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IN  MEMORIAM 
FLORIAN  CAJORI 


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*  * 


PLANE    AND    SOLID 


GEOMETEY 


BY 

WILLIAM   J.   MILNE,  Ph.D.,  LL.D. 

PRESIDENT  OP  NEW  YORK   STATE   NORMAL   COLLEGE,   ALBANY     N.T- 


n«-oo;»<€ 


NEW  YORK:.  CINCINNATI.:.  CHICAGO 

AMERICAN    BOOK    COMPANY 


MILNE'S  MATHEMATICS 


Milne's  Elements  of  Arithmetic 

Milne's  Standard  Arithmetic 

Milne's  Mental  Arithmetic 

Milne's  Elements  of  Algebra 

Milne's  Grammar  School  Algebra 

Milne's  High  School  Algebra 
Milne's  Plane  and  Solid  Geometry 
Milne's  Plane  Geometry — Separate 


COPTBIGHT,    1899,   BY 

WILLIAM   J.   MILNE. 
milnk's  obom. 

B-^      IB 


PREFACE 


It  is  generally  conceded  that  geometry  is  the  most  interesting  of  all 
the  mathematical  sciences,  yet  many  students  have  failed  to  find  either 
pleasure  or  profit  in  studying  it.  The  most  serious  hindrance  to  the 
proper  understanding  of  the  subject  has*been  the  failure  on  the  part  of 
the  student  to  grasp  the  geometrical  concept  which  he  has  been  endeav- 
oring to  establish  by  a  process  of  reasoning.  Many  attempts  have  been 
made  by  thorough  teachers  to  remedy  the  difficulty,  but  there  is  a  very 
general  agreement  that  the  most  successful  method  has  been  by  exercises 
in  "  Inventional "  Geometry.  Students  who  have  been  fortunate  enough 
to  have  the  subject  presented  in  that  way  have  usually  understood  it, 
and,  better  still,  they  have  enjoyed  it. 

While  Inventional  Geometry  has  been  full  of  interest  to  the  student, 
it  has  often  failed  to  develop  that  knowledge  of  the  science  which  is 
necessary  to  thorough  mastery,  because  it  has  not  been  progressive,  and, 
what  is  more  to  be  deplored,  it  has  failed  to  give  that  acquaintance  with 
the  forms  of  rigid  deductive  reasoning  which  is  one  of  the  chief  objects 
sought  in  the  study  of  the  science.  The  student  has  often  been  led  by 
this  objective  method  of  study  to  rely  upon  his  visual  recognition  of  the 
relations  of  lines  and  angles  in  a  drawing  rather  than  upon  the  demon- 
stration based  upon  definitions,  axioms,  and  propositions  that  have  been 
proved. 

In  this  book  the  effort  is  made  to  introduce  the  student  to  geometry 
through  the  employment  of  inventional  steps,  but  the  somewhat  frag- 
mentary and  unsatisfactory  result  of  such  teaching  is  supplemented  by 
demonstrations,  in  consecutive  order,  of  the  fundamental  propositions  of 
the  science.  The  desirability  of  training  students  to  form  proper  infer- 
ences from  the  study  of  accurately  drawn  figures  has  been  recognized  by 
the  author;  such  a  method  awakens  keen  interest  in  the  subject  and 
develops  right  habits  of  investigation,  but  there  is  necessity  also  for  the 
accuracy  of  statement  and  the  logical  training  of  the  older  methods  to 
assure  the  pleasure  and  profit  that  belong  with  both. 

Every  theorem  has  been  introduced  by  questions  designed  to  lead  the 
student  to  discover  the  geometrical  concept  clearly  and  fully  before  a 
demonstraticii  is  attempted.  They  are  not  intended  to  lead  to  a  demon- 
stration, but  rather  to  a  correct  and  definite  idea  of  what  is  to  be  proved. 


4  PREFACE, 

Many  of  the  exercises  at  the  foot  of  the  page  require  the  student  to 
infer  the  truth  involved  in  the  relations  given.  The  interrogative  form 
is  employed  for  the  purpose  of  compelling  the  student  to  obtain  the  ideas 
for  himself,  and  the  answers  he  must  give  to  the  questions  furnish  an 
admirable  training  in  accuracy  of  expression. 

A  great  abundance  of  undemonstrated  theorems  and  of  unsolved  prob- 
lems is  supplied,  and  teachers  will  find  them  quite  numerous  enough  for 
the  needs  of  any  class.  The  demonstration  of  original  theorems  and  the 
solution  of  original  problems  are  of  so  great  consequence  in  developing 
the  power  to  reason  that  every  teacher  should  insist  upon  such  w^ork. 

Much  aid  in  originating  demonstrations  may  be  obtained  from  the 
Summaries  which  follow  each  of  the  first  six  books.  These  summaries 
are  not  collections  of  propositions  that  have  been  demonstrated,  but  are 
rather  groups  of  the  truths  established  in  the  book  to  which  they  are 
appended.  If  the  student  makes  himself  thoroughly  acquainted  with 
them,  much  of  the  difl&culty  experienced  in  demonstrating  original 
theorems,  in  solving  problems,  and  in  determining  loci  will  be  removed. 

A  very  small  proportion  of  those  who  study  elementary  geometry 
expect  to  become  mathematicians  in  any  broad  sense  of  the  term,  and 
so  geometry  must  serve  to  give  them  almost  the  only  training  they  will 
get  in  formal  and  logical  argument  in  secondary  schools  and  in  colleges. 
For  this  reason  mathematical  elegance  in  demonstrations  and  in  solu- 
tions has  often  been  sacrificed  in  the  interest  of  clear  and  simple  steps, 
even  though  such  a  plan  has  required  some  expansion  of  the  text.  Ele- 
gant demonstrations  are  appreciated  by  mathematicians,  but  training  in 
formal  deductive  reasoning  is  of  more  consequence  to  most  students. 

The  author  is  indebted  to  many  authors,  both  American  and  foreign, 
who  have  preceded  him.  Their  efforts  to  present  the  subject  in  the  best 
way  have  aided  him  very  greatly  in  preparing  this  work.  He  has  selected 
large  numbers  of  supplementary  theorems  and  problems  from  several 
European  authors  of  renown,  and  yet  he  is  unable  to  give  credit  to  any 
author  in  particular,  because  they  all  seem  to  have  selected  their  exer- 
cises from  some  common  source  of  supply. 

WILLIAM  J.  MILNE. 

Albany.  N.Y. 


SUGGESTIONS   TO   TEACHERS 

1.  Thorough  teaching  and  frequent  reviews,  especially  at  the  beginning 
of  plane  and  of  solid  geometry,  will  be  rewarded  by  intelligent  progress 
and  deep  interest  on  the  part  of  the  students. 

2.  Before  the  assignment  of  any  lesson,  the  teacher  should  require  the 
students  to  draw  the  figures  and  answer  the  questions  which  are  intro- 
ductory to  the  propositions  that  are  to  be  proved  at  the  next  lesson. 

After  the  questions  have  been  answered,  require  the  students  to  express 
their  inferences  in  the  form  of  a  theorem. 

3.  While  the  students  are  answering  the  introductory  questions  or 
stating  the  inferences  suggested  by  the  exercises  at  the  bottom  of  the 
page,  the  inquiry,  "  How  do  you  know  that  this  is  true  ?  "  will  often  lead 
to  a  demonstration. 

4.  The  section  numbers  are  convenient  in  written  demonstrations,  but 
in  oral  proofs  the  reason  for  each  step  should  be  given  fully  and  accurately 
and  all  why's  should  be  answered, 

5.  Students  may  sometimes  be  allowed  to  express  definitions,  axioms, 
theorems,  etc.,  in  their  own  language,  but  as  a  general  rule  their  expres- 
sions are  inaccurate  and  faulty.  The  teacher  should  in  such  instances 
call  attention  to  the  errors  and  require  concise  and  accurate  statements. 
It  will  then  be  discovered  that  they  approximate  very  closely  those  given 
in  the  book. 

6.  The  practice  of  requiring  the  students  to  outline,  in  a  general  way, 
the  steps  they  are  to  take  in  establishing  the  truth  of  a  proposition  will 
develop  nmch  logical  power  and  cause  them  to  look  at  the  argument 
rather  than  at  its  details. 

The  following  are  suggestive  outlines  of  steps: 

Prop.  XXXVI.,  page  61. 

1.  Draw  the  diagonal  AC. 

2.  Prove  ^ABC  and  ADC  equal. 

3.  Prove^5ll2)Cand^i)ll5C. 
Then,  ABCD  is  a  parallelogram. 

6 


6  SUGGESTIONS  TO   TEACHERS, 

Prop.  XII.,  page  185. 

1.  Make  the  required  construction,  drawing  CE^  BJ^  and  CK. 

2.  Prove  A  AEC  and  AJB  equal. 

3.  Prove  AEKL  =<>  2  A  AEG,  and  AGHJ-  2  A  ^t/iS. 

4.  Frove  AEKL  o^ACHJ. 

5.  Similarly  ^Z>irX=o5(7G^i?'. 
Then,  ABDE<>  BCGF+  ACHJ. 

7.  Demonstrations  should  never  be  memorized.  If  suggestion  6  is 
observed  carefully,  students  will  not  be  likely  to  commit  to  memory  the 
words  of  the  book. 

8.  Encourage  students  to  prove  propositions  in  their  own  way,  even 
though  the  proofs  be  less  elegant  than  those  which  are  given.  Elegant 
methods  will  be  acquired  by  practice. 

9.  Written  demonstrations  should  be  required  frequently.  They 
serve  a  double  purpose,  viz. :  they  train  the  eye  and  develop  accuracy 
in  reasoning. 

All  written  work  should  be  done  neatly,  and  all  figures  should  be 
drawn  as  accurately  as  possible. 

10.  The  undemonstrated  theorems  and  unsolved  problems  are  probably 
more  numerous  than  most  classes  can  prove  or  solve  in  the  time  allotted 
to  the  subject,  consequently  teachers  are  expected  to  make  selections 
from  the  lists  given.  The  exercises  are  carefully  graded  so  that  the 
more  difficult  ones  come  at  the  end  of  each  list.  These  may  be  omitted 
at  the  first  reading  and  reserved  for  a  final  review. 

It  is  suggested  that  the  exercises  in  the  interrogative  form  at  the  foot 
of  the  page  in  Books  I  and  II,  except  the  numerical  ones,  be  employed 
at  first  only  for  the  purpose  of  developing  correct  geometrical  concepts 
and  accuracy  in  expressing  the  truth  inferred.  In  review  the  proofs  of 
the  inferences  may  be  required. 

11.  Particular  attention  should  be  given  to  the  Summary  at  the  end  of 
each  book.  The  students  should  be  required  to  state  all  the  conditions 
under  which  the  facts  given  in  black-faced  type  have  been  shown  to  be 
true.  They  will  thus  have  at  immediate  command  all  the  facts  which  can 
be  employed  in  the  demonstration  they  are  attempting. 

If  the  demonstration  of  the  inferences  and  theorems  found  at  the 
bottom  of  the  page  is  required,  the  students  should  be  referred  to  the 
summary.  They  should  understand,  however,  that  they  can  use  no  truth 
given  in  the  summary  whose  section  number  indicates  that  it  was  estab- 
lished subsequently  to  the  point  in  the  text  where  the  proposition  or 
exercise  is  found. 

The  method  of  using  the  summaries  is  illustrated  upon  page  78. 


CONTENTS 
GEOMETRY 

PAOB 

Preliminary  Definitions 9 

Lines  and  Surfaces 10 

Angles .  12 

Measurement  of  Angles  . 15 

Equality  of  Geometrical  Magnitudes 15 

Demonstration  or  Proof .        .        .        . 17 

Axioms -      ...  19 

Postulates 19 

Symbols  .        .        . •  20 

Abbreviations 20 

PLANE   GEOMETRY 

BOOK  I 

Lines  and  Rectilinear  Figures .21 

Parallel  Lines 27 

Triangles 38 

Quadrilaterals 60 

Polygons 68 

Summary         .        . ,        .        .        .74 

Supplementary  Exercises 78 

BOOK  n 

Circles     .        .        ,        . 83 

Measurement 98 

Theory  of  Limits 100 

Summary 122 

Supplementary  Exercises 124 

BOOK  III 

Ratio  and  Proportion 136 

BOOK  IV 

Proportional  Lines  and  Similar  Figures 1^7 

Summary 167 

Supplementary  Exercises .  169 

I 


CONTENTS 


BOOK  V 

PASS 

Area  and  Equivalence ,     173 

Summary 202 

Supplementary  Exercises 20^ 

BOOK  VT 

Regular  Polygons  and  Measurement  of  the  Circle 213 

Maxima  and  Minima 230 

Symmetry 234 

Nummary 237 

Supplementary  Exercises 238 


SOLID   GEOMETRY 

BOOK  VII 

Planes  and  Solid  Angles 243 

Dihedral  Angles 257 

Polyhedral  Angles .  267 

Supplementary  Exercises 272 

BOOK  VIII 

Polyhedrons 273 

Prisms 273 

Pyramids 287 

Similar  and  Regular  Polyhedrons 299 

Formulae. 305 

Supplementary  Exercises 306 

BOOK  IX 

Cylinders  and  Cones 309 

Formulae 325 

Supplementary  Exercises 325 


BOOK  X 


Spheres  ..... 
Spherical  Angles  and  Polygons 
Spherical  Measurements . 

Formulae 

Supplementary  Exercises 


327 

338 
352 
364 
364 


Exercises  for  Review 
Metric  Tables  . 


369 
383 


GEOMETRY 


PRELIMINARY  DEFINITIONS 

1.  Every  material  object  occupies  a  limited  portion  of  space 
and  is  called  a  Physical  Solid  or  Body. 

2.  The  portion  of  space  occupied  by  a  physical  solid  is  identical 
in  form  and  in  extent  with  that  solid,  and  is  called  a  Geometrical 
Solid. 

In  this  work  only  geometrical  solids  are  considered,  and  for  brevity  they 
are  called  simply  solids. 


3.  Any  limited  portion  of  space  is  called  a 
Solid. 

A  solid  has  three  dimensions,  length,  breadth, 
and  thickness. 

The  drawing  in  the  margin  is  represented  as  having 
three  dimensions.  Fig.  1. 

4.  The  limit  of  a  solid,  or  the  boundary  which  separates  it 
from  all  surrounding  space,  is  called  a  Surface. 

A  surface  has  only  two  dimensions,  length  and  breadth. 

A  page  of  a  book  is  a  surface,  but  a  leaf  of  a  book  is  a  soUd. 

5.  The  limit  or  boundary  of  a  surface  is  called  a  Line. 

A  line  has  only  one  dimension,  length.     It  has  neither  breadth 
nor  thickness. 

The  edges  of  a  cube  are  lines. 


10  GEOMETRY. 

6.  The  limits,  or  extremities,  of  a  line  are  called  Points. 

A  point  has  position  only.  It  has  neither  length,  breadth,  nor 
thickness. 

The  dots  and  lines  made  by  a  pencil  or  crayon  are  not  geometrical  points 
and  lines,  but  are  convenient  representations  of  them. 

7.  Lines,  surfaces,  and  solids  are  called  Geometrical  Magnitudes, 
or  simply  Magnitudes. 

8.  A  line  may  be  conceived  of  as  generated  by  a  point  in 
motion.  Hence  a  line  may  be  considered  as  independent  of  a 
surface,  and  it  may  be  of  unlimited  extent. 

A  surface  may  be  conceived  of  as  generated  by  a  line  in  motion. 
Hence  a  surface  may  be  considered  as  independent  of  a  solid, 
and  it  may  be  of  unlimited  extent. 

A  solid  may  be  conceived  of  as  generated  by  a  surface  in 
motion.  Hence  a  solid  may  be  considered  as  independent  of  a 
material  object. 

LINES  AND  SURFACES 

9.  1.  Select  two  points  upon  your  paper  and  draw  several 
lines  connecting  them. 

a.  Which  is  the  shortest  line  you  have  drawn  ?  If  this  line 
is  not  the  shortest  that  can  be  drawn  between  the  points,  what 
kind  of  a  line  is  the  shortest  ? 

h.  What  other  kinds  of  lines  have  you  drawn  besides  a  straight 
line? 

2.  When  a  carpenter  places  a  straightedge  upon  a  board  and 
moves  it  about  over  the  surface,  what  is  he  endeavoring  to  deter- 
mine regarding  the  surface  ? 

3.  If  the  straightedge  does  not  touch  every  point  of  the  sur- 
face of  the  board  to  which  it  is  applied,  what  has  been  discovered 
about  the  surface  ?  ^ 

4.  How  does  he  know  whether  or  not  the  surface  is  an  even 
or  a  plane  surface  ? 

5.  If  any  two  points  on  the  surface  of  a  ball  or  sphere  are 
joined  by  a  straight  line,  where  does  the  line  pass  ? 

6.  How  much  of  the  surface  of  a  perfect  sphere  is  a  plane  sur- 
face? 


GEOMETRY.  11 

10.   A  line  which  has  the  same  direction  throughout  its  whole 
extent  is  called  a  Straight  Line. 

A  straight  line  is  also  called  a  Bight  Line,  or  simply  a  Line. 
In    this   work   the    term    "line"    means   a 
straight  line  unless  otherwise  specified. 


11.  A  line  no  part  of  which  is  straight 
is  called  a  Curved  Line. 

Consequently,  a  curved  line  changes  its  direc- 
tion at  every  point. 

12.  A  line  made  up  of  several  straight 
lines  which  have  different  directions  is 
called  a  Broken  Line.  Fig.  a. 

13.  A  line  made  up  of  straight  and 
curved  lines  is  called  a  Mixed  Line. 

Any  portion  of  a  line  may  be  called  a 
segment  of  that  line. 


Fig,  4. 


14.  A  surface  such  that  a  straight  line  joining  any  two  of  its 
points  lies  wholly  in  the  surface  is  called  a  Plane  Surface,  or  a 
Plane. 

15.  A  surface,  no  part  of  which  is  plane,  is  called  a  Curved 
Surface. 

16.  Any  combination  of  points,  lines,  surfaces,  or  solids  is 
called  a  Geometrical  Figure. 

A  geometrical  figure  is  ideal,  but  it  can  be  represented  to  the  eye  by  draw- 
ings or  objects. 

17.  A  figure  formed  by  points  and  lines  in  the  same  plane  is 
called  a  Plane  Figure. 

18.  A  figure  formed  by  straight  or  right  lines  only  is  called  a 
Rectilinear  Figure. 

19.  The  science  which  treats  of  points,  lines,  surfaces,  and 
solids,  and  of  the  properties,  construction,  and  measurement  of 
geometrical  figures,  is  called  Geometry. 

20.  That  portion  of  geometry  which  treats  of  plane  figures  is 
called  Plane  Geometry. 


a  GEOMETRY. 

21.  That  portion  of  geometry  which  treats  of  figures  whose 
points  and  lines  do  not  all  lie  in  the  same  plane  is  called  Solid 
Geometry. 

ANGLES 

22.  1.  From  any  point  draw  two  straight  lines  in  different 
directions.  Draw  two  straight  lines  from  each  of  several  other 
points,  and  thus  form  several  angles. 

2.  How  does  the  angle  at  the  corner  of  this  page  compare  in 
size  with  the  angle  at  the  corner  of  the  room?  Show  your 
answer  to  be  true  by  an  actual  test. 

How  is  the  size  of  any  angle  affected  by  the  length  of  the  lines 
which  form  its  sides  9 

3.  Eorm  several  angles  at  the  same  point;  that  is,  several 
angles  having  a  common  vertex. 

4.  How  many  of  them  have  a  common  vertex  and  one  common 
side  between  them  and  are,  at  the  same  time,  on  opposite  sides  of 
the  common  side;  that  is,  how  many  angles  are  adjacent  angles? 

5.  Draw  a  straight  line  meeting  another  straight  line  so  as  to 
form  two  equal  adjacent  angles ;  that  is,  two  right  angles. 

6.  Draw  from  a  point  or  vertex  two  straight  lines  in  opposite 
directions;  that  is,  form  a  straight  angle.  How  does  a  straight 
angle  compare  in  size  with  a  right  angle  ? 

7.  Draw  several  angles,  some  greater  and  some  less,  than  a 
right  angle. 

8.  Draw  a  right  angle  and  divide  it  into  two  parts,  or  into  two 
complementary  angles. 

9.  Draw  a  straight  angle  and  divide  it  into  two  parts,  or  into 
two  supplementary  angles. 

10.  Draw  two  straiglit  lines  crossing  or  intersecting  each  other, 
thus  forming  two  pairs  of  opposite  or  vertical  angles. 

23.  The  difference  in  direction  of  two  lines  which  meet  is 
called  a  Plane  Angle,  or  simply  an  Angle. 

The  lines  are  called  the  sides  of  the  angle,    ^' 
and  the  point  where  they  meet  is  called  its 
vertex. 


The  lines  OA  and  OB  are  the  sides  of  the  angle 
formed  at  the  point  0,  and  0  is  the  vertex  of  the  angle.  Fig.  6. 


GEOMETRY.  18 

The  size  of  an  angle  does  not  depend  upon 
the  length  of  its  sides,  but  upon  the  divergence 
of  the  sides  or  upon  the  opening  between  them.  q. 

Compare  Figs.  5  and  6.  yiq.  e. 

24.  When  there  is  but  one  angle  at  a  point,  it  may  be  desig- 
nated by  the  single  letter  at  the  vertex,  or  by  three  letters. 

In  Fig.  6  the  angle  may  be  called  the  angle  A,  or  the  angle  BAG^  or  the 
angle  CAB. 

When  several  angles  have  a  common  vertex,  it  is  customary  to 
use  three  letters  in  designating  each,  placing  the  letter  at  the 
vertex  between  the  other  two. 

An  angle  is  sometimes  designated  by  a 
figure  or  small  letter  placed  in  the  opening 
of  the  angle.  ^  o" 

The  angles  formed  by  the  lines  meeting  at  O 
may  be  designated  by  AOG,  the  figure  1,  and  the  small  letter  a. 

25.  Angles  which  have  a  common  vertex  and  a  common  side, 
and  which  are  upon  opposite  sides  of  the  common  side,  are  called 
Adjacent  Angles. 

In  Fig.  7  angles  COA  and  COB  are  adjacent  angles,  having  a  common 
vertex  O,  and  a  common  side  CO  and  lying  upon  opposite  sides  of  the 
common  side.     Also  COB  and  BOD  are  adjacent  angles. 

26.  When  one  straight  line  meets  another  straight  line  so  as 
to  form  two  equal  adjacent  angles,  each  of  the  angles  is  called  a 
Right  Angle ;  and  each  line  is  said  to  be 
perpendicular  to  the  other. 

The  sides  of  a  right  angle  are  therefore 
perpendicular  to  each  other,  and  lines  per- 
pendicidar  to  each  other  form  right  angles 


with  each  other.  ^^^  ^  ^ 

27.  An  angle  whose  sides  extend  in  opposite  directions  from 
the  vertex,  thus  forming  one  straight  line,  is  called  a  Straight 
Angle. 

If  the  sides  OA  and  OB,  Fig.  9,  extend  in 

opposite  directions  from  the  vertex  O,  the  ^  yi^ ^ 

angle  AOB  is  a  straight  angle. 

A  straight  angle  is  equal  to  two  right  angles. 


14 


GEOMETRY, 


28.  An  angle  less  than  a  right  angle 
is  called  an  Acute  Angle. 

29.  An  angle  greater  than  a  right 
angle  and  less  than  a  straight  angle  is 
called  an  Obtuse  Angle. 


Fig.  10. 


Fig.  11. 


30.  An  angle  greater  than  a  straight 
angle  and  less  than  two  straight  angles 
is  called  a  Reflex  Angle. 

Acute,  obtuse,  and  reflex  angles  are 
called  oblique  angles  in  distinction  from 
right  angles  and  straight  angles. 

31.  When  two  angles  are  together 
equal  to  a  right  angle,  they  are  called 
Complementary  Angles,  and  each  is  said 
to  be  the  Complement  of  the  other. 

If  the  angle  COE  is  a  right  angle,  the  angles 
COD  and  DOE  are  complementary  angles  ;  the 
angle    COD  is  the  complement  of  the  angle  ^'^-  ^^• 

DOE  ;  and  the  angle  DOE  is  the  complement  of  the  angle  COD. 

32.  When  two  angles  are  together  equal  to  two  right  angles, 
they  are  called  Supplementary  An- 
gles, and  each  is  said  to  be  the 
Supplement  of  the  other. 

If  the  angles  AOD  and  DOB  are  to-      

gether  equal   to  two   right   angles,  the     -^ 
angles  AOD  and  DOB  are  supplemen- 
tary angles;  the  angle  AOD  is  the  supplement  of  the  angle  DOB^  and  the 
angle  DOB  is  the  supplement  of  the  angle  AOD. 

33.  When  two  lines  intersect,  the  opposite  angles  are  called 
Vertical  Angles.  a- 

The  angles  AOC  and  DOB,  and  the  angles 
AOD  and  COB  are  vertical  angles. 

34.  A  line  or  a  plane  which  divides 
any  geometrical  magnitude  into  two 
equal  parts  is  called  the  Bisector  of  that  wagnitju.dt^ 


o 

Fig.  14. 


Fro.  15. 


GEOMETRY.  16 

MEASUREMENT  OF  ANGLES 

35.  To  measure  a  magnitude  is  to  find  how  many  times  it  con- 
tains a  certain  other  magnitude  assumed  as  a  unit  of  measure. 

The  unit  of  measure  for  angles  is  sometimes  a  right  angle,  but 
very  often  it  is  a  degree. 

Suppose  the  line  OB,  having  one  of  its  extremities  fixed  at  o, 
moves  from  a  position  coincident  with  OA 
to  the  position  indicated  by  OB.     By  this 
motion  the  angle  AOB  has  been  generated. 

When  the  rotating  line  OB  has  passed 
one  half  the  distance  from  OA  around  to 
OA,  the  lines  extend  in  opposite  directions 
from  O,  and  a  straight  angle  has  been  gen- 
erated; and  since  a  straight  angle  is  equal  ^^®'  ^^' 
to  two  right  angles  (§  27),  when  the  line  has  passed  one  fourth  of 
the  distance  around  to  OA,  a  right  angle  has  been  generated,  and 
the  lines  OB  and  OA  are  perpendicular  to  each  other  (§  26). 
When  the  line  has .  rotated  entirely  around  from  OA  to  OA,  it 
has  generated  two  straight  angles,  or  four  right  angles.  Conse- 
quently :  The  total  angular  magnitude  about  a  point  in  a  plane  is 
equal  to  four  right  angles. 

Inasmuch  as  it  is  frequently  convenient  to  employ  a  smaller 
unit  of  angular  measure  than  a  right  angle,  the  entire  angular  mag- 
nitude about  a  point  has  been  divided  into  360  equal  parts,  called 
degrees;  a  degree  into  60  equal  parts,  called  minutes;  a  minute 
into  60  equal  parts,  called  seconds. 

Degrees,  minutes,  and  seconds  are  indicated  in  connection  with 
numbers  by  the  respective  symbols  °,  ',  ". 

25  degrees,  18  minutes,  34  seconds  is  written  25°  18'  34". 

A  right  angle  is  an  angle  of  90°. 

EQUALITY  OF   GEOMETRICAL  MAGNITUDES 

36  Geometrical  magnitudes  which  coincide  exactly  when  one 
is  placed  upon  or  applied  to  the  other  are  equal.  Since,  however, 
geometrical  magnitudes  are  ideal  they  are  not  actually  taken  up 
and  placed  the  one  upon  the  other,  but  this  is  conceived  to  be 
done. 


16  QEOMETRY. 

This  method  of  establishing  equality  is  called  the  Method  of 
Superposition. 

If  one  straight  line  is  conceived  to  be  placed  upon  another  straight  line 
so  that  the  extremities  of  both  coincide,  the  lines  are  equal.- 

If  an  angle  is  conceived  to  be  placed  upon  another  angle  so  that  their 
vertices  coincide  and  their  sides  take  the  same  directions,  respectively,  the 
angles  are  equal. 

If  any  figure  is  conceived  to  be  placed  upon  any  other  figure  so  that  they 
coincide  exactly  throughout  their  whole  extent,  they  are  equal. 

Figures  that  are  superposa-ble  are  sometimes  called  congruent. 


EXERCISES 

37.  Draw  as  accurately  as  possible  the  figures  which  are  sug- 
gested ;  study  them  carefully ;  infer  the  answers  to  the  questions ; 
state  your  inference  or  conclusion  in  as  accurate  form  as  possible ; 
give  the  reason  for  your  conclusion  when  you  can. 

The  student  is  asked  to  represent  by  a  drawing  any  figure  that  may  be 
required  so  that  it  may  simply  appear  to  the  eye  to  be  accurate.  Geometri- 
cal methods  of  construction  are  given  at  suitable  points  in  the  book,  but  they 
cannot  be  insisted  upon  at  this  stage. 

1.  Draw  two  straight  lines  intersecting  in  as  many  points  as 
possible.     In  how  many  points  do  they  intersect  ? 

Inference :  Two  straight  lines  can  intersect  in  only  one  point. 

2.  Draw  a  straight  line ;  draw  another  meeting  it.  How  does 
the  sum  of  the  adjacent  angles  thus  formed  compare  with  two 
right  angles  ? 

Inference :  When  one  straight  line  meets  another  straight  line,  the  sum  of 
the  adjacent  angles  is  equal  to  two  right  angles. 

3.  Draw  a  straight  line ;  from  any  point  in  it  draw  several 
lines  extending  in  different  directions.  How  does  the  sum  of  the 
consecutive  angles  formed  on  one  side  of  the  given  line  compare 
with  a  right  angle  ?     With  a  straight  angle  ? 

4.  Draw  a  straight  line ;  also  another  meeting  it  so  as  to  form 
two  adjacent  angles,  one  of  which  is  an  acute  angle.  What  kind 
of  an  angle  is  the  other  ? 


I 


GEOMETRY.  17 

5.  Draw  two  intersecting  lines.  How  many  angles  are  formed  ? 
How  do  the  opposite  or  vertical  angles  compare  in  size  ? 

6.  Draw  two  lines  intersecting  so  as  to  form  a  right  angle. 
How  does  each  of  the  other  angles  formed  compare  with  a  right 
angle  ?  How  do  right  angles  compare  in  size  ?  How  do  straight 
angles  compare  in  size  ? 

7.  Draw  two  equal  angles.  How  do  their  complements  com- 
pare ?     How  do  their  supplements  compare  ? 

8.  Draw  a  straight  line ;  select  any  point  in  that  line  and  draw 
as  many  perpendiculars  as  possible  to  the  line  at  that  point. 
How  many  such  perpendiculars  can  be  drawn  on  one  side  of  the 
line  ? 

DEMONSTRATION  OR  PROOF 

38.  The  inferences  which  the  student  has  just  made  are  proba- 
bly correct,  but  they  must  be  proved  to  be  true  before  they  can 
be  relied  upon  with  certainty  unless  their  truth  is  self-evident. 

Many  truths  have  been  inferred,  and  used  as  the  basis  of  im- 
portant enterprises  before  they  have  been  logically  demonstrated. 

Carpenters  believe  that  their  squares  are  true  if  a  line  from  the 
12-inch  mark  on  one  side  to  the  16-inch  mark  on  the  other  is 
20  inches  long ;  but  they  may  not  be  capable  of  giving  satisfac- 
tory reasons  for  their  convictions. 

Many  valuable  facts  of  geometry  may  be  inferred  by  observa- 
tion of  figures  and  objects,  but  the  value  of  the  study  to  a  student 
consists  not  so  much  in  the  knowledge  acquired  as  in  the  develop- 
ment of  the  logical  faculty  by  the  rigid  course  of  reasoning 
required  to  prove  the  truth  or  falsity  of  the  inference. 

Much  attention  must  therefore  be  given  to  the  demonstration  or 
proof  of  inferences  from  known  data,  and  of  statements  even 
though  they  may  seem  to  be  true. 

39.  A  course  of  reasoning  which  establishes  the  truth  or 
falsity  of  a  statement  is  called  a  Demonstration,  or  Proof. 

40.  A  statement  of  something  to  be  considered  or  done  is  called 
a  Proposition. 

"  All  men  are  mortal "  and  "  It  is  required  to  bisect  an  angle  "  are  propo- 
sitions. 

MILNE'S    OEOM.  2 


18  GEOMETRY, 

41.  A  proposition  so  elementary  that  its  truth  is  self-evident 
is  called  an  Axiom. 

An  axiom  is  a  self-evident  truth  to  those  only  who  understand 
the  terms  employed  in  expressing  it. 

Axioms  may  be  illustrated,  but  they  do  not  require  proof. 

Axioms  have  often  a  general  application.  Some,  however, 
apply  only  to  geometrical  magnitudes  and  relations. 

"A  whole  is  equal  to  the  sum  of  all  itg  parts"  is  a  general  axiom.  It 
can  be  employed  in  demonstrating  propositions  in  arithmetic  and  algebra  as 
well  as  in  geometry.  "  A  straight  line  is  the  shortest  distance  between  two 
points"  is  a  geometncal  axiom.  It  can  be  used  only  in  proving  propositions 
which  express  some  geometrical  truth. 

42.  A  proposition  which  requires  demonstration  or  proof  is 
called  a  Theorem. 

"  In  any  proportion  the  product  of  the  extremes  is  equal  to  the  product  of 
the  means"  is  an  algebraic  theorem, 

43.  A  theorem  whose  truth  may  be  easily  deduced  from  a 
preceding  theorem  is  often  attached  to  it,  and  called  a  Corollary. 

The  arithmetical  theorem,  "  A  number  is  divisible  by  3  when  the  sum  ot 
its  digits  is  divisible  by  3"  may  be  readily  deduced  from  the  theorem,  "A 
number  is  divisible  by  9  when  the  sum  of  its  digits  is  divisible  by  9,"  and  may 
be  attached  to  it  as  a  corollary, 

44.  A  proposition  requiring  something  to  be  done  is  called  a 
Problem.    ' 

"  Construct  an  angle  equal  to  a  given  angle  "  is  a  geometrical  problem, 

45.  A  problem  so  simple  that  its  solution  is  admitted  to  be 
possible  is  called  a  Postulate. 

*'  A  straight  line  may  be  drawn  from  one  point  to  another"  is  a  postulate. 
Postulates  are  numerous.     Some  of  those  employed  in  geometry  may  be 
found  in  §  50. 

46.  A  remark  made  upon  one  or  more  propositions,  and  show- 
ing, in  a  general  way,  their  extension  or  limitations,  their  connec- 
tion, or  their  use  is  called  a  Scholium. 

Thus,  after  the  processes  of  dividing  a  common  fraction  by  a  common 
fraction,  and  a  decimal  by  a  decimal,  have  been  taught,  a  remark  showing 
that  precisely  the  same  principles  are  involved  in  each  process  is  a  scholium. 


GEOMETRY,  19 

47.  The  enunciation  of  a  theorem  may  be  separated  into  the 
following  parts : 

1.  The  things  given,  or  granted,  called  the  Data  (singular 
datum). 

2.  A  statement  of  what  is  to  be  proved,  called  the  Conclusion. 
The  term  Hypothesis  may  be  used  instead  of  the  term  data. 

A  supposition  made  in  the  course  of  a  demonstration  is  also 
called  an  Hypothesis. 

48.  In  proofs,  or  demonstrations,  only  definitions,  axioms,  and 
propositions  which  have  been  proved  can  be  employed  to  establish  the 
truth  of  the  proposition. 

AXIOMS 

49.  1.  Things  which  are  equal  to  the  same  thing  are  equal  to  each 
other. 

2.  If  equals  are  added  to  equals,  the  sums  are  equal. 

3.  If  equals  are  taken  from  equals,  the  remainders  are  equal. 

4.  If  equals  are  added  to  unequals,  the  sums  are  unequal. 

5.  If  equals  are  taken  from  unequals,  the  remainders  are  unequal. 

6.  Things  which  are  doubles  of  equal  things  are  equal. 
1.  Things  which  are  halves  of  equal  things  are  equal. 

8.  Tlie  whole  is  greater  than  any  of  its  parts. 

9.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

10.  A  straight  line  is  the  shortest  distance  between  two  points. 

11.  If  two  straight  lines  coincide  in  two  points,  they  will  coincide 
throughout  their  whole  extent,  and  form  one  and  the  same  straight 
line. 

12.  Between  the  same  two  points  but  one  straight  line  can  be 
drawn. 

POSTULATES 

50.  1.   A  straight  line  may  be  produced  indefinitely. 

2.  A  straight  line  may  be  drawn  from  any  point  to  any  other 
point. 

3.  On  the  greater  of  two  straight  lines  a  part  can  be  laid  off  equal 
to  the  less. 

4.  A  figure  can  be  moved  unaltered  to  a  new  ptosition. 


20 


plus,  or  increased  by. 

minus,  or  diminished  by. 

multiplied  by. 

multiplied  by. 

divided  by. 
f  equals, 
I  or  is  (or  are)  equal  to. 

is  (or  are)  equivalent  to. 

is  (or  are)  greater  than. 

is  (or  are)  less  than. 

therefore,  or  hence. 


GEOMETRY, 

SYMBOLS 

A 

triangle. 

A 

triangles. 

O 

parallelogram. 

[EJ 

parallelograms. 

O 

circle. 

© 

circles. 

II  r  parallel, 
I  or  is  (or  are)  parallel  to. 


I   f  perpendicular, 
I  or  is  (or  are)  perpendicular  to. 


Js    perpendiculars. 
"    inch  or  inches. 


ABBREVIATIONS 


Adj adjacent. 

Alt alternate. 

Ax axiom. 

Circum.    .     .     .  circumference. 

Corap complement. 

Const construction. 

Cor corollary. 

Def definition. 

Ex exercise. 

Ext exterior. 

Fig figure. 

Int.       .     .     .     .  interior. 

Lat.  Surf.      .     .  lateral  surface. 


N note. 

Opp Qpposite. 

Post postulate. 

Prob problem. 

Pt point. 

Rect rectangle. 

Rt right. 

Sch scholium. 

Sect sector. 

Seg segment. 

Sim similar. 

St straight. 

Sup supplement. 


The  letters  q.e.d.  are  placed  at  the  end  of  a  proof;  they  are  the  initial 
letters  of  the  Latin  words  qtiod  erat  demonstrandum,  meaning  which  was  to 
be  proved. 

The  letters  q.e.f.  are  placed  at  tlie  end  of  a  solution  of  a, problem  for  quoa 
erat  faciendum,  meaning  which  was  to  be  done. 


PLA^E   GEOMETRY 

BOOK   I 

LINES  AND  RECTILINEAR  FIGURES 

Proposition  I 

51.  Draw  a  straight  line  and  as  many  perpendiculars  as  possible  to  the 
line  at  one  point.     How  many  can  be  drawn  ?     (§  37  ) 

Theorem.  At  any  point  in  a  straight  line  one  perperir- 
dicular  to  the  line  can  be  drawn,  and  only  one. 

Data :    Any  straight  line,  as  AB,  and 
any  point  in  that  line,  as  O.  i" 

To  prove  that  a  perpendicular  to  AB   '  \  ^.p 

can  be  drawn  at  the  point  O,  and  that  I         ,.-''' 

only  one  can  be  drawn.  j  x' 

Proof.     Suppose  a  line  DO  to  rotate    a  o  b 

about  the  point  0  as  a  pivot,  from  the  position  BO  to  AO. 

As  DO  rotates  from  the  position  BO  toward  the  position  AO,  the 
angle  DOB  will,  at  first,  be  smaller  than  the  angle  DO  A. 

As  DO  continues  to  rotate,  the  angle  DOB  will  increase  continu- 
ously, and  will  eventually  become  larger  than  angle  DOA. 

Therefore,  since  angle  DOB  is  at  first  smaller  than  angle  DOA, 
and  afterwards  larger  than  angle  DOA,  there  must  be  one  position 
of  DO,  as,  for  example,  CO,  in  which  the  two  angles  are  equal. 

By  §  26,  CO  is  then  perpendicular  to  AB. 

Since  there  is  but  one  position  in  which  the  line  DO  makes 
equal  angles  with  the  line  AB,  there  can  be  but  one  perpendicular. 

Therefore,  at  any  point  in  a  straight  line  one  perpendicula,r  to 
the  line  can  be  drawn,  and  only  one.  q.e.d. 

21 


22  PLANE   GEOMETRY.  — BOOK  L 

Proposition  II 

62.  1.  Draw  two  lines  intersecting  so  as  to  form  a  right  angle.  How 
does  each  of  the  other  angles  formed  compare  in  size  with  a  right  angle  ? 
How  do  right  angles  compare  in  size?  How'do straight  angles  compare? 
(§  37  ) 

2.  Draw  two  equal  angles  and  their  complements.  How  do  their  com- 
plements compare  in  size?    How  do  their  supplements  compare?   (§  37  ) 

Theore^n,    All  right  angles  are  equal. 

Data :  Any  right  angles,  as  ABC 
and  DEF. 

To  prove   angles  ABC  and  DBF 
equal. 


A  B  D  ■     E 

Proof.  Suppose  that  Z  DBF  is  placed  upon  Z  ABC  in  such  a 
way  that  the  point  B  falls  upon  the  point  B  and  the  line  BD  takes 
the  direction  of  the  line  BA. 

Since  by  §  26,  BC  is  perpendicular  to  BA  and  BF  is  perpendicu- 
Jar  to  BD  and  on  the  same  side  of  the  line, 

line  BF  must  take  the  same  direction  as  line  BC, 
for  otherwise  there  would  be  two  perpendiculars  to  BA  at  the  point 
B  and  by  §  51,  this  is  impossible. 

Consequently,  the  line  BF  falls  upon  the  line  BC, 

and  Z  DEF  coincides  with  Z  ABC. 
Hence,  §  36,  A  ABC  and  DEF  are  equal. 

Therefore,  all  right  angles  are  equal.  q.e.d. 

53.  Cor.  I.     All  straight  angles  are  equal. 

54.  Cor.  II.  The  complements  of  equal  angles  are  equal,  also 
the  supplements  of  equal  angles  are  equal. 

Ex.  1.    Find  the  complement  of  an  angle  of  15°  ;  27°  ;  35°  ;  49°. 

Ex.  2.    Find  the  supplement  of  an  angle  of  38°  ;  96°  ;  114°. 

Ex.  3.   The  complement  of  an  angle  is  63°.     What  is  the  angle  ? 

Ex.  4.   The  supplement  of  an  angle  is  103°.     What  is  the  angle  ? 

Ex.  5.  Find  the  complement  of  the  supplement  of  an  angle  of  165° ;  140° ; 
122°;  113°;  108°;  99°. 

Ex.  6.  Find  the  supplement  of  the  complement  of  an  angl'  of  48°;  84°; 
27°  ;  16° ;  31°  ;  64°  ;  39°. 


PLANE   GEOMETRY.  — BOOK  I.  23 

Proposition   III 

55.  1-  Draw  a  straight  line,  and  another  meeting  it.  How  does  the 
sum  of  the  adjacent  angles  thus  formed  compare  with  a  right  angle? 
With  a  straight  angle  ?     (§  37  ) 

2.  Draw  a  straight  line,  and  from  any  point  in  it  draw  several  lines 
extending  in  different  directions.  How  does  the  suin  of  the  consecutive 
angles  formed  on  one  side  of  the  line  compare  with  a  right  angle? 
With  a  straight  angle?  How  does  the  sum  of  the  consecutive  angles 
on  both  sides  of  the  line  compare  with  a  right  angle  ?  With  a  straight 
angle?     (§37) 

Theorem,  If  one  straight  line  meets  another  straight 
line,  the  sum  of  the  adjacent  angles  is  equal  to  two  right 
angles. 

Data :  Any  straight  line,  as  AB,  and  -P 

any  other  straight  line,  as  GO,  meeting 
it  in  the  point  0. 

To  prove   the  sum  of  the  adjacent 

«,ngles,  AOC  and  t,  equal  to  two  right      j 

angles. 

Proof.     When  CO  is  perpendicular  to  AB, 
by  §  26,  each  of  the  A  AOC  and  t  is  a  rt.  Z, 

and  their  sum  is  two  rt,  A. 

When  CO  is  not  perpendicular  to  AB, 

draw  DO  perpendicular  to  AB  at  the  point  O. 

Then,  by  §  26,         A  r  and  DOB  are  rt.  A, 
and  Ar-{-  A  DOB  =  2  it.  A; 

by  Ax.  9,  A  DOB  =  As-{-At, 

.'.  by  substitution, 

Ar-^As-{-At  =  2Ti  A. 

By  Ax.  9,  A  AOC  =Ar  +  As, 

and  by  substitution,    A  AOC  -[-  At  =  2  vt.  A. 

Therefore,  if  one  straight  line  meets  another  straight  line,  the  sum 
of  the  adjacent  angles  is  equal  to  two  right  angles.  q.e.d. 


