  is  equal  to  the  sum  of  twice  the  rectangle  AB,  GB, 

and  the  square  on  AC. 


Proof.     Because  AB  is  divided  into  two  points  at  (7, 
the  square  on  AB  is  equal  to  the  sum  of  the  squares  on 
AG,  CB,  and  twice  the  rectangle  AC,  CB.  (Prop.  4.) 

Add  to  each  of  these  equals  the  square  on  GB ; 
then  the  sum  of  the  squares  on  AB,  CB  is  equal  to  the 
sum  of  the  square  on  AC,  twice  the  square  on  GB,  and 
twice  the  rectangle  AC,  CB. 
But  the  sum  of  the  square  on  CB  and  the  rectangle  AC,  GB 
is  equal  to  the  rectangle  AB,  GB.  (Prop.  3.) 

Therefore  the  sum  of  the  squares  on  AB,  CB,  is  equal  to 
the  sum  of  twice  the  rectangle  AB,  CB,  and  the  square 
on  AC. 
Wherefore,  if  a  straight  line  &c. 

Corollary.  If  a  straight  line  he  divided  into  any  two 
parts,  the  square  on  one  of  the  parts  is  less  than  tlie  sum  of 
the  squares  on  the  whole  line  and  the  other  part  hy  twice  the 
rectangle  contained  hy  the  whole  line  and  the  second  part. 

*  The  algebraical  equivalent  of  this  theorem  is  the  equation 


PROPOSITION  7. 


149 


Outline  of  Alternative  Proof. 

On  AB  construct  the  square  A  DEB 
and  draw  CGF  parallel  to  AD. 

Take  BK  equal  to  BC, 
and  draw  KGH  parallel  to  BCA . 


B 


H 


Q 

E 


It  may  be  proved  that 

(1)  AX,  CBure  each  equal  to  the  rectangle  AB,  CB, 

(2)  IIF  is  equal  to  the  square  on  AC, 
and     (3)     CK  is  the  square  on  CB, 

and  hence  that  the  sum  of  the  squares  on  AB,  CB  is  equal 
to  the  sum  of  twice  the  rectangle  AB,  CB  and  the  square 
on  AC. 


EXERCISES. 


1.  ACDB  is  a  straight  line,  and  D  bisects  CB:  prove  that  the 
square  on  ^(7  is  less  than  the  sum  of  the  squares  on  AD,  DB  by  twice 
the  rectangle  AD,  DB. 

2.  If  BAG  be  an  acute  angle  and  BD,  CE  be  drawn  perpendicular 
to  CA,  AB  respectively,  then  the  rectangle  BA,  AE  is  equal  to  the 
rectangle  CA,  AD. 

3.  Shew  how  to  divide  a  given  straight  line  into  two  parts  such 
that  the  difference  of  the  squares  described  on  them  may  be  equal 
to  a  given  rectangle.     Is  a  solution  always  possible  ? 


150  BOOK  II. 

PROPOSITION  8. 

If  a  straight  line  he  divided  into  any  two  parts,  the  sum 
of  the  square  on  one  part  and  four  times  the  rectangle  con- 
tained by  the  whole  line  and  the  other  part,  is  equal  to  the 
square  on  the  straight  line  which  is  made  up  of  the  whole 
and  the  second  part  *. 

Let  the  straight  line  AB  be  divided  into  any  two  parts 
at  the  point  C  : 

it  is  required  to  prove  that  the  sum  of  four  times  the  rect- 
angle AB,  CB,  and  the  square  on  AG  is  equal  to  the 
square  on  the  straight  line  made  up  of  AB  and   CB 
together. 
Construction.     Produce  AB  to  D, 

and  make  BD  equal  to  CB.        (I.  Prop.  3.) 

A  G         B        D 

1 \ 


Proof.     Because  AD  is  divided  into  two  parts  at  B, 
the  sum  of  the  squares  on  AB,  BD  and  twice  the  rect- 
angle AB,  BD,  is  equal  to  the  square  on  AD  :    (Prop.  4.) 
and  because  AB  is  divided  into  two  parts  at  C, 
the    sum    of    the    square    on    AC    and    twice    the    rect- 
angle AB,   CB, 
is  equal  to  the  sum  of  the  squares  on  AB,  CB.     (Prop.  7.) 
Add  these  equals  together ; 
then  the  sum  of  the  squares  on  AB,  BD,  AC  and  four  times 
the  rectangle  AB,  CB, 

is  equal  to  the  sum  of  the  squares  on  AB,  CB,  AD. 
Take  away  from  these  equals,  the  equals  the  sum  of  the 
squares   on  AB,   BD  and    the  sum   of   the  squares   on 
AB,  CB; 
then  the  sum   of  the  square  on  -4C  and  four  times  the 
rectangle  AB,  CB  is  equal  to  the  square  on  AD. 
"Wherefore,  if  a  straight  line  &c. 

*  The  algebraical  equivalent  of  this  theorem  is  the  equation 
(a-&)2  +  4a6=(a  +  &)2, 
or  a^  +  i{a  +  b)b  =  {a+2b)\ 


PROPOSITION  8. 


151 


Outline  of  Alternative  Proof. 

Produce  AB  to  D  and  make  BD  equal  to  AG. 
On  AD  construct  the  square  AEFD. 


E      H 


N           P 

M 

L 

1 

Take  AG,  EH,  i'Xeach  equal  to  AG, 
and   draw  BLP,  MNH  parallel  to  AE  and  GML,  NPK 
parallel  \>o  AD. 
It  may  be  proved  that 

(1)     AL,  BK,  FN,  EM  are  each  equal  to  the  rect- 
angle AB,  AC, 
and     (2)     MP  is  equal  to  the  square  on  CB, 
and  hence  that  the  sum  of  the  square  on  ^C  and  four  times 
the  rectangle  AB,  GB  is  equal  to  the  square  on  AD. 


EXERCISES. 

1.  Prove  that  the  square  on  a  straight  Hue  is  nine  times  the 
square  on  one  third  of  the  line. 

2.  If  a  straight  line  be  bisected  and  produced  to  any  point,  the 
square  on  the  whole  line  thus  produced  is  equal  to  the  square  on  the 
part  produced  together  with  twice  the  rectangle  contained  by  the  line 
and  the  line  made  up  of  the  half  and  the  part  produced. 


162  BOOK  IL 


PROPOSITION  9. 


If  a  straight  line  he  divided  into  two  equals  and  also  into 
two  unequal  parts,  the  sum  of  the  squares  on  the  two  unequal 
parts  is  double  of  the  sum  of  the  squares  on  half  the  line  and 
on  the  line  between  the  points  of  section'^. 

Let  the  straight  line  AB  he  divided  into  two  equal 
parts  at  the  point  C,  and  into  two  unequal  parts  at  the 
point  D  : 

it  is  required  to  prove  that  the  sum  of  the  squares  on  AD^ 
DB  is  double  of  the  sum  of  the  squares  on  AC,  CD. 

A  C  D      B 

1 1 


Proof.     Because  AD  is  divided  at  C, 
the  square  on  A  JD  is  equal  to  the  sum  of  the  squares  on 
AC,  CD  and  twice  the  rectangle  AC,  CD.  (Prop.  4.) 

And  because  CB  is  divided  at  D, 
the  sum  of  the  square  on  DB  and  twice  the  rectangle  CB,  CD 
is  equal  to  the  sum  of  the  squares  on  CB,  CD. 

(Prop.  7.) 

Add  these  pairs  of  equals ; 

then  the  sum   of  the  squares  on  AD,  DB  and  twice  the 

rectangle  CB,  CD  is  equal  to  the  sum  of  the  squares  on 

AC,  CD,  CB,  CD  and  twice  the  rectangle  AC,  CD. 

Take  away  from  these  equals  twice  the  rectangle  CB,  CD, 

and  twice  the  rectangle  AC,  CD,  which  are  equal ; 
then  the  sum  of  the  squares  on  AD,  DB  is  equal  to  the 

sum  of  the  squares  on  AC,  CD,  CB,  CD, 
that  is,  is  equal  to  twice  the  sum  of  the  squares  on  AC,  CD. 
Wherefore,  if  a  straight  line  &c. 

*  The  algebraical  equivalent  of  this  theorem  is  the  equation 


PROPOSITION  9.  153 


The  theorems  of  Propositions  9  and  10  may  both  be 
included  in  one  enunciation  :  thus,  The  sum  of  the  squares 
on  the  sum  and  on  the  difference  of  two  given  straight  lines 
is  equal  to  twice  the  sum  of  the  squares  on  the  lines. 

The  straight  lines  in  both  propositions  are  A  C,  CD :  the  only 
difference  being  that  in  Prop.  9.  ^C  is  the  greater,  and  in  Prop.  10. 
CD  is  the  greater. 

For  an  outline  of  an  alternative  proof  of  Propositions  9  and  10, 
see  page  155. 


EXEKCISES. 

1.  A  straight  line  is  divided  into  two  parts,  such  that  the 
diagonal  of  the  square  on  one  of  these  parts  is  equal  to  the  whole 
line.  If  a  square  be  constructed  whose  side  is  the  difference  between 
the  aforesaid  part  and  half  the  given  line,  its  diagonal  is  equal  to  the 
other  of  the  two  parts  into  which  the  line  is  divided. 

2.  Deduce  a  proof  of  ii.  9  from  the  result  of  ii.  5. 

3.  If  a  straight  line  be  divided  into  two  equal  and  also  into  two 
unequal  parts,  the  squares  on  the  two  unequal  parts  are  equal  to 
twice  the  rectangle  contained  by  the  two  unequal  parts  together  with 
four  times  the  square  on  the  line  between  the  points  of  section. 


154  BOOK  II. 


PROPOSITION  10. 


If  a  st7'aight  line  be  bisected  and  produced  to  any  point, 
the  sum  of  the  squares  on  the  whole  line  thus  produced  and 
on  the  part  produced  is  double  of  the  sum  of  the  squares  on 
half  the  line  and  on  the  line  m^ade  up  of  the  half  and  the 
part  produced*. 

Let  the  straight  line  ^^  be  bisected  at  C,  and  produced 
toD: 

it  is  required  to  prove  that  the  sum  of  the  squares  on  AD, 
BD  is  double  of  the  sum  of  the  squares  on  AC,  CD. 

A  C  B  D 

1 1 


Proof.     Because  AD  is  divided  at  C, 
the  square  on  AD  is  equal  to  the  sum  of  the  squares  on 
AC,  CD  and  twice  the  rectangle  AC,  CD;  (Prop.  4.) 

and  because  CD  is  divided  at  B, 
the  sum  of  the  square  on  BD  and  twice  the  rectangle  CB,  CD 
is  equal  to  the  sum  of  the  squares  on  CB,  CD. 

(Prop.  7). 

Add  these  equals ; 

then  the  sum  of  the  squares  on  AD,  BD  and  twice  the 

rectangle  CB,  CD  is  equal  to  the  sum  of  the  squares  on 

AC,  CD,  CB,  CD  and  twice  the  rectangle  AC,  CD. 

Take  away  from  these  equals  twice  the  rectangle  CB,  CD, 

and  twice  the  rectangle  A  C,  CD,  which  are  equal ; 
then  the  sum  of  the  squares  on  AD,  BD  is  equal  to  the 

sum  of  the  squares  on  AC,  CD,  CB,  CD, 
that  is,  is  equal  to  twice  the  sum  of  the  squares  on  AC,  CD. 
Wherefore,  if  a  straight  line  &c. 

*  The  algebraical  equivalent  of  this  theorem  is  the  equation 


PROPOSITION  10. 


155 


Outline  of  AlterTiative  Proof  of  Propositions  9  a7ul  10. 

In  a  straight  line  A  BCD,  take  AB  equal  to  CD. 
Through  A,  B,  C,  D  draw  AJS,  BI\  CG,  DII  at  right  angles 
to  ABCD. 

Take  AK  equal  to  AB,  KL  to  BC  and  LE  to  AB. 
Draw  EFGH,  LQRS,  KMNP  parallel  to  ABCD. 

E  F  G    H 


Q               R 

M             N 

A    B 
It  may  be  proved  that 


C   D 


(1)  EQ,  ND  are  each  equal  to  the  square  on  AB, 

(2)  Z(7,  FP  are  each  equal  to  the  square  on  A  C, 

(3)  QN  is  equal  to  the  square  on  BC, 

(4)  ED  is  the  square  on  AD^ 

and     (5)     the  sum  of  ED  and  ^iV  is  equal  to  the  sum  of 

EQ,  ND,  LC,  and  FP, 
and  hence  that  the  sum  of  the  squares  on  AD,  BC  (which 
are  tlie  sum  and  the  difference  of  ^C  and  AB)  is  equal 
to  twice  the  sum  of  the  squares  on  AC,  AB. 

EXEECISES. 

1.  In  AB  the  diameter  of  a  circle  take  two  points  C  and  D 
equally  distant  from  the  centre,  and  from  any  point  E  in  the  circum- 
ference draw  EG,  ED:  shew  that  the  squares  on  EG  and  ED  are 
together  equal  to  the  squares  on  ^C  and  AD. 

2.  It  in  BG  the  base  of  a  triangle  a  point  D  be  taken  such  that 
the  squares  on  AB  and  BD  are  together  equal  to  the  squares  on  AG 
and  GD,  then  the  middle  point  of  AD  will  be  equally  distant  from 
£  and  G. 

3.  A  square  BDEG  is  described  on  the  hypotenuse  BG  of  a  right- 
angled  triangle  ABG :  shew  that  the  squares  on  DA  and  AG  are 
together  equal  to  the  squares  on  EA  and  AB. 

4.  AB  is  divided  into  any  two  parts  in  G,  and  AG,  BG  are 
bisected  in  D,  E :  prove  that  the  square  on  AE  and  three  times  the 
square  on  BE  are  equal  to  the  square  on  BD  and  three  times  the 
square  on  ^D. 


156 


BOOK  II. 


PROPOSITION  11. 

To  divide  a  given  straight  line  into  two  parts,  so  thai  tJie 
rectangle  contained  hy  the  whole  and  one  part  may  he  equal 
to  the  square  on  the  other  part^. 

Let  ABhe  the  given  straight  line  : 
it  is  required  to  divide  it  into  two  parts  in  a  point  H,  so 
that  the  rectangle  contained  by  the  whole  line  AB  and  a 
part  HB  may  be  equal  to  the  square  on  the  other  part  AH. 
CoNSTKUCTiON.     On  AB  construct  the  square  ABDC ; 

(I.  Prop.  46.) 

bisect  AC  at  E -,  (I.  Prop.  10.) 

draw  BE ;  produce  CA  to  F, 

and  make  j^i^"  equal  to  EB ;       (I.  Prop.  3.) 

and  on  AF  construct  the  square  AFGII  \    (I.  Prop.  46.) 

then  AB  \s,  divided  at  H  so  that  the  rectangle  AB,  HB  is 

equal  to  the  square  on  AH. 

Produce  GH  to  meet  CD  at  K. 


G 

H 


J^'  A  E  C 

Proof.    Because  ACis  bisected  at  E,  and  produced  to  F, 

the  sum  of  the  rectangle  FC,  FA,  and  the  square  on  AE  is 

equal  to  the  square  on  FE.  (Prop.  6.) 

But  FE  is  equal  to  EB.  (Constr.) 

Therefore  the  sum  of  the  rectangle  FC,  FA,  and  the  square 

on  AE  is  equal  to  the  square  on  EB. 

But,  because  the  angle  EAB  is  a  right  angle, 
the  square  on  EB  is  equal  to  the  sum  of  the  squares  on 
AE,  AB.  (I.  Prop.  47.) 

*  The  algebraical  equivalent  of  this  problem  is  to  find  the  smaller 
root  of  the  quadratic  equation  ax=:{a~x)^,  or  the  positive  root  of  the 
quadratic  equation  a{a-x)  —  x^. 


PROPOSITION  11.  157 

Therefore  the  sum  of  the  rectangle  FC,  FA,  and  the  square 

on  AE,  is  equal  to  the  sum  of  the  squares  on  AE,  AB. 

Take  away  from  each  of  these  equals  the  square  on  AE  ] 

then  the  rectangle  FC,  FA  is  equal  to  the  square  on  AB. 

But  the  figure  FK  is  the  rectangle  contained  by  FC,  FA, 

for  FG  is  equal  to  FA ;  (Constr.) 

and  AD  \s>  the  square  on  AB ; 

therefore  FK  is  equal  to  AD. 

Take  away  from  these  equals  the  common  part  AK; 

then  FH  is  equal  to  ED. 

But  HD  is  the  rectangle  contained  by  AB,  HB, 

for  -4^  is  equal  to  BD ;  (Constr.) 

and  FH  is  the  square  on  AH ; 
therefore  the  rectangle  AB,  HB  is  equal  to  the  square  on 
AH. 

Wherefore  the  straight  line  AB  is  divided  at  II,  so  that 
the  rectaiigle  AB,  II B  is  equal  to  the  square  on  AH. 


EXEKCISES. 

1.  Shew  that  in  a  straight  line  divided  as  in  II.  11  the  rectangle 
contained  by  the  sum  and  the  difference  of  the  parts  is  equal  to  the 
rectangle  contained  by  the  parts. 

2.  If  the  greater  segment  of  the  line  divided  in  this  proposition 
be  divided  in  the  same  manner,  the  greater  segment  of  the  greater 
segment  is  equal  to  the  smaller  segment  of  the  original  line. 

3.  Prove  that  when  a  straight  line  is  divided  as  in  this  pro- 
position the  square  on  the  line  made  up  of  the  given  line  and  the 
smaller  part  is  equal  to  five  times  the  square  on  the  larger  part. 

4.  Prove  that  in  the  figure  of  this  proposition  the  squares  on 
AB,  HB  are  together  equal  to  three  times  the  square  on  AH,  and 
that  the  difference  of  the  squares  on  AB,  AH  ia  equal  to  the  rect- 
angle AH,  AB. 

5.  Produce  a  given  straight  line,  so  that  the  rectangle  contained 
by  the  whole  line  thus  produced  and  the  given  line  may  be  equal 
to  the  square  on  the  part  produced. 


158  BOOK  II. 

PROPOSITION  12. 

In  an  obtuse-angled  triangle,  if  a  perpendicular  be  drawn 
from  either  of  the  acute  angles  to  the  opposite  side  jyroduced, 
the  square  on  the  side  opposite  the  obtuse  angle  is  greater  than 
the  sum  of  the  squares  on  tJte  other  sides,  by  twice  the  rect- 
angle contained  by  the  side  on  which,  when  produced,  the  per- 
pendicular falls,  and  the  part  of  the  produced  side  inter- 
cepted between  the  perpendicular  and  the  obtuse  angle. 

Let  ABC  be  an  obtuse-angled  triangle,  and  let  the  angle 
ACB  be  the  obtuse  angle;  from  the  point  A  let  AD  be 
drawn  perpendicular  to  BC  produced  : 

it  is  required  to  prove  that  the  square  on  AB  is  greater 
than  the  sum  of  the  squares  on  AC,  CB,  by  twice  the 
rectangle  BC,  CD. 


Proof.     Because  BD  is  divided  into  two  parts  at  C, 
the  square  on  BD  is  equal  to  the  sum  of  the  squares  on 

BC,  CD,  and  twice  the  rectangle  BC,  CD.  (Prop.  4.) 
To  each  of  these  equals  add  the  square  on  DA ; 

then  the  sum  of  the  squares  on  BD,  DA  is  equal  to 
the  sum  of  the  squares  on  BC,  CD,  DA,  and  twice  the 
rectangle  BC,  CD. 

But,  because  the  angle  at  Z)  is  a  right  angle, 
the  square  on  BA  is  equal  to  the  sum  of  the  squares  on 

BD,  DA, 

and  the  square  on  CA  is  equal  to  the  sum  of  the  squares  on 
CD,  DA.  (I.  Prop.  47.) 

Therefore  the  square  on  BA  is  equal  to  the  sum  of  the 
squares  on  BC,  CA,  and  twice  the  rectangle  BC,  CD ; 

that  is,  the  square  on  BA  is  greater  than  the  sum  of  the 
squares  on  BC,  CA  by  twice  the  rectangle  BC,  CD. 
Wherefore  in  an  obtuse-angled  triangle  &,c. 


PROPOSITION  12. 


159 


Outline  of  Alternative  Proof. 
On  the  sides  BC,  CA,  AB  of  an  obtuse-angled  triangle 
ABC^  in  which  tlie  angle   BAG  is   obtuse,  constioict  the 
squares  BBEC,  CFG  A,  AHKB. 

Draw  AL^  BAI,  CN  perpendicular  to  7iC,  CA,  AB  and 
produce  them  to  meet  the  opposite  sides  (produced  if  neces- 
sary) of  the  squares  in  P,  Q,  R. 

Draw  AD,  CK. 


D  P 

It  may  be  proved  that 

(1)     the  triangle  ABD  is  equal  to  the  triangle  KBC 
in  all  respects, 
and     (2)     the  rectangle  BP  is  equal  to  the  rectangle  BR, 

and  similarly  that  CQ  is  equal  to  CP,  and  ^i?  to  ^^, 
and    hence  that   the   square   on  BC  is    greater  than  the 
sum  of  the  squares  on  CA,  AB  by  twice  the  rectangle 
AR  or  AQ. 

EXEECISES. 

1.  The  sides  of  a  triangle  are  10,  12,  15  inches:  prove  that  it  is 
acute-angled. 

2.  On  the  side  BG  of  any  triangle  ABC,  and  on  the  side  of  BG 
remote  from  A,  a  square  BDEG  is  constructed.  Prove  that  the 
difference  of  the  squares  on  AB  and  ^C  is  equal  to  the  difference  of 
the  squares  on  AD  and  AE. 

3.  G  is  the  obtuse  angle  of  a  triangle  ABG,  and  D,  E  the  feet  of 
the  perpendiculars  drawn  from  A ,  B  respectively  to  the  opposite  sides 
produced:  prove  that  the  square  on  AB  is  equal  to  the  sum  of  the 
rectangles  contained  by  BC,  BD  and  AC,  AE. 

4.  ABG  is  a  triangle  having  the  sides  AB  and  AG  equal ;  AB  is 
produced  beyond  the  base  to  D  so  that  BD  is  equal  to  AB;  shew  that 
the  square  on  CD  is  equal  to  the  square  on  AB,  together  with  twice 
the  square  on  BG. 


160 


BOOK  II. 


PROPOSITION  13. 

I71  any  triangle^  the  square  on  a  side  subtending  an 
acute  angle,  is  less  tha7i  the  sum  of  the  squares  on  the  other 
sides,  by  twice  the  rectangle  contained  by  either  of  these  sides, 
and  the  part  of  the  side  intercepted  between  the  perpendicular 
let  fall  on  it  from  the  opposite  angle,  and  the  acute  angle. 

Let  ABC  be  a  triangle,  and  let  the  angle  ABC  be  an 
acute  angle;  let  AD  hQ  drawn  perpendicular  to  BG  and 
meet  it  (produced  if  necessary)  in  D  : 

it   is    required  to   prove  that  the  square   on    AC  is  less 
than  the  sum  of  the  squares  on  AB,  BC  by  twice  the 
rectangle  BC,  BD. 
Either  (1)  D  lies  in  BC,  or  (2)  i>  coincides  with  G,  or  (3)  D 
lies  in  BC  produced. 

(1)  (2)  (3) 


Proof.     Because  fig.  (1)  BC  is  divided  in  D,  fig.  (2)  B 
is  the  same  point  as  C,  or  fig.  (3)  BD  divided  in  C, 
the  sum  of  the  squares  on  BC,  BD  is  equal  to  the  sum  of 
the  square  on  CD,  and  twice  the  rectangle  BC,  BD. 

(I.  Prop.  47  and  II.  Prop.  7.) 
To  each  of  these  equals  add  the  square  on  DA  ; 
then  the  sum  of  the  squares  on  BC,  BD,  DA  is  equal  to 
the   sum   of   the   squares  on    CD,    DA   and   twice   the 
rectangle  BC,  BD. 

But  because  the  angle  BDA  is  a  right  angle, 
the  square  on  AB  is  equal  to  the  sum  of  the  squares  on 

BD,  DA, 
and  the  square  on  AC  is  equal  to  the  sum  of  the  squares 
on  CD,  DA.  (I.  Prop.  47.) 

Therefore  the  sum  of  the  squares  on  BG,  AB  is  equal  to  the 
sum  of  the  square  on  AC  and  twice  the  rectangle  BG,  BD : 
that  is,  the  square  on  AC  is  less  than    the   sum    of   the 
squares  on  AB,  BG  by  twice  the  rectangle  BG,  BD. 
Wherefore,  in  any  triangle  &c. 


PROPOSITION  13. 


161 


Outline  of  Alternative  Proof. 

On  the  sides  BC,  CA,  AB  of  an  acute-angled  triangle 
ABC  construct  the  squares  BDEC,  CFG  A,  AHKB. 
Draw  ALy  BM,  CN  perpendicular  to  BCy  CA,  AB,  and 
produce  tliera  to  meet  the  opposite  sides  of  the  squares 
in  P,  Q,  R. 

Draw  AD,  CK, 


It  may  be  proved  that 

(1)     the  triangle  ABD  is  equal  to  the  triangle  KBC 
in  all  respects, 
and  (2)     the  rectangle  BP  is  equal  to  the  rectangle  BR, 

and  similarly  that  CQ  is  equal  to  CP,  and  AR  to  AQ, 
and  hence  that  the  square  on  BG  is  less  than  the  sum  of 
the  squares  on  CA,  ABhy  twice  the  rectangle  AQ  or  AR. 

EXERCISES. 

1.  In  any  triangle  the  sum  of  the  squares  on  the  sides  is  equal  to 
twice  the  square  on  half  the  base  together  with  twice  the  square  on  the 
straight  line  drawn  from  the  vertex  to  the  middle  point  of  the  base. 

2.  The  base  of  a  triangle  is  given :  find  the  locus  of  the  vertex, 
when  the  sum  of  the  squares  on  the  two  sides  is  given. 

3.  The  sum  of  the  squares  on  the  sides  of  a  parallelogram  is 
equal  to  the  sum  of  the  squares  on  the  diagonals. 

4.  In  any  quadrilateral  the  squares  on  the  diagonals  are  together 
equal  to  twice  the  sum  of  the  squares  on  the  straight  lines  joining 
the  middle  points  of  opposite  sides. 

5.  The  squares  on  the  sides  of  a  quadrilateral  are  together 
greater  than  the  squares  on  its  diagonals  by  four  times  the  square 
on  the  straight  line  joining  the  middle  points  of  its  diagonals. 

T.  E.  11 


162 


BOOK  II. 


PROPOSITION  14. 

To  jind  the  side  of  a  square  equal  to  a  given  rectangle* 

Let  ABCD  be  the  given  rectangle  : 
it  is  required  to  find  the  side  of  a  square  equal  to  ABCD. 
Construction.    If  two  adjacent  sides  BA,  AD  he  equal, 
the  rectangle  is  a  square,  and  BA  or  AD  is  the  line  required. 
But  if  they  be  not  equal, 
produce  one  of  them  BA  to  E, 
and  make  AE  equal  to  ^Z>;       (I.  Prop.  3.) 
bisect  BE  at  F;  (I.  Prop.  10.) 

and  with  i^^as  centre  and  FB  as  radius, 

describe  the  circle  BGE, 

and  produce  DA  to  meet  the  circle  in  G ; 

then  AG  is  the  line  required. 

Draw  FG. 


Proof.     Because  BE  is  divided  into  two  equal  parts 
at  F,  and  into  two  unequal  parts  at  A, 
the  sum  of  the  rectangle  BA,  AE  and  the  square  on  FA  is 
equal  to  the  square  on  FE.  (Prop.  5.) 

But  FE  is  equal  to  FG. 
Therefore  the  sum  of  the  rectangle  BA,  AE  and  the  square 
on  FA  is  equal  to  the  square  on  FG. 

*  The  algebraical  equivalent  of  this  problem  is  to  find  the  positive 
root  of  the  quadratic  equation    9?  — ah. 


PROPOSITION  14.  163 

But  because  the  angles  at  A  are  right  angles,  the  square  on 
FG  is  equal  to  the  sum  of  the  squares  on  FA,  AG  ; 

(I.  Prop.  47.) 
therefore  the  sum  of  the  rectangle  BA,  AE  and  the  square 
on  FA,  is  equal  to  the  sum  of  the  squares  on  FA,  AG. 
Take  away  from  each  of  these  equals  the  square  on  FA  ; 
then  the  rectangle  BA,  AE  is  equal  to  the  square  on  AG. 
But  A  BCD  is  the  rectangle  contained  by  BA,  AE, 
since  it  is  contained  by  BA,  AD,  and  AE  \&  equal  to  AD. 
Therefore  A  BCD  is  equal  to  the  square  on  AG. 
Wherefore  the  straight  line  AG  has  been  found,  which  is 
the  side  of  a  square  equal  to  the  given  rectangle  ABCD. 

Corollary  1.  To  find  the  side  of  a  square  equal  to 
a  given  rectilineal  figure. 

Construct  a  rectangle,  i.e.  a  parallelogram,  which  has  an 
angle  equal  to  a  right  angle,  equal  to  the  given  figure, 
(I.  Prop.  45),  and  then  use  Proposition  14. 

Corollary  2.  The  square  on  the  perpendicular  from 
any  point  of  a  circle  on  any  diameter  is  equal  to  the  rectangle 
contained  by  the  parts  of  that  diameter  intercepted  between  its 
extremities  and  the  perpendicular. 


EXEKCISES. 

1.  Construct  a  rectangle  equal  to  a  given  square,  and  having 
(1)  the  sum  (2)  the  difference  of  two  of  its  adjacent  sides  equal  to 
a  given  straight  line. 

2.  The  largest  rectangle,  the  sum  of  whose  sides  is  given,  is  a 
square. 

3.  Construct  a  rectangle  equal  to  a  given  square  such  that  the 
difference  of  two  adjacent  sides  shall  be  equal  to  a  given  straight  line. 

4.  Construct  a  right-angled  triangle  equal  to  a  given  rectilineal 
figure  and  such  that  one  of  the  sides  containing  the  right  angle  is 
double  of  the  other. 

5.  Produce  a  given  straight  line  AB  both  ways  to  C  and  D  so 
that  the  rectangles  CA,  AD  and  CB,  BD  may  be  equal  respectively  to 
two  given  squares. 


164  BOOK  11. 


MISCELLANEOUS  EXERCISES. 

1.  In  a  triangle  whose  vertical  angle  is  a  right  angle  a  straight 
line  is  drawn  from  the  vertex  perpendicular  to  the  base:  shew  that 
the  square  on  either  of  the  sides  adjacent  to  the  right  angle  is  equal 
to  the  rectangle  contained  by  the  base  and  the  segment  of  it  adjacent 
to  that  side. 

2.  If  ABC  be  a  triangle  whose  angle  ^  is  a  right  angle,  and 
BE,  CF  be  drawn  bisecting  the  opposite  sides,  four  times  the  sum  of 
the  squares  on  BE  and  CF  is  equal  to  five  times  the  square  on  BG. 

3.  The  hypotenuse  AB  of  a  right-angled  triangle  ABC  19  tri- 
sected in  the  points  D,  E;  shew  that,  if  CD,  CE  be  joined,  the  sum 
of  the  squares  on  the  three  sides  of  the  triangle  CDE  is  equal  to  two 
thirds  of  the  square  on  AB. 

4.  ABCD  is  a  rectangle,  and  points  E,  F  are  taken  in  BC,  CD 
respectively.  Prove  that  twice  the  area  of  the  triangle  AEF  together 
with  the  rectangle  BE,  DF  is  equal  to  the  rectangle  AB,  BC. 

5.  On  the  outside  of  the  hypotenuse  BC,  and  the  sides  CA,  AB 
of  a  right-angled  triangle  ABC,  squares  BDEC,  AF,  and  AG  are 
described:  shew  that  the  squares  on  DG  and  EF  are  together  equal 
to  five  times  the  square  on  BC. 

6.  On  the  outside  of  the  hypotenuse  BC  of  a  right-angled  tri- 
angle ABC  and  on  the  sides  GA,  AB  squares  BDEC,  AF,  AG  are 
constructed:  prove  that  the  difference  of  the  squares  on  DG  and  EF 
is  three  times  of  the  difference  of  the  squares  onAB  and  AG. 

7.  In  the  hypotenuse  AB  of  a  right-angled  triangle  ACB,  points 
D  and  E  are  taken  such  that  AD  is  equal  to  AC  and  BE  to  BC; 
prove  that  the  square  on  DE  is  equal  to  twice  the  rectangle  BD,  AE. 

8.  One  diagonal  ^C  of  a  rhombus  ABCD  is  divided  into  any 
two  parts  at  the  point  P ;  shew  that  the  rectangle  AP,  PC  is  equal  to 
the  difference  between  the  squares  on  ^i^  &nd  PB. 

9.  In  a  given  straight  line  find  a  point  such  that  the  difference 
of  the  squares  upon  the  straight  lines  joining  it  to  two  given  points 
may  be  equal  to  a  given  rectangle.  In  what  cases  is  this  problem 
impossible? 

10.  If  a  straight  line  be  drawn  through  one  of  the  angles  of  an 
equilateral  triangle  to  meet  the  opposite  side  produced,  so  that  the 
rectangle  contained  by  the  whole  straight  line  thus  produced  and  the 
part  of  it  produced  is  equal  to  the  square  on  the  side  of  the  triangle, 
the  square  on  the  straight  line  so  drawn  will  be  double  the  square  on 
a  side  of  the  triangle. 

11.  The  square  on  any  straight  line  drawn  from  the  vertex  of  an 
isosceles  triangle  to  any  point  in  the  base  is  less  than  the  square  on  a 
side  of  the  triangle  by  the  rectangle  contained  by  the  segments  of  the 
base. 


MISCELLANEOUS  EXERCISES.  165 

12.  In  the  figure  of  II.  11,  BE  and  CH  meet  at  0 ;  shew  that 
.40  is  at  right  angles  to  CH. 

13.  Divide  a  given  straight  line  so  that  the  square  on  one  part  is 
equal  to  the  rectangle  contained  by  the  other  part  and  another  given 
straight  line. 

14.  Divide  a  given  straight  line  so  that  the  rectangle  contained 
by  the  parts  shall  be  equal  to  the  rectangle  contained  by  the  whole 
line  and  the  difference  of  the  parts. 

15.  How  many  triangles  can  be  formed  by  choosing  three  lines 
out  of  six  whose  lengths  are  3,  4,  5, 11,  12  and  13  inches?  How  many 
of  these  triangles  are  (1)  obtuse-angled,  (2)  right-angled,  (3)  acute- 
angled  ? 

16.  Prove  that  the  sum  of  the  squares  on  the  straight  lines 
drawn  from  any  point  to  the  middle  points  of  the  sides  of  a  triangle 
is  less  than  the  sum  of  the  squares  on  the  straight  lines  drawn  from 
the  same  point  to  the  angular  points  of  the  triangle  by  one  quarter 
the  sum  of  the  squares  on  the  sides  of  the  triangle. 

17.  ABBG  is  a  parallelogram  whose  diagonals  intersect  in  0. 
OL,  OM  are  drawn  at  right  angles  to  AB,  AG  meeting  them  in  L, 
M  respectively;  prove  that  the  sum  of  the  rectangles  AB,  AL  and 
AC,  AM  is  double  the  square  on  AO. 

18.  In  a  triangle  ABC  the  angles  B  and  C  are  acute:  if  E,  F 
be  the  points  where  perpendiculars  from  the  opposite  angles  meet  the 
sides  AC,  AB,  the  square  on  BC  is  equal  to  the  rectangle  AB,  BF, 
together  with  the  rectangle  A  C,  CE.  What  change  is  made  in  the 
theorem,  if  B  be  an  obtuse  angle? 

19.  Prove  that  the  locus  of  a  point,  whose  distance  from  one  of 
two  fixed  points  is  double  that  from  the  other,  is  a  circle. 

20.  At  B  in  AB  two  equal  straight  lines  BC,  BD  are  drawn 
making  equal  angles  with  AB  and  AB  produced.  Shew  that  the 
difference  of  the  squares  on  AC  and  AD  is  equal  to  twice  the  rectangle 
AB,  CD. 

21.  If  the  squares  on  the  sides  of  a  quadrilateral  be  equal  to  the 
squares  on  the  diagonals,  it  must  be  a  parallelogram. 

22.  ABC  is  a.  triangle  having  the  angle  C  greater  than  the  angle 
B,  and  D  a  point  in  the  base  BC,  such  that  the  sum  of  the  squares 
on  AB,  ^C  is  twice  the  sum  of  the  squares  on  AD,  DB.  Shew  that 
either  D  is  the  middle  point  of  the  base,  or  that  its  distance  from  the 
foot  of  the  perpendicular  from  A  on  BC  is  one  half  of  BC. 

23.  If  a  figure  be  composed  of  a  rhombus  and  the  square 
described  outwards  on  one  of  its  sides,  the  side  of  the  equivalent 
square  is  equal  to  half  the  sum  of  the  diagonals  of  the  rhombus. 

24.  ABC  is  a  triangle  in  which  C  is  a  right  angle,  and  DE  is 
drawn  from  a  point  D  in  ^C  perpendicular  to  AB;  shew  that  the 
rectangle  AB,  AE  is  equal  to  the  rectangle  AC,  AD. 


166  BOOK  11. 

25.  ABC  is  an  acute-angled  triangle  ;  perpendiculars  AFD,  BPE 
are  drawn  on  EG,  GA  from  the  opposite  angles.  Prove  that  the  rect- 
angle BD,  DG  is  equal  to  the  rectangle  AD,  PD. 

26.  A,  B,  G,  D  are  the  angular  points  of  a  parallelogram,  taken 
in  order.  If  there  be  a  point  P  such  that  the  sum  of  the  squares  on 
PA  and  PG  be  equal  to  the  sum  of  those  on  PB  and  PD,  ABGD  is  a 
rectangle. 

27.  A,  B,  G,D  are  fixed  points,  and  P  is  a  point  such  that  the  sum 
of  the  squares  on  PA,  PB,  PG,  PD  is  constant;  prove  that  P  lies  on 
a  circle,  the  centre  of  which  is  at  the  point  where  the  straight  line 
joining  the  middle  points  of  AB,  GD  cuts  the  straight  line  joining 
the  middle  points  of  AD,  BG. 

28.  A,  B,  G,  D  are  fixed  points,  and  P  is  a  point  such  that  the 
sum  of  the  squares  on  PA ,  PB  is  equal  to  the  sum  of  the  squares  on 
PG,  PD.  Prove  that  the  locus  of  P  is  a  straight  line  at  right  angles 
to  the  line  joining  the  middle  points  of  AB,  GD,  and  passes  through 
the  intersection  of  the  lines  drawn  perpendicular  to  either  of  the  pairs 
of  lines  AG,  BD  or  AD,  BG  at  their  middle  points. 

29.  In  the  plane  of  a  triangle  AEG  find  a  point  P  such  that  the 
sum  of  the  squares  on  AP,  BP  and  GP  may  be  a  minimum. 

30.  Produce  a  given  straight  line  so  that  the  rectangle  contained 
by  the  whole  line  thus  produced  and  the  produced  part  may  be  equal 
to  the  square  on  another  given  straight  line. 


CAMBRIDGE  :    PRINTBU  BY   C.   J.   CLAT,  M.A.   AND  SONS,   AT  THE   UNIVERSITY   PRESS. 


BOOK   III. 

DEFINITIONS. 

Definition  1.     Any  part  of  a  circle  is  called  an  arc. 

The  line  which  has  been  defined  (i.  Def.  22)  as  a  circle 
is  often  spoken  of  as  the  circumference  of  the  circle. 

The  reason  of  this  is  that  a  circle  is  defined  in  many  books  as  the 
part  of  the  plane  contained  by  the  line,  which  is  then  called  the 
circumference. 

Half  of  a  circle  is  called  a  semicircle. 

It  will  be  proved  hereafter  that  a  diameter  bisects  a  circle,  i.e. 
divides  it  into  two  equal  arcs.     (See  page  175.) 

Definition  2.  A  straight  line  joining  two  points  on  a 
circle  is  called  a  chord  of  the  circle. 

The  straight  line  joining  the  extremities  of  an  arc  is 
called  the  chord  of  the  arc. 

The  figure  formed  of  an  arc  and  the  chord  of  the  arc  is 
called  a  segment  of  the  circle. 

In  the  diagram  the  straight  lines  AB, 
BC,  CA  are  chords  of  the  circle  ABC  \ 
AFB,  BDC,  CEA  are  arcs. 

The  straight  line  AB  is  the  chord  of 
the  arc  AFB,  and  it  is  also  the  cliord  of 
the  arc  ACB. 

The  figure  formed'  of  the  arc  BFEC 
and  the  chord  BG  is  called  the  segment 
BFEC  or  BFAG,  or  more  often  BFC  or 

BEG  or  BAG;  and  the  figure  formed  of  the  arc  BDG  and  the  chord 
BC  is  called  the  segment  BDC. 


168 


BOOK  III. 


Definition  3.  Tke  angle  contained  by  two  chords  joining 
a  point  in  an  arc  of  a  circle  to  the  extremities  of  the  arc  is 
called  an  angle  in  the  arc,  and  the  arc  is  said  to  contain  the 
angle. 

An  angle  in  an  arc  is  often  spoken  of  as  an  angle  in  the 
segment  formed  by  the  arc  and  the  chord  of  the  arc,  and 
the  segment  is  said  to  contain  the  angle. 


The  angle  BAG  is  said  to  be  an  angle 
in  the  arc  (or  in  the  segment)  BFC  and 
the  angle  ACB  an  angle  in  the  arc  (or  in 
the  segment)  ADB ;  and  the  arc  (or  the  seg- 
ment) BFC  is  said  to  contain  the  angle 
BA  C,  and  the  arc  (or  the  segment)  ADB 
to  contain  the  angle  ACB. 


An  angle  in  an  arc  is  said  to  stand  on  the  arc  which 
forms  the  remainder  of  the  circle. 

The  angle  BAC  is  said  to  stand  on  the  arc  BDC,  and  the  angle 
ABC  on  the  arc  AEC. 

Definition  4,  Arcs,  which  contain  equal  angles,  are 
said  to  be  similar;  and  likewise  segments,  which  contain 
equal  angles,  are  said  to  be  similar. 


In  the  diagram  the  arcs  ABC,  DEF  are  said  to  be  similar,  when 
the  contained  angles  ABC,  DEF  are  equal;  and  also  the  segments 
ABC,  DEF  are  said  to  be  similar  when  the  contained  angles  ABC, 
DEF  are  equal. 


DEFINITIONS. 


169 


Definition  5.  A  jjointj  whose  distance  from  the  centre  of 
a  circle  is  less  than  the  radius  of  the  circle,  is  said  to  he 
within  the  circle ;  and  a  jjoint,  whose  distance  from  the 
centre  of  a  circle  is  greater  tJian  the  radius  of  the  circle,  is 
said  to  he  without  the  circle. 


In  the  diagram  the  point  P  is  within  the  circle,  its  distance  GP 
from  the  centre  C  being  less  than  the  radius  GA ,  and  the  point  Q,  is 
without  the  circle,  its  distance  CQ  from  the  centre  G  being  greater 
than  the  radius  GB. 

It  is  clear  that  any  line  drawn  from  a  point  P  within  a  circle  to  a 
point  Q  without  the  circle  must  intersect  the  circle  once  at  least 
(I.  Postulate  7,  page  14). 


In  the  diagram  the  straight  line  PQ 
meets  the  circle  in  the  points  R  and  S, 
and  the  straight  line  and  the  circle  inter- 
sect at  each  of  those  points. 


Definition  6.  A  sti'aight  line  and  a  circle,  which  pass 
through  a  poi7it  hut  do  not  intersect  there,  are  said  to  touch 
one  another  at  the  point.  The  sti'aight  line  is  called  a  tan- 
gent to  the  circle. 


In  the  diagram  the  circle  ABG  and 
the  straight  line  DCE  pass  through 
the  point  C,  but  do  not  intersect  there : 
they  touch  at  the  point  C,  and  DE 
is  a  tangent  to  the  circle  at  the  point 
G. 


T.  E. 


12 


170 


BOOK  III. 


In  the  diagram  the  circles  PRS, 
QRS  raeet  at  the  points  R  and  S,  and 
the  circles  intersect  at  each  of  those 
points :  for  instance,  points  on  the 
circle  PES  near  R  on  one  side  of  R 
lie  within  the  circle  QRS  and  on  the 
other  side  of  R  without  it. 


Definition  7.  Two  circles^  which  jmss  through  a  2^oint 
hut  do  not  intersect  there,  are  said  to  touch  one  another 
at  the  point. 


In  each  of  the  figures  in  the  diagram  the  circles  ABC,  BDE  pass 
through  the  point  B,  but  do  not  intersect  there:  all  points  on  the 
circle  ABC  near  B  lie  without  the  circle  BDE:  and  all  points  on 
the  circle  BDE  near  B  in  figure  (1)  lie  without  the  circle  ABC,  and 
in  figure  (2)  lie  within  the  circle  ABC. 

(See  remarks  on  the  contact  of  circles  on  pages  199  and  201.) 


Definition  8.     Circles  which  have  the  same  point  for 
a  centre  are  said  to  be  concentric. 


DEFINITIONS. 


171 


Deb'INITION  9.  The  ferpeftidicular  clrmvn  to  a  straight 
line  from  a  point  is  called  the  distance  of  the  straight  line 
from  the  point. 

If  the  distances  of  two  straight  lines  from  a  point  he  equals 
the  straight  lines  are  said  to  he  equidistant  from  the  point. 


In  the  diagram,  if  the  straight  lines  OM,  ON  be  drawn  from  the 
point  0  perpendicular  to  the  two  straight  lines  AB,  CD,  then 
OM,  ON  are  called  the  distances  of  the  two  straight  lines  AB,  CD 
from  the  point  0. 

If  OM,  ON  be  equal,  the  straight  Unes  AB,  CD  are  said  to  be 
equidistant  from  the  point  0. 

If  the  distances  of  two  straight  lines  from  a  point  he 
unequal,  the  line  the  distance  of  which  is  the  longer  is  said  to 
he  farther  from  the  point,  and  the  line  the  distance  of  which 
is  the  shorter  is  said  to  he  nearer  to  the  point. 


In  the  diagram,  if  the  straight  lines  OM,  ON  be  drawn  from  the 
point  O  perpendicular  to  the  two  straight  lines  AB,  CD,  and  if  OM  be 
less  than  ON,  then  CD  is  said  to  be  farther  from  the  point  O  than 
AB,  and  AB  is  said  to  be  nearer  to  the  point  0  than  CD. 

12—2 


172 


BOOK  III. 


Definition  10,  If  all  the  angular  points  of  a  rectilineal 
figure  lie  on  a  circle,  the  figure  is  said  to  be  inscribed  in  the 
circle,  and  the  circle  is  said  to  he  described  about  the  figure. 


In  the  diagram,  if  the  angular  points  of  the  triangle  ABC  lie 
on  the  circle  ABC,  the  triangle  ABC  is  said  to  be  inscribed  in  the 
circle  ABC,  and  the  circle  ABC  is  said  to  be  described  about  the  tri- 
angle ABC.  Similarly,  if  the  angular  points  D,  E,  F,  G  of  the 
quadrilateral  DEFG  lie  on  the  circle  DEFG,  the  quadrilateral  D£2^(r 
is  said  to  be  inscribed  in  the  circle  DEFG,  and  the  circle  DEFG  is 
said  to  be  described  about  the  quadrilateral  DEFG. 


If  all  the  sides  of  a  rectilineal  figure  touch  a  circle,  the 
figure  is  said  to  he  described  about  the  circle,  and  the  circle 
is  said  to  he  inscribed  in  the  figure. 


In  the  diagram,  if  the  sides  BC,  CA,  AB  of  the  triangle  ABC  touch 
the  circle  DEF  at  the  points  D,  E,  F  respectively,  the  triangle  ABC  is 


DEFINITIONS.  173 

said  to  be  described  about  the  circle  DEF,  and  the  circle  DEF  is  said 
to  be  inscribed  in  the  triangle  ^UC  Similarly,  if  the  sides  LG,  GH, 
HK,  KL  of  the  quadrilateral  GHKL  touch  the  circle  MNPQ  at  the 
points  M,  N,  P,  Q  respectively,  the  quadrilateral  GHKL  is  said  to  be 
described  about  the  circle  MNPQ,  and  the  circle  MNPQ  is  said  to  be 
inscribed  in  the  quadrilateral  GHKL. 


174  BOOK  III. 

PROPOSITION  1. 
A  circle  cannot  have  more  than  07ie  ce^itre. 

Let  A  be  a  centre  of  the  given  circle  BCD : 
it  is  required  to  prove  that  no  other  point  can  be  a  centre 
of  the  circle  BCD, 
Construction.     Take  any  point  E  within  the  circle  and 
draw  AE, 

and  produce  AE  both  ways  to  meet  the  circle,  beyond  A 
in  C,  and  beyond  E  in  D. 


Proof.     Because  ^  is  a  centre  of  tlie  circle, 

AC  is  equal  to  AD.  (I.  Def.  22.) 

Now  EC  is  greater  than  AC,  which  is  only  a  part  of  it, 
and  J^jD,  which  is  only  a  part  of  AD,  is  less  than  AD; 
therefore  EC  is  greater  than  ED. 
But  a  centre  of  a  circle  is  a  point  from  which  all  straight 
lines  drawn  to  the  circle  are  equal ;  (I.  Def.  22.) 
therefore  E  cannot  be  a  centre  of  the  circle. 
Similarly  it  can  be  proved  that  no  point  other  than  A  can 
be  a  centre. 
Wlierefore,  a  circle  cannot  have  &c. 

The  definition  of  a  circle  implies  that  the  figure  has  a 
centre  (I.  Def.  22):  it  is  here  proved  that  it  cannot  have 
more  tlian  one  centre:  we  shall  therefore  for  the  future 
speak  of  the  centre  of  a  circle. 


FROPOSITIOX  1.  175 


PKOPOSITION    lA. 

A  diameter  bisects  a  circle. 

Let  A  BCD  be  a    circle,    0  the 
centre  and  AOC  a  diameter: 
it  is  required  to  prove  that  the  arcs 
ABC,  ADC  are  equal. 

Construction.  Draw  any  two 
radii  OP,  OQ  making  equal  angles 
POC,  QOC  with  DC.     (I.  Prop.  23.) 

Proof.     It  is  possible  to  shift  the  figure  AOCQD  by 
turning  it  over  so  that  AOC  is  not  shifted, 
and  so  that  the  arcs  ADC,  ABC  lie  on  the  same  side  of  AC. 
If  this  be  done, 

because  the  angle  QOC  is  equal  to  the  angle  POC, 
OQ  must  coincide  in  direction  with  OP'. 
and  because  OQ  i^  equal  to  OP, 

Q  must  coincide  with  P. 

Similarly  it  can  be  proved  that  every  point  on  the  arc  ADC 

must  coincide  with  some  point  on  the  arc  ABC, 

and  every  point  on  the  arc  ABC  with  some  point  on  the 

arc  ADC. 

Therefore  the  arc  ADC  coincides  with  the  arc  ABC, 

and  is  equal  to  it. 
Wherefore,  a  dicuneter  bisects  &,c. 


EXEECISES. 

1.  Prove  by  superposition  that  circles,  which  have  equal  radii, 
are  equal. 

2.  Prove  by  superposition  that  equal  circles  have  equal  radii. 

3.  Two  circles,  which  have  a  common  centre,  but  whose  radii  are 
not  equal,  cannot  meet. 

4.  Prove  by  superposition  that  two  diameters  at  right  angles 
divide  a  circle  into  four  equal  arcs. 


176  BOOK  III. 


PROPOSITION  2. 

Jf  a  straight  line  bisect  a  chord  of  a  circle  at  right  angles, 
the  line  passes  through  the  centre. 

Let  ^^  be  a  chord  of  the  circle  ABC,  and  let  CDE  be 
the  straight  line  which  bisects  AB  at  right  angles: 
it  is  required  to  prove  that  CDIJ  passes  through  the  centre 
of  the  circle  ABC. 
Construction.     Take  any  point  G  not  in  CJS  and  on 
the  same  side  of  CJS  as  B. 

Draw  AG  cutting  CB  in  JI,  and  draw  GB,  HB. 

C 


Proof.     Because  in  the  triangles  ADH,  BDH, 
AD  is  equal  to  BJ), 
and  DH  to  BE, 
and  the  angle  ADH  is  equal  to  the  angle  BDlf, 

the  triangles  are  equal  in  all  respects;   (I.  Prop.  4.) 

therefore  HA  is  equal  to  HB. 
Therefore  GA,  which  is  the  sum  of  GH,  HA, 
is  equal  to  the  sum  of  GH,  HB. 
And  the  sum  of  GH,  HB  is  greater  than  GB;  (I.  Prop.  20.) 

therefore  GA  is  greater  than  GB. 

But  all  straight  lines  drawn  from  the  centre  to  a  circle  are 

equal.  (I.  Def.  22.) 

Hence  the  point  G  cannot  be  the  centre  of  the  circle. 

Similarly  it  can  be  proved  that  no  point  on  the  same  side 

of  GE  as  A  can  be  the  centre; 

therefore  the  centre  must  be  in  CE. 
Wherefore,  if  a  straight  line  &c. 


PROPOSITION  2. 


Corollary  1. 

Only  one  chord  drawn  through  a  point  toithin  a  circle 
which  is  7iot  the  centre  can  he  bisected  at  the  point. 

Corollary  2. 

If  two  chords  of  a  circle  bisect  each  other ^  their  point  of 
intersection  is  the  centre. 


EXEECISES. 


1.  The  straight  line,  which  joins  the  middle  points  of  two  parallel 
chords  of  a  circle,  passes  through  the  centre. 

2.  The  locus  of  the  middle  points  of  parallel  chords  in  a  circle  is  a 
straight  line. 

3.  Two  equal  parallel  chords  of  a  circle  are  equidistant  from  the 
centre. 

4.  Every  parallelogram  inscribed  in  a  circle  is  a  rectangle. 

5.  The  diagonals  of  a  rectangle  inscribed  in  a  circle  are  diameters 
of  the  circle. 

6.  If  PQ,  RS  be  two  parallel  chords  of  a  circle  and  if  PR,  QS  in- 
tersect in  U  and  if  PS,  QR  intersect  in  F,  then  UV  passes  through 
the  centre. 


178  BOOK  11 L 


PROPOSITION  3. 


If  a  straight  liyie  he  drawn  from  the  centre  of  a  circle  to 
the  7nid(Ue  point  of  a  chord,  which  is  not  a  diameter,  it  is  at 
right  angles  to  the  chord. 

Let  ABC  be  a  circle  and  F  its  centre,  and  let  D  be 
the  middle  point  oi  AB  a  chord,  Avhich  is  not  a  diameter : 
it  is  required  to  prove  that  FD  is  at  right  angles  to  AB. 

Construction.     Draw  BA,  EB. 


Proof.     Because  in  the  triangles  ADE,  BDE, 
AD  \^  equal  to  BD, 

DE  to  DE, 
and  EA  to  EB, 
the  triangles  are  equal  in  all  respects;  (I.  Prop.  8.) 
therefore  the  angle  ADE  is  equal  to  the  angle  BDE, 
and  they  are  adjacent  angles. 
Therefore  the  straight  lines  EI),  AB  are  at  right  angles  to 
each  other.  (I.  Def.  11.) 

Wherefore,  if  a  straight  line  &c. 


PROPOSITION  3.  179 


EXEKCISES. 

1.  Why  are  the  words  "which  is  not  a  diameter"  inserted  in 
enunciation  of  Proposition  3  ? 

2.  The  straight  Une,  which  joins  the  middle  points  of  two  parallel 
chords  of  a  circle,  is  at  right  angles  to  the  chords. 

3.  Circles  are  described  on  the  sides  of  a  quadrilateral  as  diame- 
ters: shew  that  the  common  chord  of  the  circles  described  on  two 
adjacent  sides  is  parallel  to  the  common  chord  of  the  other  two  circles. 

4.  A  straight  line  is  drawn  intersecting  two  concentric  circles  : 
prove  that  the  portions  of  the  straight  line  which  are  intercepted 
between  the  circles  are  equal. 

5.  A  straight  line  cuts  two  concentric  circles  in  P,  E,  and  Q,  S: 
prove  that  the  rectangle  PQ,  QR  and  the  rectangle  FQ,  PS  are  con- 
stant for  all  positions  of  the  line. 


180  BOOK  III. 

PROPOSITION  4. 

If  a  straight  liiie  he  drawn  from  the  centre  of  a  circle  at 
right  angles  to  a  chord,  it  bisects  the  chord. 

Let  A  BC  be  a  circle,  and  D  its  centre,  and  let  the  straight 
line  DE  he  drawn  at  right  angles  to  AB  a  chord,  which 
is  not  a  diameter*: 

it  is  required  to  prove  that  DE  bisects  AB,  that  is,  that 
J^  is  equal  to  BE. 

Construction.     Draw  DA,  BB. 


Proof.     Because  DB  is  equal  to  DA, 

each  being  a  radius  of  the  circle, 

the  angle  DAB  is  equal  to  the  angle  DBA.  (I.  Prop.  5.) 

And  the  angle  DEA  is  equal  to  the  angle  DEB, 

each  being  a  right  angle. 

Then  because  in  the  triangles  EAD,  EBD, 

the  angle  EAD  is  equal  to  the  angle  EBD, 

and  the  angle  DEA  to  the  angle  DEB, 

and  ED,  a  side  opposite  to  a  pair  of  equal  angles,  is  common, 

the  triangles  are  equal  in  all  respects ; 

(I.  Prop.  26,  Part  2.) 
therefore  ^^  is  equal  to  BE. 
Wherefore,  if  a  straight  line  &c. 


The  case  when  the  chord  is  a  diameter  requires  no  proof. 


PROPOSITION  4.  181 


In  Propositions  2,  3,  4  we  have  to  deal  with  tliree 
properties  of  a  straight  line : 

(a)  the  passing  through  the  centre  of  a  circle, 
(6)  the  being  at  right  angles  to  a  given  chord, 
(c)  the  bisecting  the  given  chord. 

It  is  proved  in  these  propositions  that,  if  a  straight  line 
have  any  two  of  these  three  properties,  it  necessarily  has  the 
third  property. 

Proposition  2  deduces  {a)  from  (6)  and  (c) ; 
Proposition  3  deduces  ih)  from  (c)  and  (a); 
Proposition  4  deduces  (c)  from  (a)  and  (Z>). 


EXEKCISES. 

1.  Two  chords  are  drawn  through  a  point  on  a  circle  equally 
inclined  to  the  radius  drawn  to  the  point :  prove  that  the  chords  are 
equal. 

2.  If  ABPQ,  ABBS  be  two  circles  and  BB,  QS  be  any  two 
parallel  straight  lines  drawn  through  the  points  of  section,  then 
PB,  QS  are  equal. 

3.  If  A  and  B  be  two  fixed  points  and  P  move  so  that  the  per- 
pendicular from  A  on  BP  bisects  BP,  the  locus  of  P  is  a  circle. 

4.  Draw  through  a  point  of  intersection  of  two  circles  a  straight 
line  to  make  equal  chords  in  the  two  circles. 

5.  The  locus  of  the  middle  points  of  all  chords  drawn  through  a 
fixed  point  on  a  circle  is  a  circle. 

0.  If  two  circles  PAB,  QAB  intersect  each  other  at  A,  the  locua 
of  the  middle  point  of  a  straight  line  PQ  drawn  through  J  is  a  circle. 


182  BOOK   III. 


PROPOSITION  5. 

To  find  the  centre  of  a  given  circle. 

Let  ABC  be  a  given  circle: 
it  is  required  to  find  its  centre. 

Construction.     Draw  any  two  chords  which  are  not 
parallel  and  which  cut  the  circle  in  A^  B,  and  in  C,  D. 

Bisect  AB  and  CD  at  E  and  F;  (I.  Prop.  10.) 

and  draw  EG^  EG  at  right  angles  to  AB,  CD  respectively 
meeting*  at  G :  (I.  Prop.  11.) 

then  G  is  the  centre  of  the  circle  ABC. 


Proof.     Because  the  straight  line  EG  bisects  the  chord 
AB  at  right  angles, 

EG  passes  through  the  centre;  (Prop.  2.) 

and  because  the  straight  line  EG  bisects  the  chord  CD  at 
right  angles, 

EG  passes  through  the  centre.  (Prop.  2.) 

Therefore  the  point  G,  where  the  two  lines  EG,  EG  meet,  is 

the  centre. 
"       Wherefore,  the  centre  G  of  the  given  circle  ABC  has  been 
found. 

*  The  lines  must  meet.     See  Ex.  2,  p.  51. 


PROPOSITION  5. 


183 


Outline  of  Alternative  Construction. 

Draw  any  cliord  AB,  of  the  circle  ABC. 
Bisect  AB  \n  D,  and  draw  EDF  at  right  angles  to  AB, 
meeting  the  circle  in  E  and  F. 
Bisect  EF  in  G. 


It  may  be  proved  that 

(1)  the  centre  of  the  circle  ABC  is  in  EF; 

(2)  no  other  point  but  G  can  be  the  centre. 


EXEKCISES. 

1.  Draw  all  the  lines,  which  are  wanted  to  find  the  centre  of  a 
given  circle, 

(a)  in  the  method  given  in  the  text ; 

(6)  when  the  two  chords  in  this  method  meet  on  the  circle; 

(c)  in  the  alternative  method  above. 

Which  method  requires  the  fewest  lines  ? 

2.  Draw  through  a  given  point  within  a  circle   a  chord   such 
that  it  is  bisected  at  the  point. 

3.  Describe  a  circle  with  a  given  centre  to  cut  a  given  circle  at 
the  extremities  of  a  diameter. 


184  BOOK  II L 

PROPOSITION  6. 
Every  chord  of  a  circle  lies  within  the  circle. 
Let  .4^  be  the  chord  joining  any  two  points  A,  B  on  tlie 
circle  ABC: 

it  is  required  to  prove  that  any  point  on  the  chord  AB 
between  A  and  B  is  within  the  circle. 
Construction.  Pind  the  centre  D  of  the  circle;  (Prop.  5.) 
take  any  point  E  on  AB   between  A    and  B  and  draw 
DA,  BE,  DB, 

Proof.     Because   in    the  triangle 
DAB,  DB  is  equal  to  DA, 
the  angle  DAB  is  equal  to  the  angle 
DBA;      (I.  Prop.  5.) 
but  the  exterior  angle  DEB  of  the  tri- 
angle DAE  is  greater  than  the  inte- 
rior opposite  angle  DAE;  (I.  Prop.  16.) 
therefore  the  angle   DEB  is   greater 
than  the  angle  DBE. 

And  because  in  the  triangle  DEB,  the  angle  DEB  is  greater 
than  the  angle  DBE, 
the  side  DB  is  greater  than  the  side  DE ;  (I.  Prop.  19.) 
that  is,  DE  is  less  than    DB  which  is  a  radius   of   the 
circle. 
Therefore  the  point  E  is  within  the  circle.  (Def.  5.) 

But  E  is  any  point  on  the  chord  AB  between  A  and  B,  and 
^^  is  the  chord  joining  any  two  points  on  the  circle. 
Wherefore,  every  chord  of  a  circle  &c. 

EXEKCISES. 

1.  If  a  chord  of  a  circle  be  produced,  the  produced  parts  lie 
without  the  circle. 

2.  Describe  a  circle  which  shall  pass  through  two  given  points,  and 
which  shall  have  its  radius  equal  to  a  given  straight  line  greater  in 
length  than  half  the  distance  between  the  points. 

How  many  such  circles  are  possible? 

3.  Draw  a  straight  line  to  cut  two  equal  circles  in  P,  Q  and  B,  S 
so  that  the  straight  lines  PQ,  QR,  RS  may  be  equal. 

What  condition  is  necessary  that  such  a  straight  line  can  be 
drawn  ? 


PROPOSITIONS  6  AND  7.  185 


PROPOSITION  7. 

If  two    circles  have  a  cojiimon  pointy   tliey  cannot  have 
the  same  centre. 

Let  the  two  circles  ABC,  ADE  meet  one  another  at  the 
point  A: 

it  is  required  to  prove  that  they  cannot  have  the  same 
centre. 

Construction.     Find  F  the  centre  of  one  of  the  circles 
ABC.  (Prop.  5.) 

Draw  any  straight  line  FCE  meeting  the  circles  at  two 
distinct  points  C  and  E,  and  draw  FA. 


Proof.     Because  F  is  the  centre  of  the  circle  ABC\ 

FC  is  equal  to  FA.  (I.  Def.  22.) 

But  FE  is  not  equal  to  FC ; 
therefore  FE  is  not  equal  to  FA  ; 
that  is,  two  straight  lines  FE^  FA  drawn  to  the  circle  ADE 

from  the  point  F  are  not  equal. 
Tlierefore  F  is  not  the  centre  of  the  circle  ADE.  (I.  Def.  22.) 
Wherefore,  if  two  circles  tfec. 


Corollary. 
2^wo  concentric  circles  cannot  have  a  coimrhon  point, 

T.  E.  13 


186 


BOOK  III. 


Definition.  A  point  is  said  to  rotate  about  anotJier 
point,  when  the  first  point  moves  along  a  circle,  of  which  the 
second  point  is  the  centre. 

A  finite  straight  line  is  said  to  rotate  about  a  point, 
when  each  of  its  extremities  moves  along  a  circle,  of  which 
the  point  is  the  centre,  while  the  line  remains  of  constant 
length. 

A  plane  figure  is  said  to  rotate  about  a  point,  when  each 
oftvjo  points  fixed  in  the  figure  moves  along  a  circle,  ofwhicJb 
the  point  is  the  centre,  while  the  figure  remains  unchanged  in 
shape  and  size. 


ADDITIONAL  PKOPOSITION. 


Any  finite  straight  line  may  he  shifted  from  any  one  position  in  a 
plane  to  any  other  by  rotation  about  some  point  in  the  plane. 

Let  AB,  A'B'  be  any  two  positions  of  a  finite  straight  line  in  a 
plane : 

it  is  required  to  prove  that  the  line  can  be  shifted  from  the  position 
AB  to  the  position  A'B'  hy  rotation  about  some  point  in  the  plane. 

Draw  AA'^  BB'  and  bisect  them  in  M,  N,  and  draw  MO,  NO  at 
right  angles  to  AA',BB'  meeting  in  0. 
Draw  OA,  OB,  OA',  OB'. 

Because     in     the     triangles 

AOM,   A'OM,    AM   is  equal   to 

A'M,  and  OM  to  OM, 

and  the  angle  OMA  to  the  angle 

OMA', 

the  triangles  are  equal  in  all 

respects;  (I.  Prop.  4.) 
therefore  OA  is  equal  to  OA'.  '^  B 

Similarly  it  can  be  proved  that  OB  is  equal  to  OB'. 


ROTATION.  187 

Again,  because  in  the  triangles  OAB,  OA'B', 
OA  is  equal  to  OA',  OB  to  OB',  and  AB  to  A'B', 

the  triangles  are  equal  in  all  respects ;         (I.  Prop.  8.) 
therefore  the  angle  AOB  is  equal  to  the  angle  A'OB' : 
add  to  each  the  angle  BOA' ; 
then  the  angle  AOA'  is  equal  to  the  angle  BOB'. 
It  appears  therefore  that  the  triangle  OAB  can  be  shifted  into  the 
position  OA'B'  by  being  turned  in  its  own  plane  round  the  point  O 
through  an  angle  AOA'  or  BOB' ; 
therefore  AB  can  be  shifted  to  A'B'  by  rotation  round  the  point  O. 


EXERCISES. 

1.  About  what  point  must  AB  one  side  of  a  parallelogram  ABCD 
rotate  in  order  to  take  (1)  the  position  CD,  (2)  the  position  DC? 

^  2.     Prove  that,  when  a  straight  line  rotates  about  a  point,  every 
point  in  the  line  rotates  about  the  point  through  the  same  angle. 

3.  Any  triangle  can  be  shifted  from  any  one  position  to  any 
other  position,  which  it  can  occupy  in  the  same  plane  without  being 
turned  over,  by  rotation  about  some  point  in  the  plane. 

4.  Prove  that,  when  a  plane  figure  rotates  about  a  point,  every 
point  in  the  figure  rotates  about  the  point  through  the  same  angle. 

5.  Describe  an  equilateral  triangle  of  which  one  angular  point  is 
given  and  the  others  lie  on  two  given  straight  lines. 

How  many  solutions  are  there  ? 

6.  Construct  an  equilateral  triangle,  one  of  whose  angular  points 
is  given  and  the  other  two  lie  one  on  each  of  two  given  circles. 

Find  the  limits  of  the  position  of  the  given  point  which  admit  of  a 
possible  solution. 

7.  Construct  a  square  to  have  one  vertex  at  a  fixed  point  and  two 
opposite  vertices  on  two  given  straight  lines. 


13—2 


188  BOOK  HI. 


PROPOSITION  8.     Part  1. 

Of  all  straiyht  lines  drawn  to  a  circle  from  a  2>oint  o7i  the 
circle^  the  line  ivhich  is  a  diameter  is  the  greatest;  and  of  any 
two  others,  the  one  which  subtends  the  greater  angle  at  tlie 
centre  is  the  greater. 

Let  CD^  be  a  given  circle,  A  its  centre,  and  J5  any 
point  on  the  circle;  let  BAU  be  a  diameter,  and  let  BC,  BI) 
be  any  other  two  straight  lines  drawn  from  B  to  the  circle, 
and  of  the  angles  BAG,  BAD  subtended  by  BC,  BD  at  A 
let  the  angle  BAD  be  the  greater: 

it  is  required  to  prove  that  BE  is  greater  than  BD,  and  BD 
greater  than  BC. 


Proof.     Because  AE  is  equal  to  AD, 

therefore  BE,  which  is  the  sum  of  BA,  AE, 

is  equal  to  the  sum  of  BA,  AD: 

but  the  sum  of  BA,  AD  is  greater  than  BD  ;    (I.  Prop.  20.) 

therefore  BE  is  greater  than  BD. 

'Next,  because  in  the  triangles  BAD,  BAC, 

AD  i^  equal  to  AC, 

and  BA  to  BA, 

and  the  angle  BAD  is  greater  than  the  angle  BAC, 

therefore  BD  is  greater  than  BC.     (I.  Prop.  24.) 
Wherefore,  of  all  straight  lines  &c. 

Corollary. 
A  diameter  is  the  greatest  chord  of  a  circle. 


PROPOSITION  8.     PART  1.  189 


EXERCISES. 

1.  If  two  chords  of  a  circle  subtend  equal  angles  at  the  centre, 
they  are  equal. 

2.  If  two  chords  of  a  circle  be  equal,  they  subtend  equal  angles 
at  the  centre. 

3.  Of  any  two  chords  in  a  circle  the  one  which   subtends  the 
greater  angle  at  the  centre  is  the  greater. 


190  BOOK  III. 

PROPOSITION  8.     Part  2. 

Of  all  straight  lines  drawn  to  a  circle  from  an  internal 
imint  not  the  centre^  the  one  which  passes  through  the  centre 
is  tJie  greatest^  and  the  one  which  when  produced  passes 
through  the  centre  is  the  least ;  and  of  any  two  others^  the  one 
which  subtends  the  greater  angle  at  the  centre  is  the  greater. 

Let  ODE  be  a  given  circle,  A  its  centre  and  B  any  other 
internal  point ;  let  BA  produced  beyond  A  cut  the  circle 
in  E^  and  produced  beyond  B  in  F^  and  let  BC^  BD  be  any 
other  two  straight  lines  drawn  from  B  to  the  circle,  and 
of  the  angles  BAG,  BAD  subtended  by  BG,  BD  at  A  let  the 
angle  BAD  be  the  greater: 

it  is  required  to  prove  that  BE  is  greater  than  BD^ 
BD  greater  than  BG^  and  BG  greater  than  BF. 


Proof.     Because  AE  is  equal  to  AD, 
therefore  BE^  which  is  the  sum  of  BA,  AE,  is  equal  to 

the  sum  of  BA,  AD; 
but  the  sum  of  BA,  AD  is  greater  than  BD;     (I.  Prop.  20.) 
therefore  BE  is  greater  than  BD. 
Next,  because  in  the  triangles  BAD,  BAG, 
AD  is  equal  to  AG, 
BA  to  BA, 
and  the  angle  BAD  is  greater  than  the  angle  BAG, 
therefore  BD  is  greater  than  BG.      (I.  Prop.  24.) 
Again,  because  the  sum  of  BG,  BA  is  greater  than  AG, 

(I.  Prop.  20.) 

and  AG  is,  equal  to  AF,  which  is  the  sum  of  BF,  BA, 

therefore  the  sum  of  BG,  BA  is  greater  than  the  sum  of 

BF,  BA. 

Therefore  BG  is  greater  than  BF. 

Wherefore,  of  all  straight  lines  (fee, 


PROPOSITION  8.     PART  2.  191 


EXEKCISES. 

1.  If  two  straight  lines  drawn  to  a  circle  from  an  internal  point 
not  the  centre  be  equal,  they  subtend  equal  angles  at  the  centre. 

2.  If  two  straight  lines  drawn  to  a  circle  from  an  internal  point 
not  the  centre  subtend  equal  angles  at  the  centre,  they  are  equal. 

3.  If  each  of  two  equal  straight  lines  have  one  extremity  on  one 
of  two  concentric  circles  and  the  other  extremity  on  the  other,  the 
lines  subtend  equal  angles  at  the  common  centre. 


.19:2  BOOK  III. 

PROPOSITION  8.     Part  3. 

Of  all  straight  lines  drawn  to  a  circle  from  a?i  external 
'point,  the  one  which  passes  through  the  centre  is  the  gi'eatest, 
and  the  one  which  when  produced  passes  through  the  centre 
is  the  least ;  and  of  any  two  others,  the  one  which  subtends 
the  greater  angle  at  the  centre  is  the  greater. 

Let  ODE  be  a  given  circle,  A  its  centre  and  B  a 
given  external  point;  let  BA  cut  the  circle  in  F  and  let 
BA  produced  cut  the  circle  in  E,  and  let  BD,  BG  be  any 
other  two  straight  lines  drawn  from  B  to  the  circle,  and 
of  the  angles  7i^C,  BAD  subtended  by  ^(7,  ^i>  at^  let  the 
angle  BAD  be  the  greater  : 

it  is  required  to  prove  that  BE  is  greater  than  BD, 
BD  greater  than  BG,  and  BG  greater  than  BF. 


Proof.     Because  AE  is  equal  to  AD, 

therefore  BE,  which  is  the  sum  of  BA,  AE, 

is  equal  to  the  sum  of  BA,  AD : 

but  the  sum  of  BA,  AD  is  greater  than  BD  ;    (I.  Prop.  20.) 

therefore  BE  is  greater  than  BD. 

Next,  because  in  the  triangles  BAD,  BAG, 

AD  is  equal  to  AG, 

and  BA  to  BA, 

and  the  angle  BAD  is  greater  than  the  angle  BAG, 

therefore  BD  is  greater  than  BG.    (I.  Prop.  24.) 
Again,  because  the  sum  of  BG,  GA  is  greater  than  BA, 

(I.  Prop.  20.) 
which  is  the  sum  of  BF,  FA , 
and  because  CA  is  equal  to  FA, 
therefore  BG  is  greater  than  BF. 
Wherefore,  of  all  straight  lines  &c. 


PROPOSITION  8.     PART  3.  193 

We  conclude  from  the  results  of  the  several  Parts  of  Proposition  8 
that,  if  0  be  a  fixed  point  on  the  diameter  4(7  of  a  circle  ABCD  nearer 
to  A  than  to  G  and  P  be  a  point  which 
is  capable  of  motion  along  the  circum- 
ference in  the  direction  represented  by 
the  arrow,  while  P  is  moving  along  the 
arc  ABC  from  A  to  C  the  distance  OP 
increases  continuously  from  OA  to  OG, 
and  while  P  is  moving  along  the  arc 
GDA  from  G  to  A,  the  distance  OP 
decreases  continuously  from  OG  to  OA. 

We  say  therefore  that  OG  is  a  maximum  value  of  OP,  and  OA  is  a 
minimum  value.    (See  remarks  on  page  55.) 

It  may  be  observed  here  that,  if  P  travel  round  the  circle  any 
number  of  times,  it  passes  G  and  A  alternately.  It  appears  therefore 
that  here  maximum  and  minimum  values  occur  alternately. 

The  occurrence  of  maximum  and  minimum  values  alternately  is 
true  generally  in  the  case  of  quantities  which  vary  continuously, 
i.e.  quantities  whose  magnitude  changes  without  suffering  any  abrupt 
changes. 


EXERCISES. 

1.  Find  the  shortest  distance  between  two  points  one  on  each 
of  two  circles  which  do  not  meet. 

2.  A  and  B  are  two  fixed  points;  it  is  required  to  find  a  point 
P  on  a  given  circle,  so  that  the  sum  of  the  squares  on  AP  and  BP 
may  be  the  least  possible. 

Under  what  conditions  is  the  solution  indeterminate? 


194  BOOK  III. 

PROPOSITION  9. 

From  a  point  not  tlie  centre  not  Tnore  than  two  equal 
straight  lines  can  he  drawn  to  a  circle,  one  on  each  side  of 
the  straight  line  drawn  from  that  point  to  the  centre. 

Let  J.  be  a  given  point,  and  BCD  a  given  circle, 
and  let  AB,  AD  be  two  equal  straight  lines  drawn  from  A 

to  the  circle : 
it  is  required  to  prove  that  no  other  straight  line  equal  to 

^^  or  AD  can  be  drawn  from  A  to  the  circle. 

Construction.     Find  E  the  centre  of  the  circle; 

(Prop.  5.) 
draw  EA,  EB,  ED. 


J>  D 

Proof.     Take  C  any  point  of  the  circle  on  the  same  side 
of  AE  as  AB. 

Because  B  and  C  are  equidistant  from  E, 
they  cannot  be  equidistant  from  A.     (I.  Prop.  7.) 
Therefore  there  cannot  be  another  straight  line  equal  to 
AB  drawn  from  A  to  the  circle  on  the  same  side  of  AE 
as  AB. 
Similarly  it  can  be  proved  that  there  cannot  be  another 
straight  line  equal  to  AD  drawn  from  A  to  the  circle  on 
the  same  side  of  AE  as  AD. 
Wherefore,  from  a  point  <fec. 

Corollary  1. 

If  from  a  point  three  equal  straight  lines  can  he  drawn 
to  »  circle,  that  jioiyit  is  the  centre. 

Corollary  2. 
Tvjo  circles  cannot  meet  in  more  than  two  points. 


PROPOSITION  9.  195 

If  the  straight  line  AB  drawn  to  the  circle  from  a  point  A  not 
the  centre  be  in  the  same  straight  line  as  the  centre  of  the  circle,  no 
other  straight  line  can  be  drawn  to  the  circle  from  the  point  A  equal 
to  AB.  The  line  AB  is  in  this  case  either  a  maximum  or  a  minimum 
among  the  straight  lines  drawn  from  the  point  A  to  the  circle ;  it  is  a 
maximum,  if  B  be  at  the  further  extremity  of  the  diameter  through  A, 
and  a  minimum,  if  B  be  at  the  nearer  extremity. 


EXERCISES. 

1.  If  from  any  point  within  a  circle  two  straight  lines  be  drawn 
to  the  circumference  making  equal  angles  with  the  straight  line  joining 
the  point  and  the  centre,  the  lines  are  equal  in  length. 

2.  Construct  an  equilateral  triangle,  having  two  of  its  vertices  on 
a  given  circle  and  the  third  at  a  given  point  within  the  circle. 

3.  Construct  a  square  having  one  vertex  at  a  given  point  and  two 
opposite  vertices  on  a  given  circle. 


196 


BOOK  III. 


PROPOSITION  10. 


If  two  circles  meet  at  a  point  not  in  the  same  straight  line 
as  their  centres^  the  circles  intersect  at  that  point. 

Let  ABC,  ADE  be  two  circles  meeting  at  a  point  A, 
■which  is  not  in  the  same  straight  line  as  their  centres : 
it  is  required  to  prove  that  the  circles  intersect  at  A. 


Find  F,   G  the  centres  of  the  circles 


Construction, 

ABC,  ADE;  (Prop.  5.) 

draw  AF,  FG,  GA  ;  and  through  G  draw,  on  the  same 
side  of  FG  as  GA,  two  straight  lines  Gil,  GK  to  meet 
the  circle  ADE  in  H,  K,  such  that  the  angle  FGH  is 
greater  than  the  angle  FGA,  and  the  angle  FGK  less 
than  the  angle  FGA. 
Draw  FII,  FK. 


Proof.     Because,   from  the  point  F  not  the  centre  of 
the  circle  ADE  the  straight  lines  FII,  FA,  FK  are  drawn 
to  the  circle, 
such  that  the  angle  FGH  subtended  by  FH  at  G  the  centre 

is  greater  than  the  angle  FGA  subtended  by  FA, 
and  such  that  the  angle  FGA  is  greater  than  the  angle 
FGK  subtended  by  FK, 

FH  is  greater  than  FA, 
and  FA  greater  than  FK.    (Prop.  8,  Parts  2  and  3.) 
But  FA  is  a  radius  of  the  circle  ABC : 


PROPOSITION  10.  197 

therefore  the  distance  of  the  point  H  from  the  centre  of  the 

circle  ABC  is  greater  than  the  radius,  and  //  therefore  is 

Avithout  the  circle  ABC ; 
and  the  distance  of  the  point  K  from  the  centre  of  the  circle 

ABC  is  less  than  the  radius,  and  K  therefore  is  within 

the  circle  ABC. 
Therefore  the  circles  intersect  at  the  point  A.  (Def.  5.) 

Wherefore,  if  two  circles  kc. 

Corollary. 

If  two  circles  touch,  the  jjoint  of  corUact  is  in  the  same 
straight  line  as  their  centres. 


EXEKCISES. 

1.  If  two  circles  meet  at  a  point  not  in  the  same  straight  line  as 
their  centres,  the  circles  meet  at  one  other  point. 

2.  The  straight  line  joining  the  two  points  at  which  two  circles 
meet  is  bisected  at  right  angles  by  the  straight  line  joining  the 
centres. 


198  BOOK  III. 

PROPOSITION  11. 

If  two  circles  meet  at  a  point,  vj/dch  lies  in  the  same 
straight  line  as  their  centres  and  is  between  the  centres,  the 
circles  touch  at  that  point,  and  each  circle  lies  without  the 
other. 

Let  ABC,  ADE  be  two  circles  meeting  at  a  point  A, 
which  is  in  the  same  straight  line  as  their  centres,  and  is 
between  the  centres  : 

it  is  required  to  prove  that  the  circles  touch  at  A,  and  tliat 

each  circle  lies  without  the  other. 

Construction.     Find  the  centres  F,   G  of  the  circles 

ABC,  ADE',  (Prop.  5.) 

draw  EG,  which  by  the  hypothesis  passes  through  A. 

Take  any  point  H  on  the  circle  ABC, 

and  draw  FH,  HG. 


Proof.     Because  the  sum  of  FH,  HG  is  greater  than 
EG,  (I.  Prop.  20.) 

that  is,  greater  than  the  sum  of  FA,  AG, 
and  EH  is  equal  to  FA, 
therefore  HG  is  greater  than  AG. 
But  ^6^  is  a  radius  of  the  circle  ADE; 
therefore  the  distance  of  the  point  H  from  the  centre  of 
the  circle  ADE  is  greater  than  the  radius,  and  H  there- 
fore is  without  the  circle  ADE.  (Def.  5.) 
Therefore  every  point   on   the    circle  ABC  except  A  lies 

without  the  circle  ADE. 
Therefore  the  circles  touch  at  -4.  (Def.  7.) 

Similarly  it  can  be  proved  that  every  point  on  the  circle 
ADE  except  A  lies  without  the  circle  ABC. 
Wherefore,  if  two  circles  kc. 


PROPOSITION-  11.  199 


When  one  circle  touches  another  circle  which  lies 
without  it,  the  first  circle  is  said  to  have  external  contact 
with  the  second  circle. 


In  the  diagram  each  of  the 
circles  ABC,  BDE  has  external  con- 
tact with  the  other  at  the  point  B. 


EXERCISES. 

1.  If  the  distance  between  the  centres  of  two  circles  be  greater 
than  the  sum  of  their  radii,  each  circle  lies  without  the  other. 