24 


PLANE   GEOMETRY.  — BOOK  L 


56.  Cor.  I.    The  sum  of  aU  the  consecutive  angles  which  hcuve  a 
common  vertex  in  a  line,  and  which  lie  .-^ 
on  one  side  of  it,  is  equal  to  two  right 
angles,  or  a  straight  angle. 

57.  Cor.  II.      The  sum  of  all  the 

consecutive  angles  thai  can  be  formed 
about  a  point  is  equal  to  four  right 
angles,  or  two  straight  angles. 

Ex.  7.  One  line  meets  another,  making  two  angles  with  it.  One  angle 
contains  87°.    How  many  degrees  are  there  in  the  other  ? 

Ex.  8.  Four  of  the  five  consecutive  angles  about  a  point  contain  17®,  36°, 
89°,  and  110°  respectively.    How  many  degrees  are  there  in  the  fifth  angle  ? 

Ex.  9.  K  two  lines  meet  a  third  line  at  the  same  point,  making  with  the 
third  line  angles  of  27°  and  63°  respectively,  what  is  the  angle  between  the 
two  lines? 


Proposition  IV 

58.  Constmct  two  angles  which  are  adjacent  such  that  their  sum  is 
equal  to  two  right  angles.  What  kind  of  a  line  do  their  exterior  sides 
form? 

Theoi^ew,.  If  the  sum  of  two  adjacent  angles  is  equal  to 
tivo  right  angles,  their  exterior  sides  form  one  straight 
line. 

Data :  Any  two  adjax3ent  angles,  as 
AOC  and  COB,  whose  sum  is  equal  to 
two  right  angles. 

To  prove  that  the  exterior  sides,  AO     2 

and  OB,  form  one  straight  line. 

Proof. 

From  data,  Z  AOC  -f-  Z  COB  =  2  rt.  ^ ; 

.-.  by  §  27,  Z  AOC+Z  COB  =  a  st.  Z. 

But  by  Ax.  9,      Z  AOC  +  Z  COB  =  Z  ^OF, 
.-.  by  Ax.  1,  Z  AOB  =  a  st  Z. 

Hence,  by  %  27,  AO  and  OB,  the  sides  of  Z  AOB  extending  in 
opposite  directions  from  the  point  G,  iiorm  one  straight  line. 

Therefore,  etc.  q.b.d. 


PLANE   GEOMETRY.  — BOOK  L 
Proposition  V 


25 


59.   Draw  two  intersecting  lines.    How  many  angles  are  formed?    How 
do  the  opposite  or  vertical  angles  compare  in  size  ? 

Theorem.     If  two  straight  lines  intersect,  the  vertical 
angles  are  equal. 

Data :    Any   two   intersecting    straight 
lines,  as  AB  and  CD. 

To  prove  tlie  vertical  angles,  as  v  and  t, 
equal. 


Z  r  +  Z  V  =  2  rt.  A 


Proof. 

By  §  55, 

and  Z.r  +  Z.t  =  2Tt.A) 

hence,  by  Ax.  1,  Zr-\-/.v  =  /.r-\-Zt. 

Subtracting  Z  r  from  both  sides  of  this  equality, 

by  Ax.  3,  /.v  =  Z.  t. 

In  like  manner  it  may  be  proved  that  /.r  =  Z.s. 
Therefore,  etc. 


Q.E.D. 


Ex.  10.  The  complement  of  an  angle  is  43°.  What  is  the  supplement  of 
the  angle  ? 

Ex.  11.  The  supplement  of  an  angle  is  125°.  What  is  the  complement  of 
the  angle  ? 

Ex.  12.  How  many  degrees  are  there  in  the  supplement  of  the  comple- 
ment of  an  angle  of  60°  ?    Of  43°  25'  50"  ? 

Ex.  13.  How  many  degrees  are  there  in  the  complement  of  the  supple- 
ment of  an  angle  of  159°  ?     Of  133°  15'  25"  ? 

Ex.  14.  How  many  degrees  are  there  in  the  angle  formed  by  the  bisectors 
of  two  supplementary  adjacent  angles  ? 

Ex.  15.  If  a  line  drawn  through  the  vertex  of  two  vertical  angles  bisects 
one  angle,  how  does  it  divide  the  other  ?  (§  37) 

Ex.  16.  If  one  of  the  vertical  angles  formed  by  the  intersection  of  two 
straight  lines  is  37°,  what  is  the  value  of  each  of  the  other  angles  ? 

Ex.  17.  Lines  are  drawn  to  bisect  the  two  pairs  of  vertical  angles  fonned 
by  two  intersecting  straight  lines.  What  is  the  direction  of  these  bisectors 
with  reference  to  each  other?  (§37) 


26  PLANE   GEOMETRY.^BOOK  L 

Proposition  VI 

60.  1.  Draw  a  straight  line ;  from  a  point  not  in  the  line  draw  as 
many  perpendiculars  to  the  line  as  possible.     How  many  can  be  drawn  ? 

2.  How  does  the  perpendicular  compare  in  length  with  any  other  line 
drawn  from  the  point  to  the  given  line  ? 

Theorem,  From  a  point  without  a  straight  line  only  one 
perpendicular  can  be  drawn  to  the  line. 


Data :  Any  straight  line,  as  AB,  any 
point  without  it,  as  P,  and  the  per- 
pendicular PC  drawn  from  the  point 
P  to  the  line  AB.  [1 


To  prove  that  PC  is  the  only  per- 
pendicular that  can  be   drawn  from  /       \ 
the  point  P  to  the  line  AB. 

G 

F 

Proof.     Prolong  PC  to  F,  making  CF=PC;  from  P  draw  any 
other  line  to  AB,  as  PD ;  and  draw  DF. 

Then,  PCF  is  a  straight  line. 

§  26,  ^  ^  3,nd  s  are  rt.  A, 

and,  §  52,  Zr=Zs. 

Revolve  the  figure  PCD  about  AB  as  an  axis  and  apply  it  to  the 

figure  FCD. 

DC  being  in  the  axis  remains  fixed, 

and  since  Zr=Zs, 

CP  will  take  the  direction  of  CF, 

Since,  const.,  CP  =  CF, 

P  and  F  will  coincide ; 

.*.  Ax.  11,  DP  and  DF  will  coincide, 

and,  §  36,  Zt  =  Zv. 

Revolve  PCD  back  to  its  original  position  and  prolong  PD  to  G. 

Then,  §  56,  Zt-\- Zv -\- Zw  =  2  vt  A, 


PLANE   GEOMETRY.-^ BOOK  L 


27 


and  since  /.t  =  /.Vj 

2Zt-\-Zw  =  2Yt.A', 
Z  f  +  i  Z  w  =  1  rt.  Z  ; 
that  is,  Z  ^  is  less  than  a  rt.  Z. 

PD  is  not  perpendicular  to  AB. 
But  since  PD  represents  any  line  from  P  to  AB  other  than  PC^ 
PC  is  the  only  perpendicular  that  can  be  drawn  to  AB  from  P. 
Therefore,  etc.  q.e.d. 

61.  Cor.  A  perpendicular  is  the  shortest  line  that  can  be  drawn 
from  a  j^oint  to  a  line. 

1.  Since  PCF  is  a  straight  line,  is  PDF  a  straight  line  ?  Ax.  12. 

2.  Which  line,  then,  is  the  shorter,  PCF  or  PDF?  Ax.  10. 

3.  What  part  of  PCF  is  PC  ?     Of  PDF  is  PD  ? 

4.  Then,  how  do  PC  and  PD  compare  in  length  ? 

5.  Since  PD  represents  any  line  from  P  to  AB  other  than  the 
perpendicular  PC,  what  is  the  shortest  line  that  can  be  draw» 
from  a  point  to  a  line  ? 

62.  The  distance  from  a  point  to  a  line  is  always  understood 
to  be  the  perpendicular  or  shortest  distance. 

PARALLEL  LINES 

63.  Lines  which  lie  in  the  same  plane,  and   ~ 

which  cannot  meet  however  far  they  may  be 

extended,  are  called  Parallel  Lines. 

64.  A  straight  line  which 
crosses  or  cuts  two  or  more 
straight  lines  is  called  a  Trans- 
versal. 

EF  is  a  transversal  of  AB  and  CD. 

Eight  angles  are  formed  by 
the  transversal  EF  with  the  lines 
AB  and  CD. 

65.  The  angles  above  AB  and 
those  below  CD,  or  those  without  the  two  lines  cut  by  the  trans* 
versal,  are  called  Exterior  Angles. 

Angles  r,  5,  y,  and  z  are  exterior  angles. 


28  PLANE   GEOMETRY.  — BOOK  I. 

66.  The  angles  between,  or  within  the  two  lines  cut  by  the 
transversal,  are  called  Interior  Angles. 

Angles  t,  r,  w,  and  x  are  interior  angles. 

67.  Non-adjacent  angles  without  the  two  lines,  and  on  opposite 
sides  of  the  transversal,  are  called  Alternate  Exterior  Angles. 

Angles  r  and  sr,  or  s  and  y,  are  alternate  exterior  angles. 

68.  Non-adjacent  angles  within  the  two  lines,  and  on  opposite 
sides  of  the  transversal,  are  called  Alternate  Interior  Angles. 

Angles  t  and  x,  or  v  and  w,  are  alternate  interior  angles. 

69.  Non-adjacent  angles,  which  lie  one  without  and  one  with- 
in the  two  lines,  and  on  the  same  side  of  the  transversal,  are 
called  Corresponding  Angles. 

Angles  r  and  w,  s  and  aj,  t  and  y,  or  v  and  z,  are  corresponding  angles. 
Corresponding  angles  are  also  called  Exterior  Interior  Angles. 

70.  Ax.  13.  TJirough  a  given  point  but  one  straight  line  can  be 
drawn  parallel  to  a  given  straight  line. 

Proposition  VII 

71.  Draw  a  straight  line;  also  two  other  lines  each  perpendicular  to 
the  first  line.  In  what  direction  do  the  perpendiculars  extend  with 
reference  to  each  other  ? 

Theorem,  If  two  straight  lines  are  perpendicular  to 
the  same  straight  line,  they  are  parallel. 

Data:  Any  straight  line,  as  AB, 
and  any  two  straight  lines  each  per- 
pendicular to  AB,  as  CD  and  EF. 

To  prove  CD  and  EF  parallel. 

i)       F 

Proof.  Since,  by  data,  both  CD  and  EF  are  perpendicular  to  AB, 
they  cannot  meet,  for,  if  they  should  meet,  there  would  then  be 
two  perpendiculars  from  the  same  point  to  the  line  AB,  which  is 
impossible.     §  60. 

Hence,  §  63,  CD  W  EF. 

Therefore,  etc.  q.e.d. 


PLANE   GEOMETRY.  — BOOK  I.  •         29 

Proposition  VIII 

72.  Draw  two  parallel  lines ;  also  a  transversal  perpendicular  to  one 
of  them.  What  is  the  direction  of  the  transversal  with  reference  to  the 
other  parallel  line? 

Theorem*  If  a  straight  line  is  perpendicular  to  one  of 
two  parallel  straight  lines,  it  is  perpendicular  to  the  other. 


Data:    Any    two   parallel   straight 
lines,  as  AB  and  CD,  and  any  straight     a- 
line  perpendicular  to  AB,  as  EF,  cut- 
ting CD  at  the  point  J.  q. 

To  prove  EF  perpendicular  to  CD.  G 


Proof.  If  EF  is  not  perpendicular  to  CD  at  the  point  J,  it  will 
be  perpendicular  to  some  other  line  drawn  through  that  point. 

Suppose  GH  is  that  Jine. 

Then,  hyp.,  EF  ±  GH. 

But,  data,  EF  A.  AB, 

then,  §  71,  GH  II  AB. 

But,  data,  ^  CD  W  AB, 

then,  §  70,  GH  and  CD  passing  through  J  cannot  both  be  parallel 
to  AB. 

Hence,  the  hypothesis  that  EF  is  not  perpendicular  to  CD  is 
untenable. 

Consequently,  EF  A.  CD. 

Therefore,  etc.,  q.e.d. 

Ex.  18.  Two  lines  are  drawn  each  parallel  to  AB^  and  another  line  mak- 
ing an  angle  of  90°  with  AB.  What  is  the  direction  of  this  line  with  refer- 
ence to  each  of  the  other  two  lines  ? 

Ex.  19.  State  and  illustrate  the  differences  between  a  plumb  line,  a  per- 
pendicular line,  and  a  vertical  line. 

Ex.  20.  Two  parallel  lines  are  cut  by  a  third  line  making  one  interior 
angle  35".     What  is  the  value  of  the  adjacent  interior  angle  ? 


30  •  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  IX 

73.  1.  Draw  two  parallel  lines ;  also  a  transversal.  How  many  pairs  of 
vertical  angles  are  formed?  How  many  pairs  of  supplementary  adjacent 
angles  ?  How  many  sizes  of  angles  are  formed '?  How  many  angles  of 
each  size  ?    When  may  they  all  be  of  the  same  size  ? 

2.  Name  a  pair  of  angles  whose  sum  is  equal  to  two  right  angles. 
Name  seven  other  pairs.  Name  a  set  of  four  angles  whose  sum  is  equal 
to  four  right  angles.     Name  three  other  sets. 

3.  Name  the  pairs  of  alternate  interior  angles.  How  do  the  angles  ol 
any  pair  compare  in  size  ? 

Theorem.  If  two  parallel  straight  lines  are  cut  hy  a 
transversal,  the  alternate  interior  angles  are  equal. 


Data:  Any  two  parallel  straight 
lines,  as  AB  and  CD,  cut  by  a  trans- 
versal, as  EF,  in  the  points  H  and  J. 

To  prove  the  alternate  interior  an- 
gles, as  AHJ  and  DJH,  equal. 


Proof.     Through  L,  the  middle  point  of  HJ,  draw  GK  ±  CD. 

Then,  §  72,  GK  i.  AB. 

Revolve  the  figure  JLK  about  the  point  L  and  apply  it  to  the 
figure  HLG,  so  that  LJ  coincides  with  LH. 

Then,  since,  §  59,  Z  JLK  =  Z  HLG, 

LK  takes  the  direction  of  LO. 

Const.,  JK  ±  GK  and  HG  ±  GK, 

and  since  the  point    J  falls  upon  the  point  H, 
§  60,  JK  must  fall  upon  HG. 

Since  LJ  coincides  with  LH, 

and  JK  takes  the  same  direction  as  HQ^ 

§  36,  Z  GHL  =  Akjl. 

Therefore,  etc.  q.e.d. 


PLANE   GEOMETRY.  — BOOK  I.  31 

74.  If  two  theorems  are  related  in  such  a  way  that  the  data 
and  conclusion  of  one  become  the  conclusion  and  data,  respec- 
tively, of  the  other,  the  one  is  said  to  be  the  converse  of  the  other. 

Thus,  the  converse  of  the  theorem  just  proved  is,  "  Two  straight  lines  cut 
by  a  transversal  are  parallel,  if  the  alternate  interior  angles  are  equal." 

Converse  propositions  cannot  be  assumed  to  be  true.  They 
may  be  true,  but  their  truth  must  be  established  by  proof. 

Thus,  the  truth  of  the  proposition,  "The  product  of  two  even  numbers 
is  an  even  number,"  can  be  established  readily,  but  its  converse,  "An  even 
number  is  the  prodjict  of  two  even  numbers,"  is  evidently  false. 

Proposition  X 

75.  Draw  two  lines ;  also  a  transversal.  In  what  direction  do  the  lines 
extend  with  reference  to  each  other,  if  the  alternate  interior  angles  are 
equal? 

Theorefn,  Two  straight  lines  cut  by  a  transversal  are 
parallel,  if  the  alternate  interior  angles  are  equal.  (Con- 
verse of  Prop.  IX.) 

/E 

Data :  Two  straight  lines,  as  AB  and  / 

CD,  such  that  when  cut  by  any  trans-  a --'/'" ^ 

versal,    as   EF,  the   alternate   interior  ^  / 

angles,  as  ahf  and  EJD,  are  equal.        / 

To  prove  AB  and  GB  parallel.  / 

Proof.  If  AB  is  not  parallel  to  CZ),  then  some  other  line,  as  KL, 
drawn  through  the  point  H  is  parallel  to  CD. 

Then,  hyp.  and  §  73,       Z  KUF  =  Z  EJD  ; 

but,  data,  Zahf  =  /.  ejd  ; 

hence,  Ax.  1,  Z  KHF  =  Z  AHF, 

which  is  absurd,  since  a  part  cannot  be  equal  to  the  whole. 

Hence,  the  hypothesis,  that  some  other  line,  as  KL,  drawn 
through  the  point  H  is  parallel  to  CD,  is  untenable. 

Consequently,  AB  II  CD. 

Therefore,  etc.  q.e.d. 


32 


PLANE    GEOMETRY.  —  BOOK  1. 


Proposition  XI 

76.    Draw  two  parallel  lines ;  also  a  transversal.     Name  the  pairs  of 
'corresponding  angles.     How  do  the  angles  of  any  pair  compare  in  size? 

Theorem,     If  two  parallel  straight   lines  are  cut  by  a 
transversal,  the  corresponding  angles  are  equal. 


Data:  Any  two  parallel  straight 
lines,  as  AB  and  CD,  cut  by  any- 
transversal,  as  EF. 

To  prove  the  corresponding  angles, 
as  t  and  s,  equal. 


Proof.     §  59, 

§73, 

hence.  Ax.  1, 
Therefore,  etc. 


Z.s  =  Zr; 
Zt  =  Zs. 


Q.E.D. 


Proposition  XII 

77.  Draw  two  lines;  also  a  transversal.  In  what  direction  do  the 
lines  extend  with  reference  to  each  other,  if  the  corresponding  angles  are 
equal? 

Theorem.  Two  straight  lines  cut  hy  a  transversal  are 
parallel,  if  the  corresponding  angles  are  equal.  (Converse 
of  Prop.  XI.) 


Data :  Two  straight  lines,  as  ^5 
and  CD,  such  that  when  cut  by  any 
transversal,  as  EF,  the  correspond- 
ing angles,  as  t  and  s,  are  equal. 

To  prove  AB  and  CD  parallel. 


Proof.     §  59, 

data, 

then.  Ax.  1, 
Hence,  §  75, 
Therefore,  etc. 


Zr  =  Zt, 
Zs  =  Zt; 
Zr  =  Zs. 
AB  II  CD. 


Q.E.D. 


PLANE   GEOMETRY.  — BOOK  I.  33 

Proposition  XIII 

78.  Draw  two  parallel  lines;  also  a  transversal.  How  does  the  sum 
of  the  two  interior  angles  on  the  same  side  of  the  transversal  compare 
with  a  right  angle  ? 

Theorem,  If  two  parallel  straight  lines  are  cut  hy  a 
transversal,  the  suin  of  the  two  interior  angles  on  the  same 
side  of  the  transversal  is  equal  to  two  right  angles. 

Data:  Any  two  parallel  straight 
lines,  as  AB  and  CD,  cut  by  a  trans-  / 

versal,  as  EF.  ^ 

To  prove  the  sum  of  the  two  in- 
terior angles  on  the  same  side  of  the 
transversal,  as  t  and  s,  equal  to  two  / 

right  angles.  f 

Proof.     §  73,  Z.r  =  /.s. 

Adding  Z  ^  to  each  member  of  this  equation, 
Ax.  2,  Z  r  +  Z  ^  =  Z  s  4-  Z  ^. 

But,  §  55,  Zr  +  Z^  =  2rt.  ^; 

.•.  Ax.  1,  Zs  +  Z^  =  2rt.  A. 

Therefore,  etc.  q.e.d. 

Ex.  21.  If  two  parallel  lines  are  cut  by  a  transversal,  what  is  the  sum  of 
the  two  exterior  angles  on  the  same  side  of  the  transversal  ? 

Ex.  22.  The  straight  lines  AB  and  CD  are  cut  hy  EF  in  G  and  H 
respectively ;  angle  EHD  —  38°.  What  must  he  the  value  of  the  angle 
EGB  in  order  that  AB  and  CD  may  be  parallel  ? 

Ex.  23.  A  transversal  cutting  two  parallel  lines  makes  an  interior  angle 
of  50°.  What  is  the  value  of  the  other  interior  angle  on  the  same  side  of  the 
transversal  ? 

Ex.  24.  TWO  parallel  lines  are  cut  by  a  third  line  making  one  interior 
angle  35°.  What  is  the  value  of  each  of  the  other  interior  angles  ?  How 
many  degrees  are  there  in  the  sum  of  the  interior  angles  upon  the  same  side 
of  the  transversal  ? 

Ex.  25.  How  do  lines  bisecting  any  two  alternate  interior  angles,  formed 
by  two  parallel  lines  cut  by  a  transversal,  lie  with  reference  to  each  other  ? 

Ex.  26.    The  straight  lines  AB  and  CD  are  cut  by  EF  in  G  and  H  re- 
spectively ;  angle  EHD  =  40°.     What  must  be  the  value  of  the  angle  A  GF, 
if  AB  and  CD  are  parallel  ? 
milne's  geom. — 3 


34  PLANE   GEOMETRY.  — BOOK  i. 

Proposition  XIV 

79.  Draw  two  lines;  also  a  transversal.  In  what  direction  do  th« 
lines  extend  with  reference  to  each  other,  if  the  sum  of  the  two  interior 
angles  on  the  same  side  of  the  transversal  is  equal  to  two  right  angles  ? 

Theorein,  Two  straight  lines  cut  by  a  transversal  are 
parallel,  if  the  sum  of  the  two  interior  angles  on  the  same 
side  of  the  transversal  is  equal  to  two  right  angles.  (Con- 
verse of  Prop.  XIII.) 

Data:  Two  straight  lines,  as  AB 
and  CZ),  such,  that  when  cut  by  any- 
transversal,  as  EF,  the  sum  of  the 
two  interior  angles  on  the  same 
side  of  the  transversal,  as  t  and 
s,  is  equal  to  two  right  angles. 

To  prove  AB  and  CD  parallel. 

Proof.     §55,  Zr  + Z^  =  2  rt.  zi; 

data,  Z^  +  Zs  =  2  rt.  A] 

.'.  Ax.  1,  Z.r  +  At  =  /.t  +  /.s. 

Taking  Z  t  from  each  member  of  this  equation, 

Ax.  3,  Zr  =  /.s. 

Hence,  §  75,  AB  II  CD. 

Therefore,  etc.  q.e.d. 

Ex.  27.  AB  and  CD  are  two  lines  cut  in  O  and  JST,  respectively,  by  EF-, 
A  BCF  =  123*,  and  Z  GHD  =  62°.     Are  the  lines  AB  and  CD  parallel  ? 

Ex.  28.  If  two  lines  are  cut  by  a  transversal  and  the  sum  of  the  two 
exterior  angles  on  the  same  side  of  the  transversal  is  equal  to  180°,  are  the 
lines  parallel  ? 

Ex.  29.  Two  parallel  lines  are  cut  by  a  transversal  so  that  one  exterior 
angle  is  105°.  How  many  degrees  are  there  in  the  sum  of  each  pair  of  alter- 
nate interior  angles  ? 

Ex.  30.  The  bisectors  of  two  adjacent  angles  are  perpendicular  to  each 
other.    What  is  the  relation  of  the  given  angles  to  each  other  ? 

Ex.  31.  Two  lines  are  cut  by  a  transversal.  In  what  direction  do  they 
extend  with  reference  to  each  other,  if  the  alternate  exterior  angles  are 
equal  ? 


PLANE   GEOMETRY.  — BOOK  L  35 

Proposition  XV 

80.  Draw  a  straight  line ;  also  two  other  lines  each  parallel  to  the 
given  line.  In  what  direction  do  these  two  lines  extend  with  reference 
to  each  other? 

Theorem,  Straight  lines  which  are  parallel  to  the  same 
straight  lUie  are  parallel  to  each  other. 


/K 


Data:    Any  straight  lines,  as   AB  pG" 

and   CD,   each    parallel    to    another  q 

straight  line,  as  EF. 

To  prove  AB  parallel  to  CB.  e 


ryn 


'L 


Proof.     Draw  any  transversal,  as  KL,  cutting  the  lines  AB,  CD, 
and  EF. 


Since,  data, 

CD   II  EF, 

73, 

Z.r=As. 

Since,  data, 

AB  II  EF, 

73, 

At  =  Z.s. 

Then,  Ax.  1, 

/.r  =  /.t. 

Hence,  §  77, 

AB    II   CD. 

Therefore,  etc. 

Q.E.p. 

Ex.  32.  The  straight  lines  AB,  CD,  and  EF  are  cut  in  Cr,  H,  and  J 
respectively,  by  ifZ;.  angle  KGB  =  S1°;  angle  i^JTC  =  149° ;  angle  FJL 
=  143°.     Are  the  lines  AB  and  CD  parallel  ?     AB  and  EF  ?     CD  and  EF  ? 

Ex.  33.  Can  two  intersecting  straight  lines  both  be  ptiallel  to  the  same 
straight  line  ? 

Ex.  34.  How  many  degrees  are  there  in  the  angle  formed  by  the  bisec- 
tors of  two  complementary  adjacent  angles  ? 

Ex.  35.  If  the  line  BD  bisects  the  angle  ABC,  and  EF  is  drawn  through 
B  perpendicular  to  BD,  how  do  the  angles  CBE  and  ABF  compare  in  size  ? 

Ex.  36.  If  a  straight  line  is  perpendicular  to  the  bisector  of  an  angle  at 
the  vertex,  how  does  it  divide  the  supplementary  adjacent  angle  formed  by 
producing  one  side  of  the  given  angle  through  the  vertex  ? 


36  PLANE   GEOMETRY.  — BOOK  L 

Proposition  XVI 

81.  1.  Construct  two  angles  whose  corresponding  sides  are  parallel. 
How  do  the  angles  compare  in  size,  if  both  corresponding  pairs  of  sides 
extend  in  the  same  direction  from  their  vertices  ?  If  both  pairs  extend  in 
opposite  directions  from  their  vertices  ? 

2.  Discover  whether  it  is  possible  for  the  angles  to  have  their  sides 
parallel  and  yet  not  be  equal. 

Theorem,  Angles  whose  corresponding  sides  are  parallel 
are  either  equal  or  supplementary. 

Data:  AB  parallel  to  DE,  and  BC 
parallel  to  HF,  forming  the  angles  r, 
s,  tf  s',  and  t'. 

To  prove  1.     Z.r  =  Zs,  ot  Z  s'. 

2.   Zr  and  Zi   or  ZV  supplemen-    ^        V^'  ^ 

tary.  ^  ^ 

Proof.  1.  Produce  BG  and  EB,  if  necessary,  to  intersect  as 
at  G. 

§76,  Zr  =  Zv,  and  Zs  =  Z'y;    .'.  Ax.  1,  Zr  =  Zs. 

§59,  Zs^Zs^',  .-.Ax.  1,  Zr  =  Zs'. 

2.    §  55,  Zs  +  Z^  =  2rt.Z; 

but  Zr  =  Z8\ 

Zr-\-Zt  =  2Tt.A. 

Hence,  §  32,  Zr  and  Z t  are  supplementary; 
also,  sincCj  §  59,  Z  ^  =  Z ^',  Zr  and  Z  t'  are  supplementary. 

Therefore,  etc.  q.e.d. 

82.  Scholium.  The  angles  are  equals  if  both  corresponding  pairs 
of  sides  extend  in  the  same  or  in  opposite  directions  from  their 
vertices ;  they  are  supplementary,  if  one  pair  extends  in  the  sa^ne 
and  the  other  in  opposite  directions. 

Ex.  37.  If  two  straight  lines  are  perpendicular  each  to  one  of  two  par- 
allel straight  lines,  in  what  direction  do  they  extend  with  reference  to  each 
other  ? 

Ex.  38.  How  do  lines  bisecting  any  two  corresponding  angles,  formed  by 
parallel  lines,  cut  by  a  transversal,  lie  with  reference  to  each  other  ? 


PLANE   GEOMETRY.  — BOOK  L 


87 


Proposition  XVII 

83.  1.  Construct  two  angles  whose  corresponding  sides  are  perpen- 
dicular to  each  other.  How  do  the  angles  compare  in  size,  if  both  are 
acute  ?     If  both  are  obtuse  ? 

2.  Discover  whether  it  is  possible  for  the  angles  to  have  their  sides 
perpendicular  and  yet  not  be  equal. 

Theore^n,  Angles  whose  corresponding  sides  are  perpen- 
dicular to  each  other  are  either  equal  or  supplementary. 


Data :  AB  perpendicular  to  DB,  and  CE 
perpendicular  to  FB,  forming  the  angles 
r,  s,  and  t. 

To  prove  1  /  r  =  Z  s. 

2.  Z  t  and  Z  s  supplementary. 


Proof.     1.    §  26,    Aabd  and  GBF  are  vt.  A\ 
.'.  §  52,  Z  ABD  =  Z  CBF, 

and.  Ax.  9,  Zr  +  Zv  =  Zv  +  Zs. 

Taking  Z  v  from  each  member  of  this  equation, 
Ax.  3,  Zr  =  Z  8. 

2.    §§  55,  32,     Z  t  and  Z  r  are  supplementary ; 
but  Zr  =  Zsy 

hence,  Z  ^  and  Z  s  are  supplementary. 

Therefore,  etc.  q.e.d. 

84.  Sch.  The  angles  are  equal,  if  both  are  acw^e  or  if  both  are 
obtuse  ;  they  are  supplemerdaryy  if  OTie  is  acwie  and  the  other  obtuse. 

Ex.  39.  The  bisectors  of  two  adjacent  angles  form  an  angle  of  45°. 
What  is  the  relation  of  the  given  angles  to  each  other  ? 

Ex.  40.  Two  angles  are  supplementary,  and  the  greater  is  five  times  the 
less.     How  many  degrees  are  there  in  each  angle  ? 

Ex.  41.  Two  angles  are  complementary,  and  the  greater  is  five  times  the 
less.     How  many  degrees  are  there  in  each  angle  ? 

Ex.  42.  Two  parallel  straight  lines  are  cut  by  a  transversal  so  that  one 
of  the  two  interior  angles  on  one  side  of  the  transversal  is  eleven  times  the 
other.     How  many  degrees  are  there  in  each  of  the  exterior  angles  ? 


88 


PLANE   GEOMETRY.  — BOOK  I. 


TRIANGLES 

85.  A  portion  of  a  plane  bounded  by  three  straight  lines  is 
called  a  Plane  Triangle,  or  simply  a  Triangle. 

The  straight  lines  which  bound  a  triangle  are 
called  its  sides,  their  sum  is  called  its  perimeter,  and 
the  vertices  of  the  angles  of  a  triangle  are  called 
the  vertices  of  the  triangle. 

86.  The  angle  formed  by  any  side  of  a 
triangle  and  the  prolongation  of  another 
side  is  called  an  Exterior  Angle  of  the 
triangle. 

Angle  s  is  an  exterior  angle. 

87.  An  angle  formed  within  a  triangle  by  any  two  of  its  sides 
is  called  an  Interior  Angle  of  the  triangle. 

Whenever  the  angles  of  a  triangle  or  other  enclosed  figure  are 
mentioned,  the  interior  angles  are  referred  to  unless  otherwise 
specified. 

Angles  ^,  C,  and  r  are  interior  angles. 

88.  The  interior  angles  which  are  not  adjacent  to  the  exterior 
angles  are  called  Opposite  Interior  Angles. 

Angles  A  and  C  are  opposite  interior  angles  when  s  is  the  exterior  angle. 


89.   A  triangle  whose   three   sides   are 
unequal  is  called  a  Scalene  Triangle. 


90.   A  triangle  two  of  whose  sides  are  equal  is 
called  an  Isosceles  Triangle. 


91.   A  triangle  whose  three  sides  are  equal  is 
called  an  Equilateral  Triangle. 


PLANE   GEOMETRY.  — BOOK  I. 


39 


92.  The  side  upon  which  a  triangle  is  assumed  to  stand  is 
called  the  Base  of  the  triangle. 

See  figure  accompanying  §  95. 

93.  The  angle  opposite  the  base  of  a  triangle  is  called  the 
Vertical  Angle,  and  its  vertex  is  called  the  Vertex  of  the  triangle. 


94.  The  perpendicular  distance  from  the  vertex  of  a  triangle  to 
its  base,  or  its  base  produced,  is  called  the  Altitude  of  the  triangle. 

Since  any  side  of  a  triangle  may  be  considered  as  its  base,  it  is 
evident  that  a  triangle  may  have  three  altitudes  and  that  they 
will  be  unequal,  if  the  sides  of  the  triangle  are  unequal.  If  the 
triangle  is  equilateral,  then  all  three  altitudes  will  be  equal ;  if 
the  triangle  is  isosceles,  only  two  of  the  altitudes  will  be  equal. 

95.  A  triangle,  one  of  whose  angles  is  a  right 
angle,  is  called  a  Right  Triangle. 

In  a  right  triangle,  the  side  opposite  the 
right  angle  is  called  the  hypotenuse. 

96.  A  triangle,  one  of  whose  angles  is  an 
obtuse  angle,  is  called  an  Obtuse  Triangle. 

97.  A  triangle,  each  of  whose  angles  is  an 
acute  angle,  is  called  an  Acute  Triangle. 

Obtuse  triangles  and  acute  triangles  are 
called  oblique  triangles. 

98.  A  triangle  whose  three  angles  are  equal  is  called  an  Equl 
angular  Triangle. 

See  figure  accompanying  §  91. 

99.  A  line  drawn  from  any  vertex  of  a  tri- 
angle to  the  middle  of  the  opposite  side  is 
called  a  Median,  or  Median  Line  of  the  triangle. 


40  PLANE   GEOMETRY.  —  BOOK  L 

Proposition  XVIII 

100.  1.  Make  two  triangles  such  that  two  sides  of  one,  and  the  angle 
formed  by  them,  shall  be  equal  to  the  corresponding  parts  of  the  other. 
How  do  the  triangles  compare  ?  How  do  the  third  sides  compare  ?  How 
do  the  angles  of  one  compare  with  the  corresponding  angles  of  the  other? 

2.    Under  what  conditions  are  two  triangles  equal? 

Theorem,  Two  triangles  are  equal,  if  two  sides  and  the 
included  angle  of  one  are  equal  to  two  sides  and  the  in- 
cluded angle  of  the  other,  each  to  each. 


A  B     D  E 

Data :  Any  two  triangles,  as  ABC  and  DEF,  in  which  AB  =  DE^ 
AC  =  DF,  and  angle  A  =  angle  D. 

To  prove  triangles  ABC  and  BFF  equal. 

Proof.     Place  A  ABC  upon  A  DEF,  AB  coinciding  with  DE. 

Data,  Za  —  Zb, 

hence,  AC  will  take  the  direction  of  BF; 

and  since  AC  =  BF, 

the  point  C  will  fall  upon  the  point  F. 

Since  the  point  B  falls  upon  E  and  the  point  C  upon  Fj 
BC  will  coincide  with  EF. 

Then,  A  ABC  and  BEF  coincide  in  all  their  parts. 

Hence,  §  36,  Aabc  =  A  BEF. 

Therefore,  etc.  q.e.d. 

Prove  Prop.  XVIII 

(1)  when  AB  —  BEy  BC  ~  EF,  and  angle  B  =  angle  E. 

(2)  when  AC  =  BF,  BC  =  EF,  and  angle  C  =  angle  F. 

101.  Sch.  Every  triangle  has  six  parts  or  elements ;  namely, 
three  sides  and  three  angles.  Two  equal  triangles  may  be  made 
to  coincide  in  all  their  parts.  Therefore,  each  part  of  one  is 
equal  to  the  corresponding  part  of  the  other. 


PLANE    GEOMETRY.— BOOK  I.  41 

Proposition  XIX 

102.  1.  Make  two  triangles  such  that  a  side  of  one,  and  the  angles 
formed  at  its  extremities,  shall  be  equal  to  the  corresponding  parts  of 
the  other.     How  do  the  triangles  compare?     What  parts  are  equal? 

2.   Under  what  conditions  are  two  triangles  equal  ? 

Theorem,  Two  triangles  are  equal,  if  a  side  and  two 
adjacent  angles  of  one  are  equal  to  a  side  and  two  adja- 
cent angles  of  the  other,  each  to  each. 


A  B     D  E 

Data:  Any  two  triangles,  as  ABG  and  DEF,  in  which  AB  =  DE, 
angle  A  =  angle  D,  and  angle  B  =  angle  E. 

To  prove  triangles  ABC  and  DEF  equal. 

Proof.     Place  A  ABC  upon  A  DEF,  AB  coinciding  with  DE. 

Data,  /.A=^Ad; 

hence,  AC  will  take  the  direction  of  DF, 

and  the  point  C  will  fall  upon  DF,  or  upon  DF  produced. 

Also,  data,  /.B  =  /.E\ 

hence,  BG  will  take  the  direction  of  EF, 

and  the  point  C  will  fall  upon  EF,  or  upon  EF  produced. 

Since  the  point  C  falls  upon  each  of  the  lines  DF  and  EF, 
it  must  fall  upon  their  point  of  intersection,  F. 

Then,  A  ABC  and  D^i^  coincide  in  all  their  parts. 

Hence,  §  36,  A  ^^C  =  A  DEF. 

Therefore,  etc.  q.e.d. 

Prove  Prop.  XIX 

(1)  when  angle  C  =  angle  F,  angle  B  =  angle  E,  and  BC  =  EF. 

(2)  when  angle  A  =  angle  D,  angle  C  =  angle  F,  and  AC  =  DF. 

Ex,  43.  Are  two  triangles  equal,  if  the  three  angles  of  one  are  equal  to 
the  three  angles  of  the  other,  each  to  each  ? 

Ex.  44.  Can  two  triangles,  having  two  sides  and  an  angle  of  one  respec- 
tively equal  to  two  sides  and  an  angle  of  the  other,  be  unequal  ? 


42 


PLANE  GEOMETRY.^ BOOK  I. 


Proposition  XX 

103.  1.  Draw  a  straight  line  and  a  perpendicular  to  that  line  at  its 
middle  point ;  select  any  point  in  the  perpendicular  and  from  that  point 
draw  straight  lines  to  the  extremities  of  the  given  line.  How  do  these 
lines  compare  in  length  ?  How  do  the  angles  made  by  these  lines  with 
the  perpendicular  compare  in  size  ?  How  do  the  angles  made  by  these 
lines  with  the  given  line  compare  ? 

2.  Select  any  point  not  in  the  perpendicular  and  from  that  point  draw 
straight  lines  to  the  extremities  of  the  given  line.  How  do  they  compare 
in  length  ? 

3.  Draw  a  straight  line  and  find  a  point  equidistant  from  its  ex- 
tremities; find  another  point  equidistant  from  its  extremities;  con- 
nect these  points  by  a  line  and  if  necessary  extend  it  until  it  intersects 
the  given  line.  At  what  point  does  it  intersect  the  given  line  ?  What 
kind  of  angles  does  it  make  with  the  given  line  ? 

4.  What  line  contains  every  point  that  is  equidistant  from  the  ex- 
tremities of  a  straight  line? 

Theorem,  If  a  perpendicular  is  drawn  to  a  straight  line 
at  its  middle  point, 

1,  Any  point  in  the  perpendicular  is  equidistant  from 
the  extremities  of  the  line. 

2,  Any  point  not  in  the  perpendicular  is  unequally  dis- 
tant from  the  extremities  of  the  lint* 

Data :  Any  straight  line,  as  ^5 ;  a  per- 
pendicular to  it  at  its  middle  point,  as 
CD ;  any  point  in  CD,  as  E ;  and  any  point 
not  in  CD,  as  F, 

To  prove 

1.  E  equidistant  from  A  and  5. 

2.  F  unequally  distant  from  A  and  B, 

Proof.  1.   Draw  AF,  BE,  AF,  and  BF. 

Data  and  §  26,         Z  r  and  Z  s  are  rt  z1^ 
and,  §52,  /Lr     Z«. 

In  A  ABE  and  BBE,  AD  =  J57), 

ED  is  common, 
and  Z  r  =  Z  5 ; 

.-.§100,  ^ADE=^l^BDE, 


PLANE   GEOMETRY.  — BOOK  I.  43 

and,  §  101,  AE  =  BE. 

That  is,  E  is  equidistant  from  A  and  B, 

2.   From  the  point  G,  where  AF  cuts  CD,  draw  GB, 

Ax.  10,  BF  <BG  +  GF] 

but  BG  =  AG'j  .  Why? 

.-.  substituting  AG  for  its  equal  BG, 

BF  <AG  +  GF, 

or  ^i^  <  AF. 

That  is,         2^  is  unequally  distant  from  A  and  B» 

Therefore,  etc.  q.e.d. 

104.  Cor.  I.  Every  point  that  is  equidistant  from  the  extremi- 
ties of  a  straight  line  lies  in  the  perpendicular  at  the  middle  point 
of  that  line. 

105.  Cor.  II.  If  a  perpendicular  is  erected  at  the  middle  point 
of  a  straight  line,  the  lines  joining  the  extremities  of  this  line  with 
any  point  in  the  perpendicular  make  equal  angles  with  the  line 
and  also  with  the  perpendicular.  §  101 

106.  Cor.  III.  Two  points  each  equidistant  from  the  extremities 
of  a  straight  line  determine  the  perpendicular  at  the  middle  point 
of  that  line. 

Ex.  45.  How  does  the  distance  between  two  parallel  lines  at  a  given  point 
compare  with  tlie  distance  between  them  at  any  other  point  ? 

Ex.  46.  Can  two  angles  which  are  not  adjacent  have  a  common  vertex 
and  a  common  side  ? 

Ex.  47.  If  in  an  equilateral  triangle  a  line  is  drawn  from  the  vertex  to 
the  middle  point  of  the  base,  how  do  the  triangles  thus  formed  compare 
in  size  ? 

Ex.  48.  If  two  lines  bisect  each  other,  in  what  direction  do  the  lines 
joining  their  opposite  extremities  extend  with  reference  to  each  other  ? 

Ex.  49.  If  two  sides  of  a  triangle  are  equal,  and  a  line  is  drawn  bisecting 
their  included  angle  and  intersecting  the  third  side,  how  do  the  segments 
of  the  third  side  compare  in  length  ? 

Ex.  50.  Perpendiculars  are  erected  at  the  extremities  of  a  line  and  termi- 
nate in  any  bisector  of  the  line  that  is  not  perpendicular  to  the  line.  How 
do  the  perpendiculars  compare  in  length  ? 

Ex.  51.  If  through  the  middle  point  of  a  straight  line  terminating  in  two 
parallel  lines,  a  second  straight  line  is  drawn  also  terminating  in  the  parallels, 
how  do  the  parts  of  the  second  line  compare  in  length  ? 


44 


PLANE   GEOMETRY.  — BOOK  L 


Proposition  XXI 

107.  1.  Make  two  triangles  such  that  the  sides  of  one  shall  be  equal 
to  the  corresponding  sides  of  the  other.  How  do  the  triangles  compare  ? 
How  do  the  corresponding  angles  compare? 

2.   Under  what  conditions , are  two  triangles  equal  ? 

Theorem,  Two  triangles  are  equal,  if  the  three  sides  of 
one  are  equal  to  the  three  sides  of  the  other,  each  to  each. 


Data :  Any  two  triangles,  as  ABC  and  DEF,  in  which  AB  =  DE^ 
AC  =  BF,  and  BC  =  EF. 
To  prove  triangles  ABC  and  DEF  equal. 

Proof.  Place  A  DEF  in  the  position  ABF  so  that  the  «qual 
sides,  BE  and  AB,  coincide,  and  the  vertex  F  falls  oppo&ite  C. 
Draw  CF. 

Data,  AF  =  AC,  and  BF  =  BC; 

A  and  B  are  each  equidistant  from  F  and  C; 
hence,  §  106,  AB  ±  CF  Sit  its  middle  point, 

and,  §  105,  Zr  =  Zs. 

In  A  ABC  Sind  ABF,  AC  =  AF, 

AB  is  common, 

Zr  =  Zs; 
A  ABC  =  A  ABF. 
AABC  =  ABEF, 


and 

.-.  §  100, 

That  is, 

Therefore,  etc. 

Prove  by  placing  the  triangle  so  that 

(1)  BF  will  coincide  with  AC. 

(2)  EF  will  coincide  with  BC. 


Q.E.D 


PLANE   GEOMETRY.^ BOOK  L  45 

108.  Sch.  It  is  evident  that  in  equal  triangles  the  parts  which 
are  similarly  situated  are  equal ;  that  is,  the  angles  included  by 
the  equal  sides  are  equal;  the  angles  opposite  the  equal  sides  are 
equal,  the  sides  included  between  equal  angles  are  equal,  and 
the  sides  opposite  the  equal  angles  are  equal. 

109.  In  equal  figures,  the  parts  which  are  similarly  situated 
are  called  Homologous  parts. 

Ex.  52.  Draw  two  parallel  lines  intersecting  two  parallel  lines,  and  draw 
a  line  joining  two  opposite  points  of  intersection.  How  do  the  triangles  thus 
formed  compare  ? 

Ex.  53.  Perpendiculars  are  drawn  from  the  extremities  of  a  line  to  any 
line  that  bisects  it  and  is  not  perpendicular  to  it.  How  do  the  perpendiculars 
compare  in  length  ? 

Ex.  54.  The  line  BD  is  the  bisector  of  the  angle  ABC  whose  sides  are 
equal.  Lines  are  drawn  from  any  point  of  BD^  as  ^,  to  A  and  G.  How 
do  AE  and  CE  compare  in  length  ? 

Ex.  55.  In  a  triangle  ABC  angle  A  equals  angle  B  ;  a  line  parallel  to  AB 
intersects  AC  m  D  and  BC  in  E.  How  do  the  angles  ADE  and  BED  com- 
pare ? 

Ex.  56.  If  D  is  the  middle  point  of  the  side  BC  of  the  triangle  ABC,  and 
BE  and  CF  are  perpendiculars  from  B  and  0  to  AD^  or  AD  produced,  how 
do  BE  and  GF  compare  in  length  ? 

Proposition  XXII 

110.  1.  Cut  out  a  paper  triangle  ABC.  Cut  off  the  corners  and  place 
the  vertices  A,  B,  and  C  together.  To  how  many  right  angles  is  the  sura 
of  the  three  angles  equal  ? 

2.  If  one  angle  of  a  triangle  is  a  right  angle,  how  does  the  sura  of  the 
other  two  angles  compare  with  a  right  angle? 

3.  What  is  the  greatest  nuraber  of  obtuse  angles  that  a  triangle  raay 
have?    The  greatest  nuraber  of  right  angles? 

4.  If  there  are  two  triangles  such  that  the  sum  of  two  angles  of  one  is 
equal  to  the  sum  of  two  angles  of  the  other,  how  do  the  third  angles  com- 
pare in  size  ? 

5.  If  there  are  two  right  triangles  such  that  a  side  and  an  acute  angle 
of  one  are  equal  to  the  corresponding  parts  of  the  other,  how  do  the 
triangles  compare? 

6.  Extend  one  side  of  a  triangle  through  a  vertex ;  through  the  same 
vertex  draw  a  line  parallel  to  the  opposite  side  of  the  triangle.  Since 
the  figure  thus  formed  contains  two  parallel  lines  and  a  transversal,  M'hat 
angles  of  the  figure  are  equal?  How  do3s  the  exterior  angle  of  the  tri- 
angle compare  with  the  sum  of  the  two  opposite  interior  angles? 


46  PLANE   GEOMETRY.  — BOOK  1. 

Theorem,  The  sum  of  the  angles  of  a  triangle  is  equal 
to  two  right  angles. 

o 

A  B         ~    ~~~    "d 

Datum  :     Any  triangle,  as  ABC. 

To  prove         Zr -{- Z.s-\- Zt  =  two  right  angles. 

Proof.     Produce  AB  to  D  and  draw  BE  W  AC. 

§  m,  Zr-\-Zs'  +  Zt'  =  2it.A', 

but,  §  73,  Zs'  =  Zs, 

and,  §  76,  .  Zt'  =  Zt', 

.'.  substituting  Zs  and  Zt  for  Zs'  and  Z  ^'  in  the  first  equation, 
Zr  +  Zs  +  Zt  =  2Tt.  A. 

Therefore,  etc.  q.e.d, 

111.  Cor.  I.  In  a  right  triangle  the  sum  of  the  two  acute  angles 
is  equal  to  a  nght  angle. 

112.  Cor.  II.  A  triangle  cayinot  have  more  than  one  right  angl^, 
nor  more  than  one  obtuse  angle. 

113.  Cor.  III.  If  two  angles  of  one  triangle  are  equal  to  tn'^i 
angles  of  another,  the  third  angles  are  equal. 

114.  Cor.  IV.  Two  right  triangles  are  equal,  if  a  side  and  an 
acute  angle  of  one  are  equal  to  a  side  and  an  acute  angle  of  the  of^^er, 
each  to  each. 

115.  Cor.  V.  Any  extenor  angle  of  a  triangle  is  equal  to  the 
sum  of  the  tico  opposite  interior  angles. 

1.  What  interior  angle  is  equal  to  Zt'?  Why  ? 

2.  What  interior  angle  is  equal  to  Z  s'  ?  Why  ? 

3.  To  what,  then,  is  the  whole  exterior  angle  equal  ? 

Ex.  57.  May  a  triangle  be  formed  whose  angles  are  93°,  40°,  and  OP  re- 
spectively ?    08°,  24°,  and  58°  ?     57°,  49°,  and  74°  ? 

Ex.  58.  Two  angles  of  a  triangle  are  together  equal  to  76°.  What  ia  the 
value  of  the  third  angle  ? 


PLANE   GEOMETRY.  — BOOK  I 


47 


Ex.  59.  Show  by  each  of  the-foUowing  figures  that  the  sum  of  the  three 
angles  of  a  triangle  is  equal  to  two  right  angles,  assuming  that  the  construc- 
tion lines  are  drawn  as  they  appear  to  be  drawn. 


Proposition  XXIII 

116.    1.   Draw  an  isosceles  triangle.     How  do  the  angles  opposite  the 
aqxial  sides  compare  in  size  ?. 

2.   How  do  the  angles  of  an  equilateral  triangle  compare? 

Theorem,     In  an  isosceles  triangle  the  angles  opposite 
the  equal  sides  are  equal. 


Data :  Any  isosceles  triangle,  as  ABCy  in  which 
To  prove  angle  A  =  angle  B. 

Proof.     Draw  CD  bisecting  Z  C. 
Then,  in  A  ADC  and  BDC, 


data, 

and,  const., 
.-.  §  100, 
and,  §  108, 
Therefore,  etc. 


AC=BOy 

CD  is  common, 

Aadc  =  Abdc, 
Aa  =  /.b, 


Q.E.D. 


117.   Cor.     An  equilateral  triangle  is  also  equiangular. 


48 


PLANE    GEOMETRY.  — BOOK  L 


Proposition  XXIV 

118.  1.  Draw  a  triangle  such  that  two  of  its  angles  are  equal.  How 
do  the  sides  opposite  these  angles  compare  in  length  ?  What  kind  of  a 
triangle  is  it? 

2.   How  do  the  sides  of  an  equiangular  triangle  compare  in  length  ? 

Theorein,  If  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  the  equal  angles  are  equal  and  the  triangle  is  isos- 
celes. 

c 


Data:  Any  triangle,  as  ABC,  having 

angle  A  =  angle  B. 
To  prove  AC  =  BC,  and  triangle  ABC  isosceles. 


Proof.     Draw  CD  bisecting  Z  C. 
Then,  in  A  ADC  and  BDC, 


data, 

const., 

.-.  §  113, 

and,  since 

§  102, 

and,  §  108, 
Hence,  §  90, 
Therefore,  etc. 


Za  =  Zb, 
Zr^Zs; 

Zt  =  Zv; 

CD  is  common, 

AADC  =  ABDC, 

AC  =  BC. 

A  ABC  is  isosceles. 


Q.S.P, 


119.   Cor.     An  equiangular  triangle  is  also  equilateral 


Ex.  60.  If  equal  distances  from  the  vertices  of  an  equilateral  triangle  are 
laid  off  on  its  sides  in  the  same  order,  what  kind  of  a  triangle  do  the  lines 
joining  these  points  form  ? 

Ex.  61.  From  the  extremities  of  the  base  of  an  isosceles  triangle  perpen- 
diculars are  drawn  to  the  opposite  sides ;  the  points  where  the  perpendiculars 
meet  the  opposite  sides  are  joined  by  a  straight  line.  What  is  the  direction 
of  this  hne  with  reference  to  the  base  ? 

Ex.  62.  The  base  of  an  isosceles  triangle  is  6  inches  and  the  opposite 
angle  is  60°.  How  many  degrees  are  there  in  each  of  the  base  angles? 
What  is  the  length  of  each  of  the  other  two  sides? 


PLANE   GEOMETRY.  — BOOK  I 


49 


Proposition  XXV 

120.  1.  Draw  an  isosceles  triangle  and  a  line  bisecting  its  vertical 
angle.  How  does  this  line  divide  the  base  ?  What  kind  of  angles  does 
it  form  with  the  base  ? 

2.  Draw  a  line  perpendicular  to  the  base  of  an  isosceles  triangle  at  its 
middle  point.  How  does  it  divide  the  triangle  ?  How  does  it  divide  the 
vertical  angle  ? 

3.  Draw  a  line  from  the  vertex  of  an  isosceles  triangle  and  perpen- 
dicular to  the  base.  How  does  this  line  divide  the  base  of  the  triangle  ? 
How  does  it  divide  the  vertical  angle? 

Theorem.  The  bisector  of  the  vertical  angle  of  an  isos- 
celes triangle  is  perpendicular  to  the  hose  at  its  middle 
point. 


Data:  Any  isosceles  triangle,  as  ABC^  in 
which  AG  —  BGy  and  CD  bisects  the  angle  C. 

To  prove  CB  perpendicular  to  J  5  at  its  mid- 
dle point. 


Proof.    Data,  AC—BC^ 

AA  —  ^B\ 

Aadc=i^bdc, 

AD^BD'y 
D  is  the  middle  point  of  AB  j 

CD  ±  AB, 


and,  §  116, 
.-.  §  102, 
and,  §  108, 
that  is, 
also, 
hence,  §  26, 

Therefore,  etc.  *  Q.B.D. 

121.  Cor.  I.  A  perpendicular  which  bisects  the  base  of  an  isos- 
celes triangle  bisects  the  vertical  angle, 

122.  Cor.  II.  A  line  perpendicular  to  the  base  of  an  isosceles 
triangle  and  passing  through  the  vertex  bisects  both  the  base  and  the 
vertical  angle  of  the  triangle. 

Ex.  63.  How  do  the  lines  joining  the  extremities  of  the  bases  of  two 
opposite,  or  vertical,  isosceles  triangles  compare  in  length  ? 

MILNE^S  GEOM. 4 


50 


PLANE   GEOMETRY.  — BOOK  /. 


Proposition  XXVI 

123.  Draw  two  right  triangles  such  that  the  hypotenuse  and  a  side  of 
one  shall  be  equal  to  the  corresponding  parts  of  the  other.  How  do  the 
triangles  compare  ? 

Theorem,  Two  right  triangles  are  equal,  if  the  hypote- 
nuse and  a  side  of  one  are  equal  to  the  hypotenuse  and  a 
side  of  the  other,  each  to  each. 


Data:  Any  two  right  triangles,  as  ABC  and  bef^  in  which  the 
hypotenuse  AC  =  the  hypotenuse  BF,  BC  =  EF,  and  angles  r  and 
5  are  the  right  angles. 

To  prove  triangles  ABC  and  BEF  equal. 

Proof.  Place  A  BEF  in  the  position  BBC  so  that  the  equal 
sides  EF  and  BC  coincide,  and  the  vertex  B  falls  opposite  A. 

Data,  Z  r  and  Z  s  are  rt.  A ; 

then,  Z  r  +  Z  s  =  2  rt.  ^, 

and,  §  58,  AB  and  BB  form  one  straight  lina 

Data,  AC  =  BC', 

.'.  §90,  A  ^DC  is  isosceles. 

Hence,  in  ^  ABC  and  BBC, 

AG^BCy 

Z.A=ABy 


§116, 
and,  §  113, 
.-.  §  102, 

That  is. 

Therefore,  etc. 


Aabc  =  Abbc. 
aabc  =  abef. 


Q.E.D. 


Ex.  64.   The  median  line  from  the  vertex  to  the  base  of  a  certain  triangle 
is  equal  to  one  half  the  base.     What  kind  of  an  angle  is  the  vertical  angle  ? 


PLANE  GEOMETRY.— BOOK  L  51 

Proposition  XXVII 

124.  1.   Draw  any  triangle.    How  does  any  side  compare  in  length 

with  the  sum  of  the  other  two  sides? 

2.   How  does  the  sum  of  any  two  sides  compare  with  the  third  side? 

Theorem,  Any  side  of  a  triangle  is  less  than  the  sum  of 
the  other  two  sides. 

Data :  Any  triangle,  as  ABC,  and  any  side, 

as  AC. 

To  prove  AC  less  than  AB  4-  BC. 

A  B 

Proof.  By  Ax.  10,  the  straight  line  AC,  which  is  a  side  of  the 
triangle,  is  the  shortest  distance  between  the  points  A  and  C. 

Hence,  AC  is  less  than  the  broken  line  ABC  which  joins  the 
points  A  and  C. 

That  is,  AC  is  less  than  AB  -f-  BC. 

Therefore,  etc.        «  q.e.d. 

125.  Cor.  The  sum  of  any  two  sides  of  a  triangle  is  greater  than 
the  third  side. 

Ex.  65.  May  a  triangle  he  formed  with  lines  4,  2,  and  3  inches  long  ? 
With  lines  6,  1,  and  2  inches  long  ?    5,  2,  and  3  inches  long? 

Ex.  66.  If  a  line  is  drawn  joining  the  middle  points  of  the  equal  sides  of 
an  isosceles  triangle,  what  kind  of  a  triangle  is  formed  ? 

Ex.  67.  If  the  bisectors  of  the  base  angles  of  an  isosceles  triangle  are 
produced  to  the  opposite  sides,  how  do  they  compare  in  length  ? 

Ex.  68.  The  sum  of  the  two  angles  at  the  base  of  an  isosceles  triangle  is 
64°.    What  is  the  value  of  each  angle  of  the  triangle  ? 

Ex.  69.  If  the  straight  line  which  joins  the  vertex  of  a  triangle  with  the 
middle  point  of  the  base  is  perpendicular  to  the  base,  what  liind  of  a  tri- 
angle is  it  ? 

Ex.  70.  The  perpendicular  distance  between  two  parallel  lines  is  20 
inches,  and  a  line  is  drawn  across  the  parallels  making  an  angle  of  45°  with 
the  perpendicular  at  its  upper  extremity.  What  distance  does  this  line  cut 
off  from  the  foot  of  the  perpendicular  ? 

Ex.  71.  If  from  a  point  within  a  right  angle  perpendiculars  are  drawn  to 
the  sides  containing  the  right  angle  and  each  perpendicular  is  produced  its 
own  length,  what  kind  of  a  line  will  join  the  extremities  of  the  produced 
lines  and  the  vertex  of  the  right  angle  ? 


62  PLANE   GEOMETRY.  — BOOK  L 

Proposition  XXVIII 

12^6.  Draw  a  scalene  triangle.  Where  is  the  smaller  angle  situated 
with  reference  to  the  shorter  side  ?  Where  is  the  greater  angle  situated 
with  reference  to  the  greater  side  ? 

Theorem,  If  two  sides  of  a  triangle  are  unequal,  the 
angles  opposite  are  unequal,  and  the  greater  angle  is  op- 
posite the  greater  side. 

Data:  Any  triangle,  as  ABC,  in 
which  AB  is  greater  than  BC. 

To  prove  angle  ACB,  opposite  AB,  is 
greater  than  angle  A,  opposite  BC. 


Proof.     On  BA  take  BF  equal  to  BC,  and  draw  CF. 
Ax.  8,  /.ACB  is  greater  than  Z  BCF\ 

but,  §  116,  Z  BCF=Z.  BFC; 

Z  ACB  is  greater  than  Z  BFC. 
But  since,  §  115,       Z  BFC  =  Za  +  Zacf, 

Z  BFC  is  greater  than  Z  A. 
Then,  Zacb,  the  angle  opposite  AB,  is  greater  than  Z  A,  the 
angle  opposite  BC. 

In  like  manner,  ii  AB  is  greater  than  AC,  Z  ACB  may  be  proved 
greater  than  Z  B, 

Therefore,  etc.  q.e.d. 

Prove  Z  ACB  greater  than  Z  B  when  AB  is  greater  than  AC. 

Ex.  72.    How  do  the  angles  of  a  scalene  triangle  compare  ? 

Ex.  73.  What  is  the  value  of  the  angle  formed  by  the  bisectors  of  the 
acute  angles  of  a  right  triangle  ? 

Ex.  74.    How  many  degrees  are  there  in  an  angle  of  an  equilateral  triangle  ? 

Ex.  75.  How  many  degrees  are  there  in  each  of  the  equal  angles  of  an 
isosceles  triangle,  the  angle  at  the  vertex  being  35°  50'  ? 

Ex.  76.  In  an  isosceles  triangle  one  base  angle  is  35°.  What  is  the  value 
of  the  vertical  angle  ? 

Ex.  77.  AD  is  the  bisector  of  a  base  angle  of  the  isosceles  triangle  ABC, 
the  bisector  meeting  the  side  BC  in  7> ;  the  vertical  angle  C  is  28°.  How 
many  degrees  are  there  in  angle  ADC  ? 


PLANE   GEOMETRY.— BOOK  I  53 

Proposition  XXIX 

127.  1.  Draw  any  triangle.  Where  is  the  shorter  side  situated  with 
reference  to  the  smaller  angle  ?  Where  is  the  greater  side  situated  with 
reference  to  the  greater  angle  ? 

2.   Which  side  of  a  right  triangle  is  the  greatest  ? 

Theorem,  If  two  angles  of  a  triangle  are  unequal,  the 
sides  opposite  are  unequal,  and  the  greater  side  is  opposite 
the  greater  angle.  (Converse  of  Prop.  XXVIII) 

Data:    Any   triangle,   as  ABC,  in  which  ^ 

angle  ABC  is  greater  than  angle  A. 

To  prove  AC^  opposite  angle  ABC,  greater 
than  BC,  opposite  angle  A. 

Proof.     Draw  BD  so  that  Z  ABB  =  Z  BAD. 
Then,  §  118,  AD  =  BD. 

In  A  BCD,  %  125,  BD-\-DC>BG. 

Substituting  AD  for  its  equal  BD, 

AD-\-DC>BC, 
or  AC>BC. 

That  is,  AC,  opposite  angle  ABC,  is  greater  than  BC,  opposite 
angle  A. 

In  like  manner,  if  angle  ABC  is  greater  than  angle  C,  AC  may- 
be proved  greater  than  AB. 

Therefore,  etc.  q.e.d. 

Prove  that  ^c  is  greater  than  AB  when  Zabc  is  greater  than 
Zc. 

128.  Cor.    The  hypotenuse  is  the  greatest  side  of  a  right  triangle. 

Ex.  78.  The  angles  A,  B,  and  C  of  the  triangle  ABC  &re  40°,  60°,  and 
80°  respectively.  How  do  ^O  and  AB  compare  in  length  ?  AB  and  BC? 
^C  and  50? 

Ex.  79.  CD  bisects  the  base  of  an  isosceles  triangle  ABC,  Si  base  angle  of 
which  is  55°.  How  many  degrees  are  there  in  angle  ADC  ?  In  angle  CDB  ? 
In  angle  ACD?    In  angle  DCB? 


6i 


PLANE   GEOMETRY.  — BOOK  1. 


Proposition  XXX 

129.  Construct  two  triangles  having  two  sides  of  one  equal  to  two 
sides  of  the  other,  and  the  angles  included  between  the  sides  unequal. 
How  do  the  third  sides  compare  in  length?  Which  triangle  has  the 
greater  third  side? 

Theorem.  If  two  sides  of  one  triangle  are  equal  to  two 
sides  of  another,  each  to  each,  and  the  included  angles 
are  unequal,  the  remaining  sides  are  unequal,  and  the 
greater  side  is  in  the  triangle  which  has  the  greater 
included  angle. 


Data:  Any  two  triangles,  as  ABC  and  DBF,  in  which  AC  —  DF^ 
BC  =  EF,  and  angle  ACB  is  greater  than  angle  F. 
To  prove  AB  greater  than  DE. 

Proof.     Of  the  two  sides  DF  and  EF,  suppose  that  EF  is  the 
side  which  is  not  greater.     Place  Abef  in  the  position  BBC  so 
that  the  equal  sides,  EF  and  BC,  coincide. 
Draw  CH  bisecting  Z  ACB  and  draw  BH. 
In  A  AHC  and  BHC,  AC  =  BC, 

CH  is  common, 

Zt  =  Zs; 
A  AHC  =  A  BHC, 

AH  =  BH. 
BH  +  HB  >  BB. 


and,  const., 
.-.  §  100, 
and,  §  108, 

In  A  BHB,  §  125, 


Substituting  AH  for  its  equal  BH, 

AH+  HB>BB\ 
that  is,  AB  >  BB  or  BE. 

Therefore,  etc. 


Q.E.D. 


PLANE   GEOMETRY.  — BOOK  I.  55 

Proposition  XXXI 

130.  Construct  two  triangles  that  have  two  sides  of  one  equal  respec« 
tively  to  two  sides  of  the  other,  but  the  third  sides  unequal.  How  do 
the  angles  opposite  the  third  sides  compare  in  size? 

Theorem,  If  two  sides  of  one  triangle  are  equal  to  two 
sides  of  another,  each  to  each,  and  the  third  sides  are  un- 
equal, the  angles  opposite  the  third  sides  are  unequal,  and 
the  greater  angle  is  in  the  triangle  which  has  the  greater 
third  side.  (Converse  of  Prop.  XXX.) 


Data:  Any  two  triangles,  as  ABC  and  DEF,  in  which  AC  =  DF, 
BC  =  EF,  and  AB  is  greater  than  DE. 

To  prove  angle  c  greater  than  angle  F. 

Proof.     Since  AC  =  DF, 

and  BC  =  EF, 

if  Zc  =  Zf, 

then,  §  100,  A  ABC  =  A DEFy 

and,  §  108,  AB  =  DE, 

which  is  contrary  to  data. 

If  Z  (7  is  less  than  Z  F, 

then,  Z  i^  is  greater  than  Z  C, 

and,  §  129,  DE>  AB, 

which  is  also  contrary  to  data. 

Therefore,  both  hypotheses,  namely,  that  Z.C  —  /.F,  and  that 
Z  c  is  less  than  Z  F,  are  untenable. 

Consequently,         Z  C  is  greater  than  Z  F. 

Therefore,  etc.  q.e.d. 

Ex.  80.  If  one  angle  of  a  triangle  is  equal  to  the  sum  of  the  other  two, 
what  is  the  value  of  that  angle  ?     What  kind  of  a  triangle  is  the  triangle  ? 

Ex,  81.  AT)  is  perpendicular  to  50,  one  of  the  equal  sides  of  the  isosceles 
triangle  ABC  whose  vertical  angle  is  30°.  How  many  degrees  are  there  in 
each  of  the  angles  CAB,  DAB,  and  ABC^ 


66 


PLANE   GEOMETRY.  — BOOK  L 


Proposition  XXXII 

131.  Choose  some  point  within  any  triangle  and  from  it  draw  lines 
to  the  extremities  of  one  side.  How  does  the  sum  of  these  lines  com- 
pare with  the  sum  of  the  other  two  sides  of  the  triangle? 

Theorem.,  The  sum  of  two  lines  drawn  from  a  point 
within  a  triangle  to  the  extremities  of  one  side  is  less  than 
the  sum  of  the  other  two  sides. 

Data:  Any  triangle,  as  ABC-,  any  point 
within  it,  as  D  ;  and  the  two  lines,  AD  and 
BB,  drawn  from  D  to  the  extremities  of 
AB. 

To  prove  AD  -\-  BD  less  than  AC  -\-BC. 

Proof.     Produce  AD  to  meet  BC  in  E. 

In  A  AEC,  §  124,  AE<AC+  CE. 

Adding  BE  to  both  members  of  this  inequality, 
Ax.  4,  AE  -\-BE<AC-\-CE  -\-  BE, 

or  AE  -\-BE  <AC  -^BC. 

In  A  DBEj  BD  <DE  -\-  BE. 

Adding  AD  to  both  members  of  this  inequality, 
Ax.  4,  AD  -^  BD  <  AD  -\-  DE  ■\-  BE, 

or  AD-^BD<AE-\-  BE. 

It  has  been  shown  that 

ae-\-'be<ac  +  bc\ 
hence,  AD  +  BD  <  AC -[■  BC. 

Therefore,  etc.  q.e.d. 


Why? 


Proposition  XXXIII 

132.  1.  Draw  a  straight  line  and  a  perpendicular  to  it;  select  a  point 
in  the  perpendicular,  and  from  that  point  draw  two  oblique  lines  meet- 
ing the  given  line  at  equal  distances  from  the  foot  of  the  perpendicular. 
How  do  the  oblique  lines  compare  in  length  ? 

2.  Draw  oblique  lines  from  that  point  to  points  unequally  distant  from 
the  foot  of  the  perpendicular.  How  do  they  compare  in  length  ?  Which 
is  the  greater  ? 


PLANE   GEOMETRY.  — BOOK  I.  57 

3.  Draw  two  unequal  lines  from  that  point  to  the  given  line.  Which 
one  meets  the  line  at  the  greater  distance  from  the  foot  of  the  perpen- 
dicular ? 

4.  How  many  equal  straight  lines  can  be  drawn  from  a  point  to  a 
straight  line? 

Theorem.  If  from  a  point  in  a  perpendicular  to  a  given 
straight  line,  oblique  lines  are  drawn  to  the  given  line, 

1.  The  ohlique  lines  which  meet  the  given  line  at  equal 
distances  from  the  foot  of  the  perpendicular  are  equal. 

2.  Of  oblique  lines  which  meet  the  given  line  at  unequal 
distances  from  the  foot  of  the  perpendicular  the  more 
remote  is  the  greater. 

Data :  Any  straight  line,  as  AB ; 
any  perpendicular  to  AB,  as  PB ; 
and  any  point  in  PB,  as  C,  from 

which   oblique  lines,  as    CE,   CF,  y^ 

and  (JG,  are  drawn  meeting  AB  so  // 
that  BE  =  BF,  and  /)  G  is  greater  ^     gx  '\e — 

than  BE.  ^^<\       I 

To  prove  1.  CE  =  CF.  '  <\  j 

2.  CG  greater  than  CE.  ''"-^H 

Proof.     1.   Data,  CB  X  EF  at  its  middle  point. 

Then,  §  103,  CE  =  CF. 

2.   Produce  CB  to  H,  making  BH  =  CB ;  draw  EH  and  GH. 

Then,  data  and  const.,  AB  J.  CH  Sit  its  middle  point. 

.-.  §  103,  EH=  CE,  and  GH=  CG\ 

hence.  Ax.  2,        CE  +  EH  =  2  CE,  and  CG  -\-  GH  =  2  CG. 

But,  §  131,  CE  -\-EH<CG-\-GH', 

2CE  <2  CG,  ov  CE  <CG\ 
that  is,  CG  >  CE. 

Therefore,  etc.  q.e.d. 

133.  Cor.  Only  tivo  equal  straight  lines  can  he  drawn  from  a 
point  to  a  straight  line;  and  of  two  unequal  lines  the  greater  cuts  of 
the  greater  distance  from  the  foot  of  a  perpendicular  drawn  to  the 
line  from  the  given  point. 


68  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  XXXIV 

134.   Bisect  any  angle ;  from  any  point  in  the  biseclor  draw  lines  pe? 
pendicular  to  the  sides  of  the  angle.     How  do  the  perpendiculars  com- 
pare  in  length  ?    How  do  the  distances  of  the  point  from  the  sides  ot 
the  angle  compare  ?. 

Theorem,  Every  point  in  the  bisector  of  an  angle  is 
equidistant  froin  the  sides  of  the  angle-, 

/A 

Data:  Any  angle,  as  ABC,  and  any  point  in  e/  ^ 

its  bisector  BB,  as  .F.  /       :^<p 

To  prove  F  equidistant  from  AB  and  CB.        >^^^^^      ■ 

B  Q      c 

Proof.  Draw  the  perpendiculars  fe  and  FG  representing  the 
distances  of  the  point  F  from  AB  and  CB  respectively. 

§  26,  Z  r  and  Z  s  are  rt.  A. 

Then,  in  the  rt.  A  BFE  and  BFG, 

BF  is  common, 
and,  data,  /.t  =  /.v', 

.-.  §  114,  A  BFE  =  A  BFQ, 

and,  §  108,  fe  =  fG', 

that  is,  F  is  equidistant  from  AB  and  CB. 

Therefore,  etc.  q.e.d. 

Ex.  82.  The  perpendicular  let  fall  from  the  vertex  to  the  base  of  a  tri- 
angle divides  the  vertical  angle  into  two  angles.  How  does  the  difference  of 
these  angles  compare  with  the  difference  of  the  base  angles  of  the  triangle  ? 

Ex.  83.  ^^a  is  a  triangle.  Angle  A  =  60°,  angle  B  =  40°.  The  bisector 
of  angle  A  is  produced  until  it  cuts  the  side  BC.  How  many  degrees  are 
there  in  each  angle  thus  formed  ? 

Ex.  84.  A  perpendicular  is  let  fall  from  one  end  of  the  base  of  an  isos- 
celes triangle  upon  the  opposite  side.  How  does  the  angle  formed  by  the 
perpendicular  and  the  base  compare  with  the  vertical  angle  ? 

Ex.  85.  If  an  angle  of  a  triangle  is  equal  to  half  the  sum  of  the  other  two, 
what  is  tlie  value  of  that  angle  ? 

Ex.  86.  How  does  the  sum  of  the  lines  from  a  point  within  a  triangle  to 
the  vertices  of  the  triangle  compare  with  the  sum  of  the  sides  of  the  tri- 
angle ?    With  half  that  sum  ? 


PLANE   GEOMETRY.-^BOOK  1,  69 

Proposition  XXXV 

135.  Within  an  angle  select  any  number  of  points  that  are  each  equi- 
distant from  its  sides.  Will  the  lines  joining  these  points  form  a  straight 
line?     How  will  it  divide  the  angle? 

Theorem,  Every  point  within  an  angle  and  equidistant 
from  its.  sides  lies  in  the  bisector  of  the  angle.  (Converse 
of  Prop.  XXXIV.) 

/A 

Data:  Any  angle,  as  ABC,  and  any  point  / 

within  the  angle  equidistant  from  AB  and  GB,  r^^^--  ^^^ 

as  i^.  /  ^^^ 

To  prove  F  is  in  the  bisector  of  the  angle       A>  si 

ABC.  *  °      " 

Proof.  Through  the  point  F  draw  BB\  also  draw  the  perpen- 
diculars FE  and  FG  representing  the  distances  of  the  point  F 
from  AB  and  GB  respectively. 

Then,  §  26,  Z  r  and  Z  s  are  rt.  A 

In  the  rt.  A  BFF  and  BGF, 

BF  is  common, 
and,  data,  FE  =  FG  ; 

.-.  §  123,  A  BEF  =  A  BGFf 

and,  §  108,  Zt  =  Zv', 

that  is,  ^D  is  the  bisector  of  Z  ABC. 

Hence,  F  is  in  the  bisector  of  Z  ABC. 

Therefore,  etc.  q.e.d. 

Ex.  87.  ABC  is  an  isosceles  triangle  having  a  vertical  angle  of  30°. 
From  each  extremity  of  the  base  perpendiculars  are  drawn  to  the  opposite 
sides.     What  angles  are  formed  at  the  intersection  of  these  perpendiculars  ? 

Ex.  88.  The  exterior  angle  at  the  vertex  of  an  isosceles  triangle  is  110°. 
How  many  degrees  are  there  in  each  angle  of  the  triangle  ? 

Ex.89.  The  exterior  angle  at  the  base  of  an  isosceles  triangle  is  110°. 
How  many  degrees  aa-e  there  in  each  angle  of  the  triangle  ? 

Ex.  90.  The  angle  C  at  the  vertex  of  the  isosceles  triangle  ABC  is  one 
fourth  of  the  exterior  angle  at  C.  How  many  degrees  are  there  in  angle  A  f 
In  the  exterior  angle  at  B? 

Ex.  91.  How  does  the  angle  formed  by  the  bisectors  of  the  base  angles  of 
an  isosceles  triangle  compare  with  an  exterior  angle  at  the  base  ? 


60 


PLANE   GEOMETRY.  — BOOK  L 


QUADRILATERALS 

136.   A  portion  of  a  plane  bounded  by  four  straight  lines  is 
called  a  Quadrilateral. 


137.  A  quadrilateral  which  has  no  two  sides 
parallel  is  called  a  Trapezium. 

138.  A  quadrilateral  which  has  only  two  sides 
parallel  is  called  a  Trapezoid. 

The  parallel  sides  of  a  trapezoid  are  called  its 
bases. 

139.  A  trapezoid  whose  non-parallel  sides  are 
equal  is  called  an  Isosceles  Trapezoid. 

140.  A  quadrilateral  whose  opposite  sides  are 
parallel  is  called  a  Parallelogram. 

141.  A  parallelogram  whose  angles  are  right 
angles  is  called  a  Rectangle. 

142.  A  parallelogram  whose  angles  are  oblique 
angles  is  called  a  Rhomboid. 

143.  An    equilateral    rectangle    is    called    a 
Square. 

144.  An    equilateral    rhomboid    is    called  a 
Rhombus. 


145.  The  straight  lines  which  join  the  vertices  of  the  opposite 
angles  of  a  quadrilateral  are  called  Diagonals. 

146.  The  side  upon  which  a  figure  is  assumed  to  stand  is  called 
the  Base. 

The  side  upon  which  a  trapezoid  or  a  parallelogram  is  assumed 
to  stand  is  called  its  lower  hose,  and  the  side  opposite  is  called 
its  upper  base. 

147.  The  perpendicular  distance  between  the  bases  of  a  trape- 
zoid or  of  a  parallelogram  is  called  its  Altitude. 


PLANE   GEOMETRY.  — BOOK  L  61 

Proposition  XXXVI 

148.  1.  Draw  a  quadrilateral  whose  opposite  sides  are  equal.  What 
kind  of  a  quadrilateral  is  it  ? 

2.  How  do  the  opposite  angles  of  a  parallelogram  compare  in  size? 

Theorem,  If  the  opposite  sides  of  a  quadrilateral  are 
equal,  the  figure  is  a  parallelogram. 

D 

Data:   Any  quadrilateral,  as  ABCD,  in 
which  AB  =  DC  and  AD  =  EC. 

To  prove  ABCD  di.  parallelogram. 

Proof.     Draw  AC. 

Then,  in  the  A  ABC  and  ADC,  ^ 

data,  AB  =DC,  BC  =  AD, 

and  ^(7  is  common; 

.-.§107,  AABC  =  l\ADC, 

§108,  Zr=Z^,  andZs  =  Zv; 

.-.  §  75,  AB  II  DC,  and  AD  II  BC. 

Hence,  §  140,         ABCD  is  a  parallelogram. 

Therefore,  etc.  q.e.d. 

149.  Cor.     The  opposite  angles  of  a  parallelogram  are  equal. 

Z.r=^Z.t  and  Z  <;  =  Z  s ; 

Ex.  92.  If  lines  are  drawn  joining  in  succession  the  middle  points  of  the 
sides  of  a  square,  what  figure  will  be  formed  ? 

Ex.  93.  To  how  many  right  angles  is  the  sum  of  the  angles  of  a  parallelo- 
gram equal  ?  To  what  is  the  sum  of  any  two  angles  of  a  parallelogram, 
which  'are  not  opposite,  equal  ? 

Ex.  94.  If  medians  are  drawn  from  two  vertices  of  a  triangle  and  each  is 
produced  its  own  length,  what  kind  of  a  line  will  join  the  extremities  of  the 
produced  medians  and  the  other  vertex  of  the  triangle  ? 


62  PLANE   GEOMETRY.— BOOK  /. 

Proposition  XXXVII 

150.  1.  Draw  a  quadrilateral  having  two  of  its  sides  equal  and  pa^ 
allel  to  each  other.     What  kind  of  a  quadrilateral  is  it  ? 

2.  Draw  two  parallel  lines  and  two  parallel  transversals.  How  do  the 
segments  of  the  transversals  between  the  parallel  lines  compare  in  length? 

Theorem,  If  two  sides  of  a  quadrilateral  are  equal  and 
parallel,  the  figure  is  a  parallelogram. 

Data:  Any  quadrilateral,  as  ABCD^ 
in  which  two  of  the  sides,  <is  AB  and 
DC,  are  equal  and  parallel.     . 

To  prove  ABCD  Si  parallelogram. 

Proof.     Draw  AC. 

In  the  A  ABC  and  ADC, 
data,  AB  —  DC, 

^C  is  common, 
and,  §  73,  Zr  =  /.t', 

.-.§100,  AABC  =  AADC, 

and  Zs  =  Zv. 

Hence,  §  75,  BCWAD, 

and  since,  data,  AB  \\DC, 

§  140,  ABCD  is  a  parallelogram. 

Therefore,  etc.  q.e.d. 

151.  Cor.  Parallel  lines  intercepted  betiveen  parallel  lines  are 
equal,  and  parallel  lines  are  everywhere  equally  distant. 

Ex.  95.  If  the  sides  of  a  parallelogram  are  bisected  and  these  middle 
points  joined  in  succession,  what  figure  is  formed  by  the  connecting  lines  ? 

Ex.  96.  If  the  four  interior  angles  formed  by  a  transversal  crossing  two 
parallel  lines  are  bisected  and  the  bisectors  produced  until  they  meet,  what 
figure  will  be  formed  ? 

Ex.  97.  If  a  line  is  drawn  through  the  vertices  of  two  isosceles  triangles 
on  the  same  base,  how  does  it  divide  the  base  ? 

Ex.  98.  If  two  equal  straight  lines  are  drawn  from  a  point  to  a  line,  how 
do  the  angles  formed  with  the  given  line  compare  ? 

Ex.  99.  If  lines  are  drawn  from  the  vertex  of  an  isosceles  triangle  to 
points  in  the  base  equally  distant  from  its  extremities,  how  do  they  compare 
in  length  ? 


PLAkE   GEOMETRY. —BOOK  L 


63 


Proposition  XXXVIII 

152.  1.  Draw  a  parallelogram  and  either  diagonal.     How  do  the  tri- 
angles thus  formed  compare  in  size  ? 

2.   How  do  the  opposite  sides  of  the  parallelogram  compare  in  length  ? 

Theorem.    The  diagonal  of  a  parallelogram  divides  the 
figure  into  two  equal  triangles. 

Data:  Any  parallelogram,  as  ABCD^ 
and  one  of  its  diagonals,  as  ^C. 

To   prove   triangles   ABC   and   ADC 
equal. 

Proof.     To  be  given  by  the  student. 

153.  Cor.     Tlie  opposite  sides  of  a  parallelogram  are  equal. 


Proposition  XXXIX 

154.  Draw  a  parallelogram  and  its  diagonals.  How  do  the  segments 
of  each  diagonal  compare  in  length? 

Theoretn.  The  diagonals  of  a  parallelogram  bisect  each 
other. 

•Data:  Any  parallelogram,  as  A  BCD, 
and  its  diagonals,  AO  and  BB,  inter- 
secting at  E. 

To  prove  AE  =  CE  and  BE  ==  BE. 

Proof.     In  the  A  ABE  and  CDE, 
§  153,  AB  =  DC, 

§73,  Z.r=/.t, 

and  Z  s  =  Z  v ; 

.-.  §  102,  A  ABE  =  A  CDE, 

§  108,  AE  =  CEj  and  BE  =  DE. 

Therefore,  etc.  q.e.I). 

Ex.  100.  If  the  diagonals  of  a  quadrilateral  are  equal  and  bisect  each 
other,  what  kind  of  a  figure  is  the  quadrilateral  ? 

Ex.  101.  From  a  figure  representing  a  parallelogram  and  its  diagonals, 
select  four  pairs  of  equal  triangles. 


64  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  XL 

165.  1.  Draw  two  parallelograms  such  that  two  sides  of  one,  and 
the  angle  between  them,  shall  be  equal  to  the  corresponding  parts  of  the 
other.     How  do  the  parallelograms  compare? 

2.  How  do  two  rectangles  compare,  if  the  base  and  altitude  of  one  are 
equal  to  the  corresponding  parts  of  the  other? 

Theorem,  Two  parallelograms  are  equal,  if  two  sides 
and  the  included  angle  of  one  are  equal  to  two  sides  and 
the  included  angle  of  the  other,  each  to  each. 


Data:  Any  two  parallelograms,  as  abqb  and  efoh,  in  which 
AB  =  EF,  AD  =  EH,  and  angle  A  =  angle  E. 

To  prove  parallelograms  ABCD  and  EFGH  equal. 

Proof.  Place  EJ EFGH  upon  EJ ABCD  so  that  EF  coincides 
with  its  equal  AB  and  Z  E  with  its  equal  Z  A. 

Then,  EH  coincides  with,  its  equal  AD, 

§  70,  HG  takes  the  direction  of  DC, 

and  O  falls  upon  DC,  or  upon  DC  produced. 

Also,  FG  takes  the  direction  of  BC, 

and  O  falls  upon  BC,  or  upon  BC  produced. 

Since  G  falls  upon  both  DC  and  BC,  it  must  fall  upon  their 
point  of  intersection  c,  which  is  the  only  point  common  to  DC 
and  BC. 

Hence,  §  36,  O  ABCD  =  EJefgh. 

Therefore,  etc.  q.e.d. 

156.  Cor.  Two  rectangles  are  equal,  if  the  base  and  altitude  of 
one  are  equal  to  the  base  and  altitude  of  the  other,  each  to  each. 

Ex.  102.  From  any  point  in  the  base  of  an  isosceles  triangle  lines  are 
drawn  parallel  to  the  equal  sides  and  produced  until  they  meet  the  sides  of 
the  triangle.  How  does  the  sum  of  these  two  lines  compare  with  one  of  the 
equal  sides  of  the  triangle  ? 


PLANE   GEOMETRY.— 'BOOK  I. 


65 


Proposition  XLI 

157.  Draw  three  or  more  parallel  lines  intercepting  equal  parts  on  a 
transversal;  draw  any  other  transversal.  How  do  the  parts  which  the 
parallels  intercept  on  the  second  transversal  compare  in  length  ? 

Theorefu,  If  three  or  more  parallel  lines  intercept  equal 
parts  on  any  transversal,  they  intercept  equal  parts  on 
every  transversal. 


Data :  Any  parallel  lines,  as  AH,  CM, 
EK,  and  GP,  intercepting  the  equal 
parts  AC,  CE,  and  ^G^  on  the  transver- 
sal A  G,  and  the  parts  HM,  MK,  and  KP 
on  any  other  transversal,  as  HP. 

To  prove  HM  =  MK  —  KP. 


Proof.     Draw  AB,  CD,  and  EF  each  parallel  to  HP. 
Then,  §  140,  ABMH,  CDKM,  and  EFPK  are  parallelograms, 
and,  §  153,       HM  =  AB,  MK  =  CD,  and  KP  =  EF. 

Now,  in  A  ABC,  CDE,  and  EFG, 
§  80,  AB  II  CD  WEF; 

hence,  §  76,  Zr  =  Zs  =Zt, 

Z.v  =  Zw  =  Z.X, 
AC=CE  =  EG; 
A  ABC  =  A  CDE  =  A  EFG, 
AB  =  CD  =  EF. 


and,  data, 
.-.  §  102, 
and 

But  since     HM  =  AB,  MK  =  CD,  and  KP  =  EF, 
Ax.  1,  HM  —  MK  =  KP. 

Therefore,  etc. 


Q.E.D. 


Ex.  103.  In  the  triangle  ABC  angle  A  is  double  angle  B  and  the  ex- 
terior angle  at  C  is  105°.  How  many  degrees  are  there  in  angles  A  and  B 
respectively  ? 

Ex.  104.  If  one  angle  of  a  parallelogram  is  a  right  angle,  what  is  the  value 
of  each  of  the  other  angles  ? 

Ex.  105.  One  angle  of  a  parallelogram  is  three  times  its  supplement 
What  is  the  value  of  each  angle  of  the  parallelogram  ? 


MTT.-VK'S  aHOTW- 


66 


PLANE   GEOMETRY.  — BOOK  I 


Proposition  XLII 

158.  1.  Draw  a  triangle  and  a  line  parallel  to  the  base,  bisecting  one 
of  the  sides.  How  does  it  divide  the  other  side?  How  does  the  part  of 
this  line  intercepted  by  the  sides  of  the  triangle  compare  in  length  with 
the  base  of  the  triangle  ? 

2.  Draw  a  triangle  and  a  line  connecting  the  middle  points  of  two  of 
its  sides.  What  is  the  direction  of  this  line  with  reference  to  the  third 
side  of  the  triangle? 

Theorem,  If  a  straight  line  drawn  parallel  to  the  base 
of  a  triangle  hisects  one  of  its  sides,  it  bisects  the  other  side, 
and  is  equal  to  one  half  of  the  base. 

Data:    Any    triangle,   as    ABC,   and    a 
straight  line   DE  drawn   parallel  to  AB 
bisecting  ^(7  at  Z).  d^ /^ 

To  prove  1.       BE  =^  EC. 

2.         DE  =  i  AB, 

Proof.     1.  Draw  FD  II  BC. 
Then,  in  A  DEC  and  AFD, 


data, 

DC  =  AD, 

§76, 

Zdce  =  Zadf, 

and 

Zcde  =  Zdaf', 

.-.§102, 

A  DEC  =  A  AFD, 

and 

EC  =  FD. 