2.  Prove  that  in  all  cases  the  greatest  distance  between  two  points 
one  on  each  of  two  given  circles  is  greater  than  the  distance  between 
the  centres  by  the  sum  of  the  radii. 

3.  Of  two  equal  circles  of  given  radius,  which  touch  externally  at 
P,  one  touches  OX  and  the  other  touches  OY,  where  OX,  OF  are  two 
given  straight  lines  at  right  angles  to  each  other :  prove  that  the  locus 
of  P  is  an  equal  circle. 

Shew  that  there  are  four  such  loci. 

4.  If  two  equal  circles  touch,  every  straight  line  drawn  through 
the  point  of  contact  will  make  equal  chords  in  the  two  circles. 

5.  Given  two  concentric  circles,  draw  a  chord  of  the  outer  which 
shall  be  trisected  by  the  inner  circle. 

6.  Three  circles  touch  one  another  externally  at  the  points 
A,  B,  C\  the  straight  lines  AB,  AC  are  produced  to  cut  the  circle 
JiC  at  1)  and  E:  shew  that  DE  is  a  diameter  of  BC,  and  is  parallel  to 
the  straight  line  joining  the  centres  of  the  other  circles. 


200  BOOK  III. 

PROPOSITION  12. 

If  two  circles  meet  at  a  jwint,  which  lies  in  the  same 
straight  line  as  their  centres  and  is  not  between  the  centres, 
tlie  circles  touch  at  that  imint,  and  one  of  the  circles  lies 
within  tJie  other. 

Let  ABC,  ADE  be  two  circles  meeting  at  a  point  A, 
which  is  in  the  same  straight  line  as  their  centres  and  is 
not  between  the  centres  : 

it  is  required  to  prove  that  the  circles  touch  at  A,  and  that 
one  of  the  circles  lies  within  the  other. 

Construction.     Find  the  centres  I\  G  of  the  circles 

ABC,  ADE',  (Prop.  5.) 

draw  FG,  and  produce  FG  which  by  the  hypothesis  passes 

through  A.    Let  ADE  be  the  circle  whose  centre  G  is  the 

nearer  to  A. 

Take  11  any  point  on  the  circle  ADE,  and  draw  FH,  HG. 


Proof.     Because  GA  is  equal  to  GH, 
FA,  which  is  the  sum  of  FG,  GA,  is  equal  to  the  sum  of 

FG,  GH', 
but  the  sum  of  FG,  GH  is  greater  than  FH -,   (I.  Prop.  20.) 
therefore  FA  is  greater  than  FH. 
But  FA  is  a  radius  of  the  circle  ABC ; 
therefore  the  distance  of  the  point  H  from  the  centre  of 
the  circle  ABC  is  less  than  the  radius,  and  //  therefore 
is  within  the  circle  ABC ', 
therefore  every  point  on  the  circle  ADE  except   A   lies 

within  the  circle  ABC. 
Therefore  the  circles  touch  at  A.  (Def.  7.) 

Wherefore,  if  two  circles  &c. 


PROPOSITION  12.  201 

When  one  circle  touches  another  circle,  which  lies 
within  it,  the  first  circle  is  said  to  have  internal  contact 
with  the  second  circle. 

Two  circles  can  have  external  contact  with  each  other, 
but  two  circles  cannot  have  internal  contact  with  each 
other.  If  one  circle  have  internal  contact  with  another 
circle,  the  second  circle  has  external  contact  with  the  first 
circle. 


In  the  diagram  the  circle  ABC 
internal  contact  with  the  circle  BDE 
the  circle  BDE  has  external  contact 
the  circle  ABC  at  the  point  B, 


EXERCISES. 

1.  Describe  a  circle  passing  through  a  given  point  and  touching  a 
given  circle  at  a  given  point. 

2.  If  in  any  two  given  circles  which  touch  one  another,  there  be 
drawn  two  parallel  diameters,  the  point  of  contact  and  an  extremity  of 
each  diameter,  lie  in  the  same  straight  line. 

3.  Describe  a  circle  which  shall  touch  a  given  circle,  have  its 
centre  in  a  given  straight  line,  and  pass  through  a  given  point  in  the 
given  straight  line. 

4.  Describe  a  circle  of  given  radius  to  pass  through  a  given  point 
and  to  touch  a  given  circle. 

"What  conditions  are  necessary  that  a  solution  may  be  possible  ? 


T.  E.  14 


202 


BOOK  III. 


PROPOSITION  13. 

If  two  circles  have  a  j)oint  of  contact,  they  do  not  meet  at 
any  other  point. 

Let    ABC,   ADE  be   two  circles  which    touch   at    the 
point  A  : 

it  is  required  to  prove  that  the  circles  do  not  meet  at  any 
other  point. 

Construction.     Find  F,  G  the  centres  of  the  circles 
ABC,  ADE.  (Prop.  5.) 


Proof.    Because  the  circles  ABC,  ADE  touch, 
the  point  of  contact  A  must  lie  in  the  straight  line  FG  or  in 
FG  produced.  (Prop.  10,  Coroll.) 

First  (fig.  1)  let  the  point  A  lie  in  FG  : 
then  because  the  circles  ABC,  ADE  meet  at  a  point  A  in 
the  same  straight  line  FG  as  their  centres  and  between 
the  centres, 

each  circle  lies  without  the  other.  (Prop.  11.) 

Secondly  (fig.  2)  let  the  point  A  lie  in  FG  produced  : 
then  because  the  circles  ABC,  ADE  meet  at  a  point  A  in 
the  same  straight  line  FG  as  their  centres  and  not  between 
the  centres, 

one  circle  lies  within  the  other.  (Prop.  12.) 

Therefore  in  neither  case  can  the  circles  meet  at  any  point 
other  than  A. 

Wherefore,  if  two  circles  &c. 


PROPOSITION   13.  203 

We  infer  as  a  result  of  Propositions   9 — 13   that  two 
circles  must  be  such  that  they  either 

(a)  intersect  in  two  distinct  points, 

or  (h)  touch  at  one  point,  which  is  in  the  straight 
line  joining  the  centres, 

or  (c)  do  not  meet. 


EXEECISES. 

1 .  What  is  the  greatest  number  of  contacts  which  may  exist  among 

(1)  three,  (2)  four  circles  ? 

2.  Describe  three  circles  to  have  their  centres  at  three  given 
points,  and  to  touch  each  other  in  pairs. 

3.  Into  how  many  parts  will  three  circles  divide  a  plane  ?  Dis- 
tinguish between  the  different  cases  which  may  occur,  when  the 
circles  intersect  or  touch. 


14—2 


204  BOOK  III. 

PROPOSITION  14.     Part  1. 

Chords  of  a  circle^  which  are  equal,  are  equidista7it  from 
the  centre. 

Let  AB.,  CD  be  two  equal  chords  of  the  circle  A  BCD  : 
it  is  required  to  prove  that  AB,  CD  are  equidistant  from 
the  centre. 
Construction.    Pind  U  the  centre  of  the  circle  A  BCD; 

(Prop.  5.) 
and  from  F  draw  EF^  EG  at  right  angles  to  AB,  CD. 

(I.  Prop.  12.) 
Draw  EA,  EC. 


B 

Proof.     Because  the  straight  line  EF  is  drawn  through 

the  centre  of  the  circle  at  right  angles  to  the  chord  AB,  it 

bisects  it;  (Prop.  4.) 

that  is,  AB  is  double  of  AF. 

Similarly  it  can  be  proved  that  CD  is  double  of  CG. 

But  -4^  is  equal  to  CD  ; 

therefore  AF  is  equal  to  CG. 

Next,  because  the  angles  AFE,  CGE  are  right  angles, 

the  square  on  AE  is  equal  to  the  sum  of  the  squares  on 

AF,  FE, 
and  the  square  on  CE  is  equal  to  the  sum  of  the  squares  on 
CG,  GE.  (I.  Prop.  47.) 

And  because  AE  is  equal  to  CE, 
the  square  on  ^^  is  equal  to  the  square  on  CE. 
Therefore  the  sum  of  the  squares  on  AF^  FE  is  equal  to  the 
sum  of  the  squares  on  CG,  GE. 

Because  AF  is  equal  to  CG, 
the  square  on  ^i^  is  equal  to  the  square  on  CG; 
therefore  the  square  on  FE  is  equal  to  the  square  on  GE. 


PROPOSITION   14.     PART  1.  205 

Therefore  FE  is  equal  to  GE, 
that  is,  AB,  CD  are  distant  from  the  centre. 

(Def.  9.) 
Wherefore,  chords  of  a  circle  kc. 


EXERCISES. 

1.  Chords  of  a  circle,  which  are  equal,  subtend  equal  angles  at 
the  centre. 

2.  Chords  of  a  circle,  which  subtend  equal  angles  at  the  centre, 
are  equidistant  from  the  centre. 


206  BOOK  in. 

PROPOSITION  14.    Part  2. 

Cliords  of  a  circle^  which  are  equidistant  frotn  the  centre^ 
are  equal. 

Let  ^-5,  GB  be  two  cliords  of  tlie  circle  ^^Ci>   equi- 
distant from  the  centre : 

it  is  required  to  prove  that  ^^  is  equal  to  CD. 
Construction.     Find  E  the  centre  of  the  circle  ABCD; 

(Prop.  5.) 
and  from  E  draw  EF,  EG  at  right  angles  to  A  B,  GB. 

(T.  Prop.  12.) 
Draw  EA,  EG. 


Proof.     Because  the  straight  line  EF  is  drawn  through 

the  centre  of  the  circle  at  right  angles  to  the  chord  AB^  it 

bisects  it ;  (Prop.  4.) 

that  is,  ^^  is  double  of  AF. 

Similarly  it  may  be  proved  that  GB  is  double  of  CG. 

Next,  because  the  angles  AFE^  GGE  are  right  angles, 

the  square  ot\.  AE  is  equal  to  the  sum  of  the  squares  on 

AF,  FE, 
and  the  square  on  GE  is  equal  to  the  sum  of  the  squares  on 
GG,  GE,  (I.  Prop.  47.) 

And  because  ^^  is  equal  to  GE, 
the  square  on  AE  is  equal  to  the  square  on  GE. 
Therefore  the  sum  of  the  squares  on  A  F,  FE  is  equal  to  the 
sum  of  the  squares  on  GG,  GE. 

Because  EF  is  equal  to  EG, 
the  square  on  EF  is  equal  to  the  square  on  EG; 
therefore  the  square  on  ^i^  is  equal  to  the  square  on  GG ; 
therefore  AF  \^  equal  to  GG. 


PROPOSITION  U.     PART  2.  207 

And  it  has  been  proved  that  AB  \&  double  of  AF^  and  CD 
oiCG, 

Therefore  AB  \&  equal  to  CD. 
Wherefore,  chords  of  a  circle  &c. 


Parts  1   and  2  of  Proposition  14  are  the  converses  of 
each  other. 


EXEECISES. 

1.  lu  a  circle  chords,  which  are  equidistant  from  the  centre, 
subtend  equal  angles  at  the  centre. 

2.  In  a  circle  chords,  which  subtend  equal  angles  at  the  centre, 
are  equal. 


208  BOOK  III. 


PROPOSITION  15.     Part  1. 

Of  any  two  chords  of  a  circle  the  otie  which  is  the  greater 
is  the  nearer  to  the  centre. 

Let  AB^  CD  be  two  chords  of  the  circle  ABC  I),  of  which 
-4  ^  is  greater  than  CD  : 

it  is  required  to  prove  that  ^^  is  nearer  to  the  centre  than 
CD. 
Construction.     Find  £J  the  centre  of  tlie  circle  ABCD; 

(Prop.  5.) 
and  from  E  draw  UF,  EG  at  right  angles  to  AB,  CD. 

(I.  Prop.  12.) 
Draw  EA.  EG. 


Proof.     Because  the  straight  line  EF  is  drawn  through 
the  centre  at  right  angles  to  the  chord  AB, 

AF  is  equal  to  FB,  (Prop.  4.) 

and  AB  is  double  of  AF. 
Similarly  it  can  be  proved  that  CD  is  double  of  CG. 
But  ^^  is  greater  than  CD  ; 
therefore  AF  is  greater  than  CG. 
Next,  because  the  angles  AFE,  CGE  are  right  angles, 
the  square  on  AE  is  equal  to  the  sum  of  the  squares  on 

AF,  FE, 
and  the  square  on  CE  is  equal  to  the  sum  of  the  squares  on 
CG,  GE.  (I.  Prop.  47.) 

And  because  AE  is  equal  to  CE, 
the  square  on  ^4^  is  equal  to  the  square  on  CE. 
Therefore  the  sum  of  the  squares  on  AF,  FE  is  equal  to 
the  sum  of  the  squares  on  CG,  GE ; 


PROPOSITION  15.     PART  1.  209 

Because  AF  is  greater  than  CG, 

the  square  on  AF  \^  greater  than  the  square  on  CG\ 

therefore  the  square  on  FE  is  less  than  the  square  on  GE, 

Therefore  FE  is  less  than  GE, 

that  is,  ^^  is  nearer  to  tlie  centre  than  CD. 

Wherefore,  of  any  two  chords  <kc. 


EXERCISES. 

1.  Prove  that  every  straight  line,  which  makes  equal  chords  in 
two  equal  circles,  is  parallel  to  the  straight  line  joining  the  centres  or 
passes  through  the  middle  point  of  that  line. 

2.  Find  the  shortest  chord  which  can  be  drawn  through  a  given 
point  within  a  circle. 


210  BOOK  IIL 


PROPOSITION  15.    Part  2. 

Of  any  two  chords  of  a  circle  the  one  which  is  the  nearer 
to  the  centre  is  the  greater. 

Let  AB,  CD  be  two  chords  of  the  circle  ABCD^  of  which 
^^  is  nearer  to  the  centre  than  CD  : 

it  is  required  to  prove  that -4^  is  greater  than  CD. 

Construction.     Find  E  the  centre  of  the  circle  ABCD; 

(Prop.  5.) 
and  from  E  draw  EF^  EG  at  right  angles  to  AB^  CD. 

{I.  Prop.  12.) 
Draw  EA,  EG. 


Proof.     Because  the  straight  line  EF  is  drawn  through 
the  centre  at  right  angles  to  the  chord  AB, 

AF  is  equal  to  FB,  (Prop.  4.) 

and  AB  is,  double  of  AF. 
Similarly  it  can  be  proved  that  CD  is  double  of  CG. 
Next,  because  the  angles  AFE,  CGE  are  right  angles, 
the  square  on  AE  is  equal  to  the  sum  of  the  squares  on 

AF,  FE, 
and  the  square  on  GE  is  equal  to  the  sum  of  the  squares  on 
CG,  GE.  (I.  Prop.  47.) 

And  because  AE  is  equal  to  GE, 
the  square  on  -4^  is  equal  to  the  square  on  CE. 
Therefore  the  sum  of  the  squares  on  AF,  FE  is  equal  to 
the  sum  of  the  squares  on  CG,  GE. 

Because  EF  is  less  than  EG, 
the  square  on  EF  is  less  than  the  square  on  EG ; 


PROPOSITION  15.     PART  2.  211 

therefore  the  square  on  AF  is  greater  than  the   square 
on  CG; 

therefore  AF  i&  greater  than  CG. 
Therefore  AB  i^  greater  than  CD. 
Wherefore,  of  any  two  chords  &c. 


Parts  1  and  2  of  Proposition  15  are  the  converses  of 
each  other. 


EXERCISES. 

1.  Of  any  two  chords  of  a  circle  the  nearer  to  the  centre  subtends 
the  greater  angle  at  the  centre. 

2.  Draw  through  a  given  point  a  straight  line  to  make  equal 
chords  in  two  given  equal  circles. 

Discuss  the  number  of  possible  solutions  in  tlie  dififerent  cases 
which  may  occur. 


212 


BOOK  III. 


PROPOSITION  16. 

The  straight  lioie  drawn  through  a  point  on  a  circle  at 
right  angles  to  the  radius  touches  the  circle,  and  every  other 
straight  line  drawn  through  tlie  2^oint  cuts  the  circle. 

Let  ABC  be  a  circle,  of  which  D  is  the  centre,  and  let 
^^  be  a  straight  line  drawn  through  A  at  right  angles  to 
the  radius  AD;  and  let  AF  be  any  other  straight  line 
drawn  through  A : 

it  is  required  to  prove  that  AB  touches  the  circle,  and  that 
AF  cuts  the  circle. 
Construction.     Take  any  point  F  on  AF,  and  draw 
DF,  and  from  D  draw  DG  a.t  right  angles  to  AF. 

(I.  Prop.  12.) 


Proof.     Because  in  the  triangle  DAF, 

the  angle  DAF  is  a  right  angle, 
the  angle  DFA  is  less  than  a  right  angle;  (I.  Prop.  17.) 
therefore  the  angle  DAF  is  greater  than  the  angle  DFA; 
therefore  i)^  is  greater  than  DA.  (I.  Prop.  19.) 
Therefore  the  distance  of  the  point  F  from  the  centre 
is  greater  than  the  radius,  and  F  therefore  is  without  the 
circle.  (Def.  5.) 

Similarly  it  can   be  proved   that  every   point  on  AF 
except  A  is  without  the  circle. 

Therefore  AF  touches  the  circle  ABC  at  A.        (Def.  6.) 

Next  because  in  the  triangle  DGA, 

the  angle  DGA  is  a  right  angle, 
the  angle  DAG  is  less  than  a  right  angle ;     (I.  Prop.  17.) 
therefore  the  angle  DAG  is  less  than  the  angle  DGA  ; 

therefore  DG  is  less  than  DA.     (I.  Prop.  19.) 


PROPOSITION   16.  213 

Therefore  the  distance  of  the  point  G  from  the  centre 
is  less  than  the  radius,  and  therefore  G  is  within  tlie 
circle.  (Def.  5.) 

Therefore  the  straight  line  AF  cuts  the  circle  ABC*. 

Wherefore,  tJie  straight  line  &c. 

We  infer  as  a  result  of  Proposition  16  that  a  straight 
line  and  a  circle  must  be  such  that  they  either 

(ct)  intersect  in  two  distinct  points, 
or  {b)  touch  at  one  point, 
or  (c)  do  not  meet. 


EXEKCISES. 

1.  A  point  B  is  taken  on  a  circle  whose  centre  is  C ;  PA  a  tangent 
at  any  point  P  meets  GB  produced  at  A,  and  PD  is  drawn  perpendi- 
cular to  CB :  prove  that  PB  bisects  the  angle  APD. 

2.  Describe  a  circle  to  have  its  centre  on  a  given  straight  line,  to 
pass  through  a  given  point  on  that  line  and  to  touch  another  given 
straight  line. 

3.  Describe  a  circle  to  pass  through  a  given  point  and  to  touch  a 
given  straight  line  at  a  given  point. 

4.  If  ^C  be  a  diameter  of  a  circle  ABC,  and  AP  be  drawn  per- 
pendicular to  the  tangent  at  2?,  AB  bisects  the  angle  GAP. 

5.  Prove  that  although  no  straight  line  can  be  drawn  to  pass 
between  a  circle  and  its  tangent,  yet  any  number  of  circles  can  be 
described  to  do  so. 

6.  Circles,  which  have  a  common  tangent  at  a  point,  touch  each 
other. 

7.  Prove  that  the  angle  between  a  tangent  to  a  circle  and  a  chord 
drawn  from  the  point  of  contact  is  half  of  the  angle  subtended  at  the 
centre  by  the  chord. 


A  straight  line  which  cuts  a  circle  is  often  called  a  secant. 


214 


BOOK  III. 


PROPOSITION  17. 


Through  a  given  'point   to   draw   a   tangent  to   a  given 
circle. 

Let  ABC  be  a  given  circle,  and  D  a  given  point ; 
it  is  required  through  D  to  draw  a  tangent  to  the  circle  ABC. 

First,  let  the  point  D  be  on  the  circle. 

Construction.  Find  the  centre  E  • 
(Prop.  5.) 
draw   EDy   and    draw   DF  at    right 
angles  to  DE -,  (I.  Prop.  11.) 

then  DF  is  a  tangent  drawn  as  re- 
quired. 
Proof.  Because  the  straight  line 
DF  is  drawn  through  the  point  D  on 
the  circle  ABC  at  right  angles  to 
DE  the  radius,  DF  touches  the  circle.  (Prop.  16.) 

Secondly,  let  the  point  D  be  outside  the  circle. 

Construction.    Find  the  centre  E ; 

(Prop.  5.) 

draw    EDy    cutting    the    circle    ABO 

between  E  and  D  in  B^  and  draw  BF 

at  right  angles  to  EB^  (I.  Prop.  11.) 

and  with  E  as  centre  and  ED  as 

radius  describe  a  circle  cutting  BF  in 

F.   Draw  EF^  cutting  the  circle  ABC 

between  ^  and  jF"  in  G^  and  draw  DG  : 
then  DG  is  a  tangent  drawn  as  required. 

Proof.     Because  in  the  triangles  DEG,  FEB, 
DE  is  equal  to  FE,  EG  to  EB,  and  the  angle  DEG  equal 
to  the  angle  FEB, 

the  triangles  are  equal  in  all  respects;   (I.  Prop.  4.) 
therefore  the  angle  DGE  is  equal  to  the  angle  FBE. 
But  the  angle  FBE  is  a  right  angle; 

therefore  the  angle  DGE  is  a  right  angle; 


PROPOSITION  V 


215 


and  because  the  straight  line  GD  is   drawn  through  the 

point  G  on  the  circle  ABC  at  right  angles  to  GE  the 

radius,  GD  touches  the  circle.  (Prop.  16.) 

Wherefore,  through  a  given  'point  D  a  tangent  has  been 

dratvn  to  the  circle  ABC. 


Outline  of  Alternative  Construction. 

Find  E  the  centre.  (Prop.  5.) 

Draw  ED,  and  bisect  it  in  0.      (I.  Prop.  10.) 
With  0  as  centre  and  OD  as  radius  describe 
a  circle,  cutting  the  circle  ABC  in  G. 
Draw  DG,  OG,  EG. 
It  may  be  proved  that 

(1)  the  angle  OGD  is  equal  to  the  angle  GDO, 

(2)  the  angle  OGE  is  equal  to  the  angle  GED, 
and  (3)    the  angle  EGD  is  a  right  angle, 

and  hence  that  GD  is  a  tangent  to  the  circle  ABC  at  G. 

Both  the  construction  in  the  Proposition  and  the  alternative  con- 
struction point  out  that  two  and  only  two  tangents  can  be  drawn 
to  a  circle  through  an  external  point,  one  through  a  point  on  the 
circle,  and  none  through  an  internal  point. 

When  there  is  no  danger  of  ambiguity  the  length  of  the  straight 
line  drawn  from  an  external  point  to  touch  a  circle  which  is  inter- 
cepted between  that  point  and  the  point  of  contact  is  often  spoken  of 
as  the  tangent  from  the  point  to  the  circle. 


EXEECISES. 

1.  The  two  tangents  drawn  to  a  circle  from  an  external  point 
are  equal. 

2.  Draw  a  tangent  to  a  given  circle  to  be  parallel  to  a  given 
straight  line. 

3.  Find  in  a  given  straight  line  a  point  such  that  the  tangent 
drawn  from  it  to  a  given  circle  may  be  equal  to  a  given  straight  line. 

4.  The  greater  the  distance  of  an  external  point  is  from  the 
centre  of  a  circle,  the  smaller  is  the  inclination  of  the  two  tangents 
which  can  be  drawn  from  it. 


216  BOOK  III. 


PROPOSITION  18. 


If  a  straight  line  touch  a  circle^  the  radius  draujn  to  the 
point  of  contact  is  at  right  angles  to  the  line. 

Let  the  straight  line  DE  touch  the  circle  ABC  at  the 
point  C : 

it  is  required  to  prove  that  the  radius  drawn  to  the  point  C 
is  at  right  angles  to  DOE. 

Construction.  Find  F  the  centre  of  the  circle;  (Prop.  5.) 
and  draw  FC . 


Proof.     If  DCE  were  not  at  right  angles  to  OF, 
DCE  would  cut  the  circle    (Prop.  16) ; 

but  it  does  not : 
therefore  BGE  is  at  right  angles  to  CF. 
Wherefore,  if  a  straight  line  <fec. 


PROPOSITION  18.  217 


The  Tangent  as  the  Limit  of  the  Secant. 

Let  AF  be  a  straight  line  cutting  a  given  circle  at  a  given  point  A, 
and  again  at  a  second  point  F. 

Let  D  be  the  centre  of  the   circle,  and 
AE  the  tangent  at  A.     Draw  DA,  DF. 

The  angle  FAE  is  equal  to  half  of  the  angle 
ADF.     (See  Ex.  7,  p.  213.) 

Hence  the  smaller  the  angle  ADF  is, 
or  the  smaller  the  chord  AF,  (Prop.  8,  Part  1.) 
the  smaller  is  the  angle  FAE. 

Now  because  we  can  take  the  point  F  as 
close  to  A  as  we  like,  we  can  make  the  angle 
ADF,  and  therefore  also  the  angle  FAE,  as  small  as  we  like. 
Hence  we  can  make  the  straight  line  AF  deviate  as  little  as  we  please 
from  coincidence  with  AE. 

We  express  this  fact  by  saying  that, 
the  tangent  AE  is  the  limit  of  the  secant  AF,  when  F  moves  up 
close  to  A. 

This  definition  of  a  tangent  to  a  curve  as  the  limit  of  the  secant 
through  the  point  is  one  which  admits  of  application  to  curves  of  all 
kinds. 


EXEECISES. 

1.  Through  a  given  point  draw  a  straight  line  so  that  the  chord 
which  is  intercepted  on  it  by  a  given  circle  is  equal  to  a  given  straight 
line. 

2.  Two  circles  are  concentric :  prove  that  all  chords  of  the  outer 
circle  which  touch  the  inner  are  equal. 

3.  If  two  tangents  be  drawn  to  a  circle  from  an  external  point, 
the  chord  joining  the  points  of  contact  is  bisected  at  right  angles  by 
the  straight  line  joining  the  centre  and  the  external  point. 


t.  e.  15 


218 


BOOK  III. 


PROPOSITION  19. 

If  a  straight  line  touch  a  circle,  the  straight  line  drawn 
at  right  angles  to  the  line  through  its  point  of  contact 
passes  through  the  centre. 

Let  the  straight  line  DE  touch  the  circle  AJBC  at  C, 
and  from  C  let  CA  be  drawn  at  right  angles  to  DE : 
it  is  required  to  prove  that  the  centre  of   the   circle    is 
inCA. 


Construction. 


Take  any  point  F,  not  in  CA, 
and  draw  FC. 


Proof.     Because  CA  is  at  right  angles  to  BF, 

CF  cannot  be  at  right  angles  to  DF.  (I.  Prop.  10  A.) 
But  if  a  straight  line  touch  a  circle,  the  radius  drawn  to 
the  point  of  contact  is  at  right  angles  to  the  tangent ; 

(Prop.  18.) 
therefore  the  radius  drawn  to  C  cannot  be  in  the  same 

straight  line  as  CF ; 
therefore  the  centre  cannot  lie  at  any  point  F  not  in  CA, 
that  is,  the  centre  must  lie  in  CA. 
Wherefore,  if  a  straight  line  &c, 


PROPOSITION  19.  219 

ADDITIONAL  PKOPOSITION. 
To  draio  a  comvion  tangent  to  two  given  circles. 
Let  ^,  J5  be  the  centres  of  two  given  circles,  which  we  will  call  for 
shortness  the  (A)  circle  and  the  (B)  circle  : 

Let  the  radius  of  the  (A)  circle  be  greater  than  the  radius  of  the 
(B)  circle. 

With  A  for  centre  and  the  difference  (tig.  1)  or  the  sum  (fig.  2)  of 
the  radii  for  radius  describe  a  circle, 

and  from  B  draw  BH  a  tangent  to  it.  (Prop.  17.) 

Draw  All  and  let  AH  produced  (fig.  1)  or  AH  (fig.  2)  cut  the  (A) 
circle  in  P. 

Through  P,  B  draw  PQ,  BQ  parallel  to  HB,  AH  respectively. 

(L  Prop.  31.) 
(2) 


Because  HPQB  is  a  parallelogram,  (Constr.) 

BQ  is  equal  to  HP,  (I.  Prop.  34.) 

which  is  equal  to  the  radius  of  the  (B)  circle.       (Constr.) 
Therefore  the  point  Q  is  on  the  (B)  circle. 
Again,  because  BH  is  a  tangent  at  H, 

the  angle  PHB  is  a  right  angle  ;  (Prop.  18.) 

therefore  the  parallelogram  HPQB  is  a  rectangle ;  (I.  Def .  19. ) 

therefore  the  angles  at  P  and  Q  are  right  angles,  (i.  Prop.  29,  CoroU.) 

and  PQ  is  a  tangent  to  the  (A)  and  (B)  circles  at  P  and  Q 
respectively.  (Prop.  16.) 

EXEECISES. 

1.  Prove  that  four  common  tangents  can  be  drawn  to  two  circles 
which  are  external  to  each  other. 

2.  How  many  common  tangents  can  be  drawn  to  two  intersecting 
circles? 

3.  Is  it  possible  that  two  circles  can  have  one  and  only  one 
common  tangent  ? 

4.  Draw  a  straight  line  so  that  the  chords  which  are  intercepted 
on  it  by  two  given  circles  are  equal  to  two  given  straight  lines. 

15—2 


220 


BOOK  III. 


PROPOSITION  20. 

The  angle  which  an  arc  of  a  circle  subtends  at  the  centre 
is  double  of  the  angle  which  the  arc  subtends  at  the  circum- 
ference. 

Let  ABC  \i%  a  circle,  of  which  BC  is  an  arc,  and  let 
BDC,  BAG  be  angles  subtended  by  the  arc  BG  at  the 
centre  D,  and  at  the  circumference  : 

it  is  required  to  prove  that  the  angle  BDG  is  double  of  the 
angle  BA  C. 

First,  (fig.  1)  let  the  centre  i>  lie  on  AB,  one  of  the  lines 
which  contain  the  angle  BAG. 

Construction.     Draw  DG.  - 


Proof.     Because  DA  is  equal  to  DG, 
the  angle  DGA  is  equal  to  the  angle  DAG;  (I.  Prop.  5.) 
therefore  the  sum  of  the  angles  DAG,  DGA  is  double  of 
the  angle  DAG. 
But  the  angle  BDG  is  equal  to  the  sum  of  the  angles 


DAG,  DGA; 


(I.  Prop.  32.) 


therefore  the  angle  BDG  is  double  of  the  angle  DAG. 

Next,  let  the  centre  D  lie  within  (fig.  2)  or  without 
(fig.  3)  the  angle  BAG. 

Construction.  Draw  AD  and  produce  it  to  meet  the 
circle  in  B. 

Proof.  It  follows  from  the  first  case,  that  the  angle 
UDG  is  double  of  the  angle  BAG,  and  that  the  angle 
HDB  is  double  of  the  angle  UAB ; 


PROPOSITION  20.  221 

therefore  in  (fig.  2)  the  sum  of  the  angles  EDO,  EBB  is 
double  of  the  sum  of  the  angles  EAG,  EAB, 

and  in  (fig.  3)  the  difference  of  the  angles  EDO,  EDB  is 
double  of  the  difference  of  the  angles  EAG^  EAB ; 

therefore  in  all  cases  the  angle  BDG  is  double  of  the  angle 
BAG. 
Wherefore,  the  angle  vjhich  an  arc  &c. 


In  the  diagram  of  Proposition  20  in  each  of  the  figures  the  angle 
BDC  is  double  of  the  angle  BAC.  Now  it  is  easily  seen  that  although 
in  figures  (1)  and  (3)  the  angle  BAC  is  restricted  to  values  less  than 
a  right  angle,  and  the  angle  BDG  in  consequence  to  values  less  than 
two  right  angles,  in  figure  (2)  the  angle  BAG  is  restricted  only  to 
values  less  than  two  right  angles  and  the  angle  BDG  in  consequence 
only  to  values  less  than  four  right  angles.  It  appears  therefore  that, 
if  we  wish  not  to  destroy  the  generality  of  the  theorem  of  Proposi- 
tion 20,  we  must  allow  our  definition  of  an  angle  to  include  angles 
w;hich^are  equal  to  two  or  greater  than  two  right  angles;  there  is 
nothing  inconsistent  with  a  strict  adherence  to  Euclid's  methods  in 
doing  so.  ^"~~      ■ 


EXERCISES. 

1.  Two  circles,  whose  centres  are  A  and  D,  touch  externally  at 
E :  a  third  circle,  whose  centre  is  B,  touches  them  internally  at  C 
and  F:  prove  that  the  angle  ADB  is  double  of  the  angle  EGF. 

2.  If  AB  be  a  fixed  diameter  and  DE  an  arc  of  constant  length  in 
a  fixed  circle,  and  the  straight  lines  AE,  BD  intersect  at  P,  the  angle 
APB  is  constant. 

3.  If  ^BC  be  a  triangle  inscribed  in  a  circle  and  the  angle  BAG 
be  bisected  hy  AD,  which  meets  the  circle  in  D,  then  the  diameter 
through  D  will  bisect  BG  &t  right  angles. 

4.  AB  is  a  diameter  and  PQ  any  chord  of  a  circle  cutting  AB 
within  the  circle,  and  AL  is  drawn  perpendicular  to  PQ.  Prove  that 
the  angle  LAB  is  equal  to  the  sum  of  the  angles  PAB,  QAB. 


222  BOOK  III. 

PROPOSITION  21. 

Angles  in  the  same  arc  of  a  circle  are  equal. 

Let  ABCD  be  a  circle,  and  BAC^  BBC  be  two  angles 
in  the  same  arc  BADC: 

it  is  required   to  prove  that  the  angles  BAG,  BBC   are 
equal. 

Construction.     Find  the  centre  E;  (Prop.  5.) 

and  draw  EB,  EG. 


Proof.  Because  the  angle  which  the  arc  BFG  of  the 
circle  subtends  at  the  centre  is  double  of  the  angle  which 
it  subtends  at  the  circumference, 

the  angle  BEG  is  double  of  the  angle  BAG., 
and  also  the  angle  BEG  is  double  of  the  angle  BBC ; 

*  (Prop.  20.) 

therefore  the  angle  BAG  is  equal  to  the  angle  BBG. 
Wherefore,  angles  in  the  same  arc  &c. 


Corollary.  If  a  straight  line  joining  two  points  subtend 
equal  angles  at  two  other  points  on  the  same  side  of  the  linCj 
the  four  points  lie  on  a  circle.  ' 

Let  the  straight  line  BG  subtend  equal  angles  at  the 
two  points  ^,  J9  on  the  same  side  of  BG. 

If  a  circle  be  described  about  the  triangle  BAG'^',  the 

*  That  it  is  possible  to  describe  a  circle  through  the  three  vertices 
of  a  triangle  appears  in  the  Additional  Proposition  on  page  53. 


PROPOSITION  21.  223 

circle  must  cut  BD  again  at  some  point  not  on  the  same 
side  of  ^C  as  B.  (Prop.  6.) 

Now  take  H  any  point  but  D  in  BD  or  BD  produced 
and  draw  HD. 
Then  the  angle  BHC  cannot  be  equal  to  the  angle  BDC, 

(I.  Prop.  16.) 

and  therefore  cannot  be  equal  to  the  angle  BAC. 

But  angles  in  the  same  arc  of  a  circle  are  equal.    (Prop.  21.) 

Therefore  the  circle  BAC  cannot  meet  BD  in  H ; 

that  is,  it  must  meet  it  in  D. 

In  some  books  in  the  proof  of  Proposition  21,  the  result  of  Pro- 
position 20  is  quoted  as  if  it  were  true  only  in  the  case  of  arcs  greater 
than  a  semicircle:  that  is,  as  if  the  angle  BEC,  which  the  arc  BFC 
subtends  at  the  centre,  were  restricted  to  magnitudes  less  than  two 
right  angles.  The  general  truth  of  the  theorem  is  then  deduced  as  a 
consequence. 

"We  leave  this  deduction  to  the  student  as  an  exercise. 


EXEECISES. 

1.  The  locus  of  a  point  at  which  a  given  straight  line  subtends  a 
constant  angle  is  an  arc  of  a  circle. 

2.  If  of  three  concurrent  straight  lines  inclined  at  given  angles 
to  one  another  two  pass  through  two  fixed  points,  the  third  also  passes 
through  a  third  fixed  point. 

3.  If  two  sides  of  a  triangle  of  constant  shape  and  size  pass 
through  two  fixed  points,  the  third  ahvays  touches  a  fixed  circle. 

4.  If  two  sides  of  a  triangle  of  constant  shape  and  size  always 
touch  two  fixed  circles,  the  third  side  always  touches  a  fixed  circle. 

5.  If  ABC  be  an  equilateral  triangle  described  in  a  circle  whoso 
centre  is  0,  and  if -40  produced  meet  the  circle  in  D,  then  OD,BC 
bisect  each  other. 

6.  Two  circles  ADB,  ACB  intersect  in  points^  andU.  Through 
A  any  chord  BAG  is  drawn,  and  BC,  BD  are  joined,  and  the  angle 
DBG  is  internally  bisected  by  a  line  BE  which  meets  BG  in  E. 
Shew  that  E  lies  on  a  fixed  circle. 

7.  If  ABG  be  an  isosceles  triangle  on  the  base  BG,  inscribed  in  a 
circle,  and  P,  Q  be  points  on  the  arcs  AG,AB  respectively  of  the 
circle  such  that  AQ  i^  parallel  to  BP,  then  GQ  is  parallel  to  AP. 

8.  If  the  diagonals  of  a  quadrilateral  inscribed  in  a  circle  be  at 
right  angles,  the  perpendicular  from  their  intersection  on  any  side 
bisects  the  opposite  side. 


224  BOOK  III. 

PROPOSITION  22. 

The  sum  of  two  opposite  angles  of  a  convex  quadrilateral 
inscribed  in  a  circle  is  equal  to  two  rigid  angles. 

Let  A  BCD  be  a  quadrilateral  inscribed  in  the  circle 

ABCD: 

it  is  required  to  prove  that  the  sum  of  the  angles  ABG^ 
ADC  is  equal  to  two  right  angles,  and  that  the  sum  of 
the  angles  BAD^  BCD  is  equal  to  two  right  angles. 

Construction.     Draw  AC,  BD. 


Proof.      Because  the  angles  BCA,  BDA    are   in   the 
same  arc  BCDA^ 

the  angle  BCA  is  equal  to  the  angle  ^Z>^  j  (Prop.  21.) 
and  because  the  angles  CAB,  CDB  are  in  the  same  arc 
CDAB, 

the  angle  CAB  is  equal  to  the  angle  CDB.     (Prop.  21.) 
Therefore  the  sum  of  the  angles  BCA,  CAB  is  equal  to  the 
sum  of   the   angles  BDA,    CDB,  that   is,    to  the  angle 
ADC. 
To  each  of  these  equals  add  the  angle  J  ^C  : 
then  the  sum  of  the  angles  ABC,  BCA,  CAB  is  equal  to  the 
sum  of  the  angles  ABC,  ADC. 
But  because  the  angles  ABC,  BCA,  CAB  are  the  angles 
of  a  triangle,  their  sum  is  equal  to  two  right  angles. 

(I.  Prop.  32.) 
Therefore  the  sum  of  the  angles  ABC,  ADC  is  equal  to 
two  right  angles. 

Similarly  it  can  be  proved  that  the  sum  of  the  angles 
BAD,  BCD  is  equal  to  two  right  angles. 

Wherefore,  the  sum  of  two  opposite  angles  &c. 


PROPOSITION  22.  225 

Corollary.  If  the  sum  of  two  opposite  angles  of  a  convex 
quadrilateral  he  equal  to  two  right  angles.,  the  vertices  of  the 
quadrilateral  lie  on  a  circle. 

Let  ABCD  be  a  convex  quadrilateral  in  which  the  sum 
of  the  angles  BAD^  BCD  is  equal  to  two  right  angles. 

If  a  circle  be  described  about  the  triangle  BAD^  the 

circle  must  cut  AC  again  in  some  point  not  on  the  same 

side  of  BD  as  A.  (Prop.  6.) 

Now  take  H  any  point  but  C  in  AC  ov  AC  produced 

and  draw  HB,  HD. 

Then  the  angle  BUD  cannot  be  equal  to  the  angle  BCD. 

(I.  Prop.  21.) 
Therefore  the  sum  of  the  angles  BAD.,  BHD  cannot  be  equal 

to  two  right  angles. 
But  the  sum  of  two  opposite  angles  of  a  convex  quadri- 
lateral inscribed  in  a  circle  is  equal  to  two  right  angles. 

(Prop.  22.) 
Therefore  the  circle  BAD  cannot  meet  AC  in  H ', 
that  is,  it  must  meet  it  in  C. 


EXERCISES. 

1.  If  the  sides  AB,  DC  of  a  quadrilateral  ABCD  inscribed  in  a 
circle  be  produced  to  meet  at  E,  the  triangles  AEC,  BED  are  equi- 
angular to  one  another. 

2.  A  triangle  is  inscribed  in  a  circle:  shew  that  the  sum  of  the 
angles  in  the  three  segments  exterior  to  the  triangle  is  equal  to  four 
right  angles. 

3.  If  PQRS,  pqrs  be  two  circles,  and  PprR,  QqsS  be  chords  such 
that  P,  I),  q,  Q  lie  on  a  circle,  then  R,  r,  s,  S  lie  on  a  circle. 

4.  If  any  two  consecutive  sides  of  a  convex  hexagon  inscribed  in 
a  circle  be  respectively  parallel  to  their  opposite  sides,  the  remaining 
sides  are  parallel  to  each  other. 

5.  If  any  arc  of  a  circle  described  on  the  side  JB(7  of  a  triangle 
ABC  cut  BA,  CA  produced  if  necessary  in  P  and  Q,  PQ  is  always 
parallel  to  a  fixed  straight  line. 

6.  £  is  a  point  on  one  of  the  diagonals  ^C  of  a  parallelogram 
ABCD.  Circles  are  described  about  DEA  and  BEG.  Shew  that  BD 
passes  through  the  other  point  of  intersection  of  the  circles. 


226 


BOOK  III. 


PROPOSITION  23. 

Two  arcs  of  circles,  y)hich  have  a  common  chord  and  are 
on  the  same  side  of  it,  cannot  he  similar  unless  they  are 
coincident. 

Let  ABC,  ADC  be  two  arcs  of  circles,  which  have  a 
common  chord  A  (7,  and  are  on  the  same  side  of  it : 
it  is  required  to  prove  that  ABC,  ADC  cannot  be  similar 
arcs,  unless  they  are  coincident. 

Construction.  Draw  through  A,  one  of  the  extremities 
of  the  chord  AB,  any  straight  line  ABD  to  meet  the  arcs  in 
B,  D;  and  draw  CB,  CD. 


.    G 


Proof.     If  the  points  B,  D  do  not  coincide, 
one  of  the  angles  ABC,  ADC  is  an  exterior  angle  and  the 

other  an  interior  angle  of  the  triangle  BCD ; 
therefore  the  angle  ABC  is  not  equal  to  the  angle  ADC, 

(I.  Prop.  16.) 
and  therefore  the  arc  ABC  is  not  similar  to  the  arc  ADC. 

(Def.  4.) 
It  has  now  been  proved  that,  if  any  straight  line  ABD 
meet   the  arcs  in   two   points  B  and  D   which    are   not 
coincident, 

the  arcs  cannot  be  similar. 
Therefore,  if  the  arcs  be  similar,  every  straight  line  drawn 
through  A  must  meet  the  arcs  in  two  coincident  points, 
that  is,  the  arcs  ABC,  ADC  must  coincide. 
Wherefore,  two  arcs  of  circles  &c. 


PROPOSITION  23.  227 


EXERCISES. 

1.  If  on  opposite  sides  of  the  same  straight  line  there  be  two  arcs 
of  circles,  which  contain  supplementary  angles,  the  arcs  are  parts  of 
the  same  circle. 

2.  Prove  that,  if  two  circles  have  three  points  in  common,  the 
circles  are  coincident. 


228  BOOK  III. 


PROPOSITION  24. 

Similar  arcs  of  circles^  which   have  equal  chords,   are 
equal. 

Let  ABGy  DEF  be  two  similar  arcs  of  circles,  which 
have  equal  chords  AC,  DF : 
it  is  required  to  prove  that  the  arcs  ABC,  DEF  are  equal. 


Proof.     Because  the  chords  AC,  BF  are  equal,  it  is 

possible  to  shift  the  figure  ABC,  so  that  AC  coincides  with 

DF,  A  with  D,  and  C  with  F,      (i.  Test  of  Equality,  page  5) 

and  so  that  the  arcs  ABC,  DEF  are  on  the  same  side  of 

DF. 

If  this  be  done, 

the  arc  ABC  must  coincide  with  the  arc  DEF, 
for  the  arcs  ABC,  DEF  are  similar, 
and  two  arcs  of  circles,  which  have  a  common  chord,  and 
are  on  the  same  side  of  it,  cannot  be  similar  unless  they 
coincide.  (Prop.  23.) 

Therefore  the  arcs  ABC,  DEF  are  equal. 
Wherefore,  similar  arcs  of  circles  &c. 


PROPOSITION  24.  229 


EXEBCISES. 

1.  If  D  be  a  point  in  the  side  BG  of  a  triangle  ABG  whose  sides 
AB,  AG  are  equal,  the  circles  described  about  the  triangles  ABD,  AGD 
are  equal. 

2.  Find  a  point  P  within  an  equilateral  triangle  ABG,  such  that 
the  circles  described  about  the  triangles  PBG,  PGA,  PAB  may  be  all 
equal. 

3.  Find  a  point  P  in  the  plane  of  a  triangle  ABG  such  that  the 
circles  described  about  the  triangles  PBG,  PGA,  PAB  maybe  equal. 


230 


BOOK  III. 


PROPOSITION  25. 


TofiTid  the  centre  of  the  circhy  of  which  a  given  arc  is  a 
part. 

Let  ABC  be  a  given  arc  : 
it  is  required  to  find  the  centre  of  the  circle,  of  which  the 
arc  ABC  is  a  part. 

Construction.     Draw  AC,  and  bisect  it  at  J). 

(I.  Prop.  10.) 
At  I>  draw  DB  at  right  angles  to  ^0  cutting  the  arc  at  B. 

(I.  Prop.  11.) 
Draw  ABj  bisect  it  at  B,  and  at  B  draw  BF  at  right  angles 
to  AB  meeting  BD  or  BD  produced  at  F*; 
then  F  is  the  centre  required. 

(1)  B 


(V 

B 

(s) 

B 

/e/ 

^ 

\  ^\ 

D          \ 

{/        \ 

J        A             ^ 

G 

D 


Proof.     Because  DB  bisects  the  chord  il(7  at   right 
angles, 

DB  passes  through  the  centre;         (Prop.  2.) 
and  because  EF  bisects  the  chord  J.^  at  right  angles, 

EF  passes  through  the  centre.  (Prop.  2.) 

Now  two  straight  lines  cannot  intersect  in  more  than  one 
point.  (I.  Post.  1.) 

Therefore  F,  the  point  of  intersection  of  BD  and  EF, 

is  the  centre. 
Wherefore,  the  centre  of  the  circle,  of  which  the  given  arc 
ABC  is  a  part,  has  been  found. 


*  The  lines  must  meet,  see  Ex.  2,  p.  51.     In  figure  (2)  F  coincides 
with  D, 


PROPOSITION  25.  231 


PROPOSITION   25  A. 


Equal  circles  have  equal  radii. 

If  two  circles  be  equal,  it  is  possible  to  shift  one  of  them 
so  as  to  coincide  with  the  other.  (I.  Def.  21,  page  13.) 

Let  this  be  done. 

Then,  because  a  circle  cannot  have  more  than  one  centre, 

(Prop.  1.) 
the  centres  of  the  two  coincident  circles  must  be  coincident : 
and  therefore  all  the  radii  of  both  circles  are  equal. 

Wherefore,  equal  circles  (kc. 


EXERCISES. 

1.  Having  given  two  arcs  of  circles,  shew  how  to  find  whether 
they  are  parts  of  the  same  circle. 

2.  Having  given  two  arcs  of  circles,  find  whether  they  are  parts 
of  concentric  circles. 

3.  Having  given  two  arcs  of  circles,  find  whether  one  circle  lies 
wholly  within  the  other. 


232  BOOK  III. 

PROPOSITION  26. 

In  equal  circles  the  arcs^  on  which  equal  angles  at  the 
centres  standi  are  equal ;  and  the  arcs,  on  which  equal  angles 
at  the  circumferences  stand,  are  equal. 

Let  ABCD,  EFGE  be  two  equal  circles,  and  let  AKC, 
ELG  be  two  angles  at  the  centres  standing  on  the  arcs 
ABC,  EHG,  and  let  ABC,  EFG  be  two  angles  at  the  cir- 
cumferences standing  on  the  same  arcs  :  and  let 

(1)  the  angles  A  KG,  ELG  be  equal, 

(2)  the  angles  ABC,  EFG  be  equal : 

it  is  required  in   either   case   to   prove   that   the   arcs 
ADC,  EHG  are  equal. 

Construction.     Draw  AC,  EG, 


Proof.     Because  the  angle  AKC  is  double  of  the  angle 
ABC,  (Prop.  20.) 

and  the  angle  ELG  is  double  of  the  single  EFG,   (Prop.  20.) 
in  case  (1)  because  the  angles  AKC,  ELG  are  equal, 
the  angles  ABC,  EFG  are  equal, 
and  in  case  (2)  because  the  angles  ABC,  EFG  are  equal, 
the  angles  AKC,  ELG  are  equal. 
Now  because  the  circles  are  equal, 

their  radii  AK,  KC,  EL,  LG  are  equal.    (Prop.  25  A.) 
Therefore  in  both  cases  (1)  and  (2), 

because  in  the  triangles  AKC,  ELG, 

AK  \&  equal  to  EL, 

KC  to  LG, 

and  the  angle  AKC  to  the  angle  ELG, 

the  triangles  are  equal  in  all  respects ;    (I.  Prop.  4.) 

therefore  AC  i^  equal  to  EG. 


PROPOSITION^  26.  233 

And  because  the  arcs  ABC,  EFG^   which    contain   equal 
angles,  have  equal  chords  AC,  EG, 

the  arcs  ABC,  EFG  are  equal :        (Prop.  24.) 
but  the  circles  ABCD,  EFGH  are  equal ; 
therefore  the  remaining  arcs  ADC,  EHG  are  equal. 
Wherefore,  in  equal  circles  kc. 


Corollary.  In  the  same  circle  the  arcs,  on  which  equal 
angles  at  the  centres  stand,  are  equal;  and  the  arcs,  on  which 
equal  angles  at  tlie  circumferences  stand,  are  equal. 


EXERCISES. 

1.  If  PQ,  MS  he  a  pair  of  parallel  chords  in  a  circle,  then  the  arcs 
PS,  QR  are  equal,  and  the  arcs  PR,  QS  are  equal. 

2.  A  quadrilateral  is  inscribed  in  a  circle,  and  two  opposite  angles 
are  bisected  by  straight  lines  meeting  the  circumference  in  P  and  Q  ; 
prove  that  PQ  is  a  diameter. 

3.  If  through  P  any  point  on  one  of  two  circles,  which  intersect 
in  A  and  B,  the  straight  lines  PA,  PB  be  drawn  and  produced  if 
necessary  to  cut  the  other  circle  in  Q  and  R,  the  arc  QR  is  x>i  constant 
length. 

4.  The  internal  bisectors  of  the  vertical  angles  of  all  triangles, 
on  the  same  base  and  on  the  same  side  of  it,  which  have  equal 
vertical  angles,  pass  through  one  fixed  point  and  the  external  bisectors 
through  another  fixed  point. 

5.  If  through  one  of  the  points  of  intersection  of  two  equal 
circles  a  straight  line  be  drawn  terminated  by  the  circles,  the  straight 
lines  joining  its  extremities  with  the  other  point  of  intersection  are 
equal. 


T.  K.  16 


234  BOOK  III. 

PROPOSITION  27. 

In  equal  circles,  anyles  coidaiited  hij  a7'cs,  which  are  of 
equal  length,  are  equal. 

Let  A  BCD,  EFGH  be  equal  circles,  and  let  ABC,  EFG 
be  arcs  of  equal  lengtli : 

it  is  required  to  prove  that  the  arcs  ABC,  EFG  contain 
equal  angles. 

Construction.     Find  the  centres  K,  L  of   the  circles 
ABGD,  EFGH,  (Prop.  5.) 

and  draw  AK,  KG,  EL,  LG. 


Proof.     Because  the  circles  ABCD,  EFGH  are  equal, 
their  radii  AK,  KG,  EL,  LG  are  equal.     (Prop.  25  A.) 
And  because  AK  is  equal  to  EL, 

it  is  possible  to  shift  the  fi^Mve  ABCD K  so  that  A  coincides 
with  E,  and  K  with  L,  and  so  that  the  parts  of  the  arcs 
ABC,  EFG  near  E  are  on  the  same  side  of  EL. 
If  this  be  done, 

because  the  radii  of  the  circles  are  equal  and  their  centres 
coincide, 

the  circles  must  coincide ; 
and  because  the  circles  coincide,  and  the  parts  of  the  arcs 
ABC,  EFG  near  E  are  on  the  same  side  of  EL,  those 
parts  of  the  arcs  coincide  ; 
and  because  the  arcs  are  of   equal    length   and   have  one 
extremity  common, 

therefore  the  other  extremity  must  be  common, 
that  is,  the  point  C  must  coincide  with  the  point  G, 


PROPOSITION  27.  235 

Therefore  the  arc  ABC  coincides  with  the  arc  EFG ; 
and  angles  in  the  two  arcs  are  then  angles  in  the  same  arc 
and  therefore  equal.  (Prop.  21.) 

Wherefore,  in  equal  circles  &c. 

Corollary.    In  the  same  circle^  angles  contained  by  arcs, 
which  are  of  equal  length,  are  equal. 


EXERCISES. 

1.  The  straight  lines  joining  the  extremities  of  two  equal  arcs  of 
a  circle  are  parallel  or  are  equal. 

Can  they  be  both  parallel  and  equal  ? 

2.  The  straight  lines  bisecting  any  angle  of  a  quadrilateral  in- 
scribed in  a  circle  and  the  opposite  exterior  angle,  meet  on  the  circle. 

3.  If  from  any  point  on  a  circle  a  chord  and  a  tangent  be  drawn, 
the  perpendiculars  on  them  from  the  middle  point  of  either  of  the  arcs 
subtended  by  the  chord  are  equal  to  one  another. 


16—2 


236 


BOOK  III. 


PROPOSITION  28. 

In  equal  circles^  arcs  cut  off  by  chords,  tvhich  are  equal  to 
one  another,  are  of  equal  length,  tlie  greater  equal  to  the 
greater  and  the  less  equal  to  the  less. 

Let  ABCD,  EFGH  be  equal  circles,  and  let  AC\  EG  be 
equal  chords  which  cut  off  the  two  greater  arcs  ABC,  EFG, 
and  the  two  less  arcs  ADC,  EHG  : 

it  is  required  to  prove  that  the  arcs  ABC,  EFG  are  of  equal 
length, 
and  that  the  arcs  ADC,  EHG  are  of  equal  length. 

Construction.     Find  K,  L,  the  centres  of  the  circles 
ABCD,  EFGH,  (Prop.  5.) 

and  draw  AK,  KC,  EL,  LG. 


Proof.     Because  the  circles  are  equal, 
their  radii  AK,  KC,  EL,  LG  are  equal;  (Prop.  25  A.) 
and  because  in  the  triangles  AKC,  ELG, 
KA  is  equal  to  LE, 
KC  to  LG, 
and  ^C  to  EG, 
the  triangles  are  equal  in  all  respects;  (I.  Prop.  8.) 
therefore  the  angle  AKC  is  equal  to  the  angle  ELG. 
But  in  equal  circles  the  arcs,  on  which  equal  angles  at  the 
centres  stand,  are  equal;  (Prop.  26.) 

therefore  the  arc  ADC  is  equal  to  the  arc  EHG. 
But  the  circle  ABCD  is  equal  to  the  circle  EFGH\ 

therefore  the  arc  ABC  is  equal  to  the  arc  EFG. 
Wherefore,  in  equal  circles  &c. 


PROPOSITION  28.  237 

Corollary.  In  the  scmie  circle,  arcs  cut  off  by  chords, 
which  are  equal  to  one  another,  are  of  equal  length,  the 
greater  equal  to  the  greater  a7id  the  less  equal  to  the  less. 


EXERCISES. 

1.  A  triangle  is  turned  about  its  vertex  till  one  of  the  sides 
passing  through  the  vertex  is  in  the  same  straight  line  as  the  other 
previously  was.  Prove  that  the  line  joining  the  vertex  with  the  in- 
tersection of  the  two  positions  of  the  base,  produced  if  necessary, 
bisects  the  angle  between  these  two  positions. 

2.  Find  a  point  on  one  of  two  given  equal  circles,  such  that,  if 
from  it  two  tangents  be  drawn  to  the  other  circle,  the  chord  joining 
the  points  of  contact  is  equal  to  the  chord  of  the  first  circle  formed 
by  joining  its  points  of  intersection  with  the  two  tangents  produced. 

Determine  the  conditions  of  the  possibility  of  a  solution  of  the 
problem. 


238 


BOOK  III. 


PROPOSITION  29. 

In  equal  circles,  chords,  hy  which  arcs  of  equal  length  are 
subtended,  are  equal. 

Let  ABCD,  EFGII  be  equal  circles,  and  let  AC,  EG  be 
chords  by  wliicli  ADC,  EIIG,  arcs  of  equal  length  are  sub- 
tended : 
it  is  required  to  prove  that  the  chords  AC,  EG  are  equal. 

Construction.     Find  K,  L,  the  centres  of  the  circles 
ABCD,  EFGII,  (Prop.  5.) 

and  draw  AK,  EL. 


Proof.     Because  the  circles  ABCD,  EFGH  are  equal, 
their  radii  AK,  EL  are  equal.       (Prop.  25  A.) 
And  because  AK  \%  equal  to  EL, 

it  is  possible  to  shift  the  figure  ABCDK  so  that  A  coincides 
with  E,  and  K  with  L,  and  so  that  the  parts  of  the  arcs 
ABC,  EFG  near  E  are  on  the  same  side  of  EL. 
If  this  be  done, 

because  the  radii  of  the  circles  are  equal  and  their  centres 
coincide, 

the  circles  must  coincide; 
and  because  the  circles  coincide,  and  the  parts  of  the  arcs 
ABC,  EFG  near  E  are  on  the  same  side  of  EL,  those 
parts  of  the  arcs  coincide ; 
and  because  the  arcs   are  of   equal   length  and  have  one 
extremity  common, 

therefore  the  other  extremity  must  be  common, 
that  is,  the  point  C  must  coincide  with  the  point  G. 


PROPOSITION  29.  239 

Therefore  the  chord  AC  coincides  with  the  cliord  EG 

and  is  equal  to  it. 
Wlierefore,  w  equal  circles  <fec. 


Corollary.     In  the  same  circle,  chords,  hy  ivhich  arcs  of 
equal  length  are  subtended,  are  equal. 


EXERCISES. 

1.  If  the  diagonals  of  a  quadrilateral  inscribed  in  a  circle  bisect 
one  another,  the  diagonals  are  diameters. 

2.  If  two  chords  AP,  AQ  oi  o,  circle  intersect  at  a  constant  angle 
at  a  fixed  point  A  on  the  circle,  the  chord  PQ  always  touches  a  con- 
centric circle. 

3.  Two  triangles  are  inscribed  in  a  circle  :  if  two  sides  of  the  one 
be  parallel  to  two  sides  of  the  other,  the  third  sides  are  equal. 

Is  it  necessary  that  they  are  parallel  ? 


240  BOOK  III. 

PROPOSITION  30. 

To  bisect  a  gwen  arc  of  a  circle. 

Let  ABC  be  the  given  arc  : 

it  is  required  to  bisect  it. 

Construction.     Draw  ^(7; 

bisect  it  at  i>  ;  (I.  Prop.  10.) 

at  ])  draw  DB  at  right  angles  to  AC  meeting  the  arc  at  B: 

(I.  Prop.  11.) 
the  arc  ABC  is  bisected  as  required  at  the  point  B. 
Draw  AB,  BC. 


Proof.     Because  in  the  triangles  ADB,  CDB^ 
A  J)  is  equal  to  CD, 
and  DB  to  DB, 
and  the  angle  ADB  to  the  angle  CDB, 
the  triangles  are  equal  in  all  respects;    (I.  Prop.  4.) 
therefore  ^^  is  equal  to  CB. 
But  arcs  cut  off  by  equal  chords  are  equal,  the  greater  equal 
to  the  greater,  and  the  less  equal  to  the  less; 

(Prop.  28,  Coroll.) 
and  because  BD,  if  produced,  is  a  diameter,  (Prop.  2.) 

each  of  the  arcs  AB,  CB  is  less  than  a  semicircle, 
and  therefore  the  arc  AB  is,  the  smaller  of  the  two  arcs  cut 
oif  by  the  chord  AB,  and  the  arc  CB  the  smaller  of  those 
cut  off  by  the  chord  CB  ; 

therefore  the  arc  AB  is  equal  to  the  arc  CB. 
Wherefore,  the  given  arc  ABC  is  bisected  at  B. 


PROPOSITION  30.  241 


EXERCISES. 

1.  Find  the  triangle  of  maximum  area  which  can  be  inscribed 
in  a  given  circle  having  a  given  chord  for  one  side. 

2.  Prove  that  the  triangle  of  maximum  area  inscribed  in  a  circle 
is  equilateral. 

3.  The  greatest  quadrilateral  which  can  be  inscribed  in  a  circle 
is  a  square. 

4.  Having  given  a  regular  polygon  of  any  number  of  sides  in- 
scribed in  a  circle,  inscribe  a  regular  polygon  of  double  the  number 
of  sides. 

5.  li  ABC  an  arc  of  a  circle  less  than  a  semicircle  be  bisected  in 
B,  and  AB  produced  meet  CI)  which  is  drawn  at  right  angles  io  BG  in 
J),  and  the  tangents  at  A  and  C  meet  in  A',  then  B,  C,  D,  E  lie  on  a 
circle. 


242  BOOK  III. 


PROPOSITION  31. 

An  angle  in  a  semicircle  is  a  right  angle  ;  an  angle  in  an 
arc,  which  is  greater  than  a  semicircle,  is  less  than  a  right 
angle ;  and  an  angle  in  an  arc^  which  is  less  than  a  semi- 
circle, is  greater  than  a  right  angle. 

Let  ABODE  be  a  circle,  of  which  ABD  is  a  semicircle, 
ADC  an  arc  greater  than  a  semicircle,  and  ABC  an  arc  less 
than  a  semicircle  : 

it  is  required  to  prove  that  the  angle  in  the  semicircle 
ABD  is  a  right  angle;  that  the  angle  in  the  arc  ADC 
is  less  than  a  right  angle,  and  that  the  angle  in  the 
arc  ABC  is  greater  than  a  right  angle. 

Construction.  Take  any  point  B  in  the  arc  ABC  and 
any  point  E  in  the  semicircle  AED  and  draw  AB,  BC,  CD^ 
DE,  EA,  AC,  AD. 


Proof.  Because  ACDE  is  a  quadrilateral  inscribed  in 
a  circle,  the  sum  of  the  angles  ACD,  AED  is  equal  to  two 
right  angles.  (Prop.  22.) 

But  because  each  of  the  angles  ACD,  AED  is  contained  by 
a  semicircle,  the  angle  ACD  is  equal  to  the  angle  AED  ; 

(Prop.  27,  Coroll.) 
therefore  each  of  them  is  a  right  angle. 
Next,   because  the  angle  ACD  of   the  triangle  ACD  is  a 
right  angle, 

the  angle  ADC  is  less  than  a  right  angle, 

(I.  Prop.  17.) 
and  it  is  an  angle  in  the  arc  ADC. 


PROPOSITION  31.  243 

Again,  because  ABCD  is  a  quadrilateral  inscribed  in  a 
circle,  the  sum  of  the  angles  ABC,  ADC  is  equal  to  two 
right  angles.  (Prop.  22.) 

And  the  angle  ADC  has  been  proved  to  be  less  than  a  riglit 

angle ; 
therefore  the  angle  ABC  is  greater  than  a  right  angle, 
and  it  is  an  angle  in  the  arc  ABC. 
Wherefore,  an  angle  in  a  semicircle  <fec. 


EXERCISES. 

1.  The  circles  described  on  two  equal  sides  of  a  triangle  as 
diameters  intersect  at  the  middle  point  of  the  third  side. 

2.  The  circles  described  on  any  two  sides  of  a  triangle  as 
diameters  intersect  on  the  third  side. 

3.  Construct  the  rectangle  one  of  whose  diagonals  is  a  straight 
line  given  in  magnitude  and  position  and  the  other  of  wlnose  diago- 
nals passes  through  a  given  point. 

4.  An  angle  BAG  of  constant  magnitude  turns  round  its  apex 
A  which  is  fixed.  Prove  that  the  line  joining  the  feet  of  the  perpendi- 
culars from  a  fixed  point  O  on  AB,  AC  always  touches  a  fixed  circle. 

5.  If  a  circle  A  pass  through  the  centre  of  a  circle  B,  the  tan- 
gents to  B  at  the  points  of  intersections  of  A  and  B  intersect  on  the 
circle  A. 

6.  Two  chords  AB,  CD  of  constant  length  placed  in  a  circle 
subtend  angles  at  the  centre  whose  sum  is  equal  to  two  right  angles. 
If  AC,  Bl)  intersect  in  P,  the  distances  of  P  from  the  middle  points 
of  the  chords  will  be  independent  of  their  relative  positions. 


244  BOOK  III. 

PROPOSITION  32. 

If  a  chord  he  drawn  from  the  pohit  of  contact  of  a 
tangent  to  a  circle,  each  of  the  angles  which  this  chord  makes 
with  the  tangent  is  equal  to  the  angle  iri  the  alternate  arc 
of  the  circle*. 

Let  ABCD  be  a  circle,  and  EAF  be  the  tangent  at  a 
point  A,  and  \etAC  be  a  chord  drawn  from  tlie  point  A  : 
it  is  required  to  prove  that  the  angle  CA  E  is  equal  to  the 

angle  in  the  arc  CD  A,  and  the  angle  CAF  to  the  angle  in 

the  arc  CBA. 

Construction.  At  A  draw  AB  at  right  angles  to  EAF, 
cutting  the  circle  again  at  B ;  (I.  Prop.  11.) 

take  any  point  I)  in  the  arc  ADC  and  draw  BC,  CD,  DA. 


Proof.     Because  AB  i&  drawn  at  right  angles  to  EAF, 

AB  passes  through  the  centre;         (Prop.  19.) 

therefore  BCD  A  is  a  semicircle;      (Prop.  1  A.) 

and  the  angle  BCA  is  a  right  angle;    (Prop.  31.) 

therefore  the  sum  of  the  other  two  angles  BAC,  CBA  of 

the  triangle  ABC  is  equal  to  a  right  angle.     (I.  Prop.  32.) 

And  the  sum  of  the  angles  BAC,  CAF,  that  is,  the  angle 

BAF,  is  a  right  angle ; 

therefore  the  sum  of  the  angles  BAC,  CAF  is  equal  to  the 

sum  of  the  angles  BAC,  CBA. 

*  The  alternate  arc  is  the  name  generally  given  to  the  arc  which 
lies  on  the  side  of  the  chord  opposite  to  the  angle  spoken  of. 


PROPOSITION  32.  245 

Take  away  the  common  angle  BA  C ; 

then  the  angle  OAF  is  equal  to  the  angle  CBA, 

which  is  an  angle  in  the  arc  ABC. 

Again,  because  the  sum  of  the  angles  ABC,   ADC  is 

equal  to  two  right  angles,  (Prop.  22.) 

and  the   sum  of  the    angles   CAF,   CAE   is    equal  to  two 

right  angles;  (I.  Prop.  13.) 

therefore  the  sum  of  the  angles  CAF,  CAE  is  equal  to 

the  sum  of  the  angles  ABC,  ADC ; 

and  it  has  been  proved  that  the  angle  CAF  is  equal 

to  the  angle  ABC ; 
therefore  the  angle  CAE  is  equal  to  the  angle  ADC, 

which  is  an  angle  in  the  arc  ADC. 
Wherefore,  if  a  chord  be  drawn  <fec. 

The  theorem  of  Proposition  32  follows  immediately  from  the 
theorem  of  Proposition  21,  if  we  consider  the  tangent  at  a  point 
as  the  limiting  position  of  a  chord  drawn  through  the  point.  (See 
page  217.) 

For  the  angle  between  the  tangent  AF  in  the  diagram  of  Pro- 
position 32  and  the  chord  ^C  is  the  angle  between  two  chords  of  the 
arc  ABC,  one  an  indefinitely  short  one  drawn  from  a  point  indefi- 
nitely near  ^  to  ^  and  the  other  drawn  from  the  same  point  to  C, 
and  is  therefore  an  angle  in  the  arc  ABC  and  therefore  equal  to  the 
angle  ABC.   (Prop.  21.) 

EXERCISES. 

1.  If  two  circles  touch  each  other,  any  straight  line  drawn  through 
the  point  of  contact  will  cut  off  similar  segments. 

2.  On  the  same  side  of  portions  AB,  AG  of  &  straight  line  ABC 
similar  segments  of  circles  are  described  :  prove  that  the  circles  touch 
one  another. 

3.  If  two  circles  OPQ,  Opq  touch  at  O,  and  OPp,  OQq  be  straight 
lines,  the  chords  PQ,pq  are  parallel. 

4.  If  a  straight  line  cut  two  circles  which  touch  at  0,  in  the 
points  P,  Q,  and  j^,  q,  the  angles  POp,  QOq  are  either  equal  or  supple- 
mentary. 

5.  ABC  is  a  triangle  inscribed  in  a  circle,  and  from  any  point  D 
in  BC  Q.  straight  line  DE  is  drawn  parallel  to  CA  and  meeting  the 
tangent  at^  in  E ;  shew  that  a  circle  may  be  described  round  AEBD. 


246  BOOK  III. 

PROPOSITION  33. 

To  describe  on  a  yiveyi  finite  straight  line  an  arc  of  a 
circle  containing  an  angle  equal  to  a  given  angle. 

Let  AB  be  the  given  straight  line,  and  G  the  given 
angle : 

it  is  required  to  describe  on  AB  an  arc  of  a  circle  contain- 
ing an  angle  equal  to  the  angle  C. 

Construction.     Bisect  AB  at  E.  (I.  Prop.  10.) 

If  the  angle  C  be  a  right  angle,  with  E  as  centre  and 
EA  or  EB  as  radius,  describe  a  circle : 

then  the  semicircle  on  either  side  of  AB  is  an  arc  described 
as  required.  (Prop.  31.) 

If  the  angle  C  be  not  a  right  angle,  from  A  draw  AD 
making  the  angle  BAD  equal  to  the  angle  C;  (I.  Prop.  23.) 
and  at  A^  E  draw  AF,  EF  at  right  angles  to  AD^  AB 
respectively,  (I.  Prop.  11.) 

and  let  them  meet  at  ^*. 
Draw  FB^  and  mth  F  as  centre  and  FA  as  radius  describe 

the  circle  A  GH : 
this  circle  passes  through  B  and  the  arc  AGB  on  the  side 
oi  AB  away  from  ^Z>  is  an  arc  described  as  required. 


Proof.     Because  in  the  triangles  FEA,  FEB, 

EA  is  equal  to  EB, 

and  FE  to  FE, 

and  the  angle  FEA  to  the  angle  FEB, 

the  triangles  are  equal  in  all  respects  ;         (I.  Prop.  4.) 

therefore  FA  is  equal  to  FB. 

*  The  lines  must  meet.     See  Ex.  2,  p.  51. 


PROPOSITION  33.  247 

Therefore  the  circle  AGH  passes  through  B. 
Again,  because  AD  is  drawn  from  A  at  right  angles  to 
the  radius  AF, 

AD  touches  the  circle;  (Prop.  16.) 

and  Ijecause  the  chord  AB  is  drawn  from  the  point  of 
contact  of  the  tangent  AD, 
the  angle  in  the  alternate  arc  AGB  is  equal  to  the  angle 
DAB,  (Prop.  32.) 

that  is,  to  the  angle  C. 
Wherefore,  on  the  given  straight  line  AB  the  arc  AGB 
/ms  been  described  containing  an  angle  equal  to  the  given 
angle  C. 


EXEECISES. 

1.  Find  a  point  at  which  each  of  two  given  finite  straight  lines 
subtends  a  given  angle. 

2.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle, 
and  the  foot  of  the  perpendicular  from  the  vertex  on  the  base. 

3.  Having  given  the  base  and  the  vertical  angle  of  a  triangle, 
construct  the  triangle  which  will  have  the  maximum  area. 

4.  Find  a  point  0  within  a  given  triangle  ABC,  so  that,  if  ^O, 
BO,  CO  be  joined,  the  angles  OAB,  OBC,  OCA  shall  be  all  equal. 

5.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle, 
and  the  altitude. 

6.  Find  the  locus  of  a  point  at  which  two  given  equal  straight 
lines  AB,  BG  subtend  equal  angles. 


248  BOOK  III. 


PROPOSITION  34. 

To  cut  off  from  a  given  circle  an  arc  containi^ig  an  angle 
equal  to  a  given  angle. 

Let  ABC  be  the  given  circle  and  D  the  given  angle : 
it  is  required  to  cut  off  from  the  circle  ABC  an  arc  con- 
taining an  angle  equal  to  the  angle  D. 

Construction.     Take  any  point  A   on  the  circle,  and 

through  A  draw  the  straight  line  EAF  to  touch  the  circle 

at  A.  (Prop.  17.) 

From  A   draw  AB    making  the  angle  BAE  equal  to  the 

angle  jD, 

and  cutting  the  circle  again  at  B :    (I.  Prop.  23.) 
then  the  arc  ACB,  on  the  side  oi  AB  away  from  E^  is  an  arc 
cut  off  as  required. 


Proof.     Because  the  chord  ^^  is  drawn  from  the  point 
of  contact  A  of  the  tangent  EAF, 

the  angle  EAB  is  equal  to  the  angle  in  the  alternate  arc 
ACB.  (Prop.  32.) 

And  the  angle  EAB  is  equal  to  the  angle  D  : 
therefore  the  angle  in  the  arc  ACB  is  equal  to  the  angle  D. 

"Wherefore,  the  arc  ACB  has  been  cut  off  from  the  given 
circle  ABC  containing  an  angle  equal  to  the  given  angle  D. 


PROPOSITION  34. 


249 


Through  a  given  point  two  chords  can  be  drawn  which  will  cut  off 
arcs  containing  an  angle  equal  to  a  given  angle. 


Outline  of  Alternative  Construction. 

Through  A  draw  any  chord  AP,  and  from  P  draw  PR  making  the 
angle  RPA  equal  to  the  given  angle  D,  and  cutting  the  circle  again 
&tQ. 


It  may  be  proved  that  the  arc  ABQ,  measured  from  A  on  the  side 
of  ^P  opposite  to  that  on  which  PR  is  drawn,  is  an  arc  cut  off  as 
required. 


EXERCISES. 

1.  In  a  given  circle  inscribe  an  equiangular  triangle. 

2.  Inscribe  in  a  given  circle  a  triangle,  so  that  one  angle  may  be 
a  half  of  a  second  angle  and  a  third  of  the  third  angle. 

3.  Inscribe  in  a  given  circle  a  right-angled  triangle,  so  that  one 
of  its  acute  angles  may  be  three  times  the  other, 


T.  K. 


17 


250  BOOK  III, 


PROPOSITIOISr  35. 

If  two  chords  of  a  circle  intersect  at  a  point  within  the 
circle^  the  rectangle  contained  by  the  segments  of  one  chord,  is 
equal  to  the  rectangle  contained  by  the  segynents  of  the  other 
chord. 

Let  ABGD  be  a  circle  and  AG^  BD  two  chords  inter- 
secting at  the  point  E  within  the  circle  : 
it  is  required  to  prove  that  the    rectangle  contained   by 
AEj  EC  is  equal  to  the  rectangle  contained  by  BE,  ED. 

Construction.    If  the  point  E  be  the  centre,  it  is  clear 
that  AE,  EC,  BE,  ED  are  equal,  each  being  a  radius, 
and  that  the  rectangles  AE,  EG  and  BE,  ED,  each  of  which 
is  equal  to  the  square  on  a  radius,  are  equal. 
If  E  be  not  the  centre,  find  0  the  centre ;        (Prop.  5.) 
draw  OF  &t  right  angles  to  AG,     (I.  Prop.  12.) 
and  draw  OE,  OG. 


Proof.     Because  OF  is  drawn  from  the  centre  at  right 
angles  to  the  chord  AG, 

therefore  AF  is  equal  to  FG.  (Prop.  4.) 

And  because  EC  is  the  sum  of  FG,  EF, 
and  ^^  is  the  difference  of  AF,  EF,  that  is,  of  FG,  EF, 
therefore  the  rectangle  AE,  EG  is  equal  to  the  difference 
of  the  squares  on  FG,  EF.       (II.  Prop.  5.) 
But  because  the  angles  at  F  are  right  angles, 
the  sum  of  the  squares  on  OF,  FG  is  equal  to  the  square 

on  OG, 
and  the  sum  of  the  squares  on  OF,  EF  is  equal  to  the  square 
on  OE;  (I.  Prop.  47.) 


PROPOSITION  35.  251 

therefore  the  difference  of  the  squares  on  FG^  EF  is  equal  to 

the  difference  of  the  squares  on  0C\  OE. 
Therefore  the  rectangle  AE,  EC  is  equal  to  the  difference  of 

the  squares  on  0(7,  OE. 
Similarly  it  can  be  proved  that  the  rectangle  BE,  ED  is 

equal  to  the  difference  of  the  squares  on  OB,  OE, 

that  is,  is  equal  to  the  difference  of  the  squares  on  OC,  OE, 

since  OC  is  equal  to  OB. 

Therefore  the  rectangle  AE,  EC  is  equal  to  the 

rectangle  BE,  ED. 

Wherefore,  if  two  cJwrds  of  a  circle  kc. 

There  are  two  special  cases  which  should  be  noticed  by  the  student, 
one  case,  when  the  points  E,  F  coincide,  i.e.  when  one  chord  bisects 
the  other ;  the  other  case,  when  the  points  0,  F  coincide,  i.  e.  when 
one  chord  is  a  diameter. 

In  Proposition  35  the  distances  between  the  ends  of  a  chord  and  a 
point  in  the  chord  are  spoken  of  as  segments  of  the  chord.  In  Pro- 
position 36  it  will  be  noticed  that  the  expression  segments  of  a  chord 
has  been  used  of  the  distances  between  the  ends  of  the  chord  and  a 
point  taken  in  the  chord  produced.  In  the  first  case  the  chord  is 
equal  to  the  sum  of  the  segments,  in  the  second  to  the  difference 
of  the  segments. 


EXEBCISES. 

1.  Prove  the  converse  of  Proposition  35,  i.e.  that,  if  -4C,  BD  be 
two  straight  lines  intersecting  at  E  such  that  the  rectangles  AE,  EG, 
and  BE,  ED  are  equal,  then  A,  B,  0,  D  lie  on  a  circle. 

2.  If  through  any  point  in  the  common  chord  of  two  intersecting 
circles  there  be  drawn  any  two  other  chords,  one  in  each  circle,  their 
four  extremities  all  lie  on  a  circle. 

3.  Draw  through  a  given  point  within  a  circle  a  chord,  one  of 
whose  segments  shall  be  four  times  as  long  as  the  other.  When  is 
this  possible? 

4.  Divide  a  given  straight  line  into  two  parts,  so  that  the  rect- 
angle contained  by  the  parts  may  be  equal  to  a  given  rectangle. 

5.  A,  B,  G  are  three  points  on  a  circle,  D  is  the  middle  point  of 
BG  and  AD  produced  meets  the  circle  in  E:  prove  that  the  sum  of 
the  squares  on  AB,  AG  is  double  of  the  rectangle  AD,  AE. 

17—2 


252  BOOK  III. 


PROPOSITION  36. 

If  two  chords  of  a  circle  when  produced  intersect  at  a  point 
without  the  circle,  the  rectangle  contained  by  the  segments  of 
one  chord  is  equal  to  the  rectangle  contained  by  the  segments 
of  the  other  chord. 

Let  ABDG  be  a  circle  and  let  GA^  DB  be  two  chords, 
which  intersect,  when  produced  beyond  A  and  B^  at  the 
point  E  without  the  circle  : 

it  is  required   to    prove   that  the  rectangle  contained  by 
EA,  EC  is  equal  to  the  rectangle  contained  by  EB,  ED. 

Construction.     Find  0  the  centre;  (Prop.  5.) 

drav/  OF  at  right  angles  to  ACy    (I.  Prop.  12.) 
and  draw  OE,  00. 


Proof.     Because  OF  is  drawn  from  the  centre  at  right 
angles  to  the  chord  AG, 

therefore  AF  is  equal  to  FG.  (Prop.  4.) 

And  because  EG  is  the  sum  of  EF,  FG, 
and  EA  is  the  difference  of  EF,  AF,  that  is,  of  EF,  FG, 
therefore  the  rectangle  EA,  EG  is  equal  to  the  difference 
of  the  squares  on  EF,  FG.       (II.  Prop.  6.) 
But  because  the  angles  at  F  are  right  angles, 
the  sum  of  the  squares  on  OF,  FE  is  equal  to  the  square 

on  OF, 
and  the  sum  of  the  squares  on  OF,  FG  is  equal  to  the 
square  on  OG  ;  (I.  Prop.  47.) 

therefore  the  difference  of  the  squares  on  EF,  FG  is  equal 
to  the  difference  of  the  squares  on  OE,  OG. 


PROPOSITION  36.  253 

Therefore  the  rectangle  EA,  EG  is  equal  to  the  difference 

of  the  squares  on  OE,  00. 
Similarly  it  can  be  proved  that  the  rectangle  EB^  ED  is 

equal  to  the  difference  of  the  squares  on  OE,  OD, 

that  is,  is  equal  to  the  difference  of  the  squares  on  OE^  OC^ 

since  OD  is  equal  to  OC. 

Therefore  the  rectangle  EA^  EC  is  equal  to  the 

rectangle  EB,  ED. 

Wherefore,  if  two  chords-  of  a  circle  &c. 

There  are  two  special  cases  which  should  be  noticed  by  the 
student,  one  case,  when  the  points  O,  F  coincide,  i.e.  when  one  chord 
is  a  diameter;  the  other  case,  when  the  points  JB,  D  coincide,  i.e.  when 
one  chord  is  a  tangent.  The  statement  of  the  theorem  in  the  latter 
case  appears  in  the  Corollary. 

Corollary. 

If  a  chord  of  a  circle  be  produced  to  any  point,  the 
rectangle  contained  hy  the  segments  of  the  chord  is  equal  to 
the  square  on,  the  tangerit  drawn  to  tJie  circle  from  the  point. 

This  result  is  seen  at  once  on  considering  the  tangent  as  the 
limiting  position  of  the  secant. 

EXEECISES. 

1.  Prove  the  converse  of  Proposition  36,  i.e.  that,  if  EAC,  EBD 
be  two  straight  lines  intersecting  at  E  such  that  the  rectangles  EA,EG 
and  EB,  ED  are  equal,  then  A,  B,  C,  D  lie  on  a  circle. 

2.  If  two  circles  intersect  each  other,  their  common  chord  bisects 
their  common  tangents. 

3.  From  a  given  point  as  centre  describe  a  circle  cutting  a  given 
straight  line  in  two  points,  so  that  the  rectangle  contained  by  their 
distances  from  a  given  point  in  the  straight  line  may  be  equal  to  a 
given  square. 

4.  If  yi BO  be  a  triangle  and  D  a  point  in  ^C  such  that  the  angle 
ABD  is  equal  to  the  angle  ACB,  then  the  rectangle  AC,  AJD  ia  equal 
to  the  square  on  AB. 

5.  If  from  each  of  two  given  points,  a  pair  of  tangents  be  drawn 
to  a  given  circle,  the  middle  points  of  the  chords  joining  the  points  of 
contact  of  each  pair  of  tangents  lie  on  the  circumference  of  a  circle 
passing  through  the  two  given  points. 


254 


BOOK  III. 