But,  §  151, 

BE  =  FD; 

.-.  Ax.  1, 

BE  =  EC. 

2.  §108, 

AF  =  DE, 

and,  §  151, 

FB  =DE; 

.*. 

AF  =  FB=^AB; 

but 

FB  =DE; 

hence, 

DE  =  \  AB, 

Therefore,  etc. 

Why? 


Q.E.D. 


159.   Cor.     TJie  line  joining  the  middle  points  of  two  sides  of  a 
triangle  is  parallel  to  the  third  side. 


PLANE   GEOMETRY.  — BOOK  I.  67 

For  if  the  line  is  not  parallel  to  the  third  side,  suppose  a  line 
drawn  through  2),  the  middle  point  of  ACy  parallel  to  AB.  By 
§  158, it  will  pass  through  E,  the  middle  point  of  BCj  and  we  shall 
have  two  straight  lines  drawn  between  the  same  two  points,  which 
by  Ax.  12  is  impossible.  Consequently,  the  line  joining  the  mid- 
dle points  ot  two  sides  of  a  triangle  is  parallel  to  the  third  side. 

Proposition  XLIII 

160.  Draw  a  trapezoid  and  a  line  connecting  the  middle  points  of  the 
non-parallel  sides.  What  is  the  direction  of  this  line  with  reference  to 
iihe  bases  of  the  trapezoid?  How  does  it  compare  in  length  with  the  sum 
of  the  bases  ? 

Theorem,  The  line  which  joins  the  middle  points  of  the 
non-parallel  sides  of  a  trapezoid  is  parallel  to  the  bases 
and  is  equal  to  one  half  their  sum. 

Data:  Any  trapezoid,  as  A  BOB,  and  R ^r^c 

the  line  EF  joining  the  middle  points  /              ^.'"'' 

of  the  non-parallel  sides  AB  and  B  C.  eA -^^ 

To  prove  EF  parallel  to  AB  and  BC  /.'-'' 

and  equal  to  one  half  AB  +  BC.  ^ ^ 

Proof.  Draw  AC  intersecting  EF  at  K,  and  from  H,  the  middle 
point  of  AC,  draw  HE  and  HF. 

Data,  AE  =  EBy 

and,  const.,  AH  =  HC ; 

.-.  §  159,  HE  II  BC, 

and,  §  158,  '  HE  —  ^BC, 

In  like  manner,      HF  II  AB  and  HF  =  ^AB, 

Then,  §  80,  hfWbc; 

but  HEWbC; 

:,  §  70,      EHF  is  a  straight  line  parallel  to  AB  and  DC, 

But,  data,  EKF  is  a  straight  line, 

then,  EHF  and  EKF  coincide, 

and  the  point  H  coincides  with  the  point  K, 

Now,  HE,  or  KE  =  lBC, 

and  HF,  or  KF  =  ^AB  -, 

hence.  Ax.  2,  EF  —  ^  (AB  -\-  BC). 

Therefore,  etc.  qM.D. 


68  PLANE   GEOMETRY. -BOOK  L 

POLYGONS 

161.  A  portion  of  a  plane  bounded  by  any  number  of  straight 
lines  is  called  a  Polygon. 

The  sum  of  the  straight  lines  which  bound  a  polygon  is  called 
its  perimeter. 

The  term  polygon  is  usually  applied  to  figures  of  more  than 
four  sides. 

162.  A  polygon  of  three  sides  is  called  a  trigon  or  triangle  ; 
one  of  four  sides,  a  tetragon  or  quadrilateral ;  one  of  five  sides, 
a  pentagon  ;  one  of  six  sides,  a  hexagon  ;  one  of  seven  sides,  a 
heptagon ;  one  of  eight  sides,  an  octagon ;  one  of  ten  sides,  a 
decagon;  one  of  twelve  sides,  a  dodecagon;  one  of  fifteen  sides, 
a  pentadecagon. 

163.  A  polygon  such  that  none  of  its  sides,  if 
produced,  extend  within  it  is  called  a  Convex 
Polygon. 

164.  A  polygon  such  that  two  or  more  of  its 
sides,  if  produced,  extend  within  it  is  called  a 
Concave  Polygon. 

The  reflex  angle  ABCis  called  a  re-entrant  angle. 
Unless  otherwise  stated,  polygons  considered 
hereafter  will  be  understood  to  be  convex. 

165.  A  straight  line  joining  the  vertices  of  two  non-adjacent 
angles  of  a  polygon  is  called  a  Diagonal  of  the  Polygon. 


Proposition  XLIV 

166.  1.  Draw  convex  polygons,  each  having  a  different  number  of 
sides,  and  from  any  vertex  of  each  draw  its  diagonals.  How  does  the 
number  of  triangles  into  which  each  polygon  is  divided  compare  with  the 
number  of  sides  of  the  polygon? 

To  how  many  right  angles  is  the  sum  of  the  angles  of  a  triangle  equal  ? 
To  how  many  times  two  right  angles  is  the  sum  of  the  interior  angles  of 
a  polygon  equal  ? 


PLANE   GEOMETRY.  — BOOK  I.  69 

2.   Produce  the  sides  of  any  polygon  in  succession.     To  how  many 
right  angles  is  the  sum  of  all  the  exterior  and  i:.terior  angles  equal?    To 
how  many  right  angles  is  the  sum  of  the  exterior  angles  of  a  polygon 
equal  ? 

Theorem,  The  sum  of  the  angles  of  any  eonvejc  polygon 
is  equal  to  twice  as  many  right  angles  as  the  polygon  has 
sides  less  two.    . 

D 

Data:  A  convex  polygon  of  any  number 
(n)  of  sides,  as  ABODE. 

E< 

To  prove  the  sum  of  the  angles,  A^  B,  o,  D, 
and  E  equal  to  twice  as  many  right  angles 
as  the  polygon  has  sides  less  two. 

Proof.     From  any  vertex,  as  A,  draw  the  diagonals,  AC  and  AD. 

The  number  of  triangles  thus  formed  is  two  less  than  the  num- 
ber of  sides  of  the  polygon,  or  {n  —  2)  triangles. 

By  §  110,  the  sum  of  the  angles  of  each  triangle  is  equal  to 
two  right  angles,  therefore,  the  sum  of  the  angles  of  all  the 
triangles ;  that  is,  the  sum  of  the  angles  of  the  polygon  is  equal 
to(w  — 2)2rt.  A 

Therefore,  etc,  q.e.d. 


167.  Cor.  The  sum  of  the  exterior  angles  of 
any  convex  polygon  formed  by  producing  the 
Bides  of  the  polygon  in  succession  is  equal  to 
four  right  angles. 


Ex.  106.  If  from  the  extremities  of  the  shorter  base  of  an  isosceles 
trapezoid  lines  are  drawn  parallel  to  the  equal  sides,  two  triangles  are 
formed.     How  do  they  compare  ? 

Ex.  107.  If  in  a  parallelogram  any  two  points  in  a  diagonal  equally  dis- 
tant from  its  extremities  are  joined  to  the  vertices  of  the  opposite  angles, 
what  kind  of  a  figure  is  thus  formed  ? 

Ex.  108.  How  many  degrees  are  there  in  each  angle  of  an  ecjuiangular 
polygon  of  five  sides  ? 

Ex.  109.  How  many  sides  has  a  polygon  the  sum  of  whose  mterior  angles 
^s  double  the  sum  of  its  exterior  angles  ^ 


70  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  XLV 

168.  Draw  any  triangle  and  its  three  medians.  Do  the  medians 
intersect  in  a  point  ?  Measure  the  distance  from  this  point  to  each  ver- 
tex. How  do  these  distances  compare  with  the  medians  of  which  tihey 
are  a  part  ? 

Theorem,  The  medians  of  a  triangle  pass  throiCgh  a 
point  which  is  two  thirds  of  the  distance  from  each  vertex 
to  the  middle  of  the  opposite  side. 

c 
Data:    Any   triangle,    as    ABC,    and  yf\ 

its  medians,  AD,  BE,  and  CF.  /I  \ 

To  prove  that  AB,  BE,  and  CF  pass  /     /      \ 

through  a  point,  which  is  two   thirds  ^^^^^/--'^A 

of   the   distance  from   A,  B,  and  C  to        /^Z^^^^J^^\/   \ 
the   middle   of   the  opposite   sides   re-     /^-^^        J        ^^"^^ 
spectively.  a  f  b 

Proof.  Since  two  of  the  medians  will  intersect,  if  sufficiently 
produced,  it  needs  to  be  shown  only  that  the  third  median  passes 
through  the  point  of  intersection,  to  prove  that  the  three  pass 
through  the  same  point. 

Let  any  two  of  the  medians,  as  AD  and  BE,  intersect  at  H. 

Draw  KG,  joining  K  and  G,  the  middle  points  of  AH  and  BH 
respectively  ;  also  draw  KE,  ED,  and  GD. 


Then,  §  159, 

KG  Wab,      and  ED  WAB-, 

.-.  §  80, 

KG  II  ED  ; 

also,  §  158, 

KG  =  ^AB,  and  ED^^AB 

.•. 

KG— ED', 

and,  §  150, 

KGDE  is  a  parallelogram. 

Hence,  §  154, 

KH=HD,       and   GH=HE. 

But  since,  const.. 

AK  =  KH,      and  BG  =  GH, 

AH  =  I  AD,    2^Ti^  BH=lBE, 

Then,  since  the  medians  from  any  two  vertices  intersect  in  a 
point  which  is  two  thirds  of  the  distance  from  each  vertex  to  the 
middle  of  the  opposite  side,  the  median  from  C  intersects  AD  at  if. 

That  is,  CF  passes  through  H,  and  CH  =  ^  CF. 

Therefore,  etc.  q.e.d. 


PLANE   GEOMETRY.  — BOOK  I.  71 

Proposition  XLVI 

169'.  Draw  any  triangle  and  lines  bisecting  its  angles.  Do  these  lines 
intersect  in  a  point  ?  How  do  the  distances  of  the  point  from  the  sides 
of  the  triangle  compare? 

Theorem,  The  bisectors  of  the  three  angles  of  a  triangle 
pass  through  a  point' which  is  equidistant  from  the  sides  of 
the  triangle. 

Data:  Any  triangle,  as  ABC,  and 
the  lines  AD,  BE,  and  CF,  bisecting 
the  angles  A,  B,  and  C  respectively. 

To  prove  that  AD,  BE,  and  CF  pass 
through  a  point  which  is  equidistant 
from  AB,  BC,  and  AC. 

Proof.  Since  two  of  the  bisectors  will  intersect,  if  sufficiently 
produced,  it  needs  to  be  shown  only  that  the  third  bisector  passes 
through  the  point  of  intersection  to  prove  that  the  three  pass 
through  the  same  point. 

Let  any  two  of  the  bisectors,  as  AD  and  BE,  intersect  in  H. 

Then,  §  134,  H  is  equidistant  from  AB  and  AC,  and  also  from 
AB  and  BC. 

Hence,  H  is  equidistant  from  AC  and  BC\ 

.'.  §  135,  H  lies  in  the  bisector  of  angle  C. 

That  is,  CF  passes  through  the  point  H,  which  is  equidistant 
from  AB,  BC,  and  AC. 

Therefore,  etc.  q.e.d. 

Ex.  110.  How  does  the  angle  formed  by  the  diagonals  of  a  square  com- 
pare with  a  right  angle  ? 

Ex.  111.  How  does  the  angle  formed  by  the  diagonals  of  a  rhombus 
compare  with  a  right  angle  ?     How  do  the  diagonals  divide  each  other  ? 

Ex.  112.    How  do  the  diagonals  of  a  rectangle  compare  in  length  ? 

Ex.  113.  If  from  any  point  in  the  bisector  of  an  angle  straight  lines  are 
drawn  parallel  to  the  sides  of  the  angle  and  are  produced  to  meet  the  sides, 
what  figure  is  thus  formed  ? 

Ex,  114.  The  difference  between  two  angles  of  a  parallelogram  which  have 
a  common  side  is  60°.     What  is  the  value  of  each  angle  of  the  parallelogram  ? 

Ex.  115.  If  the  middle  points  of  any  two  opposite  sides  of  a  quadrilateral 
are  joined  to  each  of  the  middle  points  of  the  diagonals,  what  kind  of  a 
figure  will  the  four  joining  lines  form  ? 


72  PLANE   GEOMETRY.  — BOOK  I. 

Proposition  XLVII 

170.  Draw  any  triangle  and  Jines  perpendicular  to  its  sides,  bisecting 
them.  Do  these  lines  intersect  in  a  point  ?  How  do  the  distances  of 
the  point  from  the  vertices  of  the  triangle  compare? 

Theorem,  The  perpendicular  bisectors  of  the  sides  of  a 
triangle  pass  through  a  point  which  is  equidistant  from 
the  vertices  of  the  triangle. 

Data:  Any  triangle,  as  ABC,  and 
FG,  DK,  and  EJ,  the  perpendicular  bi- 
sectors of  AB,  BC,  and  AC  respectively. 

To  prove  that  FG,  DK,  and  EJ  pass 
through  a  point  which  is  equidistant 
from  A,  B,  and  C.  a  f  b 

Proof.  Since  tvs^o  of  the  perpendiculars  will  intersect,  if  sufh- 
ciently  produced  (why  ?),  it  needs  to  be  shown  only  that  the  third - 
perpendicular  passes  through  the  point  of  intersection,  to  prove 
that  the  three  pass  through  the  same  point. 

Let  any  two  of  the  perpendiculars,  as  FG  and  DK,  intersect 
at  H. 

Then,  §  103,  H  is  equidistant  from  A  and  B,  and  also  from 
B  and  C. 

Hence,  H  is  equidistant  from  A  and  C ; 

.*.  §  104,      H  lies  in  the  perpendicular  bisector  of  AC. 

That  is,  EJ  passes  through  the  point  H,  which  is  equidistant 
from  A,  B,  and  C. 

Therefore,  etc.  q.e.d. 

Ex.  116.  The  middle  points  of  the  sides  of  an  equilateral  triangle  are 
joined.     What  kind  of  triangles  are  formed  ? 

Ex.  117.  How  do  the  lines  drawn  from  the  middle  points  of  the  equal 
sides  of  an  isosceles  triangle  to  the  opposite  extremities  of  the  base  compare 
in  length  ? 

Ex.  118.  Tiie  parallel  sides  of  a  trapezoid  are  35  and  55  feet  respectively. 
What  is  the  length  of  the  line  joining  the  middle  points  of  the  non-parallel 
sides  ? 

Ex.  119.  If  from  the  extremities  of  the  shorter  base  of  rn  isosceles  trape. 
zoid  perpendiculars  are  drawn  to  the  longer  base,  two  triangles  are  formed. 
How  do  they  compare  ? 


PLANE  GEOMETRY,-^ BOOK  I.  73 

Proposition  XLVIII 

171.  Draw  any  triangle  and  lines  from  the  vertices  perpendicular  to 
the  opposite  sides.     Do  these  lines  intersect  in  a  point? 

Theorem,  The  perpendiculars  from  the  vertices  of  a  tri- 
angle to  the  opposite  sides  pass  through  the  same  point. 

H  c  K 

Data:  Any  triangle,  as  ABC,     \  ^. 

and  the  lines  AD,  BE,  and  CF       \  / 

drawn  from  the  vertices  A,  B,  \      /^ 

and  C  respectively,  perpendicu-  2\     

lar  to  the  opposite  sides.  \ 

To  prove  that  AD,  BE,  and  CF 
pass  through  the  same  point. 

Q 
Proof.     Through  the  vertices  A,  B,  and  G  draw  GH,  GK,  and 
EK  parallel  to  BC^  ACy  and  AB  respectively,  and  intersecting  in 
G,  H,  and  K, 

§153,  AG=BCf 

and  AH  =  BC; 

.♦.  Ax.  1,  AG  =  AH. 

§72,  AD±GH', 

AD  is  the  perpendicular  bisector  of  GH, 
In  like  manner, 

BE  is  the  perpendicular  bisector  of  GK^ 
and  CF  is  the  perpendicular  bisector  of  HK. 

Hence,  §  170,  AD,  BE,  and  CF  pass  through  the  same  point. 
Therefore,  etc.  q.e.d. 

Ex.  120.  How  many  sides  has  a  polygon  the  sum  of  whose  exterior  angles 
is  double  the  sum  of  its  interior  angles  ? 

Ex.  121.  How  many  sides  has  a  polygon  the  sum  of  whose  interior  angles 
is  equal  to  the  sum  of  its  exterior  angles  ? 

Ex.  122.  The  perimeter  of  an  isosceles  triangle  is  176  feet,  and  the  base  is 
1^  times  one  of  the  equal  sides.  What  is  the  length  of  each  side  of  the 
triangle  ? 

Ex.  123.  How  many  sides  has  an  equiangular  polygon  which  can  be 
divided  into  equile-teral  triangles  by  lines  drawn  from  a  point  within  to  the 
vertices  of  the  polygon  ? 


T4  PLANE   GEOMETRY  -^ BOOK  L 

SUMMARY 
172.  Truths  established  in  Book  L 
1.   Two  lines  are  equal, 

0.  If  they  can  be  made  to  coincide.  §  36 
6.   If  they  are  sides  of  an  equilateral  triangle.  §  91 

c.  If  they  represent  the  distances  from  the  extremities  of  a  straight  line 
to  any  point  in  the  perpendicular  erected  at  its  middle  point.  §  103 

d.  If  they  are  homologous  sides  of  equal  triangles.  §  108 
«.  If  they  are  sides  of  a  triangle  opposite  equal  angles.  §  118 
/.  If  they  are  sides  of  an  equiangular  triangle.  §  119 
g.  If  they  are  drawn  fronj  any  point  in  a  perpendicular  to  a  line  and  cut 

off  equal  distances  on  that  line  from  the  foot  of  the  perpendicular.  §  132 

h.  If  they  represent  the  distances  of  any  point  in  the  bisector  of  an  angle 

from  its  sides.  §  134 

i.    If  they  are  the  non-parallel  sides  of  an  isosceles  trapezoid.  §  139 

j.    If  they  are  the  sides  of  a  square.  §  143 

k.   If  they  are  the  sides  of  a  rhombus.  §  144 

1.  If  they  are  parallel  and  are  intercepted  between  parallel  lines.  §  151 
w>.  If  they  are  opposite  sides  of  a  parallelogram.  §  153 
w.   If  they  are  parts  intercepted  on  one  transversal  by  parallel  lines  which 

mtercept  equal  parts  on  another  transversal.  §  157 

0.    If  one  is  half  a  side  of  a  triangle  and  the  other  is  drawn  parallel  to  it 

and  bisecting  one  of  the  other  sides.  §  158 

p.   If  one  joins  the  middle  points  of  the  non-parallel  sides  of  a  trapezoid 

and  the  other  is  equal  to  half  the  sum  of  the  parallel  sides.  §  160 

2.  Two  lines  are  parallel, 

a.   If  both  are  perpendicular  to  the  same  line.  §  71 

6.   If  when  cut  by  a  transversal  the  alternate  interior  angles  are  equal. 

§75 
c   If  when  cut  by  a  transversal  the  corresponding  angles  are  equal.     §  77 

d.  If  when  cut  by  a  transversal  the  sum  of  the  two  interior  angles  on  the 
same  side  of  the  transversal  is  equal  to  two  right  angles.  §  79 

e.  If  both  are  parallel  to  a  third  line.  »  §  80 
/.  If  they  are  the  bases  of  a  trapezoid.  §  138 
g.  If  they  are  opposite  sides  of  a  parallelogram.  §  140 
h.  If  one  is  a  side  of  a  triangle  and  the  other  joins  the  middle  points 

of  the  other  two  sides.  §  159 

i.    If  one  is  either  base  of  a  trapezoid  and  the  other  joins  the  middle 

points  of  the  non-parallel  sides.  §  160 

8.   Two  lines  are  perpendicular  to  each  other, 

a.  If  they  form  equal  adjacent  angles  with  each  other.  §  26 

^.   If  one  is  perpendicular  to  a  line  which  is  parallel  to  the  other.        §  72 


PLANE   GEOMETRY.  — BOOK  I.  75 

c.  If  any  two  or  more  points  in  one  are  each  equidistant  from  the  extremi- 
ties of  the  other.  §§  106,  104 

d.  If  one  is  the  base  of  an  isosceles  triangle  and  the  other  is  the  bisector 
of  the  vertical  angle.  §  120 

4.   Two  lines  form  one  and  the  same  straight  line, 

a.    If  they  are  the  sides  of  a  straight  angle.  §  27 

h.    If  they  are  the  exterior  sides  of  adjacent  supplementary  angles.      §  58 

6.   Two  lines  are  unequal, 

a.  If  one  is  a  perpendicular  from  a  point  to  a  straight  line  and  the  other 
is  any  other  line  from  that  point  to  the  straight  line.  §  61 

h.  If  they  represent  the  distances  from  the  extremities  of  a  straight  line 
to  any  point  without  the  perpendicular  erected  at  its  middle  point.  §  103 

c.  If  they  are  sides  of  a  triangle  and  lie  opposite  unequal  angles.        §  127 

d.  If  they  are  the  third  sides  of  two  triangles  whose  other  sides  are  equal, 
each  to  each,  and  include  unequal  angles.  '  §  129 

e.  If  they  are  drawn  from  any  point  in  a  perpendicular  to  a  line  and  cut 
off  unequal  distances  on  that  line  from  the  foot  of  the  perpendicular.      §  182 

/.  If  they  are  distances  cut  off  on  a  line  from  the  foot  of  a  perpendicular 
to  it  by  unequal  lines  from  any  point  in  the  perpendicular.  §  133 

g.  If  one  is  any  side  of  a  triangle  and  the  other  is  equal  to  the  sum  of  the 
other  two  sides.  §§  124,  125 

h.  If  one  is  equal  to  the  sum  of  two  lines  from  a  point  within  a  triangle 
to  the  extremities  of  one  side,  and  the  other  is  equal  to  the  sum  of  the  other 
two  sides.  §  131 

6.  A  line  is  bisected, 

a.  If  it  is  the  base  of  an  isosceles  triangle,  by  the  bisector  of  the  vertical 
angle.  §  120 

h.  If  it  is  the  base  of  an  isosceles  triangle,  by  a  perpendicular  from  the 
vertex.  §  122 

c.  If  it  is  either  diagonal  of  a  parallelogram,  by  the  other  diagonal.    §  154 

d.  If  it  is  the  side  of  a  triangle,  by  a  straight  line  drawn  parallel  to  the 
base  and  bisecting  the  other  side.  §  158 

7.  Lines  pass  through  the  same  point, 

a.    If  they  are  the  medians  of  a  triangle.  §  168 

6.    If  they  are  the  bisectors  of  the  three  angles  of  a  triangle.  §  169 

c.  If  they  are  the  perpendicular  bisectors  of  the  sides  of  a  triangle.   §  170 

d.  If  they  are  perpendiculars  from  the  vertices  of  a  triangle  to  the  oppo- 
site sides.  §  171 

8.  A  perpendicular,  and  only  one,  can  be  drawn  to  a  straight  line, 

a.   At  a  point  in  the  line.  §  51 

6.    From  a  point  without  the  line.  §  60 


T6  PLANE   GEOMETRY.  — BOOK  L 

9.  Two  angles  are  equal, 

a.  If  tliey  can  be  made  to  coincide.  §  36 

h.  If  they  are  right  angles.  §  52 

c.  If  they  are  straight  angles.  §  53 

d.  If  they  are  complements  of  equal  angles.  §  54 

e.  If  they  are  supplements  of  equal  angles.  §  54 
/.  If  they  are  vertical  angles.  §  59 
g.  If  they  are  alternate  interior  angles  formed  by  a  transversal  and  paral- 
lel lines.  §  73 

h.  If  they  are  corresponding  angles  formed  by  a  transversal  and  parallel 
lines.  §  76 

i.  If  their  sides  are  parallel  and  both  pairs  extend  in  the  same  or  in 
opposite  directions  from  their  vertices.  §  81 

j.  If  their  sides  are  perpendicular  to  each  other  and  both  angles  are  acute 
or  both  are  obtuse.  §  83 

k.   If  they  are  angles  of  an  equiangular  triangle.  §  98 

I.    If  they  are  formed  adjacent  to  a  straight  line  by  lines  joining  the  ex- 
tremities of  that  line  with  any  point  in  its  perpendicular  bisector.  §  105 
TO.  If  they  are  formed  by  the  perpendicular  bisector  of  a  straight  line  and 
lines  from  any  point  in  it  to  the  extremities  of  the  straight  line.  §  105 
n.   If  they  are  homologous  angles  of  equal  triangles.  §  108 
0.    If  they  are  the  third  angles  of  two  triangles  whose  other  angles  are 
equal,  each  to  each.  §  113 
p.   If  they  are  opposite  the  equal  sides  of  an  isosceles  triangle.  §  116 
q.   If  they  are  angles  of  an  equilateral  triangle.  §  117 
r.   If  they  are  the  opposite  angles  of  a  parallelogram.  §  149 

10.  Two  angles  are  supplementary, 

a.   If  their  sum  is  equal  to  two  right  angles.  §  32 

h.    If  their  corresponding  sides  are  parallel  and  one  pair  extends  in  the 

same  direction  and  the  other  in  opposite  directions  from  their  vertices.  §  81 
c.   If  their  corresponding  sides  are  perpendicular  and  one  angle  is  acute 

and  the  other  obtuse.  §  83 

11.  Two  angles  are  unequal, 

a.  If  they  are  angles  of  a  triangle  and  lie  opposite  unequal  sides.       §  126 

h.  If  they  are  the  angles  opposite  unequal  sides  of  two  triangles  whose 

other  two  sides  are  equal,  each  to  each.  §  130 

12.  An  angle  is  bisected, 

a.  If  it  is  the  vertical  angle  of  an  isosceles  triangle,  by  the  perpendicular 
bisector  of  the  base.  §  121 

h.  If  it  is  the  vertical  angle  of  an  isosceles  triangle,  by  a  line  from  the 
vertex  perpendicular  to  the  base.  §  122 

c.  By  a  line  every  point  of  which  is  equidistant  from  the  sides  of  the 
angle.  §  136 


PLANE   GEOMETRY.^ BOOK  L        .  77 

13.  An  angle  is  equal  to  the  sum  of  two  angles, 

a.  If  it  is  an  exterior  angle  of  a  triangle,  and  the  two  angles  are  the 
opposite  interior  angles.  §  115 

14.  The  sum  of  angles  is  equal  to  a  right  angle, 

a.  If  they  are  complements  of  each  other.  §  31 

h.  If  they  are  the  acute,  angles  of  a  right  triangle.  §  111 

15.  The  sum  of  angles  is  equal  to  two  right  angles, 

a.  If  they  are  supplements  of  each  other.  §  32 

6.  If  they  are   adjacent  angles  formed  by  one  straight  line  meeting 

another.  §  55 

c.  If  they  are  all  the  consecutive  angles  which  have  a  common  vertex  in 
a  line  and  lie  on  the  same  side  of  the  line.  §  56 

d.  If  they  are  the  interior  angles  formed  by  a  transversal  and  parallel 
lines  and  lie  on  the  same  side  of  the  transversal.  §  78 

e.  If  they  are  the  angles  of  a  triangle.  §  110 

16.  The  sum  of  angles  is  equal  to  four  right  angles, 

a.  If  they  are  all  the  consecutive  angles  that  can  be  formed  about  a 
point.  §  57 

6.  If  they  are  the  exterior  angles  of  any  convex  polygon  formed  by  pro- 
ducing the  sides  in  succession.  §  167 

17.  The  sum  of  angles  is  equal  to  (/;  —  2)  2  rt.  A^ 

a.  If  they  are  the  angles  of  any  convex  polygon.  §  166 

18.  Two  triangles  are  equal, 

a.  If  two  sides  and  the  included  angle  of  one  are  equal  to  the  correspond- 
ing parts  of  the  other.  §  100 

h.  If  a  side  and  two  adjacent  angles  of  one  are  equal  to  the  correspond- 
ing parts  of  the  other.  §  102 

c.  If  the  three  sides  of  one  are  equal  to  the  three  sides  of  the  other.   §  107 

d.  If  they  are  right  triangles,  and  a  side  and  an  acute  angle  of  one  are 
equal  to  the  corresponding  parts  of  the  other.  §  114 

e.  If  they  are  right  triangles,  and  the  hypotenuse  and  a  side  of  one  are 
equal  to  the  corresponding  parts  of  the  other.  §  12? 

/.    If  they  are  formed  by  dividing  a  parallelogram  by  one  of  its  diagonals. 

§152 

19.  Two  parallelograms  are  equal, 

a.   If  they  can  be  made  to  coincide.  ''    ^  §  36 

6.  If  two  sides  and  the  included  angle  of  one  are  equal  to  the  correspond- 
ing parts  of  the  other.  §  155 

20.  A  quadrilateral  is  a  parallelogram, 

a.   If  its  opposite  sides  are  parallel.  §  14C 

h.   If  its  opposite  sides  are  equal.  §  148 

c.  If  two  of  its  sides  are  equal  and  parallel.  §  150 


78  PLANE   GEOMETRY.— BOOK  L 


SUPPLEMENTARY  EXERCISES 

Ex.  124.  If  through  a  point  halfway  between  two  parallel  lines  two 
transversals  are  drawn,  they  intercept  equal  parts  on  the  parallel  lines. 

Suggestions  for  Demonstration.  1.  What  are  the  data  of  the  propo- 
sition ?  • 

2.  What  lines  and  point  in  the  figure  in  ^ ^^ ^ 

the  margin  represent  the  data  of  the  propo- 
sition ? 

3.  What  parts  of  the  figure  are  to  be    C -f ^^ 1> 

proved  equal  ? 

4.  How  may  two  lines  be  proved  equal  ?  Summary,  §  172,  1. 

5.  Since  FH  and  JG,  which  are  to  be  proved  equal,  are  parts  of  tri- 
angles, what  propositions  might  we  expect  to  employ  in  the  proof  ? 

6.  In  what  ways  may  two  triangles  be  proved  equal  ?  Summary,  §  172, 18. 

7.  What  facts  in  the  data  suggest  aid  in  determining  the  equality  of 
angles  ?  Ans.  Parallel  lines. 

8.  What  homologous  angles  in  the  two  triangles  are  equal  ? 

9.  What  other  homologous  elements  of  the  two  triangles  must  also  be 
equal  before  the  triangles  can  be  proved  to  be  equal  ? 

10.  By  careful  examination  of  the  given  figure  discover  whether  any  two 
homologous  sides  can  be  proved  equal. 

11.  Since  the  homologous  sides  cannot  be  proved  equal  from  the  given 
figure,  if  they  can  be  proved  equal  at  all,  what  expedient  must  be  resorted  to  ? 

Ans.  Construction  lines  must  be  drawn  which  will  enable  us  to  prove  a 
side  of  one  of  the  triangles  equal  to  an  homologous  side  of  the  other. 

12.  What  fact  in  the  data  has  not  yet  been  considered  which  might  suggest 
aid  in  drawing  the  construction  lines  ? 

13.  What  kind  of  a  line  measures  the  distance  between  two  parallel 
lines  ?  If  such  a  line  be  drawn  through  the  given  point,  how  is  it  divided 
at  the  given  point  ?    Then,  what  line  may  aid  in  the  proof  ? 

14.  Drawing  the  figure  as  in  the  margin, 

with  LK  perpendicular  to  the  parallel  lines,   A L  JJ ^^ 


and  passing  through  the  point  E,  which  is 
halfway  between  the  parallel  lines,  discover  . 

how  the  triangles  FEL  and  GEK  compare  ;    C J    ^ ^ 

also,  how  FE  and  GE  compare. 

15.  Since  the  homologous  angles  of  the  original  triangles  have  been  dis- 
covered to  be  equal,  and  since  the  equality  of  two  homologous  sides,  FE  and 
GE,  has  also  been  shown,  how  do  the  original  triangles  compare  ?  How 
do  the  sides  FH  and  JG  compare  ? 

Write  out  the  demonstration. 


PLANE   GEOMETRY.  — BOOK  I.  79 

General  Suggestions.  I.  Study  the  theorem  carefully  to  discover 
the  data. 

II.  Construct  a  figure,  or  figures,  to  correspond  ivith  the  data. 

III.  Discover  what  parts  of  the  figure  correspond  to  the  conclusion 
given  in  the  theorem. 

IV.  Study  the  theorem  and  the  figure  to  discover  as  many  truths 
as  possible  regarding  the  lines,  angles,  or  other  parts. 

V.  Keeping  in  mind  the  truths  just  discovered  and  the  facts  to  he 
proved,  consult  the  Summary  and  find  which  truth  will  best  aid  in 
establishing  the  proposition. 

VI.  If  none  of  the  truths  in  the  Summary  seem  to  be  directly  ap- 
plicable to  the  demonstration  sought,  draw  construction  lines  which 
may  aid  in  applying  some  one  of  the  truths. 

Ex.  125.  A  straight  line  cutting  the  sides  of  an  isosceles  triangle  and 
parallel  to  the  base  makes  equal  angles  with  the  sides. 

Ex.  126.  If  the  base  of  a  triangle  is  divided  into  two  parts  by  a  perpen- 
dicular from  the  vertex,  each  part  of  the  base  is  less  than  the  adjacent  side 
of  the  triangle. 

Ex.  127.  Any  straight  line  drawn  from  the  vertex  of  a  triangle  to  the 
base  is  bisected  by  the  straight  line  which  joins  the  middle  points  of  the 
other  sides  of  the  triangle. 

Ex.  128.  The  perpendiculars  to  the  diagonal  of  a  parallelogram  from  the 
opposite  vertices  are  equal. 

Ex.  129.  If  one  side  of  a  quadrilateral  is  extended  in  both  directions, 
the  sum  of  the  exterior  angles  formed  is  equal  to  the  sum  of  the  two  interior 
angles  opposite  the  side  produced. 

Ex.  130.  If  in  an  isosceles  triangle  perpendiculars  are  drawn  from  the 
middle  point  of  the  base  to  the  equal  sides,  the  perpendiculars  are  equal. 

Ex.  131.  A  straight  line  drawn  from  any  point  in  the  bisector  of  an  angle 
to  either  side  and  parallel  to  the  other  side  makes,  with  the  bisector  and  the 
side  it  meets,  an  isosceles  triangle. 

Ex.  132.  The  difference  between  two  sides  of  a  triangle  is  less  than  the 
third  side. 

Ex.  133.  Any  straight  line  through  the  middle  point  of  a  diagonal  of  a 
parallelogram,  and  terminated  by  the  opposite  sides,  is  bisect^d  at  that  point, 

Ex.  134.  If  either  of  the  equal  sides  of  an  isosceles  triangle  is  produced 
through  the  vertex,  the  line  bisecting  the  exterior  angle  thus  formed  is 
parallel  to  the  base  of  the  triangle.  * 

Ex.  135.  If  the  bisector  of  one  of  the  angles  of  a  triangle  meets  the 
opposite  side,  the  lines  from  the  point  of  meeting  parallel  to  the  other  sides 
and  terminated  by  them  are  equal. 


80  PLANE   GEOMETRY.  — BOOK  I. 

Ex.  136.  If  each  of  the  angles  at  the  base  of  an  isosceles  triangle  is  one 
fourth  the  vertical  angle,  every  line  perpendicular  to  the  base  forms  an  equi- 
lateral triangle  with  the  other  two  sides,  produced  when  necessary. 

Ex.  137.  If  the  straight  line  bisecting  an  exterior  angle  of  a  triangle  is 
parallel  to  a  side,  the  triangle  is  isosceles.  * 

Ex.  138.  If  the  non-parallel  sides  of  a  trapezoid  ar^  equal,  the  base 
angles  are  equal,  and  the  diagonals  are  equal. 

Suggestion^.  Through  one  extremity  of  the  shorter  parallel  side  draw  a 
line  parallel  to  the  opposite  non-parallel  side. 

Ex.  139.  If  the  angles  adjacent  to  one  base  of  a  trapezoid  are  equal, 
those  adjacent  to  the  other  base  are  also  equal. 

Suggestion.     Produce  the  non-parallel  sides. 

Ex.  140.  If  the  upper  base  of  an  isosceles  trapezoid  equals  the  sum  of 
the  non-parallel  sides,  lines  drawn  from  the  middle  point  of  the  upper  base  to 
the  extremities  of  the  lower  divide  the  figure  into  three  isosceles  triangles. 

Ex.  141.  The  opposite  angles  of  an  isosceles  trapezoid  are  supplements 
of  each  other. 

Ex.  142.  The  segments  of  the  diagonals  of  an  isosceles  trapezoid  form 
with  the  upper  and  lower  bases  two  isosceles  triangles. 

Ex.  143.  The  triangle  formed  by  joining  the  middle  points  of  the  sides 
of  an  isosceles  triangle  is  isosceles. 

Ex.  144.  If  the  two  angles  at  the  base  of  an  isosceles' triangle  are  bisected, 
the  line  which  joins  the  intersection  of  the  bisectors  with  the  vertex  of  the 
triangle  bisects  the  vertical  angle. 

Suggestion.  "  Refer  to  §  172,  9,  n. 

Ex.  145.  ABCD  is  a  parallelogram  ;  E  and  F  are  the  middle  points  of 
AD  and  BO  respectively.     Show  that  BE  and  FD  trisect  the  diagonal  AG. 

Suggestion.     Refer  to  §  172,  6,  d. 

Ex.  146.  The  exterior  angle  of  an  equiangular  hexagon  is  equal  to  the 
interior  angle  of  an  equiangular  triangle. 

Ex.  147.  If  one  diagonal  of  a  quadrilateral  bisects  two  of  the  angles,  it  is 
perpendicular  to  the  other  diagonal. 

Ex.  148.  If  one  triangle  has  two  sides  and  a  median  to  one  of  them  equal 
respectively  to  the  corresponding  parts  of  another  triangle,  the  triangles  are 
equal. 

Ex.  149.   The  diagonals  of  a  rhombus  are  unequal. 

Ex.  150.  If  one  angle  of  a  triangle  is  equal  to  the  sum  of  the  other  two, 
the  triangle  can  be  divided  into  two  isosceles  triangles. 

Suggestion.  From  the  vertex  oi  ZB  which  is  equal  to  the  sum  of  the 
other  two  angleS,  draw  BD  to  meet  AO  a,t  D,  making  ZABD  =  ZBAD. 

>Ex.  151.  If  the  diagonals  of  a  quadrilateral  bisect  each  other,  the  figure 
is  a  parallelogram. 

Ex.  152.  The  bisectors  of  two  adjacent  angles  of  a  parallelogram  inter- 
sect each  other  at  right  angles. 


PLANE   GEOMETRY.  — BOOK  I,  81 

Ex.  153.  If  the  bisectors  of  the  equal  angles  of  an  isosceles  triangle  are 
produced  until  they  meet,  they  form  with  the  base  an  isosceles  triangle. 

Ex.  154.    The  diagonals  of  a  rhombus  bisect  the  opposite  angles. 

Ex.  155.  If  two  equal  straight  lines  bisect  each  other  at  right  angles,  the 
lines  joining  their  extremities  form  a  square. 

Ex.  156.  If  the  base  of  any  triangle  is  produced  in  both  directions,  the 
sum  of  the  two  exterior  angles  diminished  by  the  vertical  angle  is  equal  to 
two  right  angles. 

Ex.  157.  In  a  quadrilateral,  if  two  opposite  sides  which  are  not  parallel 
are  produced  to  meet,  the  perimeter  of  the  greater  triangle  thus  formed  is 
greater  than  the  perimeter  of  the  quadrilateral. 

Ex.  158.  If  from  any  point  in  the  base  of  an  isosceles  triangle  lines  paral- 
lel to  the  sides  are  drawn,  a  parallelogram  is  formed  whose  perimeter  is  equal 
to  the  sum  of  the  equal  sides  of  the  triangle. 

Ex.  159.  ABCD  is  a  quadrilateral  having  angle  ABC  equal  to  angle 
ADC]  AB  and  DG  produced  meet  in  E ',  AD  and  BG  in  F.  Show  that 
angle  AED  equals  angle  AFB. 

Ex.  160.  ABO  is  an  isosceles  triangle  having  AG  equal  to  BC^  and  AG 
is' produced  through  G  its  own  length  to  D.      Then,  ABD  is  a  right  angle. 

Ex.  161.  ABG  is  a  triangle,  and  through  i>,  the  intersection  of  the  bisec- 
tors of  the  angles  B  and  C,  EDF  is  drawn  parallel  to  BG^  meeting  AB  in  E 
and  ^C  in  i^.     Then  EF  =  EB+  FG. 

Suggestion.     Prov5  ED  =  EB  and  FD  =  FG. 

Ex.  162.  ABGD  is  an  isosceles  trapezoid  and  AG  and  BD  its  diagonals 
intersecting  at  0.  Prove  that  the  following  pairs  of  triangles  are  equal: 
ABG  and  ABD  ;  ADG  and  BDO;  AOD  and  BOG. 

Ex.  163.  In  any  right  triangle  the  line  drawn  from  the  vertex  of  the  right 
angle  to  the  middle  of  the  hypotenuse  is  equal  to  one  half  the  hypotenuse. 

Suggestion.  From  the  middle  point  of  the  hypotenuse  draw  a  line  paral- 
lel to  one  of  the  other  sides. 

Ex.  164.  If  through  each  of  the  vertices  of  a  triangle  a  line  is  drawn 
parallel  to  the  opposite  side,  a  new  triangle  is  formed  equal  to  four  times 
the  given  triangle. 

Ex.  165.  Two  equal  lines,  AB  and  CZ),  intersect  at  E,  and  the  triangles 
GAE  and  BDE  are  equal.     Show  that  OB  is  parallel  to  AD. 

Ex.  166.  ABG  and  ABD  are  two  equilateral  triangles  on  opposite  sides 
of  the  same  base  ;  BE  and  BF  are  the  bisectors  respectively  of  the  angles 
ABG  and  ABD,  meeting  AG  and  AD  in  E  and  i?*  respectively.  Then,  the 
triangle  BEF  is  equilateral. 

Suggestion.     Eefer  to  §  172,  1,  f. 

Ex.  167.   ABO  is  any  triangle,  and  on  AB  and  AC  equilateral  triangles 
ADB  and  AEG  are  constructed  externally.    Show  that  GD  equals  BE. 
Suggestion.     Refer  to  §  172,  18.  a 
kilns^s  obom.  — 6 


82  PLANE   GEOMETR Y.  —  BOOK  I. 

Ex.  168.  If  through  the  extremities  of  each  diagonal  of  a  quadrilateral 
lines  parallel  to  the  other  diagonal  are  drawn,  a  parallelogram  double  the 
given  quadrilateral  will  be  formed. 

Ex.  169.  ABCD  is  a  parallelogram ;  E  and  F  are  points  on  AC,  such 
that  AE  =  FC ;  G  and  H  are  points  on  BD,  such  that  BG-  HD.  Then, 
EGFH  is  a  parallelogram. 

Ex.  170.  The  lines  joining  the  middle  points  of  the  sides  of  a  rhombus 
taken  in  order  form  a  rectangle. 

Ex.  171.  The  bisector  of  the  vertical  angle  of  a  triangle  and  the  bisectors 
of  the  exterior  angles  at  the  base  formed  by  producing  the  sides  about  the 
vertical  angle  meet  in  a  point  which  is  equidistant  from  the  base  and  the 
sides  produced. 

Suggestion.     Use  the  method  of  proof  employed  in  Prop.  XL VI. 

Ex.  172.  If  in  a  right  triangle  one  of  the  acute  angles  is  twice  the  other, 
the  hypotenuse  is  equal  to  twice  the  side  opposite  the  smaller  acute  angle. 

Suggestion.  From  the  vertex  of  the  right  angle  drav/  a  line  to  the  hy- 
potenuse, making  with  one  side  an  angle  equal  to  the  acute  angle  adjacent  to 
that  side. 

Ex.  173.  A  parallelogram  is  bisected  by  any  straight  line  passing  through 
the  middle  point  of  one  of  its  diagonals. 

Ex.  174.  If  two  quadrilaterals  have  three  sides. and  the  two  included 
angles  of  one  equal,  each  to  each,  to  three  sides  and  the  two  included  angles 
of  the  other,  the  quadrilaterals  are  equal. 

Suggestion.     Draw  homologous  diagonals. 

Ex.  175.  If  two  quadrilaterals  have  three  angles  and  the  two  included 
sides  of  one  equal,  each  to  *^ach,  to  three  angles  and  the  two  included  sides  of 
the  other,  the  quadrilaterals  are  equal. 

Ex.  176. '  The  bisectors  of  the  exterior  angles  of  a  rectangle  form  a  square. 

Ex.  177.  The  bisectors  of  the  interior  angles  of  a  parallelogram  form  a 
rectangle. 

Ex.  178.  The  bisectors  of  the  exterior  angles  of  a  quadrilateral  form  a 
quadrilateral  whose  opposite  angles  are  supplementary. 

Ex.  179.  The  bisectors  of  the  interior  angles  of  a  quadrilateral  form  a 
quadrilateral  whose  opposite  angles  are  supplementary. 

Ex.  180.  The  straight  line  drawn  from  any  vertex  of  a  triangle  to  the 
middle  point  of  the  opposite  side  is  less  than  half  the  sum  of  the  other  two 
sides. 

Suggestion.  Draw  lines  from  the  middle  point  of  the  side  opposite  the 
given  vertex,  parallel  to  each  of  the  other  two  sides. 

Ex.  181.  The  lines  which  join  the  middle  points  of  the  sides  of  a  quad- 
rilateral successively  form  a  parallelogram  whose  perimeter  is  equal  to  the 
sum  of  the  diagonals  of  the  quadrilateral. 


BOOK    II 

CIRCLES 

173.  A  plane  figure  bounded  by  a  curved  line,  every  point  ol 
which  is  equally  distant  from  a  point  within,  is  called  a  Circle ; 
the  point  within  is  called  the  Center ;  and  the  bounding  line  is 
called  the  Circumference. 

174.  Circles  which  have  the  same  center  are  called  Concentric 
Circles. 


175.  A  straight  line  from  the  center  to  the 
circumference  of  a  circle  is  called  a  Radius  of 
the  circle. 

OB,  Fig.  1,  is  a  radius  of  the  circle  AEDC. 

176.  A  "straight  line  which  passes  through 
ihe  center  of  a  circle,  and  whose  extremities 
'^xe  in  the  circumference,  is  called  a  Diameter 
of  the  circle. 

AD  and  EG,  Fig.  1,  are  diameters  of  the  circle  AEDC. 


177.  Any  part  of  a  circumference  is  called  an  Arc. 

The  curved  lines  between  A  and  B,  and  between  A  and  E,  Fig.  1,  are  arcs 
of  the  circumference  AEDC. 

178.  An  arc  which  is  one  half  a  circumference  is  called  a  Semi- 
circumference. 

179.  A  straight  line  which  joins  any  two  points  in  a  circum- 
ference is  called  a  Chord  of  the  circle. 

A  chord  which  joins  the  extremities  of  an  arc  is  said  to  subtend 
that  arc,  and 'an  arc  is  said  to  be  subtended  by  its  chord. 

83 


84 


PLANE   GEOMETRY.  — BOOK  IL 


Every  chord  of  a  circle  subtends  two  arcs. 
The  chord  AB,  Fig.  2,  subtends  the  arc  ADB,  and  also 
the  arc  ACB.  • 

When  a  chord  and  its  subtended  arc  are  men- 
tioned, the  less  arc  is  meant  unless  otherwise 
specified. 

180.  The  part  of  a  circle  included  between  an 
arc  and  its  chord  is  called  a  Segment  of  the  circle. 

ADB,  Fig.  2,  is  a  segment  of  the  circle  ADBC. 

181.  A  segment  which  is  one  half  a  circle  is  called  a  Semi- 
circle. 


182.  The  part  of  a  circle  included  between  an  arc  and  the  radii 
drawn  to  its  extremities  is  called  a  Sector  of  the  circle. 

OABB,  Fig.  2,  is  a  sector  of  the  circle  ADBC. 

183.  An  arc  which  is  one  fourth  of  a  circumference,  or  a  sector 
which  is  one  fourth  of  a  circle,  is  called  a  Quadrant. 

184.  A  straight  line  which  touches  a  circle  and  does  not  cut 
the  circumference  if  produced  is  called  a  Tangent  of  the  circle. 

The  circle  is  then  said  to  be  tangent  to  the 
line. 

The  point  where  a  tangent  touches  a  circle 
is  called  the  point  of  tangency,  or  the  point  of 
contact. 

AB,  Fig.  3,  is  a  tangent  of  the  circle  ;  and  P  is  the 
point  of  tangency. 

185.  Two  circles  are  said  to  be  tangent  to    ^' 
each  other,  if  they  are  both  tangent  to  the  fw.  8. 
same  straight  line  at  the  same  point. 

They  are  tangent  internally  or  externally  according  as  one  circle 
lies  within  or  without  the  other. 

186.  A  straight  line  which  cuts  a  circumference  in  two  points 
is  called  a  Secant  of  the  circle. 

CD,  Fig.  3,  is  a  secant  of  the  circle. 


PLANE    GEOMETRY.  — BOOK  IL 


85 


187.  An  angle  whose  vertex  is  at  the 
center  of  a  circle,  and  whose  sides  are 
radii  of  the  circle,  is  called  a  Central 
Angle. 

Angle  AOB  is  a  central  angle. 

188.  An  angle  whose  vertex  is  in  the 
circumference  of  a  circle,  and  whose  sides 
are  chords,  is  called  an  Inscribed  Angle. 

Angle  AGE  is  an  inscribed  angle. 

An  angle  whose  vertex  is  in  the  arc  of  a 
segment,  and  whose  sides  pass  through  the 
extremities  of  the  arc,  is  said  to  be  inscribed 
in  the  segment. 

189.  A  polygon  is  said  to  be  inscribed  in 
a  circle,  if  the  vertices  of  its  angles  are  in 
the  circumference. 

The  circle  is  then  said  to  be  circumscribed 
about  the  polygon. 

190.  A  polygon  is  said  to  be  circum- 
scribed about  a  circle,  if  each  side  is  tangent 
to  the  circle. 

The  circle  is  then  said  to  be  inscribed  in 

the  polygon. 

191.  Ax.  14.  All  radii  of  the  same  cirde^  or  of  equal  circles,  are 
equal. 

15.  All  diameters  of  the  same  circle,  or  of  equal  circles,  are  equal. 

16.  Turn  circles  are  equal,  if  their  radii  or  diameters  are  equal. 

17.  A  tangent  has  only  one  point  in  common  with  a  circle. 

Proposition  I 

192.  1.  Draw  a  circle  and  one  of  its  diameters;  draw  several  chords 
of  the  circle.  How  does  the  diameter  compare  in  length  with  any  other 
chord?    How  does  the  diameter  divide  the  circle?    The  circumference? 

2.  How  do  two  arcs  of  the  same  circle,  or  of  equal  circles,  compare, 
if  their  extremities  can  be  made  to  coincide? 


86  PLANE   GEOMETRY,— BOOK  II. 

Theorem.  A  diameter  of  a  circle  is  greater  than  any 
other  chord,  and  bisects  the  circle  and  its  circumference. 

'  Data :  A  circle  whose  center  is  0  ;  any 
diameter,  as  AB ;  and  any  other  chord, 
as  EF. 

To  prove  1.    AB  greater  than  EF. 

2.   AB  bisects  the  circle  and  its 
circumference. 

Proof.     1.   Draw  the  radii  OE  and  OF, 
§  125,  OE  +  OF  >  EF\ 

but.  Ax.  14,  OE  =  AO,  and  OF  =  OB] 

AO  -\-  OB  >  EFj  or  AB  >  EF. 
2.    §  176,  AB  passes  through  the  center  0. 

Eevolve  the  segment  ACB  upon  ^B  as  an  axis  until  it  comes 
into  the  plane  of  the  segment  ADB. 

Then,  the  arc  ACB  must  coincide  with  the  arc  ABB  ;  for,  if  the 
arcs  do  not  coincide,  some  points  in  the  two  arcs  are  unequally 
distant  from  the  center. 

But,  §  173,  every  point  in  the  arcs  is  equally  distant  from  0. 

Hence,  the  arcs  ACB  and  ADB  coincide  and  are  equal. 

That  is,  AB  bisects  the  circle  and  its  circumference. 

Therefore,  etc.  q.e.d." 

193.  Cor.  In  the  same  circle,  or  in  equal  circles^  arcs  whose 
extremities  can  be  made  to  coincide  are  equal. 

Proposition  II 

194.  1.  Draw  a  circle  and  divide  a  part  of  its  circumference  into  a 
number  of  equal  arcs ;  from  the  points  of  division  draw  lines  to  the 
center.     How  do  the  central  angles  thus  formed  compare  in  size? 

2.  In  a  circle  construct  two  equal  central  angles.  How  do  the  arcs 
which  subtend  them  compare  in  size? 

3.  Draw  a  circle  and  take  two  unequal  arcs ;  from  their  extremities 
draw  lines  to  the  center.  How  do  the  angles  at  the  center  compare  in 
size  ?  Which  arc  subtends  the  larger  angle  ?  Which  angle  at  the  center 
is  subtended  by  the  smaller  arc  ? 


PLANE   GEOMETRY.  — BOOK   II.  87 

Theorem,  In  the  same  circle,  or  in  equal  circles,  equal 
arcs  subtend  equal  central  angles;  conversely,  the  sides  of 
equal  central  angles  intercept  equal  arcs. 


Data:  The  equal  circles  whose  centers  are  0  and  P,  and  any 
equal  arcs,  as  AB  and  DE. 

To  prove  angle  0  =  angle  P. 

Proof.  Place  the  circle  whose  center  is  0  upon  the  equal  circle 
whose  center  is  P  so  that  arc  AB  coincides  with  the  equal  arc  DE, 
and  the  point  0  with  the  point  P, 

then,  OA  coincides  with  PD,  and  OB  with  PE. 

Hence,  §  36,  Zo  =  Z.P.  .  q.e.d. 

Conversely :  Data :  Any  equal  central  angles  in  these  circles,  as 
Z  0  and  Z  P. 

To  prove  arc  AB  =  arc  DE. 

Proof.  Place  the  circle  whose  center  is  0  upon  the  circle  whose 
center  is  P  so  that  the  point  0  coincides  with  the  point  P,  and 
OA  takes  the  direction  oi  PD. 

Data^  Zo  =:^  ZP) 

OB  takes  the  direction  of  PE) 

and,  since.  Ax.  14,    OA  =  pd,  and  OB  =  PE ; 

A  coincides  with  D,  and  B  with  E. 

Hence,  §  193,  arc  AB  =  arc  DE. 

Therefore,  etc.  q.e.d. 

195.  Cor.  In  the  same  circle,  or  in  equal  circles,  the  greater  of 
two  arcs  subtends  the  greater  central  angle;  conversely,  the  sides  of 
the  greater  of  two  central  angles  intercept  the  greater  arc. 


88  PLANE   GEOMETRY.  — BOdk  II. 

Proposition  III 

196.  1.  Draw  two  equal  circles  and  two  equal  chords,  one  in  each 
circle.   .  How  do  the  subtended  arcs  compare  in  length  ? 

2.  Draw  two  chords,  one  in  each  of  two  equal  circles,  subtending 
equal  arcs.     How  do  the  chords  compare  in  length  ? 

3.  Draw  two  equal  circles  and  two  unequal  chords,  one  in  each  circle. 
Which  chord  subtends  the  larger  arc  ?  Which  arc  is  subtended  by  the 
less  chord  ? 

Theorem,  In  the  same  circle,  or  in  equal  circles,  equal 
chords  subtend  equal  arcs;  conversely,  equal  arcs  have 
equal  chords. 


Data :  The  equal  circles  whose  centers  are  0  and  P,  and  any 
equal  chords,  as  ^B  and  DE. 

To  prove  arc  AB  =  arc  DE. 

Proof.     Draw  the  radii  OA,  OB,  PD,  and  PE, 

In  the  A  OAB  and  PDE,   AB  =  DE, 
Ax.  14,  OA  =  PD,  and  OB  =  PE', 

A  OAB  =  A  PDE,  Why  ? 

and  Zo=:Zp. 

Hence,  §  194,  arc  AB  =  arc  DE.  Q.e.d. 

Conversely :  Data :  Any  two  equal  arcs  in  these  circles,  as  AB 
and.  DE,  and  the  chords  subtending  them. 

To  prove  chord  AB  =  chord  DE. 

Proof.     By  the  student. 

197.  Cor.  In  the  same  circle,  or  in  equal  circles,  the  greater  of 
two  chords  subtends  the  greater  arc;  conversely,  the  greater  of  two 
arcs  has  the  greater  cliord. 


PLANE    GEOMETRY.  — BOOK  11. 


89 


Proposition  IV 

198.  Draw  a  circle  and  a  chord  of  the  circle ;  draw  a  radius  perpen- 
dicular to  the  chord.  How  does  this  radius  divide  the  chord?  How 
does  it  divide  the  arc  subtended  by  the  chord  ? 

Theorem.  A  radius  which  is  perpendicular  to  a  chord 
bisects  the  chord  and  its  subtended  arc. 


Data:  A  circle  whose  center  is  O;  any 
chord,  as  AB ;  and  a  radius,  as  OD,  perpen- 
dicular to  AB  at  E. 

To  prove     AE  =  BE,  and  arc  AD  =  arc  BD. 


Proof.     Di 

aw 

the  radii  OA  and  OB. 

In  the  rt.  Aaeo  and 

BEO, 

OA  —  OB, 

and 

OE  is  common ; 

.-.  §  123, 

Aaeo  =  Abeo, 

and 

AE  =  BE. 

Also, 

Z.t  =  Av', 

hence,  §  194, 

arc  AD  =  arc  BD. 

Therefore, 

etc 

/ 

Why? 


Why? 


Q.E.D. 

Ex.  182.   If  two  circumferences  intersect,  how  does  the  distance  between 
their  centers  compare  with  the  difference  of  their  radii  ? 


Proposition  V 

199.  1.  Draw  a  chord  of  any  circle  and  a  perpendicular  to  that  chord 
at  its  middle  point.  Determine  whether  the  perpendicular  passes  through 
the  center  of  the  circle. 

2.  Draw  a  line  through  the  center  of  a  circle  perpendicular  to  a  chord; 
How  does  it  divide  the  chord  ?    How  does  it  divide  the  subtended  arc  ? 

3.  Draw  a  circle  and  two  chords  which  are  not  parallel.  Erect  per- 
pendiculars at  their  middle  points  and  produce  these  perpendiculars  until 
they  intersect.  At  what  point  in  the  circle  do  the  perpendicular  bisectors 
of  the  chords  npie^t  ? 


90 


PLANE   GEOMETRY.  — BOOK  II. 


Theorem.    A  line  perpendicular  to  a  chord  at  its  mid- 
dle point  passes  through  the  center  of  the  circle. 


Data:  A  circle  whose  center  is  0;  any 
chord,  as  ^5 ;  and  CD  a  perpendicular  to 
AB  3i,t  its  middle  point. 

To  prove  that  CD  passes  through  0. 


Proof.     §  173,  A  and  B  are  equally  distant  from  0, 
data,  CD  is  the  perpendicular  bisector  of  AB ; 

hence,  §  104,  0  must  lie  in  CD  ; 

tfiat  is,  CD  passes  through  0. 

Therefore,  etc.  q.e.d. 

200.  Cor.  I.   A  line  passing  through  the  center  of  a  circle  and 
perpendicular  to  a  chord  bisects  the  chord  and  its  subtended  arc. 

201.  Cor.  II.    The  point  of  intersection  of  the  perpendicular  bi- 
sectors of  two  non-parallel  chords  is  the  center  of  the  circle. 


Proposition  VI 

202.  1.  Draw  a  circle  and  two  equal  chords.  How  do  their  distances 
from  the  center  compare  ? 

2.  Draw  a  circle  and  two  chords  equally  distant  from  the  center.  How 
do  the  chords  compare  in  length  ? 

Theorem,  In  the  same  circle,  or  in  equal  circles,  equal 
chords  are  equally  distant  from  the  center;  cotiversely^ 
chords  equally  distant  from  the  center  are  equal. 


Data :    Any  two  equal  chords,  as  ^£  and 
DEy  in  the  circle  whose  center  is  0. 

To  prove  AB  and  DE  equally  distant  from  0. 


Proof.     Draw  the  radii  OA  and  OD  ;  also  draw  the  perpendicu- 


PLANE   GEOMETRY.  — BOOK  II.  91 

lars  OG  and  OH  representing  the  distances  from  o  to  AB  and  DEj 

respectively. 

Then,  §  200,        AG  =  ^AB,  and  BH  =  ^BE ; 
.-.  Ax.  7,  AG  =BH. 

In  the  rt.  Aaog  and  pOH, 
Ax.  14,  OA  =  OBf 

AG  =  BH', 

.'.  §  123,  Aaog  =  Aboh, 

and  OG  =  OH',  .  Why? 

that  is,  AB  and  i)j&  are  equally  distant  from  O.  q.e.d. 

Conversely:  Data:    Any  two  chords,  as  AB   and  BE,  equally 
distant  from  0,  the  center  of  the  circle. 
•     To  prove  AB  =  BE. 

Proof.     In  the  rt.  ^AOG  and  BOH, 
data,  OG  =  OH, 

Ax.  14,  O^  =  02) ; 

.-.  §  123,  Aaog  =  Aboh, 

and  AG  =  BH.  Why  9 

But,  §  200,  AG  =  \AB,  and  BH  =  ^BE; 

hence.  Ax.  6,  ^£  =  D^. 

Therefore,  etc.  q.e.d 

Ex.  183.    In  how  many  points  can  a  straight  line  cut  a  circumference  ? 

Ex.  184.    How  many  centers  may  a  circle  have  ? 

Ex.  185.  If  a  straight  line  bisects  a  chord  and  its  subtended  arc,  what  is 
its  direction  with  reference  to  the  chord  ? 

Ex.  186.  Do  the  perpendicular  bisectors  of  the  sides  of  an  inscribed 
quadrilateral  meet  in  a  common  point? 

Ex.  187.  If  a  diameter  of  a  circle  bisects  a  chord,  how  does  it  divide 
the  subtended  arc  ?  In  what  direction  does  it  extend  with  reference  to  the 
chord  ? 

Ex,  188.  If  a  diameter  of  a  circle  bisects  an  arc,  how  does  it  divide  the 
chord  of  the  arc  ?    What  is  its  direction  with  reference  to  the  chord  ? 

Ex.  189.  If  the  distance  from  the  center  of  a  circle  to  a  straight  line  is 
less  than  the  radius,  will  the  line  cut  the  circumference  ?  If  the  distance 
is  greater  than  the  radius,  will  the  line  cut  the  circumference  ? 


92 


PLANE   GEOMETRY.  — BOOK  IL 


Proposition  VII 

203.  Draw  two  unequal  chords  in  the  same  circle,  or  in  equal  circles. 
Which  chord  is  nearer  the  center,  the  longer  or  the  shorter  one? 

Theorem,  In  the  same  circle,  or  in  equal  circles,  the 
less  of  two  unequal  chords  is  at  the  greater  distance  from 
the  center ^ 


Data :  In  the  equal  circles  whose  centers  are  0  and  P,  any  two 
ciiords,  as  AB  and  BE^  of  which  BE  is  the  less. 

To  prove  BE  at  a  greater  distance  from  P  than  AB  is  from  0. 

Proof.  Draw  the  radii  OA^  OB,  PB,  and  PE\  also  draw  the 
perpendiculars  OG  and  PH  representing  the  distances  from  0  to 
AB,  and  from  P  to  BE,  respectively. 

Data,  AB  >  BE, 

§200,  AG=\AB,2iTidLBH  =  ^BE\ 

AG>BH. 

Then,  take  AK  equal  to  BH,  and  draw  OK. 

In  the  isosceles  A  ABO  and  BEP, 
§  130,  /.AOBi^  greater  than  Z  BPE'^ 

Z  PBE  is  greater  than  Z  OAB, 

Then,  in  Abhp  and  AKO,  BP  =  AO, 
const.,  BH  =  AK, 

and  Z  PBH  is  greater  than  ZOAK; 

PH>OK', 
but  OK>OG', 

PH>OG', 
that  is,  BE  is  at  a  greater  distance  from  P  than  ^P  is  from  0. 

Therefore,  etc.  Q.e.d 


Why? 
Why? 


Why? 
Why? 


PLANE   GEOMETRY.  — BOOK  IL  93 

Prbposition  VIII 

204.  In  the  same  circle,  or  in  equal  circles,  draw  two  chords  unequally 
distant  from  the  center.     Which  one  is  the  shorter  ? 

Theorem.  In  the  same  circle,  or  in  equal  circles,  of  two 
chords  unequally  distant  from  the  center,  the  one  at  the 
greater  distance  is  the  less.  (Converse  of  Prop.  VII.) 

D 

Data:    In  the  circle  whose  center  is  0,        /^       \.h\ 
any  two  chords,  as  AB  and  DE,  of   which      /  /^^\\ 

DE  is  at  the  greater  distance  from  0.  of  Ne 

To  prove  DE  less  than  AB.  ^V  g~         y^ 

Proof.  Draw  the  perpendiculars  OG  and  OH  representing  the 
distances  from  0  to  AB  and  DE,  respectively. 

Now,  if  DE  =  ABj 

then,  §  202,  OH  =  OG,  which  is  contrary  to  data. 

If  DE  >  AB, 

then,  AB  <  DE, 

and,  §  203,  OG  >  OH,  which  is  also  contrary  to  data. 

Then,  since  DE  is  neither  equal  to,  nor  greater  than  AB, 

DE  is  less  than  AB. 
Therefore,  etc.  q.e.d. 

Ex.  190.  Where  does  the  line  drawn  through  the  middle  points  of  two 
parallel  chords  in  a  circle  pass  with  reference  to  the  center  ? 

Ex.  191.  If  two  circles  are  concentric,  how  do  any  two  chords  of  the 
greater,  which  are  tangent  to  the  less,  compare  in  length  ? 

Ex.  192.  If  an  isosceles  triangle  is  constructed  on  any  chord  of  a  circle 
as  its  base,  where  does  the  vertex  lie  with  reference  to  the  diameter  that  is 
perpendicular  to  the  chord,  or  to  that  diameter  produced  ? 

Ex.  193.  If  two  chords  of  a  circle  cut  each  other  and  make  equal  angles 
with  the  straight  line  which  joins  their  point  of  intersection  with  the  center, 
how  do  the  chords  compare  in  length  ? 

Ex.  194.  If  from  any  point  within  a  circle  two  equal  straight  lines  are 
drawn  to  the  circumference,  where  will  the  bisector  of  the  angle  thus  formed 
pass  with  reference  to  the  center  of  the  circle  ? 


y4  PLANE   GEOMETRY.  — BOOK  II. 

Proposition  IX 

205.  1.  Draw  a  circle  and  one  of  its  radii;  also  a  line  perpendicular 
to  the  radius  at  its  extremity.     Is  this  line  a  tangent  or  a  secant  ? 

2.  Draw  a  tangent  to  a  circle  and  a  radius  to  the  point  of  contact. 
What  kind  of  an  angle  is  formed  by  these  lines? 

Theorem,'  A  line  perpendicular  to  a  radius  at  its  ex- 
tremity is  tangent  to  the  circle;  conversely,  a  tangent  is 
perpendicular  to  the  radius  drawn  to  the  point  of  contact. 

Data  :  A  circle  whose  center  is  O ; 
any  radius,  as  0D\  and  a  line  AB 
perpendicular  to  OD  at  D. 

To  prove  AB  tangent  to  the  circle. 


Proof.     From  0  draw  any  other  line  to  AB^  as  OE. 

Then,  OD  <  OE.  Why  ? 

Since  every  point  in  the  circumference  is  at  a  distance  equal  to 
oj)  from  the  center,  and  jK  is  at  a  greater  distance,  E  is  without 
the  circumference. 

Therefore,  every  point  of  AB,  except  D,  is  without  the  circum- 
rerence. 

Hence,  §  184,  AB  is  tangent  to  the  circle  at  D.  q.e.d. 

Conversely :  Data :  Any  tangent  to  this  circle,  as  AB,  and  the 
fadius  drawn  to  the  point  of  contact,  as  OD. 

To  prove  AB  perpendicular  to  OD. 

Proof.  Ax.  17,  every  point  of  AB,  except  D,  is  without  the 
circumference. 

.*.  OD  is  the  shortest  line  that  can  be  drawn  between  0  and  AB. 

Hence,  §  61,  OD  ±  AB. 

Therefore,  etc.  q.e.d. 

Ex.  195.  If  in  a  circle  a  chord  is  perpendicular  to  a  radius  at  any  point, 
how  does  it  compare  in  length  with  any  other  chord  which  can  be  drawn 
through  that  point  ? 

Ex.  196.  If  tangents  are  drawn  through  the  extremities  of  a  diameter, 
what  is  their  direction  with  reference  to  each  other  ? 


PLANE   GEOMETRY.  — BOOK  II, 
Proposition  X 


95 


206.  1.  Draw  a  circle,  a  tangent  to  it,  and  a  chord  parallel  to  the  tan- 
gent. How  do  the  arcs  intercepted  between  the  point  of  tangency  and 
the  extremities  of  the  chord  compare? 

2.  Draw  a  circle  and  two  parallel  secants  "or  chords.  How  do  the 
intercepted  arcs  compare?  * 

Theorem.  Parallel  lines  intercept  equal  arcs  on  a  cir- 
cumference. 

Data :  A  circle  whose  center  is  O,  and  any     a 
two  parallel  lines,  as  AB  and  CDj  intercepting 
arcs  on  the  circumference. 

To  prove  that  the  arcs  intercepted  hj  AB 
and  OD  are  equal. 

Proof.     Case  I.    WJien  AB  is  a  tangent  and  CD  is  a  chord. 

Draw  to  the  point  of  tangency  the  radius  OE. 

Then,  §  205,  OE±AB; 

.'.  §  72,  oi:±CD', 

hence,  §  198,  arc  CE  =  arc  DE. 

Case  II.      When  both  AB  and  CD  are  chords. 

Draw  EF  II  AB,  and  tangent  to  the  circum-    ^. g 

ference  at  G. 

Then,  §  80,  EF  II  CD, 

Case  I,  arc  CG  =  arc  DG, 

ind  SiTC  AG  —  arc  BG ; 

.-.  Ax.  3,  arc  CA  =  arc  DB. 

Case  III.      When  AB  and  CD  are  tangents  as 
at  G  and  H  respectively. 

Draw  the  chord  EF  II  AB. 

Then,  §  80,  EF  II  CD, 

Case  I,  arc  EH  =  arc  FH, 

and  arc  Ji^G'  =  arc  FG ; 

.-.  Ax.  2,  arc  HEG  =  arc  HFG, 

Therefore,  etc.  O.B.D. 


96  PLANE   GEOMETRY.  — BOOK  II. 

Proposition  XI 

207.  Select  three  points  not  in  the  same  straight  line.  How  many 
circumferences  can  be  passed  through  them  ? 

Theorem.  Through  three  points  not  in  the  same  straight 
line  one  circumference  can  he  drawn,  and  only  one. 

Data:  Any  three  points  not  in  the  same 
straight  line,  as  A,  B,  and  C.  y^'     ~~""n 

To  prove  that  one  circumference  can  be      /  N 

drawn  through  A^  B,  and  C,  and  only  one.      ;  ^^^Jp       ^.^-fp 

Proof.    Draw  AB,  BC,  and  AC-,  and  at  their     \  j-V'' >/''' 

middle  points,  E,  F,  and  G,  respectively,  erect      \  ,.-''''  ; 
perpendiculars.  ^  ^^--~.£l""^  "^ 

§  170,  these  perpendiculars  meet  in  a  point,  as  0,  which  is 
equidistant  from  A,  B,  and  C; 

.'.  §  173,  a  circumference  described  from  O  as  a  center,  and 
with  a  radius  equal  to  the  distance  OA,  passes  through  the  points 
Ay  B,  and  C;  and,  since  the  perpendiculars  intersect  in  but  one 
point,  there  can  be  but  07ie  center,  and  consequently  but  one  cir- 
cumference passing  through  the  points  A,  B,  and  C.  q.e.d. 

208.  Cor.  I.     Circles  circumscribing  equal  triangles  are  equal. 
Cor.  II.     Two  circumferences  can  intersect  in  only  two  points.     If 

two  circumferences  have  three  points  in  common,  they  coincide 
and  form  one  circumference. 

Proposition  XII 

209.  Draw  a  circle  and  from  a  point  outside  draw  two  tangents. 
How  do  the  tangents  compare  in  length? 

Theorem.  The  tangents  drawn  to  a  circle  from  a  point 
without  are  equal. 

Data:  A  circle  whose  center  is 
0 ;  any  point  without  it,  as  A ;  and 
AB  and  AC  the  tangents  to  the 
circle  at  the  points  B  and  C,  respec- 
tively. 

To  prove     AB  =  AC. 


PLANE   GEOMETRY.  —  BOOK  II  -     97 

Proof.     Draw  OAy  OB,  and  OC. 

§  205,  A  B  and  c  are  rt.  A 

Then,  in  the  rt.  A  OAB  and  OAC, 

OB  =  0(j,  Why? 

and  .  OA  is  common; 

A  OAB  =  A  OAC,  Why? 

and  AB  =  ^C» 

Therefore,  etc.  '  q.e.d. 

210.  The  line  which  joins  the  centers  of  two  circles  is  called 
their  line  of  centers. 

211.  A  common  tangent  to  two  circles  which  cuts  their  line  of 
centers  is  called  a  common  interior  tangent;  one  which  does  not 
cut  their  line  of  centers  is  called  a  common  exterior  tangent. 

Proposition  XIII 

212.  Draw  two  intersecting  circles  and  a  chord  that  is  common  to 
both.  What  kind  of  an  angle  does  a  line  joining  their  centers  make 
with  this  common  chord  ? 

Theorem.  If  the  circumferences  of  two  circles  intersect, 
the  line  of  centers  is  perpendicular  to  the  common  chord 
at  its  middle  point. 

Data :  Two  circles  whose  centers 
are  O  and  P,  whose  circumferences 
intersect  at  A  and  5;  AB  the  common 
chord ;  and  OP  the  line  of  centers. 

To  prove  OP  perpendicular  to  AB 
at  its  middle  point. 

Proof.     §  173,     P  is  equidistant  from  A  and  B, 
and  also,  0  is  equidistant  from  A  and  B  ; 

.*.  §  104,    both  P  and  0  lie  in  the  perpendicular  bisector  of  AB. 

Hence,  Ax.  11,  OP  coincides  with  this  perpendicular  bisector  i 
that  is,  OP  ±  AB  at  its  middle  point. 

Therefore,  etc.  Q.B.D 

milne's  geom.  — 7 


98 


PLANE   GEOMETRY.  — BOOK  IL 


Proposition  XIV 

213.    Draw  two  circles  tangent  to  each  other,  and  a  line  joining  their 
centers.     Through  what  point  will  this  line  pass? 

Theorem,    If  two  circles  are  tangent  to  each  other,  their 
line  of  centers  passes  through  the  point  of  contact. 


Data :  The  two  tangent  circles 
whose  centers  are  0  and  P ;  OP 
their  line  of  centers;  and  C 
their  point  of  contact. 

To    prove    that    OP 
through  C 


Proof.     At  C  draw  the  common  tangent  AB. 


Ax.  17, 

.'.  §  205, 
also, 

hence,  §  68, 
that  is, 
Therefore,  etc. 


C  lies  in  both  circumferences; 

if  radius  PC  is  drawn,  PC  A.ABf 

if  radius  OC  is  drawn,  OC  l.AB\ 

A  OCA  +ZPCA  =2rt.  ^; 

OCP  is  a  straight  line; 

OP  passes  through  C. 

MEASUREMENT 


Why 


Q.E.D. 


214.  The  theorems  thus  far  presented  and  proved  have  usually- 
established  only  the  equality  or  inequality  of  two  magnitudes, 
but  it  is  sometimes  desirable  to  measure  accurately  the  magni- 
tudes that  are  given. 

A  magnitude  is  measured  when  we  find  how  many  times  it 
contains  another  magnitude  of  the  same  kind,  called  the  unit  of 
measure. 

The  number  which  expresses  how  many  times  a  magnitude  con- 
tains a  unit  of  measure  is  called  its  numerical  measure. 

215.  The  relation  of  two  magnitudes  which  is  determined  by 
finding  how  many  times  one  contains  the  other,  or  what  part  one 
is  of  the  other,  is  called  their  Ratio. 


PL4NE   GEOMETRY.— BOOK  11.  99 

The  ratio  of  two  magnitudes  is  the  ratio  of  their  numerical 
measures.     It  may  be  either  an  integer  or  a  fraction. 

The  ratio  of  a  line  12  ft.  long  to  one  4  ft.  long  is  3  ;  that  is,  the  12  ft.  line 
is  three  times  the  4  ft.  line.  Also  the  ratio  of  an  angle  of  10°  to  an  angle  of 
60°  is  ^ ;  that  is,  an  angle  of  10°  is  one  sixth  of  an  angle  of  60°. 

216.  Two  magnitudes  of  the  same  kind  which  contain  a  com- 
mon unit  of  measure  an  integral  number  of  times  are  called  Com- 
mensurable Magnitudes. 

217.  Two  magnitudes  of  the  same  kind  which  have  no  common 
unit  of  measure  are  called  Incommensurable  Magnitudes. 

218.  The  ratio  of  incommensurable  magnitudes  is  called  an 
mcommensurahle  ratio  ;  that  is,  it  cannot  be  expressed  exactly  by 
numbers;  and  yet  we  can  approximate  to  the  exact  numerical 
value  as  nearly  as  we  please. 

Thus,  if  the  side  of  a  square  is  1  ft.  in  length,  the  diagonal  is 
V2  ft.  in  length,  as  will  be  shown  later,  and  the  ratio  of  the 

diagonal  to  the  side  of  the  square  is  —— • 

Now,  no  integer  or  mixed  number  can  be  found  which  .is  ex- 
actly equal  to  V2^  but  by  expressing  the  square  root  of  2  in  a 
decimal  form,  the  ratio  can  be  determined  within  a  fraction  as 
small  as  we  please. 

Thus,  V2  =  1.414213+  ;  that  is,  V2  lies  between  1.414213  and 
1.414214;  therefore,  the  ratio  of  the  diagonal  of  a  square  to  its 

«5iflp  llP<5  hptwPPTl    1414213    o-pfl    1414214. 

blue  neb  uetweeu  yTycroTO^TT  '*'""•  T"o7orTyTo 

That  is,  if  the  one-millionth  part  of  a  foot  be  assumed  approxi- 
mately as  the  common  unit  of  measure,  the  side  of  the  square 
will  be  equal  to  1,000,000,  and  the  diagonal  will  be  equ^l  to  be- 
tween 1,414,213  and  1,414,214  such  units. 

By  continuing  the  process  of  finding  the  square  root  we  can 
approximate  as  closely  to  the  actual  ratio  as  we  please,  or  until 
the  fraction  contains  an  error  so  small  that  it  may  be  disregarded, 
though  it  cannot  be  eliminated. 

It  is  evident,  therefore,  that  the  ratio  of  two  magnitudes  of  the 
same  kind,  even  when  they  are  incommensurable,  may  be  obmined 
to  any  required  degree  of  precision. 


100  PLANE    GEOMETRY.  — BOOK  II. 

To  generalize :  suppose  Q  to  be  divided  into  n  equal  parts,  and 
that  P  contains  m  of  these  parts  with  a  remainder  less  than  one 

of  the  parts  ;  then,   -  =  —  within  less  than  — 
^  Q      71  n 

Since  n  may  be  taken  as  large  as  we  please,  -  may  be  made 

n 
less  than  any  assigned  measure  of  precision,  and,  consequently, 

the  value  of  —  may  be  regarded  as  the  approximate  value  of  the 

.    p     .     .^ 
ratio  —  within  any  assigned  degree  of  precision. 

THEORY   OF  LIMITS 

219.  A  magnitude  which  remains  unchanged  throughout  the 
same  discussion  is  called  a  Constant. 

220.  A  magnitude  which,  under  the  conditions  imposed  upon 
it,  may  have  an  indefinite  number  of  different  successive  dimen- 
sions is  called  a  Variable. 

221.  When  a  variable  increases  or  decreases  so  that  the  dif- 
ference between  it  and  a  constant  may  be  made  as  small  as  we 
please,  but  cannot  be  made  absolutely  equal  to  zero,  the  constant 
is  called  the  Limit  of  the  variable,  and  the  variable  is  said  to 
approach  this  limit. 

Suppose,  for  example,  that  a  point  moves  from  ^  to  P  under 

the    condition    that   it   shall    ^ , ,       \   \    b 

move   one  half  the   distance  ^  d     e  f 

during  the  first  interval  of  time ;  one  half  the  remaining  distance 
the  second  interval;  one  half  the  distance  still  remaining  the 
next  interval;  and  so  on.  The  distance  from  A  to  the  moving 
point  is  an  increasing  variable,  which  approaches  the  distance  AB 
as  a  limit,  though  it  cannot  actually  reach  it.  Also,  the  distance 
from  B  to  the  moving  point  is  a  decreasing  variable,  which  ap- 
proaches zero  as  a  limit,  though  it  cannot  actually  reach  it. 

Again,  the  sum  of  the  descending  series  1,  \,  \,  \,  -jV)  ^V'  "^V? 
etc.,  approaches  2  as  a  limit,  but  never  quite  reaches  it.  The  sum 
of  the  first  two  terms  is  1|,  of  the  first  three  terms  If,  of  the  first 
four  terms  IJ,  etc.  It  is  evident  that  the  sum  approaches  2,  and 
that,  by  taking  terms  enough,  it  may  be  made  to  differ  from  2  by 


PLANE   GEOMETRY.  — BOOK  11.  101 

as  small  a  quantity  as  we  please,  but  it  cannot  be  actually  equal 
to  2.  That  is,  the  sum  of  the  series  approaches  the  limit  2  as 
th6  number  of  terms  is  increased. 

For  further  illustration,  if,  in  the  right  triangle  ABC,  the  ver- 
tex C  indefinitely  approaches  the  vertex  B,  the 
angle  A  diminishes  and  indefinitely  approaches  zero 
as  its  limit,  and  if  it  should  actually  become  zero, 
the  triangle  would  vanish  and  become  the  straight 
line  AB.  Again,  if  the  vertex  C  indefinitely  moves 
away  from  the  vertex  B,  the  angle  A  increases 
and  indefinitely  approaches  a  right  angle  as  its 
limit,  and  if  it  should  actually  become  a  right  angle 
the  triangle  would  vanish  and  AC  and  BC  would  be  two  parallel 
lines,  each  perpendicular  to  AB.  Hence,  the  value  of  angle  A 
lies  between  the  limits  zero  and  a  right  angle,  but  it  can  never 
actually  reach  either  limit  so  long  as  the  triangle  exists. 

Proposition  XV 

222.  What  is  a  variable?  What  is  the  limit  of  a  variable?  If  two 
variables  are  always  equal,  how  do  their  limits  compare  ? 

Theorem.  If,  while  approaching  their  respective  limits, 
two  variables  are  always  equal,  their  limits  are  equal. 

Data :  Two  equal  variables, 

as  ^a;  and  Cy,  which  approach    a- +  -^      ^ 

the  limits  AB  and  CD,  respec- . 

tively.  c + ^ 

To  prove  AB  and  CD  equal. 

Proof.  Suppose  that  AB  is  greater  than  CD ;  then  some  part 
of  AB,  as  Az,  is  equal  to  CD. 

Then,  the  variable  Ax  may  have  values  between  Az  and  AB ; 
that  is,  the  variable  Ax  may  have  values  greater  than  CD,  while 
the  variable  Cy  cannot  have  values  greater  than  CD. 

This  is  contrary  to  the  condition  that  the  variables  are  equal. 

Hence,  AB  cannot  be  greater  than  CD. 

In  like  manner  it  can  be  proved  that  AB  cannot  be  less  than  CD, 

Consequently,  AB  =  CD. 

Therefore,  etc.  Q-e.d. 


102 


PLANE   GEOMETRY.^ BOOK  IL 


Proposition  XVI 

223.  In  the  same  circle,  or  in  equal  circles,  draw  two  central  angles 
such  that  the  arc  intercepted  by  the  sides  of  the  first  shall  be  three  times 
the  arc  intercepted  by  the  sides  of  the  second.  How  does  the  first  angle 
compare  in  size  with  the  second?  How  does  the  ratio  of  the  central 
angles  compare  with  the  ratio  of  the  arcs  intercepted  by  their  sides  ? 

Theorem,  In  the  same  circle,  or  in  equal  circles,  two 
central  angles  have  the  same  ratio  as  the  arcs  intercevted 
by  their  sides. 


M 


Data:  In  the  equal  circles  whose  centers  are  0  and  P,  any 
tjcntral  angles,  as  AOB  and  DPE,  whose  sides  intercept  the  arcs 
AB  and  DE  respectively. 


To  prove 


ratio 


Z  AOB 


ratio 


arc  ^5 


Z  DPE  arc  DE 

Proof.     Case  I.    When  the  arcs  are  commensurable. 
Suppose  that  M  is  the  common  unit  of  measure  for  the  two  arcs, 
and  that  it  is  contained  in  arc  AB  7  times  and  in  arc  DE  4  times. 


Then, 


tatio 


arc  AB 


ratio 


arc  DE  4 

Divide  the  arcs  AB  and  DE  into  parts  each  equal  to  the  common 
measure  M,  and  to  each  point  of  division  draw  a  radius. 

By  §  194,  each  of  these  central  angles  formed  by  any  two 
adjacent  radii  is  equal  to  every  other  central  angle  so  formed. 

The  number  of  equal  parts  into  which  the  angles  A  OB  and  DPE 
are  divided  by  these  radii  is  equal  to  the  number  of  times  M  is 
contained  in  the  arcs  AB  and  DE  respectively. 


ratio 


Z  AOB 
Z  DPE 


ratio 


Hence,  Ax.  1, 


, .     Z  AOB  ,. 

ratio  — =  ratio 

^DPE 


4 

arc  AB 

SiVC  DE 


PLANE   GEOMETRY.  — BOOK  II. 
Case  II.    When  the  arcs  are  incommensurable. 


103 


M 


bmce  arcs  AB  and  DE  are  incommensurable,  take  these  com- 
mensurable arcs  AG  and  DE,  and  suppose  that  GB  is  less  than  M. 


Then,  Case  I,      ratio 


Z  AOG 


=  ratio 


arc  A  G 


Z  DPE  arc  DE 

1  •  if  is  indefinitely  diminished,  angle  GOB  and  arc  GB  decrease, 

and  the  ratios  — and  — remain  equal,  and  indefinitely 

Z  DPE  arc  DE  "^ 

approach  the  limiting  ratios   ^  ^__  and  ^^^     _  respectively. 


Z  DPE         arc  DE 

Hence,  §  222,      ratio  ^^^  =  ratio  ^^^^. 
Z  DPE  arc  DE 


Q.E.D. 


J224.  It  was  stated  in  §  35  that  the  total  angular  magnitude  about 
A  point  in  a  plane  is  divided  into  degrees,  minutes,  and  seconds.  In 
the  same  way  the  circumference  of  a  circle  is  divided  into  360 
equal  arcs,  called  arc  degrees,  or  simply  degrees;  each  arc  degree 
is  divided  into  60  equal  parts,  called  minutes;  and  each  arc  minute 
into  60  equal  parts,  called  seconds. 

Thus  it  will  be  seen  that  the  angle  degree  and  the  arc  degree, 
the  angle  minute  and  the  arc  minute,  the  angle  second  and  the 
arc  second  correspond,  each  to  each.  The  sides  of  a  central  angle 
of  one  degree  therefore  intercept  an  arc  of  one  degree,  the  sides 
of  a  central  angle  of  ten  degrees  intercept  an  arc  of  ten  degrees, 
and,  in  general,  the  sides  of  a  central  angle  of  any  number  of 
degrees  intercept  an  arc  of  an  equal  number  of  degrees. 

Therefore,  a  central  angle  is  measured  by  its  intercepted  arc; 
that  is,  the  central  angle  contains  between  its  sides  the  same  propor- 
tion of  the  total  angular  magnitude  about  a  point  m  a  plane,  that 
the  arc  intercepted  by  its  sides  is  of  the  whole  circumference. 


104  PLANE   (JEOMETRY.-^BOOK  IL 

Proposition  XVII 

225.  1.  Draw  a  circle  and  an  inscribed  angle  one  of  whose  sides  is  a 
diameter;  draw  a  radius  to  the  extremity  of  the  other  side.  How  does 
the  inscribed  angle  compare  in  size  with  the  central  angle  subtended  by 
the  same  arc?  Since  the  central  angle  is  measured  by  the  arc  which 
subtends  it,  by  what  part  of  the  arc  is  the  inscribed  angle  measured? 

2.  Draw  other  inscribed  angles  no  one  of  whose  sides  is  a  diameter 
and  draw  a  diameter  from  the  vertex  of  each.  By  what  part  of  its  arc 
is  each  inscribed  angle  measured? 

3.  If  inscribed  angles  are  subtended  by  the  same  arc  or  chord  or  are 
inscribed  in  the  same  segment,  how  do  they  compare  in  size? 

4.  How  many  degrees  are  there  in  a  semicircumference?  Whai^ 
then,  will  be  the  size  of  all  angles  inscribed  in  a  semicircle  ? 

5.  How  does  an  angle  inscribed  in  a  segment  greater  than  a  semicircle 
compare  with  a  right  angle?    How,  if  in  one  less  than  a  semicircle? 

Theorem.  An  inscribed  angle  is  measured  hy  one  half 
the  arc  intercepted  hy  its  sides. 