PROPOSITION  37. 

If  from  an  extei'nal  point  there  he  drawn  to  a  circle  two 
straight  lines,  one-  of  which  cuts  the  circle  in  two  points  and 
the  other  meets  it,  and  if  the  recta^iyle  contained  hy  the  seg- 
ments of  the  chord  on  the  line  which  cuts  the  circle  he  equal 
to  the  square  on  the  line  which  meets  the  circle^  the  line  which 
7neets  the  circle  is  a  tangent  to  it. 

Let  ABCD  be  a  circle  and  E  an  external  point;  and 
let  EAC   be  a  straight   line   cutting   the    circle  ait  A,   C 
and  EB  a  straight  line  meeting  it  at  B,  such  that  the  rect- 
angle contained  by  EA,  EC  is  equal  to  the  square  on  EB: 
it  is  required  to  prove  that  EB  touches  the  circle. 

Construction.     Pind  the  centre  0  ;  (Prop.  5.) 

from  E  draw  ED  to  touch  the  circle  at  Z> ;  (Prop.  17.) 
and  draw  OB,  OD,  OE. 


Proof.     Because  ED  is  a  tangent,  and  OD  is  the  radius, 
the  angle  EDO  is  a  right  angle.  (Prop.  18.) 

Because  EAC  cuts  the  circle  and  ED  touches  it, 
the  rectangle  EA,  EC  is  equal  to  the  square  on  ED; 

(Prop.  36,  Coroll.) 
and  the  rectangle  EA,  EC  is  equal  to  the  square  on  EB; 
therefore  the  square  on  EB  is  equal  to  the  square  on  ED; 
therefore  EB  is  equal  to  ED. 


PROPOSITION  37.  255 

Again,  because  in  the  triangles  EBO,  EDO^ 

EB  is  equal  to  ED, 

and  OB  to  OD, 

and  OE  to  OE, 

the  triangles  are  equal  in  all  respects;  (I.  Prop.  8.) 

therefore  the  angle  EBO  is  equal  to  the  angle  EDO. 

But  EDO  is  a  right  angle ; 

therefore  the  angle  EBO  is  a  right  angle. 

And  because  BE  is  at  right  angles  to  the  radius  OB, 

BE  touches  the  circle.  (Prop.  16.) 

Wlierefore,  if  from  an  external  lyoint  &c. 


EXEKCISES. 

1.  If  three  circles  meet  two  and  two,  the  common  chords  of  each 
pair  meet  in  a  point. 

2.  If  three  circles  touch  two  and  two,  the  tangents  at  the  points 
of  contact  meet  at  a  point. 

3.  If  the  tangents  drawn  to  two  intersecting  circles  from  a  point 
be  equal,  the  common  chord  of  the  circles  passes  through  the  point. 

4.  Describe  a  circle  which  shall  touch  a  given  straight  line  at  a 
given  point,  and  shall  cut  off  from  another  given  straight  line  a  chord 
of  a  given  length. 

5.  On  OP,  the  straight  line  drawn  from  a  given  point  0  to  P  a 
point  on  a  given  straight  Hne,  a  point  Q  is  taken  such  that  the 
rectangle  OP,  OQ  is  constant:  prove  that  the  locus  of  Q  is  a  circle. 

6.  On  OP,  a  chord  of  a  given  circle  drawn  from  a  given  point  O, 
a  point  Q  is  taken  such  that  the  rectangle  OP,  OQ  is  constant:  prove 
that  the  locus  of  Q  is  a  straight  line. 


256  BOOK  III. 

PROPOSITION  37  A. 
If  two    triangles   he   equiangular    to    07ie   another,    the 
rectangle  contained  hy  any  side  of  tlie  07ie  and  any  side  of 
tlie  oilier  is  equal   to    the  rectangle    contained  by  the  cor- 
7'esj)ondi7ig  sides  *. 

Let  ABC,  DEF  be  two  triangles  which  are  equiangular 
to  one  another,  having  the  angles  at  A,  B,  C  equal  to  the 
angles  at  D,  E,  F  respectively  : 

it  is  required  to  prove  that  the  i-cctangle  AB,  EF  is  equal 
to  the  rectangle  BC,  DE. 
Construction.     In  AB,  CB  produced  beyond  B,  take 
points  G,  II  such  that  BG  is  equal  to  EF,  and  BH  to  ED; 

(I.  Prop.  3.) 
and  draw  Gil. 


Proof.     Because  the  angle  GBH  is  equal  to  the  angle 
CBA,  (I.  Prop.  15.) 

and  the  angle  FED  is  equal  to  the  angle  CBA, 
the  angle  GBH  is  equal  to  the  angle  FED. 
Because  in  the  triangles  BGII,  EFD, 

BG  is  equal  to  EF  and  BH  to  ED, 
and  the  angle  GBH  is  equal  to  the  angle  FED, 
the  triangles  are  equal  in  all  respects:     (I.  Prop.  4.) 
therefore  the  angle  BGII  is  equal  to  the  angle  EFD ; 

but  the  angle  EFD  is  equal  to  the  angle  BCA, 

therefore  the  angle  AGH  is  equal  to  the  angle  ACH; 

therefore  the  points  A,  C,  G,  II  He  on  a  circle. 

(Prop.  21,  CoroU.) 
Therefore  the  rectangle  AB,  BG  is  equal  to  the  rectangle 
CB,  BH;  (Prop.  35.) 

that  is,  the  rectangle -4^,  EF  is  equal  to  the  rectangle  BC,  DE. 
Wherefore,  if  two  triangles  &c. 

*  In  two  triangles  which  are  equiangular  to  one  another,  two 
sides  are  said  to  correspond  when  they  arc  opposite  to  equal  angles. 


PTOLEMY'S   THEOREM.  257 

PROPOSITION    37  B. 

The  rectangle  contained  hy  the  diagonals  of  a  convex 
quadrilateral  inscribed  in  a  circle  is  equal  to  the  sum  of  the 
rectangles  contained  hy  pairs  of  opposite  sides*. 

Let  ABCD  be  a  quadrilateral  inscribed  in  a  circle  and 
AC,  BD  be  its  diagonals  : 

it  is  required  to  prove  that  the  rectangle  AC,  BD  is  equal  to 
the  sum  of  the  rectangles  AB,  CD  and  BC,  AD. 

Construction.  From  B  in  BA,  on  the  same  side  of  BA 
as  CD,  draw  BE  making  the  angle 
ABE  equal  to  the  angle  CBD,  and 
meeting  AC  in  E.  (I.  Prop.  23.) 
Proof.  Because  the  angle  BAC 
is  equal  to  the  angle -Si)  C,  (Prop.  21.) 
and  the  angle  ABE  is  equal  to  the 
angle  DBC,  (Constr.) 

therefore  the  triangles  ABE,  DBC 
are  equiangular  to  one  another ; 
(I.  Prop.  32.) 
therefore  the  rectangle  AB,  CD  is  equal  to  the  rectangle 
AE,  BD.  (Prop.  37  A.) 

Again,  because  the  angle  ABE  is  equal  to  the  angle  DBC, 

the  angle  ABD  is  equal  to  the  angle  EBC ; 
and  the  angle  BDA  is  equal  to  the  angle  BCA  (i.  e.  BCE ), 

(Prop.  21.) 

therefore  the  triangles  ABD,  EBC  are  equiangular  to  one 

another;  (I.  Prop.  32.) 

therefore  the  rectangle  AD,  BC  is  equal  to  the  rectangle 

EC,  BD; 
but  it  has  been  proved  that 

the  rectangle  AB,  CD  is  equal  to  the  rectangle  AE,  BD. 
Therefore  the  sum  of  the  rectangles  AB,  CD  and  A  D,  BC  is 
equal  to  the  sum  of  the  rectangles  AE,  BD  and  EC,  BD, 
that  is,  to  the  rectangle  AC,  BD.   (II.  Prop.  1.) 
Wherefore,  the  rectangle  contained  &c. 

*  Thi.<5  theorem  is  attributed  to  Ptolemy,  a  Greek  geometer  of 
Alexandria,  who  died  about  a.d.  160. 


258 


BOOK  III. 


ADDITIONAL   PEOPOSITION. 

If  a  straight  line  be  drawn  through  a  given  point  to  cut  a  given 
circle,  the  intersection  of  the  tangents  at  the  two  points  of  section 
always  lies  on  a  fixed  straight  line*. 

Let  FRS  be  any  straight  line  drawn  through  a  given  point  P  to 
cut  a  given  circle,  whose  centre  is  O,  in  E  and  S. 

Let  QR,  QS  be  the  tangents  at  R,  S. 

Draw  OQ  intersecting  RS  in  M\  and  draw  OP,  and  draw  QH  per- 
pendicular to  OP  or  OP  produced. 


Because  the  angle  at  31  is  a  right  angle  (Ex.  3,  page  217), 
and  the  angle  at  If  is  a  right  angle, 
the  points  P,  H,  M,  Q  lie  on  a  circle ; 

fig.  1  (Prop.  21,  Coroll.)  and  fig.  2  (Prop.  22,  CoroU.) 
therefore  the  rectangle  OH,  OP  is  equal  to  the  rectangle  OM,  OQ. 

(Prop.  36.) 
But  because  QS  is  a  tangent  at  S, 

the  angle  OSQ  is  a  right  angle,  (Prop.  18.) 

and  the  angle  at  31  is  a  right  angle, 
therefore  the  rectangle  031,  OQ  is  equal  to  the  square  on  OS; 

(I.  Prop.  47.) 

Therefore  the  rectangle  OP,  OH  is  equal  to  the  square  on  OS. 

But  OP  and  OS  are  both  constants, 

therefore  OH"  is  a  constant, 

and  the  point  Q  always  lies  on  a  fixed  straight  line, 

i.e.  the  line  drawn  through  the  fixed  point  H  at  right  angles  to  OP. 

*  This  line  is  called  the  polar  of  the  given  point,  and  the  point 
is  called  the  pole  of  the  line  with  respect  to  the  circle. 


POLES  AND  POLARS. 


259 


It  has  now  been  proved  that,  if  a  point  and  a  straight  line  be  such 
that  the  straight  line  joining  the  centre  of  a  circle  to  the  point  is  at 
right  angles  to  the  line,  and  the  rectangle  contained  by  the  distances 
of  the  point  and  the  line  from  the  centre  is  equal  to  the  square  on  the 
radius  of  the  circle,  the  point  is  the  pole  of  the  line,  and  the  line  the 
polar  of  the  point  with  respect  to  the  circle. 


In  the  diagram  0  is  the  centre  of  the  circle :  if  is  a  point  in  OP 
such  that  the  rectangle  OP,  OH  is  equal  to  the  square  on  the  radius, 
and  HQ  is  at  right  angles  to  OP. 

P  is  the  pole  of  HQ,  and  HQ  is  the  polar  of  P. 
It  will  be  observed  that, 

if  P  be  without  the  circle  (fig.  1),  the  polar  cuts  the  circle : 

if  P  be  on  the  circle  (fig.  2),  the  polar  is  the  tangent  to  the  circle, 

and  if  P  be  within  the  circle  (fig.  3),  the  polar  does  not  cut  the  circle. 


260 


BOOK  III. 


ADDITIONAL  PKOPOSITION. 

If  a  quadrilateral  he  inscribed  in  a  circle,  the  square  on  the  straight 
line  joining  the  points  of  intersection  of  opposite  sides  is  less  than  the 
sum  of  the  squares  on  the  straight  lines  joining  those  points  to  the 
centre  of  the  circle  by  tiolce  the  square  on  the  radius  of  the  circle. 

Let  A  BCD  be  a  quadrilateral  inscribed  in  a  circle  whose  centre  is 
O;  and  let  the  sides  AB,  CD  meet  in  Q  and  the  sides  AD,  BC  in  JR. 

Draw  QR  and  draw  AS  making  the  angle  MAS  equal  to  the  angle 
RQD  and  meeting  RQ  in  S.  (I.  Prop.  23.) 

(1) 


Because  in  the  triangles  RAS,  RQD, 

the  angle  RAS  is  equal  to  the  angle  RQD, 
the  angle  RSA  is  equal  to  the  angle  RDQ;  (I.  Prop.  32.) 
therefore  the  points  S,  A,  D,  Q  lie  on  &  circle. 

(Prop.  21  or  20,  Coroll.) 
Therefore  the  rectangle  RS,  RQ  is  equal  to  the  rectangle  RA,  RD. 

(Prop.  36.) 
Also  it  can  be  proved  that  the  points  R,  S,  A,  B  lie  on  a  circle. 
Therefore  the  rectangle  QS,  QR  is  equal  to  the  rectangle  QA,  QB. 

(Prop.  35  or  36.) 
Therefore  the  square  on  QR, 

in  figure  (1),  being  the  sum  of  the  rectangles  RS,  RQ  and  QS,  QR, 

is  equal  to  the  sum  of  the  rectangles  RA,  RD  and  QA,  QB; 

and  in  figure  (2),  being  the  difference  of  the  rectangles  RS,  RQ,  and 

QS,  QR,  is  equal  to  the  difference  of  the  rectangles  RA,  RD 

and  QA,  QB. 

Since  the  rectangle  RA,  RD  is  equal  to  the  difference  of  the  squares  on 

RO  and  the  radius,  (Prop.  36.) 


POLES  AND  FOLARS. 


261 


and  the  rectangle  QA,  QB  in  figure  (1)  is  equal  to  the  difference  of  the 
squares  on  QO  and  the  radius,  (Prop.  36.) 

and  in  figure  (2)  is  equal  to  the  difference  of  the  squares  on  the 
radius  and  QO;  (Prop.  35.) 

it  follows  that  in  both  cases 

the  square  on  QR  is  less  than  the  sum  of  the  squares  on  QO,  RO  by 
twice  the  square  on  the  radius. 


ADDITIONAL  PEOPOSITION. 

If  one  pair  of  opposite  sides  of  a  quadrilateral  inscribed  in  a  circle 
intersect  at  a  fixed  point,  the  other  pair  of  opposite  sides  intersect  on  a 
fixed  straight  line  *. 

Let  ABGD  be  a  quadrilateral  inscribed  in  a  circle,  whose  centre  is 
0 ;  and  let  the  sides  AB,  CD  meet  in  Q  and  AD,  BG  in  R. 


Because  Q  and  R  are  the  intersections  of  opposite  sides  of  a  quadri- 
lateral inscribed  in  the  circle, 
the  square  on  QR  is  less  than  the  squares  on  OQ,  OR  by  twice  the 
square  on  the  radius;  (Add.  Prop,  page  260.) 

therefore  the  difference  of  the  squares  on  QR,  OQ  is  equal  to  the  dif- 
ference of  the  square  on  OR  and  twice  the  square  on  the  radius, 
which  is  a  constant,  if  the  point  R  be  fixed. 
Therefore  the  locus  of  the  point  Q  is  a  straight  line. 

(Ex.  2,  page  125.) 

*  We  leave  to  the  student  as  an  exercise  the  proof  that  this  line  is 
the  polar  of  the  fixed  point. 


262  BOOK  III. 


ADDITIONAL  PKOPOSITION. 

If  one  point  He  on  the  polar  of  another  point,  the  second  point  lies 
on  the  polar  of  the  first  point. 

Let  P,  Q  be  two  points  such  that  Q  lies  on  the  polar  of  P, 
i.  e.  if  QH  be  drawn  perpendicular  to  OP, 
the  rectangle  OH,  OF  is  equal  to  the  square  on  the  radius. 

Construction.    Draw  P/r  perpendicular  to  0^.  (I.  Prop.  12.) 


Proof.    Because  the  angles  at  H  and  K  are  right  angles, 

Q,  K,  P,  H  lie  on  a  circle;       (Prop.  22,  Coroll.) 
therefore  the  rectangle  OQ,  OK  is  equal  to  the  rectangle  OH,  OP, 

(Prop.  36.) 

and  therefore  to  the  square  on  the  radius ; 

and  KP  is  at  right  angles  to  0K\ 

therefore  KP  is  the  polar  of  Q, 

or,  in  other  words,  P  lies  on  the  polar  of  Q. 


POLES  AND   POLARS.  263 


EXERCISES. 

1.  Prove  that  the  polar  of  a  point  without  a  circle  is  the  straight 
line  joining  the  points  of  contact  of  tangents  drawn  from  the  point  to 
the  circle. 

2.  If  0  be  the  centre  of  a  circle,  and  the  polar  of  a  point  P  cut 
PO  in  H,  and  any  straight  line  through  P  cut  the  circle  in  B  and  S, 
then  the  polar  bisects  the  angle  RHS. 

3.  If  a  straight  line  PQR  cut  a  circle  in  Q  and  R  and  cut  the 
polar  of  P  in  K,  and  if  31  be  the  middle  point  of  QR,  then  the  rect- 
angles PQ,  PR  and  PK,  PM  are  equal. 

4.  If  P,  Q,  R,  S  he  the  points  of  contact  of  the  sides  AB,  BC, 
CD,  DA  of  a  quadrilateral -4 BCZ)  with  an  inscribed  circle,  the  straight 
lines  AC,  BD,  PR,  QS  are  concurrent. 

5.  Shew  how  to  draw  two  tangents  to  a  given  circle  from  a  given 
external  point  by  means  of  straight  lines  only. 

6.  Shew  how  to  draw  a  tangent  to  a  given  circle  at  a  given  point 
on  it  by  means  of  straight  lines  only. 


264 


BOOK  III. 


ADDITIONAL  PKOPOSITION. 

The  locus  of  a  point  from  which  tangents  drawn  to  two  given  circles 
are  equal  is  a  straight  line*. 

Let  P  be  a  point  such  that 
PQ,  PR  tangents  drawn  to  two 
given  circles  are  equal. 

Find  the  centres  A,  B  of  the 

circles;  (Prop.  5.) 

draw  AB,  AP,  AQ,BP,  BR,  and 

draw  PH  perpendicular  to  AB. 

(I.  Prop.  12.) 

Because  PQ,  PR  are  tangents 

the  angles  at  Q  and  R  are  right  angles. 
Therefore  the  sum  of  the  squares  on  PQ,  AQ  is  equal  to  the  square 
on  AP,  (I.  Prop.  47.) 

and  the  sum  of  the  squares  on  PR,  BR  is  equal  to  the  square  on  BP; 
therefore  the  difference  of  the  squares  on  AQ,  BR  is  equal  to  the 
difference  of  the  squares  on  AP,  BP, 
But  because  the  angles  at  H  are  right  angles, 

the  difference  of  the  squares  on  AP,  BP 

is  equal  to  the  difference  of  the  squares  on  AH,  HB. 

Therefore  the  difference  of  the  squares  on  AH,  HB  is  equal  to  the 

difference  of  the  squares  on  A  Q,  BR,  which  is  a  constant ; 

therefore  If  is  a  fixed  point, 

and  the  straight  line  HP  on  which  P  lies  is  drawn  through  H  at 

right  angles  to  AB  the  line  of  the  centres,  and  is  therefore  a  fixed 

straight  line. 


*  This  line  is  called  the  Radical  Axis  of  the  two  circles.  This 
name  was  given  to  the  line  by  L.  Gaultier  de  Tours,  a  French  geo- 
meter.    See  Journal  de  Vecole  Polytechnique,  torn.  ix.  p.  139  (1813). 


RADICAL   AXIS.  265 


EXERCISES. 

1.  Prove  that  the  radical  axis  of  two  intersecting  circles  passes 
through  their  points  of  intersection. 

2.  What  is  the  radical  axis  of  two  circles  which  touch  each  other? 

3.  Prove  that  the  middle  points  of  the  four  common  tangents  of 
two  circles  external  to  each  other  lie  on  a  straight  line. 

4.  Prove  that  the  radical  axes  of  three  circles  taken  two  and  two 
together  meet  in  a  point*. 

5.  Shew  how  to  draw  the  radical  axis  of  two  circles  which  do  not 
meet. 

6.  Draw  a  circle  passing  through  a  given  point  and  cutting  two 
given  circles  so  that  its  chords  of  intersection  with  the  two  circles 
may  each  pass  through  given  points. 

7.  0  is  a  fixed  point  outside  a  given  circle:  find  a  straight  line 
such  that  each  of  the  tangents  drawn  from  any  point  P  in  that  line 
to  the  circle  shall  be  equal  to  PO. 

8.  Draw  a  straight  line  in  a  given  direction  so  that  chords  cut 
from  it  by  two  given  circles  may  be  equal. 

9.  Prove  that  the  difference  of  the  squares  of  the  tangents  from 
any  point  to  two  circles  is  equal  to  twice  the  rectangle  under  the  dis- 
tance between  their  centres  and  the  distance  of  the  point  from  their 
radical  axis. 

10.  Through  two  given  points  draw  a  circle  to  cut  a  given  circle 
in  such  a  way  that  the  angle  contained  in  the  segment  cut  off  the 
given  circle  may  be  equal  to  a  given  angle. 


*  This  point  is  called  the  Radical  Centre  of  the  three  circles. 


T.  E.  18 


266  BOOK  III. 

Definition.  Two  circles  or  other  curves,  ivhich  meet  at 
a  jmint,  are  said  to  meet  at  the  angle  at  which  their  tangents 
at  the  point  meet. 

Two  circles  or  other  curves  are  said  to  he  orthogonal  or 
to  cut  orthogonally  at  a  point,  when  they  intersect  at  right 
angles  at  the  point. 


ADDITIONAL  PEOPOSITION. 

If  the  square  on  the  distance  hetioeen  the  centres  of  tico  circles  he 
equal  to  the  sum  of  the  squares  on  the  radii,  the  circles  are  orthogonal. 

liQi  A,  B  be  the  centres  of  two  circles  GPD,  EPF,  which  intersect 
at  P,  and  are  such  that  the  square  on  ^B  is  equal  to  the  sura  of  the 
squares  on  AP,  BP. 


Because  the  square  on  AB  is  equal  to  the  sum  of  the  squares  on 

AP,  BP,        . 

the  angle  .^PJB  is  a  right  angle.  (I.  Prop.  48.) 

And  BP  is  a  radius  of  the  circle  EPF ; 

therefore  AP  touches  the  circle  EPF. 

Similarly  it  can  be  proved  that  BP  touches  the  circle  CPD ; 

therefore  the  circles  GPD,  EPF  are  orthogonal. 

CoBOLLABY.     The  radius  of  each  of  two  orthogonal  circles  drawn  to 
a  point  of  intersection  is  a  tangent  to  the  other  circle. 


ORTHOGONAL   CIRCLES.  267 


EXEBCISES. 

1.  A  circle,  which  passes  through  a  given  point  and  cuts  a  given 
circle  orthogonally,  passes  through  a  second  fixed  point. 

2.  Describe  a  circle  to  cut  a  given  circle  orthogonally  at  two 
given  points. 

3.  Describe  a  circle  through  two  given  points  to  cut  a  given  circle 
orthogonally. 

4.  Two  chords  AB,  BC  of  a  circle  ACBB,  of  which  AB  is  a 
diameter,  intersect  at  E:  a.  circle  described  round  CDE  will  cut  the 
circle  ACBB  at  right  angles. 

5.  Two  circles  cut  each  other  at  right  angles  in  A,  B;  P  is  any 
point  on  one  of  the  circles,  and  the  lines  PA,  PB  cut  the  other  circle 
in  Q,  22:  shew  that  QR  is  a  diameter. 

6.  The  internal  and  external  bisectors  of  the  vertical  angle  A  of 
the  triangle  ABC  meet  the  base  in  D  and  E  respectively.  Prove  that 
the  circles  described  about  the  triangles  ABD  and  ABE  cut  at  right 
angles,  as  also  do  those  described  about  the  triangles  ACD  and  ACE. 


18—2 


268 


BOOK  III. 


ADDITIONAL  PKOPOSITION. 


Every  circle^  which  cuts  two  given  circles  orthogonally^  has  its 
centre  on  the  radical  axis  of  the  given  circles,  and  if  it  cut  the  straight 
line  joining  their  centres,  it  cuts  it  in  two  fixed  points. 

Let  ^ ,  jB  be  the  centres  of  two  given  circles  and  let  P  be  the 
centre  of  a  circle,  which  cuts  the  given  circles  orthogonally  at  Q  and  E. 
Draw  AB,  PQ,  PR,  and  draw  PH  perpendicular  to  AB. 
Because    the    circles    cut    ortho- 
gonally at  Q, 

PQ  is  a  tangent  at  Q. 
(Add.  Prop,  page  266,  Coroll.) 
Similarly  it  can  be  proved  that 

PR  is  a  tangent  at  R. 
ButPQis  equal  to  Pi?; 
therefore  P  is  a  point  on  the  radical 
axis  of  the  given  circles, 
and  therefore  H  is  a  fixed  point.  (Add.  Prop,  page  264.) 

Next,  let  the  circle  whose  centre  is  P  cut  the  line  AB  in  M,  N. 
Because  the  circles  cut  orthogonally  at  Q, 

AQ  ia  a.  tangent  to  the  circle  QMNR  at  Q, 

and  therefore  the  square  on  ^Q  is  equal  to  the  rectangle  AM,  AN; 

but  because  PH  is  at  right  angles  to  MN^ 

MH  is  equal  to  NH;  (Prop.  4.) 

and  the  rectangle  AM,  AN  is  equal  to  the  difference  of  the 

squares  on  AH,  MH;  (II.  Prop.  6.) 

therefore  the  square  on  ^Q  is  equal  to  the  difference  of  the 
squares  on  AH,  MH. 
Now  the  lines  AQ,  AH  are  of  constant  length ; 

therefore  MH  (or  NH)  is  of  constant  length. 
Therefore  the  points  M  and  N  are  at  a  constant  distance  from  H, 
which  is  a  fixed  point ; 

therefore  the  points  M  and  N  are  fixed  points. 


ORTHOGONAL  CIRCLES.  269 


We  leave  it  to  the  student  as  an  exercise  to  prove  that : 

if  the  given  circles  be  external  to  each  other,  the  points  M  and  N  are 
real,  one  within  each  of  the  given  circles ; 

if  the  circles  touch  externally,  M  and  N  coincide  with  the  point  of 
contact ; 

if  the  circles  intersect,  the  circle,  whose  centre  is  P,  does  not  intersect 
the  line  ^J5  in  real  points ; 

if  one  circle  touch  the  other  internally,  the  points  are  again  real,  and 
they  coincide  with  the  point  of  contact ; 

if  one  circle  lie  wholly  within  the  other,  the  points  M  and  N  are 
both  real,  one  within  both  circles  and  the  other  without  both 
circles. 


EXERCISES. 

1.  Draw  a  circle  to  cut  three  given  circles  orthogonally. 

2.  Prove  that  every  pair  of  circles,  which  cut  two  given  circles 
orthogonally,  has  the  same  radical  axis. 

3.  Of  four  given  circles  three  have  their  centres  in  the  same 
straight  line,  and  the  fourth  cuts  the  other  three  orthogonally;  prove 
that  the  radical  axis  of  each  pair  of  the  three  circles  is  the  same. 

4.  ABCD  is  a  quadrilateral  inscribed  in  a  circle;  the  opposite 
sides  AB  and  DC  are  produced  to  meet  at  F',  and  the  opposite  sides 
BC  and  AD  at  E:  shew  that  the  circle  described  on  EF  as  diameter 
cuts  the  circle  ABCD  a,t  right  angles. 

5.  Find  a  point  such  that  its  polar  with  respect  to  each  of  two 
given  circles  is  the  same. 


270 


BOOK  III. 


ADDITIONAL  PKOPOSITION. 

The  middle  points  of  the  sides  of  a  triangle  and  the  feet  of  the 
perpendiculars  from  the  angular  points  on  the  opposite  sides  lie  on  a 
circle. 

Let  D,  jy,  F  be  the  middle  points  of  the  sides  BC,  CA,  AB  of  a 
triangle  ABC,  and  L,  31,  N  the  feet  of  the  perpendiculars  on  them 
from  A,  B,  C.  y. 

Draw  FL,  LB,  FD,  DE.  .         _   ^  '0( 


%. 


Then  because  ALB  is  a  right-angled  triangle,  and  F  is  the  middle 
point  oi  AB, 

FL  is  equal  to  FA  ;  (Ex.  7,  page  87.) 

therefore  the  angle  FLA  is  equal  to  the  angle  FAL.  (I.  Prop.  5.) 

Similarly  it  can  be  proved  that  the  angle  A  LE  is  equal  to  the 

angle  LAE ; 

therefore  the  angle  FLE  is  equal  to  the  angle  BAG. 

Again  because  FD  is  parallel  to  AG 

and  DE  to  BA,         (Add.  Prop,  page  101.) 
FAED  is  a  parallelogram, 
and  the  angle  FDE  is  equal  to  the  angle  BA  C;  (I.  Prop.  34.) 
therefore  the  angle  FLE  is  equal  to  the  angle  FDE. 

Therefore  L,  D,  E,  F  lie  on  a  circle.     (Prop.  21,  Coroll.) 

Similarly  it  can  be  proved  that  M,  D,  E,  F  lie  on  a  circle, 

and  that  N,  D,  E,  F  lie  on  a  circle. 

But  only  one  circle  can  be  described  through  the  three  points  D,  E,  F; 

therefore  these  three  circles  are  coincident. 

Therefore  the  six  points  L,  M,  N,  D,  E,  F  lie  on  a  circle. 


THE  NINE  POINT  CIRCLE. 


271 


ADDITIONAL  PEOPOSITION. 


The  circle  through  the  middle  points  of  the  sides  of  a  triangle  passes 
through  the  middle  points  of  the  straight  lines  joining  the  angular  points 
of  the  triangle  to  the  orthocentre. 

Let  D,  E,  F  be  the  middle  points  of  the  sides  BG,  CA,  AB  of  a 
triangle  ABC. 

Draw  AL,  BM,  GN  perpendicular  to  BG,  GA,  AB,  intersecting 


at  0. 

Bisect  AO,  BO,  GO  at  a,  b,  c. 


(Add.  Prop,  page  95.) 


A 

^i 

nJ- 

Z^ 

\ 

f!/ 

//A\ 

^\^ 

/k 

In  the  triangle  OBG,  D,  c,  h  are  the  middle  points  of  the  sides, 
and  L,  M,  N  are  the  feet  of  the  perpendiculars  from  the  vertices  on 
the  opposite  sides ; 

therefore  L,  D,  c,  31,  N,  h  lie  on  a  circle.    (Add.  Prop,  page  270.) 
Similarly  it  can  be  proved  that 

L,  c,  E,  21,  a,  N  lie  on  a  circle, 

and  that  L,  M,  a,  N,  F,  b  lie  on  a  circle. 

But  only  one  circle  can  be  described  through  the  three  points  L,M,N; 

therefore  these  circles  are  coincident. 
Therefore  the  nine  points  L,  D,  c,  E,  M,  a,  N,  F,  b  lie  on  a  circle*. 


*  This  circle  is  called  the  Nine  Point  Circle  of  the  triangle. 


272  BOOK  III. 


ADDITIONAL  PKOPOSITION. 

The  feet  of  the  perpendiculars  drawn  from  any  point  on  a  circle  to 
the  three  sides  of  a  triangle  inscribed  in  the  circle  lie  on  a  straight 
line*. 

Let  ABC  be  a  triangle  and  PL,  PM,  PN 
be  the  perpendiculars  from  a  point  P  on  the 
circle  ABC  to  BG,  CA,  AB. 
Draw  LN,  NM. 

Because  PLB,  PNB  are  right  angles, 
a  circle  can  be  described  about  PNLB ; 

(III.  Prop.  21,  Coroll.) 
therefore  the  angles  PNL,  PBL  are  supple- 
mentary.     (III.  Prop.  22.) 
But  the  angles  PAC,  PBG  are  supplementary;  (III.  Prop.  22.) 
therefore  the  angle  PNL  is  equal  to  the  angle  PAC. 
Again  because  PMA,  PNA  are  right  angles, 

a  circle  can  be  described  about  PNAM;    (III.  Prop.  22,  Coroll.) 
therefore  the  angle  PN3I  is  equal  to  the  angle  PAM. 
Therefore  the  sum  of  the  angles  PNL,  PNM  is  equal  to  the  sum 
of  the  angles  PAC,  PAM, 

that  is,  to  two  right  angles.  (I.  Prop.  13.) 

Therefore  LN,  NM  are  in  the  same  straight  line.    (I.  Prop.  14.) 


EXERCISES. 

1.  If  PL,  PM,  PN  be  the  perpendiculars  drawn  from  P  a  point 
on  the  circle  ABC  to  the  sides  BC,  CA,  AB  of  an  inscribed  triangle, 
and  straight  lines  PI,  Pm,  Pn  be  drawn  to  meet  the  sides  in  I,  in,  n 
such  that  the  angles  LPl,  MPni,  NPn  are  equal  and  measured  in  the 
same  sense,  then  I,  m,  n  are  collinear. 

2.  P  is  a  point  on  the  circle  circumscribing  the  triangle  ABC. 
The  pedal  line  of  P  cuts  AG  and  BC  in  31  and  L.  Y  is  the  foot  of 
the  perpendicular  from  P  on  the  pedal  line.  Prove  that  the  rectangles 
PY,  PC,  and  PL,  PM  are  equal. 


*  This  line  is  called  the  Pedal  Line. 

Its  discovery  is  attributed  to  Dr  Robert   Simson,   and  it  is  in 
consequence  also  called  Simson's  Line. 


MISCELLANEOUS   EXERCISES. 

1.  If  any  point  P  on  a  fixed  circle  be  joined  to  a  fixed  point  J, 
and  AQ  be  drawn  equal  to  AP  at  a  constant  inclination  PAQ  to  AP, 
the  locus  of  Q  is  an  equal  circle. 

2.  Draw  a  straight  line  from  one  circle  to  another,  to  be  equal 
and  parallel  to  a  given  straight  line. 

3.  Construct  a  parallelogram,  two  of  whose  angular  points  are 
given  and  the  other  two  lie  on  two  given  circles. 

Discuss  the  number  of  possible  solutions  of  the  problem  in  the 
different  cases  which  may  occur. 

4.  Find  the  locus  of  the  centre  of  a  circle  whose  circumference 
passes  through  two  given  points. 

5.  Shew  that  the  straight  lines  drawn  at  right  angles  to  the  sides 
of  a  rectilineal  figure  inscribed  in  a  circle  from  their  middle  points 
intersect  at  a  fixed  point. 

6.  Determine  the  centre  of  a  given  circle  by  means  of  a  ruler 
with  parallel  edges  whose  breadth  is  less  than  the  diameter  of  the 
circle. 

7.  A  circle  is  described  on  the  radius  of  another  circle  as  diameter. 
Prove  that  any  chord  of  the  greatest  circle  drawn  from  the  point  of 
contact  is  bisected  by  the  lesser  circle. 

8.  Two  circles  DFG,  GEO  whose  centres  are  A  and  B  intersect  at 
C\  through  G,  DGE  and  FGG  are  drawn  equally  inclined  to  AB: 
shew  that  DE  and  FG  are  equal. 

9.  AB  and  GD  are  two  chords  of  a  circle  cutting  at  a  point  E 
within  the  circle ;  AB  is  produced  to  H  so  that  BH  is  equal  to  BE. 
The  circles  AEG  and  AGH  cut  BG  in  K  and  L;  prove  that  B  is  the 
middle  point  of  KL. 

10.  Through  either  of  the  points  of  intersection  of  two  given 
circles  draw  the  greatest  possible  straight  line  terminated  both  ways 
by  the  two  circumferences. 

11.  Through  two  points  A,  B  on  the  same  diameter  of  a  circle  and 
equidistant  from  its  centre  two  parallel  straight  lines  AP,  BQ  are 
drawn  towards  the  same  parts,  meeting  the  circle  in  P  and  Q  :  shew 
that  PQ  is  perpendicular  to  AP  and  BQ. 


274  BOOK  HI. 

12.  A  is  a  fixed  point  in  the  circumference  of  a  circle  and^BC  an 
inscribed  triangle  such  that  the  sum  of  the  squares  on  AB,AG  is 
constant ;  shew  that  the  locus  of  the  middle  point  of  BG  is  a.  straight 
line. 

13.  From  one  of  the  points  of  intersection  of  two  circles  straight 
lines  are  drawn  equally  inclined  to  the  common  chord  of  the  circles  : 
prove  that  the  portions  of  these  lines,  intercepted  between  the  other 
points  in  which  they  meet  the  circumference's  of  the  circles,  are  equal. 

14.  Describe  a  circle  of  given  radius  to  touch  two  given  circles. 
Discuss  the  number  of  possible  solutions. 

15.  Describe  three  circles  to  have  their  centres  at  three  given 
points  and  to  touch  each  other  in  pairs. 

How  many  solutions  are  there  ? 

16.  Describe  a  circle  to  touch  two  given  concentric  circles  and  to 
pass  through  a  given  point. 

Discuss  the  number  of  solutions.     Is  a  solution  always  possible? 

17.  li  AB,  CD  be  two  equal  chords  of  a  circle,  one  of  the  pairs  of 
lines  AD,  BG  and  AG,  BD  are  equal,  and  the  other  pair  parallel. 

18.  If  two  equal  chords  AB,  GD  of  a  circle  intersect  at  E,  AE  is 
equal  to  GE  or  DE. 

19.  A  quadrilateral  is  described  about  a  circle :  shew  that  two  of 
its  sides  are  together  equal  to  the  other  two  sides. 

20.  Shew  that  every  parallelogram  described  about  a  circle  is  a 
rhombus. 

21.  If  the  tangents  at  two  points  where  a  straight  line  meets  two 
circles,  one  point  being  on  each  circle,  be  parallel,  the  other  pair  also 
are  parallel. 

22.  Two  straight  lines  ABD,  ACE  touch  a  circle  at  B  and  C;  if 
DE  be  joined,  DE  is  equal  to  BD  and  GE  together:  shew  that  DE 
touches  the  circle. 

23.  When  an  equilateral  polygon  of  an  even  number  of  sides  is 
described  about  a  circle,  the  alternate  angles  are  equal. 

24.  If  a  quadrilateral  be  described  about  a  circle,  the  angles  sub- 
tended at  the  centre  of  the  circle  by  any  two  opposite  sides  of  the 
figure  are  together  equal  to  two  right  angles. 

25.  Two  radii  of  a  circle  at  right  angles  to  each  other  when  pro- 
duced are  cut  by  a  straight  line  which  touches  the  circle :  shew  that 
the  tangents  drawn  from  the  points  of  section  are  parallel  to  each 
other. 

26.  If  the  straight  line  joining  the  centres  of  two  circles,  which 
are  external  to  one  another,  cut  them  in  the  points  A,  B,  G,  D,  the 
squares  on  the  common  tangents  to  the  two  circles  are  equal  to  the 
rectangles  BD,  AG,  and  BG,  AD. 

What  is  the  corresponding  theorem  for  two  intersecting  circles  ? 


MISCELLANEOUS  EXERCISES.  275 

27.  APB  is  an  arc  of  a  circle  less  than  a  semicircle ;  tangents  are 
drawn  at  A  and  B  and  at  any  intermediate  point  P;  shew  that  the 
sum  of  the  sides  of  the  triangle  formed  by  the  three  tangents  is 
invariable  for  all  positions  of  P. 

28.  Given  a  circle,  and  a  straight  line  in  its  plane ;  draw  a  tangent 
to  the  circle,  which  shall  make  a  given  angle  with  the  given  straight 
line. 

29.  Find  the  point  in  the  circumference  of  a  given  circle,  the  sum 
of  whose  distances  from  two  given  straight  lines  at  right  angles  to 
each  other,  which  do  not  cut  the  circle,  is  the  greatest  or  least 
possible. 

30.  A  straight  line  is  drawn  touching  two  circles :  shew  that  the 
chords  are  parallel  which  join  the  points  of  contact  and  the  points 
where  the  straight  line  through  the  centres  meets  the  circumferences. 

31.  From  the  centre  C  of  a  circle,  CA  is  drawn  perpendicular  to 
a  given  straight  line  AB,  which  does  not  meet  the  circle,  and  in  ^C  a 
point  P  is  taken,  such  that  AF  is  equal  to  the  length  of  the  tangent 
li'om  A  :  prove  that,  if  Q  be  any  point  in  AB,  QF  is  equal  to  the 
length  of  the  tangent  from  Q. 

32.  If  a  quadrilateral,  having  two  of  its  sides  parallel,  be 
described  about  a  circle,  a  straight  line  drawn  through  the  centre  of 
the  circle,  parallel  to  either  of  the  two  parallel  sides,  and  terminated 
by  the  other  two  sides,  is  equal  to  a  fourth  part  of  the  perimeter  of 
the  figure. 

83.  With  a  given  point  as  centre  describe  a  circle  to  cut  a  given 
circle  at  right  angles.  How  must  the  point  be  situated  that  this  may 
be  possible  ? 

34.  If  two  circles  touch  and  a  chord  be  drawn  through  the  point 
of  contact,  the  tangents  at  the  other  points  where  the  chord  meets  the 
circles  are  parallel. 

35.  From  a  given  point  A  without  a  circle  whose  centre  is  O  draw 
a  straight  line  cutting  the  circle  at  the  points  B  and  C,  so  that  the 
area  BOG  may  be  the  greatest  possible. 

36.  Two  circles  FAB,  QAB  cut  one  another  at  A  :  it  is  required 
to  draw  through  A  a  straight  line  FQ  so  that  PQ  may  be  equal  to 
a  given  straight  line. 

37.  Two  given  circles  touch  one  another  externally  in  the  point 
P,  and  are  touched  by  the  line  AB  in  the  points  A  and  B  respectively. 
Shew  that  the  circle  described  on  AB  as  diameter  passes  through  the 
point  P,  and  touches  the  line  joining  the  centres  of  the  two  given 
circles. 

38.  From  a  point  without  a  circle  draw  a  line  such  that  the  part 
of  it  included  within  the  circle  may  be  of  a  given  length  less  than  the 
diameter  of  the  circle. 


276  BOOK  III. 

39.  When  an  equilateral  polygon  is  described  about  a  circle  it 
must  necessarily  be  equiangular  if  the  number  of  sides  be  odd,  but 
not  otherwise. 

40.  One  circle  touches  another  internally  at  the  point  A :  it  is' 
required  to  draw  from  A  a  straight  line  such  that  the  part  of  it 
between  the  circles  may  be  equal  to  a  given  straight  line,  which  is  not 
greater  than  the  difference  between  the  diameters  of  the  circles. 

41.  If  a  hexagon  circumscribe  a  circle,  the  sums  of  its  alternate 
sides  are  equal. 

42.  If  a  polygon  of  an  even  number  of  sides  circumscribe  a  circle, 
the  sum  of  the  alternate  sides  is  equal  to  half  the  perimeter  of  the 
polygon. 

43.  Under  what  condition  is  it  possible  to  describe  a  circle  to 
touch  the  two  diagonals  and  two  opposite  sides  of  a  quadrilateral  ? 

44.  If  a  parallelogram  be  circumscribed  about  a  circle  with  centre 
0,  and  a  line  be  drawn  touching  the  circle  and  cutting  the  sides  of  the 
parallelogram,  or  the  sides  produced  in  A,  B,  C,  D,  prove  that  the 
angles  AOB,  BOG,  and  COD  are,  each  of  them,  equal  to  a  half  of 
one  or  other  of  the  angles  of  the  parallelogram. 

45.  If  two  equal  circles  be  placed  at  such  a  distance  apart  that 
the  tangent  drawn  to  either  of  them  from  the  centre  of  the  other  is 
equal  to  a  diameter,  they  will  have  a  common  tangent  equal  to  the 
radius. 

46.  Draw  a  straight  line  to  touch  one  given  circle  so  that  the 
part  of  it  contained  by  another  given  circle  shall  be  equal  to  a  given 
straight  line  not  greater  than  the  diameter  of  the  latter  circle. 

47.  AB  is  the  diameter  and  G  the  centre  of  a  semicircle :  shew 
that  0  the  centre  of  any  circle  inscribed  in  the  semicircle  is  equidis- 
tant from  G  and  from  the  tangent  to  the  semicircle  parallel  to  AB. 

48.  A  circle  is  drawn  to  touch  a  given  circle  and  a  given  straight 
line.  Shew  that  the  points  of  contact  are  always  in  the  same  straight 
line  with  one  or  other  of  two  fixed  points  in  the  circumference  of  the 
given  circle. 

49.  Describe  a  circle  to  touch  a  given  circle  and  a  given  straight 
line. 

How  many  solutions  are  there  ? 

50.  A  series  of  circles  is  described  passing  through  one  angular 
point  of  a  parallelogram  and  a  fixed  point  on  the  diagonal  through 
that  angular  point.  Shew  that  the  sum  of  the  squares  on  the  tan- 
gents from  the  extremities  of  the  other  diagonal  is  the  same  for  each 
circle  of  the  series. 

51.  A  quadrilateral  is  bounded  by  a  diameter  of  a  circle,  the 
tangents  at  its  extremities,  and  a  third  tangent :  shew  that  its  area  is 
equal  to  half  that  of  the  rectangle  contained  by  the  diameter  and  the 
side  opposite  to  it. 


MISCELLANEOUS  EXERCISES  277 

52.  If  the  centres  of  two  circles  which  touch  each  other  externally 
be  fixed,  the  external  common  tangents  of  those  circles  will  touch 
another  circle  of  which  the  straight  line  joining  the  fixed  centres  is 
the  diameter. 

53.  TP^  TQ  are  tangents  from  T  to  two  circles  which  meet  in  A 
and  PQ  meets  the  circles  in  P',  Q',  and  the  angles  PAP',  QAQ'  are 
equal ;  find  the  locus  of  T. 

54.  Describe  a  circle  of  given  radius  passing  through  a  given 
point  and  touching  a  given  straight  line. 

How  many  solutions  may  there  be? 

55.  Describe  a  circle  of  given  radius  touching  a  given  straight 
line  and  a  given  circle. 

How  many  solutions  may  there  be? 

56.  Given  one  angle  of  a  triangle,  the  side  opposite  it,  and  the 
sum  of  the  other  two  sides,  construct  the  triangle. 

57.  AOB,  COD  are  two  diameters  of  a  circle  whose  centre  is  0, 
and  they  are  mutually  perpendicular.  If  P  be  any  point  on  the  cir- 
cumference, shew  that  CP  and  DP  are  the  internal  and  external 
bisectors  of  the  angle  APB. 

68.  If  AB  and  CD  be  two  perpendicular  diameters  of  a  circle  and 
P  any  point  on  the  arc  A  GB,  shew  that  D  is  equally  distant  from 
PA,  PB. 

59.  If  two  circles  intersect  each  other,  prove  that  each  common 
tangent  subtends,  at  the  two  common  points,  angles  which  are  sup- 
plementary to  each  other. 

60.  From  one  of  the  points  of  intersection  of  two  equal  circles, 
each  of  which  passes  through  the  centre  of  the  other,  a  straight  line 
is  drawn  to  intersect  the  circles  in  two  other  points :  prove  that  these 
points  form  an  equilateral  triangle  with  the  other  point  of  intersec- 
tion of  the  two  circles. 

61.  A  series  of  circles  touch  a  fixjed  straight  line  at  a  fixed  point: 
shew  that  the  tangents  at  the  points  where  they  cut  a  parallel  fixed 
straight  line  all  touch  a  fixed  circle- 

62.  Two  points  P,  Q  are  taken  in  two  arcs  described  on  the  same 
straight  line  AB,  and  on  the  same  side  of  it;  the  angles  PAQ,  PBQ 
are  bisected  by  the  straight  lines  AR,  BR  meeting  at  R:  prove  that 
the  angle  ARB  is  constant. 

63.  APB  is  a  fixed  chord  passing  through  P  a  point  of  intersec- 
tion of  two  circles  AQP,  PBR;  and  QPR  is  any  other  chord  of  the 
circles  passing  through  P:  shew  that  AQ  and  RB  when  produced 
meet  at  a  constant  angle. 

64.  A,B,C,D  are  four  points  on  a  circle.    Prove  that  the  four 

f)oints  where  the  perpendiculars  from  any  point  O  to  the  straight 
ines  AB  and  CD  meet  AC  and  BD  lie  on  a  circle. 


278  BOOK  III. 

65.  Any  number  of  triangles  are  described  on  the  same  base  BG, 
and  on  the  same  side  of  it  having  their  vertical  angles  equal,  and 
perpendiculars,  intersecting  at  D,  are  drawn  from  B  and  G  on  the 
opposite  sides;  find  the  locus  of  D. 

66.  Let  0  and  G  be  any  fixed  points  on  the  circumference  of  a 
circle,  and  OA  any  chord;  then,  if  AG  be  joined  and  produced  to  B, 
so  that  OB  is  equal  to  OA,  the  locus  of  B  is  an  equal  circle. 

67.  If  ABG  be  a  triangle,  AD  and  BE  the  perpendiculars  from 
A  and  B  upon  BG  and  AC,  DF  and  EG  the  perpendiculars  from  D 
and  E  upon  AG  and  BG,  then  FG  is  parallel  to  AB. 

68.  The  four  circles  which  pass  through  the  middle  points  of  the 
sides  of  the  four  triangles  formed  by  two  sides  of  a  quadrilateral  and 
one  of  its  two  diagonals  intersect  in  a  point. 

69.  In  a  circle  two  chords  of  given  length  are  drawn  so  as  not 
to  intersect,  and  one  of  them  is  fixed  in  position;  the  opposite 
extremities  of  the  chords  are  joined  by  straight  lines  intersecting 
within  the  circle:  shew  that  the  locus  of  the  point  of  intersection 
will  be  a  portion  of  the  circumference  of  a  circle,  passing  through 
the  extremities  of  the  fixed  chord. 

70.  The  centre  C  of  a  circle  BPQ  lies  on  another  circle  APQ  of 
which  PBA  is  a  diameter.     Prove  that  PG  is  parallel  to  BQ. 

71.  Through  one  of  the  points  of  intersection  of  two  circles, 
centres  A  and  B,  a  chord  is  drawn  meeting  the  circles  at  P  and  Q 
respectively.  The  lines  PA,  QB  intersect  in  G.  Find  the  locus 
of  C. 

72.  At  each  extremity  of  the  base  of  a  triangle  a  straight  line 
is  drawn  making  with  the  base  an  angle  equal  to  half  the  sum  of  the 
two  base  angles;  prove  that  these  lines  will  meet  on  the  external 
bisector  of  the  vertical  angle. 

73.  In  the  figure  of  Euclid  i.  47  the  feet  of  the  perpendiculars 
drawn  from  the  centre  of  the  largest  square  upon  the  three  sides  of 
the  given  right  angled  triangle  are  collinear. 

74.  Three  points  A,  B,  G  are  taken  on  a  circle.  The  tangents  at 
B  and  G  meet  in  T.  If  from  T  a  straight  line  be  drawn  parallel 
to  AB,  it  meets  ^0  in  the  diameter  perpendicular  to  AB. 

75.  If  ABG  he  a  triangle  inscribed  in  a  circle,  PQ  a  diameter, 
and  perpendiculars  be  let  fall  from  P  on  the  two  sides  meeting  in  A, 
and  from  Q  on  those  meeting  in  B,  the  lines  joining  the  feet  of  the 
two  sets  of  perpendiculars  will  intersect  at  right  angles. 

76.  Through  any  point  0  on  the  side  BG  of  an  equilateral  tri- 
angle ABG,  OK,  OL  are  drawn  parallel  to  AB,  AG  respectively  to 
meet  AG,  AB  respectively  in  K  and  L:  the  circle  through  0,  K  and  L 
cuts  AB,  AG  again  in  P  and  Q.  Prove  that  OPQ  is  an  equilateral 
triangle. 


MISCELLANEOUS  EXERCISES.  279 

77.  Two  chords  AB,  CD  of  a  circle  intersect  in  E.  From  EB,  ED, 
produced  if  necessary,  parts  EF,  EG  are  cut  off  respectively  equal  to 
ED,  EB.    Prove  that  FG  is  parallel  to  CA. 

78.  If  ABC  be  an  equilateral  triangle  and  D  be  any  point  on  the 
circumference  of  the  circle  ABC,  then  one  of  the  three  distances  DA^ 
DB,  DC  is  equal  to  the  sum  of  the  other  two. 

79.  If  through  E  a  point  of  intersection  of  two  circles  AGE,  BDE 
two  straight  lines  AB  and  GD  be  drawn  terminated  by  the  circles, 
then  A  G  and  BD  cut  one  another  at  a  constant  angle. 

80.  If  AD,  BE,  GF be  the  perpendiculars  from  the  angles  A,B,G 
of  a  triangle  on  the  opposite  sides,  these  lines  bisect  the  angles  of  the 
triangle  DEF. 

81.  AB  is  a  common  chord  of  the  segments  AGB,  ADEB  of  two 
circles,  and  through  G  any  point  on  AGB  are  drawn  the  straight 
lines  AGE,  BGD:  prove  that  the  arc  DE  is  of  invariable  length. 

82.  Divide  a,  circle  into  two  parts  so  that  the  angle  contained  in 
one  arc  shall  be  equal  to  twice  the  angle  contained  in  the  other. 

83.  A  quadrilateral  is  inscribed  in  a  circle :  shew  that  the  sum 
of  the  angles  in  the  four  arcs  of  the  circle  exterior  to  the  quadrilateral 
is  equal  to  six  right  angles. 

84.  A,  B,  G,  D  are  four  points  taken  in  order  on  the  circum- 
ference of  a  circle ;  the  straight  lines  AB,  GD  produced  intersect  at  P, 
and  AD,  BG  at  Q:  shew  that  the  straight  lines  which  bisect  the 
angles  APC,  AQC  are  perpendicular  to  each  other. 

85.  A  quadrilateral  can  have  one  circle  inscribed  in  it  and  another 
circumscribed  about  it:  shew  that  the  straight  lines  joining  the 
opposite  points  of  contact  of  the  inscribed  circle  are  perpendicular  to 
each  other. 

86.  If  D,  E,  F  be  three  points  on  the  sides  BC,  GA,  AB  of  a 
triangle,  the  perimeter  of  the  triangle  DEF  is  least  when  D,  E,  F  are 
the  feet  of  the  perpendiculars  from  A,  B,  G  on  BG,  GA,  AB. 

87.  Shew  that  the  four  straight  lines  bisecting  the  angles  of  any 
quadrilateral  form  a  quadrilateral  which  can  be  inscribed  in  a  circle. 

88.  If  a  polygon  of  six  sides  be  inscribed  in  a  circle,  the  sum  of 
three  alternate  angles  is  equal  to  four  right  angles. 

89.  AB  is  a.  diameter  of  a  circle,  GD  a  chord  perpendicular  to  it. 
A  straight  line  through  A  cuts  the  circle  in  E,  and  GD  produced  in  F: 
prove  that  the  angles  AEO,  DEF  are  equal. 

90.  If  0  be  a  point  within  a  triangle  A  BG,  and  OD,  OE,  OF  be 
drawn  perpendicular  to  BG,  GA,  AB,  respectively,  the  angle  BOG  is 
equal  to  the  sum  of  the  angles  5yi  C,  £D^.  ..    - 


280  BOOK  III. 

91.  AOB,  GOB  are  two  chords  of  a  circle  which  intersect  within 
the  circle  at  0.  Through  the  point  A  a  straight  line  ^^  is  drawn  to 
meet  the  tangent  to  the  circle  at  the  point  C,  so  that  the  angle  AFC 
is  equal  to  the  angle  BOC.    Prove  that  OF  is  parallel  to  EG. 

92.  The  sums  of  the  alternate  angles  of  any  polygon  of  an  even 
number  of  sides  inscribed  in  a  circle  are  equal. 

93.  Through  a  point  C  in  the  circumference  of  a  circle  two 
straight  lines  AOB^  DGE  are  drawn  cutting  the  circle  at  B  and  E: 
shew  that  the  straight  line  which  bisects  the  angles  ACE,  DGB  meets 
the  circle  at  a  point  equidistant  from  B  and  E. 

94.  AOB,  COB  are  two  diameters  of  the  circle  ABGB,  at  right 
angles  to  each  other.  Equal  lengths  OE,  OF,  are  measured  off  along 
OA,  OB  respectively.  Shew  that  BF  produced  cuts  BE  at  right 
angles,  and  that  these  two  straight  lines,  when  produced,  intercept 
between  them  one  fourth  of  the  circumference  of  the  circle. 

95.  If  the  extremities  of  the  chord  of  a  circle  slide  upon  two 
straight  lines,  which  intersect  on  the  circumference,  every  point  in 
the  circumference  will  trace  out  a  straight  line. 

96.  Any  point  P  is  taken  on  a  given  arc  of  a  circle  described  on  a 
line  AB,  and  perpendiculars  AG  and  BH  are  dropped  on  BP  and  AP 
respectively ;  prove  that  GH  touches  a  fixed  circle. 

97.  OA  and  OB  are  two  fixed  radii  of  a  given  circle  at  right 
angles  to  each  other  and  POQ  is  a  variable  diameter;  prove  that  the 
locus  of  the  intersection  of  PA  and  QB  is  a  circle  equal  to  the  given 
one. 

98.  Describe  a  circle  touching  a  given  straight  line  at  a  given 
point,  such  that  the  tangents  drawn  to  it  from  two  given  points  in 
the  straight  line  may  be  parallel. 

99.  Describe  a  circle  with  a  given  radius  touching  a  given  straight 
line,  such  that  the  tangents  drawn  to  it  from  two  given  points  in  the 
straight  line  may  be  parallel. 

100.  Two  circles  intersect  at  the  points  A  and  B,  from  which  are 
drawn  chords  to  a  point  G  in  one  of  the  circumferences,  and  these 
chords,  produced  if  necessary,  cut  the  other  circumference  at  B  and 
E :  shew  that  the  straight  line  BE  cuts  at  right  angles  that  diameter 
of  the  circle  ABC  which  passes  through  G. 

101.  If  squares  be  described  externally  on  the  sides  and  the  hypo- 
tenuse of  a  right-angled  triangle,  the  straight  line  joining  the  inter- 
section of  the  diagonals  of  the  latter  square  with  the  right  angle  is 
perpendicular  to  the  straight  line  joining  the  intersections  of  the 
diagonals  of  the  two  former. 

102.  G  is  the  centre  of  a  given  circle,  GA  a  straight  line  less  than 
the  radius ;  find  the  point  of  the  circumference  at  which  GA  subtends 
the  greatest  angle. 


MISCELLANEOUS  EXERGLSES.  281 

103.  If  two  chords  of  a  circle  meet  at  a  right  angle  within  or  with- 
out a  circle,  the  squares  on  their  segments  are  together  equal  to  the 
square  on  the  diameter. 

104.  On  a  given  base  BC  a  triangle  ABC  is  described  such  that 
AG  is  equal  to  the  perpendicular  from  B  upon  AG.  Find  the  locus  of 
the  vertex  A. 

105.  Draw  from  a  given  point  in  the  circumference  of  a  circle,  a 
chord  which  shall  be  bisected  by  its  point  of  intersection  with  a  given 
chord  of  the  circle. 

106.  Through  any  fixed  point  of  a  chord  of  a  circle  other  chords 
are  drawn;  shew  that  the  straight  lines  from  the  middle  point  of  the 
first  chord  to  the  middle  points  of  the  others  will  meet  them  all  at 
the  same  angle. 

107.  The  two  angles  at  the  base  of  a  triangle  are  bisected  by  two 
straight  lines  on  which  perpendiculars  are  drawn  from  the  vertex: 
shew  that  the  straight  line  which  passes  through  the  feet  of  these 
perpendiculars  will  be  parallel  to  the  base  and  will  bisect  the  sides. 

108.  A  point  P  is  taken  on  a  circle  of  which  AB  is  a  fixed  chord; 
a  parallelogram  is  described  of  which  AB  and  AP  are  adjacent  sides: 
find  the  locus  of  the  middle  points  of  the  diagonals  of  the  paral- 
lelogram. 

Find  the  maximum  length  of  the  diagonal  drawn  through  A. 

109.  ^  is  a  fixed  point  on  a  circle  whose  centre  is  0  and  BOB  is 
a  diameter.  The  tangents  at  A  and  D  meet  in  L.  Shew  that  the 
locus  of  the  intersection  of  LB  with  the  perpendicular  from  A  on  OB 
is  a  circle. 

110.  ABG  is  a  triangle  inscribed  in  a  circle ;  AB,  BE,  perpendicu- 
lars io  BG  and  AG,  meet  in  0:  AK  \b  2,  diameter  of  the  circle:  prove 
that  GK  is  equal  to  BO, 

111.  A  straight  line  touches  a  circle  at  the  point  P  and  QR  is  a 
chord  of  a  second  circle  parallel  to  this  tangent ;  PQ ,  PR  cut  the  first 
circle  in  S,  T,  and  the  second  circle  in  U,  F;  prove  that  ST,  UV  are 
parallel  to  each  other. 

112.  A,  B,  G  are  three  points  on  a  circle,  the  bisectors  of  the  angle 
BAG  and  the  angle  between  BA  produced  and  AG  meet  BG  and  BG 
produced  in  E  and  F  respectively ;  shew  that  the  tangent  at  A  bisects 
EF. 

113.  A  circle  is  described  passing  through  the  right  angle  of  a 
right-angled  triangle,  and  touching  the  hypotenuse  at  its  middle 
point :  prove  that  the  arc  of  this  circle,  cut  off  by  either  side  of  the 
triangle,  is  divided  at  the  middle  point  of  the  hypotenuse  into  two 
parts  one  of  which  is  double  of  the  other. 

114.  If  through  the  point  of  contact  P  of  two  circles  two  straight 
lines  PQq,  PRr  be  drawn  to  meet  the  circles  in  Q,  R,  and  q,  r  respec- 
tively ;  then  the  triangles  PQR,  Pqr  are  equiangular. 

T.  E.  19 


282  BOOK  III. 

115.  The  angle  between  two  chords  of  a  circle,  which  intersect 
within  the  circle,  is  equal  to  half  the  sum  of  the  angles  subtended  at 
the  centre  by  the  intercepted  arcs. 

116.  ABCD  is  a  parallelogram  and  any  straight  line  is  drawn 
cutting  the  sides  AB,  CD  in  P,  Q  respectively.  The  circle  passing 
through  A,  P,  and  Q  cuts  AD  in  S,  and  ^C  in  T.  Shew  that  the 
circles  circumscribing  the  triangle  DSQ,  CTQ  touch  one  another  at 
the  point  Q. 

117.  If  one  circle  touch  another  internally,  any  chord  of  the 
greater  circle  which  touches  the  less  is  divided  at  the  point  of  its  con- 
tact into  segments  which  subtend  equal  angles  at  the  point  of  contact 
of  the  two  circles. 

118.  Two  circles  are  drawn  touching  a  circle,  whose  centre  is  C,  in 
P  and  Q  respectively  and  intersecting  in  PQ  produced,  and  again  in 
B.    Prove  that  the  angles  CBP,  CRQ  are  equal. 

119.  The  angle  between  two  chords  of  a  circle,  which  intersect 
when  produced  without  the  circle,  is  equal  to  half  the  difference  of  the 
angles  subtended  at  the  centre  by  the  intercepted  arcs. 

120.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle, 
and  the  length  of  the  straight  line  drawn  from  the  vertex  to  the 
middle  point  of  the  base. 

121.  ^  is  a  given  point:  it  is  required  to  draw  from  A  two  straight 
lines  which  shall  contain  a  given  angle  and  intercept  on  a  given 
straight  line  a  part  of  given  length. 

122.  A  and  B  are  two  points,  XY  a  given  straight  line  of  un- 
limited length,  not  passing  through  A  or  B ;  find  the  point  (or  points) 
in  XY  at  which  the  straight  line  AB  subtends  an  angle  equal  to  a 
given  angle. 

Also  state  when  this  problem  is  impossible. 

123.  The  chords  of  two  intersecting  circles  which  are  bisected  at 
any  point  of  the  common  chord  are  equal. 

124.  Find  a  point  in  the  tangent  to  a  circle  at  one  end  of  a  given 
diameter,  such  that  when  a  straight  line  is  drawn  from  this  point  to 
the  other  extremity  of  the  diameter,  the  rectangle  contained  by  the 
part  of  it  without  the  circle  and  the  part  within  the  circle  may  be 
equal  to  a  given  square  not  greater  than  that  on  the  diameter. 

125.  If  perpendiculars  be  drawn  from  the  extremities  of  the 
diameter  of  a  circle  upon  any  chord  or  any  chord  produced,  the  rect- 
angle under  the  perpendiculars  is  equal  to  that  under  the  segments 
between  the  feet  of  the  perpendiculars  and  either  extremity  of  the 
chord. 

126.  Two  circles  intersect  and  any  straight  line  ACBD  cuts  them 
in  .4 ,  JB  and  (7,  D  respectively.  If  JB  be  a  point  on  the  line  such  that 
the  rectangles  contained  by  AC,  BE  and  J5D,  CE  are  equal,  the  locus 
of  jE  is  a  straight  line. 


MISCELLANEOUS  EXERCISES.  283 

127.  ABB  is  an  isosceles  triangle  having  the  side  AB  equal  to  the 
side  BD ;  ^C  is  drawn  at  right  angles  to  ^i?  to  meet  BD  produced  in 
<7,  and  the  bisector  of  the  angle  B  meets  AG  in  0.  Shew  that  the 
square  on  AB  is  equal  to  the  difference  of  the  rectangles  CB,  BD  and 
CA,  AO. 

128.  Through  a  given  point  without  a  circle  draw,  when  possible, 
a  straight  line  cutting  the  circle  so  that  the  part  within  the  circle 
may  be  equal  to  the  part  without  the  circle. 

What  condition  is  necessary  in  order  that  a  real  solution  may  be 
possible? 

129.  Two  equal  circles  have  their  centres  at  A  and  B:  0  is  a  fixed 
point  outside  those  circles.  A  is  the  centre  of  a  third  circle  whose 
radius  is  equal  to  OB :  prove  that  a  right-angled  triangle  can  be  con- 
structed having  its  sides  equal  to  the  tangents  from  0  to  the  three 
circles. 

130.  Prove  that,  if  a  straight  line  drawn  parallel  to  the  side  BC  of 
a  triangle  ABC  cut  the  sides  AB,  AC  in  D,  E  respectively,  the  rect- 
angle contained  by  AB^  AE  is  equal  to  the  rectangle  contained  by 
AG,  AD. 

131.  Construct  a  triangle,  having  given  the  base,  the  vertical 
angle,  and  the  length  of  the  straight  line  drawn  from  the  vertex  to  the 
base  bisecting  the  vertical  angle. 

132.  Through  two  given  points  on  the  circumference  of  a  given 
circle  draw  the  two  parallel  chords  of  the  circle  which  shall  contain 
the  greatest  rectangle. 

133.  A  straight  line  PAQ  is  drawn  through  A  one  of  the  points 
of  intersection  of  two  given  circles  to  meet  the  circles  again  in  P  and 
Q:  find  the  line  which  makes  the  rectangle  AP,  AQ  a  maximum. 

134.  Produce  a  given  straight  line  so  that  the  rectangle  contained 
by  the  whole  line  thus  produced,  and  the  part  produced,  shall  be 
equal  to  the  square  on  another  given  line. 

135.  Two  circles  intersect  in  0,  and  through  0  a  straight  line 
ORS  is  drawn  cutting  the  circles  again  in  J^  and  S.  SO  is  produced 
to  P,  so  that  the  rectangle  OP,  RS  is  constant.  Shew  that  the  locus 
of  P  is  a  straight  line. 

136.  If  from  the  foot  of  the  perpendicular  from  each  of  the 
angular  points  of  a  triangle  on  the  opposite  side  a  perpendicular  be 
drawn  to  each  of  the  other  sides,  the  feet  of  the  six  perpendiculars  so 
drawn  lie  on  a  circle. 

137.  AB,  GD  are  chords  of  a  circle  intersecting  at  0,  and  AG,  DB 
meet  at  P.  If  circles  be  described  about  the  triangles  AOG,  BOD, 
the  angles  between  their  tangents  at  0  will  be  equal  to  the  angles  at 
P,  and  their  other  common  point  will  lie  on  OP. 

19—2 


284  BOOK  III. 

138.  If  ABC  be  a  triangle,  D,  E,F  the  feet  of  the  perpendiculars 
from  A,  B,  G  on  the  opposite  sides,  0  their  point  of  intersection, 
the  rectangles  DO,  DA,  and  DE,  DF  are  equal. 

139.  A  straight  line  AD  is  drawn  bisecting  the  angle  ^1  of  a 
triangle  ABC  and  meeting  the  side  BG  in  D.  Find  a  point  E  in  BQ 
produced  either  way  such  that  the  square  on  ED  may  be  equal  to 
the  rectangle  contained  by  EB,  EG. 

140.  If  D,  D';  E,  E' ;  F,  F'  be  pairs  of  points  on  the  sides  BG, 
GA,  AB  of  a  triangle  respectively,  such  that  D,  D',  E,  E'  are  con- 
cyclic,  E,  E',  F,  F'  are  concyclic,  F,  F',  D,  D'  are  concyclic,  then 
D,  D',  E,  E'y  F,  F'  are  concyclic. 

141.  In  the  straight  line  PQ  a  point  R  is  taken,  and  circles  are 
described  on  PR,  RQ  as  diameters ;  shew  how  to  draw  a  line  through 
P  such  that  the  chords  intercepted  by  the  two  circles  may  be  equal. 

142.  If  M  be  the  middle  point  of  PQ,  where  P  and  Q  are  points 
without  a  given  circle,  the  sum  of  the  squares  on  the  tangents  to  the 
circle  from  P  and  Q  is  equal  to  twice  the  sum  of  the  square  on  the 
tangent  from  31  and  the  square  on  PM. 

143.  A  circle  FDG  touches  another  circle  BDE  in  D  and  a  chord 
^P  of  the  latter  in  F:  prove  that  the  rectangle  FA,  FB  is  equal  to 
the  rectangle  contained  by  GE  and  the  diameter  of  FDG,  where  GE  is 
drawn  perpendicular  to  AB  at  its  middle  point  G  and  on  the  same 
side  of  it  as  the  circle  FDG. 

144.  Given  the  base  and  the  vertical  angle  of  a  triangle,  prove 
that  the  locus  of  the  centre  of  the  nine-point  circle  is  a  circle. 

145.  If  a  circle  be  circumscribed  to  a  triangle,  the  middle  point 
of  the  base  is  equally  distant  from  the  orthocentre  and  the  point 
diametrically  opposite  the  vertex.  Also  these  three  points  are  in  the 
same  straight  line. 


BOOK  IV. 


DEFINITIONS. 

Definition  1. 

A  figure  of  five     sides  is  called  a    pentagon, 
one  of  six      sides  is  called  a    hexagon, 
one  of  eight  sides  is  called  an  octagon, 
one  of  ten     sides  is  called  a    decagon, 
one  of  twelve  sides  is  called  a  dodecagon*. 

Definition  2.  When  each  of  the  angular  points  of  one 
rectilineal  figure  lies  on  one  of  the  sides  of  a  second  recti- 
lineal figure^  and  each  of  the  sides  of  the  second  figure  passes 
through  one  of  the  angular  points  of  the  first  figure,  the  first 
figure  is  said  to  he  inscribed  in  the  second  figure,  and  the 
second  figure  is  said  to  he  described  about  the  first  figure. 


*  Derived  from  irhre  "five,"  ?^  "six,"  oKTiii  "eight,"  Mkol  "ten, 
ddodcKa  "twelve,"  respectively,  and  ywuia  "an  angle." 


286  BOOK  IV 


PROPOSITION  1. 

To  draw  a  chord  of  a  given  circle  equal  to  a  given  straight 
line. 

Let  ABC  be  the  given  circle,  and  D  the  given  straight 
line  : 

it  is  required  to  draw  a  chord  of  the  circle  ABG  equal 
toi>. 

Construction.     Take  any  point  A  on  the  circle  ABC, 
and  from  A  draw  AE  equal  to  D.  (I.  Prop.  2.) 

If  E  lie  on  the  circle  ABC,  what  is  required  is  done,  for  in 
the  circle  ABG  the  chord  AE  is  drawn  equal  to  B. 
But  if  E  do  not  lie  on  the  circle  ABG^ 
with  A  as  centre  and  ^^  as  radius  describe  the  circle  ECF 
cutting  the  circle  ABG  at  F. 

Draw  AF. 
^i^  is  a  chord  drawn  as  required. 


Proof.     Because  A  is  the  centre  of  the  circle  EGF, 
AFi^  equal  to  AE. 
But  AE  is  equal  to  D.  (Constr.) 

Therefore  ^i^is  equal  to  i>, 
and  it  is  a  chord  of  the  circle  ABG. 
Wherefore,  a  chord  A  F  of  the  given  circle  ABG  has  been 
drawn  equal  to  the  given  straight  line  D. 


PROPOSITION  1.  287 

It  is  clear  that  it  is  not  possible  to  draw  a  chord  of  a  given  circle 
to  be  equal  to  a  given  straight  line,  if  the  given  line  be  greater  than 
the  diameter  of  the  circle  (III.  Prop.  8,  Part  1) ;  and  further  that,  if 
a  solution  be  possible,  in  general  two  chords  can  be  drawn  from  a 
given  point  equal  to  the  given  line. 

In  the  diagram,  if  the  two  circles  intersect  in  C,  the  chord  AG 
also  is  equal  to  the  given  line. 


EXEKCISES. 

1.     In  a  given  circle  draw  a  chord  parallel  to  one  given  straight 
line  and  equal  to  another. 

.2.  On  a  given  circle  find  a  point  such  that,  if  chords  be  drawn  to 
it  from  the  extremities  of  a  given  chord,  their  sum  shall  be  equal  to  a 
given  straight  line. 

How  many  solutions  are  there  in  the  different  cases  which  may 
occur  ? 


288  BOOK  IV. 

PROPOSITION  2. 

To  inscribe  in  a  given  circle  a  triangle  equiangular  to  a 
given  triangle. 

Let  ABC   be    the  given    circle,    and   DEF  the   given 
triangle : 

it  is  required  to  inscribe  in  the  circle  ABC  a  triangle  equi- 
angular to  the  triangle  DEF. 

Construction.     Take  any  point  A  on  the  circle, 
and  through  A  draw  the  chord  ^^  to  cut  off  the  arc  AGB 

containing  an  angle  equal  to  the  angle  DFE, 
and  through  A  draw  the  chord  ^C  to  cut  off  the  arc  ABC 
containing  an  angle  equal  to  the  angle  DEF*. 

(III.  Prop.  34.) 
Draw  BC  : 
the  triangle  ABC  is  inscribed  as  required. 


Proof.     Because  the  arc  ACB  contains  an  angle  equal 
to  the  angle  DFE,  (Const r.) 

and  the  angle  ACB  is  contained  by  the  arc  ACB, 

the  angle  ACB  is  equal  to  the  angle  DFE. 
Similarly  it  can  be  proved  that 

the  angle  ABC  is  equal  to  the  angle  DEF. 
And  because  the  sum  of  three  angles  of  a  triangle  is  equal 
to  two  right  angles,  (I.  Prop.  32.) 

and  the  angles  ACB,  ABC  are  equal  to  the  angles  DFE^ 
DEF  respectively, 

*  It  must  be  noticed  that  the  arcs  AGB,  ABC  are  measured  in 
opposite  directions  along  the  circumference  from  the  point  A. 


PROPOSITION  2.  289 

the  remaining  angle  BAG  of  the  triangle  ABC  is  equal 
to  the  remaining  angle  EDF  of  the  triangle  DEF ; 

therefore  the  triangle  ABC  is  equiangular  to  the  triangle 
DEF. 
Wherefore,  a  triangle  ABC  equiangular  to  the  triangle 

DEF  has  been  inscribed  in  the  given  circle  ABC. 

Since  the  arc  ABC  may  be  measured  in  either  direction  along  the 
circumference  from  A,  we  see  that  two  triangles  equiangular  to  a 
given  triangle  can  be  inscribed  in  a  given  circle  so  as  to  have  a  vertical 
angle  equal  to  a  given  angle  of  the  triangle  at  a  given  point  on  the 
circle,  and  that  six  triangles  equiangular  to  a  given  triangle  can  be 
inscribed  in  a  given  circle,  so  as  to  have  one  of  its  vertical  angles  at 
the  given  point  on  the  circle. 


EXEKCISES. 

1.  Prove  that  all  triangles  inscribed  in  the  same  circle  equi- 
angular to  each  other  are  equal  in  all  respects. 

2.  The  altitude  of  an  equilateral  triangle  is  equal  to  a  side  of  an 
equilateral  triangle  inscribed  in  a  circle  described  on  one  of  the  sides 
of  the  original  triangle  as  diameter. 

3.  ABC,  A'B'C  are  two  triangles  equiangular  to  each  other  in- 
scribed in  a  circle  AA'BB'GC.  The  pairs  of  sides  BC,  B'C ;  GA,  C'A'; 
AB,  A'B'  intersect  in  a,  b,  c  respectively. 

Prove  that  the  triangle  abc  is  equiangular  to  the  triangle  ABC. 


290  BOOK  IV. 

PROPOSITION  3. 

To  describe  about  a  given  circle  a  triangle  equiangular  to 
a  given  triangle. 

Let  ABC    be   the   given    circle    and  DEF  the   given 
triangle : 

it  is  required  to  describe  about  the  circle  ABC  a  triangle 
equiangular  to  the  triangle  DEF. 
Construction.     Find  the  centre  G  of  the  circle  ABCj 

(III.  Prop.  5.) 
and  draw  any  diameter  HGA  meeting  the  circle  in  A. 
At  G  in   GH  draw  the  straight  lines  GB^  GC  on  opposite 
sides  of  GH  making  the  angles  BGH^  CGH  equal  to  the 
angles  EFD,  DEF,  (I.  Prop.  23.) 

meeting  the  circle  in  B,  C. 
Through  A,  B,  C  draw  MAN,  NBL,  LCM  at  right  angles 
to  GA,  GB,  GC  respectively:  (I.  Prop.  11.) 

the  triangle  LMN  is  a  triangle  described  as  required. 


Proof.     Because  the  sum  of  the  angles  of  the  quadri- 
lateral GBNA  is  equal  to  four  right  angles, 

(I.  Prop.  32,  Coroll.) 
and  two  of  the  angles  GAN,  GBN  are  right  angles, 
the  sum  of  the  angles  AGB,  ANB  is  equal  to  two  right 

angles. 
But  the  sum  of  the  angles  AGB,  IIGB  is  equal  to  two 

right  angles;  (I.  Prop.  13.) 

therefore  the  sum  of  the  angles  AGB,  ANB  is  equal  to  the 
sum  of  the  angles  AGB,  IIGB) 

therefore  the  angle  ANB  is  equal  to  the  angle  HGB, 
that  is,  to  the  angle  EFD. 


PROPOSITION  3.  291 

Similarly  it  can  be  proved  that 

the  angle  LMN  i^  equal  to  the  angle  DEF; 
therefore  the  remaining  angle  NLM  of  the  triangle  LMN 
is  equal  to  the  remaining  angle  EDF  of  the  triangle  DEF. 

(I.  Prop.  32.) 
Therefore    the    triangle   LMN  is    equiangular   to   the 
triangle  DEF. 

Again,  because  MN  is  drawn  through  A  a  point   on   the 
circle  ABC  at  right  angles  to  the  radius  AG^ 

MN  touches  the  circle.      (III.  Prop.  16.) 

Similarly  it  can  be  proved  that  NL^  LM  touch  the  circle. 

Therefore  the  triangle  LMN  is  described  about  the  circle 

ABC. 

Wherefore,  a  tricmgle  LMN  equiangular  to  the  given 

triangle  DEF  has  been  described  about  the  given  circle  ABC. 


EXERCISES. 

1.  Prove  that  all  triangles  described  about  the  same  circle  equi- 
angular to  each  other  are  equal  in  all  respects. 

2.  Describe  a  triangle  about  a  given  circle  to  have  its  sides 
parallel  to  the  sides  of  a  given  triangle. 

How  many  solutions  are  there? 

3.  The  angles  of  the  triangle  formed  by  joining  the  points  of  con- 
tact of  the  inscribed  circle  of  a  triangle  with  the  sides  are  equal  to  the 
halves  of  the  supplements  of  the  corresponding  angles  of  the  original 
triangle. 

4.  If  ABC,  A'B'C  be  two  equal  triangles  described  about  a  circle 
in  the  same  sense  and  the  pairs  of  sides  J5C,J5'C";  GA,  G'A' ;  AB,  A'B' 
meet  in  a,h,c  respectively,  a,  &,  c  are  equidistant  from  the  centre  of 
the  circle. 


292  BOOK  IV. 

PROPOSITION    4. 

To  inscribe  a  circle  in  a  given  triangle. 

Let  A  BG  be  the  given  triangle  : 
it  is  required  to  inscribe  a  circle  in  tlie  triangle  ABC. 

Construction.     Bisect  two  of  the  angles  ABC,  BCA  of 

the  triangle  ABC  by  BD,  CD  meeting  at  D,      (I.  Prop.  9.) 

and  from  D  draw  DE^  Di\  DG  perpendicular  to  BC,  CA, 

AB  respectively.  (I.  Prop.  12.) 

With  B  as  centre  and  DE^  DF,  or  DG  as  radius  describe 

a  circle : 

it  will  be  a  circle  described  as  required. 


Proof.     Because  in  the  triangles  DEB^  DGB., 
the  angle  DBE  is  equal  to  the  angle  DBG,     (Constr.) 
and  the  angle  DEB  is  equal  to  the  angle  DGB, 

(I.  Prop.  10  B.) 
and  the  side  BD  is  common, 
the  triangles  are  equal  in  all  respects; 

(I.  Prop.  26,  Part  2.) 
therefore  DE  is  equal  to  DG. 
Similarly  it  can  be  proved  that  DE  is  equal  to  DF. 
Therefore  the  three  straight  lines  DE,  DF^  DG  are  equal 
to  one  another,  and  the  circle  described  with  D  as  centre, 
and  DE,  DF,  or  DG  as  radius  passes  through  the  ex- 
tremities of  the  other  two ; 
and  touches  the  straight  lines  BC,  CA,  AB,  because  the 
angles  at  the  points  E,  F,  G  are  right  angles,  and  the 


PROPOSITION  4.  293 

straight  line  drawn  through  a  point  on  a  circle  at  right 

angles  to  the  radius  touches  the  circle.       (III.  Prop.  16.) 
Therefore  the  straight  lines  AB,  B'C,  CA  do  each  of  them 

touch  the  circle,  and  therefore  the  circle  is  inscribed  in 

the  triangle  ABC. 
Wherefore,  the  circle  EFG  has  been  imcrihed  in  the  given 
triangle  ABC. 


EXERCISES. 

1.  The  base  of  a  triauglc  is  fixed,  and  the  vertex  describes  a 
circle  passing  through  the  extremities  of  the  base:  find  the  locus  of 
the  centre  of  the  inscribed  circle. 

2.  If  a  polygon  be  described  about  a  circle,  the  bisectors  of  all 
its  angles  meet  in  a  common  point. 

3.  Describe  a  circle  to  touch  a  given  circle  and  two  given  tan- 
gents to  the  circle. 

4.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle 
and  the  radius  of  the  inscribed  circle. 

5.  Find  the  centre  of  a  circle  cutting  off  three  equal  chords  from 
the  sides  of  a  triangle. 

6.  The  triangle  whose  vertices  are  the  three  points  of  contact  of 
the  inscribed  circle  with  the  sides  of  a  triangle,  is  always  acute- 
angled. 


294 


BOOK  IV. 


It  can  be  proved  in  the  same  manner  as  in  Proposition  4 
that,  if  the  angles  at  B  and  C  of  the  triangle  be  bisected 
externally  by  BJ)^ ,  CD^ ,  meeting  at  i>, ,  and  perpendiculars, 
i)j^j,  D^F^,  Dfi^  be  drawn,  the  circle  described  with  D^  as 
centre  and  either  of  the  three  lines  i^,^,,  -^i^j,  ^fii  ^^ 
radius  will  touch  the  three  sides  of  the  triangle.  Such  a 
circle  satisfies  the  definition  (III.  Def.  10)  of  an  inscribed 
circle. 

The  circles  are  however  generally  distinguished  thus, 
the  circle  IJFG,  which  lies  wholly  within  the  triangle 
ABC,  is  called  the  inscribed  circle,  whereas  the  circle 
E^Ffi^  is  called  an  escribed  circle,  and  is  said  to  be 
escribed  beyond  the  side  BO,  to  distinguish  it  from  the 
two  other  circles  which  can,  in  a  similar  manner,  be  escribed 
beyond  CA  and  beyond  AB  respectively. 