Data :   A  circle  whose  center  is  O,  and  any- 
inscribed  angle,  as  ABO,  ^ 
To  prove  angle  ABC  measured  by  ^  arc  AC. 


Proof.     Case  I.    When  one  side  of  the  angle  is  a  diameter  of  the 
circle. 

When  AB  is  a,  diameter. 

Draw  the  radius  00, 


Then, 

OB    =    OOy 

Why? 

§90, 

A  5  oc  is  isosceles, 

and 

Zb  =  Zc. 

Why? 

§115, 

ZAOC—  Zi?  +  Zc  =  2Z5; 

or 

Zb  =  \Z  AOC. 

But,  §  224, 

Z  AOC  is  measured  by  arc  AO\ 

hence^ 

Z  jB  is  measured  by  \  arc  AC, 

PLANE   GEOMETRY.  — BOOK  IL 


106 


Case  II.    When  the  diameter  from  the  vertex  of  the  angle  lies 
between  the  sides. 

Draw  the  diameter  BB. 

Case  I.,     Z  ABB  is  measured  by  ^  arc  AD, 
and  Z  CBB  is  measured  by  i  arc  CB ; 

but,  Ax.  9,  Z  ABB  +  Z  CBB  =Z  ABC, 

and  arc  JZ)  +  arc  CB  =  arc  ^C; 

hence,  Z  ^4^8  (7  is  measured  by  |- arc  ^C. 

Case  III.    When  both  sides  of  the  angle  are  on  the  same  side  of 
the  diameter  from  the  vertex. 

Draw  the  diameter  BB. 

Case  I.,    Z  ABB  is  measured  by  ^  arc  AD, 
and  Z  CBB  is  measured  by  ^  arc  CB ; 

but  Z  ABB  —  Z  C5D  =  Z  ^5C, 

and  arc  ^D  —  arc  CD  =  arc  AC, 

hence,  Z  ^5C7  is  measured  by  ^  are  AC. 

Therefore,  etc. 


Q.E.D. 


226.  Cor.  I.  Angles  inscribed  in  the  same 
segment  of  a  circle,  or  in  equal  segments  of  the 
same  circle,  or  of  equal  circles  are  equal. 


227.   Cor.  II.    An  angle  inscribed  in  a  semi- 
circle is  a  right  angle. 


228.  Cor.  III.    An  angle  inscribed  in  a  segment  greater  than  a 
semicircle  is  less  than  a  right  angle. 

229.  Cor.  IV.    An  angle  inscribed  m  a  segment  less  than  a  semi- 
circle is  greater  than  a  right  angle. 


106  PLANE   GEOMETRY.  —  BOOK  II. 

Proposition  XVIII 

230.  Draw  a  circle  and  two  intersecting  chords ;  construct  an  inscribed 
angle  equal  to  one  of  the  vertical  angles  thus  formed,  by  drawing  from 
an  extremity  of  one  chord  a  chord  parallel  to  the  other.  How  does  the 
arc  subtending  the  inscribed  angle  compare  with  the  sum  of  the  arcs 
intercepted  by  the  sides  of  the  vertical  angles  ?  What,  then,  is  the  meas- 
ure of  the  angle  formed  by  the  two  intersecting  chords  ? 

Theorem,    An  angle  formed  hy  two  intersecting  chords 

is  measured  hy  one  half  the  sum  of  the  arc  intercepted  by 

its  sides  and  the  arc  intercepted  by  the  sides  of  its  vertical 

an^le. 

c 

Data:    Any  two   intersecting  chords,   as        X^         \    ^\ 
AB  and  CD,  forming  the  angle  r.  a/- ^ — \b 

To  prove  angle  r  measured  by                           I                      \     I 
|(arc^CH-arc£i)).  Jv i)^ 

Proof.     Draw  BE  II  AB. 

Then,  §  76,  Zs^Zr. 

§  225,  Zs  is  measured  by  ^  arc  CAE ; 

but  '  arc  CAB  =  arclc  +  arc  AB, 

and,   §  206,  arc  AB  =  arc  BB  ; 

*  arc  CAB  =  arc  ^C  +  arc  BB  ; 

hence,  Z  5  is  measured  by  ■i-(arc  ^C  -h  arc  BB). 

Consequently,  Z  r  is  measured  by  ^(arc^C  +  arc^D). 

Therefore,  etc.  ^  q.e.d. 

Ex.  197.  Prove  by  Prop.  XVII  that  the  sum  of  the  angles  of  a  triangle 
is  equal  to  two  right  angles. 

Ex.  198.  The  opposite  angles  of  an  inscribed  quadrilateral  are  supple- 
mentary. 

Ex.  199.  If  two  chords  of  a  circle  intersect  at  right  angles,  to  what  is  the 
sum  of  any  pair  of  opposite  arcs  equal  ? 

Ex.  200.  If  one  of  the  equal  sides  of  an  isosceles  triangle  is  the  diameter 
of  a  circle,  the  circumference  bisects  the  base. 

Ex.  201.  If  one  side  of  an  angle  of  a  quadrilateral  inscribed  in  a  circle  is 
produced,  the  exterior  angle  is  equal  to  the  opposite  angle  of  the  quadrilateral. 


PLANE   GEOMETRY.  — BOOK  11.  107 

Proposition  XIX 

231.  1.  At  any  point  in  the  circumference  of  a  circle  form  an  angle 
by  a  tangent  and  a  chord;  construct  vertical  angles  equal  to  this  by 
drawing  through  the  given  chord  a  chord  parallel  to  the  tangent.  How 
does  the  sum  of  the  arcs  subtending  these  vertical  angles  compare  with 
the  arc  intercepted  by  the  sicies  of  the  given  angle  ?  What,  then,  is  the 
measure  of  the  given  angle  ? 

2.  How  does  an  angle  between  a  tangent  and  a  diameter  compare  with 
a  right  angle  ?     What  is  its  arc  measure  ? 

Theorem,  An  angle  formed  hy  a  tangent  and  a  chord 
is  measured  hy  one  half  the  intercepted  arc. 

Data :  Any  tangent,  as  EB,  and  any  X"^^      \^\ 

chord,  as  AGj  forming  with  EB    the  ^/ ^__\   Xi) 

angle  r.  i  \  I 

To  prove  angle  r  measured  by  \  7^ 

iarc^C.  \  / 

Proof.  Draw  any  chord,  as  HD^  parallel  to  EB  and  cutting  AO 
in  F. 

Then,  §  76,  Z.s  =  /.r. 

§  230,         Z  s  is  measured  by  \  (arc  ^iT  +  arc  i)  (7) ; 
but,  §  206,  arc  ^jH"  =  arc  ^D  ; 

hence,    Z  s  is  measured  by  \(diVQ,  AD  +  arc  DC),  or  ^arc^C 

Consequently,     Z  r  is  measured  by  ^  arc  AC. 

Therefore,  etc.  *  q.e.d. 

232.  Cor.  A  right  angle  is  measured  by  one  half  a  semicircum- 
ference. 

Ex.  202.  The  angle  between  a  tangent  to  a  circle  and  a  chord  drawn 
from  the  point  of  contact  is  half  the  angle  at  the  center  subtended  by  that 
chord. 

Ex.  203.  A  line  which  is  tangent  to  the  inner  of  two  concentric  circles, 
and  is  a  chord  of  the  outer  circle,  is  bisected  at  the  point  of  tangency. 

Ex.  204.   The  diagonals  of  a  rectangle  inscribed  in  a  circle  are  diameters 
'     of  the  circle. 

Ex.  205.   The  bisector  of  the  vertical  angle  of  an  inscribed  isosceles 
.     triangle  passes  through  the  center  of  the  circle. 


108  PLANE   GEOMETRY.  — BOOK  IL 

Proposition  XX 

233.  At  any  point  without  the  circumference  of  a  circle  form  an 
angle  between  a  tangent  and  a  secant ;  construct  an  angle  equal  to  this 
by  drawing  from  the  point  of  tangency  a  chord  parallel  to  the  secant. 
How  does  the  arc  intercepted  by  the  sides  of  this  angle  compare  with  the 
diiference  of  the  arcs  intercepted  by  the  sides  of  the  given  angle  ?  Whatj 
then,  is  the  measure  of  the  given  angle  ? 

Theorem,  An  angle  formed  by  a  tangent  and  a  secant 
which  meet  without  a  circumference  is  measured  hy  one 
half  the  difference  of  the  intercepted  arcs. 

Data:  Any  tangent,  as  AB,  and  any- 
secant,  as  AC,  meeting  AB  without  the   ^ 
circumference  and  forming  witli  it  the 
angle  A. 

To  prove  angle  A  measured  by 
^  (arc  DF  —  arc  DE), 

c 

Proof.     From  D,  the  point  of  tangency,  draw  DM  11  AG. 
Then,  §  76,  Zr=ZA. 

§  231,  Z  r  is  measured  by  ^  arc  DH; 

but  arc  DH  =  arc  DF  —  arc  HF, 

and,  §  206,  arc  HF=SiVC  de  ; 

arc  DH  =  arc  DF  —  arc  DE ; 

hence,  Z  r  is  measured  by  |  (a^c  DF  —  arc  DE). 

Consequently,  Z  ^  is  measured  by  ^  (arc  DF  —  arc  DE), 
Therefore,  etc.  q.e.d. 

Ex.  206.  Two  chords  perpendicular  to  a  third  chord  at  its  extremities 
are  equal. 

Ex.  207.  If  a  quadrilateral  is  inscribed  in  a  circle  and  its  diagOKdls  are 
drawn,  the  diagonals  will  divide  the  angles  of  the  quadrilateral  so  that  there 
will  be  four  pairs  of  equal  angles. 

Ex.  208.  If  from  the  center  of  a  circle  a  perpendicular  is  drawn  to  either 
side  of  an  inscribed  triangle,  and  a  radius  is  drawn  to  one  end  of  this  side, 
the  angle  between  the  radius  and  the  perpendicular  is  equal  to  the  opposite 
angle  of  the  triangle. 


PLANE   GEOMETRY.  — BOOK  II. 


109 


Proposition  XXI 

234.  At  any  point  without  the  circumference  of  a  circle  form  an  angle 
between  two  secants ;  construct  an  inscribed  angle  equal  to  this  by  draw- 
ing from  the  point  of  intersection  of  either  secant  and  the  circumference 
a  chord  parallel  to  the  other  secant.  How^  does  the  arc  subtending  the 
inscribed  angle  compare  with  the  difference  of  the  arcs  intercepted  by 
the  secants  ?    What,  then,  is  the  measure  of  the  given  angle  ? 

Theorem,  An  angle  formed  hy  two  secants  which  meet 
without  a  cirjcumference  is  measured  hy  one  half  the  differ- 
ence of  the  intercepted  arcs. 


Data :  Any  two  secants,  as  AB  and 
AC,  meeting  without  the  circumference 
and  forming  the  angle  A. 

To  prove  angle  A  measured  by 
^  (arc  ^C  —  arc  DE). 


Proof.     Draw  the  chord  DF  W  AC. 

Then,  §  76,  Zr  =  /.A. 

§  225,  Z  r  is  measured  by  \  arc  BF ; 

but  arc  BF  =  arc  BC  —  arc  FC, 

and,  §  206,  arc  FC  =  arc  DE ; 

arc  5i^  =  a-rc  5(7  —  arc  Z)^ ; 
hence,  Z  r  is  measured  by  ^  (arc  BC  —  arc  DE). 

Consequently,  Z  ^  is  measured  by  ^  (arc  BC  —  arc  DE). 

Therefore,  etc.  q.e.d. 


Ex.  209.  The  tangent  at  the  vertex  of  an  inscribed  equilateral  triangle 
forms  equal  angles  with  the  adjacent  sides. 

Ex.  210.  The  angle  between  two  tangents  from  the  same  point  is  24°  15'. 
How  many  degrees  are  there  in  each  of  the  intercepted  arcs  ? 

Ex.  211.  Ond  angle  of  an  inscribed  triangle  is  38°,  and  one  of  its  sides 
subtends  an  arc  of  124°.     What  are  the  other  angles  of  the  triangle  ? 

Ex.  212.  AB^  a  chord  of  a  circle,  is  the  base  of  an  isosceles  triangle 
whose  vertex  C  is  without  the  circle,  and  whose  equal  sides  intersect  the 
circumference  at  D  and  E.    Prove  that  CD  is  equal  to  CE. 


110  PLANE   GEOMETRY.  — BOOK  IL 

Proposition  XXII 

235.  At  any  point  without  the  circumference  of  a  circle  form  an 
angle  between  two  tang'ents;  construct  an  angle  equal  to  this  by  drawing 
from  the  point  of  contact  of  either  tangent  a  chord  parallel  to  the  other. 
How  does  the  arc  intercepted  by  the  sides  of  this  angle  compare  with 
the  difference  of  the  arcs  intercepted  by  the  tangents?  What,  then,  is 
the  measure  of  the  given  angle? 

Theorem.  An  angle  formed  by  two  tangents  is  -measured 
by  one  half  the  difference  of  the  intercepted  arcs. 


Data :   Any  two  tangents,  as  AB 
and  AC,  forming  the  angle  A. 
To  prove  angle  A  measured  by 
i  (arc  DFE  —  arc  DE), 


Proof.     From  D,  the  point  of  tangency,  draw  the  chord  DF\\  AC. 

Then,  §  76,  /.r=  Z.A.     . 

§  231,  Z  r  is  measured  by  ^  arc  DF ; 

but  arc  DF  —  arc  BFE  —  arc  FE^ 

and,  §  206,  arc  FE  =  arc  DE ; 

arc  BF  —  arc  BFE  —  arc  bE\ 
hence,  Z  r  is  measured  by  J  (arc  BFE  -  sltcBE). 

Consequently,     Z  ^  is  measured  by  ^  (arc  BFE  —  arc  T)E). 

Therefore,  etc.  q.e.d. 

Ex.  213.  In  any  quadrilateral  circumscribing  a  circle  any  pair  of  opposite 
sides  is  equal  to  half  the  perimeter  of  the  quadrilateral. 

Ex.  214.  If  three  circles  touch  each  other  externally,  ahd  the  three  com- 
mon tangents  are  drawn,  these  tangents  meet  in  a  point  equidistant  from 
the  points  of  contact  of  the  circles. 

Ex.  215.  If  two  triangles  are  inscribed  in  a  circle,  and  two  sides  of  one 
are  parallel,  each  to  each,  to  two  sides  of  the  other,  the  third  sides  are  equal. 


PLANE   GEOMETRY.-- BOOK  IL  111 

Ex.  i216.    Every  parallelogram  inscribed  in  a  circle  is  a  rectangle. 

Ex.  217.  Two  sides  of  an  inscribed  triangle  subtend  \  and  \  of  the  cir« 
cumference,  respectively.     What  are  the  angles  of  the  triangle  ? 

Ex.  218.  If  a  circle  is  circumscribed  about  the  triangle  ABC,  and  a  line  is 
drawn  bisecting  angle  A  and  meeting  the  circumference  in  Z>,  angle  BCB  is 
equal  to  one  half  angle  BA  C. 

Ey  219.  AB  is  an  arc  of  65°,  DC  an  arc  of  75°  in  a  circle  whose  center 
is  0.  ^C  is  a  diameter.  How  many  degrees  are  there  in  each  angle  of  the 
triangles  A OD  and  BOG^ 

Ex  220.  If  an  equilateral  triangle  is  inscribed  in  a  circle,  and  a  diameter 
is  drawn  from  one  vertex,  the  triangle  formed  by  joining  the  other  extremity 
of  the  diameter  and  the  center  of  the  circle  with  one  of  the  other  vertices  of 
the  inscribed  triangle  is  also  equilateral. 

Ex.  221.  If  tangents  aie  drawn  to  a  circle  from  a  point  without,  the  line 
joining  that  point  with  the  center  of  the  circle  bisects  (1)  the  angle  formed 
by  the  tangents ;  (2)  the  angle  formed  by  the  radii  drawn  to  the  points  of 
tangency ;  and  (3)  the  arc  intercepted  by  these  radii. 

Proposition  XXIII 
236.  Problem,*  To  bisect  a  straight  line. 


>P 


Datum :   Any  line,  as  AB. 
Required  to  bisect  AB, 


yp 


Solution.     From  A  and  B  as  centers,  with   equal   radii   each 
^eater  than  one  half  AB,  describe  arcs  intersecting  at  C  and  D. 
Draw  CD  intersecting  AB  Sit  E. 
Then,  CD  bisects  AB  at  J^.  Q.E.F. 

Proof.     Const.,  C  and  D  are  each  equidistant  from  A  and  B, 

Hence,  §  106,     cn  is  the  perpendicular  bisector  of  AB. 

*  1.  The  student  is  urged  to  attempt  to  solve  each  problem  before  he 
studies  the  solution  given  in  the  book.  He  will  very  likely  discover  for  him- 
self the  same  method  of  solution  or  perhaps  another  one  equally  good. 

2.  The  only  implements  used  in  solving  problems  in  plane  geometry  are 
the  straightedge  and  compasses. 


112 


PLANE   GEOMETRY.-^BOOK  IL 


a- 


Proposition  XXIV 
237.  Problem,    To  bisect  an  arc  of  a  circle. 


Datum :   Any  arc  of  a  circle,  as  AB. 
Required  to  bisect  arc  AB. 


Solution.     Draw  the  chord  AB. 

From  A  and  B  as  centers,  with  equal  radii  each  greater  than 
one  half  chord  AB,  describe  arcs  intersecting  at  C  and  D. 
Draw  CD  intersecting  arc  AB  d^t  E. 
Then,  CD  bisects  arc  AB  at  Z.  q.e.p. 

Proof.     Const.,  C  and  D  are  each  equidistant  from  A  and  B ; 
.*.  §  106,     CD  is  perpendicular  to  the  chord  AB  at  its  middle  point, 
and,  §  199,  CD  passes  through  the  center  of  the  circle. 

Hence,  §  200,  CD  bisects  the  arc  AB  2i\,  E. 


Proposition  XXV 
238.  Brohlem,    To  bisect  an  angle. 


Datum:  Any  angle,  as  ^BC. 
Required  to  bisect  the  angle  ABC. 


Solution.  From  5  as  a  center  with  any  radius,  as  BE,  describe 
an  arc  intersecting  AB  'm.  D  and  CB  in  E. 

From  D  and  E  as  centers,  with  equal  radii  each  greater  than 
one  half  the  distance  DEj  describe  arcs  intersecting  at  F,  and 
draw  BF. 

Then,  BF  bisects  the  angle  ABC,  q.e.f 

Proof.     Draw  DF  and  EF. 

Then,  in  A  BDF  and  BEF^ 


PLANE   GEOMETRY.  —  BOOK  IL  113 

const.,  BD  —  BE, 

DF  =  EF, 
and  BF  is  common ; 

.-.§107,  ABBF  =  ABEFy 

and  /.'DBF  =  Zebf;  Why? 

that  is,  BF  bisects  the  angle  ABC. 

Proposition  XXVI 

239.   Problem,     At  a  given  point  in  a  line  to  erect  a 
perpendicular  to  the  line. 


% 


/o 


E:  ^        HX 


I 


Case  I.    When  the  given  point  is  between  the  extremities  of  the  line. 

Data :  Any  line,  as  AB,  and  any  point  in  AB,  as  C. 

Required  to  erect  a  perpendicular  to  AB  at  C. 

Solution.  From  C  as  a  center,  with  any  radius,  describe  arcs 
intersecting  AB  3it  D  and  E. 

From  D  and  E  as  centers,  with  equal  radii,  describe  arcs  inter- 
secting at  F.     Draw  CF. 

Then,  CF  is  the  perpendicular  required.  q.e.f. 

Proof.     By  the  student.     Suggestion.    Refer  to  §  106. 

Case  II.    When  the  given  point  is  at  the  extremity  of  the  line. 

Data :  Any  line,  as  AB,  and  the  point  at  either  extremity,  as  B. 

Required  to  erect  a  perpendicular  to  AB  at  5. 

Solution.  From  0,  any  point  without  AB,  as  a  center,  with  OB 
as  a  radius,  describe  an  arc  intersecting  AB  in  H. 

From  H,  draw  a  line  through  0  intersecting  this  arc  in  K,  and 
draw  KB. 

Then,  KB  is  the  perpendicular  required.  q.e.f. 

Proof.     By  the  student.     Suggbstion.    Refer  to  §  227. 
milne's  geom.  —  8 


114  PLANE   GEOMETRY.  — BOOK  11. 

Proposition  XXVII 

240.   Problem,    To  draw  a  perpendicular  to  a  line  from 
a  point  without  the  line. 

Data:  Any  line,  as  AB,  and  any 
point  without  the  line,  as  C.  /ff 

Required  to  draw  a  perpendicular 
from  c  to  AB. 


Solution.  From  C,  as  a  center,  with  a  radius  greater  than  the 
distance  from  C  to  AB,  describe  an  arc  intersecting  AB  at  D 
and  E. 

From  D  and  E,  as  centers,  with  equal  radii  each  greater  than 
one  half  of  DE,  describe  arcs  intersecting  at  F. 

Draw  CF  and  produce  it  to  meet  AB  as  at  G. 

Then,  CG  is  the  perpendicular  required.  q.e.f. 

Proof.     By  the  student.     Suggestion.    Refer  to  §  106L 


Proposition  XXVIII 

241.   Problem,    To  construct  an  angle  equal  to  a  given 
angle. 


Datum :  Any  angle,  2^s  ABC. 

Required  to  construct  an  angle  equal  to  ABC. 

Solution.     Draw  any  line,  as  DE. 

From  5  as  a  center,  with  any  radius,  describe  an  arc  intersect- 
ing BA  and  BC  in  F  and  G  respectively. 

From  D  as  a  center,  with  the  same  radius,  describe  an  arc  inter- 
secting DE  in  H. 


PLANE   GEOMETRY.  — BOOK  II.  115 

From  fi-  as  a  center,  with  a  radius  equal  to  the  distance  GF^ 
describe  a  second  arc  intersecting  the  first  at  J",  and  draw  DJ. 

Then,  Z  JDH  is  the  angle  required.  q.e.f. 

Proof.     Draw  GF  and  HJ. 

In  Abgf  and  DHJ,     . 
const.,  BG=  DH,  BF  =  DJ,  and  GF  =  JEfJ ; 

.-.  §  107,  Abgf  =  Abhj, 

and  Zb  =  Zb. 

Proposition  XXIX 

242.  rroblem.     Through  a  given  point  to  draw  a  line 
parallel  to  a  given  line.  ,e 


c! 


Data:  Any  line,  as  AB,  and 
any  point  not  in  AB,  as  C. 

Required     to     draw    a     line  /""\^       ' 

through  a  parallel  to  ^^.  / 

A i \ -B 

Solution.     Through  G  draw  any  line  meeting  AB,  as  EB. 
Construct  Z  ECF  equal  to  Z  EBB. 

Then,  CF  is  parallel  to  AB.  q.e.f. 

Proof.     Const.,  Z  ECF  =  Z  EBB  ; 

.-.    f  77,  CF  II  AB. 

Ex.  222.    Two  angles  of  a  triangle  being  given  to  construct  the  third. 

Ex.  223.  Two  secants  cut  each  other  without  a  circle  ;  the  intercepted 
arcs  are  72°  and  48°.     What  is  the  angle  between  the  secants  ? 

Ex.  224.  A  tangent  and  a  secant  cut  each  other  without  a  circle ;  the 
intercepted  arcs  are  94°  and  32°.  What  is  the  angle  between  the  tangent 
and  the  secant  ? 

Ex.  225.  Two  chords  of  a  circle  intersect  and  two  opposite  intercepted 
arcs  are  88°  and  26°.     What  are  the  angles  between  the  chords  ? 

Ex.  226.  A  tangent  of  a  circle  and  a  chord  from  the  point  of  contact 
intercept  an  arc  of  110°.  What  is  the  angle  between  the  tangent  and  the 
chord  ? 

Ex.  227.  If  the  radii  of  two  intersecting  circles  are  4*'"  and  9^™,  respec- 
tively, what  is  the  greatest  and  the  least  possible  distance  between  their 
centers  ? 


116  PLANE   GEOMETRY.  — BOOK  II. 

Proposition  XXX 

243.    Problem,    To  construct  a  triangle  when  two  sides 
and  the  included  angle  are  given. 


yQ 


Tb--^ 


Data :  Two  sides  of  a  triangle,  as  m  and  n,  and  the  included 
angle,  as  r. 

Required  to  construct  the  triangle. 

Solution.  Draw  any  line,  as  AD,  and  on  it  measure  AB  equal 
to  n. 

Construct  the  /,  A  equal  to  Z  r,  and  on  AG  measure  AC  equal 
to  m. 

Draw  CB. 

Then,  Aabc'ib,  the  A  required.  q.e.f. 

Proof.     By  the  student.  . 

Proposition  XXXI 

244.  Prohlem,  To  construct  a  triangle  when  a  side  and 
two  adjacent  angles  are  given. 


rj,--D 


Data :  A  side  of  a  triangle,  as  m,  and  the  angles,  as  r  and  s,  ad- 
jacent to  it. 

Required  to  construct  the  triaugle. 


PLANE   GEOMETRY.  — BOOK  II.  117 

Solution.  Draw  any  line,  as  AD,  and  on  it  measure  AB  equal 
to  m. 

Construct  Z  A  equal  to  Z  r,  and  Z  5  equal  to  Z  s. 

Produce  the'  sides  of  these  two  angles  until  they  intersect,  as 
at  C. 

Then,  AABCi^  the  A  required.  .  q.e.f. 

Proof.     By  the  student. 

245.  Sch.  The  problem  is  impossible  when  the  sum  of  the 
given  angles  equals  or  exceeds  two  right  angles.  Why  ? 


Proposition  XXXII 

246.  Problem.    To  construct  a  triangle  when  the  three 
sides  are  given. 


P 


1^ ^.. .^—-D 

Data :  The  three  sides  of  a  triangle,  as  m,  n,  and  o. 

Required  to  construct  the  triangle. 

Solution.  Draw  any  line,  as  AD,  and  on  it  measure  AB  equal 
to  0. 

From  ^  as  a  center,  with  a  radius  equal  to  n,  describe  an  arc. 

From  J5  as  a  center,  with  a  radius  equal  to  m,  describe  a  second 
arc  intersecting  the  first  in  C.     Draw  AC  and  BC. 

Then,  AABCis  the  A  required.  q.e.f. 

Proof.     By  the  student. 

247.  Sch.  The  problem  is  impossible  when  any  one  side  is 
equal  to  or  greater  than  the  sum  of  the  other  two  sides.      Why  ? 

Ex.  228.    Construct  an  equilateral  triangle. 

Ex,  229.  Prove  that  the  radius  of  a  circle  inscribed  in  an  equilateral 
triangle  is  equal  to  one  third  the  altitude  of  the  triangle. 

Ex.  230.  In  an  inscribed  trapezoid  how  do  the  non-parallel  sides  com- 
pare ?    How  do  the  diagonals  compare  ? 


118  PLANE   GEOMETRY.  — BOOK  11. 

Proposition  XXXIII 

248.  Problem,    To  construct  a  parallelogram  when  two 
sides  and  the  included  angle  are  given. 


c/. z-Ad 


'tK 


m. 


Data:  Two  sides  of  a  parallelogram,  as  m  and  n,  and  tlie 
included  angle,  as  r. 

Required  to  construct  the  parallelogram. 

Solution.  Draw  any  line,  as  AE,  and  on  it  measure  AB  equal 
to  n. 

Construct  the  Z  A  equal  to  Z  r,  and  on  AG  measure  A 0  equal 
to  m. 

From  c  as  a  center,  with  a  radius  equal  to  n,  describe  an  arc. 

From  5  as  a  center,  with  a  radius  equal  to  m,  describe  a  second 
arc  intersecting  the  first  in  D. 

Draw  CD  and  BD. 

Then,  ABDC  is  the  parallelogram  required.  q.e.f. 

Proof.     By  the  student.      Suggestion.     Refer  to  §  148. 

Proposition  XXXIV 
249.  Problem,    To  inscribe  a  circle  in  a  given  triangle. 


Datum :  Any  triangle,  as  ABC. 
Required  to  inscribe  a  circle   in 
A  ABC. 

A  E  B 

Solution.     Bisect  A  A  and  B,  produce  the  bisectors  to  intersect 
at  0,  and  draw  OE  _L  AB. 


PLANE   GEOMETRY.  — BOOK  IT.  119 

From  0  as  a  center,  with  OE  as  a  radius,  describe  the  circle  EFG. 

Then,  EFG  is  the  circle  required.  q.e.f. 

Proof.     Const.,     0  lies  in  the  bisectors  oi  Aa  and  B  ; 
.-.  §  134,  0  is  equidistant  from  AB,  AC,  and  BC. 

Hence,  a  circle  described  from  0  as  a  center,  with  a  radius 
equal  to  OE,  touches  AB,  AC,  and  BC. 

That  is,  §  190,  the  circle  EFG  is  inscribed  in  A  ABC. 

Proposition  XXXV 
250.  Problem,    To  divide  a  straight  line  into  equal  parts. 


Datum :  Any  straight  line,  as  AB. 

Required  to  divide  AB  into  equal  parts. 

Solution.  From  A  draw  a  line  of  indefinite  length,  as  AD, 
making  any  convenient  angle  with  AB. 

On  AB  measure  off  in  succession  equal  distances  corresponding 
in  number  with  the  parts  into  which  AB  is  to  be  divided. 

From  the  last  point  thus  found  on  AD,  as  C,  draw  CB,  and  from 
each  point  of  division  on  AC  draw  lines  II  CB  and  meeting  AB. 

These  lines  divid-e  AB  into  equal  parts.  q.e.f. 

Proof.     By  the  student.     Suggestion.     Refer  to  §  157. 

Ex.  231.  If  the  sides  of  a  central  angle  of  35°  intercept  an  arc  of  75"", 
what  will  be  the  length  of  an  arc  intercepted  by  the  sides  of  a  central  angle 
of  80°  in  the  same  circle  ? 

Ex.  232.  AB  and  CD  are  diameters  of  the  circle  whose  center  is  0 ;  BD 
is  an  arc  of  116°.  How  many  degrees  are  there  in  each  angle  of  the  triangles 
^OC  and  i)05? 

Ex.  '233.  If  a  circle  is  circumscribed  about  a  triangle  ABC,  and  perpen- 
diculars are  drawn  from  the  vertices  to  the  opposite  sides  and  produced  to 
meet  the  circumference  in  the  points  D,  E,  and  jP,  the  arcs  EF^  FD^  and 
DE  are  bisected  at  the  vertices. 


J>  • 


120  PLANE   GEOMETRY.  — BOOK  II. 

Proposition  XXXVI 
251.  Problem,    To  find  the  center  of  a  circle. 


Datum :  Any  circle,  as  ABD. 
Required  to  find  the  center  oi  ABD. 


Solution.     Draw  any  two  non-parallel  chords,  as  AB  and  CD. 
Draw  the  perpendicular  bisectors  of  AB  and  CD,  and  produce 
them  until  they  intersect,  as  in  0. 
Then,  0  is  the  center  of  the  circle. 
Proof.     By  the  student.     Suggestion.    Refer  to  §  201. 

Ex.  234.   To  circumscribe  a  circle  about  a  given  triangle. 

Ex.  235.  AB  is  a  chord  of  a  circle  and  J.C  is  a  tangent  at  J. ;  a  secant 
parallel  to  AB,  as  EFD,  cuts  ^C  in  ^  and  the  circumference  in  F  andD ; 
the  lines  AF,  AD,  and  BD  are  drawn.  Prove  that  tite  triangles  AEF  and 
ADB  are  mutually  equiangular. 

Proposition  XXXVII 

252.  Problem.  Through  a  given  point  to  draw  a  tangent 
to  a  given  circle. 

Data:    A  circle  whose  center  is  0,  and 
any  point,  as  Ji. 

Required  to  draw  through  A  a  tangent    [  ^/         ]    "^  j?. 

to  the  circle. 


Solution.    Case  I.      When  A  is  on  the  circumference. 

Draw  the  radius  OA. 

At  A  dT3iW  EF  ±  OA. 

Then,  EF  is  the  tangent  required. 

Proof.     Bj  the  student. 


PLANE   GEOMETRY,  — BOOK   II.  121 

Case  II.      When  A  is  without  the  circumference. 

Draw  OA. 

From  B,  the  middle  point  of  OA,  as        y^^'^^^^Tc^     ^^        \ 
a  center,  with  -B  0  as  a  radius,  describe     /         /       nT"^-^^       \ 

a  circle  intersecting  the  given  circle    /  J \  | -"--:^J^ 

at  C  and  D.  '  \  ^^  /  '^  -^^^    ' 

Drsiw  AC  Siiid  AD.  \  \dv^"""         /' 

Then,  AO  ot  AD  is  the  tangent  re-  T^^^^^^^-^-J-iL-'""' 

quired.  q.e.f. 

Proof.     By  the  student. 

Suggestion.     Draw  the  radii  OC  and  OB,  and  refer  to  §  227  and  §  205. 

Proposition  XXXVIII 

253.   Problem,    To  describe  upon  a  given  straight  line  a 
segment  of  a  circle  which  shall  contain  a  given  angle. 


Data :  Any  straight  line,  as  AB.       /    /    \         \ 

and  any  angle,  as  r.  /     /  xjq\        \ 

Required  to  describe  a  segment  \  /         j  '^^v\^    h 

of  a  circle  upon  AB  which  shall  i*r -f y^ 

contain  Z  r.  """- ^^7 


Solution.     Construct  Z  ABD  equal  to  Z  r. 

Draw  FE 1.  AB  at  its  middle  point. 

Erect  a  perpendicular  to  DB  at  B,  and  produce  it  to  intersect 
FE  at  0. 

From  0  as  a  center,  with  a  radius  equal  to  OB,  describe  a  circle. 

Then,  AMB  is  the  segment  required.  q.e.f. 

Proof.     Inscribe  any  angle  in  segment  AMB,  as  Z  s. 

Const.,  Z  ABD  =  /.r, 

§  231,  Z  ABD  is  measured  by  ^  arc  ^5  ; 

but,  §  225,  Z  s  is  measured  by  \  arc  AB, 

Z.S  —  /.  ABD  =  Zr. 

Hence,  AMB  is  the  segment  required. 


122  '       PLANE   GEOMETRY.  — BOOK  11. 

SUMMARY 
254.    Truths  established  in  Book  II. 

1.  Two  lines  are  equal, 

a.  If  they  are  radii  of  the  same  circle,  or  of  equal  circles.  Ax.  14 

b.  If  they  are  diameters  of  the  same  circle,  or  of  equal  circles.        Ax.  15 

c.  If  they  are  chords  which  subtend  equal  arcs  in  the  same  circle,  or  in 
equal  circles.  §  196 

d.  If  they  represent  the  distances  of  equal  chords  in  the  same  circle,  or  in 
equal  circles,  from  the  center.  §  202 

e.  If  they  are  chords  equally  distant  from  the  center  of  the  same  circle,  or 
of  equal  circles.  §  202 

/.     If  they  are  tangents  drawn  to  a  circle  from  a  point  without.  §  209 

g.    If  they  are  the  limits  of  two  variable  lines  which  constantly  remain 

equal  and  indefinitely  approach  their  respective  limits.  §  222 

2.  Two  lines  are  perpendicular  to  each  other, 

a.  If  one  is  a  tangent  to  a  circle  and  the  other  is  a  radius  drawn  to  the 
point  of  contact.  §  205 

b.  If  one  is  the  common  chord  of  two  intersecting  circles  and  the  other 
is  their  line  of  centers.  §  212 

3.  Two  lines  are  unequal, 

a.  If  one  is  a  diameter  of  a  circle  and  the  other  is  any  other  chord  of 
that  circle.  §  192 

b.  If  they  are  chords  of  the  same  circle,  or  of  equal  circles,  subtending 
unequal  arcs.  §  197 

c.  If  they  represent  the  distances  of  unequal  chords  in  the  same  circle, 
or  in  equal  circles,  from  the  center.  §  203 

d.  If  they  are  chords  of  the  same  circle,  or  of  equal  circles,  unequally 
distant  from  the  center.  §  204 

4.  A  line  is  bisected, 

,  a.    If  it  is  a  chord  of  a  circle,  by  a  radius  perpendicular  to  it.  §  198 

b.  If  it  is  a  chord  of  a  circle,  by  a  line  perpendicular  to  it  and  passing 
through  the  center.  §  200 

c.  If  it  is  the  common  chord  of  two  intersecting  circles,  by  their  line 
of  centers.  §  212 

5.  A  line  passes  through  a  point, 

a.  If  it  is  the  perpendicular  bisector  of  a  chord  and  the  point  is  the  center 
of  the  circle.  §  199 

b.  If  it  is  the  line  of  centers  of  two  tangent  circles,  and  the  point  is  their 
point  of  contact.  §  213 


PLANE   GEOMETRY.— BOOK  II.  123 

6.  Two  angles  are  equal, 

a.  If  they  are  central  angles  subtended  by  equal  arcs  in  the  same  circle, 
or  in  equal  circles.  §  194 

6.  If  they  are  inscribed  in  the  same  segment  of  a  circle,  or  in  equal  seg- 
ments of  the  same  circle,  or  of  equal  circles.  §  226 

7.  Two  angles  are  unequal,* 

a.  If  they  are  central  angles  subtended  by  unequal  arcs  in  the  same 
circle,  or  in  equal  circles.  §  195 

8.  An  angle  is  measured, 

a.  If  it  is  a  central  angle,  by  the  intercepted  arc.                                §  224 

h.  If  it  is  an  inscribed  angle,  by  one  half  the  intercepted  arc.           §  225 

c.  If  it  is  between  a  tangent  and  a  chord,  by  one  half  the  intercepted 
arc.  §  231 

d.  If  it  is  a  right  angle,  by  one  half  a  semicircumference.  §  232 

e.  If  it  is  between-  two  intersecting  chords,  by  one  half  the  sum  of  the 
intercepted  arcs.  §  230 

/.  If  it  is  between  a  tangent  and  a  secant,  by  one  half  the  difference  of 
the  intercepted  arcs.  §  233 

g.  If  it  is  between  two  secants  intersecting  without  the  circle,  by  one 
half  the  difference  of  the  intercepted  arcs.  §  234 

9.  Two  arcs  are  equal, 

a.  If  they  are  arcs  of  the  same  circle,  or  of  equal  circles  and  their  ex- 
tremities can  be  made  to  coincide.  §  193 

b.  If  they  subtend  equal  central  angles  in  the  same  circle,  or  in  equal 
circles.        '  §  194 

c.  If  they  are  subtended  by  equal  chords  In  the  same  circle,  or  in  equal 
circles.  §  196 

d.  If  they  are  intercepted  on  a  circumference  by  parallel  lines.         §  206 

10.  Two  arcs  are  unequal, 

a.  If  they  subtend  unequal  central  angles  in  the  same  circle,  or  in  equal 
circles.  §  195 

6.  If  they  are  subtended  by  unequal  chords  in  the  same  circle,  or  in  equal 
circles.  §  197 

11.  An  arc  is  bisected, 

a.   By  the  radius  perpendicular  to  the  chord  that  subtends  the  arc.     §  198 
6.    By  a  line  through  the  center  perpendicular  to  the  chord.  §  200 

12.  Two  circles  are  equal, 

a.    If  their  radii  or  diameters  are  equal.  Ax.  16 

h.    If  they  circumscribe  equal  triangles.  §  208 

13.  A  line  is  tangent  to  a  circle, 

a.   If  it  is  perpendicular  to  a  radius  at  its  extremity.  §  205 


124  PLANE   GEOMETRY.  — BOOK  II. 


SUPPLEMENTARY   EXERCISES 

Ex.  236.  ABC  is  an  inscribed  isosceles  triangle ;  the  vertical  angle  G  is 
three  times  each  base  angle.  How  many  degrees  are  there  in  each  of  the 
&TCSAB,  ^(7,  and  50? 

Ex.  237.  If  a  hexagon  is  circumscribed  about  a  circle,  the  sums  of  its 
alternate  sides  are  equal. 

Ex.  238.  Two  radii  of  a  circle  at  right  angles  to  each  other  are  inter- 
sected, when  produced,  by  a  line  tangent  to  the  circle.  Prove  that  the 
tangents  drawn  to  the  circle  from  the  points  of  intersection  are  parallel  to 
each  other. 

Ex.  239.  Two  circles  are  tangent  to  each  other  externally  and  each  is 
tangent  to  a  third  circle  internally.  Show  that  the  perimeter  of  the  triangle 
formed  by  joining  the  three  centers  is  equal  to  the  diameter  of  the  exterior 
circle. 

Ex.  240.  From  two  points,  A  and  B,  in  a  diameter  of  a  circle,  each  the 
same  distance  from  the  center,  two  parallel  lines  AE  and  BF  are  drawn 
toward  the  same  semicircumference,  meeting  it  in  E  and  F.  Show  that  FF 
is  perpendicular  to  AE  and  BF. 

Ex.  241.  Two  circles  are  tangent  externally  at  A  J5C  is  a  tangent  to 
the  two  circles  at  B  and  C.  Prove  that  the  circumference  of  the  circle 
described  on  50  as  a  diameter  passes  through  A. 

Ex.  242.  OA  is  a  radius  of  a  circle  whose  center  is  0  ;  -B  is  a  point  on 
a  radius  perpendicular  to  OA ;  through  B  the  chord  ^O  is  drawn  ;  at  O  a 
tangent  is  drawn  meeting  OB  produced  in  D.  Prove  that  CBD  is  an 
isosceles  triangle. 

Suggestion.     Draw  a  tangent  at  A. 

Ex.  243.  Through  a  given  point  P  without  a  circle  whose  center  is  O  two 
lines  FAB  and  PCD  are  drawn,  making  equal  angles  with  OP  and  inter- 
secting the  circumference  in  A  and  B,  C  and  Z>,  respectively.  Prove  that 
AB  equals  OD,  and  that  AP  equals  CP. 

Suggestion.  Draw  OM  and  ON  perpendicular  to  AB  and  CD,  respect- 
ively. 

Ex.  244.  If  the  angles  at  the  base  of  a  circumscribed  trapezoid  are 
equal,  each  non-parallel  side  is  equal  to  half  the  sum  of  the  parallel  sides. 

Ex.  245.  If  a  circle  is  inscribed  in  a  right  triangle,  the  sum  of  the 
diameter  and  the  hypotenuse  is  equal  to  the  sum  of  the  other  two  sides  of 
the  triangle. 

Ex.  246.  Any  parallelogram  which  can  be  circumscribed  about  a  circle 
is  equilateral, 

Ex.  247.  AB  and  CD  are  perpendicular  diameters  of  a  circle  ;  E  is  any 
point  on  the  arc  ACB.    Then,  D  is  equidistant  from  AE  and  BE. 


PLANE   GEOMETRY.  — BOOK  II.  125 

Ex.  248.  If  two  equal  chords  of  a  circle  intersect,  their  corresponding 
segments  are  equal. 

Ex.  249.  If  the  arc  cut  off  by  the  base  of  an  inscribed  triangle  is  bisected 
and  from  the  point  of  bisection  a  radius  is  drawn  and  also  a  line  to  the  oppo- 
site vertex,  the  angle  between  these  lines  is  equal  to  half  the  difference  of 
the  angles  at  the  base  of  the  triangle. 

Ex.  250.  The  two  lines  which  join  the  opposite  extremities  of  two  par- 
allel chords  intersect  at  a  point  in  that  diameter  which  is  perpendicular  to 
the  chords. 

Ex.  251.  If  a  tangent  is  drawn  to  a  circle  at  the  extremity  of  a  chord, 
the  middle  point  of  the  subtended  arc  is  equidistant  from  the  chord  and  the 
tangent. 

Ex.  252.  A  line  is  drawn  touching  two  tangent  circles.  Prove  that  the 
chords,  that  join  the  points  of  contact  with  the  points  in  which  the  line 
through  the  centers  meets  the  circumferences,  are  parallel  in  pairs. 

Ex.  253.  "Two  circles  intersect  at  the  points  A  and  B ;  through  A  a 
secant  is  drawn  intersecting  one  circumference  in  G  and  the  other  in  D ; 
through  B  a  secant  is  drawn  intersecting  the  circumference  CAB  in  E  and 
the  other  circumference  in  F.  Prove  that  the  chords  CE  and  I^F  are 
parallel. 

Suggestion.     Refer  to  Ex,  198  and  201. 

Ex.  254.  The  length  of  the  straight  line  joining  the  middle  points  of  the 
non-parallel  sides  of  a  circumscribed  trapezoid  is  equal  to  one  fourth  the 
perimeter  of  the  trapezoid. 

Ex.  255.  A  quadrilateral  is  inscribed  in  a  circle,  and  two  opposite  angles 
are  bisected  by  lines  meeting  the  circumference  in  A  and  B.  Prove  that 
AB  is  a  diameter. 

Ex.  256.  The  centers  of  the  four  circles  circumscribed  about  the  four 
triangles  formed  by  the  sides  and  diagonals  of  a  quadrilateral  lie  on  the 
vertices  of  a  parallelogram. 

Ex.  257.  If  an  equilateral  triangle  is  inscribed  in  a  circle,  the  distance  of 
each  side  from  the  center  is  equal  to  half  the  radius  of  the  circle. 

Ex.  258.  The  vertical  angle  of  an  oblique  triangle  inscribed  in  a  circle  is 
greater  or  less  than  a  right  angle  by  the  angle  contained  by  the  base  and  the 
diameter  drawn  from  the  extremity  of  the  base. 

Ex.  259.  If  from  the  extremities  of  any  diameter  of  a  given  circle  per- 
pendiculars to  any  secant  that  is  not  parallel  to  this  diameter  are  drawn, 
the  less  perpendicular  is  equal  to  that  segment  of  the  greater  which  is  con- 
tained between  the  circumference  and  the  secant. 

Ex.  260.  Two  circles  are  tangent  internally  at  J.,  and  a  chord  BC  of  the 
larger  circle  is  tangent  to  the  smaller  at  D.  Prove  that  AD  bisects  the  angle 
CAB. 

Suggestion.     Draw  AT,  the  common  tangent  of  the  circles. 


126  PLANE   GEOMETRY.^BOOK  II 

Ex.  261.  The  tangents  at  the  four  vertices  of  an  inscribed  rectangle  form 
a  rhombus. 

Ex.  262.  If  a  line  is  drawn  through  the  point  of  contact  of  two  circles 
which  are  tangent  internally,  intersecting  the  circle  whose  center  is  A  at  (7, 
and  the  circle  whose  center  is  5  at  Z>,  AC  and  BD  are  parallel. 

Ex.  263.  If  lines  are  drawn  from  the  center  of  a  circle  to  the  vertices  of 
any  circumscribed  quadrilateral,  each  angle  at  the  center  is  the  supplement 
of  the  central  angle  that  is  not  adjacent  to  it. 

Ex.  264.  Three  circles  are  tangent  to  each  other  externally  at  the  j)oints 
Af  J5,  and  C.  From  A  lines  are  drawn  through  B  and  C  meeting  the  cir- 
cumference which  passes  through  B  and  C  at  the  points  D  and  E.  Prove 
that  BE  is  a  diameter. 

Ex.  265.  If  an  angle  between  a  diagonal  and  one  side  of  a  quadrilateral 
is  equal  to  the  angle  between  the  other  diagonal  and  the  opposite  side,  the 
same  will  be  true  of  the  three  other  pairs  of  angles  corresponding  to  the 
same  description,  and  the  four  vertices  of  the  quadrilateral  lie  on  a  circum- 
ference. 

Ex.  266.  Let  the  diameter  AB  of  a  circle  be  produced  to  O,  making  BC 
equal  to  the  radius ;  through  B  draw  a  tangent,  and  from  C  draw  a  second 
tangent  to  the  circle  at  D,  intersecting  the  first  at  E ;  AD  and  BE  produced 
meet  at  F.     Prove  that  DEF  is  an  equilateral  triangle. 

Suggestion.     Draw  a  line  from  the  center  to  E. 

Ex.  267.  If  from  any  point  without  a  circle  tangents  are  drawn,  the 
angle  contained  by  the  tangents  is  double  the  angle  contained  by  the  line 
joining  the  points  of  contact  and  the  diameter  drawn  through  one  of  them. 

Ex.  268.  The  lines,  which  bisect  the  vertical  angles  of  all  triangles  on  the 
same  base  and  on  the  same  side  of  it,  and  having  equal  vertical  angles,  meet 
at  the  same  point. 

Ex.  269.  AB  and  AC  are  tangents  at  B  and  C  respectively,  to  a  circle 
whose  center  is  0;  from  D,  any  point  on  the  circumference,  a  tangent  is 
drawn,  meeting  AB  in  E  and  ^O  in  i^.  Prove  that  angle  EOF  is  equal  to 
one  half  angle  BOC. 

Ex.  270.  A  circle  whose  center  is  0  is  tangent  to  the  sides  of  an  angle 
ABC  at  A  and  C ;  through  any  point  in  the  arc  AC,  as  Z>,  a  tangent  is 
drawn,  meeting  AB  in  E,  and  CB  in  F.  Prove  (1)  that  the  perimeter  of  the 
triangle  BEF  is  constant  for  all  positions  of  Bin  AC;  (2)  that  the  angle  EOF 
is  constant. 

Ex.  271.  If  AE  and  BD  are  drawn  perpendicular  to  the  sides  BC  and 
AC,  respectively,  of  the  triangle  ABC,  and  DE  is  drawn,  the  angles  AED 
and  ABD  are  equal. 

Suggestion.     Describe  a  circle  passing  through  A,  D,  and  B. 


PLANE   GEOMETRY.  — BOOK  11.  127 

Ex.  272.  The  perimeter  of  an  inscribed  equilateral  triangle  is  equal  to 
one  half  the  perimeter  of  the  circumscribed  equilateral  triangle. 

Ex.  273.  In  the  circumscribed  quadrilateral  ABCD,  the  angles  A,  J5, 
and  Care  110°,  95°,  and  80°,  respectively,  and  the  sides  AB,  BC,  CB,  and 
DA  touch  the  circumference  at  the  points  E,  F,  G,  and  H  respectively.  Find 
the  number  of  degrees  in  each  angle  of  the  quadrilateral  EFGH. 

Ex.  274.  If  an  inscribed  isosceles  triangle  has  each  of  its  base  angles 
double  the  vertical  angle,  and  its  vertices  the  points  of  contact  of  three  tan- 
gents, these  tangents  form  an  isosceles  triangle  each  of  vv^hose  base  angles  is 
one  third  its  vertical  angle. 

Ex.  275.  ABQ  is  a  triangle  inscribed  in  a  circle  ;  the  bisectors  of  the 
angles  A,  B,  and  C  meet  in  D  ;  AD  produced  meets  the  circumference  in  E. 
Prove  that  DE  equals  CE. 

Suggestion.     Produce  CD  to  meet  the  circumference  at  F,  and  draw  AF. 

Ex.  276.  If  the  diagonals  of  a  quadrilateral  inscribed  in  a  circle  inter- 
sect at  right  angles,  the  perpendicular  from  their  intersection  upon  any  side, 
if  produced,  bisects  the  opposite  side. 

Ex.  277.  If  the  opposite  angles  of  a  quadrilateral  are  supplementary,  the 
quadrilateral  may  be  inscribed  in  a  circle. 

Suggestion.     §  207.    A  circumference  may  be  passed  through  A,  B,  and 
Z>,  and  if  it  does  not  pass  through  O,  it  will  cut  DC, 
or  DC  produced,  as  at  jE'.     Draw  ^ji5.  j),,- >.,^ 

Then,  from  data  and  the  supposition  that  the  circum-  /j         -^-.^^X 

ference  passes  through  E,  it  may  be  shown  by  raeas-         /  /  \    \ 

urement  of  inscribed  angles,  that  ZDCB  =  /.DEB.  \  /  \  j 

But,  §  11 5,  Z  DEB  =  Z  CBE  ■\-Z.DCB',  a{ ■ fB 

ZDCB=zZCBE+ZDCB,  V., ,-'"' 

which  is  absurd,  and  the  supposition  that  the  cir- 
cumference passes  through  E  and  not  through  C  is  untenable. 

Hence,  the  circumference  through  A,  B,  and  D  passes  through  C. 

Ex.  278.  The  four  lines  bisecting  the  angles  of  any  quadrilateral  form  a 
quadrilateral  which  may  be  inscribed  in  a  circle. 

Suggestion.    Kefer  to  Ex.  179  and  277. 

Ex.  279.  The  line,  drawn  from  the  center  of  the  square  described  upon 
the  hypotenuse  of  a  right  triangle  to  the  vertex  of  the  right  angle,  bisects 
the  right  angle. 

Ex.  280.  AB  and  CD  are  two  chords  of  a  circle  intersecting  at  E ; 
through  A  a  line  is  drawn  to  meet  a  line  tangent  at  C  so  that  the  angle  AFC 
equals  the  angle  BEC.     Then,  EF  is  parallel  to  BC. 

Ex.  281.  ABC  is  a  triangle  ;  AD  and  BE  are  the  perpendiculars  from  A 
and  B  upon  BC  and  J. C  respectively  ;  DF  and  EG  are  the  perpendiculars 
from  D  and  E  upon  ^C  and  BC  respectively.     Then,  FG  is  parallel  to  AB. 


128  PLANE   GEOMETRY.  — BOOK  II, 


PROBLEMS 

255.  Problems  are  valuable  for  developing  the  ingenuity  of  the 
student  in  discovering  the  auxiliary  lines  necessary  for  solution 
or  demonstration,  and  for  fixing  clearly  in  mind  the  knowledge 
previously  acquired.  They  do  not,  however,  involve  any  new 
fundamental  fact  or  principle  of  geometry,  and  may  therefore  be 
omitted  without  impairing  the  logical  development  of  the  science. 

No  definite  rules  can  be  given  for  solving  problems,  but  close 
attention  to  the  following  suggestions,  and  a  thorough  study  of 
the  Summary  will  be  of  great  assistance  in  developing  a  proper 
and  logical  method  of  procedure. 

I.  Study  carefully  the  data  of  the  problem  to  discover  every  fact 
that  is  giveuj  and  7iotice  also  what  is  required. 

II.  From  the  facts  given  deduce  all  possible  conclusions,  and  try 
to  relate  them  to  what  is  required. 

III.  If  the  2^rinciples  upon  which  the  solution  is  based  are  not 
readily  discovered  from  the  data,  try  to  get  some  clew  to  the  solution 
by  studying  the  Summary  and  by  drawing  lines  perpendicular  or 
parallel  to  other  lines;  by  forming  triangles;  by  joining  given 
points;   by  describing  circles;  etc. 

IV.  The  solution  is  often  readily  discovered  by  drawing  a  figure 
representing  the  problem  solved,  and  then  from  a  study  of  the  relor 
tions  of  the  known  and  unknown  parts  of  the  figure  discover  the  facts 
which  bear  upon  the  solution. 

Ex.  282.  Construct  the  complement  of  a  given  angle. 

Ex.  283.  Construct  the  supplement  of  a  given  angle. 

Ex.  284.  At  a  given  point  in  a  given  straight  line,  to  construct  an  angle 
of  45°. 

Ex.  285.    Divide  an  angle  into  four  equal  parts. 

Ex.  286.  At  a  given  point  in  a  given  straight  line,  to  construct  an  ang^e 
of  60°. 

Ex.^  287.   Trisect  a  right  angle. 

Ex.  288.    Construct  a  square,  having  given  one  side. 

Ex.  289.  Construct  an  isosceles  triangle,  having  given  the  base  and  the 
perpendicular  from  the  vertex  to  the  base. 

Ex.  290.    Construct  an  equilateral  triangle,  having  given  the  perimeter. 


PLANE   GEOMETRY.  — BOOK  II.  129 

Ex.  291.  Construct  an  isosceles  triangle,  having  given  the  perimeter  and 
base. 

Ex.  292.    Construct  a  rectangle,  having  given  two  adjacent  sides. 

Ex.  293.  Construct  a  rectangle,  having  given  the  shorter  side  and  the 
difference  of  two  sides. 

Ex.  294.  Construct  a  rectangle,  having  given  the  longer  side  and  the 
difference  of  two  sides. 

Ex.  295.  Construct  a  rectangle,  having  given  the  sum  and  difference  of 
two  adjacent  sides. 

Ex.  296.  Construct  a  rhombus,  having  given  one  of  its  angles  and  the 
length  of  its  side. 

Ex.  297.  Construct  an  isosceles  triangle,  having  given  the  base  and  one 
of  the  two  equal  angles  at  the  base. 

Ex.  298.  Construct  an  isosceles  right  triangle,  having  given  its  hypot- 
enuse. 

Ex.  299.  Construct  a  rhomboid,  having  given  the  perimeter,  one  side, 
and  one  angle. 

Ex.  300.  Construct  a  right  triangle,  having  given  the  hypotenuse  and 
one  side. 

Ex.  301.  Construct  a  right  triangle,  having  given  the  hypotenuse  and 
one  acute  angle. 

Ex.  302.  Construct  an  isosceles  trapezoid,  having  given  two  sides  and 
the  included  angle. 

Ex.  303.  Construct  a  trapezoid,  having  given  two  adjacent  sides,  the 
included  angle,  and  the  angle  at  the  other  extremity  of  th^  given  parallel 


Ex.  304.  From  a  given  point  without  a  given  line,  to  draw  a  line  making 
a  given  angle  with  the  given  line. 

Ex.  305.    Construct  a  square,  having  given  a  diagonal. 

Ex.  306.  Construct  a  rectangle,  having  given  one  side  and  the  angle 
included  between  it  and  a  diagonal. 

Ex.  307.  Construct  a  rectangle,  having  given  a  diagonal  and  an  angle 
between  it  and  a  side. 

Ex.  303.  Construct  a  rhombus,  having  given  its  perimeter  and  one 
diagonal. 

Ex,  309.  Construct  a  rhomboid,  having  given  two  diagonals  and  an 
angle  between  them. 

Ex.  310.  Construct  a  rectangle,  having  given  one  diagonal  and  the  angle 
included  between  the  two  diagonals. 

Ex.  311,    Construct  a  rectangle,  having  given  the  perimeter  and  one  side. 
milne's  geom.  —  9 


130  PLANE   GEOMETRY.  — BOOK  II. 

Ex.  312.  Construct  a  trapezoid,  having  given  two  sides,  tlie  included 
angle,  and  the  difference  between  the  two  parallel  sides. 

Ex.  313.  Construct  a  trapezium,  having  given  three  consecutive  sides 
and  the  two  included  angles. 

Ex.  314.  Construct  a  trapezium,  having  given  two  adjacent  sides  and 
the  three  angles  adjacent  to  these  sides. 

Ex.  315.  Construct  an  isosceles  triangle,  having  given  the  base  and  the 
vertical  angle. 

Ex.  316.    Construct  an  equilateral  triangle,  having  given  its  altitude. 

Ex.  317.  Construct  a  triangle,  having  given  two  sides  and  the  angle 
opposite  one. 

Ex.  318.  Construct  a  triangle,  having  given  its  base,  the  median  to  the 
base,  and  the  angle  included  between  them. 

Ex.  319.  Construct  a  right  triangle,  having  given  its  hypotenuse  and 
having  one  of  its  acute  angles  double  the  other, 

Ex.  320.  Construct  a  trapezoid,  having  given  the  sum  and  difference  of 
the  parallel  sides,  and  the  sum  and  difference  of  the  angles  at  the  base. 

Ex.  321.   Construct  a  rhombus,  having  given  its  diagonals. 

Ex.  322.  Construct  a  rhomboid,  having  given  two  adjacent  sides  and  an 
angle  not  included  by  them. 

Ex.  323.  Construct  a  rhomboid,  having  given  one  side  and  the  angles 
included  between  it  and  the  diagonals. 

Ex.  324.  Construct  an  isosceles  trapezoid,  having  given  the  bases  and 
the  altitude. 

Ex.  325.  Construct  an  isosceles  trapezoid,  having  given  the  altitude  and 
the  sum  and  difference  of  the  parallel  sides. 

Ex.  326.  Construct  a  triangle,  having  given  two  angles  and  a  side  oppo- 
site one. 

Ex.  327.  Draw  a  line  which  shall  pass  through  a  given  point  and  make 
equal  angles  with  two  given  intersecting  lines. 

Ex.  328.  Construct  a  right  triangle,  having  given  one  side  and  the  angle 
opposite. 

Ex.  329.  Construct  an  isosceles  trapezoid,  having  given  the  bases  and 
a  diagonal. 

Ex.  330.  Construct  an  isosceles  trapezoid,  having  given  the  bases  and 
one  angle. 

Ex.  331.  From  two  given  points,  draw  two  equal  straight  lines  which 
shall  meet  in  the  same  point  of  a  line  given  in  position. 

Ex.  332.  ABC  is  an  isosceles  triangle.  Draw  a  straight  line  parallel  to 
the  base  AB  and  meeting  the  equal  sides  in  E  and  F,  so  that  BF^  EF,  and 
EA  are  all  equal. 


PLANE   GEOMETRY.  — BOOK  II.  131 

Ex.  333.  Given  two  straight  lines  which  cannot  be  produced  to  their 
intersection,  to  draw  a  third  line  which  would  pass  through  their  intersec- 
tion and  bisect  their  contained  angle. 

'     Ex.  334.    Constrnct  a  triangle,  having  given  the  altitude  and  the  angles 
at  the  base. 

Ex.  335.  Given  the  middlQ  point  of  a  chord  in  a  given  circle,  to  draw 
the  chord. 

Ex.  336.  Construct  a  circle  to  pass  through  two  given  points  and  have 
its  center  on  a  given  straight  line.     When  is  this  impossible  ? 

Ex.  337.    Draw  a  tangent  to  a  circle  parallel  to  a  given  straight  line. 

Ex.  338.    Draw  a  tangent  to  a  circle  perpendicular  to  a  given  straight  line. 

Ex.  339.  Draw  a  straight  line  tangent  to  a  given  circle  and  making  with 
a  given  line  a  given  angle. 

Ex.  340.  Construct  a  circle  of  given  radius  to  pass  through  two  given 
points.     When  is  this  impossible  ? 

Ex.  341.  Construct  a  circle  tangent  to  two  intersecting  lines  with  its 
center  at  a  given  distance  from  their  intersection.  How  many  such  circles 
can  be  drawn  ? 

Ex.  342.  From  a  given  point  as  a  center,  to  describe  a  circle  tangent  to 
a  given  circle.     How  many  solutions  may  there  be  ? 

Ex.  343.  Construct  a  circle  of  given  radius  tangent  to  a  given  circle  at 
a  given  point.    How  many  solutions  may  there  be  ? 

Ex.  344.  Draw  a  common  tangent  to  two  given  circles.  How  many 
solutions  may  there  be  ? 

Ex.  345.  In  a  given  circle,  to  inscribe  a  triangle  equiangular  to  a  given 
triangle. 

Ex.  346.  About  a  given  circle,  to  circumscribe  a  triangle  equiangular  to 
a  given  triangle. 

Ex.  347.  Construct  a  triangle,  having  given  the  vertical  angle,  one  of  the 
sides  containing  it,  and  the  altitude. 

Ex.  348.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle, 
and  one  other  side. 

Suggestion.  On  the  given  base  construct  a  segment  that  will  contain  an 
angle  equal  to  the  given  angle. 

Ex.  349.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle, 
and  the  foot  of  the  perpendicular  from  the  vertex  to  the  base. 

Ex.  350.  Construct  a  triangle  whose  vertex  is  on  a  given  straight  line, 
and  having  given  its  base  and  vertical  angle. 

Ex.  351.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle, 
and  the  altitude. 

Ex.  352.  J^escribe  a  circle  with  a  given  center  to  intersect  a  given  circle 
at  the  extremities  of  a  diameter.     Is  this  ever  impossible  ? 


132  PLANE  GEOMETRY, --BOOK  IL 

Ex.  353.  Construct  a  circle  to  pass  through  a  given  point  and  be  tangent 
to  a  given  circle  at  a  given  poiut.     When  is  this  impossible  ? 

Ex.  354.  Construct  a  circle  to  pass  through  a  given  point  and  touch  l» 
given  straight  line  at  a  given  point. 

Ex.  355.    Construct  a  circle  to  touch  three  given  straight  lines. 

Ex.  356.  Within  an  equilateral  triangle,  to  describe  three  circles  each 
tangent  to  the  other  two  and  to  two  sides  of  the  triangle. 

Ex.  357.  Construct  a  circle  of  given  radius  to  touch  two  given  straight 
lines. 

Ex.  358.  Construct  a  circle  of  given  radius,  having  its  center  on  a  given 
straight  line  and  touching  ano£her  given  straight  line.  How  many  solutions 
may  there  be  ? 

Ex.  359.  Construct  a  right  triangle,  having  given  the  radius  of  the 
inscribed  circle  and  one  of  the  sides  containing  the  right  angle. 

Ex.  360.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle, 
and  the  length  of  the  median  to  the  base. 

Ex.  361.  Construct  a  triangle,  having  given  the  three  middle  points  of 
its  sides. 

Ex.  362.  Construct  a  circle  of  given  radius  to  pass  through  a  given  point 
and  touch  a  given  straight  line. 

Ex.  363.  From  the  vertices  of  a  triangle  as  centers,  to  describe  three 
circles  which  shall  be  tangent  to  each  other. 

Ex.  364.  Construct  a  triangle,  having  given  the  base,  the  altitude,  afld 
the  radius  of  the  circumscribed  circle. 

Ex.  365.  Three  given  straight  lines  meet  at  a  point ;  draw  another  straight 
line  so  that  the  two  portions  of  it  intercepted  between  the  given  lines  are 
equal.    How  many  solutions  may  there  be  ? 

Suggestion.    Form  a  parallelogram. 

Ex.  366.  Through  a  given  point,  between  two  intersecting  straight  lines, 
to  draw  a  line  terminated  by  the  given  lines  and  bisected  at  the  given  point. 

Ex.  367.  Construct  a  circle  to  intercept  equal  chords  of  given  length  on 
three  given  straight  lines. 

Ex.  368.  Construct  a  triangle,  having  given  one  angle,  the  opposite  side, 
and  the  sum  of  the  other  two  sides. 

The  Locus  of  a  Point 

256.  When  a  point  equidistant  from  the  extremities  of  a 
straight  line  is  to  be  found,  the  middle  point  of  the  line  meets 
the  conditions.  But  there  are  other  points  which  also  fulfill  the 
required  conditions,  for  every  point  in  the  perpendicular  to  the 
given  line  at  its  middle  point  is  equidistant  from  the  extremities 
of  the  line. 


PLANE   GEOMETRY.  — BOOK  11.  133 

Such  a  perpendicular  is  called  the  locals  of  the  points  which  are 
equidistant  from  the  extremities  of  the  line. 

The  *line,  or  system  of  lines,  containing  every  point  which 
satisfies  certain  given  conditions,  and  no  other  points,  is  called 
the  Locus  of  those  points. 

A  locus  may  also  be  described  as  a  line,  or  the  lines,  traced  by 
a  point  which  moves  in  accordance  with  given  conditions. 

To  prove  that  a  certain  line,  or  system  of  lines,  is  the  required 
locus,  it  must  be  shown : 

1.  That  every  point  in  the  lines  satisfies  the  given  conditions. 

2.  That  any  point  not  in  the  lines  cannot  satisfy  the  given  con- 
ditions. 

Ex.  369.  Find  the  locus  of  a  point  which  is  equidistant  from  two  inter- 
secting straight  lines. 

Data  :  Any  two  straight  lines,  as  AB  and  CD,  ^ 

intersecting  at  the  point  K.  /|\  ! 

Required  to  find  the  locus  of  a  point  equidistant  /  •  -"j^j 

from  AB  and  CD.  ^^^^  /    •  A\        ^^ 

Solution.  A  point  equidistant  from  two  inter-  •  -^^^"""<^^!i>^^^  xr 
secting  straight  lines  suggests  a  point  in  the  bisec-  3^'lS^'"^^ 

tor  of  an  angle.  a^^'^^'^        i       '  ^"^^D 

Draw  EF  bisecting  A  CKB  and  AKD.,  and  also  i 

GH  bisecting  A  AKC  and  BKD. 

Then,  §  134,  every  point  in  EF  is  equidistant  from  AB  and  CD,  and 
every  point  in  GH  is  equidistant  from  AB  and  CD. 

If  all  other  points  are  unequally  distant  from  AB  and  CD.,  then  EF  and 
GH  is  the  required  locus. 

From  P  any  point  without  the  bisectors  draw  TM  ±  CD,^  and  BE  L  AB., 
intersecting  EF  in  J.    From  J  draw  JL  ±  CD,  and  also  draw  BL. 

Then,  § 61,        PL>  PM,  and,  §  125,  P./  ^JL>PL\ 
PJ^JL>  PM. 

But  .JLz=JB\  Why? 

PJ+JB>  PM,  or  PB  >  PM. 

That  is,  the  point  P  is  unequally  distant  from  AB  and  CD. 
Hence  EF  and  GH  is  the  required  locus. 

Ex.  370.    Find  the  locus  of  "a  point  at  a  given  distance  from  a  given  point. 

Ex.  371.  Find  the  locus  of  a  point  equidistant  from  two  parallel  straight 
lines. 

*  In  this  and  the  next  four  paragraphs  the  lines  mentioned  may  be  either 
straight  or  curved. 


134  PLANE  GEOMETRY.  — BOOK  II, 

Ex.  372.  Find  the  locus  of  a  point  at  a  given  distance  from  a  given 
straight  line. 

Ex.  373.  Find  a  point  vv^hich  is  equidistant  from  three  given  points  not 
in  the  same  straight  line. 

Ex.  374.  Find  the  locus  of  a  point  equidistant  from  the  circumferences 
of  two  concentric  circles. 

Ex.  375.  Find  a  point  in  a  given  straight  line  wliich  is  equidistant  from 
two  given  points. 

Ex.  376.  Find  the  locus  of  the  center  of  a  circle  tangent  to  each  of  two 
parallel  lines. 

Ex.  377.  Find  the  locus  of  the  center  of  a  circle  which  touches  a  given 
line  at  a  given  point. 

Ex.  378.  Find  the  locus  of  the  center  of  a  circle  of  given  radius  that 
passes  through  a  given  point. 

Ex.  379.  Find  the  locus  of  the  center  of  a  circle  which  is  tangent  to  a 
given  circle  at  a  given  point. 

Ex.  380.  Find  the  locus  of  the  center  of  a  circle  of  given  radius  and 
tangent  to  a  given  circle. 

Ex.  381.  Find  the  locus  of  the  center  of  a  circle  passing  through  two 
given  points. 

Ex.  382.  Find  the  locus  of  the  center  of  a  circle  of  given  radius  and 
tangent  to  a  given  line. 

Ex.  383.  Find  the  locus  of  the  center  of  a  circle  tangent  to  each  of  two 
intersecting  lines. 

Ex.  384.  Find  the  locus  of  the  middle  points  of  a  system  of  parallel 
chords  drawn  in  a  circle. 

Ex.  385.  Find  the  locus  of  the  middle  points  of  equal  chords  of  a  given 
circle. 

Ex.  386.  Find  the  locus  of  the  extremities  of  tangents  of  fixed  length 
drawn  to  a  given  circle. 

Ex.  387.  Find  the  locus  of  the  middle  points  of  straight  lines  drawn 
from  a  given  point  to  meet  a  given  straight  line. 

Ex.  388.  Find  the  locus  of  the  vertex  of  a  right  triangle  on  a  given  base 
as  hypotenuse. 

Ex.  389.  Find  the  locus  of  the  middle  points  of  all  chords  of  a  circle 
drawn  from  a  fixed  point  in  the  circumference. 

Ex.  390.  Find  the  locus  of  the  middle  point  of  a  straight  line  moving 
between  the  sides  of  a  right  angle. 

Ex.  391.  Find  the  locus  of  the  points  of  contact  of  tangents  from  a  fixed 
point  to  a  system  of  concentric  circles. 

Ex.  392.  Find  the  locus  of  the  middle  points  of  secants  drawn  from  a 
given  point  to  a  given  circle. 


BOOK  III 


RATIO  AND  PROPORTION 

257.  1.   How  is  a  magnitude  measured  ? 

2.  What  is  the  numerical  measure  of  a  magnitude  ? 

3.  What  is  the  common  measure  of  two  or  more  magnitudes  ? 

4.  What  is  meant  by  the  ratio  of  two  magnitudes  ? 

5.  How  may  the  ratio  of  two  magnitudes  be  determined  ? 

6.  Since  the  ratio  of  two  magnitudes  is  the  ratio  of  their 
numerical  measures,  what  is  the  relation  of  two  magnitudes 
whose  numerical  measures  are  8  and  16  respectively?  5  and 
10?    12  and  36?     15  and  45? 

7.  How  does  8  compare  with  2  ?  What  is  the  relation  of  3 
to  9?  Of  12  to  4?  Of  18  to  3?  Of  20  to  40?  Of  25  to  75? 
Of  35  to  70  ? 

8.  What  is  the  ratio  of  1  ft.  to  1  yd.  ?  3  in.  to  1  ft.  ?  2*=«»  to 
jdm  9    5dm  to  2^"  ?    2  sq.  ft.  to  2  sq.  yd.  ?    3  cu.  ft.  to  1  cu.  yd.  ? 

258.  The  quantities  compared  are  called  the  Terms  of  the  ratio. 
A  ratio  is  denoted  by  a  colon  placed  between  the  terms. 

The  ratio  of  2  to  5  is  expressed  2  : 5. 

259.  The  first  term  of  a  ratio  is  called  the  Antecedent  of  the 
ratio.  The  second  term  of  a  ratio  is  called  the  Consequent  of  the 
ratio. 

260.  The  antecedent  and  consequent  together  form  a  Couplet. 

261.  Since  the  ratio  of  two  quantities  may  be  expressed  by  a 

fraction,  as  -,  it  follows  that: 
b 

The  changes  which  may  be  made  upon  the  terms  of  a  fraction 

ivithout  altering  its  value  may  be  made  upon  the  terms  of  a  ratio 

without  altering  the  ratio. 


136  PLANE   GEOMETRY.  — BOOK  III. 

262.  1.  What  two  numbers  have  the  same  relation  to  each 
other  as  3  has  to  6  ?     2  to  8  ?     5  to  15  ?     8  to  4? 

2.  What  numbers  have  the  same  relation  to  each  other  that 
4  in.  has  to  2  f t.  ?     5  ft.  to  2  yd.  ?     5^"^  to  1™  ?     S'^"  to  8^'"  ? 

3.  What  number  has  the  same  relation  to  6  that  2  has  to  4  ? 

4.  What  number  has  the  same  relation  to  12  that  3  has  to  9  ? 

5.  What  number  has  the  same  ratio  to  8  that  5  has  to  15  ? 

263.  An  equality  of  ratios  is  called  a  Proportion. 

The  sign  of  equality  is  written  between  the  equal  ratios. 

a:h  =  c:d  is  di>  proportion^  and  is  read :  the  ratio  of  a  to  6  is  equal  to  the 
ratio  of  c  to  d,  or  a  is  to  6  as  c  is  to  d. 

The  double  colon,  : :,  is  frequently  used  instead  of  the  sign  of 
equality. 

264.  The  antecedents  of  the  ratios  which  form  a  proportion 
are  called  the  Antecedents  of  the  proportion,  and  the  consequents 
of  those  ratios  are  called  the  Consequents  of  the  proportion. 

In  the  proportion  a:h  —  c-.d,  a  and  c  are  the  antecedents^  and  h  and  d 
are  the  consequents  of  the  proportion. 

265.  The  first  and  fourth  terms  of  a  proportion  are  called  the 
Extremes  and  the  second  and  third  terms  are  called  the  Means  of 
the  proportion. 

In  the  proportion  a :  6  =  c :  d,  a  and  d  are  the  extremes^  and  6  and  c  are 
the  means. 

266.  A  quantity  which  serves  as  both  means  of  a  proportion 
IS  called  a  Mean  Proportional. 

In  the  proportion  a :  6  =  6  :  c,  6  is  a  mean  proportional.  * 

267.  Since  a  proportion  is  an  equality  of  ratios,  and  the  ratio 
of  one  quantity  to  another  is  found  by  dividing  the  antecedent 
by  the  consequent,  it  follows  that : 

A  proportion  may  he  expressed  as  an  equation  in  which  both  mem- 
bers are  fractions. 

The  proportion  a:b=.C'.d  may  be  written  -  =  -. 

Such  an  expression  is  to  be  read  as  the  ordinary  form  of  a  proportion  is 
read. 


PLANE  GEOMETRY.-rBOOK  III,  137 

268.  Since  a  proportion  may  be  regarded  as  an  equation  in 
whicli  both  members  are  fractions,  it  follows  that: 

1.  The  changes  which  may  be  made  upon  the  members  of  an  equor 
Hon  without  destroying  the  equality  may  be  made  upon  the  couplets 
of  a  proportion  without  destroying  the  equality  of  the  ratios. 

2.  Tlie  changes  which  may  be  made  upon  the  terms  of  a  fraction 
without  altering  the  value  of  the  fraction  may  be  made  upon  the 
terms  of  each  ratio  of  a  proportion  without  destroying  the  propor- 
tion. 

Proposition  I 

269.  1.  Form  several  proportions,  as  3 : 5  =  9 :  15,  and  discover  how 
the  product  of  the  extremes  compares  with  the  product  of  the  means  in 
each. 

2.  If  the  means  in  any  proportion  are  the  same,  how  may  the  means 
be  found  from  the  product  of  the  extremes? 

3.  Form  a  proportion  whose  consequents  are  equal.  How  do  the 
antecedents  compare? 

4.  Form  a  proportion  in  which  either  antecedent  is  equal  to  its  conse- 
quent.    How  does  the  other  antecedent  compare  with  its  consequent? 

Theorem,  In  any  proportion,  the  product  of  the  extremes 
is  equal  to  the  product  of  the  means. 

Data :  a :  d  =  c :  cL 

To  prove  cm?  =  6c 

Proof.    From  data,  §  267,    %  =  %; 

0     a 

Multiplying  each  member  of  this  equation  by  6d, 

Therefore,  etc  q.b.d. 

270.  Cor.  I.  A  mean  proportional  between  two  quantities  is 
equal  to  the  square  root  of  their  product. 

Lf  Jib  =  b:c,  find  the  value  of  b. 

271.  Cor.  II.  If  in  any  proportion  any  antecedent  is  equal  U* 
its  consequent,  the  other  antecedent  is  equal  to  its  consequent. 


138  PLANE   GEOMETRY.  — BOOK  III. 

272.  Cor.  III.  If  the  consequents  of  any  proportion  are  equaly 
the  antecedents  are  equal,  and  conversely. 

For,  if  a:h  =  c:hf 

a     c 

1=1' 

and  multiplying  by  6,  a  =  c. 

Proposition  II 

273.  1.  If  the  product  of  the  extremes  of  a  proportion  is  48,  -what 
may  the  extremes  be?     If  72?  If  30?  If  36?  If  6a2?  If  12  aft?  If  ahcl 

2.  If  the  product  of  the  means  is  48,  what  may  the  means  be  ?  If 
96?  If  108?  If  Qhcdl   If  ahcdl   If  a%2?  If  ahc'i 

3.  Form  a  proportion  the  product  of  whose,  extremes  or  means  is  60 ; 
72;  84;  80;  64;  144;  x^y'^ ,  xyz;  xyzv. 

Theorem,  If  the  product  of  two  quantities  is  equal  to  the 
-product  of  two  others,  eithsr  two  -may  he  D%ade  the  extremes 
of  a  proportion  of  which  the  other  two  are  the  means. 

Data :  ad  —  be. 

To  prove  a:h  =  c:d. 

Proof.     Data,  ad  =  be. 

Dividing  each  member  of  this  equation  by  bd, 


a     c 
b~d' 

at  is, 

a:b  =  c:d. 

Therefore,  etc. 

Q.E.D. 

Ex.  393.  If  the  vertical  angle  of  an  isosceles  triangle  is  30°,  what  is  its 
ratio  to  each  of  the  base  angles  ? 

Ex.  394.  If  the  exterior  angle  at  the  base  of  an  isosceles  triangle  is  100°, 
what  is  its  ratio  to  each  angle  of  the  triangle  ? 

Ex.  395.  If  one  of  the  acute  angles  of  a  right  triangle  is  40°,  what  is  its 
ratio  to  the  other  acute  angle  ?    To  the  right  angle  ? 

Ex.  396.  The  interior  angles  on  the  same  side  of  a  transversal  cutting 
two  parallel  lines  are  to  each  other  as  8  to  2.  How  many  degrees  are  there 
in  each  angle  ? 

Ex.  397.  The  vertical  angle  of  an  isosceles  triangle  has  the  same  ratio  to 
a  right  angle  that  an  angle  of  40*^  has  to  an  angle  of  an  equilateral  triangle. 
How  many  degrees  are  there  in  each  angle  of  the  isosceles  triangle  ? 


PLANE   GEOMETRY.  — BOOR  til.  139 

Proposition  III 

274.  1.  Form  a  proportion  and  transpose  the  means.  How  do  the 
resulting  ratios  compare? 

2.  Transpose  the  extremes.     How  do  the  resulting  ratios  compare? 

3.  Transform  similarly  and  investigate  other  proportions. 

Theorem,  In  any  proportion,  the  first  term  is  to  the  third 
as  the  second  is  to  the  fourth;  that  is,  tl%e  terms  are  in  pro- 
portion by  alternation. 

Data :  a'.h  =  c:d. 

To  prove  a  :  c  =  6  :  c?. 

Proof.     From  data,  §  267,    ?  =  -• 

0     a 

Multiplying  each  member  of  this  equation  by  -, 

c 

that  is,  a :  c  =  6  :  d. 

Therefore,  etc.  q.e.d. 

Proposition  IV 

275.  1.  Form  a  proportion.  If  the  antecedent  of  each  ratio  becomes 
the  consequent,  and  the  consequent  the  antecedent,  how  do  the  resulting 
ratios  compare  ? 

2.   Transform  similarly  and  investigate  other  proportions. 

Theorem,  In  any  proportion,  the  ratio  of  the  second  term 
to  the  first  is  equal  to  the  ratio  of  the  fourth  term  to  the 
third;  that  is,  the  terms  are  in  proportion  hy  inversion. 

Data :  a-.h  —  cd. 

To  prove  6  :  a  =  c? :  c. 

Proof.     From  data,  §  269,   6c  =  ad. 

Dividing  each  member  of  this  equation  by  o/a^ 

b^d, 

a     c ' 
that  is,  b:a  =  d:c. 

'  Therefore,  etc.  q.e.d. 


140  PLANE   GEOMETRY.  — BOOK  III. 

Proposition  V 

276.  1.  Forni  a  proportion.  How  does  the  ratio  of  the  sum  of  the 
first  two  terms  to  either  term  compare  with  the  ratio  of  the  sum  of  the 
last  two  terms  to  the  corresponding  term  ? 

2.   Transform  similarly  and  investigate  other  proportions. 

Theorem.  In  any  -proportion,  the  ratio  of  tlw  sum  of  the 
first  two  terms  to  either  term  is  equal  to  the  ratio  of  the  sum 
of  the  last  two  terms  to  the  eorresponding  term;  that  is,  the 
terms  are  in  proportion  hy  composition. 

Data :  a\h  —  c.d. 


or 


To  prove        a  +  bi 

6  =  c-hd: 

d,  and  a  +  b 

Proof.     §  267, 

a_ 
b^ 

_c 
~~d 

Adding  1  to  each  member  of  this  equation. 

i^'- 

=1+1' 

a-\-b_ 
b 

c  +  d, 

-    d    ' 

that  is,  a  +  b:b  —  c-\-d:d. 

In  like  manner  it  may  be  shown  that  a  -\-b:a  =  c  +  d:c. 
Therefore,  etc.  q.e.d. 

Proposition  VI 

277.  1.  Form  a  proportion.  How  does  the  ratio  of  the  difference  of 
the  first  two  terms  to  either  term  compare  with  ratio  of  the  difference  of 
the  last  two  terms  to  the  corresponding  term? 

2.   Transform  similarly  and  investigate  other  proportions. 

Theoretn,  In  any  proportion,  the  ratio  of  tJie  difference 
between  the  first  two  terms  to  either  terin  is  equal  to  tJie 
ratio  of  the  difference  between  the  last  two  terms  to  the  cor- 
responding term;  that  is,  the  term^s  are  in  proportion  hy 
division. 

Data :  a:b  —  c'.d. 

To  prove        a  —  6:6  =  c  —  d:c?,  and  a  —  b:a  =  c  —  d'.c. 


PLANE   GEOMETRY.  — BOOK  III.  143 

Proof.     §267,  ?  =  ^. 

h     d 

Subtracting  1  from  each  member  of  this  equation, 

h  ^^~d     ^' 
a  —  h     c  —  d 

that  is,  a  —  b:b  =  c  —  d:d. 

In  like  manner  it  may  be  shown  that  a  —  b:a=c  —  d:c. 
Therefore,  etc.  q.e.d. 

Proposition  VII 

278.  1.  Form  a  proportion.  How  does  the  ratio  of  the  sum  of  the 
first  two  terms  to  their  difference  compare  with  the  ratio  of  the  sum  of 
the  last  two  terms  to  their  difference? 

Theorem,  In  amj  proportion,  the  ratio  of  the  sum  of 
the  first  two  terirvs  to  their  difference  is  equal  to  the  ratio 
of  the  sum  of  ths  last  two  terms  to  their  difference;  that 
is,  the  terms  are  in  proportion  by  composition  and  division* 

Data :  a  :  ft  =  c :  c?. 

To  prove  a  -{-  b  i  a  —  b  ==  c  -\-  d :  c  —  d. 

Proof.     §§276,267,      <^l±A=,^±A,  (1) 

b  d 

and,  §§  277,  267,  ^^^  =  ^-^.  (2) 

b  d 


Dividing  (1)  by  (2), 


a-{-b  _c-{-d^ 
a  —  b     G  —  d^ 


that  is,  a-\-b:a  —  b  —  c-\-d'.Q  —  d. 

Therefore,  etc.  q.e.d. 


142  PLANE  GEOMETRY,^ BOOK  IIL 

Proposition  VIII 

279.  1.  Form  a  series  of  equal  ratios,  as  2 :  3  =  4  :  6  =  8  :  12  =  10 :  15. 
How  does  the  ratio  of  the  sum  of  the  antecedents  to  the  sum  of  the  con- 
sequents compare  with  the  ratio  of  any  antecedent  to  its  consequent? 

2.   Transform  similarly  and  investigate  other  series. 

Theorem.  In  a  series  of  equal  ratios,  the  sum  of  the 
antecedents  is  to  the  sum  of  the  consequents  as  any  ante- 
cedent is  to  its  consequent. 

Data :    Any  series  of  equal  ratios,  as  a:b  =  C'.d  =  e  :/=  g :  h. 
To  prove  a-f  c  ^  e -^g-.h-^-d  +/+  A  =  a  :  6,  or  c :  d,  etc. 

Proof.     Denoting  each  ratio  by  r,  "  =  -  =  -  =  ^  =  r.  (1) 

b     d    f    h 

From  (1),       a^hr,  c^dr,  e^ftf  and  g^hr.  (2) 

Adding  equations  (2), 

a-hc-he4-g'  =  (&4-d  4-/-I-  h)r.  (3) 

Dividing  (3)  by(b^d  +/+  h), 

g-f  c4-e4-9^y 
b-^d  4-/+  h 
Substituting  the  value  of  r  in  (1), 

b  +  d+f+h-b'  ""'  d'  ^' 
that  is,    a-^c-^e-^g:b  +  d  +/+  fe  =  a :  6,  or  c :  d,  etc. 
Therefore,  etc.  q.e.to. 

•  Ex.  398.   It  a:b  =  c:d,  prove  that  a:a  +  6  =  C:c  +  d. 

Ex.  399.    liaib  =  b  ic,  prove  that  a  -\- b  :  b  +  c  =  a  :b. 

Ex.  400.  AD  bisects  angle  A  at  the  base  of  the  isosceles  triangle  ABC^ 
and  meets  the  side  BO  in  D.  If  angle  0  is  68°,  what  is  its  ratio  to  angle 
ADB? 

Ex.  401.  The  sura  of  the  angles  of  a  polygon  expressed  in  right  angles  is 
to  the  number  of  its  sides  as  4  is  to  3.     How  many  sides  has  the  polygon  ? 

Ex.  402.  If  the  angle  formed  by  two  secants  intersecting  without  a  circle 
is  30°  and  the  smaller  of  the  intercepted  arcs  is  20°,  what  is  the  ratio  of  the 
smaller  arc  to  the  larger  ? 

Ex.  403.  If  the  angle  formed  by  two  intersecting  chords  of  a  circle  is  40° 
and  one  of  the  intercepted  arcs  is  30°,  what  is  the  ratio  of  that  arc  to  the 
opposite  intercepted  arc  ? 


PLANE   GEOMETRY.  — BOOK  III.  143 

Proposition  IX 

280.  Form  two  or  more  proportions  in  which  the  corresponding 
consequents  are  equal,  as  2:3  =  6:9  and  5  :  3  =  15  :  9.  How  does  the 
ratio  of  the  sum  of  the  antecedents  of  the  first  couplets  to  their  common 
consequent  compare  with  the  ratio  of  the  sum  of  the  antecedents  of  the 
second  couplets  to  their  common  consequent? 

Theorem,  When  two  or  more  proportions  have  the  same 
quantity  as  the  consequents  of  the  first  couplets  and  an- 
other quantity  as  the  consequents  of  the  second  couplets, 
the  sum  of  the  antecedents  of  the  first  couplets  is  to  their 
common  consequent  as  the  sum  of  the  antecedents  of  the 
second  couplets  is  to  their  common  consequent. 