PROPOSITION  L  295 


EXERCISES. 

1.  Prove  that  the  radius  of  the  inscribed  circle  of  a  triangle  is 
less  than  the  radius  of  any  one  of  the  escribed  circles. 

2.  Prove  that  the  greatest  of  the  escribed  circles  of  a  triangle  is 
that  which  is  escribed  beyond  the  greatest  side,  and  the  least,  beyond 
the  least  side. 

3.  If  the  centres  of  the  escribed  circles  of  a  triangle  be  joined,  and 
the  points  of  contact  of  the  inscribed  circle  be  joined,  the  two  tri- 
angles so  formed  are  equiangular  to  each  other. 

4.  A  circle  touches  the  side  BC  of  a  triangle  ABC  and  the  other 
two  sides  produced :  shew  that  the  distance  between  the  points  of  con- 
tact of  the  side  BC  with  this  circle  and  with  the  inscribed  circle,  is 
equal  to  the  difference  between  the  sides  AB  and  AG. 

5.  Construct  a  triangle,  having  given  its  base,  one  of  the  angles 
at  the  base,  and  the  distance  between  the  centre  of  the  inscribed  circle 
and  the  centre  of  the  circle  touching  the  base  and  the  sides  produced. 

6.  Prove  that,  ii  A,B  he  two  fixed  points  on  a  circle  and  P  a 
variable  point,  the  locus  of  the  centre  of  each  of  the  escribed  circles 
of  the  triangle  APB  is  a  circle. 

7.  The  centre  of  the  inscribed  circle  of  a  triangle  is  the  ortho- 
centre  of  the  triangle  formed  by  the  centres  of  the  escribed  circles. 


296 


BOOK  IV. 


PROPOSITION   5. 

2^0  describe  a  circle  about  a  given  triangle. 

Let  ABC  be  the  given  triangle  : 
it  is  required  to  describe  a  circle  about  the  triangle  ABG. 

Construction.     Bisect  two  of  the  sides  AB^  AG  oi  the 
triangle  ABG,  at  B,  E,  (I.  Prop.  10.) 

and  draw  DF^  EF  at  right  angles  to  AB,  AG  meeting  at  F. 

(I.  Prop.  12.) 

Draw  FA  J  and  with  F  as  centre  and  FA  as  radius  describe 

a  circle: 

this  is  a  circle  described  as  required. 

F  must  lie  either  in  BG  (fig.  2)  or  not  in  BG  (figs.  1  and  3). 

If  F  do  not  lie  in  BG,  draw  FB,  FG. 


Proof.     Because  in  the  triangles  FDA,  FDB, 

AD  \^  equal  to  BD,  (Constr.) 

and  DF  is  equal  to  DF, 
and  the  angle  ADF  is  equal  to  the  angle  BDF,  (Constr.) 
the  triangles  are  equal  in  all  respects;     (I.  Prop.  4.) 
therefore  FA  is  equal  to  FB. 
Similarly  it  can  be  proved  that  FA  is  equal  to  FG. 
Therefore  the  circle  described  with  F  as  centre  and  FA  as 
radius  passes  through  the  points  B  and  G,  and  is  de- 
scribed about  the  triangle  ABG. 
"Wherefore,    a  circle  ABG  has  been  described  about  the 
given  triangle  ABG* 


PROPOSITION  5.  297 

The  construction  of  Proposition  5  shews  that  only  one 
circle  can  be  described  about  a  given  triangle,  a  theorem 
which  has  already  been  established  otherwise. 

(III.  Prop.  9,  CorolL  2.) 

The  circle  ABC  is  often  spoken  of  as  the  circumscribed 
circle  of  the  triangle  ABC. 

Propositions  4  and  5  solve  problems  of  the  same  nature ;  each 
shews  how  to  describe  a  circle  to  satisfy  three  given  conditions.  The 
problem  of  Proposition  4  to  describe  a  circle  to  touch  three  given 
straight  lines,  admits  of  4  solutions;  Proposition  5,  to  describe  a  circle 
to  pass  through  three  given  points,  admits  of  but  a  single  solution. 

A  circle  can  be  described  to  satisfy  three  (and  not  more  than 
three)  independent  conditions,  but  it  will  be  found  that  the  solution 
is  not  always  unique :  if  the  problem  be  one  which  can  be  solved  by 
geometrical  methods,  the  number  of  solutions  will  be  found  to  be  1  or 
2or4  =  2x2or8  =  2x2x2or  some  higher  power  of  2. 

The  number  2  occurs  in  one  of  its  powers  from  the  fact  that  at 
each  step  of  the  solution  where  choice  is  possible,  the  choice  lies 
between  the  two  intersections  of  a  circle  and  a  straight  line  or  the 
two  intersections  of  two  circles. 

If  it  be  required  to  describe  a  circle  to  touch  four  or  more  given 
straight  lines,  or  to  pass  through  four  or  more  given  points,  relations 
of  some  kind  must  exist  between  the  positions  of  the  lines  or  of  the 
points  in  order  that  a  solution  may  be  possible. 


EXEECISES. 

1.  Inscribe  in  an  equilateral  triangle  another  equilateral  triangle 
having  each  side  equal  to  a  given  straight  line. 

2.  Shew  how  to  cut  off  the  corners  of  an  equilateral  triangle,  so 
as  to  leave  a  regular  hexagon. 

3.  The  sides  AB,  AG  oi  a.  triangle  are  produced  and  the  exterior 
angles  are  bisected  by  straight  Hues  meeting  in  0:  if  a  circle  be 
described  about  the  triangle  BOG,  its  centre  will  be  on  the  circle 
described  about  the  triangle  ABG. 


T.  E.  20 


298  BOOK  IV. 

PROPOSITION   6. 

To  inscribe  a  square  in  a  given  circle. 

Let  ABCD  be  the  given  circle : 
it  is  required  to  inscribe  a  square  in  the  circle  ABCD. 

CpNSTRUCTiON.     Find  the  centre  E  of  the  circle  ABCD, 

(III.  Prop.  5.) 
and  draw  two  diameters  AEC,  BED  at  right  angles  to  one 
another.  (I.  Prop.  11.) 

Draw  ^5,  BG,  CD,  DA: 
the  quadrilateral  ABCD  is  a  square  inscribed  as  required. 


Proof.     Because  the  angle  BEC  is  double  of  the  angle 
BAG, 

and  the  angle  AED  is  double  of  the  angle  ACD, 

(III.  Prop.  20.) 
and  the  angle  BEC  is  equal  to  the  angle  AED, 

(I.  Prop.  15.) 

therefore  the  angle  BAG  is  equal  to  the  angle  AGD. 

And  because  AC  meeting  AB,   CD  makes  the   alternate 

angles  BA  C,  A  CD  equal, 

AB,  CD  are  parallel.  (I.  Prop.  27.) 

Similarly  it  can  be  proved  that  AD,  BG  are  parallel. 

Therefore  the  quadrilateral  ABCD  is  a  parallelogram. 

Again,  because  ABC  is  an  angle  in  a  semicircle  ABC, 

the  angle  ABC  is  a  right  angle.   (III.  Prop.  31.) 
Therefore  the  parallelogram  ABCD  is  a  rectangle. 

(I.  Def.  19.) 


PROPOSITION  6.  299 

Again,  because  in  the  triangles  AEB,  CEB^ 
AE  is  equal  to  CE, 
BE  to  BE, 
and  the  angle  AEB  to  the  angle  GEB, 
the  triangles  are  equal  in  all  respects;  (I.  Prop.  4.) 

therefore  BA  is  equal  to  BC. 
Therefore  the  rectangle  ABCD  is  a  square. 

(I.  Def.  20.) 
Wherefore,  a  square  ABCD  has  been  inscribed  iJi  the 
given  circle  ABCD. 


EXERCISES. 

1.  Inscribe  a  regular  octagon  in  a  given  circle. 

2.  Shew  how  to  cut  off  the  comers  of  a  square  so  as  to  leave  a 
regular  octagon. 

3.  Inscribe  in  a  given  square,  a  square  to  have  its  sides  equal 
to  a  given  straight  line. 


20—2 


300 


BOOK  IV. 


PROPOSITION  7. 

To  describe  a  square  about  a  given  circle. 

Let  ABCD  be  the  given  circle  : 

it  is  required  to  describe  a  square  about  it. 
Construction.     Find  E  the  centre  of  the  circle  ABCD^ 

(III.  Prop.  5.) 

and  draw  two  diameters  AEC,  BED  at  right  angles  to  one 

another.  (I.  Prop.  11.) 

Draw  GAF,  HGK  parallel  to  BD,  and  GBH,  FDK  parallel 

to  AG: 
the  quadrilateral  PGHK  is  a  square  described  as  required. 

Q  A  F 


B 


r 

V 

H 


K 


Proof.     Because  GF,  HK  are  each  parallel  to  BD, 

GF,  HK  are  parallel  to  each  other.     (I.  Prop.  30.) 
Similarly  it  can  be  proved  that 

Gil  J  FK  are  parallel  to  each  other ; 
therefore  FGHK  is  a  parallelogram. 
Again,  because  GAEB  is  a  parallelogram,  (Constr.) 

the  angle  AGB  is  equal  to  the  angle  AEB, 

(I.  Prop.  34.) 
which  is  a  right  angle;  (Constr.) 

therefore  the  parallelogram  FGHK  is  a  rectangle. 

(I.  Def.  19.) 
Again,  because  GBDF  is  a  parallelogram, 

GF  is  equal  to  BD,  a  diameter  of  the  circle. 
Similarly  it  can  be  proved  that 

GH  is  equal  to  AG,  a  diameter  of  the  circle; 
therefore  GF  is  equal  to  GH. 
Therefore  the  rectangle  FGHK  is  a  square.  (I.  Def.  20.) 


PROPOSITION  7.  301 

Again,  because  A£)  intersects  the  parallel  lines  GF,  BD, 
the  angle  GAB  is  equal  to  the  alternate  angle  AED^ 

(I.  Prop.  29.) 

which  is  a  right  angle ;  (Constr.) 

therefore  6^^^  touches  the  circle.   (III.  Prop.  16.) 

Similarly  it  can  be  proved  that  GBH^  HCK,  KDF  touch 

tlie  circle. 

Wherefore,  a  square  FGHK  has  been  described  about  the 
given  circle  A  BCD. 


EXEBCISES. 

1.  Describe  a  regular  octagon  about  a  given  circle. 

2.  Prove  that  the  area  of  a  circumscribed  square  of  a  circle  is 
double  that  of  an  inscribed  square. 

3.  If  two  circles  be  such  that  the  same  square  can  be  inscribed 
in  one  and  described  about  the  other,  the  circles  must  be  concentric. 

Is  any  other  condition  necessary  ? 

4.  If  a  parallelogram  admit  of  a  circle  being  inscribed  in  it  and 
another  circle  being  described  about  it,  the  parallelogram  must  be  a 
square. 


302 


BOOK  IV. 


PROPOSITION   8. 

To  iyiscrihe  a  circle  in  a  given  square. 

Let  A  BCD  be  the  given  square  : 

it  is  required  to  inscribe  a  circle  in  it. 

Construction.     Draw  AC,  BD  intersecting  in  E. 

From  E  draw  EF,  EG  perpendicular  to  J) A,  AB  two 

of  the  sides  of  the  square,  (I.  Prop.  12.) 

and  with  E  as  centre  and  EF  or  EG  as  radius  describe 

a  circle : 

it  is  a  circle  inscribed  as  required. 


Proof.     Because  CD  is  equal  to  AD, 

(1.  Prop.  34,  Coroll.  1.) 
the  angle  CAD  is  equal  to  the  angle  ^  C/>.  (I.  Prop.  5.) 

And  because  AG  meets  the  parallels  AB,  DC, 
the  angle  BA  C  is  equal  to  the  alternate  angle  A  CD  ; 

(I.  Prop.  29.) 

therefore  the  angle  BAG  is  equal  to  the  angle  DAG. 

And  because  in  the  triangles  GAE,  FAE, 

the  angle  GAE  is  equal  to  the  angle  FAE, 

and  the  angle  AGE  to  the  angle  AFE, 

and  AE  equal  to  AE, 
the  triangles  are  equal  in  all  respects  ; 

(I.  Prop.  26,  Part  2.) 

therefore  EG  is  equal  to  EF, 

and  therefore  the  circle  described  with  E  as  centre  and 

EF  or  EG  as  radius  passes  through  the  extremity  of  the 

other,  and  touches  the  two  sides  DAy  A  B.  (III.  Prop.  16.) 


PROPOSITION  8.  303 

Similarly  it  can   be  proved  that  this  circle  touches  each 
of  the  sides  ^(7,  CX): 

it  is  therefore  inscribed  in  the  square  ABCD. 
Wherefore,  a  circle  FGH  has  been  inscribed  in  the  given 
square  ABCD. 


EXEECISES. 


1.  Prove  that  a  circle  can  be  inscribed  in  any  rhombus. 

2.  Two  opposite  sides  of  a  convex  quadrilateral  are  together 
equal  to  the  other  two.  Shew  that  a  circle  can  be  inscribed  in  the 
quadrilateral. 

3.  AD,  BE  are  common  tangents  to  two  circles  ABC,  DEC, 
that  touch  each  other ;  shew  that  a  circle  may  be  inscribed  in  the 
quadrilateral  ABED,  and  a  circle  may  be  described  about  it. 


304 


BOOK  IV. 


PROPOSITION   9. 
To  describe  a  circle  about  a  given  square. 

Let  ABCD  be  the  given  square: 

it  is  required  to  describe  a  circle  about  it. 
Construction.     Draw  AC,  BD  intersecting  at  E; 
and  with  E  as  centre  and  EA^  EB^  EC  or  ED  as  radius 
describe  a  circle : 
it  is  a  circle  described  as  required. 


Proof.     Because  in  the  triangles  BAC,  DAC, 

BA  is  equal  to  EA,  (I.  Prop.  34,  Coroll.  1.) 
and  BC  to  EC, 
and  AC  to  AC, 
the  triangles  are  equal  in  all  respects;  (I.  Prop.  8.) 
therefore  the  angle  BAC  is  equal  to  the  angle  BAC, 

or  the  angle  BAC  is  half  of  the  angle  BAD. 
Similarly  it  can  be  proved  that 

the  angle  ABD  is  half  of  the  angle  ABC. 
But  the  angle  BAD  is  equal  to  the  angle  ABC', 

(I.  Prop.  29,  Coroll.  and  I.  Prop.  10  B.) 
therefore  the  angle  BAE  is  equal  to  the  angle  ABE, 

Therefore  BE  is  equal  to  AE.         (I.  Prop.  6.) 
Similarly  it  can  be  proved  that  CE  and  DE  are  each  of 

them  equal  to  AE  or  BE. 
Therefore  the  circle  described  with  E  as  centre  and  one  of 
four  lines  EA,  EB,  EC,  or  ED  as  radius  passes  through 
the  extremities  of  the  other  three,  and  is  described  about 
the  square  ABCD. 
Wherefore,  a  circle  ABCD  has  been  described  about  the 
given  square  ABCD. 


PROPOSITION  9.  305 


EXEECISES. 

1.  A  point  is  taken  without  a  square  such  that  the  angles  sub- 
tended at  it  by  three  sides  of  the  square  are  equal :  shew  that  the 
locus  of  the  point  is  the  circumference  of  the  circle  circumscribing 
the  square. 

2.  Find  the  locus  of  a  point  at  which  two  given  sides  of  a  square 
subtend  equal  angles. 

3.  If  a  quadrilateral  be  capable  of  having  a  quadrilateral  of 
minimum  perimeter  inscribed  in  it,  it  must  admit  of  a  circle  being 
inscribed  in  it. 

4.  ABCD  is  a  quadrilateral  inscribed  in  a  circle,  and  its 
diagonals,  AC,  BD  intersect  at  right  angles  in  E;  K,  L,  M,  N  are  the 
feet  of  the  perpendiculars  from  E  on  the  sides  of  the  quadrilateral. 
Shew  that  KLMN  can  have  circles  inscribed  in  it  and  described 
about  it. 


306  BOOK  IV. 

PROPOSITION    10. 

To  construct  a  triangle  having  each  of  two  angles  double 
of  the  third  angle. 

Construction'.     Take  any  straight  line  AB, 
and  divide  it  at  (7,  so  that  the  rectangle  AB,  BC  may  be 
equal  to  the  square  on  AC.     (II.  Prop.  11.) 
With  centre  ^  and  radius  AB  describe  the  circle  BDE, 
and  draw  a  chord  BD  equal  to  AG,         (Prop.  1.) 
and  draw  DA : 
the  triangle  ABD  is  a  triangle  constructed  as  required. 
Draw  DC,  and  about  the  triangle  ACD  describe  the  circle 
ACD.  (Prop.  5.) 


Peoof.     Because  the  rectangle  AB,  BC  is  equal  to  the 
square  on  AC,  and  AC  is  equal  to  BD, 

the  rectangle  A  B,  BC  is  equal  to  the  square  on  BD. 

And  because  from  the  point  B  without  the  circle  ACD, 

BCA  is  drawn  cutting  it  in  C  and  A,  and  BD  is  drawn  " 

meeting  it  in  D, 

and  the  rectangle  AB,  BC  is  equal  to  the  square  on  BD, 

therefore^/)  touches  the  circle  ACD.     (III.  Prop.  37.) 

And  because  BD  touches  the  circle  ACD,  and  DC  is,  drawn 

from  the  point  of  contact  D, 
the  angle  BDC  is  equal  to  the  angle  DAC.    (III.  Prop.  32.) 

To  each  of  these  add  the  angle  CDA ; 
then  the  whole  angle  BDA  is  equal  to  the  sum  of  the  angles 

CDA,  DAC. 
But  the  angle  BCD  is  equal  to  the  sum  of  the  angles  CDA, 
DAG;  (I.  Prop.  32.) 

therefore  the  angle  BDA  is  equal  to  the  angle  BCD. 


PROPOSITIOF  10.  307 

But  because  AD  is  equal  to  AB, 
the  angle  ABD  is  equal  to  the  angle  ADB\    (I.  Prop.  5.) 
therefore  the  angle  DBA  is  equal  to  the  angle  BCD. 
And  because  the  angle  DBO  is  equal  to  the  angle  BCD, 
DC  is  equal  to  DB.  (I.  Prop.  6.) 

But  DB  is  equal  to  CA',  (Constr.) 

therefore  CA  is  equal  to  CD; 
therefore  the  angle  CD  A  is  equal  to  the  angle  CAD. 

(I.  Prop.  5.) 
Therefore  the  sum  of  the  angles  CA  D,  CD  A  is  double  of  the 

angle  CAD. 

But   the  angle  BCD  is    equal  to   the  sum  of  the  angles 

CAD,  CDA',  (I.  Prop.  32.) 

therefore  the  angle  BCD  is  double  of  the  angle  CAD. 

And  the  angle  BCD  lias  been  proved  equal  to  each  of  the 

angles  BDA,  DBA ; 

therefore  each  of  the  angles  BDA,  DBA  is  double  of  the 

angle  BAD. 

Wherefore,  a  triangle  ABD  has  been  constructed  havirtg 

each  of  two  angles  ABD,  ADB  double  of  the  third  angle 

BAD. 

It  will  be  observed  that  the  smaller  angle  of  the  triangle  constructed 
in  this  proposition  is  equal  to  a  fifth  of  two  right  angles. 


EXERCISES. 

1.  Describe  an  isosceles  triangle  having  each  of  the  angles  at 
the  base  one  third  of  the  vertical  angle. 

2.  Divide  a  right  angle  into  five  equal  parts. 

3.  In  the  figure  of  Proposition  10,  if  the  two  circles  cut  again  at 
E,  then  BE  is  equal  to  DC. 

4.  In  the  figure  of  Proposition  10,  the  circle  AC  J)  is  equal  to  the 
circle  described  about  the  triangle  ABD. 

5.  In  the  figure  of  Proposition  10,  if  AF  be  the  diameter  of  the 
smaller  circle,  DF  is  equal  to  a  radius  of  the  circle  which  circum- 
scribes the  triangle  BCD. 

6.  If  in  the  figure  of  Proposition  10,  the  circles  meet  again  in 
E,  then  GE  is  parallel  to  BD. 


308  BOOK  IV. 

PROPOSITION    11. 

To  inscribe  a  regular  pentagon  in  a  given  circle. 

Let  ABODE  be  the  given  circle  : 
it  is  required  to  inscribe  a  regular  pentagon  in  the  circle 
ABODE. 
Construction.  Construct  a  triangle  FGH  having  each 
of  the  angles  at  G^  H  double  of  the  angle  at  F;  (Prop.  10.) 
in  the  circle  ABODE,  inscribe  the  triangle  AOD^  equi- 
angular to  the  triangle  FGH,  so  that  the  angles  OAD,  ADC, 
DO  A  may  be  equal  to  the  angles  GFH,  FHG,  EOF  re- 
spectively. (Prop.  2.) 
Bisect  the  angles  AOD,  ADO  by  the  straight  lines  OE,  DB, 

(I.  Prop.  9.) 
and  draw  OB,  BA,  AE,  ED : 
then  ABODE  is  a  pentagon  inscribed  as  required. 


Proof.     Because  each   of   the   angles  AOD,  ADO  is 

double  of  the  angle  OAD,  (Constr.) 

and  they  are  bisected  by  OE,  DB, 

the  five  angles  ADB,  BDO,  OAD,  DOE,  EOA  are  equal. 

But  equal  angles  at  the  circumference  of  a  circle  stand  on 

equal  arcs;     (III.  Prop.  26,  Coroll.) 

therefore  the  five  arcs  AB,  BO,  OD,  DE,  EA  are  equal. 

And  the  chords,  by  which  equal  arcs  are  subtended,  are  equal ; 

(III.  Prop.  29,  Coroll.) 
therefore  the  five  straight  lines  AB,  BO,  OD,  DE,  EA  are 

equal ; 
therefore  the  pentagon  ABODE  is  equilateral. 
Again,  the  arc  ED  is  equal  to  the  arc  BA ; 
to  each  of  these  add  the  arc  AE; 


PROPOSITION  11.  309 

then  the  whole  arc  AED  is  equal  to  the  whole  arc  BAE. 

And  the  angle  AED  is  contained  by  the  arc  AED,  and 

the  angle  BAE  by  the  arc  BAE ; 

therefore  the  angle  AED  is  equal  to  the  angle  BAE. 

(III.  Prop.  27,  CoroU.) 
Similarly  it  can  be  proved  that  each  of  the  angles  ABC, 
BCD,  CDE  is  equal  to  the  angle  AED  or  BAE; 
therefore  the  pentagon  ABODE  is  equiangular. 
Therefore  ABODE  is  a  regular  pentagon. 
"Wherefore,  a  regular  pentagon  ABODE  has  been  inscribed 
in  the  given  circle  ABODE. 

The  following  is  a  complete  Geometrical  construction  for  inscribing 
a  regular  decagon  or  pentagon  in  a  given  circle. 

Find  0  the  centre. 

Draw  two  diameters  AOC,  BOB  at 
right  angles  to  one  another. 

Bisect  OB  in  E. 

Draw  AE  and  cut  off  EF  equal  to 
OE. 

Place  round  the  circle  ten  chords 
equal  to  AF. 

These  chords  will  be  the  sides  of 
a  regular  decagon.      Draw  the  chords 
joining  five  alternate  vertices  of  the  decagon  ;  they  will  be  the  sides  of 
a  regular  pentagon. 

We  leave  the  proof  of  this  as  an  exercise  for  the  student. 

EXERCISES. 

1.  A  regular  pentagon  is  inscribed  in  a  circle,  and  alternate 
angular  points  are  joined  by  straight  lines.  Prove  that  these  lines 
will  form  by  their  intersections  a  regular  pentagon. 

2.  If  ABCBE  be  a  regular  pentagon,  AC  EBB  is  a  regular  star 
shaped  pentagon,  each  of  whose  angles  is  equal  to  two-fifths  of  a  right 
angle. 

3.  ABCBE  is  a  regular  pentagon;  draw  ^C7  and  BB,  and  let  BB 
meet  AG  sX  F\  shew  that  AC  is  equal  to  the  sum  of  AB  and  BF. 

4.  li  AB,  BC,  and  CB  be  sides  of  a  regular  pentagon,  the  circle 
which  touches  AB  and  CB  at  B  and  C  passes  through  the  centre  of 
the  circle  inscribed  in  the  pentagon. 


310 


BOOK  IV. 


PROPOSITION    12. 

To  describe  a  regular  pentagon  about  a  given  circle. 

Let  ABCDE  be  the  given  circle  : 
it  is  required  to  describe  a   regular  pentagon   about  the 
circle  ABCDE. 
Construction.    Let  A^  B,C,  D,  E  be  the  angular  points 
of  a  regular  pentagon  inscribed  in  the  circle  ;       (Prop.  11.) 
so  that  AB,  BC,  CD,  DE,  EA  are  equal  arcs. 

Find  the  centre  F ;  (III.  Prop.  5.) 

draw  FA,  FB,  EC,  FD,  FE, 

and  draw  GAH,  HBK,  KCL,  LDM,  MEG  at  right  angles 

to  FA,  FB,  EC,  ED,  EE  respectively;   (I.  Prop.  11.) 

then  GHKLM  is  a  pentagon  described  as  required. 

Draw  FK,  EL, 


Proof.     Because  in  the  triangles  FBK,  ECK, 
FB  is  equal  to  EC,  and  FK  to  FK, 
and  the  angle  FBK  equal  to  the  angle  ECK, 

each  being  a  right  angle,     (I.  Prop.  10  B.) 
and  each  of  the  angles  FKB,  EEC  therefore  is  less  than 
a  right  angle,  (T.  Prop.  17.) 

the  triangles  are  equal  in  all  respects ; 

(I.  Prop.  26  A,  CoroU.) 

therefore  KB  is  equal  to  KG, 

and  the  angle  BEK  equal  to  the  angle  CEK, 

and  the  angle  BKE  to  the  angle  CKF; 

therefore  KF  bisects  each  of  the  angles  BEG,  BKC. 

Similarly  it  can  be  proved  that  LF  bisects  each  of  the 

angles  CED,  CLD. 


PROPOSITION  12.  311 

Again,  because  the  arc  BG  is  equal  to  the  arc  CD^ 
the  angle  BFC  is  equal  to  the  angle  CFD. 

(III.  Prop.  27,  Coroll.) 

And  because  the  angle  KFC  is  half  of  the  angle  BFC, 

and  the  angle  LFC  is  half  of  the  angle  CFD; 

therefore  the  angle  KFC  is  equal  to  the  angle  LFC. 

Now  because  in  the  triangles  KFC,  LFC, 

the  angle  KFC  is  equal  to  the  angle  LFC, 

and  the  angle  KCF  to  the  angle  LCF, 

and  FC  to  FC, 

the  triangles  are  equal  in  all  respects;    (T.  Prop.  26,  Part  1.) 

therefore  KC  is  equal  to  LC, 

and  the  angle  FKC  equal  to  the  angle  FLC. 

Now  it  has  been  proved  that  KB  is  equal  to  KC, 

and  that  KL  is  double  of  KC ; 

and  it  can  similarly  be  proved  that  KII  is  double  of  KB ; 

therefore  UK  is  equal  to  KL. 

Similarly  it  can  be  proved  that  any  two  consecutive  sides 

of  GHKLM  are  equal ; 

therefore  the  pentagon  GHKLM  is  equilateral. 

And  because  it  has  been  proved  that  the  angles  FKC,  FLC 

are  equal, 

and  that  the  angle  BKC  is  double  of  the  angle  FKC, 

and  the  angle  CLD  double  of  the  angle  FLC, 

therefore  the  angle  BKC  is  equal  to  the  angle  CLD. 

Similarly  it  can  be  proved  that  any  two  consecutive  angles 

of  GHKLM  are  equal. 

Therefore  the  pentagon  GHKLM  is  equiangular. 

The  pentagon  is  therefore  regular. 

And  because  each  side  is  drawn  at  right  angles  to  a  radius 

of  the  circle  at  its  extremity,  it  touches  the  circle; 

(III.  Prop.  16.) 
therefore  the  pentagon  is  described  about  the  circle  ABODE. 
Wherefore,  a  regular  pentagon  GHKLM  has  been  de- 
scribed about  the  given  circle  ABODE. 


312 


BOOK  IV. 


PROPOSITION    13. 

To  irtscribe  a  circle  in  a  given  regular  pentagon. 

Let  ABODE  be  the  given  regular  pentagon  : 
it  is  required  to  inscribe  a  circle  in  ABODE. 

Construction.     Bisect  any  two  consecutive  angles  of 
the  pentagon  ABC,  BCD  by  BE,  OE  (I.  Prop.  9.) 

meeting  at  E; 

draw  EG,  EH,  EK,  EL,  EM  perpendicular  to  AB,  BO,  CD, 

DE,  EA  respectively.  (I.  Prop.  12.) 

With  E  as  centre  and  EG,  EH,  EK,  EL  or  EM  as  radius 

describe  a  circle: 

it  is  a  circle  inscribed  as  required. 

Draw  AE. 


G  K  D 

Proof.     Because  in  the  triangles  ABE,  OBE, 

AB  is  equal  to  CB^  (Hypothesis.) 

and  BE  to  BE, 
and  the  angle  ABE  to  the  angle  OBE,    (Constr.) 
the  triangles  are  equal  in  all  respects ; 

(I.  Prop.  4.) 
therefore  FA  is  equal  to  EC, 
and  the  angle  BAE  is  equal  to  the  angle  BOE. 

Again,  because  the  angle  BAE  is  equal  to  the  angle  BOD, 

(Hypothesis.) 
and  the  angle  BAE  has  been  proved  equal  to  the  angle 

BOE, 
and  the  angle  BOE  is  half  of  the  angle  BOD,    (Constr.) 
therefore  the  angle  BAE  is  half  of  the  angle  BAE, 
ov  AE  bisects  the  angle  BAE. 


PROPOSITION  13.  313 

Similarly  it  can  Ije  pi'oved  that  EF,  DF  bisect  the  angles 

AED,  EDO; 

therefore  the  bisectors  of  all  the  angles  of  the  pentagon 

meet  in  a  point. 

Again,  because  in  the  triangles  FGH,  FCK^ 
the  angle  FUG  is  equal  to  the  angle  FKC^ 
and  the  angle  FOE  to  the  angle  FGK^    (Constr.) 

and  FG  to  FG, 
the  triangles  are  equal  in  all  respects; 

(I.  Prop.  26,  Part  2.) 
therefore  FH  is  equal  to  FK. 
Similarly  it  can  be  proved  that  the  perpendiculars  on  any 

two  consecutive  sides  are  equal  to  one  another  : 
therefore  FG^  FH^  FK,  FL,  FM  are  equal,  and  the  circle 
described  with  F  as  centre  and  one  of  the  five  lines  FG, 
FH,  FK,  FL  or  FM  as  radius  passes  through  the  ex- 
tremities of  the  other  four  ;  and  because  the  angles  at 
G,  II,  K,  L,  M  are  right  angles,  it  touches  AB,  EG,  CD, 
BE,  EA.  (III.  Prop.  16.) 

Wherefore,  a  circle  GHKLM  has  been  inscribed  hi  the 
given  regular  pentagon  ABODE. 


EXERCISES. 

1.  How  many  conditions  are   necessary  in  order  that  a  given 
pentagon  may  admit  of  a  circle  being  inscribed  in  it  ? 

2.  Prove  that  the  bisectors  of  all  the  angles   of   any   regular 
polygon  meet  in  a  point. 


T.  12.  21 


314  BOOK  IV. 

PROPOSITION    14. 
To  describe  a  circle  about  a  given  regular  pentagon. 

Let  ABODE  be  tbe  given  regular  pentagon  : 
it  is  required  to  describe  a  circle  about  ABODE. 

Construction.     Bisect  any  two  consecutive  angles  of 

the  pentagon  ABO,  BOD,  by  BF,  OF    (I.  Prop.  9.) 

meeting  at  F,  and  draw  FA,  FE,  FD ; 

with  F  as  centre  and  FA,  FB,  FO,  FD  or  FE  as  radius 

describe  a  circle : 

it  will  be  a  circle  described  as  required. 


Proof.     Because  the  angle  ABO  is  equal  to  the  angle 
BOD,  (Hypothesis.) 

and  the  angle  FBO  is  half  of  the  angle  ABO, 
and  the  angle  FOB  is  half  of  the  angle  BOD, 
therefore  the  angle  FBO  is  equal  to  the  angle  FOB; 

therefore  FG  is  equal  to  FB.         (I.  Prop.  6.) 

Again,  because  in  the  triangles  ABF,  OBF, 

AB  is  equal  to  OB,  (Hypothesis.) 

BE  to  BF, 
and  the  angle  ABF  to  the  angle  OBF,      (Gonstr.) 
the  triangles  are  equal  in  all  respects ; 

(I.  Prop.  4.) 
therefore  FA  is  equal  to  FG, 
and  the  angle  FAB  to  the  angle  FOB. 
And  because  the  angle  FAB  is  equal  to  the  angle  FOB, 
and  the  angle  BAE  to  the  angle  BOD, 

(Hypothesis.) 


PROPOSITION  U.  315 

and  the  angle  FOB  is  half  of  the  angle  BCD^ 

(Constr.) 
therefore  the  angle  FAB  is  half  of  the  angle  BAE; 

and  FA  bisects  the  angle  BAF. 
Similarly  it  can  be  proved  that  FD,  FE   bisect  the 
angles  CDE,  DEA  respectively, 

and  that  FD  and  FE  are  each  of  them  equal  to  FA  or  FC', 

therefore  FA,   FB,   FC,   FD,  FE  are   all  equal,  and   the 

circle  described  with  F  as  centre  and  one  of   the  five 

lines  FA,  FB,  FC,  FD,  FE  as  radius  passes  through  the 

extremities  of  the  other  four,  and  is  described  about  the 

pentagon  ABCDE. 

Wherefore,  a  circle  ABCDE  has  been  described  about  the 

given  regular  pentagon  ABCDE. 


EXERCISES. 

1.  Describe  a  regular  decagon  to  have  five  of  its  vertices  coin- 
cident with  those  of  a  given  regular  pentagon. 

2.  How  many  conditions  are  necessary  in  order  that  a  given 
pentagon  may  admit  of  a  circle  being  described  about  it  ?  State  the 
conditions. 

3.  Shew  how  to  cut  off  the  corners  of  a  regular  pentagon  so  as 
to  leave  a  regular  decagon. 


21— -2 


S16  BOOK  IV. 

PROPOSITION    15. 

2^0  iyiscribe  a  regular  hexagon  in  a  given  circle. 

Let  ABC  DBF  be  the  given  circle  : 
it  is  required  to  inscribe  a  regular  hexagon  in  the  circle 
ABGDEF. 

Construction.     Find  G  the  centre  of  the  circle ; 

(III.  Prop.  6.) 
draw  any  diameter  AGD^ 
and  with  A  as  centre  and  AG  as  radius  describe  the  circle 

6?^i^  intersecting  the  circle  ABC  DBF  in  B  and  F. 

Draw  BG,  FG  and  produce  them  to  meet  the  circle  again 

in  E  and  C,  and  draw  AB,  BC,  CD,  DE,  EF,  FA : 

then  ABODE F  is  a  hexagon  inscribed  as  required. 


Proof.     Because  G  is  the  centre  of  the  circle  ABCDEF, 

GB  is  equal  to  GA. 

And  because  A  is  the  centre  of  the  circle  BGF, 

AB  i&  equal  to  AG. 

Therefore  AB,  BG,  GA  are  all  equal. 

Therefore  the  angles  AGB,  BAG^  GBA  are  all  equal. 

(I.  Prop.  5,  CoroU.  1.) 

But  the  sum  of  these  three  angles  is  equal  to  two  right 

angles;  (I.  Prop.  32.) 

therefore  the  angle  AGB  is  equal  to  one-third  of  two  right 

angles. 
Similarly  it  can  be  proved  that  the  angle  EGA  is  equal  to 
one-third  of  two  right  angles. 


PROPOSITION  15.  317 

But  the  sum  of  the  three  angles  BGA^  AGF,  FOE  is  equal 

to  two  right  angles;.  (I,  Prop.  13.) 

therefore  the  angle  FGE  is  one-third  of  two  right  angles, 

and  therefore  the  angles  EGA,  AGF,  FGE  are  equal. 

And  because  opposite  vertical  angles  are  equal, 

(I.  Prop.  15.) 

the  angles  opposite  to  these  are  equal ; 

therefore  all  the  angles  AGE,  FGE,  EGD,  DGC,  CGB,  EGA 

are  equal. 

Therefore  the  arcs  AF,  FE,  ED,  DC,  CB,  BA  are  equal. 

(III.  Prop.  26,  Coroll.) 
Therefore  the  chords  AF,  FE,  ED,  DC,  CB,  BA  are  equal. 

(HI.  Prop.  29,  Coroll.) 
Therefore  the  liexagon  ABC  DEE  is  equilateral. 
Again,  because  the  arcs  BAF,  AFE,  FED,  EDC,  DCB, 
CBA  are  equal, 

the  angles  BAF,  AFE,  FED,  EDC,  DCB,  CBA   in  those 
arcs  are  equal.  (III.  Prop.  27,  Coroll.) 

Therefore  the  hexagon  ABC  DEE  is  equiangular  : 
it  is  therefore  regular,   and  it  is  inscribed  in  the  circle 
ABCDEF. 
Wherefore,   a  regular  liexagon  ABCDEF  has  been  in- 
scribed  in  the  given  circle  ABCDEF. 


EXEECISES. 

1.  Inscribe  a  regular  dodecagon  in  a  given  circle. 

2.  If  ABCDEF  be  a  regular  hexagon,  and  AC,  BD,  CE,  DF, 
EA,  FB  be  drawn,  they  will  form  another  regular  hexagon  of  one 
third  the  area. 

3.  The  perimeter  of  the  inscribed  equilateral  triangle  of  a  circle  is 
three  quarters  the  perimeter  of  the  circumscribed  regular  hexagon. 

4.  Six  equal  circles  can  be  described  each  touching  a  given  circle 
and  two  of  the  others. 


318 


BOOK  TV. 


PROPOSITION   16. 

To  inscribe  a  regular  polygon  of  fifteen  sides  in  a  given 
circle. 

Let  A  BCD  be  the  given  circle  : 
it  is  required  to  inscribe  a  regular  polygon  of  fifteen  sides 
in  the  circle  A  BCD. 

Construction.     Let  ^,  (7  be  two  angular  points  of  an 

equilateral  triangle  inscribed  in  the  circle,  (Prop.  2.) 

and  let  A^B,DhQ  three  angular  points  of  a  regular  pentagon 

inscribed  in  the  circle.  (Prop.  1 1 . 

Draw  CD, 

and  place  round  the  circle  fifteen  chords  AL,  LM...  each 

equal  to  CD. 

The  figure  ALM ...  is  a  polygon  inscribed  as  required. 


Proof.     If  the  whole  circle  contain  fifteen  equal  parts, 
the  arc  ABC,  which  is  a  third  of  the  circle,  contains  five 

such  parts, 
and  the  arc  ABD,  which  is  made  up  of  the  arcs  AB,  BD, 
each  of  which  is  a  fifth  of  the  circle,  contains  six  such 
parts ; 
therefore  the  arc  CD^  which  is  the  difference  of  the  arcs 
ABD,  ABC,  consists  of  one  such  part; 

therefore  the  arc  CD  is  one-fifteenth  of  the  circle. 

And  because  the  arcs  AL,  LM,  .,.,  which  are  the  shorter 

arcs  cut  off* by  equal  chords  AL,  LM,  ...,  are  equal, 

(III.  Prop.  28,  Coroll.) 
each  of  the  arcs  AL,  LM,  ...,  is  one  fifteenth  of  the  circle. 


PROPOSITION  16.  319 

Therefore  the  extremity  of  the  last  chord  coincides  with 
the  point  A,  and  the  extremities  of  the  chords  which  have 
been  placed  round  the  circle  exactly  divide  the  circle  into 
fifteen  equal  parts. 

The  figure  ALM...  therefore  is  equilateral. 
And  as  each  of  the  angles  is  contained  by  an  arc  made  up 
of  two  of  the  fifteen  equal  parts, 
all  the  angles  are  equal ; 

(III.  Prop.  27,  Coroll.) 
therefore  the  figure  ALM...  is  equiangular. 

Wherefore,  a  regular  polygon  of  fifteen  sides  han  been  in- 
scribed in  the  given  circle  AJiCD. 


EXEECISES. 

1.  Prove  that  ia  the  figure  of  Proposition  16  a  vertex  of  the 
inscribed  regular  polygon  of  fifteen  sides  coincides  with  each  of  the 
vertices  of  the  regular  figures  used  in  the  construction. 

2.  Prove  that,  if  all  but  one  of  the  bisectors  of  the  angles  of  a 
polygon  meet  in  a  point,  they  all  do  so,  and  a  circle  can  be  inscribed 
in  the  polygon. 

3.  Prove  that,  if  all  but  one  of  the  rectangular  bisectors  of  the 
sides  of  a  polygon  meet  in  a  point,  they  all  do  so,  and  a  circle  can  be 
described  about  the  polygon. 


320  BOOK  IV. 

When  a  regular  polygon  of  any  number  of  sides  is  given,  we  can 
inscribe  a  regular  polygon  of  the  same  number  of  sides  in  a  given 
circle,  and  we  can  also  describe  a  regular  polygon  of  the  same 
number  of  sides  about  a  given  circle. 

Moreover  we  can  always  inscribe  a  circle  in  a  given  regular 
polygon  and  describe  a  circle  about  it. 

Methods  have  now  been  given  for  the  construction  of  regular 
figures  of  3,  4,  5,  6,  and  15  sides.  When  any  regular  polygon  is  given 
we  can  construct  a  regular  polygon  of  double  the  number  of  sides 
by  describing  a  circle  about  the  polygon  and  bisecting  the  smaller 
arcs  subtended  by  the  sides  of  the  given  polygon,  and  so  on  in 
succession  for  each  duplication  of  the  number  of  sides.  Thus  we  see 
that  we  can  by  EucUd's  methods  construct  regular  polygons  of  3  x  2"', 
4x2"',  5x2"  and  15  x  2'^  sides,  where  «  is  any  positive  integer,  in- 
cluding zero.  It  was  proved  by  Gauss*  in  the  year  1801  that  by 
purely  geometrical  methods  those  regular  polygons  can  be  con- 
structed, the  number  of  whose  sides  is  a  prime  f  number  of  the  form 
2'*  +  !.  This  general  law  relating  to  the  number  of  sides  includes  the 
case  of  the  triangle  {n  =  l)  and  the  pentagon  (n=2);  the  next  two 
cases  are  those  of  the  polygons  which  have  17  sides  {n  =  4)  and 
257  sides  {n  =  7)  respectively. 


*  Disquisitiones  Arithmeticce  (sectio  septima). 
t  A  prime  number  is  an  integer  which  is  not  divisible  without 
remainder  by  any  integer  except  itself  and  unity. 


MISCELLANEOUS   EXERCISES. 

1.  To  inscribe  a  square  in  a  given  parallelogram. 

2.  On  a  given  circle  find  a  point  such  that,  if  chords  be  drawn  to 
it  from  the  extremities  of  a  given  chord  of  the  circle,  their  difference 
shall  be  equal  to  a  given  straight  line  less  than  the  given  chord. 

3.  AB  is  a  fixed  chord  of  a  circle  whose  centre  is  O,  and  CD  is 
any  other  chord  equal  to  AB.  The  extremities  of  these  chords  are 
joined  by  straight  lines.  Prove  that,  if  the  joining  lines  meet  each 
other,  their  point  of  intersection  lies  on  the  circle  which  circumscribes 
the  triangle  AOB. 

4.  Through  each  angular  point  of  a  triangle  two  straight  lines 
are  drawn  parallel  to  the  straight  lines  joining  the  centre  of  the  cir- 
cumscribed circle  to  the  other  angular  points  of  the  triangle.  Prove 
that  these  six  straight  lines  form  an  equilateral  hexagon,  which  has 
three  pairs  of  equal  angles.. 

5.  If  two  equilateral  triangles  be  described  about  the  same  circle, 
they  will  form  an  equilateral  hexagon  whose  alternate  angles  are 
equal. 

6.  Describe  about  a  given  circle  a  quadrilateral  equiangular  to  a 
given  quadrilateral. 

7.  If  the  diameter  of  one  of  the  escribed  circles  of  a  triangle  be 
equal  to  the  perimeter  of  the  triangle,  the  triangle  is  right-angled. 

8.  The  diameter  of  the  inscribed  circle  of  a  right-angled  triangle 
is  equal  to  the  difference  between  the  sum  of  the  two  smaller  sides 
and  the  hypotenuse. 

9.  Construct  a  triangle  having  given  the  centres  of  its  inscribed 
circle  and  of  two  of  its  escribed  circles. 

10.  The  side  AB  of  a  triangle  ABC  touches  the  escribed  circles 
at  (xj,  Gg,  Gg:  prove  that  G^G^  is  equal  to  BC,  and  G^G^  to  CA. 

11.  If  a  circle  be  inscribed  in  a  right-angled  isosceles  triangle, 
the  distance  from  the  centre  of  the  circle  to  the  right  angle  will  be 
equal  to  the  difference  between  the  hypotenuse  and  a  side. 


322  BOOK  IV. 

12.  Describe  a  circle  to  touch  each  of  two  given  straight  lines 
and  to  have  its  centre  at  a  given  distance  from  a  third  given  straight 
line. 

13.  Three  circles  are  described,  each  of  which  touches  one  side  of 
a  triangle  ABC,  and  the  other  two  sides  produced.  If  D  be  the  point 
of  contact  of  the  side  BC,  E  that  of  AC,  and  F  that  of  AB,  then  AE 
is  equal  to  BD,  BF  to  CE,  and  CB  to  AF. 

14.  If  the  diagonals  of  the  quadrilateral  ABCD  intersect  at  right 
angles  at  0,  the  sum  of  the  radii  of  the  inscribed  circles  of  the  tri- 
angles A  OB,  BOC,  COD,  DO  A  is  equal  to  the  difference  between  the 
sum  of  the  diagonals  and  the  semi-perimeter  of  the  quadrilateral. 

15.  Having  given  the  hypotenuse  of  a  right-angled  triangle  and 
the  radius  of  the  inscribed  circle,  construct  the  triangle. 

16.  If  the  inscribed  circle  of  a  triangle  ABC  touch  the  sides  AB, 
AC  qX  the  points  D,  E,  and  a  straight  line  be  drawn  from  A  to  the 
centre  of  the  circle  meeting  the  circumference  at  G,  the  point  G  is  the 
centre  of  the  inscribed  circle  of  the  triangle  ADE. 

17.  Two  sides  of  a  triangle  whose  perimeter  is  constant  are  given 
in  position :  shew  that  the  third  side  always  touches  a  fixed  chcle. 

18.  The  points  of  contact  of  the  inscribed  circle  of  a  triangle  are 
joined ;  and  from  the  angular  points  of  the  triangle  so  formed  perpen- 
diculars are  drawn  to  the  opposite  sides :  shew  that  the  triangle  of 
which  the  feet  of  these  perpendiculars  are  the  angular  points  has  its 
sides  parallel  to  the  sides  of  the  original  triangle. 

19.  Four  triangles  are  formed  by  three  out  of  four  given  points 
on  a  given  circle :  shew  that  a  circle  may  be  described  so  as  to  pass 
through  the  centres  of  the  inscribed  circles  of  the  four  triangles. 

20.  The  rectangle  of  the  segments  of  the  hypotenuse  of  a  right- 
angled  triangle  made  by  the  point  of  contact  of  the  inscribed  circle  is 
equal  to  the  area  of  the  triangle. 

21.  If  on  the  sides  of  any  triangle  three  equilateral  triangles  be 
constructed,  the  centres  of  the  inscribed  circles  of  these  triangles  are 
the  vertices  of  an  equilateral  triangle. 

22.  Describe  three  equal  circles  to  touch  each  other  and  a  given 
circle. 

23.  Construct  an  isosceles  triangle,  having  its  base  equal  to  the 
greater,  and  the  diameter  of  its  inscribed  circle  equal  to  the  less  of 
two  given  straight  lines. 

24.  In  a  given  right-angled  triangle,  the  lengths  of  the  sides  con- 
taining the  right  angle  are  6  and  8  feet  respectively.  Find  the  lengths 
of  the  segments  into  which  the  hypotenuse  is  divided  by  the  circle 
inscribed  in  the  triangle. 


MISCELLANEOUS  EXAMPLES.  323 

25.  Two  triangles  ABC,  DEF  are  inscribed  in  the  same  circle  so 
that  AB,  BE,  GF  meet  in  one  point  0 ;  prove  that,  if  O  be  the  centre 
of  the  inscribed  circle  of  one  of  the  triangles,  it  will  be  the  ortho- 
centre  of  the  other. 

26.  A  circle  B  passes  through  the  centre  of  another  circle  A ;  a 
triangle  is  described  circumscribing  A  and  having  two  angular  points 
on  B  :  prove  that  the  third  angular  point  is  on  the  line  of  centres. 

27.  A  circle  is  escribed  to  the  side  i3(7  of  a  triangle  ABC  touching 
the  other  sides  in  F  and  (r.  A  tangent  DE  is  drawn  parallel  to  UC 
meeting  the  sides  in  D,  E.  DE  is  found  to  be  three  times  BC  in 
length.     Shew  that  DE  is  twice  AF. 

28.  The  sum  of  the  diameters  of  the  inscribed  and  the  circum- 
scribed circles  of  a  right-angled  triangle  is  equal  to  the  sum  of  the 
sides  containing  the  right  angle. 

29.  Prove  that  two  circles  can  be  described  with  the  middle  point 
of  the  hypotenuse  of  a  right-angled  triangle  as  centre  to  touch  the 
two  circles  described  on  the  two  sides  as  diameters. 

30.  The  perpendicular  from  A  on  the  opposite  side  i>C  of  a 
triangle  ABC,  meets  the  circumference  of  the  circumscribed  circle  in 
G.  If  P  be  the  point  in  which  the  perpendiculars  from  the  angles 
upon  the  opposite  sides  intersect,  then  PG  is  bisected  by  BC. 

31.  Through  G,  the  middle  point  of  the  arc  ACB  of  a  circle,  any 
chord  GP  is  drawn,  cutting  the  straight  line  AB  in  Q.  Shew  that  the 
locus  of  the  centre  of  the  circle  circumscribing  the  triangle  BQP  is  a 
straight  line. 

32.  The  distance  of  the  orthocentre  of  a  triangle  from  any 
vertex  is  double  of  the  distance  of  the  centre  of  the  circumscribed 
circle  from  the  opposite  side. 

33.  Two  equilateral  triangles  ABC,  DEF  are  inscribed  in  a  circle 
whose  centre  is  0.  AC,  DF  intersect  in  P,  and  AB,  DE  in  Q.  Prove 
that  either  POQ  is  a  straight  line,  or  a  circle  can  be  described  about 
APOQD. 

34.  Given  the  base,  the  difference  of  the  angles  at  the  base  and 
the  radius  of  the  circumscribing  circle  of  a  triangle,  shew  how  to  con- 
struct the  triangle. 

35.  From  the  vertices  of  a  triangle  draw  straight  lines  which 
shall  form  an  equilateral  hexagon  whose  area  is  double  that  of  the 
triangle. 

36.  If  on  each  side  of  an  acute-angled  triangle  as  base,  an 
isosceles  triangle  be  constructed  the  sides  of  each  being  equal  to  the 
radius  of  the  circumscribed  circle,  the  vertices  of  these  triangles  form 
the  vertices  of  a  triangle  equal  in  all  respects  to  the  original  triangle. 

37.  If  O  be  the  orthocentre  of  the  triangle  ABC,  the  triangle 
formed  by  the  centres  of  the  circles  OBG,  OCA,  GAB  is  equal  to  the 
triangle  ABC  in  all  respects. 


324  BOOK  IV. 

38.  The  ends  of  a  straight  line  AB  move  along  two  fixed  straight 
lines  in  a  plane ;  prove  that  there  is  a  point,  in  rigid  connection  with 
AB,  which  describes  a  circle. 

39.  The  circle  through  B,  C  and  the  centre  of  the  circle  inscribed 
in  the  triangle  ABC  meets  the  sides  AB,  AC  again  in  E,  F;  prove  that 
EF  touches  the  inscribed  circle. 

40.  Let  ABC  be  a  triangle,  0  the  centre  of  the  inscribed  circle, 
and  0',  0",  0'",  the  centres  of  the  escribed  circles  situated  in  the 
angles  A,  B,  C,  respectively;  prove  (1)  that  the  circumscribed  circle 
passes  through  the  middle  points  of  the  lines  00',  00",  00'"; 
(2)  that  the  four  points  0,  B,  C,  0'  lie  on  a  circle  which  has  its 
centre  on  the  circumscribed  circle;  (3)  that  the  points  0",  B,  C,  0'" 
lie  on  a  circle  whose  centre  is  on  the  circumscribed  circle. 

41.  The  triangle  of  least  perimeter  which  can  be  inscribed  in  a 
given  acute-angled  triangle  is  the  triangle  formed  by  joining  the  feet 
of  the  perpendiculars  from  the  angular  points  on  the  opposite  sides. 

42.  Describe  a  circle  to  touch  a  given  straight  line,  and  pass 
through  two  given  points. 

43.  Describe  a  circle  to  pass  through  two  given  points  and  cut 
off  from  a  given  straight  line  a  chord  of  given  length. 

44.  Describe  a  circle  to  pass  through  two  given  points,  so  that 
the  tangent  drawn  to  it  from  another  given  point  may  be  of  a  given 
length. 

45.  If  I,  0  be  the  centres  of  the  inscribed  and  the  circumscribed 
circles  of  a  triangle  ABC,  and  if  AI  be  produced  to  meet  the  circum- 
scribed circle  in  F,  then  OF  bisects  BC. 

46.  If  I  be  the  centre  of  the  inscribed  circle  of  the  triangle  ABC, 
and  AI  produced  meet  the  circumscribed  circle  at  F,  FB,  FI,  and  FG 
are  all  equal. 

47.  Construct  a  triangle  having  given  one  angular  point  and  the 
centres  of  the  inscribed  and  the  circumscribed  circles. 

48.  0  is  the  centre  of  the  circumscribed  circle  of  a  triangle  ABC ; 
D,  E,  F  are  the  feet  of  the  perpendiculars  from  A,  B,  C,  on  the  op- 
posite sides:  shew  that  OA,  OB,  OC  are  respectively  perpendicular 
to  EF,  FD,  BE. 

49.  The  four  circles  each  of  which  passes  through  the  centres  of 
three  of  the  four  circles  touching  the  sides  of  a  triangle  are  equal  to 
one  another. 

50.  Construct  a  triangle  having  given  an  angle  and  the  radii  of 
the  inscribed  and  the  circumscribed  circles. 

51.  From  the  vertex  of  a  triangle  draw  a  straight  line  to  the  base 
so  that  the  square  on  the  straight  line  may  be  equal  to  the  rectangle 
contained  by  the  segments  of  the  base. 


MISCELLANEOUS  EXAMPLES.  325 

52.  Four  triangles  are  formed  by  three  out  of  four  given  straight 
lines ;  shew  that  the  circumscribed  circles  of  these  triangles  all  pass 
through  a  common  point. 

53.  The  straight  line  joining  the  middle  points  of  the  arcs  of  a 
circle  cut  off  by  two  sides  of  an  inscribed  equilateral  triangle  is 
trisected  by  those  sides. 

54.  The  perpendicular  from  an  angle  of  an  equilateral  triangle 
on  the  opposite  side  is  equal  to  three  quarters  of  the  diameter  of  the 
circumscribed  circle. 

55.  If  the  inscribed  and  the  circumscribed  circles  of  a  triangle  be 
concentric,  the  triangle  is  equilateral. 

56.  The  angle  C  of  the  triangle  ABC  is  a  right  angle.  P  is  the 
intersection  of  the  diagonals  of  a  square  on  AC,  and  Q  of  those  of  a 
square  on  BC.  Prove  that  the  circumscribed  circle  of  the  triangle 
ABC  passes  through  the  intersection  of  PQ  with  a  perpendicular  to 
AB  drawn  through  the  middle  point  of  AB. 

57.  Describe  three  circles  to  touch  in  pairs  at  three  given  points. 

68.  A  rhombus  is  described  about  a  given  rectangle.  Prove  that 
its  centre  coincides  with  that  of  the  rectangle  and  that  each  of  its 
angular  points  lies  either  on  a  fixed  straight  line  or  on  a  fixed  circle. 

59.  Describe  an  isosceles  triangle  such  that  thi'ee  times  the  verti- 
cal angle  shall  be  four  times  either  of  the  other  angles. 

60.  li  ABCDE  be  a  regular  pentagon,  and  AC,  BD  intersect  at 
O,  then  ^0  is  equal  to  DO,  and  the  rectangle  AC,  CO  is  equal  to  the 
square  on  BC. 

61.  Prove  that  the  difference  of  the  squares  on  a  diagonal  and  on 
a  side  of  a  regular  pentagon  is  equal  to  the  rectangle  contained  by 
them. 

62.  If  with  one  of  the  angular  points  of  a  regular  pentagon  as 
centre  and  one  of  its  diagonals  as  radius  a  circle  be  described,  a  side 
of  the  pentagon  will  be  equal  to  a  side  of  the  regular  decagon  inscribed 
in  the  circle. 

63.  If  ABCDE  be  a  pentagon  described  about  a  circle,  and  F  be 
the  point  of  contact  of  AB,  then  twice  AF  is  equal  to  the  difference  of 
the  sum  of  AB,  AE,  CD  and  the  sum  of  BC,  DE. 

64.  AOA',  BOB'  are  two  diameters  of  a  circle  at  right  angles  to 
one  another.  £0  is  bisected  at  C  and  AC  cuts  the  circle  on  BO  as 
diameter  in  D  and  E.  Circles  having  A  as  centre  and  AD,  AE  as 
radii  are  described  cutting  the  original  circle  in  jP  and  F^,  G  and  G' : 
prove  that  A'GFF'G'  is  a  regular  pentagon. 

65.  A  regular  hexagon  ABCDEF  is  inscribed  in  a  circle.  A 
second  circle  is  described  through  A  and  B  to  cut  the  first  circle  at 
right  angles :  and  a  third  circle  is  described  through  A  and  C  to  cut 
the  first  circle  at  right  angles.  Prove  that  the  diameter  of  the  third 
circle  is  three  times  that  of  the  second. 


326  BOOK  IV. 

66.  If  the  alternate  angles  of  an  equilateral  hexagon  be  equal  to 
one  another,  a  drcle  can  be  inscribed  in  the  hexagon. 

67.  Construct  a  regular  polygon  of  2n  sides  of  equal  perimeter 
with  a  given  regular  polygon  of  n  sides. 

68.  Any  equilateral  figure  which  is  inscribed  in  a  circle  is  also 
equiangular. 

69.  Prove  that  a  polygon  which  is  described  about  one  and 
inscribed  in  another  of  two  concentric  circles  must  be  regular. 

70.  Shew  that  it  is  always  possible  to  describe  about  a  circle  a 
polygon  equiangular  to  any  given  polygon.  Will  the  two  polygons  be 
necessarily  similar? 

71.  Find  the  locus  of  the  centre  of  the  circumscribing  circle  of  a 
triangle,  when  the  vertical  angle  and  the  sum  of  the  sides  containing 
it  are  given. 

72.  Given  the  circumscribed  circle,  an  escribed  circle  and  the 
centre  of  the  inscribed  circle,  construct  the  triangle. 

73.  Given  an  angular  point,  the  circumscribing  circle  and  the 
orthocentre,  construct  the  triangle. 

74.  Describe  a  square  about  a  given  quadrilateral. 
How  many  solutions  are  there? 

75.  Construct  a  square  so  that  each  side  shall  touch  one  of  four 
given  circles. 

How  many  solutions  are  there  ? 


CAMBBinaE:  PBIITTBD  BY  C.  J.  CLAY,  M.A.  &SON8,  AT  THE  UNIVEBSITY  PEB8S. 


BOOK    V. 


DEFINITIONS. 

Definition  1.  If  one  magnitude  he  equal  to  another 
magnitude  of  the  same  kind  repeated  twice,  thrice  or  any 
number  of  times,  the  first  is  said  to  be  a  multiple  of  the 
second,  and  the  second  is  said  to  be  an  aliquot  part  or  a 
measure  of  the  first. 

If  one  magnitude  A  be  equal  to  m  times  another  magnitude  B 
of  the  same  kind  {m  being  an  integer,  i.e.  a  whole  number),  A  is  said 
to  be  the  m^''  multiple  of  B,  and  B  the  m*  part  of  A. 

If  A  be  any  multiple  of  B  and  if  C  be  the  same  multiple  of  D,  then 
A  and  C  are  said  to  be  equimultiples  of  B  and  D. 

The  magnitudes  treated  of  in  Book  V.  are  not  necessarily  Geome- 
trical magnitudes :  but  they  are  assumed  to  be  such  that  any  magnitude 
can  be  supposed  to  be  repeated  as  often  as  desired,  in  other  words, 
that  any  multiple  we  please  of  a  magnitude  can  be  taken.  They  are 
assumed  also  to  be  such  that  any  one  taken  twice  is  greater  than  it 
is  alone ;  such  quantities  as  those  which  are  called  in  Algebra  either 
negative  or  imaginary  are  excluded  from  consideration. 

The  capital  letters  A,  B,  C,D  &c.  will  be  used  to  denote  magnitudes, 
and  the  small  letters  m,  n,  p,  q  &c.  to  denote  whole  numbers. 

When  a  magnitude  A  is  spoken  of,  the  letter  A  is  supposed  to  repre- 
sent the  magnitude  itself. 

Definition  2.  The  relatioyi  of  one  magnitude  to  another 
of  the  same  kind  with  respect  to  the  multiples  of  the  seco7id  or 
of  aliquot  parts  of  the  second,  ivhich  the  first  is  greater  than, 
equal  to,  or  less  than,  is  called  the  ratio  of  the  first  to  the 
second. 

It  is  difficult  to  convey  a  precise  idea  of  "ratio"  by  a  definition. 
The  student  will  gradually  acquire  a  firmer  grasp  of  the  meaning  of 
the  term  as  he  proceeds.  It  is  important  to  bear  in  mind  that 
the  difference  between  two  magnitudes  is  not  their  ratio. 


328  BOOK    V. 

Whenever  the  ratio  of  one  magnitude  to  another  is  spoken  of,  it 
is  necessarily  implied,  although  it  may  not  always  be  express?7  stated, 
that  the  two  magnitudes  are  of  the  same  kind. 

In  the  ratio  of  one  magnitude  to  another,  the  first  is 
called  the  antecedent  and  the  second  the  consequent  of  the 
ratio. 

In  the  ratio  of  A  to  B,  A  is  the  antecedent  and  B  the  consequent. 

The  ratio  of  a  magnitude  to  an  equal  magnitude  is 
called  a  ratio  of  equality,  and  is  said  to  be  equal  to  unity; 

the  ratio  of  a  magnitude  to  a  less  magnitude  is  called 
&  ratio  of  greater  inequality,  and  is  said  to  be  greater 
than  unity; 

the  ratio  of  a  magnitude  to  a  greater  magnitude  is 
called  a  ratio  of  less  inequality,  and  is  said  to  be  less  than 
unity. 

The  ratio  of  one  diameter  to  another  diameter  of  the  same  circle 
is  a  ratio  of  equality : 

the  ratio  of  a  diagonal  to  a  side  of  a  square  is  a  ratio  of  greater 
inequality : 

the  ratio  of  the  area  of  a  circle  to  the  area  of  a  square  described 
about  the  circle  is  a  ratio  of  less  inequality. 

Definition  3.  If  one  magnitude  repeated  any  number 
of  times  be  greater  than,  equal  to,  or  less  than  a  second  7nagni- 
tude  repeated  any  other  number  of  times,  the  ratio  of  the 
first  magnitude  to  the  second  magnitude  is  said  to  be 
greater  than,  equal  to,  or  less  than  the  ratio  of  the  second 
number  to  the  first  number. 

li  A,  B  he  two  magnitudes  such  that  m  times  A  is  equal  to  n 
times  B,  the  ratio  of  ^  to  £  is  equal  to  the  ratio  of  n  to  m. 

Similarly,  if  m  times  A  be  greater  than  n  times  B,  the  ratio  of 
^  to  iJ  is  greater  than  that  of  n  to  m ;  and  if  m  times  A  be  less  than 
n  times  J5,  the  ratio  of  ^  to  B  is  less  than  that  of  n  to  m. 

Definition  4.  When  two  magnitudes  of  the  same  kind 
are  such  that  some  measure  of  the  first  is  equal  to  some 
measure  of  the  second,  the  two  magnitudes  are  said  to  be 
commensurable. 

Two  magnitudes  of  the  same  kind,  which  are  not  com- 
mensurable, are  said  to  be  incommensurable. 


DEFINITIONS.  329 

li  AyByChe  three  magnitudes  of  the  same  kind  such  that  C  is  the 
w*^  part  of  A  and  G  is  the  n*'^  part  of  B,  or,  in  other  words,  such  that 
A  is  equal  to  m  times  C  and  B  is  equal  to  7i  times  C,  then  ^  and  B 
have  a  common  measure  C,  and  therefore  are  commensurable. 

If  the  ratio  of  ^  to  5  be  equal  to  the  ratio  of  one  integer  to 
another,  say  that  of  n  to  w,  and  the  ratio  of  C  to  D  be  also  equal  to 
that  of  n  to  m,  the  ratio  of  ^  to  B  is  equal  to  that  of  C  to  D  :  and 
similarly,  if  the  ratio  of  ^  to  5  be  equal  to  that  of  n  to  m,  and  the 
ratio  of  C  to  D  be  greater  or  less  than  that  of  n  to  m,  the  ratio  of 
il  to  £  is  less  or  greater  respectively  than  that  of  C  to  D. 

A  complete  method  is  thus  afforded  of  testing  the  equality  or  the 
inequality  of  the  ratios  of  pairs  of  commensurable  magnitudes:  but 
the  same  method  is  not  applicable  to  incommensurable  magnitudes. 

Now  it  will  be  manifest  from  what  has  been  said  that,  if  we  have 
four  magnitudes  A,  B,  C,  D,  of  which  A  and  B  are  incommen- 
surable, the  ratio  of  G  to  D  cannot  be  equal  to  that  of  A  to  B,  unless 
G  and  D  be  also  incommensurable. 

It  is  possible  to  find  two  magnitudes  of  the  same  kind  that  are 
not  commensurable.  It  can  be  proved  that  a  diagonal  and  a  side  of 
the  same  square  are  such  a  pair  of  magnitudes,  and  also  that  the 
circumference  and  a  diameter  of  the  same  circle  are  another  such  pair 
of  magnitudes.  The  question  arises  how  the  ratios  of  two  pairs  of 
such  incommensurable  magnitudes  are  to  be  compared. 

It  is  easy  to  prove  that  a  diagonal  of  a  square  is  greater  than  once 
and  less  than  twice  a  side ;  these  inequalities  give  a  very  rough 
comparison  of  the  lengths  of  the  two  lines.  It  can  be  proved  that 
10  times  a  diagonal  is  greater  than  14  times  and  less  than  15  times 
a  side :  these  inequalities  give  a  less  rough  comparison  of  the  lengths. 
Again,  it  can  be  proved  that  100  times  a  diagonal  is  greater  than 
141  times  and  less  than  142  times  a  side:  these  inequalities  give 
a  still  less  rough  comparison  of  the  lengths  of  the  two  lines. 

These  facts  are  represented  by  saying  that  the  ratio  of  a  diagonal 
to  a  side  is  greater  than  the  ratio  of  1  to  1  and  less  than  that  of  2  to  1 : 
greater  than  the  ratio  of  14  to  10  and  less  than  that  of  15  to  10  : 
greater  than  the  ratio  of  141  to  100  and  less  than  that  of  142 
to  100. 

Pairs  of  ratios  of  greater  and  greater  numbers  might  be  quoted, 
between  which  the  ratio  of  a  diagonal  to  a  side  always  lies:  but  no 
two  numbers  can  be  found  such  that  the  ratio  in  question  is  equal  to 
the  ratio  of  the  numbers. 

T.  E.  22 


330  BOOK    V. 

In  the  following  definition  of  the  equality  of  two  ratios,  the  case 
of  incommensurable  magnitudes  is  included  as  well  as  that  of  com- 
mensurable magnitudes. 


Definition  5.  If  four  rtuignitudes  he  such  that,  when 
any  equimultiples  whatever  of  the  first  and  the  third  are 
taken,  and  also  any  equimultiples  whatever  of  the  second  and 
the  fourth,  the  multiples  of  the  first  and  the  third  are  simul- 
taneously either  both  greater  than,  or  both  equal  to,  or  both 
less  than  the  multiples  of  the  second  and  the  fourth  respec- 
tively, the  ratio  of  the  first  magnitude  to  the  second  is  said  to 
he  equal  to  the  ratio  of  the  third  to  the  fourth. 

When  the  ratio  of  the  first  of  four  magnitudes  to  the 
second  is  equal  to  that  of  the  third  to  the  fourth,  the  mag- 
nitudes are  said  to  be  proportionals  or  in  proportion. 

When  four  magnitudes  are  proportionals,  the  first  is  said 
to  he  to  the  second  as  the  third  to  the  fourth. 

Let  A,  B,  G,  D  he  four  magnitudes,  of  which  A  and  B  are  of  the 
same  kind,  and  C  and  D  are  of  the  same  kind,  and  let  any  equimul- 
tiples whatever  of  A  and  C,  say  m  times  A  and  m  times  C,  be  taken,  and 
any  equimultiples  whatever  of  B  and  Z),  say  n  times  B  and  n  times  D ; 
then,  if  m  times  A  be  greater  than  n  times  B  and  also  m  times  G 
greater  than  n  times  D,  or  else  m  times  A  be  equal  to  n  times  B  and 
also  w  times  G  equal  to  n  times  D,  or  else  m  times  A  be  less  than  n 
times  B  and  also  m  times  G  less  than  n  times  D,  for  every  possible 
pair  of  whole  numbers  m  and  n,  the  ratio  of  ^  to  J5  is  equal  to  the 
ratio  of  G  to  D,  and  A,  B,  G,  D  are  proportionals. 

The  fact  that  four  magnitudes  A,  B,  G,  D  are  in  proportion  is 
denoted  by  saying  that  A  has  to  B  the  same  ratio  that  G  has  to  D,  or 
that  the  ratio  of  .4  to  B  is  equal  to  that  of  G  to  D,  or  that  A  is 
to  i?  as  C  to  D :  it  is  expressed  still  more  concisely  by  the  notation 

A:B  =  G:D. 
Note.    It  will  be  observed  that  when  four  magnitudes  A ,  B,  G,  D 


DEFINITIONS.  331 

are  defined  in  order  as  proportionals,  i.e.  ^1  is  to  JB  as  C  to  D,  they 
are  at  the  same  time  defined  as  proportionals  also  in  the  three  several 
orders  5,  A,  D,  C\  C,  D,  A,  B ;  and  D,  C,  B,  A;  that  is  to  say,  it 
follows  from  the  definition  that,  if  any  one  of  the  four  proportions 

A  :  B  =  C  :  D,     B  :  A=D  :  C,     C  :  D  =  A  :  B,     D  :  C  =  B  :  A 

exist,  the  other  three  exist  also. 

It  follows  at  once  from  Definition  5  that  if,  of  A,  B,  C,  D,  four 
magnitudes,  A  be  equal  to  C  and  B  be  equal  to  D,  the  ratio  of  J^  to  i? 
is  equal  to  the  ratio  of  C  to  D  ;  and  further  that  if,  of  ^4,  7?,  C,  three 
magnitudes,  A  be  equal  to  B,  the  ratio  of  ^  to  C  is  equal  to  the  ratio 
of  B  to  C,  and  also  the  ratio  of  C  to  J  is  equal  to  the  ratio  of  C  to  B. 

Definition  6.  When /our  magnitudes  are  proportionals, 
the  first  and  the  third,  the  antecedents,  are  said  to  he 
homologous  to  one  another,  and  the  second  and  the  fourth, 
the  consequents,  are  also  said  to  he  homologous  to  one  an- 
other. 

In  the  proportion  4  is  to  £  as  O  to  D,  the  antecedents  A  and  G 
are  homologous  to  one  another  and  the  consequents  B  and  D  are 
homologous  to  one  another. 

Definition  7.  If  it  he  possible  to  take  equimultiples  of 
the  first  and  the  third  of  four  magnitudes  and  equimultiples 
of  the  second  and  the  fourth,  such  that  the  multiple  of 
the  first  is  greater  than  that  of  the  second  and  the  multiple 
of  the  third  not  greater  than  that  of  the  fourth,  the  ratio  of 
the  first  to  the  second  is  said  to  he  greater  than  that  of  the 
third  to  the  fourth  ;  and  the  ratio  of  the  third  to  the  fourth 
is  said  to  he  less  than  the  ratio  of  the  first  to  the  second. 

As  an  example  the  ratios  of  the  numbers  2  to  3  and  5  to  8  may 
be  taken. 

If  the  5^^  multiples  of  the  first  and  the  third  be  taken  and  the  S"^ 
multiples  of  the  second  and  the  fourth,  the  multiples  in  order  are 
10,  9,  25,  24 :  here  10  is  greater  than  9  and  25  greater  than  24 :  but 
equality  between  the  ratios  is  not  thereby  established. 

22—2 


332  BOOK    V. 

If  the  ll**^  multiples  of  the  first  and  the  third  be  taken,  and  the  7*** 
multiples  of  the  second  and  the  fourth,  the  multiples  in  order  are 
22,  21,  56,  56 :  here  22  is  greater  than  21  and  55  not  greater  than  56 : 
and  the  fact  is  established  that  the  ratio  of  2  to  3  is  greater  than  that 
of  5  to  8. 

Definition  8.  The  ratio  of  the  first  of  three  magnitudes 
of  the  same  kind  to  the  third  is  said  to  be  compounded  of  the 
ratio  of  the  first  to  the  second  ayid  the  ratio  of  the  second  to 
the  third. 

The  ratio  of  the  first  of  three  magnitudes  of  the  same  kind 
to  the  third  is  also  said  to  he  the  ratio  of  the  ratio  of  the  first 
magnitude  to  the  second  to  the  ratio  of  the  third  magnitude 
to  the  second. 

If  A,  B,  C  be  three  magnitudes  of  the  same  kind,  the  ratio  of 
Aio  G  \B  compounded  of  the  ratio  ^  to  J5  and  the  ratio  B  to  C. 

Further,  the  ratio  of  A  to  C  is  said  to  be  the  ratio  of  the  ratio  A 
to  B  to  the  ratio  C  to  B. 

Definition  9.  If  the  first  of  a  number  of  magnitudes 
of  the  same  kind  be  to  the  second  as  the  second  to  the  third 
and  as  the  third  to  the  fourth  and  so  on,  the  magnitudes  are 
said  to  be  in  continued  proportion.  .        ^  i 

If  a  number  of  magnitudes  be  in  continued  proportion, 
the  ratio  of  the  first  to  the  third  is  said  to  be  duplicate  of 
the  ratio  of  the  first  to  the  second,  and  the  ratio  of  the  first  to 
the  fourth  is  said  to  he  triplicate  of  the  ratio  of  the  first 
to  the  second. 

If  four  magnitudes  be  in  proportion,  the  first  and  the 
fourth  are  called  the  extremes  and  the  second  and  the  fourth 
the  means  of  the  proportion. 

If  three  magnitudes  he  iii  continued  proportion,  the  first 
and  the  third  are  called  the  extremes  and  the  second  the 
mean  of  the  proportion ;  also  the  second  is  called  a  mean 
proportional  between  the  first  and  the  third,  and  the  third  is 
called  a  third  proportional  to  the  first  and  tJie  second. 


PROPOSITION  1.  333 


PROPOSITION  1.     [Euclid  Elem.  Prop.  4.] 

If  four  magnitudes  in  order  he  proportionals  and  any 
equimultiples  of  the  antecedents  he  taken,  and  any  equi- 
multiples of  the  consequents,  the  four  multiples  are  pro- 
portionals in  the  same  order  as  the  magnitudes* * 

Let  the  magnitudes  A,  B,  C,  D  be  proportionals,  and 
let  m,  n  be  two  given  numbers : 

it  is  required  to  prove  that  m  times  A,  n  times  B,  tn  times  (7, 
n  times  D  are  proportionals. 

Construction.     Let  p,  q  be  any  two  numbers,  and  let 
m  times  A  be  E,  ?*  times  B  be  F,  m  times  C  he  G  and  n         -^ 
times  i>  be /^  1' >  A  .^    13  ^^G-        "^  J 

'>>  f+ 

Proof.     Because  E^  G  are  equimultiples  of  A,  (7,     "^ 

p  times  E,  p  times  6^  are  equimultiples  of  ^,  (7, 

no  matter  what  number  p  may  be ; 

and  because  F,  H  are  equimultiples  of  B,  D, 

q  times  F,  q  times  H  are  equimultiples  of  B,  D, 

no  matter  what  number  q  may  be  ; 

and  because  A  is  to  B  a,s  C  to  i>, 

p  times  j^  and  ^  times  G 

are  both  greater  than,  both   equal  to  or  both   less  than 

q  times  F  and  q  times  if  respectively, 

for  all  values  of  p  and  q;  (Def.  5.) 

therefore  ^  is  to  i^  as  6^  to  if ;  (Def.  5.) 

that  is,  m  times  A  is  to  n  times  B  as  m  times  C  to  ?i  times  i>. 

Wherefore,  if  four  magnitudes,  &c. 

*  Algebraically.    If  a  :  &=c  :  d,  then  wia  -.nb-mc:  nd. 


334  BOOK   V. 


PROPOSITION  2.     [Euclid  Elem.  Prop.  8.] 

The  greater  of  two  magnitudes  has  to  a  third  magnUude 
a  greater  ratio  than  the  less  has;  and  a  third  magnitude 
has  to  the  less  of  two  other  magnitudes  a  greater  ratio  than 
it  has  to  the  greater*. 

Let  A,  B,  C  be  three  magnitudes  of  the  same  kind,  of 
which  A  is  greater  than  B : 
it  is  required  to  prove  that  A  has  to  (7  a  greater  ratio  than 

B  has  to  0,  and  that  C  has  to  ^  a  greater  ratio  tlian  G 

has  to  A. 

Construction.     Let  the  excess  of  A  over  ^  be  i>; 
take   the   m^^   equimultiples    of   B,  D,  such  that  each  is 
greater  than  (7,  and  of  the  multiples  of  C  let  the  p^"^ 
multiple  be  the  first  which  is  greater  than  m  times  B^ 
and  let  n  be  the  number  next  less  than  ^9. 

Proof.    Because  m  times  B  is  not  less  than  n  times  G, 
and  7n  times  D  is  greater  than  C ',        (Constr.) 
therefore  the  sum  of  m  times  B  and  m  times  D  is  greater 
than  the  sum  of  n  times  G  and  G, 
that  is,  m  times  A  is  greater  than  p  times  G ; 

and  m  times  B  is  less  than  p  times  G;       (Constr.) 
therefore  A  has  to  C  a  greater  ratio  than  B  has  to  G. 

(Del  7.) 
Next,  because  p  times  G  is  less  than  m  times  Aj 
and  p  times  C  is  greater  than  in  times  B, 
therefore  G  has  to  -ff  a  greater  ratio  than  C  has  to  A. 

(Def.  7.) 
"Wherefore,  the  greater,  &c. 

*  Algebraically.    If  a > 6,  then  a  i  o-b  :  c  and  c  :  b>c  :  a. 


FUOFOSITIONS  2,  3.  335 


PROPOSITION  3.     [Euclid  Elem.  Prop.  9.] 

If  the  ratio  of  the  first  of  three  magnitudes  to  the  third  he 
equal  to  the  ratio  of  the  second  to  the  thirds  the  first  magni- 
tude is  equal  to  the  second** 

Let  yl,  ^,  (7  be  three  magnitudes  of  the  same  kind  such 
that  ^  is  to  C  as  i?  is  to  C : 

it  is  required  to  prove  that  A  is  equal  to  B. 

Proof.  Because,  any  magnitude  greater  than  B  has 
to  C  a  greater  ratio  than  B  has  to  C,  (Prop.  2.) 

and  A  has  to  C  the  same  ratio  as  B  to  C, 
A  cannot  be  greater  than  B. 

Again,  because  any  magnitude  less  than  B  has  to  C  a 

less  ratio  than  B  to  C,  (Prop.  2.) 

and  A  has  to  C  the  same  ratio  as  B  to  C, 

A  cannot  be  less  than  B. 

Therefore  A  must  be  equal  to  B. 

Wherefore,  if  the  ratio  of  the  first  ^  &c. 

*  Algebraically,     li  a  :  c  =  h  :  c,  then  a  =  6. 


336  BOOK   F. 

PROPOSITION  4.     Part  1.     [Euclid  Elem.  Prop.  10.] 

If  the  ratio  of  the  first  of  three  magnitudes  to  the  third 
he  greater  than  the  ratio  of  the  second  to  the  third,  the  fi^'st 
magnitude  is  greater  than  the  second*. 

Let  A,  B,  G  be  three  magnitudes  of  the  same  kind,  such 
that  A  has  to  C  a  greater  ratio  than  B  to  C : 
it  is  required  to  prove  that  A  is  greater  than  B. 

Proof.  Because  the  ratio  of  ^  to  C  is  greater  than 
that  of  B  to  (7, 

there  are  some  equimultiples,  say  the  m*'^  multiples,  of 
A,  B  and  some  multiple,  say  the  n^^  multiple,  of  (7,  such 
that  m  times  A  is  greater  than  n  times  C, 
and  m  times  B  not  greater  than  n  times  (7;  (Def.  7.) 

therefore  there  is  some  number  m  such  that  m  times  A  is 
greater  than  m  times  B', 

therefore  A  is  greater  than  B. 

Wherefore,  if  the  ratio  of  the  first  ^  &c. 

PROPOSITION  L     Part  2.     [Euclid  Elem.  Prop.  10.] 

If  the  ratio  of  the  first  of  three  m>agnitudes  to  the  second 
he  greater  than  the  ratio  of  the  first  to  the  third,  the  second 
magnitude  is  less  than  the  thirdj. 

Let  A,  B,  (7  be  three  magnitudes  of  the  same  kind,  such 
that  A  has  to  B  a,  greater  ratio  than  ^  to  C: 
it  is  required  to  prove  that  B  is  less  than  G. 

Proof.  Because  the  ratio  of  ^  to  J5  is  greater  than  that 
of  A  to  (7,  there  is  some  multiple,  say  the  m^^  multiple,  of 
A,  and  there  are  some  equimultiples,  say  the  n^^  multiples, 
of  B,  G,  such  that  m  times  A  is  greater  than  n  times  B, 

and  m  times  A  not  greater  than  n  times  G;  (Def.  7.) 
therefore  there  is  some  number  n  such  that  n  times  G  is 
greater  than  n  times  B ; 

therefore  G  is  greater  than  B. 
Wherefore,  if  the  ratio  of  the  first,  &c. 

*  Algebraically.     If  a  :  ob  :  c,  then  a>b. 
+  Algebraically.    If  a  :  6>a  :  c,  then  6<c. 


FMOFOSITIONS  4,  5.  337 


PROPOSITION  5.     [Euclid  Elem.  Prop.  11.] 

Ratios,  which  are  equal  to  the  same  ratio,  are  equal  to 
one  another*. 

Let  A,  B,  6',  Z),  E,  F  be  six  magnitudes,  such  that 
^  is  to  J?  as  C  to  B, 
and  also  £"  is  to  ^  as  C  to  i> : 
it  is  required  to  prove  that 

^  is  to  ^  as  E  to  F. 

Proof.     Because  ^  is  to  ^  as  (7  to  D, 
in  times  A  and  rti  times  G 
are  both  greater  than,   both  equal  to  or  both  less  than 
n  times  B  and  n  times  D  respectively, 
for  all  values  of  m  and  n\  (Def.  5.) 

and  because  jE^  is  to  ^  as  C  to  D, 
m  times  E  and  7?i  times  C 
are  both  greater  than,  both  equal  to,  or  both  less  than 
71  times  F  and  n  times  D  respectively, 
for  all  values  of  m  and  n', 

therefore  m  times  A  and  Tti  times  E 
are  both  greater  than,  both  equal  to  or  both   less  than 
n  times  B  and  n  times  F  respectively, 
for  all  values  of  7n  and  n; 

therefore  ^  is  to  i?  as  ^  to  F.  (Def.  5.) 

Wherefore,  ratios,  which  are  equal,  &c. 

*  Algebraically,     li  a  :h=^c  -.  dy  and  e  •.f=c\d, 
then  a  :  h  =  e  :/. 


338  BOOK    V, 


PROPOSITION  6.     [Euclid  Elem.  Prop.  12.] 

If  any  riumber  of  ratios  be  eqical,  each  ratio  is  equal  to 
the  ratio  of  the  sum  of  the  antecedents  to  the  sum  of  the  con- 
sequents^* 

Let  A^  -S,  C,  i),  E^  F  be  any  number  of  magnitudes 
of  the  same  kind,  such  that  the  ratios  of  A  to  B,  C  to  J), 
E  to  F  are  equal : 

it  is  required  to  prove  that 

-4  is  to  ^  as  the  sum  of  A,  C,  E  to  the  sum  of  B,  I),  F. 

Construction.  Take  any  equimultiples,  say  the  m^^ 
multiples,  oi  A,  C,  E^  and  any  equimultiples,  say  the  n^^ 
multiples,  of  B,  i>,  F. 

Proof.     Because  ^  is  to  ^  as  C  to  D,  and  also  as  E  to  F^ 
therefore  m  times  A,  tn  times  C  and  m  times  E 
are  simultaneously  all  greater  than,  all  equal  to  or  all  less 
than  n  times  B^  n  times  D  and  n  times  F  respectively, 

for  all  values  of  in  and  n\  (Def.  5.) 

therefore  m  times  A  and  m  times  the  sum  of  A,  C,  E 
are  simultaneously  all  greater  than,  all  equal  to  or  all  less 
than  n  times  B  and  n  times  the  sum  of  B,  D,  F  respectively, 
for  all  values  of  m  and  n. 
Therefore 
A  is  to  B  as  the  sum  of  A,  (7,  E  to  the  sum  of  B,  I),  F. 

(Def.  5.) 
Wherefore,  if  any  number  of  ratios,  &c. 

Corollary.  The  ratio  of  two  mojgnitudes  is  equal  to  tlie 
ratio  of  any  two  equimultiples  ofthem^. 

*  Algebraically,    li  a:h  =  c  :  d=e  :f, 
then  a  :  h  —  aJrC-\-e  :  &  +  flf+/. 

t  Algebraically,    a  :h=ma  imh. 


PROPOSITIONS  6,  7.  339 


PROPOSITION  7.     [Euclid  Elem.  Prop.  13.] 

If  the  first  of  three  ratios  be  equal  to  the  second  and  the 
second  greater  than  tJie  third,  the  first  is  greater  than  the  third*- 

Let  A,  Bj  Cy  />,  E,  F  be  six  magnitudes,  such  that  the 
ratio  of  ^  to  ^  is  equal  to  that  of  C  to  Z>,  and  the  ratio  of 
(7  to  Z>  is  greater  than  that  of  US'  to  F: 

it  is  required  to  prove  that  the  ratio  of  J  to  -5  is  greater 
than  that  of  E  to  F. 

Proof.  Because  the  ratio  of  C  to  i>  is  greater  than 
that  of  E  to  F,  it  is  possible  to  find  some  equimultiples,  say 
the  m^^  multiples,  of  C  and  E,  and  some  equimultiples,  say 
the  ri*'^  multiples,  of  D  and  F,  such  that 

m  times  C  is  greater  than  n  times  D 
and  m  times  E  not  greater  than  n  times  F.     (Def.  7.) 

Again,  because  ^  is  to  -6  as  C  to  i>, 

m  times  A  and  m  times  C 
are  simultaneously  both  greater  than,  both  equal  to,  or  both 
less  than  n  times  B  and  n  times  D  respectively, 

for  all  values  of  m  and  n.  (Def.  5.) 

Therefore  for  some  values  of  m  and  n 

m  times  A  is  greater  than  n  times  B 
and  m  times  E  not  greater  than  n  times  F, 
Therefore  the  ratio  of  ^  to  i?  is  greater  than  that  of  E 
to  F.  (Def.  7.) 

Wherefore,  if  the  first  of  three  ratios,  tfec. 

*  Algebraically.    If  a\h  =  C'.d,   and  c'.d>e:f, 
then  a  :  h>e  :/. 


340  BOOK    V. 


PROPOSITION  8.     [Euclid  Elem.  Prop.  14.] 

If  the  first  of  four  'proportionals  of  the  same  kind  he 
greater  than  the  third,  the  second  is  greater  than  the  fourth; 
if  the  first  be  equal  to  tlie  second,  the  third  is  equal  to  the 
fourth;  if  the  first  he  less  than  the  second,  the  third  is  less 
than  the  fourth*. 

Let  A,  B,  C,  D  be  four  magnitudes  of  the  same  kind 
such  that  A  is  to  B  as  C  to  D: 

it  is  required  to  prove  that,  if  A  be  greater  than  C,  B  is 
greater  than  i>,  and,  if  A  be  equal  to  C,  B  is  equal  to  D, 
and,  if  A  be  less  than  G,  B  is  less  than  D. 

Proof.     First.     Let  A    be  greater  than   C.     Because 
A,  B,  C  are  three  magnitudes  and  A  is  greater  than  C, 
the  ratio  of  ^  to  ^  is  greater  than  that  of  C  to  ^ ; 

(Prop.  2.) 
but  the  ratio  of  ^  to  ^  is  equal  to  that  oi  C  to  D; 
therefore  the  ratio  of  C  to  Z>  is  greater  than  that  of  C  to  ^; 

(Prop.  7.) 
therefore  B  is  greater  than  D.    (Prop.  4,  Part  2.) 

Next.     Let  A  be  equal  to  C. 

Because  A  is  to  B  a^s  G  to  JJ, 

Bisto  A  a.s  Bto  G;         (Del  5.  Note.) 
and  A  is  equal  to  G; 
therefore  ^  is  to  C  as  i)  to  (7; 

therefore  B  is  equal  to  D.  (Prop.  3.) 

Lastly.     Let  A  be  less  than  G. 

Because  A  is  to  B  as  G  to  D, 

G  is  to  D  as  A  to  B;         (Del  5.  Note.) 
therefore,  by  the  first  case, 

if  G  be  greater  than  A,  D  is  greater  than  B, 
that  is,  if  A  be  less  than  G,  B  is  less  than  D. 
Wherefore,  if  the  first,  &c. 

*  Algebraically.     If         a  i  b  =  c  :  d, 
then  a>  =  <:c  according  as  b>  =  <.d. 


PROPOSITIONS  8,9.  341 


PROPOSITION  9.     [Euclid  Elem.  Prop.  16.] 

If  the  Jirst  of  four  magnitudes  of  the  same  kind  he  to 
the  second  as  the  third  to  the  fourth^  then  also  the  first  is  to 
the  third  as  the  second  to  the  fourth*. 

Let  A,  B,  C,  D  be  four  magnitudes  of  the  same  kind 
such  that  A  is  to  ^  as  (7  to  /) : 

it  is  required  to  prove  that  ^  is  to  C  as  J5  to  D. 

Construction.  Take  any  equimultiples,  say  the  m^^ 
multiples,  of  A,  B,  and  any  equimultiples,  say  the  n^^ 
multiples,  of  C,  D. 

Proof.  Because  m  times  ^  is  to  m  times  -6  as  ^1  to  B^ 
and  because  n  times  C  is  to  n  times  i>  as  C  to  ^, 

(Prop.  6.     CoroU.) 

and  A  is  to  B  as  C  to  B:        (Hypothesis.) 

therefore  m  times  ^  is  to  m  times  B  as  n  times  C  to 

n  times  B.  (Prop.  5.) 

Therefore  m  times  A  and  m  times  B 
are  both  greater  than,  both  equal  to  or  both  less  than 

n  times  C  and  n  times  D  respectively, 
for  all  values  of  ??^  and  n  ;  (Prop.  8.) 

therefore  ^  is  to  C  as  ^  to  B.  (Dei.  5.) 

Wherefore,  if  the  first,  &c. 

*  Algebraically.     If  a  :  b  =  c  :  d,   then  a  :  c  =  b  i  d. 


342  BOOK   V. 


PROPOSITION  10.     [Euclid  Elem.  Prop.  17.] 

If  the  suin  of  the  first  and  the  second  of  four  magnitudes 
he  to  the  second  as  the  sum  of  the  third  and  the  fourth  to  the 
fourth^  the  first  is  to  the  second  as  the  third  to  the  fourth*. 

Let  A,  B,  0,  D  be  four  magnitudes,  A  and  B  being  of 
the  same  kind  and  C  and  D  of  the  same  kind,  such  that 

the  sum  of  A  and  B  is  to  B  as  the  sum  of  C  and  D  to  D: 
it  is  required  to  prove  that 

^  is  to  -5  as  C  to  D. 


th 


Construction.  Take  any  equimultiples,  say  the  m* 
multiples,  oi  A,  B^  C,  D  and  any  equimultiples,  say  the  n^ 
multiples,  of  ^,  i>; 

then  the  sums  of  m  times  B  and  n  times  B  and  of  m  times 
D  and  n  times  D  are  equimultiples  of  B  and  D  respectively. 

Proof.     Because  the  sum  of  A  and  ^  is  to  ^  as  the  sum 
of  C  and  D  to  Z>,  (Hypothesis.) 

therefore  m  times  the  sum  of  A  and  B 

and  m  times  the  sum  of  C  and  D  are  simultaneously 
both  greater  than,  both  equal  to  or  both  less  than 
the  sum  of  m  and  n  times  B 
and  the  sum  of  m  and  n  times  D  respectively, 

for  all  values  of  m  and  n:  (Def.  5.) 

therefore  m  times  A  and  m  times  G 
are  simultaneously  both  greater  than,  both  equal  to  or  both 
less  than  n  times  B  and  n  times  D  respectively, 
for  all  values  of  m  and  n ; 
therefore  J^  is  to  ^  as  C  to  D.  (Del  5.) 

Wherefore,  if  the  su7n,  &c. 

*  Algebraically.    If  a  +  b  :  b=c  +  d  :  d,  then  a  :  b  =  c  :  d. 


PROPOSITIONS   10,  11.  343 

PROPOSITION  11.     [Euclid  Elem.  Prop.  18.] 

If  the  first  of  four  magnitudes  he  to  the  second  as  the 
third  to  tJie  fourth,  then  the  sum  of  the  first  and  the  second  is 
to  t/ie  second  as  the  sum  of  the  third  and  tlie  fourth  to  the 
fourth*. 

Let  -4,  j8,  C,  i>  be  four  magnitudes,  A  and  B  being  of 
the  same  kind,  and  C  and  D  of  the  same  kind,  such  that 
.4  is  to  ^  as  (7  to  Z> : 

it  is  required  to  prove  that 
the  sum  of  A  and  ^  is  to  J5  as  the  sum  of  C  and  D  to  D. 
Construction.      Take   any  equimultiples,  say  the  mP^ 
multiples,   of    A^    G   and    any    equimultiples,    say    the    n^^ 
multiples,  of  B,  D. 

Proof.     Because  ^  is  to  -6  as  C  to  Z),        (Hypothesis.) 
therefore  m  times  A  and  m  times  G 
are  simultaneously  both  greater  than,  both  equal  to  or  both 
less  than  n  times  B  and  n  times  B  respectively, 
for  all  values  of  m  and  n\  (Def.  5.) 

therefore  m  times  the  sum  of  A  and  B 
and  m  times  the  sum  of  G  and  D 
are  simultaneously  both  greater  than,  both  equal  to  or  both 
less  than  the  sum  of  m  and  n  times  B 

and  the  sum  of  m  and  n  times  D  respectively, 
for  all  values  of  m  and  n. 
And  it  is  manifest  that 

m  times  the  sum  of  A  and  B 
and  m  times  the  sum  of  G  and  D 
are  simultaneously  both  greater  than 

])  times  B  and  'p  times  D  respectively, 
for  all  values  of  m  and  />,  where  m  is  not  less  than  p. 
Therefore  m  times  the  sum  of  A  and  B 
and  m  times  the  sum  of  G  and  D 
are  simultaneously  both  greater  than,  both  equal  to  or  both 
less  than  p  times  B  and  p  times  I)  respectively, 
for  all  values  of  m  and  jo; 
therefore  the  sum  of  A  and  jB  is  to  ^  as  the  sum  of  G 
and  D  to  D.  (Def.  5.) 

Wherefore,  if  tJie  first,  kc. 

*  Algebraically.     If  a  :  &  =  c  :  d,  then  a  +  6  :  &=c  +  d  :  d. 


344  BOOK    V. 


PROPOSITION    12.     [Euclid  Elem.  Prop.  19.] 

If  the  sum  of  the  first  and  the  second  of  four  magnitudes 
he  to  the  sum  of  the  third  and  the  fourth  as  the  second  to  the 
fourth^  the  first  is  to  the  second  as  the  third  to  the  fourth'*' . 


Let  A^  B,  C,  D  he  four  magnitudes  of  the  same  kind, 
such  that 
the  sum  of  A  and  B  is  to  the  sum  of  C  and  D  a.s  B  to  D  : 
it  is  required  to  prove  that  A  is  to  ^  as  (7  to  Z>. 

Proof.     Because 
the  sum  of  A  and  B  is  to  the  sum  of  C  and  i)  as  i5  to  J), 
the  sum  of  A  and  -5  is  to  ^  as  the  sum  of  C  and  D  to  B. 

(Prop.  9.) 
Therefore  ^  is  to  ^  as  C  is  to  B.      (Prop.  10.) 

Wherefore,  if  the  sum,  &c. 


Algebraically.     If   a  +  b  :  c  +  d  =  b  :  d,  then  a  ;  h  =  c  :  d. 


PROPOSITIONS   12,   13.  345 


PROPOSITION  13.     [Euclid  Elem.  Prop.  20.] 

If  the  first  of  six  magnitudes  be  to  the  second  as  the  fourth 
to  the  fifths  and  the  second  he  to  the  third  as  the  fifth  to  the 
sixth,  then  the  first  and  the  fourth  are  hoth  greater  than, 
both  equal  to,  or  both  less  tJian  the  third  and  the  sixth 
respectively"^. 

Let  A,  B,  G,  D,  E,  i^be  six  magnitudes,  A,  B,  C  being 
of  the  same  kind  and  D,  E,  F  of  the  same  kind,  such  that 

A  is  to  ^  as  i)  to  E,  and  ^  is  to  (7  as  ^  to  i^ : 
it  is  required  to  prove  that  A  and  D  are  both  greater  than, 
both  equal  to  or  both  less  than  C  and  F  respectively. 

Proof.     First,  let  A  be  greater  than  C. 

Because  ^  is  to  ^  as  i>  to  E,      (Hypothesis.) 
and  the  ratio  of  ^  to  ^  is  greater  than  that  of  G  to  B, 

(Prop.  2.) 

the  ratio  oi  Dto  E  is  greater  than  that  of  G  to  B;  (Prop.  7.) 

and  because  C  is  to  ^  as  2^  to  E;  (Dei.  5,  Note.) 

the  ratio  oi  D  to  E  is  greater  than  that  of  i^to  E;  (Prop.  7.) 

therefore  D  is  greater  than  F.  (Prop.  4.) 

Secondly,    because   the    magnitudes    are   proportionals 

when  taken  in  the  orders  A,  B,  D,  E ;  B,  G,  E,  F,  they 

are  also  proportionals  when  taken  in  the  orders  E,  E,  A,  B ; 

E,  F,B,G;  (Def.  5,  Note.) 

therefore  by  the  first  case, 

if  D  be  greater  than  F,  A  is  greater  than  G. 
Lastly.     The  magnitudes  are  also  proportionals  when 
taken  in  the  orders  G,  B,  F,E;  B,  A,  E,  D ;  (Def.  5,  Note.) 
therefore  by  the  first  and  second  cases, 

if  G  be  greater  than  A,  F  \^  greater  than  D, 
and  if  F  be  greater  than  D,  G  is  greater  than  A  ; 
therefore  A  and  D  are  both  greater  than,  both  equal  to  or 
both  less  than  G  and  F  respectively. 
Wherefore,  if  the  first,  tfec. 

*  AlgebraicaUy.     If  a  :  h  =  d  :  e  and  b  :  c  =  e  :/,         ^^^^  ^  -    ^ 
then  a>  =  <c  according  as  d>  =  </. 

T.  E.  23 


346  BOOK    V, 


PROPOSITION  14.     [Euclid  Elem.  Prop.  22.] 

If  the  first  of  six  magnitudes  be  to  tlie  second  as  thefourtK 
to  the  fifth,  and  the  second  he  to  the  third  as  the  fifth  to  tJie 
sixth,  then  the  first  is  to  the  third  as  the  fourth  to  the  sixth*. 

Let  A,  B,  C,  D,  E,  F  be  six  magnitudes,  A,  B,  C  being 
of  the  same  kind  and  D,  E^  F  of  the  same  kind,  such  that 

-4  is  to  ^  as  Z)  to  E,  and  B  is  to  C  as  E  to  F: 
it  is  required  to  prove  that  ^  is  to  C  as  i>  to  F. 

Construction.     Take   any  equimultiples,  say  the   m^^ 
multiples,  of  ^,  D, 
and  any  equimultiples,  say  the  n^^^  multiples,  of  B,  E, 
and  any  equimultiples,  say  the  p^^  multiples,  of  C,  F. 

Proof.     Because  ^  is  to  jB  as  />  to  E, 
and  ^  is  to  C  as  ^  to  F; 
therefore  m  times  A  is  to  n  times  B&sm  times  Dion  times  E, 
and  n  times  -5  is  to  p  times  C  as  n  times  E  to  p  times  F. 

(Prop.  1.) 
Therefore  m  times  A  and  m  times  D 
are  both  greater  than,   both  equal  to   or   both  less  than 
]}  times  C  and  p  times  F  respectively, 

for  all  values  of  m  and  jo.  (Prop.  13.) 

Therefore  A  is  to  C  as  J)  to  F.  (Del  5.) 

Wherefore,  if  the  first,  &g. 

Corollary.  Ratios  which  are  duplicate  of  equal  ratios 
are  equal  f. 


*  Algebraically.     It  a  :  b  —  d  :  e  and  b  :  c  =  e  :  f,  then  a  :  c  =  d  :f. 
t  Algebraically.    If  a  :  b  =  b  :  c  and  d  :  e  =  e  :/  and  a  :  b==d  :  e, 
then  a  :  c  =  d  :  /. 


PROPOSITION  U,    15.  347 

PROPOSITION  15. 

If  the  first  of  six  magnitudes  have  to  the  second  a  greater 
ratio  than  the  fourth  to  the  fifth,  and  the  second  have  to  the 
third  a  greater  ratio  than  the  fifth  to  the  sixth,  then  the  first 
lias  to  the  third  a  greater  ratio  than  the  fourth  to  the  sixth'''. 

Let  A,  B,  C,  D,  E,  F  be  six  magnitudes,  A,  B,  C  being 
of  the  same  kind,  and  D,  E,  F  of  the  same  kind,  such  that 
A  has  to  j5  a  greater  ratio  than  D  to  E, 
and  B  has  to  6'  a  greater  ratio  than  E  to  F: 
it  is  required  to  prove  that 

A  has  to  (7  a  greater  ratio  than  D  to  F. 
Construction.     Because  the  ratio  of  ^  to  -5  is  greater 
than    that   of   D  to  E,   it  is  possible  to  find  some  equi- 
multiples, say  the  m*^  multiples,  of  A  and  i>,  and  some  equi- 
multiples, say  the  n^^  multiples,  of  B  and  E,  such  that 
m  times  A  is  greater  than  n  times  B, 
and  7Ai  times  D  not  greater  than  n  times  E:  (Def.  7.) 
and  because  the  ratio  oi  B  to  C  is  greater  than  that  of  E 
to  F,   it  is  possible  to  find  some   equimultiples,  say  the 
p^^  multiples,  of  B  and  E,  and  some  equimultiples,  say  the 
q^^  multiples,  of  C  and  F,  such  that 

2)  times  B  is  greater  than  q  times  (7, 
and  p  times  E  not  greater  than  q  times  F.    (Def.  7.) 
Let  p  times  m  be  r  and  n  times  q  be  s, 
and  let  n  times  B  he  H  and  p  times  B  be  K. 
Proof.     Because  m  times  A  is  greater  than  n  times  B, 
and  p  times  m  is  r,  and  n  times  B  is  ^, 
therefore  r  times  A  is  greater  than  p  times  ZT ; 
and  because  p  times  ^  is  greater  than  q  times  C, 
and  p  times  B  is  K  and  n  times  g'  is  s, 
therefore  n  times  ^  is  greater  than  s  times  C ; 
and  because  n  times  B  is  ^,  and  />  times  B  is  ^, 
jo  times  H  is  equal  to  n  times  ^; 
therefore  r  times  A  is  greater  than  s  times  (7. 
Similarly  it  can  be  proved  that 

r  times  D  is  not  greater  than  s  times  /'^ ; 
therefore  A  has  to  C  a  greater  ratio  than  Z>  to  F.     (Def.  7.) 
Wherefore,  if  the  first,  &c. 

*  Algebraically.     If  a  :  bxl :  e  and  b  :  oe  :  f, 
then  a  :  c> d  :  /. 

23—2 


348  BOOK 


PROPOSITION  16. 
Ratios,  of  which  equal  ratios  are  duplicate,  are  equal*. 

Let  A,  B,  C,  D,  E,  F  be  six  magnitudes,  A,  B,  C  being 
of  the  same  kind,  and  D,  E,  F  of  the  same  kind,  such  that 
^  is  to  i?  as  ^  to  C, 
and  D  is  to  E  as  E  to  F, 
and  also  ^  is  to  C  as  i)  to  F\ 
it  is  required  to  prove  that 

^  is  to  ^  as  />  to  E. 

Proof.     If  the  ratio  of  ^  to  5  were  greater  than  that 
of  D  to  E, 

then  also,  since  ^  is  to  j5  as  ^  to  C, 
and  Z>  is  to  ^  as  J^"  to  F, 
the  ratio  oi  B  to  C  would  be  greater  than  that  of  E  to  F; 

(Prop.  7.) 

therefore  the  ratio  of  ^  to  C  would  be  greater  than  that  of 

DtoF.  (Prop.  15.) 

Similarly,  if  the  ratio  of  ^  to  ^  were  less  than  D  to  E, 

then  the  ratio  of  A  to  C  would  be  less  than  that  of  D  to  F; 

therefore  ^  is  to  ^  as  i>  to  E. 

Wherefore,  ratios,  of  which  &c. 


*  Algebraically.     U  a  :  c  =  d  :f  and  a  :  b  =  b  :  c  and  d  :  e  =  e  :  /, 
then  a  :  b  =  d  :  e, 

negative  quantities  being  excluded. 


BOOK    YL 


DEFINITIONS. 

It  is  often  convenient  to  speak  of  closed  rectilineal  figures  as  a 
class.  The  wording  of  definition  15  of  Book  i.  (page  11)  implies 
that  the  term  polygon  does  not  include  a  triangle  or  a  quadrilateral. 
This  restriction  for  the  future  will  not  be  maintained,  and  any  closed 
rectilineal  figure,  no  matter  what  the  number  of  its  sides  may  be, 
will  be  included  under  the  term  polygon. 

Definition  1.  When  the  angles  of  one  polygon  taken  in 
order  are  equal  to  the  angles  of  another  taken  in  order,  the 
two  polygons  are  said  to  be  equiangular  to  one  another. 


F 
A 

The  polygons  ABODE,  FGHKL  are  equiangular  to  one  another, 
if  the  angles  at  ^,  5,  C,  D,  E  be  equal  to  the  angles  at  F,  G,  H,  K, 
L  respectively. 

Pairs  of  vertices  A,  F;  B,  G;  &c.,  at  which  the  angles  are  equal,  are 
corresponding  vertices:  and  pairs  of  sides  AB,  FG;  BC,  GH;  &c. 
joining  corresponding  pairs  of  vertices  are  corresponding  sides. 

In  this  definition  there  is  one  more  condition  of  equality  than  is 
necessary.  If  n  - 1  of  the  angles  of  a  polygon  of  n  sides  be  equal  to 
w  -  1  of  the  angles  of  another  polygon  of  n  sides,  the  remaining  angles 
must  be  equal.     (See  I.  Prop.  32,  Coroll.) 

Definition  2.  When  the  ratio  of  a  side  of  one  of  two 
polygo7is,  which  are  equiangular  to  one  another,  to  the  cor- 
responding side  of  the  other  is  the  same  for  all  pairs  of  cor- 
responding sides,  the  polygons  are  said  to  he  similar  to 
one  another. 


350  BOOK   VL 

The  polygons  ABODE,  FGHKL  are  similar  to  one  another,  if 
the  angles  &t  A,  B,  C,  D,  E  be  equal  to  the  angles  at  F,  G,  H,  K,  L 


respectively,  and  if  also  all  the  ratios  of  AB  to  FG,  BG  to   GH, 
CD  to  HK,  DE  to  KL,  EA  to  LF  be  equal  to  one  another. 

It  will  be  seen  hereafter  that  there  are  in  this  definition  three 
more  conditions  of  equality  than  are  necessary.  One  unnecessary 
condition  is  contained  in  the  statement  that  the  polygons  are  equi- 
angular to  one  another.  Other  two  unnecessary  conditions  are  con- 
tained in  the  statement  that  all  the  ratios  are  equal.  For  it  can  be 
proved  that  in  general,  if  in  two  equiangular  polygons  all  but  two  of 
the  ratios  of  corresponding  sides  be  equal,  all  the  ratios  are  equal. 

Definition  3.  If  in  each  of  two  given  finite  straight 
lines  a  point  he  taken  such  that  the  segments  of  the  first  line 
are  in  the  same  ratio  as  the  segments  of  the  second  line^  the 
two  lines  are  said  to  he  cut  proportionally  hy  the  points. 

^'^     P  ^^'    B 

.B  A ^ P 


■i D 


D 

In  the  diagram  (figure  1)  the  points  P,  Q  cut  the  straight  lines 
AB,  CD  proportionally,  if  AP  be  to  PB  as  CQ  to  QD. 

This  definition  is  extended  also  to  the  case,  where  the  points  P 
and  Q  are  in  the  lines  AB,  CD  produced  (figure  2).  It  must  however 
be  noticed  that  both  points  P,  Q  must  be  in  the  lines  themselves,  or 
both  points  in  the  lines  produced :  otherwise  the  lines  are  not  said  to 
be  cut  proportionally. 

In  figure  1  the  points  P,  Q  are  said  to  cut  the  lines  AB,  CD 
intemaUy;  in  figure  2  the  points  P,  Q  are  said  to  cut  the  lines 
AB,  CD  extemaUy. 


DEFINITIONS.  351 

Definition  4.  In  some  cases,  where  one  side  of  a  tri- 
angle is  specially  distinguished  from  the  other  tivo  sides,  that 
side  is  called  the  base  of  the  triangle  aoid  the  perpendicular* 
upon  that  side  from  the  ojyposite  vertex  is  called  the  altitude 
of  the  triangle. 

Similarly,  one  side  of  a  parallelogram  is  sometimes  called 
the  base  and  tlie  perpendicular  distance  between  it  and  the 
opposite  side  the  altitude  of  the  parallelogram. 

Definition  5.  When  a  straight  line  is  divided  into  tivo 
parts,  so  that  the  whole  is  to  one  part  as  that  part  to  the 
other  part,  the  line  is  said  to  be  divided  in  extreme  and 
mean  ratio. 

Definition  6.  The  figure  formed  of  an  arc  of  a  circle 
and  the  radii  drawn  to  its  extremities  is  called  a  sector  of 
the  circle. 

The  angle  between  the  radii,  which  is  subtended  by  the 
arc,  is  called  the  angle  of  the  sector. 

B 

(i)  ^ 


G  A 

If  0  be  the  centre  of  the  circle,  of  which  the  arc  ABC  is  a  part, 
and  OA,  OC  be  radii,  the  figure  OABC  is  a  sector. 

In  figure  1  the  angle  of  the  sector  is  less  than  two  right  angles, 
in  figure  2  the  angle  of  the  sector  is  greater  than  two  right  angles. 

Definition  7.  Points  lying  on  a  straight  line  are  said 
to  be  collinear.     A  set  of  such  p)oints  is  called  a  range. 

Straight  lines  passing  through  a  point  are  said  to  be 
concurrent.  A  set  of  such  lines  is  called  a  pencil.  T]ie 
lines  are  called  the  rays  of  the  pencil  and  the  ])oint  is  called 
the  vertex  of  the  pencil. 

A  set  of  points  ABCD...  lying  on  a  straight  line  is  called  the 
range  ABCD.... 

*  (Seel.  Def.  11,  p.  9.) 


352  BOOK    VL 

A  set  of  straight  lines  drawn  from  a  point  0  to  a  series  of  points 
A,  B,  C,  D...  is  called  the  pencil  0  (ABCD...). 

Definition  8.  Four  points  on  a  straight  line^  such  that 
one  pair  divide  the  straight  line  joining  the  other  pair 
internally  and  externally  in  the  same  ratio,  are  called 
a  harmonic  range. 

In  the  diagram,  if  AP  be  to  PB  as  AQ  to  QB,  then  A,  P,  B,  Q 
is  a  harmonic  range. 

P  O 


B 


Because  AP  is  to  PB  b,9  AQ  to  QB,  therefore  QB  is  to  BP  as  QA  to 
AP  (V.  Def.  5,  Note,  and  V.  Prop.  9),  that  is,  the  points  B,  A  divide 
the  distance  QP  internally  and  externally  in  the  same  ratio. 

The  two  points  which  form  either  pair  of  a  harmonic 
range  are  said  to  be  conjugate  to  one  another. 

The  points  A,  B  and  P,  Q  are  two  pairs  of  conjugate  points. 

Definition  9.  A  pencil  of  four  rays  passing  through 
the  four  points  of  a  harmonic  range  is  called  a  harmonic 
pencil. 

Two  rays,  which  pass  through  a  pair  of  conjugate  points 
of  a  harmonic  range,  are  called  conjugate  rays  of  the  pencil. 

Definition  10.  Four  points  on  a  straight  line,  such  that 
one  pair  divide  the  straight  line  joining  the  other  pair 
internally  and  externally  in  different  ratios,  are  called 
an  anharmonic  range. 

The  ratio  of  the  ratio  of  internal  division  to  the  ratio  of 
external  division  is  called  the  ratio  of  the  anharmonic  range. 

If  two  anharmonic  ranges  have  equal  ratios,  they  are 
called  like  anharmonic  ranges. 

Definition  11.  A  pencil  of  four  rays  passing  through 
the  four  points  of  an  anharmonic  range  is  called  an  anhar- 
monic pencil. 


DEFINITIONS. 


353 


The  ratio  of  the  range  is  also  said  to  he  the  ratio  of  the 
pencil. 

If  two  pencils  have  equal  ratios,  they  are  called  like  an- 
harmonic  pencils. 

Quadrilaterals  are  often  divided  into  three  classes  (1)  convex, 
(2)  re-entrant,  (3)  cross,  the  natures  of  which  appear  from  the  adjoin- 
ing diagrams. 


If  the  sides  of  a  quadrilateral  be  produced  both  ways,  the  character 
of  the  complete  figure  which  is  so  formed  is  independent  of  the  class 
to  which  the  quadrilateral  belongs,  as  is  evident  from  the  adjoining 
diagram. 


Such  a  figure  is  called  a  complete  quadrilateral,  of   which  the 
following  is  a  definition. 

Definition    12.     The  figure  formed   hy  four   infinite 
straight  lines  is  called  a  complete  ciuadrilateral. 

The  straight  line  joining  the  intersection  of  one  pair  of 
lines  to  the  intersection  of  the  other  pair  is  called  a  diagonal. 

There  are  three  such  diagonals.     In  the  last  diagram  PP\  QQ', 
RR'  are  diagonals  of  the  complete  quadrilateral  there  represented. 


354  BOOK    VL 

PROPOSITION  1. 

The  ratio  of  two  triangles  of  the  same  altitude  is  equal 
to  the  ratio  of  tlieir  bases. 

Let  the  triangles  ABC,  ADE  be  two  triangles  of  the 
same  altitude,  that  is,  having  a  common  vertex  A  and  their 
bases  BC,  BE  in  a  straight  line : 

it  is  required  to  prove  that  the  triangle  ABC  is  to  the 
triangle  ADE  as  BC  to  DE. 

Construction.     In  CB  produced,  take  any  number  of 
straight  lines  BE,  EG  each  equal  to  BC,  and  in  DE  pro- 
duced, any  number  EU,  HK,  KL  each  equal  to  DE\ 
and  draw  AE,  AG,  AH,  AK,  AL. 


H       K 


Proof.     Because  BC,  EB,  GE  are  equal, 
the  triangles  ABC,  AEB,  A  GEare  equal.  (I.  Prop.  38,  Coroll.) 
Therefore  the  triangle  AGC   is  the  same  multiple  of   the 
triangle  ABC,  that  GC  is  of  BC. 
Similarly  it  can  be  proved  that 
the  triangle  AD  Lis  the  same  multiple  of  the  triangle  ADE^.. 
that  DL  is  of  DE. 

Again,  if  GC  be  equal  to  DL, 

the  triangle  AGC  is  equal  to  the  triangle  ADL, 

(I.  Prop.  38,  Coroll.) 

and  if  GC  be  greater  or  less  than  DL,  the  triangle  AGC  is 

greater  or  less  respectively  than  the  triangle  ADL. 

Therefore  since  of   the  four  magnitudes  the  triangles 

ABC,  ADE  and  the  lines  BC,  DE,  the  triangle  AGC  and 

the  line  GC  are  any  equimultiples  wliatever  of  the  first  and 


PROPOSITION   1.  355 

the  third,  and  the  triangle  ADL  and  the  liiie  DL  are  any 
equimultiples  whatever  of  the  second  and  tlie  fourth,  and 
it  has  been  proved  that 
the  triangle  AGC   and  GC  are  both  greater  than,   both 

equal  to  or  both  less  than  the  triangle  ADL  and  DL 

respectively ; 

therefore 
the  triangle  ABC  is  to  the  triangle  ADE  as  BG  to  DE. 

(V.  Def.  5.) 
Wherefore,  the  ratio  of  two  triangles,  &c. 

Corollary  1.  TJie  ratio  of  two  triangles  of  equal  alti- 
tudes is  equal  to  the  ratio  of  their  hoses. 

Corollary  2.  The  ratio  of  two  triangles  of  equal  bases 
is  equal  to  the  ratio  of  their  altitudes. 

Corollary  3.  The  ratio  of  two  parallelograms  of  equal 
altitudes  is  equal  to  the  ratio  of  their  ba^es. 

Each  parallelogram  is  double  of  the  triangle  on  the 
same  base  and  of  the  same  altitude.  The  ratio  of  the 
parallelograms  therefore  is  equal  to  the  ratio  of  the  triangles. 

(V.  Prop.  6,  Coroll.) 


EXEECISES. 

1.  The  diagonals  of  a  convex  quadrilateral,  two  of  whose  sides  are 
parallel  and  one  of  them  double  of  the  other,  cut  one  another  at  a 
point  of  trisection. 

2.  The  sum  of  the  perpendiculars  on  the  two  sides  of  an  isosceles 
triangle  from  any  point  of  the  base  is  constant. 

3.  If  straight  hues  AO,  BO,  CO  be  drawn  from  the  vertices  of  a 
triangle  ABC,  and  AO  produced  cut  BC  in  D,  the  triangles  AOB, 
AOC  have  the  same  ratio  a.a  BD,  DC. 

4.  If  in  the  sides  BC,  CA  of  a  triangle  points  D,  E  be  taken,  such 
that  BD  is  twice  DC,  and  CE  twice  EA,  and  the  straight  lines  AD, 
BE  intersect  in  O,  then  the  areas  of  the  triangles  EOA,  AOB,  BODy 
ABC  are  in  the  ratios  of  the  numbers  1,  6,  8,  21. 


356 


BOOK  VI, 


PROPOSITION  2.     Part  1. 

If  a  straight  line  he  parallel  to  one  side  of  a  triangle^ 
it  cuts  the  other  sides  proportionally. 

Let  the  straight  line  DE  be  parallel  to  the  side  BO  of 
the  triangle  ABC,  and  cut  the  sides  AB,  AC  or  these 
sides  produced  in  i),  E  respectively : 

it  is  required  to  prove  that  BD  is  to  DA  as  CE  to  EA. 

Construction.     Draw  BE,  CD. 


Proof.     Because  the  two  triangles  BDE,  CDE  have 
the  side  DE  common,  and  BC  is  parallel  to  DE, 

the  triangles  BDE,  CDE  are  equal.  (I.  Prop.  37.) 

Therefore  the  triangle  BDE  is  to  the  triangle  ADE  as 

the  triangle  CDE  to  the  triangle  ADE.    (V.  Def.  5,  Note.) 

But  the  triangle  BDE  is  to  the  triangle  ADE  as  BD  to 

DA;  (Prop.  1.) 

and  the  triangle  CDE  is  to  the  triangle  ADE  as  CE  to 

EA  ;  (Prop.  1.) 

therefore  BD  is  to  DA  as  CE  to  EA.  (V.  Prop.  5.) 

Wherefore,  if  a  straight  line  &c. 

Corollary.     Because  BC  is  parallel  to  DE  a  side  of 
the  triangle  ADE,  it  follows  that 

DB  is  to  BA  as  EC  to  CA ; 
therefore  also  A B  ifi  to  AD  r^  AC  to  AE. 

(V.  Props.  10,  and  11.) 


PROPOSITION  2.     PART  1.  357 


EXEECISES. 

1.  A  straight  line  drawn  parallel  to  BC,  one  of  the  sides  of 
a  triangle  ABC,  meets  AB  at  D  and  AC  at  E;  if  BE  and  CD  meet  at 
F,  then  the  triangle  ADF  is  equal  to  the  triangle  AEF. 

2.  If  in  a  triangle  ABC  a  straight  line  parallel  to  BC  meet  AB  at 
D  and  ^C  at  E,  and  if  i?E  and  CD  meet  at  jP:  then  AF  produced 
if  necessary  will  bisect  BC  and  DE. 

3.  Through  D,  any  point  in  the  base  of  a  triangle  ABC,  straight 
lines  DE,  DF  are  drawn  parallel  to  the  sides  AB,  AC,  and  meet  the 
sides  at  E,  F:  shew  that  the  triangle  AEF  is  a  mean  proportional 
between  the  triangles  FBD,  EDC. 

4.  If  two  sides  of  a  quadrilateral  be  parallel,  any  straight  line 
drawn  parallel  to  them  will  cut  the  other  sides  proportionally. 

5.  If  a  straight  line  EF,  drawn  parallel  to  the  diagonal  ^  C  of  a 
parallelogram  ABGD,  meet  AD,  DC,  or  those  sides  produced,  in  E 
and  F  respectively,  then  the  triangle  ABE  is  equal  in  area  to  the 
triangle  BCF. 

6.  ABC  is  a  triangle,  and  through  D,  a  point  in  AB,  DE  is  drawn 
parallel  to  BC  meeting  ^C  in  ^.  Through  C  a  line  CF  is  drawn 
parallel  to  BE,  meeting  AB  produced  in  F.  Prove  that  AB  is  a 
mean  proportional  between  AD  and  AF. 

7.  Through  a  given  point  within  a  given  angle  draw  a  straight 
line  such  that  the  segments  intercepted  between  the  point  and  the 
lines  which  form  the  angle  may  have  to  one  another  a  given  ratio. 

8.  Find  a  point  D  in  the  side  ^B  of  a  triangle  ABC  such  that 
the  square  on  CD  is  in  a  given  ratio  to  the  rectangle  AD,  DB. 


358 


J300K   VI. 


PROPOSITION  2.     Part  2. 

If  a  straight  line  cut  two  sides  of  a  triangle  proportion- 
all't/j  it  is  parallel  to  the  third  side. 

Let  the  straiglit  line  DE  cut  the  sides  AB^  AC  oi  the 
triangle  ABC,   or  these  sides  produced,  proportionally  in 
i),  E  respectively,  so  that  BB  is  to  DA  as  CE  to  EA : 
it  is  required  to  prove  that  DE  is  parallel  to  BC. 


Draw  BE,  CD. 


Proof.     Because  BD  is  to  DA  as  CE  to  EA, 

(Hypothesis.) 
and  as  BD  to  DA  so  is  the  triangle  BDE  to  the  triangle 
ADE,  (Prop.  1.) 

and  as  CE  to  EA  so  is  the  triangle  CDE  to  the  triangle 
ADE',  (Prop.  1.) 

therefore  the  triangle  BDE  is  to  the  triangle  A  DE 
as  the  triangle  CDE  to  the  triangle  ADE;      (V.  Prop.  5.) 
i.e.  the  triangles  BDE,  CDE  have  the  same  ratio  to  the 
triangle  ADE ; 
therefore  the  triangles  BDE,  CDE  are  equal. 

(V.  Prop.  3.) 
But  these  triangles  have  a  common  side  DE  and  lie  on  the 
same  side  of  it ; 

therefore  BC  is  parallel  to  DE.    (I.  Prop.  39.) 

Wherefore,  if  a  straight  line  &c. 


PROPOSITION  2.     PART  2.  359 

Corollary.     If  AB  be  to  AD  as  AC  to  AE^ 
it  follows  that  BD  is  to  DA  as  C^  to  EA 

(V.  Props.  10  and  11.) 
and  therefore  BG  is  parallel  to  AD. 


EXEECISES. 

1.  Prove  that  there  is  only  one  point  which  divides  a  given 
straight  line  internally  in  a  given  ratio,  and  only  one  point  which 
divides  a  given  straight  line  externally  in  a  given  ratio. 

2.  If  DBF  be  a  triangle  inscribed  in  a  triangle  ABC  and  have  its 
sides  parallel  to  those  of  ABC,  then  D,  E,  F  must  be  the  middle 
points  of  £C,  CA,  AB. 

3.  From  a  point  Fj  in  the  common  base  of  two  triangles  ACB, 
ADB,  straight  lines  are  drawn  parallel  to  AC,  AD,  meeting  BC,  BD 
&t  F,  G:  shew  that  FG  is  parallel  to  CD. 

4.  If  two  given  distances  PQ,  RS  be  measured  off  on  two  fixed 
parallel  straight  lines,  then  the  locus  of  the  intersection  of  each  of 
the  pairs  PS,  QR  and  PR,  QS  is  a  parallel  straight  line. 

5.  On  three  straight  lines  OAP,  OBQ,  OCR  the  points  are  chosen 
so  that  AB,  PQ  are  parallel  and  BC,  QR  are  parallel ;  prove  that 
AC,  PR  also  are  parallel. 

6.  If  two  opposite  sides  AB,  DC  of  a  quadrilateral  A  BCD  be 
parallel,  any  straight  line  PQ  which  cuts  AD,  BC  proportionally  must 
be  parallel  to  AB  and  DC. 

7.  Take  D,  E,  the  middle  points  of  the  sides  CA,  CB  of  a  triangle ; 
join  D  and  E,  and  draw  AE,  BD,  intersecting  in  0 ;  then  the  areas  of 
the  triangles  DOE,  EOB,  BOA  are  in  continued  proportion. 


360  BOOK   VI 


PROPOSITION  3.     Part  1. 

If  an  angle  of  a  triangle  he  bisected  internally  or 
externally  hy  a  straight  line  which  cuts  the  opposite  side  or 
that  side  produced,  the  ratio  of  the  segmeiits  of  that  side  is 
equal  to  the  ratio  of  the  other  sides  of  the  triangle. 

Let  the  angle  BAG  of  the  triangle  ABC  be  bisected 
internally  or  externally  by  the  straight  line  AD  which  cuts 
in  D  the  opposite  side  BC  (fig.  1)  or  BC  produced  (fig.  2): 
it  is  required  to  prove  that  BD  is  to  DC  as  BA  to  AC. 

Construction.  Through  C  draw  CE  parallel  to  DA  to 
meet  BA  produced  or  BA  in  E;  and  take  in  BA  or  BA 
produced  a  point  F  on  the  side  of  A  away  from  E, 


Proof.     Because  AC  intersects  the  parallels  AD,  EC, 
the  angle  DAC  is  equal  to  the  angle  ACE;  (T.  Prop.  29.) 

and  because  FAE  intersects  the  parallels  AD,  EG, 
the  angle  FAD  is  equal  to  the  angle  AEG.  (I.  Prop.  29.) 
And  the  angle  DAG  is  equal  to  the  angle  FAD; 

(Hypothesis.) 
therefore  the  angle  AEG  is  equal  to  the  angle  AGE; 

therefore  ^C  is  equal  to  AE.      (I.  Prop.  6.) 
Now  because  AD  is  parallel  to  EC,  one  of  the  sides  of 
the  triangle  BEG, 

BD  is  to  DC  as  BA  to  AE ;  (Prop.  2.) 

and  AG  has  been  proved  equal  to  AE; 

therefore  BD  is  to  DC  as  BA  to  AC. 

Wherefore,  if  an  angle  &c. 


PROPOSITION  3.     PART  I.  361 


EXEECISES. 

1.  ^ BO  is  a  triangle  which  has  its  base  BG  bisected  in  D.  BE, 
DF  bisect  the  angles  ADC,  ADB  meeting  AG,  AB  in  E,  F.  Prove 
that  EF  is  parallel  to  BC. 

2.  If  AD  bisect  the  angle  BAG,  and  meet  BG  in  D,  and  DE,  DF 
bisect  the  angles  ADG,  ADB  and  meet  AG,  AB  in  E,  F  respectively, 
then  the  triangle  BEF  is  to  the  triangle  GEF  as  BA  is  to  A  G. 

3.  An  internal  point  0  is  joined  to  the  vertices  of  a  triangle  ABG. 
The  bisectors  of  the  angles  BOG,  GOA,  AOB  meet  BC,  GA,  AB 
respectively  in  D,  E,  F:  prove  that  the  ratio  compounded  of  the 
ratios  AE  to  EG,  CD  to  DB,  and  BF  to  FA  is  unity. 

4.  One  circle  touches  another  internally  at  O.  A  straight  line 
touches  the  inner  circle  at  G,  and  meets  the  outer  one  in  A,  B:  prove 
that  OA  is  to  OjB  as  ^0  to  GB. 

5.  The  angle  ^  of  a  triangle  ABG  is  bisected  by  AD  which  cuts 
the  base  at  D,  and  0  is  the  middle  point  of  BC :  shew  that  CD 
has  the  same  ratio  to  OB  that  the  difference  of  the  sides  has  to 
their  sum. 

6.  AD  and  AE  bisect  the  interior  and  the  exterior  angles  at  A  of 
a  triangle  ABG,  and  meet  the  base  at  D  and  E  ;  and  O  is  the  middle 
point  of  BG:  shew  that  OB  is  a  mean  proportional  between  OD 
and  OE. 

7.  If  A,  B,  G  be  three  points  in  a  straight  line,  and  D  a  point 
at  which  AB  and  BG  subtend  equal  angles,  then  the  locus  of  D  is 
a  circle. 


T.  B.  24 


362  .  BOOK    Vl. 


PROPOSITION  3.     Part  2. 

If  a  straight  line  drawn  through  a  vertex  of  a  triangle 
cut  the  opposite  side  or  that  side  produced,  so  that  the  ratio 
of  the  segments  of  that  side  is  equal  to  the  ratio  of  the  other 
sides  of  the  triangle,  tJie  straight  line  bisects  the  vertical  angle 
internally  or  externally. 

Let  the  straight  line  AD  drawn  through  A  one  of  the 
vertices  of  the  triangle  ^i5C  cut  the  opposite  side  BC  or  BG 
produced  in  D,  so  that  BD  is  to  DC  as  BA  to  AG: 

it  is  required  to  prove  that  AD  bisects  the  angle  at  A 
internally  or  externally. 

Construction.  Through  G  draw  GE  parallel  to  DA  to 
meet  BA  produced  (fig.  1)  or  BA  (fig.  2)  in  E;  and  take  in 
BA  or  BA  produced  a  point  F  on  the  side  of  A  away  from  E. 


Proof.     Because  DA  is  parallel  to  GE  one  of  the  sides 
of  the  triangle  BEG, 

BD  is  to  DG  as  BA  to  AE.  (Prop.  2.) 

And  BD  is  to  DG  as  BA  to  AG',  (Hypothesis.) 

therefore  BA  i^  to  AG  as  BA  to  AE',  (Y.  Prop.  5.) 

therefore  ^^  is  equal  to  AG ',     (Y.  Prop.  3.) 

therefore  the  angle  AGE  is  equal  to  the  angle  AEG. 

(I.  Prop.  5.) 

Again,  because  AG  intersects  the  parallels  AD,  EG, 
the  angle  DAG  is  equal  to  the  angle  AGE;  (I.  Prop.  29.) 

and  because  FAE  intersects  the  parallels  AD,  EC, 
the  angle  FAD  is  equal  to  the  angle  AEG',  (I.  Prop.  29.) 

therefore  the  angle  DAG  is  equal  to  the  angle  FAD. 

Wherefore,  if  a  straight  line  &c. 


PROPOSITION  3.     PART  2.  363 


EXERCISES. 


1.  The  bisector  of  the  angle  BAG  of  a  triangle  ABC  meets  BG  in 
D;  a  straight  line  EGF  parallel  to  BG  meets  AB,  AD,  AG  in  E,  G,  F 
respectively;  prove  that  EG  is  to  GF  as  BD  to  DG. 

2.  The  sides  AB,  AG  oi  a.  given  triangle  ABG  are  produced  to 
any  points  D,  E,  so  that  DE  is  parallel  to  BG.  The  straight  line  DE 
is  divided  at  F  so  that  DF  is  to  jPE  as  BD  is  to  C£  :  shew  that  the 
locus  of  i^  is  a  straight  line. 

3.  If  a  chord  of  a  circle  AB  be  divided  at  G  so  that  ^  C  is  to  GB  as 
AP  to  PB,  where  P  is  a  point  on  the  circle  :  then  a  circle  can  be  de- 
scribed to  touch  AB  at  G  and  the  given  circle  at  P. 

4.  ABGD  is  a  quadrilateral :  if  the  bisectors  of  the  angles  at  A 
and  G  meet  in  jBD,  then  the  bisectors  of  the  angles  at  B  and  D  meet 
in^C. 

5.  If  ^,  B,  C,  D  be  four  points  in  order  on  a  straight  line,  such 
that  AB  is  to  BG  as  AD  to  DG,  and  P  be  any  point  on  the  circle 
described  on  BD  as  diameter,  then  PB,  PD  are  the  bisectors  of  the 
angle  APG. 


24—2 


364 


BOOK   VI. 


PROPOSITION  4. 

If  two  triangles  be  equiangular  to  one  another,  they  are 
similar. 

Let  tlie  triangles  ABC,  DEF  be  equiangular  to  one  an- 
other : 
it  is  required  to  prove  that  the  ratios 

AB  to  DE,  BG  to  EF  and  CA  to  FD  are  equal. 

Construction.  Of  the  two  lines  BA,  ED  let  BA  be  the 
greater*.  In  BA  take  BG  equal  to  ED,  and  in  BG  take 
BII  equal  to  EF;  and  draw  GH. 


Proof.     Because  in  the  triangles  GBH,  DEF, 

BG  is  equal  to  ED,  and  BH  to  EF, 

and  the  angle  GBU  to  the  angle  DEF, 

the  triangles  are  equal  in  all  respects;  (I.  Prop.  4.) 

therefore  the  angle  BGH  is  equal  to  the  angle  EDF] 

and  the  angle  EDF  is  equal  to  the  angle  BAG; 

(Hypothesis.) 
Hence  the  angle  BGH  is  equal  to  the  angle  BAG, 

and  AC  is  parallel  to  GH ;  (I.  Prop.  28.) 
therefore  BA  is  to  BG  as  BG  to  BH -, 

(Prop,  2,  Part  1,  Coroll.) 
and  GB  is  equal  to  DE,  and  BH  to  EF-, 
therefore  AB  is  to  DE  as  BG  to  ^i^. 
Similarly  it  can  be  proved  that  either  of   these  ratios  is 
equal  to  the  ratio  GA  to  FD. 

Wherefore,  if  two  triangles  &c. 


*  The  case  when  BA  is  equal  to  ED  has  already  been  dealt  with. 
(I.  Prop.  26.) 


PROPOSITION  4.  365 


EXEKCISES. 

1.  A  common  tangent  to  two  circles  cuts  the  straight  line  joining 
the  centres  externally  or  internally  in  the  ratio  of  the  radii. 

2.  If  AB,  CD,  two  parallel  straight  lines,  be  divided  proportionally 
by  P,  g,  so  that  AP  is  to  PB  as  GQ  to  QD,  then  AG,  PQ,  BD  meet 
in  a  point. 

3.  ABGD  is  a  parallelogram ;  P  and  Q  are  points  in  a  straight 
line  parallel  to  AB;  PA  and  QB  meet  at  R,  and  PD  and  QG  meet  at 
S ;  shew  that  RS  is  parallel  to  AD. 

4.  The  tangents  at  the  points  P  and  Q  of  a  circle  intersect  in  T: 
if  from  any  other  point  R  of  the  circle  the  perpendiculars  RM,  RN 
be  drawn  to  the  tangents  TP  and  TQ,  and  the  perpendicular  RL  be 
drawn  to  the  chord  PQ,  then  RL  is  a  mean  proportional  between 
RM  and  RN. 

5.  A  straight  line,  parallel  to  the  side  BG  of  a  triangle  ABG, 
meets  the  sides  A,B,  AG  (or  those  sides  produced)  at  D  and  E.  On 
DE  is  constructed  a  parallelogram  DEFG,  and  the  straight  lines 
BG,  GF  (produced  if  necessary)  meet  each  other  at  S.  Prove  that 
AS  is  parallel  to  DG  or  EF. 

6.  Inscribe  an  equilateral  triangle  in  a  given  triangle,  so  as  to 
have  one  side  parallel  to  a  side  of  the  given  triangle. 

7.  If  two  triangles  have  their  bases  equal  and  in  the  same 
straight  line,  and  also  have  their  vertices  on  a  parallel  straight 
line,  any  straight  line  parallel  to  their  bases  will  cut  off  equal  areas 
from  the  two  triangles. 

8.  In  a  given  triangle  ABG  draw  a  straight  line  PQ  parallel  to 
AB  meeting  AG,  BG  in  P,  Q,  so  that  PQ  may  be  a  mean  proportional 
between  BQ,  QG. 

9.  Two  circles  intersect  at  A,  and  a  straight  line  is  drawn  bisect- 
ing the  angle  between  the  tangents  at  A.  Prove  that  the  segments  of 
the  line  cut  off  by  the  circles  are  proportional  to  the  radii. 

10.  If  AGB,  BGD  be  equal  angles,  and  DB  be  perpendicular  to  BG 
and  BA  to  AG,  then  the  triangle  DBG  is  to  the  triangle  ABG  as  DG 
is  to  GA . 


366  BOOK   VI. 

PROPOSITION  5. 

If  the  ratios  of  the  three  sides  of  one  triangle  to  the  three 
sides  of  another  triangle  he  equals  the  triangles  are  equi- 
angular to  one  another. 

Let  the  given  triangles  ABC,  DEF  be  such  that  the 
ratios  AB  to  DE,  BG  to  EF  and  CA  to  FD  are  equal: 
it  is  required  to  prove  that  the  triangles  ABC,  DEF  are 
equiangular  to  one  another. 

Construction.  On  the  side  of  EF  away  from  D  draw 
EG  J  FG  making  the  angles  FEG,  EFG  equal  to  the  angles 
GBA,  BOA  respectively.  (I.  Prop.  23.) 


Proof.     Because  in  the  triangles  ABC,  GEF, 
the  angles  ABC,  BOA  are  equal  to  the  angles  GEF,  EFG, 
the  triangles  ABC,  GEF  are  equiangular  to  one  another, 

(I.  Prop.  32.) 

and  therefore  AB  is  to  GE  as  BC  to  EF.  (Prop.  4.) 

And  AB  is  to  DE  as  BC  to  EF;    (Hypothesis.) 

therefore  AB  is  to  GE  as  AB  to  DE;  (V.  Prop.  5.) 

therefore  GE  is  equal  to  DE.       (Y.  Prop.  3.) 

Similarly  it  can  be  proved  that  GF  is  equal  to  DF. 

Then  because  in  the  triangles  DEF,  GEF, 
DE,  EF,  FD  are  equal  to  GE,  EF,  FG  respectively, 

the  triangles  are  equal  in  all  respects;  (I.  Prop.  8.) 
therefore  the  triangles  DEF,  GEF  are  equiangular  to  one 

another ; 
and  the  triangle  GEF  was  constructed  so  as  to  be  equi- 
angular to  the  triangle  ABC ; 
therefore   the   triangles   ABC,    DEF   are   equiangular   to 
one  another. 
Wherefore,  if  the  ratios  &c. 


PROPOSITION  5.  367 


PROPOSITION  5.  A. 


If  one  pair  of  angles  of  two  triangles  he  equal  and  another 
pair  of  angles  he  supplementary,  tJie  ratios  of  the  sides  opposite 
to  these  pairs  of  angles  are  eqvxil. 

Let  ABC,  DBF  be  two  triangles  in  which  the  angles 
ABC,  DBF  are  equal,  and  the  angles  ACB,  DFE  are  sup- 
plementary : 

it  is  required  to  prove  that  AB  is  to  DE  as  ^  C  to  DF. 

Construction.  Of  the  two  angles  ACB,  DFE,  let  ACB 
be  the  less.  With  A  as  centre  and  AC  as  radius  describe  a 
circle  cutting  BC  in  G ;  and  draw  AG. 


Proof.     Because  ^C  is  equal  to  AG, 
the  angle  AGC  is  equal  to  the  angle  ACG ;  (I.  Prop.  5.) 
and  the  angle  AGB  is  the  supplement  of  the  angle  AGC, 
and  the  angle  DFE  is  the  supplement  of  the  angle  ACB ; 

(Hypothesis.) 
therefore  the  angle  AGB  is  equal  to  the  angle  DFE; 
and  the  angle  ABG  is  equal  to  the  angle  DEF ; 

(Hypothesis.) 

therefore  the  triangles  ABG,  DEF  are  equiangular  to  one 

another;  (I.  Prop.  32.) 

therefore  AB  is  to  DE  as  AG  to  DF;  (Prop.  4.) 

and  ^C  is  equal  to  AG ;  (Constr.) 

therefore  AB  is  to  DE  as  AC  to  DF. 

Wherefore,  if  one  pair  of  angles,  kc. 


368  BOOK   VI. 


PEOPOSITION  6. 

Jf  the  ratios  of  two  sides  of  one  triangle  to  two  sides  of 
another  triangle  he  equal,  and  also  the  angles  contahied  hy 
those  sides  he  equal,  the  triangles  are  equiangular  to  one 
another. 

Let  ABC,  DEF  be  two  triangles  in  which  ^i?  is  to  DE 
as  BG  to  EF, 

and  the  angle  ABC  is  equal  to  the  angle  DEF: 
it  is  required  to  prove  that  the  triangles  ABC,  DEF  are 
equiangular  to  one  another. 

Construction.  Of  the  two  lines  BA,  ED  let  BA  be  the 
greater*.  In  BA  take  BG  equal  to  ED,  and  in  BG  take 
BU  equ9,l  to  EF;  and  draw  GH. 


Proof.     Because  BA  is  to  ED  as  BG  to  EF^ 
and  BG  is  equal  to  ED,  and  BH  to  EF, 
therefore  BA  is  to  BG  as  BG  to  BH ; 
therefore  GH  i&  parallel  to  AC,  .(Prop.  2,  Part  2,  Coroll.) 
and  the  angle  BGH  is  equal  to  the  angle  BAG. 

(I.  Prop.  29.) 

Again,  because  in  the  triangles  DEF,  GBH, 
ED  is  equal  to  BG  and  EF  to  BH, 
and  the  angle  DEF  to  the  angle  GBH, 
the  triangles  are  equal  in  all  respects ;  (I.  Prop.  4.) 

*  The  case  when  BA  is  equal  to  ED  has  already  been  dealt  with, 
(I.  Prop.  4.) 


PROPOSITION  6.  369 

therefore  the  angle  EDF  is  equal  to  the  angle  BGH, 
and  therefore  to  the  angle  BAC. 
And  the  angles  at  B  and  E  are  equal ;  (Hypothesis.) 
therefore  the  triangles  ABC,  DEF  are  equiangular  to  one 
another.  (I.  Prop.  32.) 

Wherefore,  if  the  ratios  kc. 


EXERCISES. 

1.  Shew  that  the  locus  of  the  middle  points  of  straight  lines 
parallel  to  the  base  of  a  triangle  and  terminated  by  its  sides  is  a 
straight  line. 

2.  CAB,  CEB  are  two  triangles  having  the  angle  B  common  and 
the  sides  CA,  CE  equal ;  if  BAE  be  produced  to  D  and  ED  be  taken 
a  third  proportional  to  BA,AC,  then  the  triangle  BDG  is  similar  to  the 
triangle  BAG. 

3.  From  a  point  E  in  the  common  base  of  the  triangles  ACB, 
ADB,  straight  lines  are  drawn  parallel  to  AC,  AD,  meeting  BC,  BD  in 
F  and  G  ;  shew  that  FG,  CD  are  parallel. 

4.  C  is  a  point  in  a  given  straight  line  AB,  and  AB  is  produced 
to  0,  so  that  CO  is  a  mean  proportional  between  AO  and  BO.  If  P 
be  any  point  on  a  circle  described  with  centre  0  and  radius  OC,  then 
the  angles  APG,  BPC  are  equal. 

6.  If  a  point  0  be  taken  within  a  parallelogram  ABCD,  such  that 
the  angles  OBA,  ODA  are  equal,  then  the  angles  OAD,  OCD  are 
equal. 

6.  If  two  points  P,  Q,  be  such  that  when  four  perpendiculars  PM, 
Pm,  QN,  Qn  are  dropped  upon  the  straight  lines  AMN,  Amn,  PM  is  to 
Pm  as  QN  to  Qn,  then  P  and  Q  lie  on  a  straight  line  through  A. 

7.  If  on  the  three  sides  of  any  triangle,  equilateral  triangles  be 
described  either  all  externally  or  all  internally,  the  centres  of  the 
circles  inscribed  in  these  triangles  are  the  vertices  of  an  equilateral 
triangle. 

8.  The  straight  line  OP  joining  a  fixed  point  O  to  a  variable 
point  P  on  a  fixed  circle  is  divided  in  Q  in  a  constant  ratio ;  prove 
that  the  locus  of  Q  is  a  circle. 

9.  Given  the  base  and  the  vertical  angle  of  a  triangle,  find  the 
locus  of  the  intersection  of  bisectors  of  sides. 


370  BOOK   VI. 


PROPOSITION  7. 

If  the  ratios  of  two  sides  of  one  triangle  to  two  sides  of 
another  triangle  he  equal,  and  also  the  angles  opposite  to 
one  pair  of  these  sides  he  equal,  the  angles  opposite  to  the 
other  pair  of  sides  are  equal  or  supplementary. 

Let  ABG,  DEF  be  two  triangles,  in  which 
AB  is  to  DE  as  BC  to  EF, 
and  the  angle  BAG  is  equal  to  the  angle  EDF: 
it  is  required  to  prove  that  the  angles  ACB,  DFE  are  either 
equal  or  supplementary. 

Construction.     On  the  side  of  EF  2,yf2iy  from  D, 
draw  EG  making  the  angle  FEG  equal  to  the  angle  GBA, 
and  draw  FG  making  the  angle  EFG  equal  to  the  angle 
EGA.  (I.  Prop.  23.) 


Proof.     Because   the  triangles  ABG,   GEF  are  equi- 
angular to  one  another,  (I.  Prop.  32.) 
AB  is  to  GE  as  AG  to  GF;  (Prop.  4.) 
and  AB  is  to  DE  as  BG  to  EF-,  (Hypothesis.) 
therefore  AB  is  to  GE  as  AB  to  DE,  (Y.  Prop.  5.) 
and  GE  is  equal  to  DE.           (V.  Prop.  3.) 
Now  because  in  the  triangles  GEF,  DEF, 
GE  is  equal  to  DE  and  EF  to  EF, 
and  the  angle  EGF  to  the  angle  EDF', 

(for  each  is  equal  to  the  angle  at  A) 


PROPOSITION  7.  371 

therefore  the  angles  GFU,  DFE  are  either  equal  or  supple- 
mentary; (I.  Prop.  26,  A.) 
and  the  angle  GFE  is  equal  to  the  angle  ACB ; 

(Constr.) 
therefore  the  angles  ACB^  DFE  are  either  equal  or  sup- 
plementary. 

Wherefore,  if  the  ratios  ttc. 

Corollary.  When  two  of  the  ratios  of  a  side  of  one 
triangle  to  the  corresponding  side  of  another  triangle  W^e 
equal,  and  also  the  angles  opposite  to  one  pair  of  these  sides 
equal,  the  triangles  are  equiangular  to  one  another,  provided 
that  of  the  angles  opposite  to  the  second  pair  of  sides, 

(1)  each  be  less  than  a  right  angle, 

(2)  each  be  greater  than  a  right  angle, 
or   (3)  one  of  them  be  a  right  angle. 

(I.  Prop.  2Q  A,  Coroll.) 


EXEKCISE. 

Prove  that,  if  ABCD,  EFGH  be  two  quadrilaterals,  such  that  the 
angles  ABC,  ABC  are  equal  to  the  angles  EFG,  ERG  respectively, 
and  the  ratios  AB  to  EF,  BC  to  EG,  CD  to  GH  are  equal,  and  if  the 
angles  BAD,  FEH  be  both  acute  angles,  then  the  quadrilaterals  are 
similar. 


372  BOOK    VI. 

PROPOSITION  8. 

In  a  right-angled  triangle,  if  a  perpendicular  be  drawn 
froTYi  the  opposite  vertex  to  the  hypotenuse,  the  perpendicular 
is  a  mean  proportional  between  the  segments  of  the  hypo- 
tenuse, and  each  of  the  sides  of  the  triangle  is  a  mean  pro- 
portional between  the  hypotenuse  and  the  segment  of  it 
adjacent  to  that  side. 

*  Let  ABC   be  a  right-angled  triangle,  and   let  AD    be 
drawn  perpendicular  to  the  hypotenuse  BC : 
it  is  required  to  prove  that  BD  is  to  DA  as  AD  to  DC, 
that  BC  is  to  BA  as  BA  to  BD,  and  that  BC  is  to  CA  as 
CA  to  CD. 


Proof.     Because  in  the  triangles  ABC,  DBA, 

the  right  angle  BAC  is  equal  to  the  right  angle  BDA, 

and  the  angle  ABC  is  equal  to  the  angle  DBA, 

therefore  the  triangles  ABC,  DBA  are  equiangular  to  one 

another.  (I.  Prop.  32.) 

Similarly  it  can  be  proved  that  the  triangles  DAC,  ABC  are 

equiangular  to  one  another. 
Therefore  the  triangles  DBA,  DAC  are  equiangular  to  one 
another. 
Now,  because  the  triangles  DBA,  DAC  are  equiangular 
to  one  another, 

BD  is  to  DA  as  AD  to  DC  ]  (Prop.  4.) 

and  because  the  triangles  ABC,  DBA  are  equiangular  to  one 
another, 

BC  is  to  BA  as  BA  to  BD',  (Prop.  4.) 

and  because  the  triangles  ABC,  DAC  are  equiangular  to 
one  another, 

BC  is  to  CA^^AC  to  CD.  (Prop.  4.) 

Wherefore,  in  a  right-angled  triangle  &c. 


PEOFOSITION  8.  373 


EXEECISES. 

1.  If  the  perpendicular  drawn  from  the  vertex  of  a  triangle  to  the 
base  be  a  mean  proportional  between  the  segments  of  the  base,  the 
triangle  is  right-angled. 

2.  If  a  triangle  whose  sides  are  unequal  can  be  divided  into  two 
similar  triangles  by  a  straight  line  joining  the  vertex  to  a  point  in  the 
base,  the  vertical  angle  must  be  a  right  angle. 

3.  If  CD,  CE,  the  internal  and  the  external  bisectors  of  the  angle 
at  C  in  a  triangle  ABC  having  a  right  angle  at  A,  cut  BA  in  D  and 
jK  respectively,  then  AC  is  a  mean  proportional  between  AD,  AE. 

4.  A  perpendicular  AD  is  drawn  to  the  hypotenuse  BC  of  a  right- 
angled  triangle  from  the  opposite  vertex  A :  and  perpendiculars 
DE,  DF  are  drawn  from  D  to  the  sides  AB,  AC ;  prove  that  a  circle 
will  pass  through  the  four  points  B,  E,  F,  G. 

5.  On  the  tangent  to  a  circle  at  A  two  points  C  and  B  are  taken 
such  that  AC  IB  equal  to  CB  :  the  straight  lines  joining  B,  C  to  F,  the 
opposite  extremity  of  the  diameter  through  A,  cut  the  circle  in  D,  E 
respectively ;  prove  that  AE  is  to  ED  as  FA  to  FD. 

6.  A  chord  CD  is  drawn  parallel  to  a  diameter  AB  of  a  circle,  and 
AC,  AD  are  produced  to  cut  the  tangent  a.t  B  in  E,  F  respectively  ; 
prove  that  the  sum  of  the  rectangles  AC,  CE  and  AD,  DF  is  equal  to 
the  square  on  AB. 

7.  If  ^  be  a  point  outside  a  circle  and  B  be  the  middle  point  of 
the  chord  of  contact  of  tangents  drawn  from  A,  and  P,  Q  be  any 
two  points  on  the  circle,  then  PA  is  to  QA  as  PB  to  QB. 

8.  Two  circles  intersect  in  A,  B;  from  B  perpendiculars  BE,  BF 
are  drawn  to  their  diameters  AC,  AD;  prove  that  C,  E,  F,  D  lie  on  a 
circle,  which  is  cut  at  right  angles  by  the  circle  whose  centre  is  A  and 
radius  AB. 

9.  The  circumference  of  one  circle  passes  through  the  centre  of 
another  circle.  If  from  any  point  of  the  former  circle  two  straight 
lines  be  drawn  to  touch  the  latter  circle,  the  straight  line  joining  the 
points  of  contact  is  bisected  by  the  common  chord  of  the  two  circles. 


374  BOOK  VI. 


PROPOSITION  9. 


From  a  given  finite  straight  line  to  cut  off  any  aliqtwt 
part  required. 

Let  AB  be  the  given  finite  straight  line : 
it  is  required  to  cut  off  from  AB  a  given  aliquot  part,  say 
the  ?^*^  part. 

Construction.  From  A  draw  any  straight  line  AC 
making  an  angle  with  AB,  and  in  it  take  any  point  D,  and 
cut  off  ^^  the  same  multiple  oi  AB  that  ^^  is  of  the  part 
to  be  cut  off,  i.  e.  take  AU  equal  to  n  times  AD. 

Draw  BB,  and  draw  DF  parallel  to  it  meeting  AB  inF: 
then  AF  is  the  part  required. 


Proof.     Because  FD  is  parallel  to  BF,  one  of  the  sides 
of  the  triangle  ABE, 

AB  is  to  AF  as  AE  to  AD ;  (Prop.  2,  Part  1,  Coroll.) 
and  AE  is  equal  to  n  times  AD; 
therefore  -4^  is  equal  to  n  times  AF. 
Therefore  AF  is  the  n*^  part  of  AB. 

Wherefore,  yVom  the  given  straight  line  AB,  AF  the  part 
required  has  been  cut  off. 


PROPOSITION  10.  375 


PROPOSITION  10. 

To  divide  a  given  finite  straight  line  similarly  to  a  given 
divided  straight  line. 

Let  ^^  be  a  given  straight  line  and  CD  another  given 
straight  line  divided  in  E : 
it  is  required  to  divide  AB  similarly  to  CD. 

Construction.     Draw  AF  making  an  angle  with  AB\ 
cut  off  A  G,  GH  equal  to  CE^  ED  respectively. 

Draw  HB,  and  draw  GK  parallel  to  HB  meeting  ABinK: 
then  ^^  is  divided  at  K  similarly  to  CD  at  E. 


Proof.  Because  GK  is  parallel  to  HB  one  of  the  sides 
of  the  triangle  AHB^ 

AK  is  to  KB  as  AG  to  GH;  (Prop.  2.) 

and  AG  is  equal  to  CE,  and  GH  to  ED.       (Constr.) 

Therefore  ^Z  is  to  KB  as  CE  to  ED. 
Wherefore,  the  straight  line  AB  has  been  divided  at  K 
similarly  to  the  straight  line  CD  at  E. 

EXERCISES. 

1.  If  three  straight  lines  passing  through  a  point  0  cut  two 
parallel  straight  lines  ABC,  PQR  in  A,  P;  B,  Q;  G,  R,  then  the 
lines  A  0,  PR  are  similarly  divided  in  B,  Q. 

2.  Draw  a  straight  line  through  a  given  point  A,  so  that  the 
perpendiculars  upon  it  from  two  other  given  points  B  and  G  may 
be  in  a  given  ratio. 

3.  Draw  through  two  given  points  on  a  circle  two  parallel  chords 
which  shall  have  a  given  ratio  to  one  another. 


376 


BOOK   VI. 


PROPOSITION  11. 


lines 


To  find  a  third  proportional  to  two  given  finite  straight 


Let  AB,  CD  be  two  given  straight  lines : 
it  is  required  to  find  a  third  proportional  to  AB^  CD. 

Construction.  Draw  from  any  point  P  a  pair  of  straight 
lines  PE,  PF  making  an  angle  with  one  another,  and  from 
PE  cut  off  PG,  GH  equal  to  AB^  CD  respectively  and  from 
PF  cut  off  PK  equal  to  CD. 

Draw  GK  and  draw  HL  parallel  to  GK  meeting  PF  in  L : 
then  KL  is  a  third  proportional  to  AB^  CD. 


Proof.     Because  GK  is  parallel  to  HL  one  of  the  sides 
of  the  triangle  PHL, 

PG  is  to  GH  as  PK  to  KL;  (Prop.  2.) 

and  PG  is  equal  to  AB  and  GH  and  PK  are  each  equal  to 
CD; 

therefore  AB  is  to  CD  as  CD  to  KL. 

Wherefore  to  the  two  given  straight  lines  AB^  CD  a  third 
proportional  KL  has  been  found. 


PROPOSITION  12.  377 


PKOPOSITION  12. 

To  find  a  fourth  proportional  to  three  given  straight  lines. 

Let  AB,  CD,  UFhe  three  given  straight  lines : 
it  is  required  to  find  a  fourth  proportional  to  AB,  CD,  EF. 

Construction.  Draw  from  any  point  P  a  pair  of 
straight  lines  PG,  PH  making  an  angle  with  one  another; 
and  from  PG  cut  off  PK,  KL  equal  to  AB,  CD  respectively, 

and  from  PH  cut  off  PM  equal  to  EF. 

Draw  KM,  and  draw  LN  parallel  to  KM  meeting  PH  in  N: 

then  MN  is  a  fourth  proportional  to  AB^  CD,  EF. 


Proof.     Because  KM  is  parallel  to  LN  one  of  the  sides 
of  the  triangle  PLN, 

PK  is  to  KL  as  PM  to  MN -,  (Prop.  2.) 

and  PK  is  equal  to  AB,  KL  to  CD,  and  PM  to  EF-, 
therefore  ^^  is  to  CD  as  EF  to  MN. 

Wherefore  to  the  three  given  straight  lines  AB,  CD,  EF, 
a  fourth  proportional  MN  has  been  found. 


EXEKCISE. 

1.     0  is  a  point  on  a  straight  line  AB\  find  a  point  B  in  AB 
produced,  such  that  BA  is  to  BB  as  CA  to  GB. 


T.  E.  25 


378 


BOOK   VI. 


PROPOSITION  13. 

To  find  a  mean  proportional  between  two  given  straight 
lines. 

Let  AB,  CD  be  two  given  straight  lines: 
it  is  required  to  find  a  mean  proportional  between  J^  and 
CD. 

Construction.     Draw  any  straight  line   and   from    it 
cut  oQ  EF,  FG  equal  to  AB^  CD  respectively. 

Describe  a  circle  on  EG  as  diameter  and  draw  FII  at 
right  angles  to  EG  meeting  the  circle  in  H: 

then  FH  is  a  mean  proportional  between  AB  and  CD. 
Draw  EH,  HG. 


Proof.     Because  EHG  is  a  semicircle, 

the  angle  EHG  is  a  right  angle;  (III.  Prop.  31.) 
and  because  HF  is  the  perpendicular  from  H  on  the  hypo- 
tenuse of  the  right-angled  triangle  EHG, 

EF  is  to  FH  as  FH  to  FG',  (Prop.  8.) 

and  EF  is  equal  to  ^^  and  FG  to  CD; 
therefore  AB  is  to  FH  as  FH  to  CD, 

Wherefore,  between  the  two  given  straight  lines  AB,  CD 
a  mean  proportional  FH  has  been  found. 


PROPOSITION  13.  379 


EXEKCISES. 

1.  Find  a  mean  proportional  between  two  given  straight  lines 
by  the  use  of  the  theorem  of  Proposition  37  of  Book  iii. 

2.  Divide  a  given  finite  straight  line  into  two  parts,  so  that  their 
mean  proportional  may  be  of  given  length. 

3.  Construct  an  isosceles  triangle  equal  to  a  given  triangle  and 
having  the  vertical  angle  equal  to  one  of  the  angles  of  the  given 
triangle. 

4.  Find  a  third  proportional  to  two  given  straight  lines  by  a 
method  similar  to  that  of  Proposition  13. 


25—2 


380  BOOK   VI. 

Definition.  If  the  ratio  of  a  side  of  one  polygon  to  a 
side  of  another  'polygon  he  equal  to  the  ratio  of  an  adjacent 
side  of  the  second  to  an  adjacent  side  of  the  Jtrst^  those  sides 
are  said  to  he  reciprocally  proportional. 

PROPOSITION  14.     Part  1. 

If  two  parallelograms,  which  have  a  pair  of  equal  angles, 
he  equal  in  area,  their  sides  ahout  the  equal  angles  are  reci- 
procally proportional. 

Let  ABGD,  EFGH  be  two  parallelograms,  which  have 
the  angles  at  B  and  H  equal,  and  which  are  equal  in  area : 
it  is  required  to  prove  that  AB  is  to  HG  as  EH  to  BC. 

Construction.  From  AB,  GB  produced  cut  oSBN,  BL 
equal  to  HG,  HE,  and  complete  the  parallelograms  AL,  LN. 

L  M 


Proof.     Because  in  the  parallelograms  LN,  EG, 
LB  is  equal  to  EH,  and  BN  to  HG, 
and  the  angle  LBN  to  the  angle  EHG, 
therefore  the  parallelograms  LN,  EG  are  equal  in  area. 

(I.  Props.  4  and  34.) 
And  the  area  of  EG  is  equal  to  the  area  of  ^C; 
therefore  the  area  oi  AL  is  to  the  area  of  LN  as  the  area 
oi  AL  to  the  area  of  AC. 
And  AB  is  to  BN  as  the  area  of  AL  to  the  area  of  LN, 
and  LB  is  to  BC  as  the  area  of  ^Z  to  the  area  oi  AC  ; 

(Prop.  1,  Coroll.  3.) 
therefore  AB  is  to  BN  as  LB  to  BC,  (V.  Prop.  5.) 
that  is,  AB  is  to  HG  as  EH  to  BC. 

Wherefore,  if  two  parallelograms  &c. 


PROPOSITION  U.  381 

PROPOSITION  14.     Part  2. 

If  two  parallelograms^  which  have  a  pair  of  equal  angles, 
have  their  sides  about  the  equal  angles  reciprocally  propor- 
tional, the  parallelograms  are  equal  in  area. 

Let  ABCD,  EFGH  be  two  parallelograms,  which  have 
the  angles  at  B,  H  equal  and  in  which  ^^  is  to  EG  as  EH 
to  PC: 

it  is  required    to   prove   that   the   parallelograms  A  PCD, 
EFGH  are  equal  in  area. 

Construction.  From  AP,  CB  produced  cut  off  PN,  PL 
equal  to  HG,  HE  and  complete  the  parallelograms  AL,  LN. 


Proof.     Because  in  the  parallelograms  LN,  EG, 
LP  is  equal  to  EH,  and  PN  to  HG, 
and  the  angle  LPN  to  the  angle  EHG, 
therefore  the  parallelograms  LN,  EG  are  equal  in  area. 

(I.  Props.  4  and  34.) 

Because  AP  is  to  HG  as  EH  to  PC, 

and  PN  is  equal  to  HG  and  LP  to  EH, 

therefore  ^^  is  to  PN  as  LP  to  PC. 

And  ^i5  is  to  BN  as  the  area  of  ^Z  to  the  area  of  LN, 

and  LB  is  to  BC  as  the  area  of  AL  to  the  area  oi  AC; 

(Prop.  1,  Coroll.  3.) 
therefore  the  area  of  AL  is  to  the  area  of  LN  as  the  area 
of  ^Z  to  the  area  of  AG.  (V.  Prop.  5.) 

Therefore  the  area  of  LN  is  equal  to  the  area  of  AG. 
And  the  parallelograms  LN,  EG  are  equal  in  area ; 

therefore  the  parallelograms  AC,  EG  are  equal  in  area. 
Wherefore,  if  two  parallelograms  <fec. 


382  BOOK   VI. 


PROPOSITION  15.     Part  1. 

If  two  triangles,  which  have  a  pair  of  equal  angles,  be 
equal  in  area,  their  sides  about  the  equal  angles  are  recipro- 
cally proportional. 

Let  ABC,  DEF  be  two  triangles,  which  have  the  angles 
at  A,  D  equal  and  which  are  equal  in  area : 
it  is  required  to  prove  that  BA  is  to  DF  as  ED  to  AC. 

Construction.  From  BA,  CA  produced  cut  off  AH, 
AG  equal  to  DF,  DE  respectively,  and  draw  BG,  GH. 

E 
D 


Proof.     Because  in  the  triangles  AGH,  DEF, 
GA  is  equal  to  ED,  and  AH  to  DF, 
and  the  angle  GAH  to  the  angle  EDF, 
the  triangles  are  equal  in  all  respects.  (I.  Prop.  4.) 
And  the  area  of  DEF  is  equal  to  the  area  of  ABC ; 
therefore  the  area  of  AGB  is  to  the  area  of  AGH  as  the 

area  of  ABG  to  the  area  of  ABC. 
And  BA  is  to  AH  as  the  area  oi  ABG  to  the  area  of  GAH, 
and  GA  is  to  AC  as  the  area  of  ABG  to  the  area  of  ABC; 

(Prop.  1.) 
therefore  BA  is  to  AH  a,s  G A  to  AC ;  (Y.  Prop.  5.) 
and  AH  is  equal  to  DF  and  GA  to  ED ;  (Constr.) 
therefore  BA  is  to  DF  as  ED  to  AC. 

Wherefore,  if  two  triangles  &c. 


PROPOi:HTION  15.  383 


PROPOSITION  15.     Part  2. 

If  two  triangles,  which  have  a  pair  of  equal  angles,  have 
their  sides  about  the  equal  angles  reciprocally  proportional, 
the  triangles  are  equal  in  area. 

Let  ABC,  DEF  be  two  triangles,  which  have  the  angles 
2X  A,  D  equal  and  in  which  BA  is  to  DF  as  ED  to  AC'. 
it  is  required  to  prove  that  the  triangles  ABC,  DEF  are 
equal  in  area. 

Construction.  From  BA,  CA  produced  cut  o^  AH,  AG 
equal  to  DF,  DE  respectively,  and  draw  BG,  GH. 


Proof.     Because  in  the  triangles  AGH,  DEF, 
GA  is  equal  to  ED,  and  AH  to  DF, 
and  the  angle  GAH  to  the  angle  EDF, 
the  triangles  are  equal  in  all  respects.   (I.  Prop.  4.) 

Because  BA  is  to  DF  as  ED  to  A  C, 

and  AHi^  equal  to  DF  and  GA  to  ED] 

therefore  BA  is  to  AH  as  GA  to  AC. 

And  BA  is  to  AH  as  the  area  of  GBA  to  the  area  of  AGH, 

and  GA  is  to  ^C  as  the  area  of  GBA  to  the  area  of  ABC ; 

(Prop.  1.) 
therefore  the  area  of  GBA  is  to  the  area  of  AGH  as  the 
area  of  GBA  to  the  area  of  ABC;  (Y.  Prop.  5.) 

therefore  the  area  of  AGH  is  equal  to  the  area  of  ABC ; 
and  the  triangles  AGH,  DEF,  being  equal  in  all  respects, 
are  equal  in  area ; 
therefore  the  triangles  ABC,  DEF  are  equal  in  area. 

Wherefore,  if  two  triangles  &c. 


384 


BOOK    VL 


PROPOSITION  16.     Part  1. 


If  four  straight  lines  be  proportionals,  the  rectayigle  con- 
tained hy  the  extremes  is  equal  to  the  rectangle  contained  by 
the  means. 

Let  the  straight  lines ^-J^,  CD,  EF,  GH  be  proportionals, 
so  that  ^5  is  to  CD  as  EF  to  GH: 
it  is  required  to  prove  that  the  rectangle  contained  by  ^^ 

and  GH  is  equal  to  the  rectangle  contained  by  CD  and 

EF. 

Construction.  From  any  point  P  draw  two  straight 
lines  at  right  angles  and  on  one  cut  off  PK,  PL  equal  to 
AB,  CD  respectively;  and  on  the  other  cut  off  PM,  PN 
equal  to  EF^  GH  respectively ;  and  complete  the  rectangles 
ML,  NK. 


G 


L     K 


Proof.     Because  ^^  is  to  CD  as  EF  to  GH, 
and  PK,  PL,  PM,  PJST,  are  equal   to  AB,  CD,  EF,  GH 

respectively, 

therefore  PK  is  to  PL  as  PM  to  PK 
And  the  angle  at  P  is  common  to  the  two  rectangles  NK, 

ML; 
therefore  the  rectangle  KK  is  equal  to  the  rectangle  ML ; 

(Prop.  14,  Part  2.) 
and  KK  is  contained  by  PK,  PN  and  ML  is  contained  by 

PL,  PM. 
Therefore  the  rectangle  contained  by  AB,  GH  is  equal  to 

the  rectangle  contained  by  CD,  EF. 

Wherefore,  if  four  straight  lines  &c. 


PROPOSITION  16. 


385 


PROPOSITION  16.     Part  2. 

If  the  rectangle  co7itained  by  the  first  and  the  fourth  of 
four  given  straight  lines  he  equal  to  the  rectangle  contained 
by  the  second  and  the  third,  the  four  lines  are  proportionals. 

Let  AB,  CD,  EF,  GU  be  four  straight  lines,  such  that 
the  rectangle  contained  hj  AB  and  GH  is  equal  to  the  rect- 
angle contained  by  CD  and  EF: 
it  is  required  to  prove  that  ^^  is  to  CD  as  EF  to  GH. 

Construction.  From  any  point  P  draw  two  straight 
lines  at  right  angles,  and  on  one  cut  off  PK,  PL  equal  to 
AB,  CD  respectively;  and  on  the  other  cut  off  PM,  PN 
equal  to  EF,  6^^  respectively ;  and  complete  the  rectangles 
ML,  NK. 


M 

N 


A- 

n 

B 

E 

F 

G 

H 

L     K 


Proof.     Because  the  rectangle  contained  hj  AB   and 
GH  is  equal  to  the  rectangle  contained  by  CD  and  EF, 
and  PK,  PL,  PM,  PN  are  equal  to  AB,  CD,  EF,  GH  re- 
spectively, 
the  rectangle  contained  by  PK  and  PN  is  equal  to  the 
rectangle  contained  by  PL  and  PM, 
that  is,  the  rectangle  NK  is  equal  to  the  rectangle  ML ; 
therefore  PK  is  to  PL  as  PM  to  PN; 

(Prop.  14,  Part  1.) 
and  PK  is  equal  to  AB,  PL  to  CD,  PM  to  EF  and  PN  to 
GH;  (Constr.) 

therefore  AB  is  to  CD  as  EF  to  GH, 

Wherefore,  if  the  rectangle  &c. 


386  BOOK   VI. 


PROPOSITION  17.     Part  1. 

If  three  straight  lines  be  proportionals,  the  rectangle  con- 
tained by  the  extreynes  is  equal  to  the  square  on  the  mean. 

Let  the  three  straight  lines  AB,  CD,  EF  be  propor- 
tionals, so  that  AB  is  to  CD  as  CD  to  EF: 
it  is  required  to  prove  that  the  rectangle  contained  by  AB 
and  EF  is  equal  to  the  square  on  CD. 

Construction.     Draw  a  straight  line  GH  equal  to  CD. 


G D 

G H 

E F 


Proof.   Because  ^5  is  to  CD  as  CD  to  EF,  (Hypothesis.) 
and  GR  is  equal  to  CD;  (Constr.) 

therefore  AB  is  to  CD  as  GH  to  EF; 
therefore  the  rectangle  contained  by  ^^  and  EF  is  equal 
to  the  rectangle  contained  by  CD  and  GH, 

(Prop.  16,  Part  1.) 
which  is  equal  to  the  square  on  CD,  for  GH  is  equal  to  CD. 

Wherefore,  if  three  straight  lines  &c. 

EXEKCISE. 

1.  A  square  is  inscribed  in  a  right-angled  triangle  ABC,  so  that 
two  corners  D,  E  lie  on  the  hypotenuse  AB  and  the  other  two  on 
the  sides  BC,  CA\  prove  that  the  square  is  equal  to  the  rectangle 
AD,  EB. 


PROPOSITION  17.  387 


PROPOSITION  17.     Part  2. 

If  the  rectangle  contained  by  the  first  and  the  third  of 
three  given  straight  lines  he  equal  to  the  square  on  the  second, 
the  three  straight  lines  are  proportionals. 

Let  AB,  CD,  ^^  be  three  given  straight  lines,  such  that 
the  rectangle  contained  hj  AB  and  EF  is   equal    to   the 
square  on  CD: 
it  is  required  to  prove  that  AB  is  to  CD  as  CD  to  EF. 

Construction.     Draw  a  straight  line  Gil  equal  to  CD. 

A B 

G D 

G R 

E F 


Proof.     Because  GH  is  equal  to  CD, 
the  square  on  CD  is  equal  to  the  rectangle  contained  by 

CD  and  GH; 
therefore  the  rectangle  contained  by  ^^  and  EF,  which  is 
equal  to  the  square  on  CD,  is  equal  to  the  rectangle  con- 
tained by  CD  and  GH; 

therefore  AB  is  to  CD  as  GH  to  EF;  (Prop.  16,  Part  2.) 
and  GH  is  equal  to  CD; 
therefore  AB  is  to  CD  as  CD  to  EF. 

Wherefore,  if  the  rectangle  &c. 


388  BOOK    VL 

PROPOSITION  18. 

On  a  given  straight  line  to  construct  a  polygon  similar 
to  a  given  polygon  so  that  the  given  straight  line  may  cor- 
respond to  a  given  side  of  the  given  polygon. 

Let  AB  be  the  given  straight  line,  and  PQRST  the 
given  polygon: 

it  is  required  to  construct   on  -4^   a   polygon   similar   to 
PQR^T  so  that  AB^  PQ  may  be  corresponding  sides. 

Construction.  From  P,  Q  the  extremities  of  PQ^ 
draw  the  diagonals  PR,  PS,  QS,  QT  of  the  polygon 
PQRST.  At  A,  B  on  the  same  side  oi  AB  make  the  angles 
BAG,  BAD,  BAE  equal  to  the  angles  QPR,  QPS,  QPT 
respectively,  and  the  angles  ABC,  ABD,  ABE  to  the  angles 
PQR,  PQS,  PQT  respectively,  and  draw  CD,  DE: 

then  ABODE  is  a  polygon  constructed  as  required. 


Proof.     Because   the   triangles  ABC,  PQR   are   equi- 
angular to  one  another,  (Constr.) 
^(7  is  to  PR  as  AB  to  PQ;            (Prop.  4.) 
and  because  the  triangles  ABD,  PQS  are  equiangular  to 
one  another,  (Constr.) 
AD  is  to  PS  as  AB  to  PQ  ;            (Prop.  4.) 
therefore  AD  i^  to  PS  s>^  AG  to  PR.    (V.  Prop.  5.) 
Again,  because  in  the  triangles  DAG,  SPR, 
the  ratios  of  -4i>  to  PS  and  AG  to  PR  are  equal, 

and  the  angles  DAG,  SPR  are  equal;     (Constr.) 

therefore  the  triangles  DAG,  SPR  are  equiangular  to  one 

another,  and  GD  is  to  RS  as  AG  to  PR;  (Prop.  6.) 

therefore  GD  is  to  RS  as  AB  to  PQ.    (V.  Prop.  5.) 


PROPOSITION  18.  389 

Again  because  the  triangles  ABC^  PQR  are  equiangular 
to  one  another,  (Const r.) 

the  angle  ACB  is  equal  to  the  angle  PRQ ; 
and  because  the  triangles  DAC,  SPR  have  been  proved 
equiangular  to  one  another, 

the  angle  ACD  is  equal  to  the  angle  PRS ; 
therefore  the  angle  BCD  is  equal  to  the  angle  QRS, 

Similarly  it  can  be  proved  that  the  ratio  of  any  other 
corresponding  pair  of  sides  of  the  polygons  ABODE,  PQRST 
is  equal  to  that  of  AB  to  PQ,  and  that  any  corresponding 
pair  of  angles  are  equal. 

Wherefore,  on  the  given  straight  line  AB,  the  polygon 
ABODE  has  been  constructed  similar  to  the  given  polygon 
PQRST,  so  that  AB,  PQ  are  corresponding  sides. 


EXEECISES. 

1.  Given  the  length  of  the  line  joining  the  middle  point  of  a 
side  of  a  square  with  an  end  of  the  opposite  side;  determine,  by 
any  method,  the  length  of  a  diagonal  of  the  square. 

2.  Inscribe  in  a  given  triangle  a  second  triangle  so  that  its  sides 
may  be  parallel  to  three  given  straight  lines. 

In  how  many  ways  can  this  be  done  ? 

3.  In  a  triangle  ABC  inscribe  a  square  so  that  two  of  its  vertices 
may  be  on  BC  and  the  other  two  on  AB,  AC. 

4.  In  a  semicircle  inscribe  a  square,  so  that  two  corners  may  lie 
in  the  diameter  and  two  on  the  circumference. 

5.  In  a  given  sector  of  a  circle  inscribe  a  square  so  that  two 
corners  may  lie  on  the  arc  and  one  on  each  of  the  bounding  radii. 

6.  In  a  given  sector  inscribe  a  square  so  that  two  corners  may  be 
on  one  of  the  bounding  radii,  one  on  the  other  bounding  radius  and 
one  on  the  arc. 


390 


BOOK    VI. 


PROPOSITION   19. 


Similar  triangles  are  to  one  another  in  the  ratio  duplicate 
of  the  ratio  of  two  corresponding  sides. 

Let  ABC,  BEF  be  similar  triangles  and  BG,  EF  be 
corresponding  sides: 
it  is  required  to  prove  that  the    triangle  ABC   is    to  the 

triangle  DEF,  in  the  ratio  duplicate  of  the  ratio  of  BC 

to  EF. 

Construction.     Find  a  third  proportional  to  BC,  EF 
and  from  BC  cut  off  BG  equal  to  it.     Draw  A  G. 


Proof.     Because  the  triangles  ABC,  BEF  are  similar, 

AB  is  to  BE  as  BC  to  EF ;  (Prop.  4.) 

and  BG  is  to  EF  as  EF  to  BG;  (Constr.) 

therefore  AB  is  to  BE  as  EF  to  BG\  (V.  Prop.  5.) 

and  the  angle  ABG  is  equal  to  the  angle  BEF \ 

therefore  the  triangles  ABG,  BEF  are  equal  in  area. 

(Prop.  15,  Part  2.) 

And  the  triangle  ^^C  is  to  the  triangle  ^^6r  as  BC  to  BG ; 

therefore  the  triangle  ABC  is  to  the  triangle  BEF  as  BC 

to  BG; 

And  because  BC  is  to  EF  as  EF  to  BG, 

BC  has  to  BG  the  ratio  duplicate  of  the  ratio  of  BC  to  EF. 

(V.  Def.  9.) 
Therefore  the  triangle  ABG  has  to  the  triangle  BEF  the 
ratio  duplicate  of  the  ratio  of  BC  to  EF. 
Wherefore,  similar  triangles  &c. 

Corollary.  If  three  straight  lines  be  proportionals,  the 
first  is  to  the  third  as  any  triangle  on  the  first  to  a  similar 
triangle  on  the  second. 


PROPOSITION  19.  391 


EXEKCISES. 

1.  Through  a  point  within  a  triangle  three  straight  lines  are  drawn 
parallel  to  the  sides,  dividing  the  triangle  into  three  triangles  and 
three  parallelograms:  if  the  three  triangles  be  equal  to  each  other 
in  area,  each  is  one-ninth  of  the  original  triangle. 

2.  An  isosceles  triangle  is  described  having  each  of  the  angles  at 
the  base  double  of  the  third  angle:  if  the  angles  at  the  base  be 
bisected,  and  the  points  where  the  lines  bisecting  them  meet  the 
opposite  sides  be  joined,  the  triangle  will  be  divided  into  two  parts 
having  the  same  ratio  as  the  base  to  the  side  of  the  triangle. 

3.  ABC  is  a  triangle,  the  angle  A  being  greater  than  the  angle  B : 
a  point  D  is  taken  in  BG,  such  that  the  angle  CAD  is  equal  to  B. 
Prove  that  CD  is  to  CB  in  the  ratio  duplicate  of  the  ratio  of  AD  to 
AB. 

4.  The  sides  BC,  CA,  AB  of  an  equilateral  triangle  ABC  are 
divided  in  the  points  D,  E,  F  so  that  the  ratios  BD  to  DC,  CE  to  EA 
and  ^i^  to  FB  are  each  equal  to  2  to  1.  Find  the  ratio  of  the  triangle 
DEF  to  the  triangle  ABC. 

6.  If  a  straight  line  AB  be  produced  to  a  point  G  so  that  AB  is 
a  mean  proportional  between  AG  and  GB,  then  the  square  on  AB  is 
to  the  square  on  BG  as  AB  to  the  excess  of  AB  over  BG. 

6.  Find  a  mean  proportional  between  the  areas  of  two  similar 
right-angled  triangles  which  have  one  of  the  sides  containing  the 
right  angle  common. 

7.  Bisect  a  given  triangle  by  a  line  parallel  to  its  base. 

8.  Bisect  a  given  triangle  by  a  line  drawn  perpendicular  to  its 


9.  Divide  a  given  triangle  into  two  parts,  having  a  given  ratio  to 
one  another,  by  a  straight  line  parallel  to  one  of  its  sides. 

10.  ABG  is  a  triangle;  AB  is  produced  to  E:  AD  is  a  straight 
line  meeting  BG  in  D :  BF  is  parallel  to  ED  and  meets  AD  in  F: 
construct  a  triangle  similar  to  ABG  and  equal  to  AEF, 


392  BOOK   VI. 

PROPOSITION  20. 

A  pair  of  similar  polygons  may  he  divided  into  pairs  of 
similar  triangles^  each  pair  having  the  same  ratio  as  the 
polygons. 

Let  ABC  BE,  PQRST  be  a  pair  of  similar  polygons: 
it  is  required  to  prove  that  the  polygons  can  be  divided 
into  pairs  of  similar  triangles. 
Construction.     Take  any  point  L  within  the  polygon 
ABODE,  and  draw  LA,  LB,  LC,  LB,  LE. 

Within  the  polygon  PQRST,  draw  PX,  QX  making  the 
angles  QPX,  PQX.  equal  to  the  angles  BAI-j,  ABL  re- 
spectively, and  draw  XR,  XS,  XT. 


Proof.     Because   the  triangles  LAB,  XPQ   are   equi- 
angular to  one  another,  (Constr.) 
LB  is  to  XQ  as  AB  to  PQ ;            (Prop.  4.) 
and  because  the  polygons  ABGBE,  PQRST  are  similar, 

AB  is  to  PQ  as  BO  to  QR ;  (Def.  2.) 

therefore  LB  is  to  XQ  as  BC  to  QR.  (V.  Prop.  5.) 
Again   because   the   polygons    ABCBE,    PQRST    are 
similar, 

the  angle  ABC  is  equal  to  the  angle  PQR ; 
and  the  angle  ABL  is  equal  to  the  angle  PQX ;    (Constr.) 
therefore  the  angle  LBC  is  equal  to  the  angle  XQR. 
Therefore   the   triangles  LBC,    XQR   are    equiangular   to 
one  another,  (Prop.  6.) 

and  therefore  similar.  (Prop.  4.) 

Similarly  it  can  be  proved  that  the  triangles  LOB, 
LBE,  LEA  are  similar  to  the  triangles  XRS,  XST,  XTP 
respectively. 


PROPOSITIOJS'  20.  393 

Again,  because  ^5  is  to  PQ  as  BG  to  QR; 
and  because  the  triangle  LAB  is  to  the  triangle  XPQ  in 

the  ratio  duplicate  of  the  ratio  of  ^j5  to  PQ, 
and  the  triangle  LBC  is  to  the  triangle  XQR  in  the  ratio 

duplicate  of  the  ratio  of  BG  to  QR,  (Prop.  19.) 

therefore  the  triangle  LAB  is  to  the  triangle  XPQ  as  the 

triangle  LBG  to  the  triangle  XQR.  (V.  Prop.  14.  Coroll.) 
Similarly  it  can  be  proved  that  each  of  the  ratios  of  the 
triangles  LGD,  LDE,  LEA  to  the  triangles  XRS,  XST, 
XTP  respectively  is  equal  to  the  ratio  of  the  triangle  LAB 
to  the  triangle  XPQ. 
Therefore  the  polygon  ABGDE  is  to  the  polygon  PQRST 

as  the  triangle  LAB  to  the  triangle  XPQ.      (V.  Prop.  6.) 

Wherefore,  a  pair  of  similar  polygons  <fec. 

Corollary.     Similar  p>olygon8  are  to  one  anotlter  in  the 
ratio  duplicate  oftlie  ratio  of  two  corresponding  sides. 


EXERCISES. 

1.  If  ABC  be  a  right-angled  triangle  and  CD  be  drawn  perpen- 
dicular to  the  hypotenuse,  then  AD  is  to  DB  as  the  square  on  ^C  to 
the  square  on  CB. 

2.  If  a  straight  line  be  drawn  from  each  corner  of  a  square  to 
the  nearer  point  of  trisection  of  the  next  side  of  the  square  in  order, 
so  as  to  form  a  square,  this  square  will  be  two-fifths  of  the  original 
square.  What  will  be  the  area  of  the  new  square,  if  the  lines  be 
drawn  to  the  further  point  of  trisection  ? 


T.  E.  26 


394 


BOOK   VI 


PROPOSITION  21. 

Polygons   which   are   similar   to  the  same  polygon   are 
similar  to  07ie  another. 

Let  each  of  the  polygons  ABC...,  FGH...^  be  similar  to 
the  polygon  PQR... : 

it  is  required  to  prove  that  ABG...^  FGH...  are  similar  to 
one  another. 


Proof.      Because   the   polygons   ABC...,   PQR...    are 
similar, 

the  angle  ABC  is  equal  to  the  angle  PQR, 

and  AB  is  to  PQ  as  BC  to  QR',  (Def.  2.) 

and  because  the  polygons  FGH...^  PQR...  are  similar, 
the  angle  FGH  is  equal  to  the  angle  PQR, 
and  PQ  is  to  FG  as  QR  to  GH. 
Therefore  the  angle  ABC  is  equal  to  the  angle  FGH, 

and  AB  is  to  FG  as  BC  to  GH.  (V.  Prop.  14.) 
Similarly  it  can  be  proved  that  every  pair  of    corre- 
sponding angles  of  the  polygons  ABC...,  FGH...  are  equal 
and  that  the  ratios  of  all  pairs  of  corresponding  sides  are 
equal. 

Therefore  the  polygons  ABC...,  FGH...  are  similar. 

"Wherefore,  polygons  which  are  similar  <fec. 


PEOFOSITION  21.  395 


EXERCISES. 


1.  Prove  that,  if  ABCD,  EFGH  be  two  quadrilaterals  which  are 
equiangular  to  one  another  and  are  such  that  the  ratios  AB  to  EF, 
and  BC  to  FG  are  equal,  the  quadrilaterals  are  similar. 

2.  Prove  that,  if  ABCD,  EFGH  be  two  quadrilaterals,  which  are 
equiangular  to  one  another  and  are  such  that  the  ratios  of  AB  to  EF 
and  CD  to  GH  are  equal,  the  quadrilaterals  are  similar. 

What  exceptional  case  may  occur  ? 

3.  Prove  that,  if  ABCD,  EFGH  be  two  quadrilaterals  such  that 
the  angles  ABC,  BCD  are  equal  to  the  angles  EFG,  FGH  respectively, 
and  the  ratios  of  AB  to  EF,  BC  to  FG  and  CD  to  GH  are  all  equal, 
the  quadrilaterals  are  similar. 


26—2 


396 


BOOK   VI. 


PROPOSITION  22.     Part  1. 

If /our  straight  lines  be  proportionals,  the  ratio  of  two 
similar  polygons  similarly  described  on  the  first  pair  is  equal 
to  the  ratio  of  two  similar  polygons  similarly  described  on 
the  second  pair. 

Let  the  four  straight  lines  AB,  CD,  EF,  GH  be  propor- 
tionals, and  let  AKLB,  CMND  be  two  similar  polygons 
similarly  described  on  AB,  CD,  and  EPQRF,  GSTUH  be 
two  similar  polygons  similarly  described  on  EF,  GH-. 
it  is  required  to  prove  that  AKLB  is  to  CMND  as  EPQRF 
to  GSTUH, 


Proof.     Because  AB  is  to  Ci>  as  EF  to  GH, 
and  AKLB  has  to  CMND  the  ratio  duplicate  of  the  ratio 
of  AB  to  CD,  (Prop.  20,  Coroll.) 

and  EPQRF  has  to  GSTUH  the  ratio  duplicate  of  the  ratio 
oi  EF  to  GH, 
therefore  AKLB  is  to  CMND  as  EPQRF  to  GSTUH. 

(V.  Prop.  14,  Coroll.) 
Wherefore,  if  four  straight  lilies  &c. 


EXERCISE. 


1.  Perpendiculars  are  let  fall  from  two  opposite  angles  of  a  rect- 
angle on  a  diagonal :  shew  that  they  will  divide  the  diagonal  into 
equal  parts,  if  the  square  on  one  side  of  the  rectangle  be  double  that 
on  the  other. 


PEOPOSITION  22.  397 


PROPOSITION  22.     Part  2. 

If  the  ratio  of  two  similar  polygons  similarly  described 
on  the  first  and  the  second  of  four  straight  lines  he  equal  to 
the  ratio  of  two  similar  polygons  similarly  described  on  the 
third  and  the  fourth^  the  four  straight  lines  are  propor- 
tionals. 

Let  AB,  CD,  EF,  GUhe  four  given  straight  lines,  and 
let  AKLB,  CMND  be  two  similar  polygons  similarly  de- 
scribed on  AB,  CD,  and  EPQRF,  GSTUH  be  two  similar 
polygons  similarly  described  on  EF,  GH,  and  let  AKLB, 
CMND,  EPQRF,  GSTUH  be  proportionals : 
it  is  required  to  prove  that  -4^  is  to  CD  as  EF  to  GH. 


Proof.     Because  AKLB  has  to  CMND  the  ratio  dupli- 
cate of  the  ratio  of  AB  to  CD,  (Prop.  20,  Coroll.) 
and  EPQRF  has  to  GSTUH  the   ratio  duplicate  of  the 
ratio  of  EF  to  GH, 

and  AKLB  is  to  CMND  as  EPQRF  to  GSTUH, 
therefore  AB  is  to  CD  as  EF  to  GH.  (V.  Prop.  1 6.) 

Wherefore,  if  the  ratio  <fec. 


398  BOOK    VI. 

PROPOSITION  23. 

If  two  triangles  have  an  angle  of  the  one  equal  to  an 
angle  of  the  other^  the  ratio  of  the  areas  of  the  triangles  is 
equal  to  the  ratio  compounded  of  the  ratios  of  the  sides  about 
the  equal  angles. 

Let  the  triangles  ABC,  DEF  have  the  angles  at  B  and 
E  equal : 
it  is  required  to  prove  that  the  ratio  of  the  triangle  ABC 

to  the  triangle  DEF  is  equal  to  the  ratio  compounded  of 

the  ratios  AB  to  DE  and  BC  to  EF. 

Construction.  In  AB,  CB  produced  cut  off  BG,  BH 
equal  to  ED,  EF,  and  draw  CG,  GH. 


Proof.     Because  in  the  triangles  GBH,  DEF, 
BG  is  equal  to  ED,  and  BH  to  EF, 
and  the  angle  GBH  to  the  angle  DEF, 
the  triangles  are  equal  in  all  respects.  (I.  Prop.  4.) 
And  because  the  triangle  ABC  is  to  the  triangle  BGC  as 
AB  to  BG,  (Prop.  1.) 

and  the  triangle  GCB  is  to  the  triangle  GBH  as  CB  to  BH; 
therefore  the  triangle  ABC  has  to  the  triangle  GBH  the 
ratio  compounded  of  the  ratios  AB  to  BG  and  CB  to  BH; 

(V.  Def.  8.) 
therefore  the  triangle  ABC  has  to  the  triangle  DEF  the 
ratio  compounded  of  the  ratios  AB  to  DE  and  BC  to  EF. 
Wherefore,  if  two  triangles  &c. 

Corollary.  If  two  parallelograms  have  an  angle  of  the 
one  equal  to  an  angle  of  the  other,  the  ratio  of  the  areas  of 
the  parallelograms  is  equal  to  the  ratio  compounded  of  the 
ratios  of  the  sides  about  the  eq^Mcl  angles. 


PROPOSITION  23.  399 

It  is  proved  in  Proposition  23  that  the  ratio  of  the  triangle  ABC 
to  the  triangle  DEF  is  equal  to  the  ratio  compounded  of  the  ratios 
AB  to  BE  and  BC  to  EF.  Similarly  it  can  be  proved  that  the  ratio 
of  the  triangle  ABC  to  the  triangle  DEF  is  equal  to  the  ratio  com- 
pounded of  the  ratios  BC  to  EF  and  AB  to  BE.  And  since  any  two 
ratios  can  be  represented  by  the  ratios  AB  to  BE  and  BC  io  EF^ 
if  the  lines  be  chosen  of  proper  lengths,  it  follows  that  the  magnitude 
of  the  ratio  compounded  of  two  given  ratios  is  independent  of  the 
order  in  tvhich  tliey  are  compounded. 

Again,  because  the  proof  of  Proposition  23  is  applicable  to  two 
right-angled  triangles,  we  may  assume  the  equal  angles  at  B  and  E  to 
be  right  angles,  in  which  case  the  triangle  ^2^C  is  equal  to  half  the 
rectangle  AB,  BC  and  the  triangle  BEF  is  equal  to  half  the  rectangle 
BE,  EF.  It  follows  that  the  ratio  compounded  of  AB  to  BE,  and 
BC  to  EF  is  equal  to  the  ratio  of  the  rectangle  AB,  BC  to  the  rectangle 
BE,  EFy  or,  in  other  words,  the  ratio  compounded  of  the  ratios  of  two 
pairs  of  lines  is  equal  to  the  ratio  of  the  rectangle  contained  by  the 
antecedents  to  the  rectangle  contained  by  the  consequents. 


EXEECISES. 

1.  A  and  B  are  two  given  points;  AC  and  BB  are  perpendicular 
to  a  given  straight  line  CB:  AB  and  BC  intersect  at  E,  and  EF  is 
perpendicular  to  CB :  shew  that  AF  and  BE  make  equal  angles  with 
CB. 

2.  If  a  triangle  inscribed  in  another  have  one  side  parallel  to  a 
side  of  the  other,  its  area  is  to  that  of  the  larger  triangle  as  the  rect- 
angle contained  by  the  segments  of  either  of  the  other  sides  of  the 
original  triangle  is  to  the  square  on  that  side. 

3.  If  on  two  straight  lines  OABC,  OFEB,  the  points  be  so 
chosen  that  AE  is  parallel  to  BB,  and  AF  parallel  to  CB,  then  also 
BE  is  parallel  to  CE. 

4.  Find  the  greatest  triangle  which  can  be  inscribed  in  a  given 
triangle  so  as  to  have  one  side  parallel  to  one  of  the  sides  of  the 
given  triangle. 

5.  Find  the  least  triangle  which  can  be  described  about  a  given 
triangle. 


400  BOOK  VL 


PROPOSITION  24. 


A  parallelogram  about  a  diagonal  of  another  parallelo- 
gram is  similar  to  it. 

Let  the  parallelogram  AEFG  be  about  the  diagonal  AC 
of  the  parallelogram  ABCD : 
it  is  required  to  prove  that  AEFG  is  similar  to  ABCD. 


Proof.     Because  EF  is  parallel  to  BC, 
the  angles  AEF,  AFE  are  equal  to  the  angles  ABC,  ACB 
respectively,  (I.  Prop.  29.) 

and  therefore  the  triangles  AEF,  ABC  are  equiangular  to 

one  another; 
therefore  the  parallelograms  AEFG,  ABCD  are  equiangular 
to  one  another. 
And  because  the  triangles  AEF,  ABC  are  equiangular 
to  one  another, 

^  J'  is  to  AB  as  EF  to  BC;  (Prop.  4.) 

and  ^i^is  equal  to  AG,  and  BC  to  AD;  (I.  Prop.  34.) 

therefore  also  ^^  is  to  ^jB  as  ^6^  to  ^i>. 

Similarly  it  can  be  proved  that  the  ratios  of  all  pairs  of 

corresponding  sides  of  the  parallelograms  AEFG,  ABCD 

are  equal. 

Therefore  the  parallelograms  are  similar. 

Wherefore,  a  parallelogram,  &c. 


PROPOSITION  24.  401 


EXERCISES. 


1.  Prove  that,  in  the  figure  of  VI.  24,  EG  and  BD  are  parallel. 

2.  Prove  that,  if  in  the  figure  of  VI.  24,  EF,  OF  produced  cut 
CD,  CB  in  fl,  K,  then  HG,  GA,  KE  meet  in  a  point. 

3.  Prove  that,  if  two  similar  quadrilaterals  ABCD,  AEFG  be  so 
placed  that  ABE,  ADG  are  straight  lines,  then  the  points  A,  F,  C  lie 
on  a  straight  hne. 

4.  In  a  given  triangle  inscribe  a  rhombus  which  shall  have  one 
of  its  angular  points  at  a  given  point  in  the  base,  and  a  side  on  that 
base. 

5.  Construct  a  parallelogram  similar  to  a  given  parallelogram,  so 
that  two  of  its  vertices  are  on  one  side  of  a  given  triangle  and  the 
other  vertices  on  the  other  two  sides. 


402 


BOOK   VI, 


PROPOSITION  25. 

To  construct  a  polygon  shnila?'  to  a  given  polygon  and 
equal  to  another  given  polygon. 

Let  ABODE  be  one  given  polygon,  and  FGffX smother: 
it  is  required  to  construct  a  polygon  similar  to  ABODE  and 
equal  to  FGHK. 
Construction.     Construct  on  AB  a  rectangle  AL  equal 
to  ABODE, 

and  on  BL  construct  a  rectangle  LM  equal  to  FGHK. 

(I.  Prop.  45.) 
Find  FQ  a  mean  proportional  between  AB  and  BM, 

(Prop.  13.) 
and  on  FQ  construct  a  polygon  PQRST  similar  to  ABODE, 
so  that  FQ,  AB  are  corresponding  sides:  (Prop.  18.) 

then  FQRST  is  a  polygon  constructed  as  required. 

D 

K 


B 
L 

M 

P  Q 

Proof.     Because  ABODE  is  to  FQRST  in  the  ratio 
duplicate  of  the  ratio  of  ^^  to  FQ, 

and  J  ^  is  to  BM  in  the  ratio  duplicate  of  the  ratio  of  ^^ 
toP^; 

therefore  ABODE  is  to  FQRST  as  AB  to  BM; 

and  ^^  is  to  BAI  as  the  rectangle  AL  to  the  rectangle  LM, 

that  is,  as  ABODE  to  FGHK 

Therefore  FQRST  is  equal  to  FGHK; 

and  it  was  constructed  similar  to  ABODE. 

"Wherefore,  apolygoji  FQRST  has  been  constructed  similar 
to  the  polygon  ABODE  and  equal  to  the  polygon  FGHK. 


EXEECISES. 

1.  Construct  a  sqimre  equal  to  a  given  equilateral  triangle. 

2.  Construct  an  equilateral  triangle  equal  to  a  given  rectangle. 


PMOPOSITION  25.  403 

In  Proposition  4  it  was  proved  that,  if  two  tHangles  be  equiangular 
to  one  another,  they  are  similar.  Hence  the  condition  of  the  equality 
of  the  ratios  of  corresponding  sides,  which  appears  in  Definition  2, 
is  unnecessary  in  the  case  of  two  triangles,  which  are  equiangular  to 
one  another. 

If  we  take  the  case  of  two  polygons  ABODE,  PQRST  of  more 
than  three  sides,  which  are  equiangular  to  one  another,  and  which 
are  such  that  all  but  two  of  the  ratios  of  pairs  of  corresponding  sides 
are  equal,  say  AB  to  PQ,  BC  to  QR,  CD  to  RS,  where  two  adjacent 
sides  are  omitted,  we  can  prove  that  the  polygons  are  similar. 


Take  any  point  L  within  ABCD,  and  draw  LA,  LB,  LC,  LD, 
DA.  Within  the  polygon  PQRS  draw  PX,  QX,  making  the  angles 
QPX,  PQX  equal  to  the  angles  BAL,  ABL  respectively,  and  draw 
XR,  XS,  SP. 

It  can  be  proved,  as  in  Proposition  20,  that  the  triangles  ALB, 

BLC,  CLD  are  similar  to  the  triangles  PXQ,  QXR,  RXS  respectively ; 

therefore  the  angles  ALB,  BLC,  CLD  are  equal  to  the  angles 

PXQ,  QXR,  RXS  respectively,  and 

therefore  the  angle  ALD  is  equal  to  the  angle  PXS; 

also  each  of  the  ratios  LA  to  XP^  LB  to  XQ,  LC  to  XR  and  LD  to  XS 

is  equal  to  the  ratio  of  AB  to  PQ,  and 

therefore  ^L  is  to  PX  as  LD  to  XS. 

Therefore  tTie  triangles  ALD,  PXS  are  similar,  (Prop.  6.) 

and  AD  is  to  PS  as  LA  to  XP. 

Hence  the  two  triangles  AED,  PTS  are  equiangular  to  one  another ; 
therefore  they  are  similar,  and  each  of  the  ratios  DE  to  ST,  EA  to 
TP  is  equal  to  the  ratio  of  AD  to  PS,  (Prop.  4)  which  is  equal  to  the 
ratio  of  LA  to  XP,  and  therefore  to  the  ratio  AB  to  PQ. 

In  this  case  therefore  the  two  polygons  are  similar. 

This  method  reduces  the  case,  where  the  two  sides  whose  ratios 
are  omitted  are  adjacent,  to  the  similar  case  of  quadrilaterals  (Ex.  1, 
page  395).  A  similar  method  will  reduce  the  case,  where  the  two 
sides  whose  ratios  are  omitted  are  not  adjacent,  to  the  similar  case  of 
quadrilaterals  (Ex.  2,  page  395). 

The  two  cases  together  justify  the  remark  on  Definition  2,  page  850. 


404  BOOK   VI. 


PROPOSITION  26. 

If  two  sirnilar  parallelograms  have  a  comynon  angle  and 
be  similarly  jjlaced,  one  is  about  the  diagonal  of  the  other. 

Let  the  parallelograms  ABCD,  AEFG  be  similar  and 
similarly  placed  and  have  a  common  angle  at  A : 
it  is  required  to  prove  that  the  points  A,   F,  C  lie  on  a 
straight  line. 

Construction.     Draw  AF  and  AG. 

A  E 


Proof.     Because  the  parallelograms  AEFG,  ABGD  are 
similar, 

AG  is  to  AB  as  GF  to  DO;  (Def.  2.) 

and  the  angle  AGF  is  equal  to  the  angle  ADC; 

therefore  the  triangles  AGF,  ADC  are  equiangular  to  one 

another.  (Prop.  6.) 

Therefore  the  angle  GAF  is  equal  to  the  angle  DAC, 

i.e.  the  three  points  A,  F,  G  lie  on  a  straight  line. 

Wherefore,  if  two  similar  parallelograms  &c. 


EXEEGISE. 

1.  Inscribe  in  a  given  triangle  a  parallelogram  similar  to  a  given 
parallelogram  so  as  to  have  two  corners  on  one  side  and  one  on  each 
of  the  other  sides  of  the  triangle. 


PEOPOSITION  30.  405 


PROPOSITION  30. 

To  divide  a  given  straight  line  in  extreme  and  mean  ratio. 

Let  AB  be  the  given  straight  line: 
it  is  required  to  divide  it  in  extreme  and  mean  ratio. 

Construction.  Divide  AB  at  the  point  C  into  two 
parts  so  that  the  rectangle  AB^  BC  may  be  equal  to  the 
square  on  AC.  (II.  Prop.  11.) 


B 


Proof.     Because  the  rectangle  AB,  BG  is  equal  to  the 
square  on  AC, 

AB  is  to  AC  as  AC  to  BC.  (Prop.  17.) 

Wherefore,  the  given  straight  line  AB  has  been  divided 
at  C  in  extreine  and  mean  ratio. 


EXERCISES. 

1.  Two  diagonals  of  a  regular  pentagon  which  meet  within  the 
figure  divide  each  other  in  extreme  and  mean  ratio. 

2.  Divide  a  given  straight  line  into  two  parts  so  that  any  triangle 
described  on  the  first  part  may  have  to  a  similar  and  similarly  de- 
scribed triangle  on  the  second  part  the  ratio  which  the  whole  has  to 
the  second  part. 


406 


BOOK  Vl. 


PKOPOSITION  31. 


A  polygon  on  the  hypotenuse  of  a  right-angled  triangle 
is  equal  to  the  sum  of  the  polygons  siTriilarly  described  on  the 
other  sides. 

Let  ABC  be  a  right-angled  triangle  having  the  right 

angle  BAG: 

and  let  BDC,  GEA^  AFB  be  similar  polygons  similarly- 
described  on  BG,  GA,  AB  respectively : 

it  is  required  to  prove  that  the  polygon  BBG  is  equal  to 
the  sum  of  the  polygons  CUA^  AFB. 


Proof.     Because  BBG  has  to  GBA  the  ratio  duplicate 
of  the  ratio  of  BG  to  GA, 
and  the  square  on  BG  has  to  the  square  on  GA  the  ratio 

duplicate  of  the  ratio  of  BG  to  GA;       (Prop.  20,  Coroll.) 
therefore  BBG  is  to  GBA  as  the  square  on  BG  to  the  square 

on  GA; 
therefore  BBG  is  to  the  square  on  BG  as  GBA  to  tlie  square 

on  GA.  (V.  Prop.  9.) 


PROPOSITIOir  31.  4or 

Similarly  it  can  be  proved  that 
BDC  is  to  the  square  on  BC  as  AFB  to  the  square  on  AB. 
Therefore  BDC  is  to  the  square  on  BC 
as  the  sum  of  CEA^  AFB  to  the  sum  of  the  squares  on  CA, 
AB;  (V.  Prop.  6.) 

and  the  square  on  BC  is  equal  to  the  sum  of  the  squares 
on  CA,  AB;  (I.  Prop.  47.) 

therefore  BDC  is  equal  to  the  sum  of  CEA,  AFB. 
Wherefore,  a  polygon  tfec. 


EXERCISES. 


1.  Divide  a  given  finite  straight  line  into  two  parts  so  that  the 
squares  on  them  shall  be  to  one  another  in  a  given  ratio. 

2.  Construct  an  equilateral  triangle  equal   to   the   sum  of  two 
given  equilateral  triangles. 

3.  On  two  given  lines  similar  triangles  are  described ;  construct 
a  similar  triangle  equal  to  the  difference  of  the  given  triangles. 

4.  Construct  a  triangle  equal  to  the  sum  of  three  given  similar 
triangles  and  similar  to  them. 

6.     Construct  a  polygon  equal  to  the  sum   of  any  number  of 
similar  polygons  and  similar  to  them. 


408  BOOK  VI. 


PROPOSITION  32. 

If  two  triangles  have  sides  parallel  in  pairs,  the  straight 
lines  joining  the  corresponding  vertices  meet  in  a  point. 

Let  ABC,  DEF  be  two  triangles  such  that  the  sides 
BC,  GA,  AB  are  parallel  to  the  sides  EF,  FD,  DE  respec- 
tively : 

it  is  required  to  prove  that  the  straight  lines  joining  the 
pairs  of  points  A,  Dy  B,  E;  C,  F  meet  in  a  point. 

Construction.  Draw  AD,  BE  and  let  them  meet  in  G; 
and  draw  GC,  GF. 


Proof.     Because  AB  is  parallel  to  DE, 

the  angles  GAB,  GBA  are  equal  to  the  angles  GDE,  GED 

respectively  ;  (I.  Prop.  29.) 

therefore  the  triangles  GAB,  GDE  are  equiangular  to  one 

another;  (I.  Prop.  32.) 

therefore  GB  is  to  GE  as  BA  to  ED;    (Prop.  4.) 

and  because  the  triangles  ABC,  DEF  are  equiangular  to 

one  another,  (I.  Prop.  34,  Coroll.  2.) 

BA  is  to  ED  as  BC  to  EF;  (Prop.  4.) 

therefore  GB  is  to  GE  as  BC  to  EF;  (V.  Prop.  5.) 

and  the  angle  GBC  is  equal  to  the  angle  GEF  \ 

(I.  Prop.  29.) 
therefore  the  triangles  GBC,  GEE  are  equiangular  to  one 
another;  (Prop.  6.) 

therefore  the  angles  BGC,  EGF  are  equal, 
that  is,  the  points  C,  F,  G  lie  on  a  straight  line, 
or,  in  other  words,  AD,  BE,  GF  meet  in  a  point. 
Wherefore,  if  two  triangles  &c. 


PROPOSITION  32.  409 

It  will  be  seen  at  once  that,  if  in  the  diagram  of  Proposition  32 
AB  be  equal  to  BE,  then  the  straight  lines  AD,  BE  do  not  meet  at 
any  point  at  a  finite  distance,  in  other  words,  they  are  parallel.  Also, 
because  the  triangles  ABC,  DEF  are  similar,  if  AB  be  equal  to  DE, 
then  also  BG  is  equal  to  EF,  and  therefore  BE  and  CF  are  parallel. 

Hence  we  must  consider  the  case  when  the  two  triangles  are 
similarly  placed  and  equal  as  a  special  case  in  which  the  point  of 
intersection  of  the  lines  joining  the  corresponding  vertices  is  at  an 
infinite  distance. 


EXEKCISES. 

1.  If  two  similar  triangles  be  similarly  placed  on  two  parallel 
straight  lines,  the  lines  joining  corresponding  vertices  meet  in  a 
point. 

2.  If  any  two  similar  polygons  have  three  pairs  of  corresponding 
sides  parallel,  the  straight  lines  joining  the  corresponding  vertices 
meet  in  a  point. 

3.  AB  is  a  fixed  diameter  of  a  circle  ABC:  PQ  is  a  straight  line 
parallel  to  AB  and  of  constant  length,  which  moves  so  that  its  middle 
point  traces  out  the  circle  ABC;  find  the  locus  of  the  intersection  of 
AP,  BQ  and  of  AQ,  BP. 

4.  Prove  that,  if  the  corresponding  sides  of  ABCD,  EFGH  two 
squares  be  parallel,  the  straight  lines  AE,  BF,  CG,  DH  pass  through 
a  point,  and  AG,  BH,  CE,  DF  pass  through  another  point. 


T.  E.  27 


410 


BOOK    VI. 


PROPOSITION  33.     Part  1. 

In  equal  circles  angles  at  the  centres  have  the  same  ratio 
as  the  arcs  on  which  they  stand. 

Let  BCD,  MNO  be  two  given  equal    circles,  and    let 
BA  6',  MLN  be  two  angles  at  their  centres : 
it  is  required  to  prove  that  the  angle  BAG  is  to  the  angle 
MLN  as  the  arc  BC  to  the  arc  MN. 

Construction.  From  A  draw  any  number  of  radii  AD, 
AE,  AF  making  the  angles  CAD,  DAE,  EAF  each  equal 
to  the  angle  BAC;  and  from  L  draw  any  number  of  radii 
LO,  LP,  LQ,  LR  making  the  angles  NLO,  OLP,  PLQ,  QLR 
each  equal  to  the  angle  MLN. 


Proof.     Because  the  angles  BAC,  CAD,  DAE,  EAF 
are  all  equal, 

the  arcs  BG,  CD,  DE,  EF  are  all  equal;  (III.  Prop.  26.) 
therefore  the  angle  BAF  and  the  arc  BF  are  equimultiples 
of  the  angle  BAC  and  the  arc  BC. 
Similarly  it  can  be  proved  that 
the  angle  MLR  and  the  arc  MR  are  equimultiples  of  the 
angle  MLN  and  the  arc  MN. 
And,  because  the  circles  are  equal,  if  the  angle  BAF  be 
equal  to  the  angle  MLR, 

the  arc  ^i^is  equal  to  the  arc  MR,  (III.  Prop.  26.) 

and  if  the  angle  BAF  he  greater  or  less  than  the  angle  MLR, 

the  arc  BF  is  greater  or  less  respectively  than  the  arc  MR. 

Therefore  the  angle  BAC  is  to  the  angle  MLN  as  the  arc 

BC  to  the  arc  MN.  (V.  Def.  5.) 

Wherefore,  in  equal  circles  &c. 


PROPOSITION  33.     PART  1.  411 

Corollary.  In  equal  circles  angles  at  the  circumferences 
have  the  same  ratio  as  the  arcs  on  which  they  stand. 

The  angles  at  the  centres  are  double  of  the  angles  at 
the  circumferences,  and  therefore  have  the  same  ratio. 

(V.  Prop.  6,  Coroll.) 

In  the  construction  of  Proposition  33  there  is  nothing  to  limit  the 
magnitude  of  the  multiple  angles  BAF,  31 LR ;  they  may  be  greater 
than  two  right  angles,  greater  than  four  right  angles,  or  greater  than 
any  multiple  of  four  right  angles,  and  at  the  same  time  the  multiple 
arcs  BF,  MR  will  be  greater  than  half  the  circumference  of  the  circle, 
greater  than  the  circumference,  or  greater  than  any  multiple  of  the 
circumference. 

In  the  Third  Book  (page  221)  we  had  occasion  to  remark  that 
the  admittance  of  angles  equal  to  or  greater  than  two  right  angles 
was  not  inconsistent  with  Euclid's  methods.  We  may  now  go 
further  and  say  that  the  admittance  of  angles  without  any  restric- 
tion whatever  on  their  magnitude  is  essential  to  his  method.  The 
validity  of  the  proof  of  this  Proposition  depends  on  the  possibility  of 
choosing  any  multiples  we  please  of  the  angles  BAG,  MLN,  that  is, 
of  taking  the  multiple  angles  BAF^  MLR  as  large  as  we  please. 


27—2 


412  BOOK   VI. 


PROPOSITION  33.     Part  2. 

In  equal  circles^  the  areas  of  sectors  have  the  same  ratio 
as  their  angles. 

Let  BCD,  MNO  be  two  given  equal  circles,  and  let  BAC, 
MLN  be  two  angles  at  their  centres : 

it  is  required  to  prove  that  the  angle  BAG  is  to  the  angle 
MLN  as  the  sector  BAG  to  the  sector  MLN. 

Construction.  From  A  draw  any  number  of  radii  AD, 
AE,  J/' making  the  angles  CAD,  DAE,  EAF  ea>c\\  equal  to 
the  angle  BAC ;  and  from  L  draw  any  number  of  radii 
LO,  LP,  LQ,  LR  making  the  angles  NLO,  OLP,  PLQ, 
QLR  each  equal  to  the  angle  MLN. 


Proof.  Because  the  angle  CAD  is  equal  to  the  angle 
BAC,  it  is  possible  to  shift  the  figure  CAD  so  that  AC  will 
be  on  AB,  and  AD  on  AC  ]  if  this  be  done,  then  the  point 
C  will  coincide  with  B  and  D  with  C, 

and  therefore  the  arc  CD  with  the  arc  BC.  (III.  Prop.  23.) 
Therefore  the  sector  CAD  coincides  with  the  sector  BAC 
and  is  equal  to  it  in  all  respects. 
Similarly  it  can  be  proved  that  the  sectors  DAE,  EAF 
are  each  equal  to  the  sector  BAC. 

Therefore  the  angle  BAF  and  the  sector  BAF  are   equi- 
multiples of  the  angle  BAC  and  the  sector  BAC. 
Similarly  it  can  be  proved  that 
the  angle  MLR  and  the  sector  MLR  are  equimultiples  of 
the  angle  MLN  and  the  sector  MLN. 


FBOPOSITIOK  33.     PART  2.  413 

And  it  can  be  proved  as  before  that,  if  the  angle  BAF  be 
equal  to  the  angle  MLR, 

the  sector  BAF  in  equal  to  the  sector  MLR; 

and,  if  the  angle  BAF  he  greater  or  less  than  the  angle  MLR, 

the  sector  is  greater  or  less  respectively  than  the  sector  MLR; 

therefore  the  angle  BAG  is  to  the  angle  MLN  as  the  sector 

BAC  to  the  sector  MLN.  (V.  Def.  5.) 

Wherefore,  in  equal  circles  tkc. 

Corollary.     In  equal  circles  the  areas  of  sectors  have  the 
same  ratio  as  the  arcs  on  which  they  stand. 


414  BOOK    VI. 

PROPOSITION  34. 

If  an  angle  of  a  triangle  he  bisected  by  a  straight  line 
which  cuts  the  opposite  side,  the  rectangle  contained  by  the 
segments  of  that  side  is  less  than  the  rectangle  contained  by 
the  other  sides  by  the  square  on  the  line. 

Let  the  angle  BAC  of  the  triangle  ABC  be  bisected  by 
the  straight  line  AD,  which  cuts  BC  at  D : 
it  is  required  to  prove  that  the  rectangle  BD,  BC  is  less 
than  the  rectangle  BA,  AC  hy  the  square  on  AD. 

Construction.     Describe  the  circle  ABC;  (IV.  Prop.  5.) 
produce  AD  to  meet  the  circle  at  E,  and  draw  EG. 


Proof.     Because  in  the  triangles  BAD,  BAC, 
the  angle  BAD  is  equal  to  the  angle  EAC, 

(Hypothesis.) 
and  the  angle  ABD  to  the  angle  AEC, 

(III.  Prop.  21.) 

therefore  the  triangles  are  equiangular  to  one  another ; 

therefore  BA  is  to  EA  ^^  AD  to  AC \    (Prop.  4.) 

therefore  the  rectangle  BA,  AC  \s,  equal  to  the  rectangle 

EA,  AD,  (Prop.  16.) 

that  is,  to  the  rectangle  ED,  DA  together  with  the  square 

on  AD.  (II.  Prop.  3.) 

And  the  rectangle  ED,  DA  is  equal  to  the  rectangle  BD,  DC', 

(III.  Prop.  35.) 
therefore  the  rectangle  BD,  DC  is  less  than  the  rectangle 
BA,  AC  hy  the  square  on  AD. 
Wherefore,  if  an  angle  &c. 


PROPOSITION  34.  415 


EXERCISES. 

1.  If  an  angle  of  a  triangle  be  bisected  externally  by  a  straight 
line  which  cuts  the  opposite  side  produced,  the  rectangle  contained 
by  the  segments  of  that  side  is  greater  than  the  rectangle  contained 
by  the  other  sides  by  the  square  on  the  line. 

2.  Prove  that,  if  the  internal  and  the  external  bisectors  of  the 
vertical  angle  of  a  triangle  ABC  cut  BC  in  D  and  E,  then  the  square 
on  BE  is  equal  to  the  difference  of  the  rectangles  EB,  EC  and 
BE,  BG. 

3.  If  I  be  the  centre  of  the  inscribed  circle  of  a  triangle  ABC 
and  AI  produced  cut  the  circumscribed  circle  ABC  in  E,  then  the 
rectangle  contained  by  AI,  IE  is  equal  to  twice  the  rectangle  con- 
tained by  the  radii  of  the  circumscribed  and  the  inscribed  circles, 

(See  Ex.  46,  page  324.) 

4.  If  Ij  be  the  centre  of  the  circle  of  the  triangle  ABC  escribed 
beyond  BC  and  AI^  cut  the  circumscribed  circle  ABC  in  E,  then  the 
rectangle  contained  by  AI-^,  I^E  is  equal  to  twice  the  rectangle  con- 
tained by  the  radii  of  the  circumscribed  and  the  escribed  circles. 

5.  ^^  is  the  base  of  a  triangle  ABC  whose  sides  are  segments 
of  a  line  divided  in  extreme  and  mean  ratio.  CP  the  bisector  of 
the  angle  C,  and  CQ  the  perpendicular  from  C  on  AB  meet  AB  in  P 
and  Q.  Prove  that  the  square  on  CP  is  equal  to  twice  the  rectangle 
contained  by  PQ  and  AB. 


416  BOOK    VI. 


PROPOSITION  35. 


If  a  perpendicular  he  drawn  from  a  vertex  of  a  triangle 
to  tJie  opposite  side,  the  rectangle  contained  hy  the  oilier  sides 
of  the  triangle  is  equal  to  the  rectangle  contained  hy  the 
perpendicular  and  the  diameter  of  the  circle  described  about 
the  triangle. 

Let  AD  be  the  perpendicular  drawn  from  the  vertex  A 
of  the  given  triangle  ABC  to  the  opposite  side  BG \ 
it  is   required  to  prove  that  the  rectangle  contained  by 
AB,  AG  is  equal  to  the  rectangle  contained  by  AD  and 
the  diameter  of  the  circle  described  about  ABG. 

Construction.     Describe  the  circle  ABC;  (IV.  Prop.  5.) 
draw  the  diameter  AB  and  draw  jEC. 


Proof.     Because  in  the  triangles  BAD,  EAC, 
the  angle  ABD  is  equal  to  the  angle  AEG,  (III.  Prop.  21.) 
and  the  angle  ADB  to  the  angle  AGE;  (III.  Prop.  31.) 
therefore  the  triangles  are  equiangular  to  one  another, 

(I.  Prop.  32.) 

and  BA  is  to  EA  as  AD  to  AG;         (Prop.  4.) 

therefore  the  rectangle  BA,  AG  is  equal  to  the  rectangle 

EA,  AD.  (Prop.  16.) 

Wherefore,  if  a  perpendicular  (fee. 


PROPOSITION  35  A.  417 


PROPOSITION  35  A. 


The  ratio  of  twice  the  area  of  a  triangle  to  the  rectangle 
contained  hy  two  of  the  sides  is  equal  to  the  ratio  of  the  thir-d 
side  to  the  diameter  of  the  circumscribed  circle  of  tlie 
triangle. 

Let  ABC  be  a  triangle  : 

it  is  required  to  prove  that  twice  the  area  of  ABC  is  to 
the  rectangle  contained  by  AC,  BC  as  AB  to  the  diameter 
of  the  circle  described  about  ABC. 

Construction.  Describe  the  circle  ABC;  draw  the 
diameter  AJE,  draw  AD  perpendicular  to  BC  and  draw  JSC. 


Proof.     Because  in  the  triangles  BAD,  EAC, 
the  angle  ABD  is  equal  to  the  angle  A  EC,    (III.  Prop.  21.) 
and  the  angle  ADB  to  the  angle  ACE ;  (III.  Prop.  31.) 
therefore  the  triangles  are  equiangular  to  one  another, 

(I.  Prop.  32.) 

and  yli)  is  to  ^(7  as  ^^  to  AE.  (Prop.  4.) 

Therefore  the  rectangle  AD,  BC  is  to  the  rectangle  A  C,  BC 

as  AB  to  AE;  (Prop.  1.) 

and  the  rectangle  AD,  BC  is  equal  to  twice  the  area  of  the 

triangle  ABC ; 
therefore   twice  the  area   of  the  triangle  ABC  is  to  the 
rectangle  AC,  BCas  AB  to  the  diameter  of  the  circle  ABC. 
Wherefore,  the  ratio  &,c. 


418  BOOK   VI. 


ADDITIONAL  PKOPOSITION   1. 

If  a  straight  line  cut  the  three  sides  of  a  triangle  produced  if 
necessary,  the  ratio  compounded  of  the  ratios  of  the  segments  of  the 
sides  taken  in  order  is  equal  to  unity.* 

Let  the  sides  BC,  CA,  AB  ot  the  triangle  ABC  he  cut  by  the 
straight  line  LMN  in  L,  M,  N  respectively. 

Through  C  draw  CZ  parallel  to  LMN  to  meet  ABN  in  Z. 


Because  ZG,  NML  are  parallel, 

AM  is  to  MC  as  AN  to  NZ, 
and  CL  is  to  LB  as  ZN  to  NB ;  (Prop.  2.) 

therefore  the  ratio  compounded  of 

the  ratios  AM  to  MC  and  CL  to  LB  is  equal  to  the  ratio  compounded 
of  the  ratios  AN  to  NZ  and  ZN  to  NB, 

i.e.  the  ratio  AN  to  NB ;  (V.  Def.  8.) 

therefore  the  ratio  compounded  of  the  ratios  AM  to  MC,  CL  to  LB 
and  BN  to  NA  is  equal  the  ratio  compounded  of  the  ratios  AN  to 
NB  and  NB  to  AN,  that  is,  to  the  ratio  AN  to  AN,  i.e.  to  unity. 

(V.  Def.  2.) 


*  This  theorem  is  attribute  1  to  Menelaus,  a  Greek  Geometer,  who 
lived  in  the  latter  part  of  the  first  century  a.d. 


THEOREM  OF  MENELAUS.  419 

Definition.     A  straight  line  drawn  to  cut  a  series  of 
lines  is  often  called  a  transversal. 

The  straight  line  LMN  in  Additional  Proposition  1  is  a  transversal 
of  the  triangle  ABC. 


EXEECISES. 

1.  Points  E,  F  are  taken  in  the  sides  AC,  AB  of  a  triangle  such 
that  AE  is  twice  EC  and  BF  is  twice  FA  ;  FE  produced  cuts  BG  in 
D  ;  find  the  ratio  BD  to  DC. 

2.  If  the  bisectors  of  the  angles  B,  C  of  a  triangle  ABC  meet 
the  opposite  sides  in  D  and  E,  and  if  the  straight  line  DE  produced 
meet  BC  produced  in  F,  then  the  external  angle  at  A  is  bisected 
by^F. 

3.  BB  is  the  perpendicular  let  fall  from  one  end  of  the  base 
upon  the  straight  line  bisecting  the  vertical  angle  BAC  of  a  tri- 
angle. liBA  be  three  times  as  long  as  AC,  AD  will  be  bisected  at 
the  point  E,  where  it  cuts  the  base. 

4.  If  a  side  J?C  of  a  triangle  ABC  be  bisected  by  a  straight  line 
which  meets  the  sides  AB,  AC,  produced  if  necessary,  in  D  and  E 
respectively,  then  AE  is  io  EC  b.b  AD  to  DB. 

5.  If  one  side  of  a  given  triangle  be  produced  and  the  other 
shortened  by  equal  quantities,  the  line  joining  the  points  of  section 
will  be  divided  by  the  base  in  the  inverse  ratio  of  the  sides. 

6.  In  the  sides  AB,  AC  oi  o,  triangle  ABC  two  points  D,  E  are 
taken  such  that  /)D  is  equal  to  CE ;  DE,  BC  are  produced  to  meet  at 
F:  shew  that  AB  ib  to  AC  as  EF  to  DF. 


420  BOOK   VI. 

The  converse  of  the  theorem  on  page  418  may  be  stated  as 
follows : — 

ADDITIONAL   PROPOSITION  2. 

If  three  points  be  taken  on  the  sides  of  a  triangle  (either  one  on 
a  side  produced  and  the  other  two  on  sides,  or  else  all  three  on  sides 
produced),  such  that  the  ratio  compounded  of  the  ratios  of  the  segments 
of  the  sides  taken  in  order  is  equal  to  unity,  the  three  points  lie 
on  a  straight  line. 

Let  three  points  X,  M,  N  be  taken  on  the  sides  BG,  CA,  AB 
of  a  triangle  ABC,  either  all  on  sides  produced  (fig.  2)  or  one  L  on 
a  side  produced,  and  the  others  M,  N  on  sides  (fig.  1)  such  that  the 
ratio  compounded  of  the  ratios  AM  to  MG,  GL  to  LB  and  BN  to  NA 
is  equal  to  unity. 


Draw  LM  and  let  it  produced  cut  AB  in  P. 
Then  the  ratio  compounded  of  the  ratios 

AM  to  MG,  GL  to  LB  and  BP  to  PA  is  equal  to  unity; 

(Add.  Prop.  1.) 
and  the  ratio  compounded  of  the  ratios 

AM  to  MG,  GL  to  LB  and  BN  to  NA  is  equal  to  unity; 

(Hypothesis.) 
therefore  the  ratio  BP  to  PA  is  equal  to  the  ratio  BN  to  NA ; 

therefore  BP  is  to  BA  as  BN  to  BA ;  (V.  Prop.  10  or  11) 
therefore  BP  is  equal  to  BN;  (V.  Prop.  3.) 

that  is,  P  coincides  with  N, 
or,  in  other  words,  L,  M,  N  are  in  a  straight  line. 


THEOREM  OF  MENELAUS,  421 


EXEKCISES. 


1.  The  inscribed  circle  of  a  triangle  ABC  touches  the  sides 
BC,  CA,  AB  at  D,  E,  F;  EF,  FD,  DE  produced  meet  BC,  CA,  AB 
in  L,  M,  N :  prove  that  L,  M,  N  are  coUinear. 

2.  An  escribed  circle  of  a  triangle  ABC  touches  the  side  BC  at 
D  and  the  sides  AC,  AB  produced  &t  E,  F;  ED,  FD  produced  cut 
AB,  AC  in  K,  H  respectively;  prove  that  FE,  BC,  KH  meet  in  a 
point. 

3.  If  AB,  CD,  EF  be  three  parallel  straight  Unes,  and  AC,  BD 
meet  in  N,  CE,  DF  meet  in  L,  and  EA,  FB  meet  in  M,  then  L,  M,  N 
lie  on  a  straight  line. 

4.  If  D,  E,  F  be  the  points  of  contact  with  BC,  CA,  AB  of  the 
inscribed  circle,  or  of  any  one  of  the  escribed  circles  of  the  triangle 
ABC,  the  lines  AD,  BE,  CF  pass  through  a  point. 

5.  If  D  be  the  point  of  contact  of  the  inscribed  circle  of  a  triangle 
ABC  with  BC,  and  E,  F  the  points  of  contact  of  escribed  circles  with 
CA  produced  and  BA  produced  respectively,  then  AD,  BE,  CF  meet 
in  a  point. 

6.  If  one  escribed  circle  of  a  triangle  ABC  touch  AC  in  F  &nd 
BA  produced  in  G  and  another  escribed  circle  touch  AB  in  H  and 
CA  produced  in  K,  then  FH,  KG  produced  cut  BC  produced  in  points 
equidistant  from  the  middle  point  of  BC. 


422  BOOK   VI. 


ADDITIONAL  PROPOSITION   3. 

If  three  straight  lines  be  drawn  from  the  vertices  of  a  triangle  meet- 
ing in  a  point  and  cutting  the  opposite  sides  or  the  sides  produced,  the 
ratio  compounded  of  the  ratio  of  the  segments  of  the  sides  taken  in 
order  is  equal  to  unity* 

Let  the  straight  lines  A  0,  £0,  CO  be  drawn  from  the  vertices  of 
the  triangle  ABC  meeting  in  0,  and  cutting  BC,  CA,  AB  in  D,  E,  F 
respectively. 


Through  C  draw  HCG  parallel  to  AB  to  meet  BO,  AO  produced 
in  H,  G. 

Then  because  the  triangles  A  OF,  GOG  are  equiangular  to  one  another, 

AF  is  to  GC  as  FO  to  CO ;  (Prop.  4.) 

and  because  the  triangles  FOB,  COH  are  equiangular  to  one  another, 

FB  is  to  CH  as  FO  to  CO ; 

therefore  AF  is  to  GC  as  FB  to  CH,        (V.  Prop.  5.) 

and  therefore  AF  is  to  FB  as  GC  to  CH.    (V.  Prop.  9.) 

And  because  the  triangles  CEH,  AEB  are  equiangular  to  one  another, 

CEi^ioAE  as  CH  io  AB-, 
and  because  the  triangles  BDA,  CDG  are  equiangular  to  one  another, 

BD  i^  to  CD  BLBBAioCG',    \ 
therefore  the  ratio  compounded  of  the  ratios 

AF  to  FB,  CE  to  EA,  and  BD  to  DC 
is  equal  to  the  ratio  compounded  of  the  ratios 

GC  to  CH,  CH  to  AB,  and  AB  to  CG, 
which  is  equal  to  unity. 

*  This  theorem  was  first  published  in  the  year  1678  by  Giovanni 
Ceva,  an  Italian. 


CEVA'S  THEOREM.  ^23 


In  Additional  Proposition  3  it  has  been  proved  that,  if  through 
the  vertices  A,  B,  C  of  a  triangle  three  concurrent  straight  lines  ADy 
BE,  CF  be  drawn  meeting  the  sides  BC,  CA,  AB  in  D,  E,  F,  the  ratio 
of  BD  to  DC  is  equal  to  the  ratio  compounded  of  the  ratios  J^i^'to  FA 
and  AE  to  EC. 

In  Additional  Proposition  1  it  has  been  proved  that,  if  the  straight 
line  FE  be  drawn  and  produced  to  meet  BC  produced  in  L,  the  ratio 
BL  to  LC  is  equal  to  the  ratio  compounded  of  the  ratios  BE  to  FA 
and  AE  to  EC. 

Therefore  BD  is  to  DC  as  BL  to  LC, 
or,  in  other  words,  BDCL  is  a  harmonic  range. 

It  is  a  remarkable  fact  that,  although  the  theorem  on  page  418 
had  been  known  as  early  as  the  1st  century,  the  theorem  on  page  422, 
which  seems  a  very  natural  complement  to  the  other,  should  not  have 
been  discovered  until  the  17th  century. 


EXERCISES. 

1.  Prove  Ceva's  Theorem  for  a  triangle  ABC  and  a  point  O, 
(1)  when  0  lies  between  AB  produced  and  AC  produced,  (2)  when  O 
lies  between  BA  produced  and  CA  produced. 

2.  Prove  Ceva's  Theorem  by  using  the  result  of  Ex.  3,  page  355. 

3.  Prove  Ceva's  Theorem  by  the  use  of  Menelaus'  Theorem, 
considering  in  the  figure  of  Add.  Prop.  3  COF  a  transversal  of  the 
triangle  ABD  and  BOE  a  transversal  of  the  triangle  ADC. 

4.  D,  E,  F  are  the  points  in  which  the  bisectors  of  the  angles 
A,B,  C  of  &  triangle  cut  the  opposite  sides ;  prove  that,  if  BC  be  equal 
to  half  the  sum  of  the  sides  AB,  AC,  then  EF  bisects  AD. 


424  BOOK   VL 


The  converse  of  the  theorem  on  page  422  may  be  stated  as 
follows ; — 

ADDITIONAL  PKOPOSITION  4. 

If  three  straight  lines  he  drawn  through  the  vertices  of  a  triangle 
cutting  the  opposite  sides  {either  all  three  sides,  or  else  one  side  and 
the  other  two  sides  produced)  so  that  the  ratio  compounded  of  the 
ratios  of  the  segments  of  the  sides  taken  in  order  is  equal  to  unity^ 
the  three  straight  lines  meet  in  a  point. 

Let  three  straight  lines  AD,  BE,  CF  be  drawn  from  the  vertices 
A,  B,  C  of  a  triangle  ABC  to  cut  the  opposite  sides  in  D,  E,  F 
respectively,  so  that  the  ratio  compounded  of  the  ratios  AF  to 
FB,  BD  to  DC  and  CE  to  EA  is  equal  to  unity. 


Let  AD,  CF  meet  in  0 :   draw  BO  and  produce  it  to  meet  CA 
in  P. 

Then  because  the  ratio  compounded  of  the  ratios  AF  to  FB, 
BD  to  DC  and  CP  to  PA  is  equal  to  unity,  (Add.  Prop.  3.) 

and  also  the  ratio  compounded  of  the  ratios 

AF  to  FB,  BD  to  DC  and  CE  to  EA  is  equal  to  unity;  (Hypothesis.) 

therefore  the  ratio  CP  to  PA  is  equal  to  the  ratio  CE  to  EA ; 

therefore  CP  is  to  CA  as  CE  to  CA ;  (V.  Prop.  10  or  11.) 

therefore  CP  is  equal  to  CE  ;  (V.  Prop.  3.) 

therefore  P  coincides  with  E, 

or,  in  other  words,  AD,  BE,  CF  meet  in  a  point. 


CENTROID   OF    WEIGHTS.  425 

If  in  the  sides  BG,  GA,  AB  of  a  triangle,  points  D,  E,  F  be  taken 
such  that  BD  is  to  D  (7  as  w  to  m,  GE  is  to  EA  as  Z  to  n,  and  AF  is 
to  FB  as  m  to  I,  where  Z,  m,  n  are  any  three  integers, 
the  straight  lines  AD,  BE,  GF  meet  in  a  point,  say  0.  (Add.  Prop.  4.) 

It  is  proved  by  Add.  Prop.  1  that  the  ratio  of  AO  to  OD  is 
equal  to  the  ratio  compounded  of  the  ratios  AE  to  EG  and  GB  to 
BD,  that  is,  of  the  ratios  n  to  I  and  m  +  7i  to  n\ 

therefore  ^0  is  to  OD  as  m-\-n  to  I. 

Similarly  it  appears  that  BO  is  to  OE  as  n  +  l  to  ?«  and  that 
GO  is  to  Oi^  as  Z  +  m  to  n. 


It  follows  that,  if  we  divide  BG  in  D  so  that  J5D  is  to  DG  as  n  to 
w,  and  then  divide  D^  in  0  so  that  DO  is  to  OA  as  Z  to  m  +  n,  we 
arrive  at  the  same  point,  as  if  we  divide  GA  in  E  so  that  GE  is  to 
EA  as  Z  to  w,  and  then  divide  EB  in  0  so  that  jBO  is  to  OS  as  m  to 
w  +  Z,  or  as  if  we  divide  AB  in  F  so  that  ^jP  is  to  FB  as  m  to  I,  and 
then  divide  EG  in  0  so  that  FO  is  to  OC  as  n  to  Z  +  m. 

This  point  0  is  called  the  centroid  of  weights  I,  in,  n  bX  A,  B,  G 
respectively.  It  appears  that  the  position  of  the  centroid  of  three 
weights  is  independent  of  the  order  in  which  the  weights  are  taken, 
or,  in  other  words,  the  centroid  of  three  weights  is  a  unique  point. 

It  is  not  difl&cult  to  see  that  this  proposition  can  be  extended  to 
any  number  of  weights,  so  that  we  may  state  the  proposition  in  the 
general  form,  the  centroid  of  a  number  of  given  weights  is  a  unique 
point. 

EXERCISE. 

1.  From  the  vertex  A  oi  a.  triangle  ABG  a  straight  line  is  drawn 
cutting  BG  in  D,  and  the  angles  BDA,  GDA  are  bisected  by  straight 
lines  cutting  AB,  AG  in  F,  E  respectively:  prove  that  AD,  BE,  GF 
intersect  in  a  point. 

T.  E.  28 


426  BOOK   VI. 


ADDITIONAL  PBOPOSITION   5. 

The  locus  of  a  point,  the  ratio  of  whose  distances  from  two  given 
points  is  constant,  is  a  circle*. 

Let  A,  B  he  two  given  points  and  P  a  point  such  that  the  ratio  of 
AP  to  BP  is  equal  to  the  given  ratio. 

Draw  PA,  PB ;  and  draw  PC,  PD  the  internal  and  the  external 
bisectors  of  the  angle  APB  meeting  AB  in  C  and  AB  produced  in  D. 


G     B 


Because  PC,  PB  are  the  bisectors  of  the  angle  APB, 
therefore  the  ratios  of  ^C  to  GB  and  AD  to  BB  are  equal  to  the  ratio 
of  AP  to  PB ;  and  the  ratio  of  AP  to  PB  is  constant ; 

therefore  G  and  D  are  two  fixed  points.    (Ex.  1,  page  359.) 
And  because  PG,  PD  are  the  bisectors  of  the  angle  APB, 

the  angle  GPD  is  a  right  angle.      (Ex.5,  page  43. ) 
Therefore  every  point  on  the  locus  of  P  must  lie  on  the  circle  upon 
the  fixed  line  GD  as  diameter. 

Next  we  will  prove  that  every  point  of  the  circle  belongs  to  the 
locus. 

Let  P  be  any  point  on  the  circle  described  on  GD  as  diameter. 
Draw  PA,  PG ;  and  draw  PE  making  the  angle  GPE  equal  to  the 
angle  GPA  and  meeting  GD  at  E ; 

then  AP  is  to  PE  aq  AG  to  GE.   (Prop.  3,  Part  1.) 
Again,  because  GPD  is  a  right  angle, 

PD  is  the  external  bisector  of  the  angle  APE ; 
therefore  AP  is  to  PE  as  AD  to  DE. 

*  This  theorem  is  attributed  to  Apollonius  of  Perga,  a  Greek 
geometer,  who  lived  in  the  latter  part  of  the  third  century  b.c. 


CIRCLE   OF  APOLLONIUS.  427 

Therefore  ^C  is  to  CE  as  AB  to  BE  ;  (V.  Prop.  5.) 

and  ^C  is  to  GB  as  AB  to  BE; 

therefore  CE  is  to  EB  as  CB  to  BB, 

and  iJ  coincides  with  B^  which  is  a  fixed  point.    (Ex.  1,  page  359.) 

Therefore  AP  is  to  PB  in  the  fixed  ratio  AG  io  CB  for  every  point  P 

on  the  circle. 

We  may  state  the  result  of  this  proposition  thus: — If  a  circle  he 
described  upon  the  straight  line  joining  two  conjugate  points  of  a 
harmonic  range  as  a  diameter,  the  ratio  of  the  distances  of  a  point  on 
the  circle  from  the  other  pair  of  conjugate  points  is  constant. 


EXEKCISES. 

1.  Prove  that,  if  A,  B  he  two  fixed  points  and  P  be  a  point  such 
that  PA  is  equal  to  m  times  PB, 

(1)  if  m  vanish,  the  locus  reduces  to  the  point  A ; 

(2)  if  m  be  equal  to  unity,  the  locus  is  the  straight  line  which 
bisects  AB  &t  right  angles ; 

(3)  if  m  be  infinitely  great,  the  locus  reduces  to  the  point  B; 

(4)  if  m  be  greater  than  unity,  the  locus  is  a  circle  excluding  A 
and  including  B ; 

(5)  if  m  be  less  than  unity,  the  locus  is  a  circle  including  A  and 
excluding  B ; 

(6)  if  m  be  greater  than  unity,  the  greater  the  value  of  m,  the 
less  the  circle; 

(7)  if  m  be  less  than  unity,  the  less  the  value  of  m,  the  less  the 
circle ; 

(8)  the  loci  for  two  different  values  of  m  do  not  intersect. 

2.  A  and  B  are  the  centres  of  two  circles.  A  straight  line  PQ 
parallel  to  AB  meets  the  circles  in  P  and  Q :  find  the  locus  of  the 
point  of  intersection  of  AP  and  BQ. 

3.  Find  a  point  such  that  its  distances  from  three  given  points 
may  be  in  given  ratios. 

4.  Prove  that,  if  a  map  be  laid  flat  on  another  map  of  the  same 
district  on  a  larger  scale,  there  is  one  place  in  the  district  which  is 
represented  in  the  two  maps  by  points  which  are  superposed  one  on 
the  other. 

28—2 


428  BOOK    VI. 


ADDITIONAL  PKOPOSITION   6. 

If  the  line  hetioeen  one  pair  of  conjugates  of  a  harmonic  range  be 
bisected^  the  square  on  half  the  line  is  equal  to  the  rectangle  contained 
by  the  segments  of  the  line  hetioeen  the  other  pair  of  conjugates  made  by 
the  point  of  bisection. 

Let  AGBB  be  a  harmonic  range,  such  that  ^0  is  to  CB  as  AD  to 
BB,  so  that  A,  B  are  one  pair  of  conjugates  and  C,  D  the  other. 

First,  let  0  be  the  middle  point  of  CD,  the  line  between  one  pair 
of  conjugates. 

Describe  the  circle  on  CD  as  diameter. 

Take  any  point  P  on  the  circle,  and  draw  PA,  PC,  PB,  PO. 


Because  the  angle  OPG  is  equal  to  the  angle  OGP, 

the  sum  of  the  angles  OPB,  BPG  is  equal  to  the  sum  of  the  angles 

GAP,  GPA  ;  (I.  Prop.  32.) 

and  because  AP  is  to  PB  as  ^C  to  CB,  (see  page  427.) 

the  angle  BPG  is  equal  to  the  angle  GPA ; 

therefore  the  angle  OPB  is  equal  to  the  angle  GAP; 

therefore  OP  touches  the  circle  described  about  APB; 

(Converse  of  III.  Prop.  32.) 

therefore  the  square  on  OP,  which  is  equal  to  the  square  on  OG,  is 

equal  to  the  rectangle  OA,  OB.  (III.  Prop.  36,  Coroll.) 


HARMONIC  RANGE.  429 


Secondly,  let  C/  be  the  middle  point  of  AB,  the  line  between  the 
second  pair  of  conjugates. 


The  rectangle  OA,  OB  is  equal  to  the  difference  between  the 
squares  on  00'  and  O'B,  (II.  Prop.  10.) 

and  the  rectangle  OA,  OB  is  equal  to  the  square  on  0C\ 
therefore  the  square  on  OG  is  equal  to  the  difference  between  the 
squares  on  00'  and  O'B. 
Therefore  the  square  on  O'B  is  equal  to  the  difference  between  the 
squares  on  00'  and  00,  which  is  equal  to  the  rectangle  O'G,  O'D. 

(H.  Prop.  10.) 

Note.  All  the  circles,  which  are  the  loci  of  the  point  P  for 
different  values  of  the  ratio  AP  to  BP,  have  their  centres  in  the  line 
AB,  and,  since  the  rectangle  O'G,  O'D  is  equal  to  the  square  on  the 
tangent  from  0'  to  the  circle  GPD,  the  straight  line  which  bisects  AB 
at  right  angles  is  the  radical  axis  of  every  pair  of  such  circles.  Such 
a  series  of  circles  is  called  Coaxial. 

The  points  A,  B  are  called  the  limiting  points  of  the  series  of 
circles. 


EXEECISES. 

1.  If  two  circles  be  described  upon  the  straight  lines  joining  the 
two  pairs  of  conjugate  points  of  a  harmonic  range  as  diameters,  the 
circles  cut  orthogonally. 

2.  A  common  tangent  to  two  given  circles  is  divided  harmonically 
by  any  circle  which  is  coaxial  with  the  given  circles. 


430 


BOOK   VI. 


ADDITIONAL  PEOPOSITION   7. 

A  chord  of  a  circle  is  divided  harmonically  by  any  point  on  it  and 
the  polar  of  the  point. 

Let  PQ  be  a  chord  of  the  given  circle  CQPD :  take  A  any  point 
on  PQ  produced  and  draw  the  diameter  ACD. 

Let  B  be  the  point  such  that  A  GBD  is  a  harmonic  range. 

Draw  PB,  BQ  and  draw  BR  at  right  angles  to  AB  meeting  PQ 


Because  A  CBD  is  a  harmonic  range  and  0  is  the  middle  point  of  CD, 
the  rectangle  OA,  OB  is  equal  to  the  square  on  OC;    (Add.  Prop.  6.) 
therefore  BR  is  the  polar  of  A.  (see  page  259.) 

Because  AP  is  to  PB  as  AG  to  CB, 

and  AQ  is  to  QB  as  AG  to  GB ;         (Add.  Prop.  5.) 
therefore  AP  is  to  PB  as  AQ  to  QB ; 
and  PB  is  to  PA  as  QB  to  AQ  ;  (V.  Def.  5  note.) 

therefore  PB  is  to  BQ  a.s  PA  to  AQ;  (V.  Prop.  9.) 

therefore  BA  is  the  external  bisector  of  the  angle  PBQ ; 

(Prop.  3,  Part  2.) 
therefore  BR  is  the  internal  bisector, 

and  PB  is  to  RQ  as  PB  to  BQ,    (Prop.  3,  Part  1.) 
and  therefore  as  PA  to  AQ; 
therefore  AQRP  is  a  harmonic  range. 


HARMONIC  PROPERTY  OF  POLAR.       431 


It  may  be  remarked  that  in  the  diagram  on  page  430  the  point  A 
is  taken  outside  the  circle.  If  the  point  were  inside  the  circle,  say  R, 
its  polar  would  intersect  PQB  in  A.  (Add.  Prop,  on  page  262.) 

Hence  the  theorem  is  established  generally. 


EXEKCISES. 

1.  Prove  that,  if  AGBD  be  a  harmonic  range,  and  if  0  be  the 
middle  point  of  CD,  then  ^C  is  to  Ci?  as  ^0  to  OC. 

2.  Establish  the  theorem  of  page  426  by  proving  that  in  the 
figure  of  page  430  the  triangles  ABQ,  APO  are  similar  and  also  that 
the  triangles  ABP,  AQO  are  similar. 

3.  If  any  straight  line  PQR  be  drawn  touching  one  given  circle 
at  Q  and  cutting  another  at  P,  R,  the  segments  PQ,  QR  subtend 
equal  or  supplementary  angles  at  either  of  the  limiting  points  of  the 
coaxial  system  to  which  the  given  circles  belong. 

4.  If  any  straight  line  PQES  be  drawn  cutting  two  given  circles 
of  a  coaxial  system  in  P,  S  and  Q,  R,  the  segments  PQ,  RS  subtend 
equal  or  supplementary  angles  at  either  of  the  limiting  points. 


432  BOOK   VI. 

ADDITIONAL   PEOPOSITION   8. 

If  a  pencil  he  drawn  from  a  point  to  the  four  points  of  a  harmonic 
range  and  if  a  straight  line  he  drawn  through  one  of  the  points  parallel 
to  the  ray  which  passes  through  the  conjugate  point,  the  part  of  the  line 
intercepted  hetween  the  rays  through  the  other  pair  of  points  is  hisected 
at  the  point. 

Let  ABGD  be  a  harmonic  range,  and  0  be  any  point  not  in  the 
straight  line  AT).  Let  OA,  OB,  OD  be  drawn*,  and  let  FCH  be  drawn 
parallel  to  ^0,  meeting  OB,  OD,  produced  if  necessary,  in  F,  H. 


Because  the  triangles  OAB,  FOB  are  equiangular  to  one  another, 
OA  is  to  FC  as  AB  to  BG;  (Prop.  4.) 

and  because  the  triangles  OAD,  HCD  are  equiangular  to  one  another, 
OA  is  to  HC  as  AD  to  DC. 
And  because  ABCD  is  a  harmonic  range, 

AB  ia  to  BC  &s  AD  to  DC; 
therefore  OA  is  to  FC  as  OA  to  HG;        (V.  Prop.  5.) 
therefore  FG  is  equal  to  HG.  (V.  Prop.  3.) 

Note.    The  converse  of  this  proposition  is  true,  viz.  if  the  line  FH 
be  bisected  at  C,  then  ABGD  is  a  harmonic  range. 


EXEECISE. 

1.  Give  a  construction  to  find  the  fourth  point  of  a  harmonic 
range  when  three  points  are  given. 

*  The  ray  OG  of  the  pencil  0  {ABGD)  is  omitted  in  the  figure,  as 
it  is  not  wanted  in  the  proof.  Similar  omissions  will  be  met  with 
elsewhere. 


HARMONIC   RANGES  AND   PENCILS.      433 


ADDITIONAL   PEOPOSITION   9. 


The  points^  in  which  a  harmonic  pencil 
form  a  harmonic  range. 


cut  by  any  straight  line^ 


Let  0  {A BCD)  be  a  pencil  drawn  through  the  points  of  the  har- 
monic range  ABCD :  let  abed  be  any  other  straight  line  cutting  the 
rays  OA,  OB,  OC,  OD  in  a,  b,  c,  d  respectively. 

Through  C,  c  draw  GCF,  gcf  parallel  to  OA  cutting  the  rays  OB, 
OD,  produced  if  necessary,  in  F,  G  and  /,  g. 


Because  ABCD  is  a  harmonic  range,  and  GCF  is  parallel  to  OA, 
therefore  FC  is  equal  to  CG.         (Add.  Prop.  8.) 
And  because /c^r  is  parallel  to  FCG, 

therefore  fc  is  equal  to  eg.        (Ex.  1,  page  369.) 
And  because  fcg  is  parallel  to  Oa  and  fc  is  equal  to  eg, 

abed  is  a  harmonic  range.  (Note,  p.  432.) 


EXEKCISES. 

1.  The  pencil  formed  by  joining  the  four  angular  points  of  a 
square  to  any  point  on  the  circumscribing  circle  of  the  square  is  a 
harmonic  pencil. 

2.  Give  a  construction  for  drawing  the  fourth  ray  of  a  harmonic 
pencil,  when  three  rays  are  given. 

3.  Draw  a  harmonic  pencil  of  which  the  rays  pass  through  the 
angular  points  of  a  rectangle,  and  one  of  which  is  given  in  direction. 

4.  CA,  CB  are  two  tangents  to  a  circle;  E  is  the  foot  of  the 
perpendicular  from  B  on  AD  the  diameter  through  A ;  prove  that  CD 
bisects  BE. 


434  BOOK    VI. 


ADDITIONAL   PEOPOSITION   10. 

If  two  harmonic  ranges  have  two  corresponding  points,  one  in 
each  range,  coincident,  the  straight  lines  joining  the  other  pairs  of 
corresponding  points  pass  through  a  point. 

Let  AGED,  Achd  be  two  harmonic  ranges,  of  which  the  point  A 
is  a  common  point. 

Draw  Gc,  Bb  and  let  them,  produced  if  necessary,  meet  in  0,  and 
draw  OD. 


Because  O  {AGBD)  is  a  harmonic  pencil, 
if  Acb  cut  OD  in  d', 
then  Acbd'  is  a  harmonic  range ;        (Add.  Prop.  9.) 
therefore  Ac  is  to  cb  as  Ad'  to  d'b ; 

but  Acbd  is  a  harmonic  range ; 
therefore  Ac  is  to  cb  as  Ad  to  db; 
therefore  Ad'  is  to  d'b  as  Ad  to  db,         (V.  Prop.  5.) 
and  d'  coincides  with  d;  (Ex.  1,  page  359.) 

i.e.  Gc,  Bb,  Dd  meet  in  a  point. 


EXEECISE. 

1.  Prove  that  the  intersections  of  the  pairs  of  straight  lines  (76, 
Be;  Bd,  Db;  Dc,  Gd  in  the  above  figure  lie  on  a  straight  line  which 
passes  through  A. 


HARMONIC  RANGES  AND   PENCILS.      435 


ADDITIONAL  PKOPOSITION   11. 

If  two  harmonic  pencils  have  two  corresponding  rays,  one  of  each 
pencil,  coincident,  the  intersections  of  the  other  three  pairs  of 
corresponding  rays  lie  on  a  straight  line. 

Let  0  (ABCD),  0'  (abed)  be  two  harmonic  pencils,  of  which  OAaO' 
is  a  common  ray. 

Let  OB,  O'b  meet  in  Q,  and  OC,  O'c  in  R ;  draw  QR  and  let  it 
meet  00'  in  P,  OD  in  S,  and  O'd  in  s. 


Then  because  O  (ABCD)  is  a  harmonic  pencil, 

PQRS  is  a  harmonic  range;  (Add.  Prop.  9.) 

therefore  PQ  is  to  QR  as  PS  to  SR ; 
and  because  0'  (abed)  is  a  harmonic  pencil, 

PQRs  is  a  harmonic  range ; 
therefore  PQ  is  to  QR  as  Ps  to  sR  ; 
therefore  Ps  is  to  sR  as  PS  to  SR,  (V.  Prop.  5.) 

and  the  points  S,  s  coincide;      (Ex.  1,  page  359.) 
i.e.  the  intersections  of  OB,  O'bi  OC,  O'c,  and  OD,  O'd  are  collinear. 


EXEECISE. 

1.  Prove  that  the  straight  lines  joining  the  intersections  of  the 
pairs  of  straight  lines  OB,  O'c  and  OC,  O'b;  OC,  O'd  and  OD,  O'c; 
OD,  O'b  and  OjB,  O'd  intersect  in  a  point  which  lies  on  the  line  00'. 


436  BOOK   VI. 


ADDITIONAL  PKOPOSITION   12. 

If  a  pencil  be  drawn  from  a  point  to  the  four  points  of  an  anharmonic 
range  and  if  a  straight  line  be  drawn  through  one  of  the  points  parallel  to 
the  ray  which  passes  through  a  second  point,  the  part  of  it  intercepted 
between  the  rays  through  the  other  pair  of  points  will  be  divided  in  a 
constant  ratio  at  the  first  point. 

Let  ABCD  be  an  anharmonic  range  and  0  be  any  point  not  in  the 
straight  line  AB. 

Let  OA,  OB,  OD  be  drawn  and  FCH  be  drawn  parallel  to  ^O 
meeting  OB,  OD,  produced  if  necessary,  in  F,  H, 


Because  the  triangles  OAB,  FOB  are  equiangular  to  one  another, 
OA  is  to  FC  a&  AB  to  BG ;  (Prop.  4.) 

and  because  the  triangles  OAD,  HCD  are  equiangular  to  one  another, 

OA  is  to  HC  as  AD  to  DC ; 

therefore  the  ratio  of  the  ratio  OA  to  FC  to  the  ratio  OA  to  HC  is 

equal  to  the  ratio  of  the  ratio  AB  to  BC  to  the  ratio  AD  to  DC, 

which  is  constant ; 

i.  e.  the  ratio  HC  to  FC  is  constant  and  is  equal  to  the  ratio  of  the 

range  ABCD.  (Def.  10.) 


EXEKCISES. 

1.  If  ABCD,  ABCE  be  two  like  anharmonic  ranges,  then  the 
points  D,  E  coincide. 

2.  If  the  ratio  of  the  range  ABCD  be  equal  to  the  ratio  of  the 
range  ADGB,  the  range  ABCD  is  harmonic. 


ANHARMONIC  RANGES  AND   PENCILS.    437 

ADDITIONAL  PKOPOSITION   13. 

The  points  in  which  an  anharmonic  pencil  is  cut  by  any  straight 
line  form  an  anharmonic  range  of  constant  ratio. 

Let  0  {ABCD)  be  a  pencil  drawn  through  the  points  of  the  an- 
harmonic range  ABCD. 

Let  ahcd  be  any  other  straight  line  cutting  the  rays  OA,  OB,  OC, 
OD  in  a,  b,  c,  d  respectively. 

Through  C,  c  draw  GCH,  gch  parallel  to  OA  cutting  the  rays  OD, 
OB  in  G,  H  and  g,  h. 


Because  ABCD  is  an  anharmonic  range, 

and  GCH  is  parallel  to  OA, 
therefore  the  ratio  of  GC  to  CH  is  the  ratio  of  the  range  ABCD ; 

(Add.  Prop.  12.) 
and  because  abed  is  an  anharmonic  range  and  gch  is  parallel  to  Oa, 
therefore  the  ratio  of  ^c  to  ch  is  the  ratio  of  the  range  abed ; 

(Add.  Prop.  12.) 
and  because  gch  is  parallel  to  GCH, 

gc  is  to  ch&B  GC  to  CH;         (Ex.  2,  page  365.) 
therefore  abed  is  an  anharmonic  range  of  ratio  equal  to  that  of  the 
range  ABCD. 

EXEKCISES. 

1.  Find  a  point  on  a  given  straight  line  such  that  lines  drawn 
from  it  to  three  given  points  shall  intercept  on  any  parallel  to  the 
given  line  lengths  having  a  given  ratio. 

2.  Three  points  F,  G,  H  are  taken  on  the  side  BC  of  a  triangle 
ABC :  through  G  any  line  is  drawn  cutting  AB  and  AC  in  L  and  31 
respectively;  FL  and  HM  intersect  in  K;  prove  that  K  lies  on  a 
fixed  straight  line  passing  through  A. 


438  BOOK    VI. 


ADDITIONAL  PEOPOSITION   14. 

If  two  like  anharmonic  ranges  have  two  corresponding  points^  one 
in  each  range,  coincident,  the  straight  lines  joining  the  other  pairs 
of  corresponding  points  pass  through  a  point. 

Let  ACBD,  Acbd,  be  two  like  anharmonic  ranges,  of  which  the 
point  ^  is  a  common  point. 

Draw  Cc,  Bb,  and  let  them,  produced  if  necessary,  meet  in  0 ;  and 
draw  OD. 


Because  0{ACBD)  is  an  anharmonic  pencil, 
if  Acb  cut  OD  in  d', 
then  Acbd'  is  an  anharmonic  range  of  ratio  equal  to  that  of  the  pencil; 

(Add.  Prop.  13.) 
and  Acbd  is  a  like  anharmonic  range; 
therefore  the  ratio  of  the  ratio  Ac  to  cb  to  the  ratio  Ad'  to  d'b  is  equal 
to  the  ratio  of  the  ratio  Ac  to  cb  to  the  ratio  Ad  to  db ; 
therefore  the  ratio  Ad'  to  d'b  is  equal  to  the  ratio  Ad  to  db, 

and  d'  coincides  with  d,  (Ex.  1,  page  359.) 

i.e.  Cc,  Bb,  Dd  meet  in  a  point. 


EXEECISE. 

1.     Prove  that  the  intersections  of  the  pairs  of  straight  lines  Cb, 
Be ;  Bd,  Db ;  Dc,  Cd  in  the  above  figure  lie  on  a  straight  line  through  A, 


ANHARMONIC  RANGES  AND   PENCILS.     439 


ADDITIONAL  PKOPOSITION  15. 

If  two  like  anharmonic  pencils  have  two  corresponding  rays,  one 
in  each  pencil,  coincident,  the  intersections  of  the  other  three  pairs  of 
corresponding  rays  lie  on  a  straight  line. 

Let  0  (ABCD),  0'  (abed)  be  two  like  anharmonic  pencils,  of  which 
OAaO'  is  a  common  ray.  Let  OB,  O'b  meet  in  Q,  and  OC,  O'c  in  R; 
draw  QR  and  let  it  meet  00'  in  P,  OD  in  S  and  O'd  in  s. 


Because  PQRS  is  a  transversal  of  the  anharmonic  pencil  0  {ABCD), 

PQRS  is  a  range  of  ratio  equal  to  that  of  the  pencil;  (Add.  Prop.  13) 

and  because  PQRs  is  a  transversal  of  the  anharmonic  pencil  0'  {abed), 

PQRs  is  a  range  of  ratio  equal  to  that  of  the  pencil ; 

and  because  O  {ABCD),  0'  {abed)  are  like  anharmonic  pencils, 

(Hypothesis) 
therefore  PQRS,  PQRs  are  two  like  anharmonic  ranges ; 

therefore  the  points  S,  s  coincide;  (Ex.  1,  page  436) 
i.e.  the  intersections  of  OB,  O'b;  OC,  O'c;  and  OD,  O'd  are  coUinear. 


EXEKCISE. 

1.  Prove  that  the  straight  lines  joining  the  intersections  of  the 
pairs  of  straight  lines  OB,  O'c  and  OC,  O'b;  OC,  O'd  and  OD,  O'c; 
OD,  O'b  and  OB,  O'd  intersect  in  a  point  which  lies  on  the  straight 
line  00'. 


440  BOOK    VI, 


ADDITIONAL  PKOPOSITION  16. 

The  anharmonic  ratio  of  the  pencil  formed  by  joining  four  given 
points  on  a  circle  to  any  fifth  point  on  the  same  circle  is  constant. 

Let  A,  B,  G,  B  he  four  given  points  on  a  circle,  and  let  0  be  any 
fifth  point  on  the  circle,  and  let  the  pencil  O  (ABCD)  be  drawn. 

Take  0'  any  other  point  in  the  same  arc  AD  as  0 ; 
then  the  angles  AO'B,  BO'C,  CO'D  are  equal  to  the  angles  AOB, 
BOG,  COD  respectively; 
and  the  pencil  0' {ABGD)  is  equal*  to  the  pencil  O  (ABGD). 


Next  take  0'  any  point  in  the  arc  AB. 

Then  the  angles  BO'G,  GO'D  are  equal  to  the  angles  BOG,  GOD 
respectively, 
and  the  angle  between  O'B  and  AO'  produced,  say  O'A^,  is  equal  to 
the  angle  AOB, 

and  the  pencil  O'  (A-^BGD)  is  equal  to  the  pencil  0  (ABGD). 
Similarly  it  can  be  proved  that  the  pencil  is  the  same  for  all 
positions  of  O'  on  the  circle. 


EXEECISE. 

1.  The  locus  of  the  vertex  of  a  harmonic  pencil,  whose  rays  pass 
through  the  angular  points  of  a  square,  is  the  circumscribed  circle 
of  the  square. 


*    In  the  sense  that  one  pencil  can  be  shifted  so  that  its  rays 
coincide  with  the  rays  of  the  other  pencil.  (I.  Def.  21.) 


ANHARMONIG  PROPERTIES  OF  A  CIRCLE.     441 

ADDITIONAL  PEOPOSITION  17. 

The  anhai-nwnic  ratio  of  the  range  formed  by  the  intersections  of 
four  given  tangents  to  a  circle  by  any  fifth  tangent  to  the  same  circle 
is  constant. 

Let  Aa,  Bh,  Cc,  Dd  be  the  tangents  at  four  given  points,  A,  B,  C, 
D  on  a  circle,  and  let  them  cut  the  tangent  at  any  fifth  point  T  on  the 
circle  in  a,  b,  c,  d. 

Find  P  the  centre,  and  take  any  point  0  on  the  circle. 

Draw  PA,  PB,  PT,  Pa,  Pb. 


Because  the  angle  APT  is  equal  to  twice  the  angle  AOT, 

(in.  Prop.  20) 
and  also  equal  to  twice  the  angle  aPT, 
the  angle  ^OT  is  equal  to  the  angle  aPT. 
Similarly  it  can  be  proved  that 

the  angle  BOT  is  equal  to  the  angle  bPT ; 
therefore  the  angle  AOB  is  equal  to  the  angle  aPb. 
Similarly  it  can  be  proved  that  the  angles  BOG,  COD  are  equal  to 
the  angles  bPc,  cPd ; 

therefore  the  pencil  0  (ABCD)  is  equal  to  the  pencil  P  (abed), 
and  the  pencil  0  {ABCD)  has  a  constant  ratio;   (Add.  Prop.  16) 
therefore  the  pencil  P  (abed)  has  a  constant  ratio  ; 
and  therefore  the  range  abed  has  a  constant  ratio.  (Add.  Prop.  13) 


EXEECISE. 

1.     If  a  straight  line  cut  the  four  sides  of  a  square  in  a  harmonic 
range,  it  touches  the  inscribed  circle  of  the  square. 

T.  E.  29 


442  BOOK    VI. 


ADDITIONAL  PKOPOSITION  18. 

The  anharmonic  ratio  of  the  pencil  formed  by  joining  four  points 
on  a  circle  to  any  fifth  point  on  the  circle  is  the  same  as  the  ratio  of 
the  rectangles  contained  by  the  chords  that  join  the  points. 

Let  A,  B,  G,  Dhe  four  given  points  on  a  circle.  Draw  AB,  AC, 
AD,  BC,  CD. 

In  AD  take  Ab  equal  to  AB;  draw  Bb  and  let  it  be  produced  to 
meet  the  circle  in  0;  draw  OA,  OC,  OD  and  let  OG  cut  AD  in  c; 
draw  be  parallel  to  CD  to  meet  OG  in  e,  and  draw  Ae. 


Because  be  is  parallel  to  CD, 

the  angle  bee  is  equal  to  the  angle  OGD, 

which  is  equal  to  the  angle  OAc ;  (III.  Prop.  21) 

therefore  A,  b,  e,  0  lie  on  a  circle. 

Therefore  the  angle  Aeb  is  equal  to  the  angle  AOB, 

which  is  equal  to  the  angle  A  CB  ; 

and  the  angle  Abe  is  equal  to  the  supplement  of  the  angle  AOe, 

which  supplement  is  equal  to  the  angle  ABC ; 
therefore  the  triangles  Abe,  ABC  are  equiangular  to  one  another; 
and  ^6  is  equal  to  AB  ;  (Constr.) 

therefore  be  is  equal  to  BC. 
Now  the  anharmonic  ratio  of  the  pencil  0  {ABCD)  is  equal  to  that 
of  the  range  AbcD,  (Add.  Prop.  13) 

which  is  equal  to  the  ratio  of  the  ratio  Ab  to  be  to  the  ratio  AD  to  Dc, 
that  is,  to  the  ratio  compounded  of  the  ratios  Ab  to  be  and  Do  to  AD, 
which  is  the  ratio  of  the  rectangle  Ab,  cD  to  the  rectangle  AD,  be. 

(See  page  399.) 


THE  SIX  AN  HARMONIC  RATIOS.  443 

And  because  he  is  parallel  to  CD, 

the  triangles  DcG,  bee  are  equiangular  to  one  another  ; 

therefore  cD  is  to  be  as  CD  to  be ;  (Prop.  4) 

therefore  the  ratio  of  the  rectangle  Ab,  eD  to  the  rectangle  AD,  be 
is  equal  to  the  ratio  of  the  rectangle  AB,  CD  to  the  rectangle  AD,  be. 
Therefore  the  anharmonic  ratio  of  the  pencil  O  (ABCD)  is  equal 
to  the  ratio  of  the  rectangle  AB,  CD  to  the  rectangle  AD,  BC. 


The  anharmonic  ratio  of  a  range  ABCD  is  defined  to  be 
(Definition  10)  the  ratio  of  the  ratio  AB  to  BC  to  the  ratio  AD  to  DC, 
which,  by  Definition  8  of  Book  V.,  is  equal  to  the  ratio  compounded  of 
the  ratios  AB  to  BC  and  DC  to  AD ;  and  this  last  ratio  has,  on 
page  399,  been  shewn  to  be  equal  to  the  ratio  of  the  rectangle 
AB,  CD  to  the  rectangle  AD,  BC. 

Now  it  can  be  proved  (Ex.  1,  page  137)  that,  if  ABCD  be  a  range, 
the  sum  of  the  rectangles  AB,  CD  and  AD,  BC  ia  equal  to  the 
rectangle  AC,  BD. 

If  therefore  any  two  of  these  three  rectangles  be  given,  the  third 
is  at  once  found.  There  are  six  ratios,  of  which  one  of  these  rectangles 
is  the  antecedent  and  another  the  consequent;  if  any  one  of  these 
ratios  be  given,  the  other  five  ratios  are  at  once  found.  Now  in 
the  definition  the  ratio  of  the  rectangle  AB,  CD  to  the  rectangle 
AD,  BC  is  defined  as  the  anharmonic  ratio  of  the  range  ABCD. 
There  is  no  reason  against  adopting  any  other  of  the  six  ratios  as  the 
ratio  of  the  range,  but  it  is  important  strictly  to  adhere  throughout  an 
investigation  to  one  and  the  same  ratio. 


EXEKCISES. 

1.  The  anharmonic  ratio  of  the  range  formed  by  the  intersections 
of  four  given  tangents  to  a  circle  with  any  fifth  tangent  is  equal  to  the 
ratio  of  the  rectangles  contained  by  the  chords  which  join  the  points 
of  contact  of  the  given  tangents. 

2.  Two  fixed  points  D,  E  are  taken  on  the  diameter  AB  of  a 
circle,  and  P  any  point  on  the  circumference;  perpendiculars  AM, 
BN  are  let  fall  on  PD,  PE ;  prove  that  the  ratio  of  the  rectangle 
PM,  PN  to  the  rectangle  AM,  BN  is  constant. 

29—2 


444  BOOK    VI. 


ADDITIONAL  PEOPOSITION  19. 

If  two  tnangles  he  such  tJiat  the  straight  lines  joining  their  vertices 
in  pairs  pass  through  a  point,  the  intersections  of  pairs  of  correspond- 
ing sides  lie  on  a  straight  line. 

Let  ABC,  abc  be  two  given  triangles  such  that  the  straight  lines 
Aa,  Bb,  Cc  meet  in  a  point  0. 

Let  the  pairs  of  sides  BG,  be;  CA,  ca;  AB,  ah,  produced  if 
necessary,  meet  in  D,  E,  F  respectively,  and  let  OBh  cut  AC,  ac  in 
H,h. 


Because  EAHC,  EaJic  cut  the  same  pencil  0{EAHG), 

EAHC,  Eahc  are  like  anharmonic  ranges ;  (Add.  Prop.  13), 
therefore  the  pencils  B  (EAHC),  h  (Eahc)  are  like  anharmonic  pencils; 

and  they  have  a  common  ray  BHhb ; 
therefore  the  intersections  of  the  pairs  of  rays  BC,  he ;   BE,  bE ; 
BA,  ba  lie  on  a  straight  line;  (Add.  Prop.  15); 

that  is,  D,  E,  F  lie  on  a  straight  line. 

Note.  The  point  0  is  often  called  the  pole  of  the  triangles  ABC, 
abc,  and  the  straight  Hne  DEE  the  axis  of  the  triangles. 

This  theorem  may  then  be  enunciated  thus,  compolar  triangles  are 
coaxial. 

Compolar  triangles  are  often  said  to  be  in  perspective. 


TRIANGLES   IN  PERSPECTIVE. 


445 


ADDITIONAL  PEOPOSITION  20. 

If  two  triangles  he  such  that  the  intersections  of  their  sides  taken  in 
pairs  lie  on  a  straight  line,  the  straight  lines  joining  pairs  of  corre- 
sponding vertices  meet  in  a  point.* 

Let  ABC,  abc  be  two  given  triangles  such  that  the  pairs  of  sides 
BC,  be;  CA,  ca;  AB,  ab  intersect  in  three  points  D,  E,  F  lying  on  a 
straight  line. 

Let  the  straight  lines  Aa,  Bb,  Cc  be  drawn  and  let  Bb  cut  AC,  ac, 
DEF  in  H,  h,  K  respectively. 


Because  the  pencils  B  {EFKD),  b  (EFKD)  have  a  common  trans- 
versal EFKD, 

they  are  like  anharmonic  pencils ; 

and  because  EAHC  is  a  transversal  of  the  pencil  B  (EFKD), 

and  Eahc  is  a  transversal  of  the  pencil  b  {EFKD), 

EAHC,  Eahc  are  like  anharmonic  ranges ;  (Add.  Prop.  13) 

and  they  have  a  common  point  E ; 

therefore  the  lines  Aa,  Hh,  Cc  meet  in  a  point,  (Add.  Prop.  14) 

that  is,  Aa,  Bb,  Cc  meet  in  a  point. 

Note.     The  above  theorem  may  be  enunciated  thus,  coaxial  tri- 
angles are  compolar. 


*  The  theorems  of  this  and  the  preceding  pages  are  attributed  to 
Gerard  Desargues  (born  at  Lyons  1593,  died  1662)x 


446  BOOK   VL 


ADDITIONAL  PEOPOSITION  21. 

If  a  hexagon  he  inscribed  in  a  circle,  the  intersections  of  -pairs  of 
opposite  sides  lie  on  a  straight  line.  * 

Let  ABCDEF  be  a  hexagon  inscribed  in  a  given  circle. 

Let  the  pairs  of  sides  AB,  DE ;  BG,  EF -,  CD,  FA  meet  in  L,  iH, 
N  respectively,  and  let  AB,  CD  meet  in  P,  and  BC,  DE  in  Q. 


^  L  M 

Because  A,  B,  G,D,  E,  F  are  points  on  a  circle, 
A  (BCDF),  E  (BCDF)  are  like  anharmonic  pencils;  (Add.  Prop.  16) 
and  because  PCDN  is  a  transversal  of  the  pencil  A  (BCDF), 
and  BCQM  is  a  transversal  of  the  pencil  E  (BCDF), 
PCDN,  BCQM  are  like  anharmonic  ranges ;  (Add.  Prop.  13) 
and  they  have  a  common  point  C ; 
therefore  the  lines  PB,  DQ,  NM  meet  in  a  point,  (Add.  Prop.  14) 

that  is,  AB,  DE,  NM  meet  in  a  point, 
or,  in  other  words,  the  points  L,  M,  N  lie  on  a  straight  line. 


EXEECISES. 

1.  If  the  tangents  to  the  circumscribed  circle  of  a  triangle  ABG 
at  A,  B,  C  meet  the  sides  BC,  CA,  AD,  in  L,  M,  N,  then  L,  M,  N 
lie  on  a  straight  line. 

2.  If  ACE,  BDF  be  two  ranges,  then  the  intersections  of  the 
pairs. of  straight  lines  AB,  DE  ;  BC,  EF ;  CD,  FA  are  coUinear. 

*  This  theorem  is  true  for  any  conic.  It  was  discovered  at  the 
age  of  sixteen  by  Blaise  Pascal  (born  at  Clermont  1623,  died  at  Paris, 
1662). 


PASCAL'S  AND   BBIANCHON'S  THEOREMS.     447 


ADDITIONAL  PEOPOSITION  22. 

If  a  hexagon  be  described  about  a  circle,  the  straight  lines  joining 
opposite  vertices  pass  through  a  point.* 

Let  ABCDEF  be  a  hexagon  described  about  a  given  circle. 
Let  AB  meet  CD,  DE  in  L,  M  and  EF  meet  BC,  CD  in  N,  P. 


Because  the  six  sides  of  the  hexagon  are  tangents  to  a  circle,  the 
points  where  the  sides  AB,  EF  are  cut  by  the  remaining  sides  of  the 
hexagon  form  like  anharmonic  ranges,  (Add.  Prop.  17) 

that  is,  ABLM,  FNPE  are  like  anharmonic  ranges ; 
therefore  D  {ABLM),  C{FNPE)  are  like  anharmonic  pencils; 
and  they  have  a  common  ray  LGDP-, 
therefore  the  pairs  of  rays  DA,  GF;  DB,  CN;  DM,  CE  intersect  on 
a  straight  line,  (Add.  Prop.  15) 

that  is,  DA ,  CF  meet  in  a  point  on  the  line  BE, 
or,  in  other  words,  AD,  BE,  GF  meet  in  a  point. 


EXERCISE. 

1.     If  the  inscribed  circle  of  a  triangle  ABG  touch  the  sides  BG, 
GA,  AB  in  D,  E,  F,  then  AD,  BE,  GF  meet  in  a  point. 

*  This  theorem  also  is  true  for  any  conic.    It  was  discovered  by 
Charles  Julien  Brianchon  (born  at  Sevres,  1785). 


448  BOOK   VI. 


SIMILAE  FIGUKES. 

If  two  similar  figures  be  placed  so  that  their  corresponding  sides 
are  parallel  in  pairs,  the  figures  are  then  said  to  be  similar  and 
similarly  situate. 

It  appears  from  Proposition  32  that,  if  two  triangles  be  similar 
and  similarly  situate,  then  the  lines  joining  pairs  of  corresponding 
vertices  meet  in  a  point.  It  is  easily  proved  that,  if  two  polygons  of 
any  number  of  sides  be  similar  and  similarly  situate,  then  the  lines 
joining  pairs  of  corresponding  vertices  meet  in  a  point,  and  further 
that  the  ratio  of  the  distances  of  this  point  from  any  pair  of  cor- 
responding points  of  the  two  figures  is  constant,  and  is  equal  to  the 
ratio  of  a  pair  of  corresponding  sides  of  the  two  figures.  Such  a 
point  is  called  in  consequence  a  centre  of  similitude  of  the  two 
figures.  -- 

It  is  also  easily  seen  that,  if  a  point  be  a  centre  of  similitude  of 
two  figures,  it  is  also  a  centre  of  similitude  of  any  two  figures 
similarly  described  with  reference  to  the  first  pair  of  similar  figures  ; 
for  instance,  if  a  point  be  a  centre  of  similitude  of  two  triangles,  it  is 
also  a  centre  of  similitude  of  the  circumscribed  circles  of  the  triangles, 
and  also  a  centre  of  similitude  of  the  inscribed  circles  and  so  on. 

If  a  pair  of  corresponding  points  lie  on  the  same  side  of  the 
centre  of  similitude,  it  is  called  a  centre  of  direct  similitude:  if  a 
pair  of  corresponding  points  lie  on  opposite  sides  of  the  centre  of 
similitude,  it  is  called  a  centre  of  inverse  similitude. 

It  must  be  noticed  that  the  centre  of  direct  similitude  is  at  an 
infinite  distance  in  the  case  when  the  two  similar  figures  are  equal. 

As  an  example  we  will  prove  an  important  theorem  with  reference 
to  the  centres  of  similitude  of  the  circumscribed  circle  and  the  Nine 
Point  circle  of  a  triangle  : 

The  orthocentre  and  the  centroid  of  a  triangle  are  the  centres 
of  direct  and  inverse  similitude  of  the  circumscribed  circle  and  the 
Nine  Point  circle  of  the  triangle. 


CENTRES   OF  SIMILITUDE. 


449 


If  ABC  be  a  triangle,  and  D,  E,  Fhe  the  middle  points  of  its  sides, 
the  triangles  ABC,  DEF  are  equiangular  to  one  another, 
and  the  ratio  of  AB  to  DE  is  equal  to  the  ratio  2  to  1. 

(Add  Prop.,  page  101.) 
Also  the  radii  of  the  circles  ABC,  DEF  are  in  the  ratio  of  2  to  1. 
Because  the  lines  AD,  BE,  CF  meet  in  a  point  G, 
and  the  ratios  ^G  to  GD,  BG  to  GE,  CG  to  GF  are  each  equal  to  2  to  1, 

(Add  Prop.,  page  103.) 

A 


G  is  the  centre  of  inverse  similitude  of  the  two  circles  ABC,  DEF. 
Again,  if  P  be  the  orthocentre  of  the  triangle  ABC, 

and  a,  6,  c  be  the  middle  points  of  PA,  PB,  PC, 
the  ratios  PA  to  Pa,  PB  to  Pb,  PC  to  Pc  are  each  equal  to  2  to  1. 
Therefore  P  is  the  centre  of  direct  similitude   of  the   triangles 
ABC,  abc,  and  therefore  of  the  circles  ABC,  abc. 

Now  the  circles  abc,  DEF  are  identical.       (Add.  Prop.,  page  271.) 
Therefore  P  is  the  centre  of  direct  simihtude  of  the  circles  ABC, 
DEF. 

Now,  if  0,  0'  be  the  centres  of  the  circles  ABC,  DEF, 
since  the  radii  of  the  circles  are  in  the  ratio  2  to  1, 
their  centres  of  similitude  lie  in  the  straight  line  00', 
and  divide  the  distance  internally  and  externally  in  the  ratio  2  to  1. 
That  is,  P,  0',  G,  0  lie  on  ar  straight  line, 

and  PO  is  equal  to  twice  PO', 
and  GO  is  equal  to  twice  GO'. 


450  BOOK   VI. 


ADDITIONAL  PEOPOSITION  23. 

Every  straight  line  which  passes  through  the  extremities  of  two 
parallel  radii  of  tivo  fixed  circles  passes  through  a  centre  of  similitude 
of  the  circles. 

Let  AP,  BQ  be  two  parallel  radii  of  two  given  circles,  whose 
centres  are  A,  B. 

Draw  ABy  PQ  and  let  them,  produced  if  necessary,  meet  at  S. 

(2) 


Because  the  triangles  APS,  BQS  are  equiangular  to  one  another, 
AS  is  to  BS  as  AP  to  BQ ;  (Prop.  4) 

that  is,   S  is  a  point,   which  divides  the  distance  AB  externally 
(fig.  1)  or  internally  (fig.  2)  in  the  ratio  of  the  radii; 

therefore  ^S  is  a  fixed  point.       (Ex.  1,  page  359.) 
Again,  because  the  triangles  APS,  BQS  are  equiangular  to  one 
another,  SP  is  to  SQ  as  AP  to  BQ ;  (Prop.  4) 

that  is,  the  ratio  of  the  distances  SP  to  SQ  is  a  constant  ratio ; 
therefore  S  ib  &  centre  of  similitude  of  the  circles. 

In  figure  1,  where  the  two  radii  AP,  BQ  are  drawn  in  the  same 
sense,  Sis  in  the  line  of  centres  AB  produced,  and  the  two  distances  SP, 
SQ  are  drawn  in  the  same  direction.    Sis  a.  centre  of  direct  similitude. 

In  figure  2,  where  AP,  BQ  are  drawn  in  opposite  senses,  S  is  in 
the  line  of  centres  AB,  and  the  two  distances  SP,  SQ  are  drawn  in 
opposite  directions.    Sis  a.  centre  of  inverse  similitude. 


CENTRES  OF  SIMILITUDE.  451 


ADDITIONAL  PROPOSITION  24. 

If  a  circle  he  drawn  to  touch  two  given  circles,  the  straight  line 
which  passes  through  the  points  of  contact  passes  through  one  of  the 
centres  of  similitude  of  the  given  circles. 

Let  a  circle  be  drawn  to  touch  two  given  circles,  whose  centres  are 
A,  B,  at  P,  Q. 

Draw  -4P,  BQ  and  let  them,  produced  if  necessary,  meet  at  0  *. 

Draw  PQ  and  let  it,  produced  if  necessary,  meet  the  circles  again 
atP',  Q':  draw^Q'. 


Because  the  circles  touch  at  P,  the  centre  of  the  circle  which 
is  described  to  touch  the  given  circles  must  lie  in  ^P ; 

(III.  Prop.  10,  Coroll.) 
similarly  it  must  lie  in  PQ ; 
therefore  0  is  the  centre ; 
therefore  the  angle  OPQ  is  equal  to  the  angle  OQP, 
which  is  equal  to  the  angle  BQQ'  and  therefore  to  the  angle  BQ'Q\ 

therefore  APO,  BQ'  are  parallel ;  (I.  Prop.  28) 

therefore  PQ  passes  through  a  centre  of  simiUtude.  (Add.  Prop.  23.) 

*  If  ^P,  PQ  be  parallel,  so  that  the  point  0  is  infinitely  distant, 
and  the  radius  of  the  circle  which  touches  the  given  circles  at  P,  Q 
is  infinitely  large,  PQ  becomes  one  of  the  common  tangents  of  the 
given  circles,  which  common  tangents  pass  through  S. 


452  BOOK   VL 


ADDITIONAL  PKOPOSITION  25. 

If  two  straight  lines  he  drawn  through  a  centre  of  similitude  of  two 
given  circles  to  cut  the  circles,  a  pair  of  chords  of  the  two  circles  join- 
ing pairs  of  inverse*  points  intersect  on  the  radical  axis  of  the  given 
circles,  and  a  pair  of  chords  of  the  two  circles  joining  pairs  of  cor- 
responding points  are  parallel. 

Let  >S  be  a  centre  of  similitude  of  two  given  circles  PQqp,  P'Q'q'p' 
and  let  SPQP'Q',  Spqp'q'  be  any  two  straight  lines  drawn  through 
S  cutting  the  circles ;  P,  Q' ;  Q,  P' ;  p,  q' ;  q,  p'  being  pairs  of 
inverse  points  and  P,  P'\  Q,  Q'  ;p,  p'\  q,  q'  being  pairs  of  correspond- 
ing points. 

Draw  Pq,  Q'p',  and  let  them,  produced  if  necessary,  meet  at  R. 


Because  S  ia  a.  centre  of  similitude,  Sp  is  to  8p'  as  SQ  to  SQ'; 

therefore  Qp,  Q'p'  are  parallel ;      (Prop.  2,  Part  2) 
therefore  the  angles  PQp,  PQ'p'  are  equal  j   (I.  Prop.  29) 
and  the  angles  PQp,  Pqp  are  equal;      (III.  Prop.  21) 
therefore  the  angles  PQ'p'-,  Pqp  are  equal. 
Therefore  the  four  points  P,  q,  p',  Q'  lie  on  a  circle, 

(III.  Prop.  22,  Coroll.) 

and  the  rectangle  RP,  Rq  is  equal  to  the  rectangle  RQ',  Rp'; 

(III.  Prop.  36) 

therefore  the  squares  on  the  tangents  drawn  from  R  to  the  two  circles 
are  equal, 

and  consequently  R  lies  on  the  radical  axis  of  the  given  circles. 

(Add.  Prop,  page  264.) 

*  For  this  name  see  page  460. 


CENTRES  OF  SIMILITUDE.  453 

ADDITIONAL  PEOPOSITION  26. 

IJ  a  straight  line  he  drawn  through  a  centre  of  similitude  of  two 
given  circles,  the  rectangle  contained  by  the  distances  of  two  inverse 
points  is  constant. 

Let  /S  be  a  centre  of  similitude  of  two  given  circles,  whose  centres 
are  A,  B. 

Draw  MNS  a  common  tangent  through  S  (see  note  on  page  451), 
and  let  SQ'QP'P  be  any  straight  line  through  S  cutting  the  circles 
in  P,  P'  and  Q,  Q'. 

Draw  AP,  BQ,  AM,  BN. 


Because  the  rectangle  SP,  SP'  is  equal  to  the  square  on  SM, 

SP  is  to  SM  as  SM  to  SP'.      (Prop.  17,  Part  2.) 
And  because  S  ia  a,  centre  of  similitude, 

SP  is  to  SQ  as  SM  to  SN; 
and  therefore  SP  is  to  SM  a.a  SQ  to  SN;     (V.  Prop.  9) 
therefore  SQ  is  to  SN  as  SM  to  SP';         (V.  Prop.  5) 
therefore  the  rectangle  SQ,  SP'  is  equal  to  the  rectangle  SM,  SN. 

(Prop.  16,  Part  1.) 
Similarly  it  can  be  proved  that  the  rectangle  SP,  SQ'  is  equal 
to  the  rectangle  SM,  SN. 

EXEBCISE. 

1.    Prove  that  in  the  above  figure  the  rectangle  PQ,  P'Q'  is  equal 
to  the  square  on  MN, 


454  BOOK   VI. 

ADDITIONAL   PEOPOSITION  27. 

The  six  centres  of  similitude  of  three  given  circles  taken  in  pairs  lie 
three  by  three  on  four  straight  lines  *. 

Let  A,  B,  C  he  the  centres  of  three  given  circles;  let  a,  b,  c  be  the 
radii  of  the  A,  B^  G  circles,  and  let  D,  D'  be  the  centres  of  direct  and 
of  inverse  similitude  of  the^,  G  circles,  E,  E'  those  of  the  G,  A  circles, 
and  F,  F'  those  of  the  ^,  J5  circles. 


Because  BD  is  to  DO  as  b  to  c, 
and  GE  is  to  EA  as  e  to  a, 
and  AF  is  to  FB  as  a  to  b,       (Add.  Prop.  23) 
therefore  the  ratio  compounded  of  the  ratios 

BB  to  DG,  GE  io  EA  and  AF  to  FB, 
is  equal  to  the  ratio  compounded  of  the  ratios 

6  to  c,  c  to  a  and  a  to  b,  that  is  to  unity ; 

therefore  DEF  is  a  straight  line.       (Add.  Prop.  2.) 
Similarly  it  can  be  proved  that  DE'F',  D'EF',  D'E'F  are  straight 
lines, 

and  also  that  the  lines  of  each  of  the  sets  AD',  BE',  GF';  AD',  BE,  GF ; 
AD,  BE',  GF;  AD,  BE,  GF'  meet  in  a  point.  (Add.  Prop.  4.) 

*  These  lines  are  called  the  axes  of  similitude  of  the  three  circles. 


CIRCLES  TOUCHING  TWO   CIRCLES.       455 

ADDITIONAL  PEOPOSITION  28. 

If  a  point  on  the  radical  axis  of  two  given  circles  he  joined  to  the 
points  of  contact  of  a  circle,  which  touches  both  the  given  circles,  by 
straight  lines  which  cut  the  circles  again,  another  circle  can  be  de- 
scribed to  touch  the  given  circles  at  the  points  of  section. 

Let  0  be  a  point  on  the  radical  axis  of  the  given  circles  A,  B; 
and  let  a  circle  touch  them  at  P,  Q. 

Draw  OP,  OQ  and  let  them  meet  the  A,  B  circles  in  p,  q.  Draw 
PT,  QS,  pt,  qs  the  tangents  to  the  circles  at  P,  Q,  jp,  q  and  draw 

8 


Because  0  is  on  the  radical  axis  of  the  circles  A,  B, 

the  rectangle  OP,  Op  is  equal  to  the  rectangle  OQ,  Oq; 

therefore  P,  Q,  q,  p  lie  on  a  circle.  (Ex.  1,  page  253.) 
The  difference  of  the  angles  tpO,  sqO  is  equal  to 

the  difference  of  the  angles  TPO,  SQO, 
which  is  equal  to  the  difference  of  the  angles  PQO,  QPO, 
which  is  equal  to  the  difference  of  the  angles  qpO,  pqO ; 

(in.  Prop.  22) 
therefore  the  angle  tpq  is  equal  to  the  angle  sqp. 
It  is  therefore  possible  to  describe  a  circle  touching  the  circles 
A,  B  atp,  q. 

CoROLLABY.  If  the  radical  centre  of  three  given  circles  he  joined 
to  the  points  of  contact  of  a  circle,  which  touches  all  the  three  given 
circles,  by  straight  lines  which  cut  the  circles  again,  another  circle  can 
be  described  to  touch  the  given  circles  at  the  points  of  section. 

(See  Ex.  140,  page  284.) 


456  BOOK   VL 


ADDITIONAL  PEOPOSITION  29. 

If  two  circles  he  drawn  to  touch  three  given  circles,  so  that  the 
straight  line  joining  the  tioo  points  of  contact  on  each  of  the  given  circles 
passes  through  the  radical  centre  of  the  given  circles,  the  radical  axis 
of  the  pair  of  circles  is  one  of  the  axes  of  similitude  of  the  three  given 
circles. 

Let  PQR,  pqr  be  a  pair  of  circles  touching  three  given  circles  A, 
By  G  &t  P,  p;  Q,  q;  R,  r,  so  that  Pp,  Qq,  Rr  pass  through  0  the 
radical  centre  of  the  circles  A,B,C.  (Add.  Prop.  28,  Coroll.)' 


Draw  PQi  pq  and  let  them,  produced  if  necessary,  meet  in  F. 
Because  the  circles  PQR,  pqr  touch  the  circles  A,  B, 
the  lines  PQ,  pq  pass  through  one  of  the  centres  of  similitude  of  the 
circles  A  and  B  ;  (Add.  Prop.  24) 

therefore  the  rectangle  FP,  FQ  is  equal  to  the  rectangle  Fp,  Fq ; 

(Add.  Prop.  26.) 
therefore  i^  is  a  point  on  the  radical  axis  of  the  circles  PQR,  pqr. 
Similarly  it  can  be  proved  that  a  centre  of  similitude  of  each  of 
the  pairs  of  circles  B,  C  and  C,  A  lies  on  the  radical  axis  of  PQR, 
pqr. 

Therefore  the  radical  axis  of  the  circles  PQR,  pqr  is  an  axis  of 
similitude  of  the  circles  A,  B,  G. 


CIRCLES  TOUCHING   THREE  CIRCLES.     457 


ADDITIONAL  PKOPOSITION  30. 

If  a  pair  of  chords  of  two  given  circles  intersect  in  the  radical 
axis,  and  if  the  chords  intersect  the  polars  of  one  of  the  centres  of 
similitude  in  two  points  collinear  loith  the  centre  of  similitude,  then 
the  points  of  intersection  of  the  chords  ivith  the  circles  lie  two  by  two 
on  two  straight  lines  through  the  centre  of  similitude. 

Let  S  be  one  of  the  centres  of  similitude  of  two  given  circles  A,  J5; 
let  L,  M  be  two  points  collinear  with  S,  on  the  polars  of  S  with 
respect  to  the  circles  A,  B\  and  let  0  be  a  point  on  the  radical  axis. 

Draw  OL,  and  let  it  intersect  the  circle  A  in  P,  p. 

Draw  SP,  Sp,  and  let  them  meet  the  circle  A  in  P',  p',  and  let  the 
corresponding  points  to  P,  P',  p,  p\  where  the  lines  meet  the  circle 
B  be  Q',  Q,  q\  q. 


Because  L  is  on  the  polar  of  S,  the  line  P^  passes  through  L. 

(Add.  Prop,  page  261.) 
Because  S  is  the  centre  of  similitude,  and  the  intersections  of 
Pp,  P'p\  and  of  Q'q',  Qq  are  corresponding  points,  and  Pp,  P'p'  inter- 
sect at  L, 

and  L,  31  are  corresponding  points, 

therefore  Qq,  Q'q'  intersect  at  M; 

and  because  S  is  the  centre  of  similitude, 

Pp,  Qq  intersect  on  the  radical  axis,    (Add.  Prop.  25) 

that  is,  Qq  must  pass  through  0  ;  and  it  also  passes  through  M. 

Therefore  the  points  Q,  q  obtained  by  the  above  construction  are 

the  points  where  the  straight  line  OM  cuts  the  circle  B,  that  is,  if  OL 

cut  the  circle  ^  in  P,  ^  and  OM  cut  the  circle  B  in  Q,q,  then  PQ,  pq 

pass  through  S. 

T.  E.  30 


458 


BOOK    VI. 


ADDITIONAL  PBOPOSITION  31. 

To  describe  a  circle  to  touch  three  given  circles.* 

Let  A,  B,  G  be  three  given  circles.  Find  0  the  radical  centre  of 
the  circles  A,  B,  C,  and  draw  DEF  one  of  their  axes  of  similitude. 

Find  the  poles  L,  M,  N  of  the  line  DEF  with  respect  to  the  circles 
A,B,C  respectively.  Draw  OL,  OM,  ON  and  let  them,  produced  if 
necessary,  cut  the  circles  A,  B,  C  respectively  in  P,  ^;  Q,  q;  R,  r. 


Because  L,  M  are  the  poles  of  DEF  with  respect  to  the  circles 
A,  J5,  therefore  L,  31  are  on  the  polars  of  F  with  respect  to  the  circles 
A,Bi 

and  because  F  is  a  centre  of  similitude  of  the  two  circles, 
therefore  LMF  is  a  straight  line. 
Therefore  the  line  PQ  passes  through  F.  (Add.  Prop.  30.) 


*  This  solution  of  the  problem  is  due  to  Joseph  Diez  Gergonne 
(born  at  Nancy  1771,  died  at  Montpellier  1859). 


CIRCLES  TOUCHING   THREE  CIRCLES.     459 


Therefore  a  circle  can  be  described  through  P,  Q  to  touch  the 
given  circles  at  P,  Q.  (Add.  Prop.  24.) 

Similarly  it  can  be  proved  that  circles  can  be  described  through 
Q,  P,  and  through  P,  P  to  touch  at  each  pair  of  points. 
Therefore  these  three  circles  are  identical  (see  Ex.  140,  page  284), 
that  is,  the  circle  PQR  touches  the  given  circles  at  P,  Q,  P. 
Similarly  the  circle  pqr  touches  the  given  circles  at  p,  q,  r. 

Since  there  are  four  axes  of  similitude,  and  since  two  circles  can 
be  obtained  by  the  foregoing  construction  from  each  axis  of  similitude, 
there  are  2  x  4  or  8  circles  which  can  be  described  to  touch  three  given 
circles. 

It  is  readily  seen  that,  if  three  circles  touch  a  fourth  circle,  there 
are  two  distinct  possible  types  of  configuration  of  the  three  circles 
relatively  to  the  circle  which  they  touch  : 

(1)  the  three  circles  may  lie  on  the  same  side  of  the  fourth  circle, 

(2)  two  of  the  three  circles  may  lie  on  one  side  and  the  third  on 
the  other  side  of  the  fourth  circle. 

Since  any  one  of  the  three  given  circles  may  be  the  one  which  lies 
by  itself  on  the  one  side  of  the  fourth  circle,  the  type  (2)  may  be  sub- 
divided into  three  different  groups. 

It  can  be  proved  without  much  difficulty  that  of  the  eight  circles 
which  can  be  described  to  touch  three  given  circles  A,  B,  C,  there 
are  four  pairs  of  circles,  such  that 

^,  P,  0  lie  on  the  same  side  of  each  circle  of  one  pair ; 

A  lies  on  one  side,  and  P,  C  on  the  other,  of  each  circle  of  a 
second  pair ; 

P  lies  on  one  side,  and  C,  A  on  the  other,  of  each  circle  of  a  third 
pair ; 

and  C  lies  on  one  side,  and  -4,  P  on  the  other,  of  each  circle  of  a 
fourth  pair. 

Each  of  these  four  pairs  of  circles  is  obtained  by  the  above  con- 
struction from  one  of  the  axes  of  similitude  of  the  given  circles. 


30—2 


400  BOOK   VI. 


INVEKSION. 

1.  We  will  now  proceed  to  give  an  account  of  a  geometrical  Method 
called  Inversion,  and  we  will  do  so  without  stating  the  theorems 
which  we  are  about  to  establish  in  the  formal  way  in  which  theorems 
have  been  stated  heretofore.  We  will  further  depart  from  our  former 
method  by  making  use  of  the  arithmetical  method  of  representing 
geometrical  magnitudes  (see  page  135),  so  that  we  shall  not  be 
debarred  from  using  fractions  to  represent  the  ratios  of  geometrical 
quantities,  and  if  necessary  we  shall  use  the  signs  ordinarily  used  in 
Algebra  to  signify  addition,  subtraction,  and  the  other  elementary 
operations. 

2.  Definition.  If  0  be  a  fixed  point  and  P  any  other  point,  and  if 
on  the  straight  line  OP  (produced  if  necessary)  we  take  a  point  P'  such 
that 

OP  .OP'  =  a\ 
where  a  is  a  constant,  then  each  of  the  points  P,  P'  is  called  the 
inverse  of  the  other  with  respect  to  the  circle  whose  centre  is  0  and 
radius  a. 


The  straight  line  OP  is  often  called  the  radius  vector  of  the 
point  P. 

The  point  0  is  called  the  pole  of  inversion  and  a  the  radius  of 
inversion. 

If  the  point  P  trace  out  a  curve,  the  curve  which  is  the  locus  of 
P'  is  called  the  inverse  of  the  curve  which  is  the  locus  of  P. 


INVERSION.  461 

3.  If  we  consider  the  points  P',  P",  which  are  the  inverses  of  P 
with  respect  to  the  same  pole  and  different  radii  of  inversion  a,  b, 
since  OP.  OP'=a?  and  OP  .  OP"  =  h^, 

therefore  OP'  :  OP"  =  a?  :  62. 

It  follows  that  since  the  points  P',  P"  lie  on  the  same  radius  vector, 
and  OP'  :  OP"  is  a  constant  ratio,  the  loci  of  P',  P"  are  two  similar 
curves  of  which  0  is  a  centre  of  similitude. 

4.  If  P,  P',  and  Q,  Q!  be  two  pairs  of  inverse  points,  then  since 

OP  .  OP'=a^  and  OQ  .  OQ'  =  a\ 
therefore  OP  .  0F=  OQ  .  OQ' ; 

it  follows  that  P,  Q,  Q',  P'  lie  on  a  circle,  and  the  angle  OPQ  is  equal 
to  the  angle  OQ'P'. 

Again,  if  the  line  OQQ'  approach  nearer  and  nearer  to  the  posi- 
tion of  OPP',  the  lines  QP,  Q'F  in  the  limit  are  the  tangents  to  the 
curves  which  are  the  loci  of  P  and  P'  at  P  and  P'  respectively. 

(See  page  217.) 

We  may  state  this  result  in  the  form  : 

Two  inverse  curves  at  two  inverse  points  cut  the  radium  vector 
through  the  pole  of  inversion  at  the  same  angle  on  opposite  sides. 


It  follows  that  any  two  curves  cut  at  the  same  angle  as  their 
inverse  curves  at  the  inverse  point.     (See  Definition  on  page  266.) 

5.  From  the  definition  of  inversion  it  follows  at  once  that  a 
straight  line  through  the  pole  inverts  into  itself. 

Three  points  A,  B,  C  in  the  same  radius  vector,  whose  distances 
OA,  OB,  00  are  such  that  0A  +  0C=20B,  invert  into  three  points 
A',  B'y  C\  whose  distances  OA',  OB',  OC  are  such  that 

1       _1___2^ 
OA''^  0C~  OB'' 


462 


BOOK    VL 


Three  magnitudes  such  as  OA,  OB,  OG  are  said  to  be  in  Arith- 
metical Progression,  and  three  magnitudes  such  as  0A\  0B\  OC  are 
said  to  be  in  Harmonical  Progression. 

Because  _L+^^^A^, 

therefore  OB'  .  00' +0 A'  .  OB' =  20 A'  :  OC  ; 
whence  OA'  .B'C'  =  OC'  .  A'B', 
andOA'  :A'B'  =  OC'  :  G'B'; 
therefore  OA'B'C  is  a  harmonic  range. 

Also  three  points  A,  B,  C  in  the  same  radius  vector,  whose  distances 
OA,  OB,  OC  are  such  that  OA  .  OC=OB^,  invert  into  three  points 
A',  B',  C,  whose  distances  are  such  that  OA' .  OC'=OB'^. 

Three  magnitudes  such  as  OA,  OB,  OG  are  said  to  be  in 
Geometrical  Progression. 

It  may  be  noticed  that  three  magnitudes  a,  b,  c  are  in  Arithmetical, 
Geometrical,  or  Harmonical  Progression  according  as 
a  —  b  :  b  —  c  =  a  :  a, 
a-b  :  b-c  =  a  :  b, 
or  a-b  :b-c  =  a:  c  respectively. 

6.  Let  P  be  a  point  on  a  fixed  straight  line ;  draw  OA  perpendicular 
to  the  line,  and  on  OA,  produced  if  necessary,  take  the  point  A'  such 
that  OA.OA'  =  a'^. 


Let  P'  be  the  inverse  of  P,  so  that  OP .  OP'  =  a^ :  then 
OP.OP'=.OA.OA';  therefore  P,  A,  A',  P'  lie  on  a  circle,  and  the 
angle  0P'^'=  the  angle  O^P  =  a  right  angle;  therefore  the  locus  of 
P'  is  a  circle  whose  diameter  is  OA'.     Hence  the  theorem  : 

A  straight  line  which  does  not  pass  through  the  pole  of  inversion 
inverts  into  a  circle  which  passes  through  the  pole  and  touches  there  a 
parallel  to  the  given  line. 


INVERSION. 


463 


7.    Again  let  P  be  any  point  on  a  circle,  of  which  OA  is  a  diameter. 
Take  in   OA,  produced  if   necessary,  the   point  A'  such  that 
OA.OA'=a?. 


Let  F  be  the  inverse  of  P,  so  that  OP.  OP'=a^; 
then  OP.OP'=OA.OA'; 
therefore  P,  A,  A\  P',  lie  on  a  circle, 

and  the  angle  P'A'0=the  angle  OPA=a,  right  angle,  i.e.  the  locus 
of  P'  is  a  straight  line  through  A'  the  inverse  of  A  drawn  at  right 
angles  to  OA'.    Hence  the  theorem  : 

A  circle  which  passes  through  the  pole  of  inversion  inverts  into  a 
straight  line  parallel  to  the  tangent  at  the  pole. 

8.  Again,  let  P  be  any  point  on  a  circle,  which  does  not  pass  through 
the  pole,  and  P'  be  the  inverse  point  so  that  OP .  OP'  =  a^ :  let  OP,  pro- 
duced if  necessary,  cut  the  circle  in  Q ;  find  A  the  centre  and  draw  QA , 
and  let  OA,  produced  if  necessary,  cut  P'B  drawn  parallel  to  QA  in  B. 


Because  OP .  OQ=t%  where  t  represents  the  length  of  the  tangent 
from  0  to  the  given  circle,  and  OP  .  OP'  =  a\ 
therefore  OP' :  OQ  =  a^:  t'^=OB  :  OA=BP' :  AQ  ; 

therefore  the  locus  of  P'  is  a  circle  such  that  B  is  its  centre  and  0  is 
one  of  the  centres  of  similitude  of  it  and  the  given  circle. 

Hence  the  theorem :  A  circle,  which  does  not  pass  through  the  pole 
of  inversion,  inverts  into  a  circle  which  does  not  pass  through  the  pole, 
and  is  sv^h  that  the  pole  is  a  centre  of  similitude  of  the  two  circles. 


464 


BOOK   VI. 


9.    Again,  if  P,  P'  and  Q,  Q'  be  two  pairs  of  inverse  points, 
OP.  OP'=OQ.OQ'  =  a\ 
the  triangles  OPQ,  OQ'P'  are  similar. 

P'Q'^OQ:  ^OQ'.OQ_       a^ 
PQ~OP~OP.  OQ  ~  OP.OQ' 


Therefore 


P'Q:=a 


OP.OQ 


'OP.OQ' 

Thus  the  distance  between  two  points  in  a  figure  is  expressed  in 
terms  of  the  distance  between  their  inverse  points  and  the  distances 
of  the  inverse  points  from  the  pole. 

10.  Now  let  us  take  the  ordinary  definition  of  a  circle,  i.e.  the 
locus  of  a  point  P  such  that  its  distance  PC  from  a  fixed  point 
G  is  constant. 

Let  F  be  the  inverse  point  of  P ;  take  C  the  inverse  point  of  C. 

If  we  please  we  may  in  this  investigation  choose  the  radius  of 
inversion  (see  page  461)  equal  to  the  tangent  drawn  from  0  to  the 
original  circle ;  then  P'  lies  on  the  same  circle  as  P, 

and  OG.OO'=OT^--^OP.  OP'. 


Therefore  the  triangles  OP'G',  OOP  are  similar, 
and  OP':P'C'=OG:CP. 

Hence  the  theorem  (which  has  been  already  proved  otherwise. 
Add.  Prop.  5) : 

The  locus  of  a  point  the  distances  of  which  from  two  fixed  points 
are  in  a  constant  ratio  is  a  circle. 

It  will  be  seen  that  the  straight  line  TT\  which  is  the  polar  of  O, 
and  the  circle  on  00  as  diameter  are  inverses  of  each  other. 


INVERSION. 


465 


11.  Again,  it  is  known  from  Book  III.  that  all  straight  lines  which 
cut  a  circle  at  right  angles  pass  through  the  centre,  and  also  that  all 
straight  lines  through  the  centre  cut  the  circle  at  right  angles. 

If  therefore  we  invert  a  figure  consisting  of  a  series  of  straight 
lines  cutting  a  series  of  concentric  circles  (centre  C)  at  right  angles, 
we  shall  obtain  a  series  of  coaxial  circles  passing  through  0  the  pole 
of  inversion  and  through  C  the  inverse  point  of  G,  and  cutting  each 
of  a  second  series  of  circles  at  right  angles.     (Add.  Prop,  page  268.) 


Hence  the  theorem : 

A  system  of  concentric  circles  inverts  into  a  system  of  coaxial  circles. 

In  the  above  diagram  0  is  the  pole  of  inversion,  and  the  tangent 
from  0  to  the  circle  PQR  (or  P'Q'R')  is  taken  as  the  radius  of  inver- 
sion so  that  the  circle  inverts  into  itself.  The  points  0,  A,  B  in  the 
left-hand  figure  are  supposed  to  coincide  with  0,  B',  ^i'  respectively  in 
the  right-hand  figure :  the  figures  are  drawn  apart  merely  for  the  sake 
of  clearness. 

Further,  if  we  invert  the  last  system  with  respect  to  any  point,  we 
shall  get  a  system  of  exactly  the  same  nature ;  viz.  a  system  of  circles 
passing  through  two  fixed  points  and  cutting  each  of  another  system 
of  circles  at  right  angles. 

Hence  the  theorem : 

A  system  of  coaxial  circles  inverts  into  another  system  of  coaxial 
circles,  and  the  limiting  points  into  the  limiting  points. 


466  BOOK   VI. 


12.  A  circle  can  be  inverted  into  itself  with  respect  to  any  point 
as  pole  of  inversion,  if  the  tangent  to  the  circle  from  the  point  be 
taken  as  the  radius  of  inversion. 

Any  two  circles  can  be  inverted  into  themselves  with  respect  to 
any  point  on  their  radical  axis  as  the  pole  of  inversion. 

Any  three  circles  can  be  inverted  into  themselves  with  respect  to 
their  radical  centre  as  the  pole  of  inversion. 

Hence  we  conclude  that  when  one  circle  is  drawn  to  touch  two 
given  circles  at  P  and  Q,  if  0  be  any  point  on  the  radical  axis 
of  the  given  circles,  the  lines  OP,  OQ  will  cut  the  circles  again  in 
two  points  P',  Q\  such  that  another  circle  can  be  described  to  touch 
the  given  circles  at  P',  Q'. 

We  also  conclude  that  when  one  circle  is  drawn  to  touch  three 
given  circles  at  P,  Q,  B,  if  0  be  the  radical  centre,  the  lines  OP, 
OQ,  OB  cut  the  circles  again  in  three  points  P',  Q\  B'  such  that  a 
circle  described  through  them  wiU  touch  the  circles  at  P',  Q\  B'. 

These  two  theorems  have  been  proved  before  (page  455). 

13.  If  two  circles  PQO,  pqO  be  described  to  touch  two  given  circles 
A,  B  &t  P,  p;  Q,  q,  and  to  touch  each  other  at  0,  then  the  centre  of 
similitude  F,  which  is  the  point  of  intersection  of  the  straight  lines 
PQ,  pq,  must  lie  on  the  common  tangent  to  the  circles  at  0. 

(Add.  Prop.  29.) 

Hence,  if  0  be  taken  as  the  pole  of  inversion,  the  two  circles 
OPQ,  Opq  will  invert  into  two  parallel  common  tangents  to  the 
inverse  circles  (page  423),  and  therefore  the  given  circles  will  invert 
into  a  pair  of  equal  circles. 

And  since  FO^=FP  .FQ=FM  .  FN,  if  FMN  be  a  common  tangent 
to  the  given  circles,  (Add.  Prop.  26) 

we  see  that  the  pole  0  may  be  chosen  anywhere  on  the  circle,  whose 
centre  is  F  and  whose  radius  is  a  mean  proportional  between  the 
tangents  from  the  point  F  to  the  circles  A,  B. 

Hence  with  any  point  on  two  definite  circles  as  pole  of  inversion 
we  can  invert  two  given  circles  into  two  equal  circles. 

It  follows  that  with  any  one  of  certain  definite  points  as  pole  of 
inversion,  three  given  circles  can  he  inverted  into  three  equal  circles. 


INVERSION. 


467 


14.     If  we  take  A\  B',  C  three  points  in  order  on  a  straight  line, 
the  relation  A'B'  +  B'G'=A'C'  exists  between  the  segments. 


If  we  invert  with  respect  to  any  pole  0,  the  three  inverse  points 
A,  B,  C  will  lie  on  a  circle  through  O,  and  the  chords  will  satisfy  the 
relation 

AB  BC  AG  ,  .^.^ 

oaTob  +  obTog  =  oaTog  '      ^'''  P^^"  ^^^)' 

or  AB.OG  +  BG  .  OA  =  AG  .  OB, 

which  is  Ptolemy's  Theorem.  (III.  Prop.  37  B.) 

Again  if  we  take  A\  B\  G',  D',  four  points  in  order  on  a  straight 
line,  the  relation 

A'B' .  G'D'  +  A'D' .  B'G'=A'G' .  B'D' 
exists  between  the  segments.  (Ex.  1,  page  137.) 

If  we  invert  with  respect  to  any  pole  0,  the  four  inverse  points 
A,  B,  G,  D  will  he  on  a  circle  through  0,  and  the  chords  will  satisfy 
the  relation, 

AB  GD  AD  BG     _      AG  BD 

OA  .  OB  '  OG  .  OD'^  OA  .  OD  '  OB  .  0G~  OA  .  OG  '  OB  .  OD  * 
or  AB.GD  +  AD.BG=AC.BD, 

which  also  is  a  form  of  Ptolemy's  Theorem. 


im  BOOK   VI. 


PEAUCELLIEE'S   CELL. 

15.  Before  we  leave  the  subject  of  inversion,  it  will  be  as  well  to 
explain  the  nature  of  a  simple  piece  of  mechanism,  by  the  use  of 
which  the  inverse  of  any  given  curve  can  be  drawn. 

The  figure  represents  such  an  instrument ;  OA,  OB  are  two  equal 
rods  and  AP,  PB,  BP\  P'A,  four  other  equal  rods,  all  six  being  freely 
hinged  together  at  the  points  0,  A,  B,  P,  P\ 


This  instrument  is  generally  called  a  Peaucellier's  Cell*. 
Because  APBP'  is  a  rhombus, 
its  diagonals  bisect  each  other  at  right  angles  at  M;  (Ex.  1,  page  39)- 
and  because  OAB  is  an  isosceles  triangle, 
and  M  is  the  middle  point  of  AB, 
OM  is  at  right  angles  to  AB. 
Therefore  OPP'  is  a  straight  Hne, 

and  the  rectangle  OP  .  OP'=OAP-MP^  (II.  Prop.  6) 
=  OA^-AP^ 
=  a  constant. 
If  therefore  the  point  0  be  fixed,  and  P  be  made  to  trace  out  any 
curve,  P'  will  trace  out  the  inverse  curve. 

*  This  mechanical  invention  is  due  to  A.  Peaucellier,  Capitaine 
du  G^nie  (a  Nice),  who  proposed  the  design  of  such  an  instrument  as 
a  question  for  solution  in  the  Nouvelles  Annales  1864  (p.  414).  His 
solution  was  published  in  the  Nouvelles  Annales  1873  (pp.  71 — 8). 


CASETS  THEOREM.  469 


We  will  now  proceed  to  prove  a  theorem  which  establishes  a 
relation  between  six  of  the  common  tangents  of  paii-s  of  four  circles 
which  touch  a  fifth  given  circle,  of  a  kind  similar  to  that  which 
Ptolemy's  Theorem  establishes  between  the  chords  joining  the  points 
of  contact. 

Let  a  circle  whose  centre  is  0  touch  two  given  circles  whose 
centres  are  ^,  B  at  P,  Q  ;  let  PQ,  produced  if  necessary,  cut  the 
circles  A,  B  again  in  P',  Q'  and  pass  through  their  centre  of  similitude 
F ;  and  let  pqF  be  one  of  their  common  tangents  through  F. 

(Add.  Prop.  24  note.) 


Because  P  is  a  centre  of  similitude  of  the  circles, 

pP'  is  parallel  to  qQ,  and  pP  to  qQ' ',  (Add.  Prop.  25) 

therefore  ^2  :  FQ  =Fq  :  FQ, 
&ndLpq  :PQ'  =  Fq  :  FQ'; 
therefore  pq^  :  P'Q  .  PQ'  =  Fq'^ :  FQ  .  FQ' ; 
and  PgS^PQ  .  FQ' ; 
therefore  pq^= P'Q  .  PQ'.         (Ex.  1,  page  453.) 
Again  because  P  is  a  centre  of  similitude, 

OPA  is  parallel  to  Q'B,  and  OQB  to  P'A  ; 
therefore  OA  :  OP  =  P'Q  :  PQ, 
and  OB  :  OQ=PQ'  :  PQ; 
therefore  OA  .  OB  :  OP  .  OQ  =  PQ'  .  P'Q  :  PQ^, 
or  OA  .OB  :  OP^=pq^  :  PQ^; 
therefore,  if  OA  .  OB=OL\ 
then  OL  :  OP=pq  :  PQ.  (V.  Prop.  16.) 


470  BOOK  VI. 


Similarly,  if  two  other  circles,  whose  centres  are  C,  D,  touch  the 
same  circle,  centre  0,  at  R,  S,  and  rs  be  their  common  tangent, 
OC  .  OD  :  OP^~  =  rs^:  RS^ ; 
therefore,  if  OC  .  OD  =  OM^, 
then  OM  :OP=rs  :BS;  (V.  Prop.  16) 

therefore  OL  .  OM :  OP^=pq  .rs:  PQ.  RS, 
If  we  denote  the  common  tangent  to  the  two  circles  which  touch  the 
fifth  circle  at  P,  Q  by  {PQ),  and  so  on  for  the  other  pairs  of  circles, 
we  may  write  this  proportion, 

OL  .  OM:  OP^={PQ)  .  (RS)  :  PQ  .  RS. 
It  can  be  proved  in  a  similar  manner  that, 
if  OL"^=OA  .  OC  and  OM'^=OB  .  OD, 

then  OL' .  OM'  :  OP^={PR)  .  (QS)  :  PR  .  QS. 

Now  0L2  :  OL'^=OA  .  OB  :  OA  .  OC=OB  :  OC, 

and  OM'^  :  OM^=OB  .  OD  :  OC  .  OD  =  OB  :  OC ■ 

therefore  OL^  :  OL'^=OM'^  :  0M% 

and  0L'.0L'=0M'  '.OM;                    (V.  Prop.  16) 

therefore  OL  .  0M=0L' .  OM'.                       (Prop.  16.) 

Hence  we  have 

(PQ)  .  (RS)  :  PQ  .  RS={PR)  .  {QS)  :  PR  .  QS 
and  similarly  =  {PS)  .  {QR)  :  PS  .  QR. 

And  because,  if  PQRS  be  a  convex  quadrilateral, 

PR  .  QS=PQ  .  RS  +  PS  .  QR  (Ptolemy's  theorem), 
therefore  (PR)  .  {QS)  =  {PQ)  .  {RS)  +  {PS)  .  {QR}. 

This  theorem  is  generally  known  as  Casey's  Theorem*. 

It  may  be  observed  that  the  common  tangent  of  each  pair  of 
circles,  which  appears  in  the  equation,  is  that  tangent  which  passes 
through  the  same  centre  of  similitude  of  the  circles  as  the  chord 
joining  the  points  of  contact  of  the  circles  with  the  fifth  circle. 


*  This  theorem  was  discovered  by  John  Casey  (born  at  Kilbenny, 
County  Cork,  1820,  died  at  Dublin  1890). 


CASETS  THEOREM.  471 


It  is  readily  seen  that,  if  four  circles  touch  a  fifth  circle,  there  are 
three  distinct  possible  types  of  configuration  of  the  four  circles 
relatively  to  the  circle  which  they  touch  ; 

(1)  the  four  circles  may  lie  on  the  same  side  of  the  fifth  circle, 

(2)  three  of  the  four  circles  may  lie  on  one  side  and  the  fourth 
circle  on  the  other  side, 

(3)  two  of  the  four  circles  may  lie  on  one  side  and  the  other  two 
on  the  other  side. 

The  converse  of  Casey's  Theorem  may  be  stated  in  the  following 
manner :  if  an  equation  of  the  form  of  the  equation  in  Casey's  Theorem 
exist  between  the  common  tangents  of  four  circles  taken  in  pairs,  the 
common  tangents  being  chosen  in  accordance  with  one  of  the  three 
possible  types  of  configuration,  then  the  four  circles  touch  a  fifth  circle. 

The  truth  of  this  converse  theorem  which  is  often  assumed  without 
any  attempt  at  proof  can  be  proved,  but  the  proof  of  it  is  thought  to 
be  beyond  the  scope  of  this  work. 


472  BOOK   ri. 


CONTINUITY. 

Let  us  consider  a  variable  point  P  on  a  given  straight  line,  on 
which  A,  B  are  two  fixed  points.     It  is  seen  at  once  that, 

(1)  if  P  be  outside  AB  beyond  A ,  then  the  excess  of  PB  over  PA 
is  equal  to  ^7? ; 

(2)  if  P  be  in  AB,  then  the  sum  of  AP  and  PB  is  equal  to  AB, 
and 

(3)  if  P  be  outside  AB  beyond  B,  then  the  excess  of  ^P  overPP 
is  equal  to  AB. 


(1)  (2)  (3) 


We  can  write  these  results  in  the  forms 

(1)  PA+AB  =  PB, 

(2)  AB=AP  +  PB, 

(3)  AB  +  BP=AP. 

Here  we  observe  that,  while  P  changes  from  one  side  of  A  to  the 
other,  the  distance  PA,  which  vanishes  when  P  coincides  with  A, 
changes  sides  in  the  equation,  which  otherwise  remains  unchanged; 
and  again  that  while  P  changes  from  one  side  of  B  to  the  other,  the 
distance  PB  similarly  changes  sides  in  the  equation. 

A  geometrical  theorem  consists,  in  many  cases,  of  a  proof  that  a 
certain  equation  exists  between  a  number  of  geometrical  magnitudes, 
such  equation  remaining  unchanged  in  form  for  variations  in  the 
geometrical  magnitudes  involved,  consistent  with  the  conditions  to 
which  they  are  subject. 


CONTINUITY.  473 

It  is  found  in  many  of  such  theorems,  as  in  the  illustration 
which  we  have  just  given,  that,  if  subject  to  continuous  variation  of 
some  chosen  geometrical  magnitude  some  other  magnitude  con- 
tinuously diminish  and  vanish,  then  in  the  equation  which  applies  to 
the  configuration  determined  by  the  next  succeeding  values  of  the 
chosen  variable  magnitude,  the  magnitude  which  has  vanished 
appears  on  the  opposite  side  of  the  equation.  This  fact  is  due  to  the 
absence  of  any  sudden  changes  in  the  magnitudes  under  considera- 
tion.    The  general  law  that  no  sudden  change  occurs  is  often  spoken 

of  as  THE  PRINCIPLE  OF  CONTINUITY. 

Let  as  consider  a  variable  point  P  on  a  given  straight  line,  on 
which  ^  is  a  fixed  point ;  and  let  us  consider  any  equation  between 
variable  geometrical  magnitudes,  one  of  which  is  PA  the  distance 
between  P  and  A.  The  principle  of  continuity  leads  us  to  expect 
that,  if  P  in  the  variation  of  its  position  pass  from  one  side  to 
the  other  of  A,  the  sign  of  PA  in  the  equation  will  change.  In  other 
words,  we  may  consider  the  equation  to  remain  unchanged  in  form,  if 
we  resolve  to  represent  by  the  expression  PA  not  only  the  distance 
between  P  and  A,  but  also  the  fact  that  the  distance  is  measured 
from  P  towards  A.  This  result  is  at  once  obtained  by  resolving 
that  -PA  shall  represent  a  distance  equal  to  PA  and  measured  in 
the  opposite  direction;  in  other  words,  that  AP=  -PA. 

Let  us  return  to  the  consideration  of  a  variable  point  P  on  a  given 
straight  line,  on  which  A,  6  are  two  fixed  points.  It  appears  that 
the  equation  which  exists  between  the  distances  between  the  points 
takes  different  forms  according  as  P  is  (1)  in  BA  produced,  (2)  in 
AB,  or  (3)  in  AB  produced. 

If  we  allow  the  use  of  the  minus  sign,  we  may  write  these  equations, 

(1)  AB  +  PA-PB  =  0, 

(2)  AB-AP-PB  =  0, 

(3)  AB-AP  +  BP=0, 

where  each  symbol  such  &s  AB  represents  merely  the  length  of  a 
line  measured  in  the  same  direction  as  AB. 

It  is  at  once  seen  that,  if  we  adopt  the  convention  that 
PQ  =  -QP, 
all  these  equations  are  the  same ;  each  may  be  written 

AB  =  AP+PB, 
or  AB  +  BP  +  PA=:0. 

T.  E.  31 


474  BOOK   VL 

The  first  form  expresses  that  the  operation  of  passing  from  ^  to  £  is 

the  same  as  passing  from  ^  to  P  and  then  from  P  to  J5 ; 
the  second  form  expresses  that  the  aggregate  result  of  the  operations 

of  passing  from  Aio  B^  and  then  from  5  to  P  and  then  from  P 

to  A  is  to  arrive  at  the  point  A  of  starting ; 
both   of  which  facts  are  true  for  all  combinations  of  three  points 

A,B^  P  ons.  straight  line. 

The  results  of  the  theorems  contained  in  Propositions  5  and  6  of 
Book  II.  become  the  same,  if  we  take  into  account  the  fact  that  the 
distance  BD  is  measured  in  opposite  directions  in  the  two  figures  :  and 
similarly  the  results  of  Propositions  12  and  13  of  Book  II.  become 
the  same,  if  we  take  into  account  the  sign  of  CD. 

As  a  further  illustration  of  the  Principle  of  Continuity  we  will 
take  Ptolemy's  Theorem. 

Let  us  consider  a  variable  point  P  on  a  circle,  on  which  A,  B,  G 
are  three  fixed  points.     It  is  proved  in  III.  Prop.  37  B,  that 

(1)  if  P  be  in  the  arc  ^P, 

AB.PC=BC.PA  +  CA.PB; 

(2)  if  P  be  in  the  arc  BC, 

BG.PA  =  CA,PB  +  AB.PC; 
and 

(3)  if  P  be  in  the  arc  CA, 

GA  .  PB=AB  .  PG  +  BG .  PA. 
These  equations  may  be  written 

(1)  AB.PG-BG.PA-GA,PB  =  0, 

(2)  AB.PG-BG.PA  +  GA.PB  =  0, 

(3)  AB.  PG  +  BG.  PA -GA.PB  =  0. 

Hence,  while  P  passes  along  the  arc  from  one  side  of  B  to  the 
other,  the  sign  of  PB,  which  vanishes  when  P  coincides  with  P, 
changes  sign  in  the  equation,  which  otherwise  remains  unchanged, 
and  so  on  for  passage  through  C  or  ^. 


PORISMS.  475 


POEISMATIC  PEOBLEMS. 

In  some  cases,  when  a  Geometrical  problem  is  submitted  for 
solution,  it  is  found  that  some  relation  between  the  geometrical 
magnitudes  or  figures  which  are  given  is  necessary  in  order  that 
a  solution  may  be  possible,  and  that,  if  one  solution  be  possible, 
and  therefore  the  relation  exist,  the  number  of  possible  solutions 
is  infinite.     The  solution  is  then  said  to  be  indeterminate. 

Such  a  problem  is  called  a  porism. 

"We  might  take  as  an  illustration  of  such  a  problem  the  problem, 
to  construct  a  triangle  which  shall  be  inscribed  in  one  and  described 
about  another  of  two  given  concentric  circles.  It  is  easily  seen  that, 
if  one  such  triangle  exist,  the  radius  of  the  first  circle  must  be  equal 
to  twice  the  radius  of  the  second,  and  that  then  every  equilateral 
triangle  which  is  inscribed  in  the  first  circle  is  also  described  about 
the  second  circle. 

Again  we  might  take  the  problem,  to  construct  a  triangle  so  that 
each  vertex  shall  lie  on  one  of  three  given  concentric  circles,  and  that 
each  side  shall  touch  a  fourth  given  concentric  circle.  It  is  easily 
proved  by  means  of  the  method  of  rotation  (page  186)  that,  if  such  a 
triangle  be  possible,  an  infinite  number  of  such  triangles  are  possible. 

We  will  now  proceed  to  find  the  relation  which  must  exist  between 
two  circles  which  are  not  concentric,  in  order  that  it  may  be  possible 
to  construct  a  triangle  which  shall  be  inscribed  in  one  and  described 
about  the  other,  and  we  shall  prove  that,  if  it  be  possible  to  construct 
one  such  triangle,  it  is  possible  to  construct  an  infinite  number. 


31—2 


476  BOOK   VI. 


The  square  on  the  straight  line  joining  the  centres  of  the  circum- 
scribed circle  and  the  inscribed  circle  of  a  triangle  is  less  than  the 
square  on  the  radius  of  the  circumscribed  circle  by  twice  the  rectangle 
contained  by  the  radii  of  the  two  circles. 

Let  ABC  be  the  circumscribed  circle  of  the  triangle  ABC  and  DEF 
the  inscribed  circle  touching  AB  at  F. 

Find  0, 1  the  centres  of  the  circles  ABC,  DEF.  Draw  AI,  IB, 
BO,  IF  and  let  AI,  BO  produced  meet  the  circle  ABC  at  H,  K. 

Draw  BH,  HK. 


Because  the  angle  FAI  (or  BAH)  is  equal  to  the  angle  HKB, 
and  the  angle  AFI  is  equal  to  the  angle  KHB, 
the  triangles  FAI,  HKB  are  equiangular  to  one  another; 
therefore  the  rectangle  AI,  BH  is  equal  to  the  rectangle  KB,  IF. 

(III.  Prop.  37  A.) 

Because  the  angle  BIH  is  equal  to  the  sum  of  the  angles  BAI,  ABI, 

which  is  equal  to  the  sum  of  the  angles  HBC,  CBI, 

that  is,  to  the  angle  IBH ; 

therefore  BH  is  equal  to  IH ; 

therefore  the  rectangle  AI,  BH  is  equal  to  the  rectangle  AI,  IH, 

which  is  equal  to  the  difference  of  the  squares  on  OB,  01; 

(III.  Prop.  35) 
therefore  the  difference  of  the  squares  on  OB,  01  is  equal  to  the  rect- 
angle KB,  IF, 

that  is,  to  twice  the  rectangle  contained  by  the  radii. 


PORISM  OF  TWO   CIRCLES.  ill 

If  two  circles  he  such  that  one  triangle  can  he  constructed  that  is 
inscribed  in  one  circle  and  circumscrihed  ahout  the  other  circle,  an 
infinite  number  of  such  triangles  can  he  constructed. 

Let  ABC  be  a  triangle:  and  let  its  circumscribed  and  inscribed 
circles  be  described,  and  let  I  be  the  centre  of  the  inscribed  circle. 


Take  any  other  point  A'  on  the  cir- 
cumscribed circle  ABC;  draw  AI,  A' I  and 
produce  them  to  cut  the  circle  ABC  oX 
H,H'. 

With  centre  H'  and  radius  H'l  draw 
a  circle  cutting  the  circle  ABC  in  B\  C. 

Draw  A'B\  B'C,  C'A'. 


Because  H'B'  is  equal  to  H'l, 

the  angle  H'lB'  is  equal  to  the  angle  H'B' I; 
therefore  the  sum  of  the  angles  IB' A',  lA'B'  is  equal  to  the  sum  of  the 
angles  H'B'C,  C'B'I.  (I.  Prop.  32.) 

And  because  the  chords  H'B',  H'C  are  equal, 

the  angle  lA'B'  is  equal  to  the  angle  H'B'C; 
therefore  the  angles  IB' A',  C'B'I  are  equal, 
that  is,  B'l  is  the  bisector  of  the  angle  A' B'C. 
And  A'l  is  the  bisector  of  the  angle  B'A'C  ; 
therefore  I  is  the  centre  of  the  inscribed  circle  of  the  triangle  A'B'C. 
Because  AIH,  A'lH'  are  two  chords  of  the  circle  ABC, 
the  rectangle  AI,  III  is  equal  to  the  rectangle  A'l,  IH'\ 
and  the  rectangle  AI,  IH  is  equal  to  twice  the  rectangle  contained  by 
the  radius  of  the  circle  ABC  and  the  radius  of  the  inscribed  circle 
of  the  triangle  ABC,  (page  476) 

and  the  rectangle  A'l,  IH'  is  equal  to  twice  the  rectangle  contained  by 
the  radius  of  the  circle  ABC  and  the  radius  of  the  inscribed  circle 
of  the  triangle  A'B'C.  (page  476.) 

Therefore  the  radii  of  the  inscribed  circles  of  the  triangles  ABC, 
A'B'C  are  equal: 

and  the  circles  have  a  common  centre  I. 
Therefore  the  triangle  A'B'C  has  the  same  circumscribed  circle  and 
the  same  inscribed  circle  as  the  triangle  ABC. 


478 


BOOK   VL 


We  now  proceed  to  prove  a  theorem  which  might  fairly  have  been 
included  at  an  earlier  stage. 

Let  PAB  be  any  triangle  and  ^  be  a  point  in  AB,  such  that 
mAE=nEB.  Here  E  is  the  centroid  of  weights  m  and  ns^i  A  and  B. 
(See  page  425. ) 

Draw  PM  perpendicular  to  -4ij  and  draw  PE. 


Because  PA^=PE^  +  EA^  -  2ME  .  AE ;  (IL  Prop.  13) 

and  PB^  =  PE^  +  EB^  +  2ME  .  EB,  (II.  Prop.  12) 

and  because  mAE  =  nEBy 

and  therefore  mME  .  AE  =  nME  .EB; 

therefore        mPA^  +  nPB^  ={m  +  n)  PE^  +  mEA  2  +  nEB^. 

Hence  the  theorem : 

The  sum  of  any  multiples  of  the  squares  on  the  distances  of  any 
point  from  two  given  points  is  equal  to  the  sum  of  the  same  multiples  of 
the  squares  on  the  distances  of  the  given  points  from  the  centroid  of 
weights  at  the  given  points  proportional  to  those  multiples,  together 
with  the  sum  of  the  multiples  of  the  square  on  the  distance  of  the  point 
from  the  centroid. 

Because  mAE=nEB, 

therefore  mAE^  =  nAE  .  EB, 

and  mAE^  +  nEB^ =nAE  .EB  +  nEB^ 

=  nAB.EB. 

We  may  therefore  write  the  result  of  this  theorem  in  the  form 
mPA^  +  nPJ52  =  (m  +  n)  PE^  +  nBE  .  BA 
or  nPB^  -  nBE  .  BA  =  (m  +  n)  PE^  -  mPA^ 


COAXIAL  CIRCLES. 


479 


Next,  let  us  assume  P  a  point  such  that  the  tangents  P'l\  PS 
drawn  from  it  to  two  given  circles,  whose  centres  are  A,  B,  satisfy 
the  equation  mPT=nPS. 

Take  two  points  0,  0'  in  AB,  such  that 

AO .  AO'=AT^  and  BO  .  BO'  =  BS^'. 

(Add.  Prop,  page  268.) 


Because  viPT=nPS, 

mWr-  =  n''PS^; 
therefore  m"  {PA^-A  T^)  =  n^  {PB"^  -  BS^) , 

or  m^{PA'^-AO.AO')  =  n'^{PB^-BO.BO'). 

Now  by  the  last  theorem,  if  qAO=pOO'  and  rOO'  =  qO'B, 
q{PB^-BO.BO')  =  {r  +  q)PO'^-rPO^, 
and  q  {PA^  -  AO  .  AO')  =  {p  +  q)  PO"^  -pPO'\ 

Therefore 

n2  {q  +  r)  PO'^  -  n^rPO^=m^  {p  +  q)  PO^  -  m^pPO'^, 
or  {m2  {p  +  q)  +  nV}  P0^=  {n^  {q  +  r)+  m'^p}  PO'^. 

Therefore  PO  is  to  PO'  in  a  constant  ratio,  and  therefore  the 
locus  of  P  is  one  of  the  system  of  coaxial  circles,  of  which  0,  0'  are  the 
limiting  points.  (See  Note  on  page  429.) 

In  consequence  of  the  very  algebraical  character  of  the  proof 
which  has  just  been  given,  we  will  give  another  proof  of  the  same 
theorem  depending  in  a  great  measure  on  the  theory  of  similitude. 


480 


BOOK  VI. 


Let  us  again  assume  P  is  a  point  such  that  the  tangents  PT^  PS 
drawn  to  two  given  circles,  whose  centres  are  A,  B,  satisfy  the 
equation 

mPT=nPS. 

There  must  be  some  point  Q  in  AB,  such  that,  if  QH,  QK  be  the 
tangents  to  the  circles, 

'mQH  =  nQK\ 
therefore  m^PT^  =  'nPPS^, 

and  m^QH^=n^QK^; 

therefore  m^{PT^-  QH  2)  =  n^  (P^^  -  QK^) , 

or  m^  (P^2 -AQ^)=  r?  {PB^  - BQ^). 


Now  draw  two  circles,  with  centres  A  and  B,  and  radii  AQ,  BQ, 
and  let  PQ  cut  these  circles  in  L,  M  respectively;  then 

PA^-AQ^  =  PQ.PL, 
and  PP2  -  QB^  =  PQ .  PM;  (III.  Prop.  36) 

therefore  rn^PQ  .  PL  =  n^PQ  .  PM, 

or  mPPL  =  n^PM. 

Therefore  m^  {PQ  +  QL)  =  n^  {PQ  -  QM), 

or  {n^-vi^)PQ  =  m^QL  +  n^QM. 

Therefore  {n^  -  m^)  PQ  :  QM=  m^QL  +  n^QM  :  QM 

=  mPQA+n^QB  :  QB, 
since  QL  :  QM=  QA  :  QB. 

From  this  it  follows  that  the  ratio  QP  :  Q3I  is  constant ;  therefore 
the  locus  of  P  is  a  circle  which  passes  through  Q,  and  has  its  centre 
in  the  line  AB.  (See  page  450.) 

Hence  the  theorem :  the  locus  of  a  point,  such  that  the  tangents 
drawn  from  it  to  two  given  circles  are  in  a  constant  ratio,  is  a  circle. 


COAXIAL   CIRCLES.  481 


Next,  let  P  be  a  point  on  the  locus  such  that  FT,  Ft,  the  tangents 
drawn  to  two  given  circles,  are  in  a  given  ratio. 

Draw  Ss  a  common  tangent  to  the  two  given  circles. 

Let  Ss  cut  the  radical  axis  of  the  given  circles  in  M,  and  the 
circle  which  is  the  locus  of  P  in  ^  and  R. 


Because  both  Q  and  R  are  points  on  the  locus  of  P, 
QS  is  to  Qs  as  RS  to  Rs ; 
therefore  SQsR  is  a  harmonic  range. 
And  because  M  is  the  middle  point  of  Ss, 
MS^  =  Ms^  =  MQ.MR, 
and  MQ  .  MR  is  equal  to  the  square  on  the  tangent  from  M  to  the 
locus  of  P. 
It  follows  therefore  that  M  is  a  point  on  the  radical  axis  of  each 
of  the  pairs  of  circles;    and  since  their  centres  are  collinear,  the 
circles  have  a  common  radical  axis. 

Hence  the  theorem : 

The  circle  which  is  the  locus  of  a  point,  such  that  the  tangents  draion 
from  it  to  tioo  given  circles  are  in  a  constant  ratio,  belongs  to  the  same 
coaxial  system  as  the  given  circles. 


482  BOOK   VL 

We  will  now  prove  the  theorem  : 

If  two  opposite  sides  of  a  quadrilateral,  which  is  inscribed  in  a 
circle,  touch  another  circle,  the  other  sides  of  the  quadrilateral  touch  a 
third  circle  coaxial  with  the  other  two. 

Let  ABCB  be  a  quadrilateral  inscribed  in  a  given  circle ; 

and  let  AB,  CD  touch  another  given  circle  at  P,  Q. 
Draw  PQ  and  let  it  be  produced  to  meet  AD,  BCin  It,  S. 


Because  the  angles  BPS,  DQR  are  equal 

and  the  angles  PBS,  QDR  are  equal,  (III.  Prop.  21) 

therefore  the  triangles  BPS,  DQR  are  equiangular  to  one  another; 
therefore  BP  is  to  BS  as  DQ  to  DR.  (Prop.  4.) 

Similarly  it  can  be  proved  that  the  triangles  APR,  GQS  are  equi- 
angular to  one  another  and  that  AP  is  to  AR  as  CQ  to  CS. 
Again,  because  the  angles  BSP,  DRQ  are  equal, 

a  circle  can  be  described  to  touch  BG,  DA  at  S,  R. 

Next,  because  in  the  two  triangles  ARP,  BSP, 
the  angles  APR,  BPS  are  equal, 
and  the  angles  ARP,  BSP  are  supplementary; 
therefore  AP  is  to  AR  as  BP  to  BS.  (Prop.  5  A.) 

Therefore  the  ratios  of  the  tangents  drawn  from  the  four  points 
A,  B,  G,  D  to  the  two  circles  PQ,  RS  are  equal. 

Therefore  the  four  points  A,  B,  G,  D  aM  lie  on  a  circle  coaxial 
with  the  two  circles  PQ,  RS  ;  (page  481) 

that  is,  the  circle  ABGD  is  coaxial  with  the  circles  PQ,  RS, 
or,  in  other  words,  the  circle  RS  is  coaxial  with  the  circles  ABGD 
and  PQ. 


FONCELET'S   THEOREMS.  483 


Next,  let  ABC,  A'B'C  be  two  triangles  inscribed  in  a  circle,  such 
that  AB,  A'B'  touch  a  second  circle  and  J5C,  B'C  a  third  circle 
belonging  to  the  same  coaxial  system. 


Because  ABB' A'  is  a  quadrilateral  inscribed  in  the  circle  ABC, 

and  AB,  B'A'  touch  another  circle, 

therefore  AA',  BB'  touch  a  circle  of  the  same  coaxial  system ; 

and  because  BCG'B'  is  a  quadrilateral  inscribed  in  the  circle  ABC, 

and  BG,  C'B'  touch  another  circle  of  the  system ; 

therefore  BB\  CC  touch  a  circle  of  the  system. 

Therefore  AA',  BB',  CC  touch  the  same  circle  of  the  system. 

And  because  AA'C'C  is  a  quadrilateral  inscribed  in  the  circle  ABC, 

and  AA',  CC  touch  another  circle  of  the  system, 

therefore  AC,  A'C  touch  a  circle  of  the  system; 

that  is,  A'C  always  touches  that  circle  of  the  system  which  retouches, 

or,  in  other  words,  A'C  touches  a  definite  circle  of  the  system. 


Hence  the  theorem : 

If  two  sides  of  a  triangle  inscribed  in  a  given  circle  touch  given 
circles  of  the  same  coaxial  system  as  the  first  circle,  the  third  side 
touches  a  fourth  fixed  circle  of  the  system.* 

Corollary.  If  all  the  sides  but  one  of  a  polygon  inscribed  in  a 
given  circle  touch  given  circles  of  the  same  coaxial  system  as  the  first 
circle,  the  reviaining  side  touches  another  fixed  circle  of  the  system.* 

*  These  theorems  are  due  to  Jean  Victor  Poncelet  (born  at  Metz 
1788,  died  at  Paris  18G7). 


484  BOOK   VI. 


MISCELLANEOUS   EXERCISES. 

1.  Two  triangles  ABC,  BCD  have  the  side  BC  common,  the 
angles  at  B  equal,  and  the  angles  ACB,  BDC  right  angles.  Shew 
that  the  triangle  ABC  is  to  the  triangle  BCD  as  AB  to  BD. 

2.  The  rectangle  contained  by  two  straight  lines  is  a  mean  pro- 
portional between  the  squares  described  upon  them. 

3.  Any  polygons  whatsoever  described  about  a  circle  are  to  one 
another  as  their  perimeters. 

4.  The  sum  of  the  perpendiculars  drawn  from  any  point  within 
an  equilateral  triangle  on  the  three  sides  is  invariable. 

5.  In  a  parallelogram  E,  F,  G,  H  are  the  middle  points  of  the 
Bides  AB,  BC,  CD,  DA  ;  if  AF,  AG,  CE,  CH  be  drawn,  the  parallelo- 
gram formed  by  them  is  one-third  of  the  parallelogram  ABCD. 

6.  ABC  is  a  triangle,  D  any  point  in  AB  produced ;  E  a  point  in 
BC,  such  that  CE  is  to  EB  as  AD  to  BD.  Prove  that  DE  produced 
bisects  A  C. 

7.  ABC,  ABD  are  triangles  on  the  same  base,  and  CD  meets  the 
base  in  E  ;  then  CE  is  to  DE  as  the  triangle  ABC  to  the  triangle  ABD. 

8.  Triangles  of  unequal  altitudes  are  to  each  other  in  the  ratio 
compounded  of  the  ratios  of  their  altitudes  and  their  bases. 

9.  If  triangles  ABC,  AEF  have  a  common  angle  A,  the  triangle 
ABC  is  to  the  triangle  AEF  as  the  rectangle  AB,  AC  to  the  rectangle 
AE,  AF. 

10.  O  is  the  centre  of  the  circle  inscribed  in  a  triangle  ABC,  and 
BO,  CO  meet  the  opposite  sides  in  D,  E  respectively.  Prove  that  the 
triangles  BOE,  COD  are  to  one  another  in  the  ratio  of  the  rectangles 
AE,AB;  AD,  AC. 

11.  If  in  the  sides  of  a  triangle  BC,  CA,  AB,  points  D,  U,  F  be 
taken  such  that  BD  is  twice  DC,  CE  twice  EA,  and  AF  twice  FB,  and 
AD,  BE,  CF  intersect  in  pairs  in  P,  Q,  R,  then  the  areas  of  the 
triangles  PQR,  ABC  are  in  the  ratio  of  1  to  7. 

12.  A  point  and  a  straight  line  being  given,  to  draw  a  line  parallel 
to  the  given  line  such  that  all  the  lines  drawn  through  the  point  may 
be  cut  by  the  parallels  in  a  given  ratio. 

13.  If  from  a  point  0  in  the  base  BC  of  a  triangle  OM  and 
ON  he  drawn  parallel  to  the  sides  AB  and  AC  respectively,  then  the 
area  of  the  triangle  AMN  is  a  mean  proportional  between  the  areas  of 
BNO  and  CMO. 

14.  If  P,  Q  be  two  points  within  a  parallelogram  ABCD,  and  if 
PA,  QB  meet  in  R,  and  PD,  QC  meet  in  S,  and  if  PQ  be  parallel  to 
AB^  then  RS  is  parallel  to  AD. 


MISCELLANEOUS  EXERCISES.  485 

15.  A  point  P  is  taken  on  the  bisector  of  the  angle  BAG  of  the 
triangle  ABC  between  A  and  the  base;  prove  that,  if  ^C  be  greater 
than  AB,  the  ratio  PC  to  PB  is  greater  than  the  ratio  ^C  to  CB. 

16.  On  a  circle  of  which  AB  is  a  diameter  take  any  point  P. 
Draw  PC,  PD  on  opposite  sides  of  AP,  and  equally  inclined  to  it, 
meeting  AB  at  C  and  D :  prove  that  AC  is  to  BC  as  AD  is  to  BD. 

17.  Apply  VI.  3  to  solve  the  problem  of  the  trisection  of  a  finite 
straight  line. 

18.  AB  is  a  diameter  of  a  circle,  CD  is  a  chord  at  right  angles  to 
it,  and  E  is  any  point  in  CD ;  AE  and  BE  are  drawn  and  produced 
to  cut  the  circle  at  F  and  G  :  shew  that  the  quadrilateral  GFDG 
has  any  two  of  its  adjacent  sides  in  the  same  ratio  as  the  remaining 
two. 

19.  ABCD  is  a  quadrilateral  having  the  opposite  sides  AD,  BG 
parallel;  ^  is  a  point  in  AB,  and  the  straight  lines  EC,  ED  are 
drawn;  AF  is  drawn  parallel  to  EC  meeting  CD  in  F;  shew  that 
BF  is  parallel  to  ED. 

20.  If  a  straight  line  AD  be  drawn  bisecting  the  angle  BAG  of 
the  triangle  BAG,  and  meeting  BC  in  D,  and  FDC  be  drawn  perpen- 
dicular to  AD,  to  meet  AB  and  AG  produced  if  necessary,  in  F  and  E 
respectively,  and  EG  he  drawn  parallel  to  BC,  meeting  AB  in  G, 
then  BG  is  equal  to  BF. 

21.  The  side  AB  oi  the  triangle  ABC  is  produced  to  G ,  and  the  angle 
CBG  bisected  by  BE  meeting  AC  in  E:  the  angle  EBG  is  bisected  by 
BL  and  the  exterior  angle  of  the  triangle  EBG  got  by  producing  EB 
is  bisected  by  BF.  Shew  that,  if  BL  be  parallel  to  ^  C  and  BF  meet 
AG  in  F,  CE  is  a  mean  proportional  between  GA  and  GF. 

22.  Each  acute  angle  of  a  right-angled  triangle  and  its  cor- 
responding exterior  angle  are  bisected  by  straight  lines  meeting  the 
opposite  sides ;  prove  that  the  rectangle  contained  by  the  portions  of 
those  sides  intercepted  between  the  bisecting  lines  is  four  times  the 
square  on  the  hypotenuse. 

23.  A  and  B  are  fixed  points,  and  AP,  BQ  are  parallel  chords  of 
a  variable  circle,  which  passes  through  the  fixed  points.  If  the  ratio 
of  AP  to  BQ  be  constant,  then  the  loci  of  P  and  Q  are  each  a  pair  of 
circles,  and  the  sum  of  the  radii  of  two  of  these  circles,  and  the 
difference  of  the  radii  of  the  other  two,  are  independent  of  the  magni- 
tude of  the  ratio. 

24.  If  two  chords  AB,  AC,  drawn  from  a  point  A  in  the  circum- 
ference of  the  circle  ABC,  be  produced  to  meet  the  tangent  at  the 
other  extremity  of  the  diameter  through  A  in  D,  E  respectively,  then 
the  triangle  AED  is  similar  to  the  triangle  ABC. 

25.  AB  is  the  diameter  of  a  circle,  E  the  middle  point  of  the 
radius  OB ;  on  AE,  EB  as  diameters  circles  are  described ;  PQL  is  a 
common  tangent  meeting  the  circles  at  P  and  Q,  and  AB  produced  at 
L :  shew  that  BL  is  equal  to  the  radius  of  the  smaller  circle. 


4.SQ  BOOK   VL 

26.  From  B  the  right  angle  of  a  right-angled  triangle  ABC,  Bp  is 
let  fall  perpendicular  to  AC ;  from  p, pq  is  let  fall  perpendicular  to  BA  ; 
from  q,  qr  is  let  fall  perpendicular  to  AC,  and  so  on;  prove  that 

Bp+pq  +  &c.  :  AB  =  AB  +  AC  :  BC. 

27.  An  isosceles  triangle  is  described  on  a  side  of  a  square  and 
the  vertex  joined  with  the  opposite  angles :  the  middle  segment  of  the 
side  has  to  either  of  the  outside  segments  double  of  the  ratio  of  the 
altitude  of  the  triangle  to  its  base. 

28.  A  straight  line  BE  is  drawn  parallel  to  the  side  BC  oi  a, 
triangle  ABC.  Q  is  a  point  in  BC  such  that  the  rectangle  BC,  CQ  is 
equal  to  the  square  on  DE,  and  CR  is  taken  equal  to  DE  in  BC  pro- 
duced.    Prove  that  AR  is  parallel  to  DQ. 

29.  ABC,  DEF  are  triangles,  having  the  angle  A  equal  to  the 
angle  D ;  and  AB  is  equal  to  DF :  shew  that  the  areas  of  the  triangles 
are  as  ^C  to  DE. 

30.  CA,  CB  are  diameters  of  two  circles  which  touch  each  other 
externally  at  C;  a  chord  AD  of  the  former  circle,  when  produced, 
touches  the  latter  at  E,  while  a  chord  BF  of  the  latter,  when  pro- 
duced, touches  the  former  at  G:  shew  that  the  rectangle  contained 
by  AD  and  BF  is  four  times  that  contained  by  DE  and  FG. 

31.  If  AA'  B'B,  BB'  C'C,  GC  A' A  be  three  circles,  and  the 
straight  lines  AA',  BB',  CC  cut  the  circle  A'B'C  again  in  a,  /3,  y  re- 
spectively, the  triangle  a^-y  will  be  similar  to  the  triangle  ABC. 

32.  On  the  two  sides  of  a  right-angled  triangle  squares  are 
described :  shew  that  the  straight  lines  joining  the  acute  angles  of  the 
triangle  and  the  opposite  angles  of  the  squares  cut  off  equal  segments 
from  the  sides,  and  that  each  of  these  equal  segments  is  a  mean 
proportional  between  the  remaining  segments. 

33.  ABA'B'  is  a  rectangle  inscribed  in  a  circle  and  AB  is  twice 
A'B  ;  ^C  is  a  chord  equal  to  AB  and  meeting  B'A'  in  F  and  BA'  in  E ; 
prove  that  AF:AE  =  CF :  CA. 

34.  If  BD,  CD  are  perpendicular  to  the  sides  AB,  AC  ot  a  triangle 
ABC  and  CE  is  drawn  perpendicular  to  AD,  meeting  AB  in  E,  then 
the  triangles  ABC,  ACE  are  similar. 

35.  Describe  a  circle  touching  the  side  BC  of  the  triangle  ABC 
and  the  other  two  sides  produced;  and  shew  that  the  distance  between 
the  points  of  contact  oi  BC  with  this  circle  and  the  inscribed  circle  is 
equal  to  the  difference  between  AB  and  AC. 

36.  A  straight  line  AB  is  divided  into  any  two  parts  at  C,  and  on 
the  whole  straight  line  and  on  the  two  parts  of  it  equilateral  triangles 
ADB,  ACE,  BCF  are  described,  the  two  latter  being  on  the  same  side 
of  the  straight  line,  and  the  former  on  the  opposite  side;  G,  H,  K are 
the  centres  of  the  circles  inscribed  in  these  triangles :  shew  that  the 
angles  AGH,  BGK  are  respectively  equal  to  the  angles  ADC,  BDCt 
and  that  GHK  is  an  equilateral  triangle. 


MISCELLANEOUS  EXERCISES  487 

37.  Two  circles,  centres  A  and  B,  touch  one  another  at  G.  A 
straight  line  is  drawn  cutting  one  circle  in  P  and  Q  and  the  other 
circle  in  R  and  S.  Prove  that  the  ratio  of  the  rectangle  PR,  PS  to 
the  square  on  CP  is  constant. 

38.  AB  is  the  diameter  of  a  circle,  E  the  middle  point  of  the 
radius  OB ;  on  AE,  EB  as  diameters  circles  are  described.  PQL  is  a 
common  tangent,  meeting  the  circles  in  P  and  Q,  and  AB  produced 
in  L.     Shew  that  BL  equals  the  radius  of  the  lesser  circle. 

39.  If  through  the  vertex,  and  the  extremities  of  the  base  of  a 
triangle,  two  circles  be  described,  intersecting  one  another  in  the  base, 
or  the  base  produced,  their  diameters  are  proportional  to  the  sides  of 
the  triangle. 

40.  D  is  the  middle  point  of  the  base  BC  of  an  isosceles  triangle, 
OF  perpendicular  to  AB,  DE  perpendicular  to  CF,  EG  parallel  to  the 
base  meets  AD  in  G;  prove  that  EG  is  to  GA  in  the  triplicate  ratio 
ofBD  to  DA. 

41.  Two  straight  lines  and  a  point  between  them  are  given  in 
position;  draw  two  straight  lines  from  the  given  point  to  terminate 
in  the  given  straight  lines,  so  that  they  shall  contain  a  given  angle 
and  have  a  given  ratio. 

42.  Four  lines,  AB,  CD,  EF,  GH,  drawn  in  any  directions,  inter- 
sect in  the  same  point  P ;  then  if  from  any  point  m  in  one  of  these 
lines,  another  be  drawn  parallel  to  the  next  in  order,  cutting  the  re- 
maining two  in  p  and  q ;  the  ratio  mp  :pq  is  the  same  in  whichever 
line  the  point  m  is  taken. 

43.  If  P,  Q  be  the  points  of  intersection  of  a  variable  circle  drawn 
through  two  given  points  A,  B  with  a  fixed  circle,  prove  that  the 
ratio  AP  .  AQ  :  BP  .  BQ  ia  constant. 

44.  A  quadrilateral  is  divided  into  four  triangles  by  its  diagonals; 
shew  that  if  two  of  these  triangles  are  equal,  the  remaining  two  are 
either  equal  or  similar. 

45.  ABCD  is  a  quadrilateral  inscribed  in  a  circle,  E,  F,  G  are  the 
points  of  intersection  of  AB  and  CD,  AC  and  BD,  AD  and  BC  re- 
spectively. K  is  the  foot  of  the  perpendicular  let  fall  from  F  on  EG. 
Prove  that  KA  :  KB=zFA  :  FB. 

46.  Divide  a  given  finite  straight  line  similarly  to  a  given  divided 
straight  line  parallel  to  the  first  line. 

47.  If  two  parallel  straight  lines  AB,  CD  be  divided  proportionally 
at  P,  Q,  so  that  AP  is  to  PB  as  CQ  to  QD,  then  the  straight  lines  AC, 
PQ,  BD  meet  in  a  point. 

48.  BA  C,  DAE  are  similar  equal  triangles,  BAD  and  CAE  being 
straight  lines;  and  the  parallelograms  of  which  BC  and  DE  are 
diagonals  are  completed.  Prove  that  the  lines  drawn  to  complete  the 
parallelograms  themselves  form  a  parallelogram  whose  diagonal  passes 
through  A. 


488  BOOK   VI. 

49.  If,  in  similar  triangles,  any  two  equal  angles  be  joined  to  the 
opposite  sides  by  straight  lines  making  equal  angles  with  homologous 
sides;  these  lines  shall  have  the  same  ratio  as  the  sides  on  which 
they  fall,  and  shall  divide  those  sides  proportionally. 

50.  APB,  CQD  are  parallel  straight  lines,  and  ^P  is  to  PB  as 
DQ  to  QC,  prove  that  the  straight  lines  PQ,  AD,  BG  meet  in  a  point. 

51.  Describe  a  circle  which  shall  pass  through  a  given  point  and 
touch  two  given  straight  lines. 

52.  AD  the  bisector  of  the  base  of  the  triangle  ABC  is  bisected 
in  E,  BE  cuts  ^C  in  i^,  prove  that  AF  :FC '.:1:2. 

53.  A  straight  line  drawn  through  the  middle  point  of  a  side  of  a 
triangle  divides  the  other  sides,  one  internally,  the  other  externally  in 
the  same  ratio. 

54.  In  the  triangle  ABC  there  are  drawn  AD  bisecting  BC,  and 
EF  parallel  to  J5C  and  cutting  AB,  AC  in  E,  F.  Shew  that  BF  and 
CE  intersect  in  AD. 

55.  In  the  triangle  ACB,  having  C  a  right  angle,  AD  bisecting 
the  angle  A  meets  CB  in  D,  prove  that  the  square  on  ^C  is  to  the 
square  on  AD  as  £C  to  2BD. 

56.  AB  and  CD  are  two  parallel  straight  lines ;  E  is  the  middle 
point  of  CD ;  AC  and  BE  meet  at  F,  and  AE  and  BD  meet  at  G :  shew 
that  FG  is  parallel  to  AB. 

57.  A,  B,  C  are  three  fixed  points  in  a  straight  line ;  any  straight 
line  is  drawn  through  C;  shew  that  the  perpendiculars  on  it  from  A 
and  B  are  in  a  constant  ratio. 

58.  If  the  perpendiculars  from  two  fixed  points  on  a  straight 
line  passing  between  them  be  in  a  given  ratio,  the  straight  line  must 
pass  through  a  third  fixed  point. 

59.  Through  a  given  point  draw  a  straight  line,  so  that  the  parts 
of  it  intercepted  between  that  point  and  perpendiculars  drawn  to  the 
straight  line  from  two  other  given  points  may  have  a  given  ratio. 

60.  A  tangent  to  a  circle  at  the  point  A  intersects  two  parallel 
tangents  at  B,  C,  the  points  of  contact  of  which  with  the  circle  are 
D,  E  respectively ;  and  BE,  CD  intersect  at  F :  shew  that  AF  is 
parallel  to  the  tangents  BD,  CE. 

61.  P  and  Q  are  fixed  points ;  AB  and  CD  are  fixed  parallel 
straight  lines ;  any  straight  line  is  drawn  from  P  to  meet  AB  at  M, 
and  a  straight  line  is  drawn  from  Q  parallel  to  P3I  meeting  CD  at 
N :  shew  that  the  ratio  of  PM  to  QN  is  constant,  and  thence  shew 
that  the  straight  hne  through  31  and  N  passes  through  a  fixed  point. 

62.  If  two  circles  touch  each  other,  and  also  touch  a  given 
straight  line,  the  part  of  the  straight  line  between  the  points  of  con- 
tact is  a  mean  proportional  between  the  diameters  of  the  circles. 


MISCELLANEOUS  EXERCISES.  489 

63.  If  at  a  given  point  two  circles  intersect,  and  their  centres  lie 
on  two  fixed  straight  lines  which  pass  through  that  point,  shew  that 
whatever  be  the  magnitude  of  the  circles  their  common  tangents  will 
always  meet  in  one  of  two  fixed  straight  lines  which  pass  through 
the  given  point. 

64.  From  the  angular  points  of  a  parallelogram  ABCD  perpen- 
diculars are  drawn  on  the  diagonals  meeting  them  at  E,  F,  G,  H 
respectively:  shew  that  EFGH is  a  parallelogram  similar  to  ABCD. 

65.  ABODE  is  a  regular  pentagon,  and  AD,  BE  intersect  at  O : 
shew  that  a  side  of  the  pentagon  is  a  mean  proportional  between  AO 
and  AD. 

66.  ACB  is  a  triangle,  and  the  side  ^C  is  produced  to  D  so  that 
CD  is  equal  to  AC,  and  BD  is  joined:  if  any  straight  line  drawn 
parallel  io  AB  cuts  the  sides  AC,  CB,  and  from  the  points  of  section 
straight  lines  be  drawn  parallel  to  DB,  shew  that  these  straight  lines 
will  meet  AB  at  points  equidistant  from  its  extremities. 

67.  If  a  circle  be  described  touching  externally  two  given  circles, 
the  straight  line  passing  through  the  points  of  contact  will  intersect 
the  straight  line  passing  through  the  centres  of  the  given  circles  at  a 
fixed  point. 

68.  A  and  B  are  two  points  on  the  circumference  of  a  circle  of 
which  C  is  the  centre ;  draw  tangents  at  A  and  B  meeting  at  T ;  and 
from  A  draw  AN  perpendicular  to  CB  :  shew  that  BT  is  to  BC  as  BN 
is  to  NA. 

69.  Find  a  point  the  perpendiculars  from  which  on  the  sides  of  a 
given  triangle  shall  be  in  a  given  ratio. 

70.  A  quadrilateral  ABCD  is  inscribed  in  a  circle  and  its  dia- 
gonals AC,  BD  meet  at  0.  Points  P,  Q  are  taken  in  AB,  CD  such 
that  AP  is  to  PB  &a  AO  to  OB,  and  CQ  is  to  QD  as  CO  to  OD ;  prove 
that  a  circle  can  be  described  to  touch  AB,  CD  at  P,  Q. 

71.  Prove  that  the  diagonals  of  the  complements  of  parallelo- 
grams about  a  diagonal  of  a  parallelogram  meet  in  the  diagonal  of 
the  parallelogram. 

72.  Through  a  point  G  of  the  side  CD  of  a  parallelogram  ABCD 
are  drawn  AG  and  BG  meeting  the  sides  in  E  and  F;  and  GH  is 
drawn  parallel  to  EF,  meeting  AF  in  H ;  prove  that  FH  is  equal 
to^D. 

73.  Any  regular  polygon  inscribed  in  a  circle  is  a  mean  propor- 
tional between  the  inscribed  and  circumscribed  regular  polygons  of 
half  the  number  of  sides. 

74.  If  two  sides  of  a  parallelogram  inscribed  in  a  quadrilateral  be 
parallel  to  one  of  the  diagonals  of  the  quadrilateral,  then  the  other 
sides  of  the  parallelogram  are  parallel  to  the  other  diagonal. 

75.  A  circle  is  described  round  an  equilateral  triangle,  and  from 
any  point  in  the  circumference  straight  lines  are  drawn  to  the  angular 
points  of  the  triangle  :  shew  that  one  of  these  straight  lines  is  equal 
to  the  other  two  together. 

T.  E.  32 


490  BOOK    VI. 

76.  ABC  is  a  triangle.  At  ^  a  straight  line  JD  is  drawn  making 
the  angle  CAD  equal  to  CBA,  and  at  C  the  straight  line  CD  is  drawn 
making  the  angle  ACD  equal  to  BAC.  Shew  that  AD  is  a  fourth 
proportional  to  AB,  BG  and  CA. 

77.  If  bd  be  drawn  cutting  the  sides  AB,  AD  and  the  diagonal  AG 
of  the  parallelogram  ABCD  in  b,  d,  and  c  respectively,  so  that  Ab  is 
equal  to  Ad,  then  the  sum  of  AB,  AD  is  to  the  sum  of  Ab,  Ad  as  AG 
to  twice  Ac. 

78.  Having  given  the  base  of  a  triangle  and  the  opposite  angle, 
find  that  triangle  for  which  the  rectangle  contained  by  the  perpen- 
diculars from  the  ends  of  the  base  on  the  opposite  sides  is  greater 
than  for  any  other. 

79.  Through  each  angular  point  of  a  quadrilateral  a  straight  line 
is  drawn  perpendicular  to  the  diagonal  which  does  not  pass  through 
that  point,  shew  that  the  parallelogram  thus  formed  is  similar  to  the 
parallelogram  formed  by  joining  the  middle  points  of  the  sides  of  the 
given  quadrilateral. 

80.  ABCD  is  a  quadrilateral  inscribed  in  a  circle;  BA,  CD  pro- 
duced meet  in  P,  and  AD,  BG  produced  in  Q.  Prove  that  PC  is  to 
PB  as  QA  to  QB. 

81.  Through  D,  any  point  in  the  base  of  a  given  triangle  ABC, 
straight  lines  DE,  DF  are  drawn  parallel  to  the  sides  AB,  AC  and 
meeting  the  sides  in  E,  F  and  EF  is  drawn ;  shew  that  the  triangle 
AEF  is  a  mean  proportional  between  the  triangles  FBD,  EDG. 

82.  If  through  the  vertex  and  the  extremities  of  the  base  of  a 
triangle  two  circles  be  described  intersecting  each  other  in  the  base  or 
the  base  produced,  their  diameters  are  proportional  to  the  sides  of 
the  triangle. 

83.  Draw  a  straight  line  such  that  the  perpendiculars  let  fall 
from  any  point  in  it  on  two  given  straight  lines  may  be  in  a  given 
ratio. 

84.  In  any  right-angled  triangle,  one  side  is  to  the  other,  as  the 
excess  of  the  hypotenuse  above  the  second,  to  the  line  cut  off  from 
the  first  between  the  right  angle  and  the  line  bisecting  the  opposite 
angle. 

85.  AB  is  a  fixed  straight  line,  C  a  moving  point,  and  CD  a  line 
parallel  to  AB  ;  a  variable  line  PQR  is  drawn  cutting  AC  in  P,  BG  in 
Q  and  CDin  R',  prove  that  if  the  ratios  AP  to  PC,  and  BQ  to  QC  he 
constant,  GR  is  of  constant  length. 

86.  If  I,  Ii  be  the  centres  of  the  inscribed  circle  of  a  triangle 
ABC  and  of  the  circle  escribed  beyond  BG,  the  rectangle  AI,  AI^  is 
equal  to  the  rectangle  AB,  AG. 

87.  If  I,  Ij  be  the  centres  of  the  inscribed  circle  of  a  triangle 
ABC  and  of  the  circle  escribed  beyond  BG,  and  Z>,  E  be  the  points  of 
contact  of  those  circles  with  AB,  then  ID  is  to  DB  as  EB  to  EI^. 


MISCELLANEOUS  EXERCISES.  491 

88.  If  Ij,  I2  ^6  *h®  centres  of  the  circles  of  a  triangle  ABC 
escribed  beyond  BG,  CA  respectively  and  E,  F  be  their  points  of 
contact  with  AB,  then  I^E  is  to  EB  as  BF  to  I^F. 

89.  0  is  a  fixed  point  in  a  given  straight  line  OA,  and  a  circle  of 
given  radius  moves  so  as  always  to  be  touched  by  OA ;  a  tangent  OP 
is  drawn  from  O  to  the  circle,  and  in  OP  produced  PQ  is  taken  a 
third  proportional  to  OP  and  the  radius :  shew  that  as  the  circle 
moves  along  OA,  the  point  Q  will  move  in  a  straight  line. 

90.  On  AB,  AC,  two  adjacent  sides  of  a  rectangle,  two  similar 
triangles  are  constructed,  and  perpendiculars  are  drawn  to  AB,  AC 
from  the  angles  which  they  subtend,  intersecting  at  the  point  P.  If 
AB,  AC  he  homologous  sides,  shew  that  P  is  in  all  cases  in  one  of 
the  diagonals  of  the  rectangle. 

91.  If  at  any  two  points  A,  B  ;  AC,  BD  be  drawn  at  right  angles 
to  AB  on  the  same  side  of  it,  so  that  AB  is  &  mean  proportional 
between  AG  and  BD;  the  circles  described  on  AC,  BD  as  diameters 
will  touch  each  other. 

92.  One  circle  touches  another  internally  at  A,  and  from  two 
points  in  the  line  joining  their  centres,  perpendiculars  are  drawn 
intersecting  the  outer  circle  in  the  points  B,  C,  and  the  inner  in  D,  E; 
shew  that  AB  :  AG  =  AD  :  AE. 

93.  Find  a  straight  line  such  that  the  perpendiculars  on  it  from 
three  given  points  shall  be  in  given  ratios  to  each  other. 

94.  Divide  a  given  arc  of  a  circle  into  two  parts,  so  that  the 
chords  of  these  parts  shall  be  to  each  other  in  a  given  ratio. 

95.  CAB,  GEB  are  two  triangles  having  a  common  angle  GBA, 
and  the  sides  opposite  to  it  CA,  GE  equal.  If  BA  be  produced  to  D, 
and  ED  be  taken  a  third  proportional  to  BA,  AC,  then  the  triangle 
BDG  is  similar  to  the  triangle  BAG. 

96.  One  side  of  a  triangle  is  given,  and  also  its  points  of  inter- 
section with  the  bisector  of  the  opposite  angle  and  the  perpendicular 
from  the  opposite  vertex  ;  construct  the  triangle. 

97.  The  diameter  of  a  circle  is  a  mean  proportional  between  the 
sides  of  an  equilateral  triangle  and  a  regular  hexagon  which  are 
described  about  the  circle. 

98.  If  two  regular  polygons  of  the  same  number  of  sides  be  con- 
structed, one  inscribed  in  and  the  other  described  about  a  given  circle, 
and  a  third  of  double  the  number  of  sides  be  inscribed  in  the  circle, 
this  last  is  a  mean  proportional  between  the  other  two. 

99.  A  triangle  DEF  is  inscribed  in  a  triangle  ABC  so  that  DE, 
DF  are  parallel  to  BA,  CA  respectively;  prove  that  the  triangle 
DEF  is  to  the  triangle  ABC  as  the  rectangle  BD,  DC  to  the  square 
on  BC. 

32—2 


492  BOOK   VI. 

100.  The  vertical  angle  0  of  a  triangle  is  bisected  by  a  straight 
line  which  meets  the  base  at  D,  and  is  produced  to  meet  the  circle 
ABC  at  E ;  prove  that  the  rectangle  contained  by  CD  and  CE  is  equal 
to  the  rectangle  contained  by  ^C  and  CB. 

101.  A  straight  line  is  divided  in  two  given  points,  determine 
a  third  point,  such  that  its  distances  from  the  two  given  points  may 
be  proportional  to  its  distances  from  the  ends  of  the  line. 

102.  AB'iBQ.  diameter  of  a  circle,  PQ  a  chord  perpendicular  to  AB, 
O  any  point  on  the  circle;  OP,  OQ  meet  AB  in  R  and  S;  prove  that 
the  rectangle  AR  .  BS  is  equal  to  the  rectangle  A  S  .  BR. 

103.  A,  B  are  two  fixed  points  and  P  a  variable  point  on  a  circle, 
AA',  BB'  are  drawn  parallel  to  a  fixed  line  to  meet  the  circle  in  A',  B' : 
the  fixed  line  meets  AB'  in  D,  A'B  in  D\  AP  in  E,  BP  in  E' \  prove 
that  BE  .  B'E'  is  constant. 

104.  Prove  that,  if  ABCD  be  a  quadrilateral  not  inscriptible  in  a 
circle,  a  point  E  exists  such  that  the  rectangle  AB,  CD  is  equal  to  the 
rectangle  AE,  BD  and  the  rectangle  AD,  BC  is  equal  to  the  rectangle 
CE,  BD.     Hence  prove  the  converse  of  Ptolemy's  Theorem. 

105.  BE  and  CF  are  perpendiculars  upon  AD  the  bisector  of  the 
angle  ^  of  a  triangle  ABC.  The  area  of  the  triangle  is  equal  to  either 
of  the  rectangles  AE,  CF  or  AF,  BE. 

106.  If  the  exterior  angle  CAE  of  a  triangle  be  bisected  by  the 
straight  line  AD  which  likewise  cuts  the  base  produced  in  D ;  then 
BA  .  AC,  the  rectangle  of  the  sides,  is  less  than  the  rectangle  BD .  DC 
by  the  square  on  AD. 

107.  ABC  is  an  isosceles  triangle,  the  side  AB  being  equal  to  ^C; 
F  is  the  middle  point  of  BC;  on  any  straight  line  through  A  perpen- 
diculars EG  and  CE  are  drawn:  shew  that  the  rectangle  AC,  EF 
is  equal  to  the  sum  of  the  rectangles  EC,  EG  and  FA,  EG. 

108.  Describe  a  circle  which  shall  pass  through  a  given  point  and 
touch  a  given  straight  line  and  a  given  circle. 

109.  Divide  a  triangle  into  two  equal  parts  by  a  straight  line  at 
right  angles  to  one  of  the  sides. 

110.  If  a  straight  line  drawn  from  the  vertex  of  an  isosceles  triangle 
to  the  base,  be  produced  to  meet  the  circumference  of  a  circle  de- 
scribed about  the  triangle,  the  rectangle  contained  by  the  whole  line 
so  produced  and  the  part  of  it  between  the  vertex  and  the  base  shall 
be  equal  to  the  square  on  either  of  the  equal  sides  of  the  triangle. 

111.  Two  straight  lines  are  drawn  from  a  point  A  to  touch  a 
circle  of  which  the  centre  is  E  ;  the  points  of  contact  are  joined  by  a 
straight  line  which  cuts  EA  at  H;  and  on  HA  as  diameter  a  circle  is 
described :  shew  that  the  straight  lines  drawn  through  E  to  touch  this 
circle  will  meet  it  on  the  circumference  of  the  given  circle. 


MISCELLANEOUS  EXERCLSES  493 

112.  Two  triangles  BAD,  BAG  have  the  side  BA  and  the  angle  A 
common:  moreover  the  angle  ABB  is  equal  to  the  angle  ACB:  shew 
that  the  rectangle  contained  by  AC,  BD  is  equal  to  that  contained  by 
AB,  BC. 

113.  ABCD  is  a  quadrilateral  in  a  circle;  the  straight  lines  CE, 
DE  which  bisect  the  angles  ACB,  ABB,  cut  BD  and  AC  d.t  F  and  G 
respectively.-  shew  that  EF  is  to  EG  as  ED  is  to  EC. 

114.  A  Bqn&reDEFG  is  inscribed  in  a  right-angled  triangle  ABC, 
so  that  D,  G  are  in  the  hypotenuse  AB  oi  the  triangle  E  in  AC,  and  F 
in  CB :  prove  that  the  area  of  the  square  is  equal  to  the  rectangle 
AD,  BG. 

115.  A,  B,  C  are  three  points  in  order  in  a  straight  line:  find  a 
point  P  in  the  straight  line  so  that  PB  may  be  a  mean  proportional 
between  PA  and  PC. 

116.  AB  is  a  diameter,  and  P  any  point  in  the  circumference  of  a 
circle;  AP  and  BP  are  joined  and  produced  if  necessary;  from  any 
point  G  in  AB  a  straight  line  is  drawn  at  right  angles  to  AB  meeting 
AP  at  D  and  BP  at  E,  and  the  circumference  of  the  circle  at  F:  shew 
that  CD  is  a  third  proportional  to  CE  and  CF. 

117.  If  F  be  a  point  in  the  side  CB  of  a  right-angled  triangle  and 
CD,  FE  be  perpendiculars  on  the  hypotenuse  AB,  then  the  sum  of  the 
rectangles  AD,  AE  and  CD,  EF  is  equal  to  the  square  on  AC. 

118.  In  the  figure  of  II.  11  shew  that  four  other  straight  lines 
besides  the  given  straight  line  are  divided  in  the  required  manner. 

119.  A  straight  line  CD  bisects  the  vertical  angle  C  of  a  triangle 
ABC,  and  cuts  the  base  in  D,  on  AB  produced  a  point  E  is  taken 
equidistant  from  C  and  D :  prove  that  the  rectangle  AE  .  BE  is  equal 
to  the  square  on  DE. 

120.  If  the  perpendicular  in  a  right-angled  triangle  divide  the 
hypotenuse  in  extreme  and  mean  ratio,  the  less  side  is  equal  to  the 
alternate  segment. 

121.  ABC  is  a  right-angled  triangle,  CD  a  perpendicular  from  the 
right  angle  upon  AB;  shew  that  if  ^C  is  double  of  BC,  BD  is  one- 
fifth  of  AB. 

122.  Through  a  given  point  draw  a  chord  in  a  given  circle  so  that 
it  shall  be  divided  at  the  point  in  a  given  ratio.  Find  the  limiting 
value  of  the  ratio. 

123.  ABCD  is  a  parallelogram;  from  B  a  straight  line  is  drawn 
cutting  the  diagonal  ^C  at  F,  the  side  DC  at  G,  and  the  side  AD  pro- 
duced at  E :  shew  that  the  rectangle  EF,  EG  is  equal  to  the  square 
on  BF. 

124.  Find  a  point  in  a  side  of  a  triangle  from  which  two  straight 
lines  drawn,  one  to  the  opposite  angle,  and  the  other  parallel  to  the 
base,  shall  cut  off  towards  the  vertex  and  towards  the  base,  equal 
triangles. 


494  BOOK   VI. 

125.  On  a  chord  AB  of  a  circle  any  point  P  is  taken :  on  AP,  PB 
any  two  similar  and  similarly  situated  triangles  APE,  PBF  are  con- 
structed, and  the  straight  line  £i^  joining  the  vertices  of  these  triangles 
is  produced  to  meet  AB  produced  in  Q.  If  any  circle  be  described 
touching  AB  at  P  the  common  chord  of  these  two  circles  passes 
through  Q. 

126.  With  a  point  A  in  the  circumference  of  a  circle  ABC  as 
centre,  a  circle  PBG  is  described  cutting  the  former  circle  at  the 
points  B  and  C;  any  chord  AD  of  the  former  meets  the  common 
chord  BC  at  E,  and  the  circumference  of  the  other  circle  at  0 :  shew 
that  the  angles  EPO  and  DPO  are  equal  for  all  positions  of  P. 

127.  It  is  required  to  cut  off  from  one  given  line  a  part  such  that 
it  may  be  a  mean  proportional  between  the  remainder  and  another 
given  line. 

128.  Construct  a  square  so  that  its  vertices  shall  lie  on  four  of  the 
sides  of  a  regular  pentagon. 

129.  Shew  how  to  divide  a  given  triangle  into  any  number  of 
equal  parts  by  lines  parallel  to  the  base. 

130.  Divide  a  given  triangle  by  a  straight  line  drawn  in  a  given 
direction  into  two  parts  whose  areas  shall  be  in  a  given  ratio. 

131.  If  E  be  the  intersection  of  the  diagonals  of  a  quadrilateral 
ABCD,  which  has  the  sides  AB  and  CD  parallel,  then 

(i)  the  straight  line  joining  the  middle  points  oi  AB  and  CD 
passes  through  E ; 

(ii)  if  P  be  any  point  in  DB  produced,  the  rectangles  PB,  EC  and 
PD,  EA  are  together  equal  to  the  rectangle  PE,  AC. 

132.  Two  quadrilaterals  ABCD,  ABEF  in  which  BC,  CD,  DA 
are  equal  to  BE,  EF,  FA  respectively,  are  on  the  same  side  of  AB. 
Prove  that  if  the  rectangles  OA,  OD  and  OB,  OC  be  equal,  where  O 
is  the  point  of  intersection  of  the  bisectors  of  the  angles,  DAF, 
CBE,  then  the  quadrilaterals  are  equal  in  area. 

133.  Prove  that  the  area  of  a  quadrilateral,  whose  sides  are  all  of 
given  lengths,  is  a  maximum  when  two  opposite  angles  of  the  quadri- 
lateral are  supplementary. 

134.  Having  four  given  finite  straight  lines,  construct  the  quadri- 
lateral of  maximum  area  which  can  be  formed  with  them  taken  in  a 
given  order  for  sides. 

135.  Either  of  the  complements  is  a  mean  proportional  between 
the  parallelograms  about  the  diameter  of  a  parallelogram. 

136.  Shew  that  one  of  the  triangles  in  the  figure  of  iv.  10  is  a 
mean  proportional  between  the  other  two. 

137.  The  sides  AB  and  ^C  of  a  triangle  ABC  are  produced  to  D 
and  E  respectively,  and  DE  is  joined.  A  point  F  is  taken  in  J5C  such 
that  BF :  FC  =  AB.AE  :  AC.  AD,  prove  that,  if  AF  be  joined  and 
produced,  it  will  pass  through  the  middle  point  of  DE. 


MISCELLANEOUS  EXERCISES.  495 

138.  In  any  triangle  ABC,  if  BB  be  taken  equal  to  one-fourth  of 
BG,  and  CE  one-fourth  of  ^C,  the  straight  line  drawn  from  C  through 
the  intersection  of  BE  and  AD  will  divide  AB  into  two  parts,  which 
are  in  the  ratio  of  nine  to  one. 

139.  Lines  drawn  from  the  extremities  of  the  base  of  a  triangle 
intersecting  on  the  line  joining  the  vertex  with  the  middle  point  of  the 
base,  cut  the  sides  proportionally ;  and  conversely. 

140.  D  is  the  middle  point  of  the  side  BC  of  a  triangle  ABC,  and 
P  is  any  point  in  AD\  through  P  the  straight  lines  BPE,  CPF  are 
drawn  meeting  the  other  sides  at  E,  F:  shew  that  EF  is  parallel 
to  BC\ 

141.  ABC  is  a  triangle  and  D,  E,  F  points  in  the  sides  BC,  GA, 
AB  respectively  such  that  AD,  BE,  GF  meet  in  0 ;  prove  that  the 
ratio  ^0  to  OD  is  equal  to  the  sum  of  the  ratios  AF  to  FB  and  AE 
to  EC. 

142.  Through  any  point  O  within  triangle  ABC  straight  lines 
AO,  BO,  CO  are  drawn  cutting  the  opposite  sides  in  D,  E,  F.  EF, 
FD,  DE  are  produced  to  meet  the  sides  again  inG,  H,  K.  Prove  that 
circles  on  DG,  EH,  FK  as  diameters  pass  through  the  same  two 
points. 

143.  Find  a  point  without  two  circles  such  that  the  tangents 
drawn  therefrom  to  the  circles  shall  contain  equal  angles. 

144.  Prove  that  the  locus  of  a  point,  at  which  two  given  parts 
of  the  same  straight  line  subtend  equal  angles  is  two  circles. 

145.  Find  on  a  given  straight  line  AB  two  points  P,  Q  such  that 
APQB  is  a  harmonic  range,  and  the  ratio  AP  to  PQ  is  equal  to  a  given 
ratio. 

146.  P  is  a  point  on  the  same  straight  line  as  the  harmonic 
range  ABCD  ;  prove  that 

o^_PB     PD 
AG~  BC'^  DC' 

147.  A  chord  AB  and  a  diameter  CD  of  a  circle  cut  at  right 
angles.  If  P  be  any  other  point  on  the  circle,  P  {A  CBD)  is  a  harmonic 
pencil. 

148.  If  two  circles  touch  one  another  externally  and  from  the 
centre  of  one  tangents  be  drawn  to  the  other,  the  chord  joining  the 
points  in  which  the  first  circle  is  cut  by  the  tangents,  will  be  an  har- 
monic mean  between  the  radii. 

149.  A,  B,  G,  A',  B',  C  are  two  sets  of  three  points  lying  on  two 
parallel  straight  lines  ;  prove  that  the  intersections  of  the  three  pairs 
of  hnes  AA',  B'C;  BB',  C'A  ;  CC,  A'B  lie  on  a  straight  line. 

150.  P  and  Q  are  any  two  points  in  AD,  BC  two  opposite  sides 
of  a  parallelogram ;  X  and  Y  are  the  respective  intersections  of 
AQ,  BP,  and  DQ,  GP',  prove  that  XY,  produced,  bisects  the 
parallelogram. 


496  BOOK   VI. 

151.  If  the  sides  of  a  quadrilateral  circumscribing  a  circle  touch 
at  the  angular  points  of  an  inscribed  quadrilateral,  all  the  diagonals 
meet  in  a  point. 

152.  The  square  inscribed  in  a  circle  is  to  the  square  inscribed  in 
the  semicircle  as  5  to  2. 

153.  Describe  a  rectangle  which  shall  be  equal  to  a  given  square 
and  have  its  sides  in  a  given  ratio. 

154.  In  a  given  rectangle  inscribe  a  rectangle  whose  sides  shall 
have  to  one  another  a  given  ratio. 

155.  Describe  a  triangle  similar  to  a  given  triangle,  one  angular 
point  being  on  the  bounding  diameter  of  a  given  semicircle,  and  one 
of  the  sides  perpendicular  to  this  diameter,  and  the  other  two  angular 
points  on  the  arc  of  the  semicircle. 

156.  If  M,  N  be  the  points  at  which  the  inscribed  and  an 
escribed  circle  touch  the  side  ^C  of  a  triangle  ABC  and  if  BM 
be  produced  to  cut  the  escribed  circle  again  at  P,  then  NP  is  a 
diameter. 

157.  Shew  that  in  general  two  circles  can  be  described  to  cut  two 
lines  AB,  AC  &\,  given  angles  and  to  pass  through  a  fixed  point  P.  If 
T,  T'  be  the  centres  of  these  circles,  then  PA  bisects  the  exterior  angle 

Tcr. 

158.  From  a  given  point  outside  two  given  circles  which  do  not 
meet,  draw  a  straight  line  such  that  the  portions  of  it  intercepted  by 
the  circles  shall  be  in  the  same  ratio  as  their  radii. 

159.  A\  B',  C  are  the  middle  points  of  the  sides  of  the  triangle 
ABC,  and  P  is  the  orthocentre  and  PA',  PB',  PC  produced  meet  the 
circumscribing  circle  in  A",  B'\  C";  prove  that  the  triangle  A"B"G"  is 
equal  in  all  respects  to  the  triangle  ABC. 

160.  If  through  any  point  in  the  arc  of  a  quadrant,  whose  radius 
is  R,  two  circles  be  drawn  touching  the  bounding  radii  of  the 
quadrant,  and  r,  r'  be  the  radii  of  these  circles,  then  ri-'—BP'. 

161.  Let  two  circles  touch  one  another,  and  a  common  tangent  be 
drawn  to  them  touching  them  in  P,  Q.  If  a  pair  of  parallel  tangents 
be  drawn  to  the  two  circles  meeting  PQ  in  A,  B,  and  if  the  line  joining 
their  points  of  contact  meet  PQ  in  G,  then  the  ratio  of  AP  to  BQ  is 
either  equal  to  or  duplicate  of  the  ratio  of  PC  to  QC,  according  as  one 
or  another  pair  of  parallel  tangents  is  taken. 

162.  If  three  circles  touch  a  straight  line  one  of  the  circles  which 
touches  the  three  circles  passes  through  their  radical  centre. 

163.  Two  circles  cut  in  the  points  A,  B;  any  chord  through  A 
cuts  the  circles  again  at  P  and  Q ;  shew  that  the  locus  of  the 
point  dividing  PQ  in  a  constant  ratio  is  a  circle  passing  through  A 
and  B. 


MISCELLANEOUS  EXERCISES  497 

164.  AB  and  AC  are  two  fixed  straight  lines,  and  0  is  a  fixed 
point.  Two  circles  are  drawn  through  0,  one  of  which  touches  AB 
and  ^C  at  D  and  E  respectively,  and  the  other  touches  them  at  F 
and  G  respectively.  Prove  that  the  circles  passing  round  OBF  and 
OEG  touch  one  another  at  0. 

165.  Prove  that  the  locus  of  the  middle  points  of  the  sides  of  all 
triangles  which  have  a  given  orthocentre  and  are  inscribed  in  a  given 
circle  is  another  circle. 

166.  From  any  point  P  on  the  circle  ABC  a  pair  of  tangents 
PQ,  PR  are  drawn  to  the  circle  DEF,  and  the  chord  QR  is  bisected 
in  S.  Shew  that  the  locus  of  /S  is  a  circle;  except  when  the  circle 
ABC  passes  through  the  centre  of  the  circle  DEF,  when  the  locus  of 
S  is  a  straight  line. 

167.  It  is  required  to  describe  a  circle  through  two  given  points 
A,  B  and  to  touch  a  given  circle  which  touches  AB  at  D.  Find  C  the 
centre  of  the  circle :  draw  CA,  CB,  and  draw  AO,  BO  at  right  angles 
to  CA,  CB  respectively.  Prove  that  CD  produced  will  cut  the  circle 
in  a  point  P,  such  that  the  circle  APB  will  touch  the  given  circle 
at  P. 

168.  O  is  the  radical  centre  of  three  circles.  Points  A,  B,  C  are 
taken  on  the  radical  axes  and  AB,  BC,  CA  are  drawn.  Prove  that 
the  six  points  in  which  these  meet  the  three  given  circles  lie  on  a 
circle.  If  radii  vectores  are  drawn  from  O  to  these  six  points,  they 
meet  the  three  given  circles  in  six  points  on  a  circle  and  its  common 
chords  with  the  three  circles  meet  in  pairs  on  OA,  OB,  OC. 

169.  If  from  a  given  point  S,  a  perpendicular  SY  he  drawn  to 
the  tangent  PF  at  any  point  P  of  a  circle,  centre  C,  and  in  the  line 
MP  drawn  perpendicular  to  CS,  or  in  MP  produced,  a  point  Q  be 
taken  so  that  MQ  =  SY,  Q  will  lie  on  one  of  two  fixed  straight  lines. 

170.  The  diagonals  AC,  BD,  of  a  quadrilateral  inscribed  in  a 
circle  cut  each  other  in  E.  Shew  that  the  rectangle  AB,  BC  is  to 
the  rectangle  AD,  DC  as  BE  to  ED. 

171.  The  square  on  the  straight  line  joining  the  centres  of  the 
circumscribed  circle  and  an  escribed  circle  of  a  triangle  is  greater  than 
the  square  on  the  radius  of  the  circumscribed  circle  by  twice  the 
rectangle  contained  by  the  radii  of  the  circles.     (See  p.  476.) 

172.  If  one  triangle  can  be  constructed  such  that  one  of  two 
given  circles  is  the  circumscribed  circle  and  the  other  of  the  given 
circles  is  one  of  its  escribed  circles,  an  infinite  number  of  such  triangles 
can  be  constructed.     (See  p.  477.) 

173.  A,  B,  C  are  three  circles:  prove  that,  if  the  common 
tangent  of  A  and  C  be  equal  to  the  sum  of  the  common  tangents  of  A 
and  B  and  of  B  and  C,  the  three  circles  touch  a  straight  line. 

174.  Three  circles  A,  B,  C  touch  a  fourth  circle  :  prove  that  the 
ratio  of  the  common  tangent  of  A  and  B  to  the  common  tangent  of 
A  and  C  is  independent  of  the  radius  of  A. 


498  BOOK   VI. 

175.  If  ABGD  be  a  quadrilateral  inscribed  in  a  circle  and  a 
second  circle  touch  this  at  A ,  and  the  tangents  to  it  from  B,  C,  D  be 
Bb,  Gc,  Dd,  then  the  rectangle  BD,  Cc  is  equal  to  the  sum  of  the 
rectangles  BC,  Dd  and  CI),  Bb. 

176.  ABCD  is  a  quadrilateral  inscribed  in  a  circle,  F  the  inter- 
section of  the  diagonals:  shew  that  the  rectangle  AF,  FD  is  to  the 
rectangle  BF,  FG  as  the  square  on  ^D  to  the  square  on  BG. 

177.  The  diagonals  of  a  quadrilateral,  inscribed  in  a  circle,  are 
to  one  another  in  the  same  ratio  as  the  sums  of  the  rectangles  con- 
tained by  the  sides  which  meet  their  extremities. 

178.  The  sides  of  a  quadrilateral  ABGD  produced  meet  in  P 
and  Q.  Prove  that  the  rectangles  PD,  DQ  ;  AD,  DG ;  PB,  BQ ; 
AB,  BG  are  proportionals. 

179.  Prove  that  the  locus  of  a  point  such  that  the  square  on  its 
distance  from  a  fixed  point  varies  as  its  distance  from  a  fixed  straight 
line  is  a  circle. 

180.  A  quadrilateral  circumscribes  a  circle.  Shew  that  the 
rectangles  contained  by  perpendiculars  from  opposite  angles  upon 
any  tangent  are  to  one  another  in  a  constant  ratio. 


INDEX. 


Acute-angled  triangle,  12. 
Algebra  and  Geometry  connected, 

134. 
Aliquot  part,  definition  of,  327. 
Alternate  angles,  definition  of,  75. 

—  arc,  definition  of,  244. 
Altitude,  351. 
Ambiguous  case,  preface,  x. 

—  extension  of,  367. 
Angle,  acute,  9. 

—  in  an  arc,  definition  of,  168. 

—  in  a  segment,  168. 

—  obtuse,  9. 

—  of  intersection  of  curves,  266. 

—  not  limited  to  2  rt.  z  s  (ex- 
tension of  application  of  the 
term),  221. 

—  plane,  7. 

—  right,  9. 

—  of  a  sector,  351. 
Angles,  addition  of,  8. 

—  of  any  magnitude,  essential 
to  Euclid's  method,  411. 

—  all  right  angles  equal,  37. 

—  test  of  equality  of,  8. 
Anharmonic  pencils,  having  com- 
mon ray,  439. 

transversals  of,  487. 

—  property  of  four  tangents  to 
circle,  441. 

—  property  of  four  points  on 
circle,  440. 

—  ratio  of  four  points  on  circle, 
442. 

—  ranges,  having  common  point, 
438. 

property  of,  436. 

Antecedent,  definition  of,  328. 
Apollonius,  circle  of,  426. 
Arc,  definition  of,  167. 
Arcs,  direction  of  measurement 

of,  289. 
Area,  definition  of,  10. 
Arithmetical  Progression,  462. 
Away  from,  why  used,  33,  41. 
Axiom,  use  of  term,  preface,  vi. 


Axioms,  definition  of,  15. 

—  examples  of,  15. 

—  the  list  of,  imperfect,  preface, 
vii. 

Axis  of  similitude  defined,  454. 

—  of  triangles,  444. 

—  radical,  264. 

of  pair  of  circles  touching 

three  circles,  456. 

Base  of  a  parallelogram,  351. 

—  of  a  triangle,  351. 
Brianchon's  theorem,  447. 

Casey's  theorem,  469 — 471. 
Centre,  definition  of,  13.    Cf.  174. 

—  of  direct  similitude,  when 
infinitely  distant,  454. 

—  radical,  265. 

—  of  similitude,  (a)  direct,  {h) 
inverse,  455. 

condition  that  chords  in- 
tersecting on  radical  axis  should 
pass  through  inverse  points 
with  respect  to,  457. 

rectangle  under  distances 

to  inverse  points  on  two  circles 
from,  453. 

Centroid,  centre  of  gravity,  de- 
finition of,  103. 

—  of  weights,  425. 
Ceva,  theorem  of,  422. 

Ceva's  theorem,  converse  of,  424. 
Chord,  definition  of,  167. 

—  of  an  arc,  167. 
a  circle,  167. 

contact  of  circle  touching 

two  circles,  451. 
Chords  of  two  circles  intersecting 

on  radical  axis,  452. 
Circle,  definition  of,  13. 

—  escribed,  definition  of,  294. 

—  inscribed,  definition  of,  172, 
294. 

—  of  Apollonius,  426. 

—  Nine-point,  271. 


500 


BOOK   VI. 


Circles,    coaxial,     definition     of, 
429. 

—  construction  for  circles  to  touch 
three,  458. 

—  equal,  have  equal  radii,  231. 

—  four  pairs  ofcircles  touch  three, 
459. 

—  orthogonal,  266. 

—  pair  of,  touching  two  circles, 
455. 

touching  three  circles,  455. 

—  touching  three  circles,  radical 
axis  of  pair  of,  456. 

Circumference,  definition  of,  167. 
Circumscribed  circle,  definition  of, 

297.     Cf.  172. 
Coaxial  circles,  properties  of,  268, 

465,  479,  481,  482,  483. 

—  triangles,  definition  of,  444. 

are  compolar,  445. 

CoUinear  points,  351. 
Commensurable,    definition     of, 

328. 
Complete  quadrilateral,  353. 
Complement.        Complementary 

angles,  definitions  of,  45. 
Complements  of  parallelograms, 

&c..  111. 
Compolar  triangles,  definition  of, 

444. 

are  coaxial,  444. 

Concentric,  definition  of,  170. 
Concurrent  lines,   definition   of, 

351. 
examples  of,   53,   71,   95, 

103,  422,  424. 
Configuration    of    three    circles 

touching  a  fourth  circle,  459. 

—  of  four  circles  touching  a  fifth 
circle,  471. 

Conjugate  points,  352. 

—  rays,  352. 

Consequent,  definition  of,  328. 
Contact,  external,  defined,  199. 

—  internal,  definition  of,  201. 
Contain.     Arcs  contain  angles, 

168. 
Contained,  a  rectangle  is,  134. 
Continuity,    principle    of,     472, 

473,  474. 


Converse  propositions,  27. 

not  necessarily  true,  27. 

Convex  figure,  definition  of,  10. 
Corollary,  =a  further  inference 

from  facts  proved. 
Correspondence  of  sides  of  AS, 256. 
Corresponding  angles,  definition 

of,  75. 

—  vertices,  349. 
Curve,  inverse  of  a,  460. 
Curves,  angle  between,  266. 

Data,  definition  of,  61. 

—  often  conditioned,  61. 
Decagon,  definition  of,  285. 
Desargues,  theorems  due  to,  444, 

445. 
Diagonals,  of  a  complete  quadri- 
lateral, 353. 

—  definition  of,  10. 
Diameter,  definition  of,  13. 

—  bisects  circle,  175. 
Direction  of  measurement  of  arcs, 

289. 
Distance  from  a  point,  definition 
of,  171. 

—  between  centres  of  inscribed 
and  circumscribed  circles  of 
triangle,  476. 

Distances  of  point  from  two 
points,  sum  of  multiples  of 
squares  on,  478. 

Dodecagon,  definition  of,  285. 

DupHcate  ratio,  definition  of,  332. 

Each  to  each,  why  discarded,  pi-e- 

face,  ix. 
Enunciation,   joint  for  Props.  5 

and  6,  Book  II.,  145. 

—  joint  for  Props.  9  and  10,  Book 
II.,  153. 

Equal  in  all  respects,  definition 

of,  13. 
EquaUty  of  ratios,  definition  of, 

330. 
remarks  on,  329. 

—  ratio  of,  328. 
Equiangular  triangles,  property 

of  sides  of  pair  of,  256. 
Equilateral  figure,  11. 


INDEX. 


501 


Equimultiples,  definition  of,  327. 

Euclid,  by  his  Postulates  restrict- 
ed himself  in  his  use  of  instru- 
ments, 21. 

—  in  Prop.  4,  Book  I.,  assumes 
that  two  lines  cannot  have  a 
common  part,  23. 

Example  of  the  principle  of  con- 
tinuity, 474. 
Exterior  angles,  definition  of,  75. 
Extreme  and  mean  ratio,  351. 
Extremes,  definition  of,  832. 

Figure,  definition  of,  preface,  vi. 

—  circumscribed,  172,  285. 

—  convex,  definition  of,  10. 

—  equiangular,  11. 

—  equilateral,  11. 

—  inscribed,  172,  285. 

—  plane,  6. 

—  rectilinear,  9. 

—  regular,  11. 

Figures,  impossible,  preface,  ix. 

Gauss,  what  regular  polygons  can 
be  inscribed  in  a  circle,  320. 

Geometrical  Progression,  462. 

Gergonne's  construction  for  cir- 
cles touching  three  circles,  458. 

Gnomon,  why  discarded,  preface, 
vi. 

Harmonic  pencils,   having  com- 
mon ray,  435. 
transversals  of,  433. 

—  property  of  polar,  430. 

—  range,  defined,  352. 
property  of,  428,  432. 

—  ranges,  having  common  point, 
434. 

Harmonical  Progression,  462. 
Hexagon,  definition  of,  285. 
Homologous,  definition  of,  331. 
Hypotenuse,  definition  of,  11. 

Impossible  figures,  preface,  ix. 

Incommensurable,  definition  of, 
328. 

Inequality,  ratio  of  greater  in- 
equality, 328. 

—  ratio  of  less  inequality,  328. 


Interior  angles,  definition  of,  75. 
Inverse  curves,  definition  of,  460. 
mechanical  mode  of  draw- 
ing, 468. 

—  of  circle  through  pole,  463. 

—  of  circle  not  through  pole,  463. 

—  of  straight  hne,  462. 
Inversion,  definition  of,  460. 

—  angle  between  two  curves  not 
altered  by,  461. 

—  distance  between  two  points 
in  terms  of  inverse  points, 
464. 

—  coaxial  circles  invert  into  co- 
axial circles,  465. 

—  concentric  circles  invert  into 
coaxial  circles,  465. 

—  limiting  points  invert  into 
limiting  points,  465. 

—  locus  of  point  P,  where 

mP.4  =  ?iP^,  464. 

—  one  circle  may  be  its  own  in- 
verse, 466. 

—  pole  of,  460. 

—  Ptolemy's  theorem  proved  by, 
467. 

—  radius  of,  definition  of,  460. 

—  two  circles  may  be  their  own 
inverses,  466. 

—  two  circles,  how  inverted  into 
equal  circles,  466. 

—  two  points  and  their  inverses 
lie  on  a  circle,  461. 

—  three  circles  may  be  their  own 
inverses,  466. 

—  three  circles,  how  inverted  into 
equal  circles,  466, 

Isosceles,  definition  of,  11. 

Letters,  used  to  represent  magni- 
tudes, 327. 
Like  anharmonic  pencils,  353. 

—  anharmonic  ranges,  352. 
Limit,  explained,  217. 
Limiting  points  of  a  series  of  co- 
axial circles,  429. 

Line,  definition  of,  2. 

—  pedal,  definition  of,  272. 

—  Simson's,  272. 

—  straight,  definition  of,  4. 


502 


BOOK   71, 


Lines,  addition  of,  6. 

—  cut  proportionally  (1)  extern- 
ally, (2)  internally,  350. 

Locus,  definition  of,  and  example, 
39. 

—  of  points,  distances  from  which 
to  two  fixed  points  are  equal, 
39. 

distances  from  which  to  two 

fixed  points  are  in  a  fixed  ratio, 

426,  464. 
tangents  from  which  to  two 

circles  are  equal,  264. 
tangents  from  which    to 

two  circles  are  in  a  fixed  ratio, 

479,  480,  481. 

Magnitude  of  angles  not  limited 
by  Euclid's  method,  411. 

Magnitudes  represented  by  letters, 
327. 

Maximum  value,  definition  of,  55. 

example  of,  190,  192,  193, 

195. 

Maximum  and  minimum,  illus- 
tration of,  55. 

—  and  minimum  values  occur 
alternately,  193. 

Mean  proportional,  definition  of, 

332. 
Means,  definition  of,  332. 
Measure,  definition  of,  327. 
Menelaus,  theorem  of,  418. 

converse  of,  420. 

Method  of  Superposition,  5. 
Minimum  value,  definition  of,  55. 

example  of,  55,  57,  193, 195. 

Multiple,  definition  of,  327. 

—  definition  of  m'^^,  327. 

Nine-point   circle,  definition   of, 

271. 
properties    of,    270,    271, 

448,  449. 
Notation  of  proportion,  330. 

Obtuse-angled  triangle,  definition 

of,  12. 
Octagon,  definition  of,  285. 
Orthocentre,  definition  of,  95. 


Orthogonal  circles,  266—268. 
Orthogonally,  system   of  circles 
cutting  two  circles,  268. 

Parallel,  definition  of,  7. 

Parallelogram,  definition  of,  12. 

Parallelograms  about  the  dia- 
gonal of  a  parallelogram.  111, 

Part,  definition  of  w^^^  327. 

Parts  of  a  triangle,  definition  of, 
74. 

Pascal's  theorem,  446. 

Peaucellier's  cell,  468. 

Pedal  line,  272. 

Pencil,  anharmonic,  352, 

—  definition  of,  351. 

—  how  denoted,  351. 

—  harmonic,  352. 

—  like  anharmonic,  353. 
Pencils,  equality  of,  test  of,  440. 
Pentagon,  definition  of,  285. 

—  regular,  construction  for,  309. 
Perimeter,  definition  of,  9. 
Perpendicular,  definition  of,  9. 
Perspective,  triangles  in,  444. 
Plane,  definition  of,  6. 

Point,  angular,  9. 

—  definition  of,  2. 

Points,  conjugate,  defined,  352. 

—  limiting,  of  a  series  of  coaxial 
circles,  429. 

Polar,  definition  of,  258. 
Pole,  definition  of,  258. 

—  of  inversion,  460. 

—  of  triangles,  444, 
Polygon,  definition  of,  11. 

—  the  term  extended  to  include 
triangles  and  quadrilaterals, 
349. 

Polygons,  equiangular,  349. 

—  number  of  conditions  of  equi- 
angularity  of,  349. 

—  number  of  necessary  conditions 
of  similarity  of,  350. 

—  regular,  which  can  be  inscribed 
in  a  circle,  320. 

—  similar,  349. 
Poncelet's  theorems,  483. 
Porism,  definition  of,  475. 

—  examples  of,  475 — 477. 


INDEX. 


503 


Porism  of  inscribed  and  circum- 
scribed circles,  477. 
— condition  for,  476. 

—  of  set  of  coaxial  circles, 
483. 

Postulate,  definition  of,  'preface, 
viii,  4. 

—  I.  Two  straight  lines  cannot 
enclose  a  space,  4. 

—  II.  Two  straight  lines  cannot 
have  a  common  part,  4, 

— •  III.  A  straight  line  may  be 
drawn  from  any  point  to  any 
other  point,  5. 

—  IV.  A  finite  straight  line  may 
be  produced  to  any  length,  5. 

—  V.  All  right  angles  equal, 
preface,  viii,  9. 

—  "VI.  A  circle  may  be  described 
with  any  centre,  and  with  any 
radius,  preface,  ix,  14. 

—  VII.  Any  straight  line  drawn 
through  a  point  within  a  closed 
figure  must,  if  produced  far 
enough,  intersect  the  figure  in 
two  points  at  least,  14. 

—  VIII.  Any  line  joining  two 
points  one  within  and  the  other 
without  a  closed  figure  must 
intersect  the  figure  in  one  point 
at  least,  14. 

—  IX.  If  the  sum  of  the  two  in- 
terior angles,  which  two  straight 
lines  make  with  a  given  straight 
line  on  the  same  side  of  it,  be 
not  equal  to  two  right  angles, 
the  two  straight  lines  are  not 
parallel,  51. 

Principle  of  continuity,  definition 
of,  472. 

examples  of,  474. 

Problem,  definition  oi,  preface,  xi. 

Problems  often  admit  of  several 
solutions,  17,  19,  249,  287. 

Progression,  Arithmetical,  Geo- 
metrical, Harmonical,  462. 

Proportion,  proportionals,  defini- 
tion of,  330. 

—  continued,  definition  of,  332. 

—  notation  of,  330. 


Proportional,  reciprocally,  defini- 
tion of,  380. 

Proportionally  cut,  externally, 
350. 

internally,  350. 

Ptolemy's  Theorem,  property  of 
chords  joining  four  points  on 
circle,  257. 

Casey's  extension  of,  469» 

470,  471. 

Pythagoras,  Theorem  of,  120. 

Quadrilateral,  complete,  353. 

harmonic    properties    of, 

423. 

—  convex,  353. 

—  cross,  353. 

—  definition  of,  11. 

—  re-entrant,  353. 

Kadical  axis,  264. 

—  centre,  265. 
Kadius,  definition  of,  13. 

—  of  inversion,  460. 

—  vector,  definition  of,  460. 
Eange,  anharmonic,  352. 

—  definition  of,  351. 

—  how  denoted,  351. 
Eanges,  like  anharmonic,  352. 
Eatio  of  anharmonic  range,  352, 

443. 

—  compounded,  definition  of,  332. 
independent    of    order  of 

composition,  399. 

—  definition  of,  327. 

—  of  equality,  328. 

—  of  greater  inequality,  328. 

—  of  less  inequality,  328. 

—  of  a  pencil,  353. 

—  of  ratios,  332. 
Eatios  compounded,  332. 

—  equality  of,  defined,  330. 
Eays,  conjugate,  352. 

—  of  a  pencil  defined,  351. 
Eeciprocally  proportional,  379 
Eectangle,  definition  of,  12,  134. 

—  how  denominated,  134. 
Ee-entrant  quadrilaterals,  353. 
Eelations  between  a  line  and  a 

circle,  213. 


504 


BOOK  VI. 


Belations  between  two  circles, 
203. 

BespectfuUy,  how  used,  preface, 
ix. 

Ehomboid,  why  discarded,  pre- 
face, vi. 

Ehombus,  definition  of,  12. 

Bight  angles  are  equal,  37.  Cf. 
preface,  viii. 

Eight-angled  triangle,  definition 
of,  11. 

Eotation,  point  about  point,  186. 

—  line  about  point,  186. 

—  plane  figure  about  point,  186. 

Secant,  definition  of,  213. 
Sector,  angle  of  a,  351. 

—  of  a  circle,  351. 
Segment,  definition  of,  167. 
Segments  of  a  chord,  251. 
Semicircle,  definition  of,  167. 
Sides,  corresponding,  256,  349. 

definition  of,  9. 

Similar  arcs,  definition  of,  168. 

—  figures,  similar  and  similarly 
situate,  448. 

—  segments,  definition  of,  168. 
Similarity  of  figures,   condition 

of,  403. 
Similitude,  axis  of,  defined,  454. 

—  centre  of,  defined,  448. 

direct,  definition  of,  448. 

at  infinite  distance,  409, 

451. 
inverse,  definition  of,  448. 

—  centres  of,  of  two  circles,  450. 

of  three  circles,  454. 

of    circle    of    nine  points 

and  circumscribed  circle,  449. 
Simson's  Hue,  272. 
Solid,  definition  of,  3. 
Square,  definition  of,  12. 

—  ordinary  definition,  why  dis- 
carded, preface,  vi. 

Straight  lines,  test   of   equality 

of,  6. 
Superposition,  method  of,  5. 
note  on,  preface,  viii. 


Supplement,  supplementary  an- 
gles, definitions  of,  45. 
Surface,  definition  of,  3, 

Tangent,  definition  of,  169. 

—  common  to  two  circles,  219. 

—  limit  of,  a  secant,  217,  245, 
253. 

examples  of,  245,  253. 

Test  of  equality  of  angles,  8. 

geometrical  figures,  5. 

straight  lines,  5. 

Third  Proportional,  definition  of, 

332. 
Theorem  of  Apollonius,  426. 

—  Brianchon's,  447. 

—  Casey's,  469,  470,  471. 

—  Ceva's,  422. 

—  of  Desargues,  444. 

—  of  Menelaus,  418. 

—  Pascal's,  446. 

—  Poncelet's,  482,  483. 

—  Ptolemy's,  257. 

—  of  Pythagoras,  120. 
Touch,  meaning  of,  169. 
Transversal,  definition  of,  419. 
Trapezium,  why  discarded,  pre- 
face, vi. 

Triangles,  coaxial,  444. 

—  compolar,  444.  ♦ 

—  definition  of,  11. 

—  equal,  note  on,  74. 

—  inscribed  in  circle,  416. 

—  missing  case  of  equality  of, 
preface,  x. 

Triplicate  ratio,  definition  of,  332. 

Units  of  length  and  of  area,  135. 
Unique  point,  centroid  of  weights 
is  a,  422. 

Vertex,  7. 

—  definition  of,  9. 

—  of  a  pencil  defined,  351. 

Weights,  centroid  of,  425. 
Within  a  circle,  169. 
Without  a  circle,  169. 


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Kohlrausch.  DasJahri8i3.  By  Wilhelm  Wagner,  Ph.D.  2.r. 
Lessing  and  Gellert.    Selected  Fables.    By  Karl  Hermann 

Breul,  M.A.,  Ph.D.     3^. 

Mendelssohn's  Letters.  Selections  from.   By  J.  Sime,  M.A.  3^. 
Raumer.    Der  erste  Kreuzzug  (1095 — 1099).    By  Wilhelm 

Wagner,  Ph.D.     is. 

Riehl.    Culturgeschichtliche  Novellen.     By  H.  J.  Wolsten- 

HOLME,  B.A.  (Lond.).     3^-.  td. 

Schiller.    Maria  Stuart.    By  Karl  Hermann  Breul,  M.A. 

PhD.     -is.bd. 

Wilhelm  Tell.    By  the  same  Editor,    ns.  6d.    Abridged 

Edition,     is.  6d. 

Geschichte  des  Dreissigjahrigen  Kriegs.    Buch  III. 


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XJhland.    Ernst,  Herzog  von  Schwaben.    By  H.  J.  Wolsten- 

HOLME,  B.A.     3J.  6d. 

V.     ENGLISH. 
Bacon's  History  of  the  Reign  of  King  Henry  VII.     By 

the  Rev.  Professor  Lumby,  D.D.     3^. 

Cowley's  Essays.     By  the  Rev.  Professor  Lumby,  D.D.    4^. 
Discourse  of  the  Commonwealf  of  thys  Realme  of  Englande. 

First  printed  in  1581,  and  commonly  attributed  to  W.  S.    Edited  from  the  MSS. 
by  the  late  Elizabeth  Lamond.  [In  the  Press. 

Milton's  Comus  and  Arcades.     By  A.  W.  Verity,   M.A., 

sometime  Scholar  of  Trinity  College,     -^s. 

Milton's  Ode  on  the  Morning  of  Christ's  Nativity,  L' Allegro, 

II  Penseroso  and  Lycidas.     By  the  same  Editor,     is.  6d. 

Milton's  Samson  Agonistes.    By  the  same  Editor.    2s.  6d. 
Milton's  Paradise  Lost.   Books  T.  II.   By  the  same  Editor.   2s. 

BkS.  III.  IV.    By  the  same.    [Preparing. 

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More's  History  of  King  Richard  III.  By  J.  R.  Lumby,D.D.  3^-.  6^. 
More's  Utopia.     By  Rev.  Prof.  Lumby,  D.D.     3^.  (^d. 
Sidney,  Sir  Philip.    An  Apologie  for  Poetrie.     By  E.  S. 

Shuckburgh,  M.  a.    The  Text  is  a  revision  of  that  of  the  first  edition  of  1595.    Si'. 

Thales  to  Cicero,  A  Sketch  of  Ancient  Philosophy.    By 

Joseph  B.  Mayor,  M.A.     3^.  6d. 

The  Two  Noble  Kinsmen.     By  the   Rev.    Professor   Skeat, 

Litt.D.     2>^.tii. 

VI.     EDUCATIONAL  SCIENCE. 

Comenius,  John  Amos,  Bishop  of  the  Moravians.     His  Life 

and  Educational  Worlcs,  by  S.   S.   Laurie,  LL.  D.,    F.R.S.E.     3J.  dd. 

Education,  Three  Lectures  on  the  Practice  of.  L  On  Mark- 
ing, by  h.w.  eve,m.  a.  II.  On  Stimulus,  hy  A.  SiDGwicK,  M.A.  III.  On 
the  Teaching  of  Latin  Verse  Composition,  by  E.  A.  Abbott,  D.D.     is. 

Stimulus.     A    Lecture  delivered   for  the   Teachers'   Training 

Syndicate,  May,  1882,  by  A.  Sidgwick,  M.A.     \s. 

Locke  on  Education.    By  the  Rev.  R.  H.  Quick,  M.A.    3^.  dd. 
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the  Edition  of  1673.     By  O.  Browning,  M.A.     2J. 

Modern  Languages,  Lectures  on  the  Teaching  of.    By  C. 

Coi.beck:,  M.A.     is. 

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Lectures  delivered  in  the  University  of  Cambridge  in  the  Lent  Term,  1883,  by 
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late  Head  Master  of  Exeter  Grammar  School,     is. 

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hiodities  and  the  Countries  of  the  World.     By  H.  R.  Mill,  D.Sc,  F.R.S.E.    i.r. 

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