Data :  a:b  =  c:d,  (1) 

e:b=f:d,  (2) 

and  g:b  =  h:d.  (3) 

To  prove  a-\-e-{-g:b  =  c  -|-/+  h  :  d, 

(4) 

.      (6) 
(6) 


Proof. 

From  (1) 

a      c 

I'd' 

from  (2) 

h      d' 

from  (3) 

9.  =  ^., 
b     d' 

...(4) +  (5) +  (6),         « 

+  e  +  ^     c-hf+h, 
b          ~        d        ' 

that  is, 

a  +  ' 

e+9 

■.b  =  c+f+h:d. 

Therefore,  etc. 

Q.E.D. 

281.  Cor.  When  two  or  more  proportions  have  the  same  quan- 
tity as  the  antecedents  of  the  first  couplets,  and  another  quantity  as 
the  antecedents  of  the  second  couplets,  the  common  antecedent  of  the 
first  couplets  is  to  the  sum  of  their  consequents  as  the  common  ante- 
cedent  of  the  second  couplets  is  to  the  sum  of  their  consequents. 

}lx.  404.  It  a:b  =  c:d,  prove  that  2  a  +  h  :  b  =2  c  +  d:  d. 

Ex.  405.  It  a:b  =  c  :d,  prove  that  a  :S  a  +  b  =  c  :3  c  +  d, 

Ex.  406.  If  a  :b  =  b  :c,  prove  that  2  a  —  b  :  a  =  2b  —  c  :b, 

Ex.  407.  Ua:b  =  b:c,  prove  that  a  +  36:d  =  6  +  3c:c 


144  PLANE   GEOMETRY.  — BOOK  III, 

Proposition  X 

282.    1.  Form  a  proportion ;  multiply  or  divide  the  terms  of  either 
ratio  by  any  number.     How  do  the  resulting  ratios  compare  ? 
2.   Transform  similarly  and  investigate  other  proportions. 

Theorem.  If  in  a  proportion  the  terms  of  either  coup- 
let are  inultiplied  hy  any  quantity,  the  resulting  ratios 
forin  a  proportion. 

Data :  a:h  =C'.d. 


To  prove 

ma 

:7nb  = 

=  c:d. 

Proof. 

a 

c 
d 

Multiplying  both  terms  of  the  first  fraction 

by 

ma 
mb 

c 

^d' 

that  is, 

ma 

:mb  = 

=  c:d. 

Therefore,  etc. 

Q.E.D. 

283.  Cor.  If  in  a  proportion  the  terms  of  either  couplet  are 
divided  by  any  quantity,  the  resulting  ratios  form  a  proportion. 

Proposition  XI 

284.  1.  Form  a  proportion;  multiply  or  divide  the  antecedents  or 
the  consequents  by  any  number.     How  do  the  resulting  ratios  compare? 

Theorem,  If  in  any  proportion  the  antecedents  or  the 
consequents  are  multiplied  hy  the  same  quantity,  the  re- 
sulting ratios  are  in  proportion. 

Data :  a\b  =  c:d. 

To  prove  ma:b  =  mc:  d, 

and  a:nb  =  c:  nd. 

M-  (1) 

Multiplying  (1)  by  m,        ^  =  ^ ; 
that  is,  ma:b  =  mc:d. 


PLANE   GEOMETRY.  — BOOK  IIL  145 


a        c 


Dividing  (1)  by  71,  ^  =  ^; 


that  is,  a:nh  =  c'.  nd. 

Therefore,  etc.  q.e.d. 

285.  Cor.  If  in  any  proportion  the  antecedents  or  the  conae" 
quents  are  divided  by  the  same'  quantity,  the  resulting  ratios  are  in 
proportion. 

Proposition  XII 

286.  1.  Form  several  proportions.  Multiply  together  their  correspond- 
ing terms,  and  discover  whether  the  resulting  quantities  form  a  pro- 
portion. 

2.  If  there  is  an  equal  antecedent  and  consequent  in  the  same  couplet, 
or  in  corresponding  couplets,  cancel  them  from  the  products  of  the  corre- 
sponding terms.     Do  the  resulting  quantities  form  a  proportion  ? 

Theorem.  The  products  of  the  corresponding  terms  of 
any  number  of  proportions  are  in  proportion. 

Data :         a:b  =  c:d,  e  :f=  g  :  h,  and  k:  l  =  m:o. 

To  prove  aek :  bji  =  cgm  :  dho. 

^     ^  a      c    e     q         ^    h     m 

Proof.  T  =  :5'   >=??   ^^d   y  =  — 

b     d    f    h  I      o 

Multiplying  these  equations  together, 

aek  _  cgm 

'bfl~~dho'^ 
that  is,  aek :  bjl  =  cgm :  dho. 

Therefore,  etc.  q.e.d. 

287.  Cor.  In  finding  the  proportion  formed  by  the  products  of 
the  corresponding  terms  of  any  number  of  proportions,  an  equal 
antecedent  and  consequent  in  the  same  couplet,  or  in  corresponding 
couplets,  may  be  dropped, 

Por,  if  a:b  =  ci  d, 

and  b:e—f:c, 

§  286,  ab:be  =  cf:  do. 

Dividing  the  terms  of  the  first  couplet  by  b  and  the  terms  of  the 
second  by  c,  §  283,  a :  e  =/;  d, 

milne's  geom.  —  to 


146  PLANE   GEOMETRY,  — BOOK  III, 

Proposition  XIII 

288.   1.   Form  a  proportion;  raise  the  terms  of  both  ratios  to  the 
same  power.     How  do  the  resulting  ratios  compare  ? 

2.  Extract  the  same  root  of  the  terms  of  both  ratios  in  a  proportion, 
as  4  :  9  =  16  :  36.     How  do  the  resulting  ratios  compare  ? 

3.  Transform  similarly  and  investigate  other  proportions. 

Theorem.    In  any  proportion,  like  powers  or  like  roots 
of  the  terms  are  in  proportion. 

Data :  a  :  6  =  c    d 

Li        II 
To  prove  i"  :  &"  s=.  c"  •  d",  and  a** :  &"  =s  c"  :  d*. 

Proof.  J  =  |.  (1) 

Raising  both  fractions  in  (1)  to  the  wth  power, 

that  is,  a''\b^=:c''i  <t. 

Extracting  the  nth  root  of  both  fractions  io  (1), 

n — i» 

that  13,  or" ;  ft'' » (f* :  d*. 

Therefore,  etc.  Q.I5.D. 

Ex.  408.  Make  the  changes  that  may  be  made  upon  the  following  propor- 
tion without  destroying  the  equality  of  the  ratios  :  16  :  36  :a  4  : 9. 
Ex.  409.    If  a  :  6  =  c  :  df,  prove  that  ma  inh  =  71(10'.  nd. 
Ex.  410.    If  a:6  =  c:(f,  prove  that  a -h^b  :b  =  c  +  id:d, 
Ex.  411.  If  a  :  6  =  6  : c,  prove  that  a^  -{•  ab  :  b^  +  be  =  a  iC 
Ex.  412.   If  a  :  6  =  6  :  c,  prove  that  a  :  c  =  (a  +  6)2  :  (6  +  c)*. 
Ex.  413.    If  a:  b  =  m:n,  and  b:c  =  n:Of  prove  that  aic^mti^ 
Ex.  414.    If  a  :  6  =  c  :  ^,  prove  that 

ma  —  nb :  ma  •{■  nb  =  mc  -~  nd  :mc  ^  nd 
Ex.  415.   Ua:b  =  c:d,  prove  that 

So  +  4(>:4o-6&  =  8c  +  4d:4c-6<i 


BOOK  IV 

PROPORTIONAL  LINES   AND  SIMILAR  FIGURES 

Proposition  I 

289.  1.  Draw  a  line  parallel  to  the  base  of  a  triangle  through  the 
middle  point  of  one  side  and  cutting  the  other  side.  How  do  the  seg- 
ments of  the  other  side  compare  in  length  ? 

2.  Draw  a  line  parallel  to  the  base  one  fourth,  one  sixth,  or  any  part 
of  the  distance  from  the  extremity  of  the  base  to  the  vertex.  How  do 
the  segments  of  the  other  side  compare  ? 

3.  How  does  the  ratio  of  one  of  these  sides  to  either  of  its  segments 
compare  with  the  ratio  of  the  other  to  its  corresponding  segment  ? 

Theorem,  A  line  which  is  parallel  to  one  side  of  a  tri- 
angle and  meets  the  other  two  sides  divides  those  sides 
proportionally. 

c 
Data:  Any  triangle,  as  ABC,  and  any  /\ 

line  parallel  to  AB,  as  de,  meeting  AC  /......X^^ 

and  BC  in  D  and  E,  respectively.  /      ]7Z    ^v 

To  prove   CD  \DA  =  CE  :EB.  ^Z '— —^5    -M 

Proof.     Case  I.    When  CD  and  DA  are  commensurable. 
Suppose  that  Jif  is  a  unit  of  measure  common  to  CD  and  DA,  and 
that  M  is  contained  in  CD  3  times  and  in  D^  2  times. 

Then,  hyp.,  CD  \DA  =  Z:2. 

Divide  CD  and  DA  into  parts  each  equal  to  the  common  measure 
if,  and  from  each  point  of  division  draw  lines  parallel  to  AB. 
§  157,  these  lines  divide  CE  into  3  and  EB  into  2  equal  parts ; 

.-.  CE:EB  =  ^.2, 

and,  Ax.  1,  CD  :  DA  =  CE  :  EB. 

147 


148  PLANE   GEOMETRY.  — BOOK  IV. 

Case  II.    When  CD  and  DA  are  incommensurable. 

Since  CD  and  DA  are  incommensurable, 

c 
suppose  that  CD  and  DF  are  commensur-  A. 

able  and  that  FA  <  M.  /     ^\ 

Draw                 FG  II DE.  ^/       ~       \.  ^ 

Case  I,      CD  :DF  =  CE  :  EG.  ^  ~~ ^-^ 

If  M  is  indefinitely  diminished,  the  ratios  CD  :  DF  and  CE  :  EG 

remain  equal,   and   indefinitely   approach  their  limiting  ratios 

CD  :  DA  and  CE  :  EB,  respectively. 
Hence,  §  222,  CD  :  DA  =  CE  :  EB. 

Therefore,  etc.  q.e.d. 

290.  Cor.  '  A  line  which  is  parallel  to  one  side  of  a  triangle  and 
meets  the  other  two  sides  divides  them  so  that  one  side  is  to  either  of 
its  segments  as  the  other  side  is  to  its  corresponding  segment. 

Proposition  II 

291.  Draw  a  line  dividing  each  of  two  sides  of  a  triangle  into  halves, 
or  into  other  proportional  parts.  What  is  the  direction  of  this  line  with 
reference  to  the  third  side  ? 

Theorem.  A  lUie  which  divides  two  sides  of  a  triangle 
proportionally  is  parallel  to  the  third  side.  (Converse  of 
Prop.  I.) 

Q 

Data:    Any  triangle,  as  ABC,  and                      /\ 
the  line  DE  dividing  AC  and  ^C  so                 /      ^vs^ 
that  CA  :  CD  =  CB  :  CE.  r/- ^^ 

To  prove      DE  II AB.  /  \. 

A^ ^J5 

Proof.  If  DE  is  not  parallel  to  AB,  some  other  line  drawn 
through  D  will  be  parallel  to  AB. 

Suppose  that  DF  is  that  line. 

Then,  §  290,  CA\CD  =  CB -.  CF\ 

but,  data,  CA  :CD  =  CB:GE', 

CB  :  CF  =  CB  :  CE  j 
hence,  §  272,  OF  =  CE. 


PLANE   GEOMETRY.— BOOK  IV.  149 

But  this  is  impossible  unless  F  coincides  with  E ; 
that  is,  Ax.  11,  unless  DF  coincides  with  DE. 

Therefore,  the  hypothesis,  that  some  line  other  than  DE  drawn 
through  D  is  parallel  to  AB,  is  untenable. 

Hence,  DE  II  AB. 

Therefore,  etc.  ^  q.e.d. 

Proposition  III 

292.  Draw  a  triangle  whose  sides  are  6",  5",  and  3",  or  any  other 
dimensions ;  bisect  any  one  of  its  angles  and  produce  the  bisector  to 
meet  the  opposite  side. 

How  does  the  ratio  of  the  segments  of  this  side  made  by  the  bisector 
compare  with  the  ratio  of  the  sides  of  the  triangle  adjacent  to  these 
segments  ? 

Theorem.  The  bisector  of  an  angle  of  a  triangle  divides 
the  opposite  side  into  segments  which  are  proportional  to 
the  adjacent  sides. 


Data:  Any  triangle,  as  ABC,  and  CD 

the  bisector  of  one  of  its  angles,  ACB. 

To  prove     AD  :  DB  =  AC  :  CB. 


D 


Proof.     From  B  draw  a  line  parallel  to  CD  and  meeting  AC  pro- 
duced in  E. 


Then,  §  289, 

AD 

:  DB  =  AC  :  CE, 

also. 

Zr  =  Zs, 

Why? 

and 

Zt  =  Zv', 

Why? 

but,  data, 

Zr  =  Zt; 

.-. 

Z  s  =  Zv, 

Why? 

and,  §  118, 

CB  =  CE. 

Substituting  CB 

in  the  proportion  for  its  equal  CE, 

AD 

:  DB  :=AC:  CB. 

Therefore,  etc. 

Q.E.D. 

150 


PLANE  GEOMETRY,-^ BOOK  IV. 


Proposition  IV 

293.    Draw  a  triangle  whose  sides  are  6",  5",  and  3",  or  any  other 

dimensions;  bisect  an  exterior  angle  at  any  vertex  and  produce  the 
bisector  to  meet  the  opposite  side  produced.  How  does  the  ratio  of  the 
distances  from  the  point  of  meeting  to  each  extremity  of  the  opposite 
side  compare  with  the  ratio  of  the  other  sides  of  the  triangle  ? 

Theorem,  The  bisector  of  an  exterior  angle  of  a  triangle 
meets  the  opposite  side  produced  at  a  point  the  distances  of 
which  from  the  extremMies  of  this  side  are  proportional  to 
the  other  two  sides. 

Data:  A  triangle,  as  ABO\ 
an  exterior  angle,  as  BC7i> ;  and 
its  bisector  CE  meeting  AB  pro- 
duced in  E. 

To  prove      AE :  BE=AC .  BO. 

Proof.    Draw  BF II  CE. 


Then,  §  290, 
also, 
and 
but,  lata, 

and,  §  118, 


AE:BE  =  AO:  FO, 

Z  5  as  Z  Vy 

FC  =  BQ. 

Substituting  BQ  in  the  proportion  for  its  equal  FC^ 

AE '.  BE  =  AC  :  BC. 

Therefore,  etc. 


Why? 
Why? 

Why? 


Q.B.D. 

294.  Sch.  This  proposition  is  not  true,  if  the  triangle  is  equi- 
lateral. •  Why  ? 

Ex.  416.  The  base  of  a  triangle  is  10  ft.  and  the  other  sides  8  ft  and 
12  ft.  Find  the  segments  of  the  base  made  by  the  bisector  of  the  vertical 
angle. 

Ex.  417.  The  sides  ^C  and  5(7  of  the  triangle  ABC  dire  5  ft.  and  8  ft. 
respectively.  If  a  line  drawn  parallel  to  the  base  divides  AC  into  segments 
of  2  ft.  and  3  ft.,  what  are  the  segments  into  which  it  divides  BC2 


PLANE   GEOMETRY.  — BOOK  IV.  151 

295.  If  a  straight  line  is  divided  at  a  point  between  its  extrem- 
ities, it  is  said  to  be  divided  internally.  The  line  is  equal  to  the 
sum  of  the  internal  segments. 

If  a  straight  line  is  produced  and  divided  at  a  point  on  the 
part  produced,  it  is  said  to  be  divided  externally.  The  line  is 
equal  to  the  difference  between  the  external  segments. 


c 

Fig.l. 

In  Fig.  1,  AB  is  divided  internally  at  C,  and 
AB  =  AG+  GB. 


Fig.  2. 

In  Fig.  2,  AB  is  produced  and  divided  externally  at  E^  and 
AB  =  BE-  AE. 

296.  If  a  line  is  divided  at  a  given  point  so  that  one  segment 
is  a  mean  proportional  between  the  whole  line  and  the  other  seg- 
ment, it  is  said  to  be  divided  in  extreme  and  mean  ratio. 

In  Fig.  1,  if  AB'.AC^AC:  CB, 

4.J5  is  divided  internally  in  extreme  and  mean  ratio  at  the  point  C. 

In  Fig.  2,  if  AB  :  AE  =  AE  :  BE, 

AB  is  divided  externally  in  extreme  and  mean  ratio  at  the  point  E. 

297.  If  a  line  is  divided  internally  and  externally  into  seg- 
ments which  have  the  same  ratio,  it  is  said  to  be  divided  har- 
monically. 


-E 


c       B 
Fig.  8. 

InFig.  3,  if  ^C:C5  =  6:3, 

and  if  AE:BE  =  Q'.^, 

AC :  CB  =  AE :  BE, 

and  AB  is  divided  harmonically  at  the  points  C  and  E. 

Ex.  418.  A  line  drawn  parallel  to  the  base  of  triangle  ^BC  divides  AC 
into  segments  of  3^'"  and  8<^">  respectively,  and  the  segment  of  J5C,  corre- 
sponding to  S*!™,  is  6^"".     What  is  the  length  ot  BC? 


152  PLANE   GEOMETRY.  — BOOK  IV. 


Proposition  V 

298.  Draw  a  triangle  whose  sides  are  6",  5",  and  3",  or  any  other 
dimensions ;  bisect  an  interior  and  an  exterior  angle  at  one  vertex  and 
produce  the  bisectors  to  meet  the  opposite  side  and  the  opposite  side 
produced  respectively.  How  do  the  ratios  of  the  internal  and  external 
segments  of  the  opposite  side  compare  ? 

Theorem,  The  bisectors  of  an  interior  and  of  an  exterior 
angle  at  one  vertex  of  a  triangle  divide  the  opposite  side 
harmonically. 

Data:  Any  triangle,  as  ABC,  ^ 

and  the  bisectors  CE  and  CF  of 
the  interior  and  an  exterior 
angle  at  C,  respectively. 

To  prove   AE  :  EB  ==  AF  :  BF. 

Proof.      §  292,  AE  :  EB  =z  AG  \  BC, 

§293,  AF:BF=:AC:BC', 

hence,  AE  :  EB  =  AF  :  BF.  Why  ? 

Therefore,  etc.  q.e.d. 

299.  Polygons  whose  homologous  angles  are  equal,  and  whose 
homologous  sides  are  proportional,  are  called  Similar  Polygons. 

In  similar  polygons,  points,  lines,  and  angles  that  are  similarly 
situated  are  called  homologous  points,  lines,  and  angles. 

Equal  polygons  are  similar,  but  similar  polygons  are  not 
necessarily  equal. 

Thus,  all  equilateral  triangles  are  similar,  but  not  all  equilateral  triangles 
are  equal. 

Ex.  419.  The  sides  AB,  BC,  and  AC  of  the  triangle  ABC  are  respec- 
tively 8  in.,  5  in.,  and  10  in.  The  bisector  of  the  exterior  angle  at  C  meets 
AB  produced  in  E.     What  is  the  length  of  BE  ? 

Ex.  420.  In  triangle  ABC,  AC  is  16""  and  J?C  is  5>n.  The  bisector  of 
the  exterior  angle  at  C  meets  AB  produced  in  E.  If  AE  is  21'",  what  is  the 
length  of  the  side  AB  ? 

Ex.  421.  The  bisectors  of  an  interior  and  exterior  angle  at  C  of  the 
triangle  ABC  meet  the  opposite  side  and  the  opposite  side  produced  in  E 
and  F,  respectively.  If  AB  is  14  in.  and  EB  is  4  in.,  what  are  the  interna' 
and  external  segments  of  AB  ? 


PLANE   GEOMETRY.  — BOOK  IV,  153 

Proposition  VI 

300.  Draw  a  triangle ;  draw  another  whose  angles  are  equal,  each  to 
each,  to  the  angles  of  the  first  and  one  of  whose  sides  is  double,  or  any 
other  number  of  times,  the  homologous  side  of  the  first.  How  do  the 
ratios  of  any  two  pairs  of  homologous  sides  jcompare?  What  kind  of 
triangles  are  they?     Why? 

Theorem,  Two  triangles  are  similar,  if  the  angles  of  one 
are  equal  to  the  angles  of  the  other,  each  to  each. 

Data:  Any  two  trian- 
gles, as  ABO  and  BEF,  in 
which  angle  A  —  angle  D, 
angle  B  =  angle  E,  and 
angle  c  =  angle  F. 

To   prove   A  ABC    and    ^' 

DEF  similar. 

Proof.  In- the  greater  triangle,  ABC,  measure  CG  equal  to  DF, 
CH  equal  to  EF,  and  draw  GH. 

Then,  §  100,  A  GHC  =  A  DEF, 

and  /.CGH  =  Z.D  =  ZA',  Why? 

GH  W  AB.  Why? 

Hence,  §  290,  AC:GC  =  BC:  HC, 

or,  substituting  DF  for  its  equal  GC,  and  EF  for  its  equal  HC, 
AC:BF  =  BC:EF. 

In  like  manner,  AB  :  DE  =  BC  :  EF. 

Since,  in  the  A  ABC  and  DEF  the  homologous  sides  are  pro- 
portional, and,  from  data,  the  homologous  angles  are  equal, 
§  299,  A  ABC  and  DEF  are  similar. 

Therefore,  etc.  q.e.d. 

301.  Cor.  I.  Two  triangles  are  similar ,  if  two  angles  of  one  are 
equal  to  ttuo  angles  of  the  other,  each  to  each. 

302.  Cor.  II.  Two  right  triangles  are  similar,  if  an  acute  angle 
of  one  is  equal  to  an  acute  angle  of  the  other. 

Ex.  422.  The  sides  of  a  triangle  are  5,  7,  and  9.  If  the  side  of  a  similar 
triangle  homologous  to  7  is  8,  what  are  the  other  sides  of  the  triangle  ? 


154 


PLANE   GEOMETRY.  —  BOOK  IV, 


Proposition  VII 

303.  Draw  any  two  triangles  such  that  the  sides  of  one  are  propor- 
tional to  the  sides  of  the  other ;  for  example,  draw  one  triangle  whose 
sides  are  3",  4",  and  5",  and  another  whose  sides  are  6",  8",  and  10". 
How  do  the  homologous  angles  compare  in  size  ?  What  kind  of  triangles 
are  they?     Why? 

TTieorem,  Two  triangles  are  similar,  if  the  sides  of  one 
are  proportional  to  the  sides  of  the  other,  ea^ch  to  eaeh. 

Data:  Any  two  tri- 
angles, as  ABC  and  DEF^ 
such  that 

AC :DF  = BC:EF 
—  AB  :  DE. 

To    prove    A  ABC    and.    ^ 
^EF  similar. 


Proof.     In  the  greater  triangle,  ABC,  measure  CG  equal  to  DF^ 
GH  equal  to  EF,  and  draw  GH. 


Data, 

Then,  §  291, 
hence, 
and,  §  301, 
.-.  §  299, 
that  is, 

But,  dat^, 

whence,  §  272, 
and,  §  107, 

But 
hence. 


AC iBF  =  BC lEFj 
AC:  GC  =  BC\HC. 
GH  II  AB  ; 
Z.A=/.  CGH,  /.B  =iZ.  CHG, 
A  ABC  and  GHC  are  similar; 
AC:  GC  =  AB:  GH\ 
AC  :  DF  =  AB  :  GH. 
AC:DF  =  AB:DE; 
AB  :  GH  =  AB  :  BE ; 
GH  =  BE, 
A  GHC  =  A  DEF. 
A  ABC  and  GHC  are  similar ; 
A  ABC  and  BEF  are  similar. 


Why? 


Q.E.D. 


304.    Sch.      In  §  299  the  characteristics  of  similar  polygons 
were  defined  as : 

1.  Their  homologous  angles  are  equal. 

2.  Their  homologous  sides  are  proportional. 

From  §  300  and  ^  303  it  is  seen  that  in  the  case  of  triancjles, 


PLANE   GEOMETRY.  — BOOK  IV. 


166 


either  condition  involves  the  other;  that  is,  if  the  homologous 
angles  of  two  triangles  are  equal,  the  homologous  sides  are  pro- 
portional, and  conversely;  hence,  triangles  are  similar,  if  theii 
homologous  angles  are  equal  or  if  their  homologous  sides  are 
proportional.  In  the  case  of  polygons  of  more  than  three  sides 
either  condition  may  exist  without  involving  the  other. 

Thus,  a  square  and  a  rhombus  may  Rave  their  sides  all  equal  and,  conse- 
quently, proportional,  but  the  angles  of  the  square  are  right  angles,  and 
'those  of  the  rhombus  are  oblique ;  therefore,  the  figures  are  not  similar. 
Also  a  square  and  a  rectangle  have  their  angles  all  equal,  but  their  sides  may 
not  be  proportional ;  consequently,  the  figures  are  not  similar. 


Proposition  VIII 

305.  Draw  two  similar  triangles  whose  sides  are  3",  4",  and  5",  and 
6",  8",  and  10"  respectively,  or  any  two  similar  triangles;  draw  lines 
representing  their  altitudes.  How  does  the  ratio  of  their  altitudes  com- 
pare with  the  ratio  of  any  two  homologous  sides  ? 

Theorem.  The  altitudes  of  similar  triangles  are  to  each 
other  as  any  two  homologous  sides. 

Data:  Any  two  similar 
triangles,  as  ABC  and  DEF, 
and  their  altitudes,  as  CQ 
and  FH,  respectively. 

To  prove  CG  :  FH  =  AC : 
DF  =  BC:FF  =  AB  :  DE, 


Proof.    Data, 
.-.  §  299, 
§94, 
.-.  §  302, 
and,  §  299, 


A  ABC  Sind  DEF  are  similar; 
ZA  =  ZDy 

Aagc  and  DHF  are  rt.  A\ 
rt.  Aagc  and  DHF  are  similar, 
CG  :  FH  =  AC  :  DF. 


In  like  manner  it  may  be  shown  that 

CG:FH  =  BC:EF. 
But,  §  299,  BC'.EF  =  AB\DE\ 


hence, 


CG  :  FH  =  AC  :  DF  =  BC  I  EF  ==  AB  :  DE. 


Therefore,  etc. 


Why? 

Q.E.D. 


166  PLANE  GEOMETRY.^ BOOK  IV. 

Proposition  IX 

306.  Draw  two  triangles  such  that  an  angle  of  one  is  equal  to  an 
angle  of  the  other,  and  the  including  sides  in  the  first  triangle  are  3" 
and  5",  and  in  the  second  6"  and  10".  How  do  the  homologous  angles 
compare?  How  do  the  ratios  of  any  two  pairs  of  homologous  sides 
compare?  What  name  is  given  to  triangles  that  have  such  relations  to 
3ach  other? 

Theorem.  Two  triangles  are  similar,  if  an  angle  of  one  is 
equal  to  an  angle  of  the  other,  and  the  sides  about  these 
angles  are  in  proportUm. 

Data:  Any  two  trian- 
gles,  as  ABC  and  VEF,  in 
which  angle  C  =  angle  F, 
and  AO:BO  =  DF:  EF, 

To  prove  A  ABC  and 
DEF  similar. 

Proof.  In  the  greater  triangle,  ABC,  measure  CQ  equal  to  BFf 
CH  equal  to  EF,  and  draw  QH. 

Then,  since,  da.ta,  Z.C  =  Af, 

§100,  AOHC^^ADEF. 

Data,  AC:BC  =  J)F:EF; 

AC:BC=GC:HC.  Why? 

Hence,  §  291,  GH  II  AB, 

ZA^Z  CQH,  and  Z  B  =  Z  CHO ;  Why  ? 

hence,  §  301,  A  ABC  and  GHC  are  similar; 

that  is,  A  ABC  and  DEF  are  similar. 

Therefore,  etc.  q.e.d. 

Ex.  423.  The  sides  of  a  triangle  are  8^™,  10<*™,  and  12^™  in  length  respec- 
tively. If  a  line  9'^"*  long,  parallel  to  the  longest  side,  terminates  in  tlie  other 
two,  what  are  the  segments  into  which  it  divides  them  ? 

Ex.  424.  If 'the  bisector  of  an  interior  angle  of  a  triangle  divides  the  side 
opposite  the  angle  into  two  segments  which  are  6  ft.  and  8  ft.  respectively, 
and  if  the  side  of  the  triangle  adjacent  to  the  8  ft.  segment  is  20  ft.,  what  is 
the  length  of  the  other  side  of  the  triangle  f 


PLANE   GEOMETRY.  — BOOK  IV. 


157 


Proposition  X 

307.  Draw  two  triangles  whose  sides  are  parallel,  each  to  each,  or  per- 
pendicular, each  to  each.  What  may  be  inferred  regarding  the  relative 
size  of  the  homologous  angles?     Then,  what  kind  of  triangles  are  they? 

Theorem.  Two  triangles  are  similar,  if  their  sides  are 
parallel,  each  to  eojch,  or  are  perpendicular,  each  to  each. 

Data :  Any  two  tri- 
angles, as  ABC  and 
DBF,  in  which  AB, 
AC,  and  BC  are  par- 
allel or  perpendicular  ^z i ^ ^^       j.  ^^ 

respectively    to    DE, 
DFy  and  EF. 

To  prove  A  ABC 
and  DEF  similar. 

Proof.  By  §§  81,  83,  angles  which  have  their  sides  either  paraL 
lei  or  perpendicular  are  either  equal  or  supplementary. 

1.  Suppose  that  each  of  the  angles  of  one  triangle  is  supple- 
mentary to  the  corresponding  angles  of  the  other ;  that  is,  suppose 
/.A-\-/:d=2  rt.  Zs;  Zj5 +Z^  =  2  rt.  Zs;  Zc  +  Zi^  =  2  rt.  a 

Then,  the  sum  of  the  interior  angles  of  the  two  triangles  is 
equal  to  6  rt.  A,  which  is  impossible. 

2.  Suppose  that  one  angle  of  one  triangle  is  equal  to  the  cor- 
responding angle  of  the  other,  and  that  the  other  two  angles  of 
the  triangles  are  supplementary,  each  to  each;  that  is,  suppose 
/La  =  /.D;  Z^  +  Z^=:2rt.  Z;  Z(7  +  Zi^  =  2rt.  Z. 

Then,  the  sum  of  the  angles  of  the  two  triangles  exceeds 
4  rt.  Z,  which  is  impossible. 

3.  Suppose  that  two  angles  of  one  triangle  are  equal  to  the 
corresponding  angles  of  the  other,  each  to  each ;  then  the  third 
angles  must  be  equal ; 

that  is,  suppose  Z.  A  =  Z  D\  /.B  =  Z.E, 

then,  /.C  =  /.F; 

that  is,  A  ABC  and  DEF  are  mutually  equiangular. 

Hence,  §  300,      A  ABC  and  DEF  are  similar. 

Therefore,  etc.  Q.E.D. 


158  PLANE   GEOMETRY.  — BOOK  IV, 

Proposition  XI 

308.  1.  Draw  three  or  more  lines  which  meet  in  a  point  and  two 
parallel  lines  cutting  them.  Discover  whether  any  pairs  of  triangles  thus 
formed  are  similar.     Are  the  pairs  of  bases  proportional? 

2.  If  three  non-parallel  lines  intersect  two  parallel  lines,  making  the 
intercepted  segments  4"  and  6"  on  one  side  of  the  middle  line  and  8"  and 
12"  on  the  other  side,  will  the  non-parallel  lines  meet  in  the  same  point 
if  produced  ? 

Theorem,  Lines  which  meet  in  a  point  intercept  propor- 
tional segments  upon  two  parallel  lines;  conversely,  non- 
parallel  lines  which  intercept  proportional  segments  upon 
two  parallel  lines  meet  in  a  point. 

Data:  Any  lines,  as  AH^  BH,  and 
CH,  which  meet  at  a  point,  as  H,  and 
intercept  the  segments  AB,  BC,  DE, 
and  EF  upon  two  parallel  lines,  AC 
and  DF. 

To  prove  ABiDE  =  BC:  EF. 

Proof.     Zr  =  Zs,Zt  =  Zv,Ziv  =  Zx,  and  Z  y 

:  §  301,  Aabh  and  BEH  are  similar, 

and      ,  A  BCH  and  EFH  are  similar. 

Then,  §  299,  AB  :DE  =BH:  EH, 

*».nd  BC:  EF  =  BH:EHj 

AB  :  BE  =  BC  :  EF.         *'  Q.E.D. 

Conversely :  Data :  Non-parallel  lines,  as  AB,  BE,  and  CF,  inter- 
secting parallel  lines  AC  and  DF,  so  that  AB  :  BE  =  BC  :  EF. 

To  prove  that  AD,  BE,  and  CF,  if  produced,  meet  in  a  point. 

Proof.     Produce  AD  and  BE  to  meet  in  H  and  draw  CH. 

Suppose  that  J  is  the  point  in  which  DF  intersects  CH. 

Then,  AB:  DE  =  BC  :  EJ,  Why  ? 

but,  since,  data,  ab  :  de  =  BC  :  EF, 

this  is  impossible,  unless       EJ  =  EF, 
and  J  and  F  coincide. 

Then,  CF  passes  through  H. 

Consequently,  AD,  BE,  and  CF  meet  in  a  point. 

Therefore,  etc.  q.b.d. 


PLANE   GEOMETRY.  —  BOOK  IV.  159 


Proposition  XII 

t 

309.  Draw  two  polygons  such  that  they  may  be  divided  into  the  same 
number  of  triangles,  similar,  each  to  each,  and  similarly  situated.  How 
do  the  homologous  angles  of  these  polygons  compare  in  size  ?  How  do 
the  ratios  of  any  two  pairs  of  homologous  sides  compare  ?  What  kind  ot 
polygons  are  they  ?    Why  ? 

Theorem,  Two  polygons  are  similar,  if  each  is  composed 
of  the  same  number  of  triangles  which  are  simAlar,  each  to 
eoAih,  and  similarly  placed. 


I^ata:  Any  two  polygons,  as  ABODE  and  FGHJK,  composed  of 
triangles  ABC,  ACD,  and  ABE ;  and  FGH,  FHJ,  and  FJK,  respec- 
tively, which  are  similar,  each  to  each,  and  are  similarly  placed. 

To  prove  ABODE  and  FGHJK  similar. 

Proof.  /Lb  =Zg.  Why  ? 

Also,  '  Z.r  =  /.Sy 

and  At  =  /.v, 

/.BOD  =  Z.GHJ.  Why? 

In  like  manner  it  may  be  shown  that  Z  ODE  =  Z  HJK,  etc. 

Hence,  the  homologous  angles  of  the  polygons  are  equal. 

Again,  §  299,  AB  :  FG  =  BO  :  GH  =  AC  :  FH  =  OD  :  HJ  =  etc., 
or  AB  :  FG  =  BC :  GH  =  OD  :  HJ  =  etc. ; 

that  is,  the  homologous  sides  of  the  polygons  are  proportional. 

Hence,  §  299,  ABODE  and  FGHJK  are  similar. 

Therefore,  etc.  q.b.d. 

Ex.  425.  If  a  stick  .3  ft.  long,  in  a  vertical  position,  casts  a  shadow 
1  ft.  7^  in.  long,  how  high  is  a  church  steeple  which  at  the  same  time  cast* 
a  shadow  78  ft.  in  length  ? 


160 


PLANE   GEOMETRY.  — BOOK  IV. 


Proposition  XIII 

• 
310.  Draw  two  similar  polygons  and  from  two  homologous  vertices 
draw  diagonals  dividing  the  polygons  into  triangles.  How  many  tri- 
angles are  there  in  each  polygon  ?  How  do  the  homologous'^  angles  of  the 
corresponding  triangles  compare  in  size?  How  do  the  ratios  of  their 
sides  compare  ?     Then,  what  kind  of  triangles  are  they  ? 

Theorem,  If  two  polygons  are  similar,  they  may  he 
divided  by  diagonals  into  the  same  numher  of  triangles 
which  are  similar,  each  to  each,  and  similarly  placed. 
(Converse  of  Prop.  XII.) 


Data:  Any  two  similar  polygons,  as  ABCDE  and  FGHJK.     ' 
To  prove  that  the  polygons  ABODE  and  FGHJK  may  be  divided 

by  diagonals  into  the  same  number  of  triangles  which  are  similar, 

each,  to  each,  and  are  similarly  placed. 

Proof.     From  any  two  homologous  vertices,  as  A  and  F,  draw 

the  diagonals  AC,  AD,  FH,  and  FJ. 

.   In  A  ABC  and  FGH,      Z.  B  =  Z.  G, 


and  AB  :FG  =BC:GH', 

.'.  §  306,  A  ABC  and  FGH  are  similar, 

Z  r  =  Zs] 

Zbcd  =  Z  ghj'j 

Zt  =  Zv', 

BC:  GH=  ACiFH, 
BC:  GH=  CDiHJj 
AC'.FH  =  CD  :HJ, 
and,  §  306,  A  ACD  and  FHJ  are  similar. 

In  like  manner,   A  ADE  and  FJK  are  similar. 

Therefore,  etc. 


and 
but 


and 


Why? 


Why? 
Why? 
Why? 
Why? 
Why? 
Why? 


Q.E.D. 


PLANE   GEOMETRY.  — BOOK  IV.  161 

Proposition  XIV    . 

311.  Draw  two  similar  polygons;  measure  the  sides  of  each.  How 
does  the  ratio  of  the  perimeters,  or  the  sums  of  the  homologous  sides, 
compare  with  the  ratio  of  any  two  homologous  sides? 

Theorem,  The  perimeters  of  similar  -polygons  are  to  each 
other  as  any  two  homologous  sides. 


Data:   Any  two  similar  polygons,  as  ABODE  and  FGHJK, 
Denote  their  perimeters  by  P  and  Q  respectively. 

To  prove  P  :  Q  =  AB  :  FG  =  etc. 

Proof.  AB  :  FG  =  BC  :  GH  =  CD  :  HJ  =  etc. ;  Why  ? 

.-.  §  279,         AB  -i-BC-^-  etc. :  FG  +  GH-{-  etc.  =  AB:FG  =  etc. ; 
that  is,  P  :  Q  =  AB  :  FG  =  etc. 

Therefore,  etc.  q.e.d. 

Proposition  XV 

312.  1 .  Draw  a  right  triangle  whose  sides  are  3",  4^',  and  5",  or  any 
other  right  triangle ;  from  the  vertex  of  the  right  angle  draw  a  perpen- 
dicular to  the  hypotenuse.  How  do  the  angles  of  each  of  the  triangles 
thus  formed  compare  in  size  with  the  homologous  angles  of  the  original 
triangle"?  How  does  the  ratio  of  the  longer  segment  of  the  hypotenuse 
to  the  perpendicular  compare  with  the  ratio  of  the  perpendicular  to  the 
shorter  segment  ? 

2.  How  does  the  ratio  of  the  hypotenuse  to  either  side  about  the  right 
angle  compare  with  the  ratio  of  the  same  side  to  the  segment  of  the 
hypotenuse  adjacent  to  it  ? 

3.  Draw  a  circle  and  its  diameter ;  from  any  point  in  the  circumfer- 
ence draw  a  perpendicular  to  the  diameter.  How  does  the  ratio  of  the 
longer  segment  of  the  diameter  to  the  perpendicular  compare  with  the 
ratio  of  the  perpendicular  to  the  shorter  segment  ? 

MILNfi's   OEOM. — 11 


162  PLANE    GEOMETRY.  —  BOOK  IV. 

Theorem,  If  in  a  right  triangle  a  perpendicular  is 
drawn  from  the  vertex  of  the  right  angle  to  the  hypotenuse, 
the  perpendicular  is  a  mean  proportional  between  the  seg- 
ments of  the  hypotenuse. 

Data:  Any  right  triangle,  as  ABC,  and 
the  perpendicular  CD,  from  the  vertex  of 
the  right  angle  C,  upon  AB. 

To  prove       AD  :  CD  =  CD  :  BD. 

Proof.     In  the  rt.  A  ABC  and  ACDy 

Za  is  common ; 

.*.  §  302,  A  ABC  and  ACD  are  similar. 

In  like  manner  it  may  be  shown  that  A  ABC  and  CBD  are 
similar ; 
hence,  A  ^  CD  and  C^i)  are  similar.  Why? 

Now,  AC,  AD,  and  CD  are  the  sides  of  A  ACD  homologous 
respectively  to  BC,  CD,  and  BD  of  A  CBD  ; 

hence,  §  299,  AD  :  CD  =z  CD  :BD. 

Therefore,  etc.  q.e.d. 

313.  Cor.  I.  Each  side  about  the  right  angle  is  a  mean  propor- 
iional  between  the  hypotenuse  and  the  adjacent  segment. 

314.  Cor.  II.  The  perpendicular  to  a  diameter  from  any  point  in 
the  circumference  of  a  circle  is  a  mean  proportional  between  the 
segments  of  the  diameter. 

Proposition  XVI 

315.  From  a  point  without  a  circumference  draw  a  tangent  and  a 
secant;  from  the  point  of  tangency  draw  chords  to  the  points  at  which 
the  secant  intersects  the  circumference.  What  angles  of  the  figure  are 
equal?  What  two  triangles  are  similar?  Then,  how  does  the  ratio  of 
the  secant  to  the  tangent  compare  with  the  ratio  of  the  tangent  to  the 
external  segment  of  the  secant  ? 

Theorem.  If  from  a  point  without  a  circle  a  secant  and 
a  tangent  are  drawn,  the  tangent  is  a  mean  proportional 
between  the  whole  secant  and  the  external  segment. 


PLANE   GEOMETRY.  — BOOK  IV, 


Data:  Any  circle,  as  BCD', 
any  point  without,  as  A ;  any 
secant  from  A,  as  ADB  ;  and  the 
tangent  from  ^,  as  ^C. 

To  prove 

AB  :  AC  =  AC:  AD, 

Proof.     Draw  BC  and  DC. 

In  A  ABC  and  ADC,  Z  A  is  common, 
§  225,  Z  ^  is  measured  by  |  arc  DC, 

and,  §  231,  Z  ACD  is  measured  by  ^  arc  DC; 

.'.  ■  Zb  ==Zacd. 

Hence,  §  301,         A  ABC  and  ADC  are  similar, 
and,  §299,  AB  :AC  =  AC:AD. 

Therefore,  etc. 

Proposition  XVII 


Why? 


Q.E.D. 


316.  JProhlem,    To  divide  a  straight  line  into  parts  pro- 
portional to  any  number  of  given  lines. 

n  ,  H  G  r. 


D^' 


^-0 


Data :  Any  straight  line,  as  JIS  ;  also  the  lines  I,  m,  and  n. 

Required  to  divide  AB  into  parts  proportional  to  I,  m,  and  n. 

Solution.  From  one  extremity  of  AB,  as  A,  draw  a  line,  as  AC, 
making  with  AB  any  convenient  angle. 

On  ^C  measure  AD,  DE,  and  EF  equal  to  I,  m,  and  n  respec- 
tively.    Draw  FB. 

Through  D  and  E  draw  lines  parallel  to  FB,  meeting  AB  in  H 
and  G  respectively. 

Then,  AH,  HG,  and  GB  are  the  parts  required.  q.e.f. 

Proof.     By  the  student.     Suggestion.    Refer  to  §  289. 


164  PLANE   GEOMETRY.  — BOOK  IV. 


Proposition  XVIII 

317.  Problem.     To  find  a  fourth*  proportional  to  three 
£iven  lines. 


A-^-- 

Y 


F' 


"0 

Data :  Any  three  lines,  as  I,  m,  and  n. 
Required  a  fourth,  proportional  to  I,  m,  and  n. 
Solution.     Draw  any  two  lines,  as  AB  and  ACj  forming  any 
convenient  angle  at  ^. 
On  AB  take  AD  =  m. 
On  AC  take  AE  =  I,  and  EF  =  n. 
Draw  ED. 

From  F  draw  a  line  parallel  to  ED  meeting  AB  in  G. 
Then,  DG  is  the  proportional  required.  q.e.f. 

Proof.      By  the  student.      Suggestion.     Refer  to  §  289. 

Proposition  XIX 

318.  Problem,    To  find  a  third  "t  proportional  to  two  given 
lines. 

I  .  m  D  G  „ 


Data :  Any  two  lines,  as  I  and  m. 
Required  a  third  proportional  to  I  and  m. 

Solution.     Draw  any  two  lines,  as  AB  and  AC,  forming  any 
convenient  angle  at  ^. 
On  ^j5  take  AD  =  m. 
On  ^C  take  AE  =  l,  and  EF  =  m. 

*  When  a  :  6  =  c  ;  d,  d  is  termed  the  fourth  proportional  to  a,  ft,  and  c. 
t  When  a  :  6  =  6  :  c,  c  is  termed  the  third  proportional  to  a  and  6. 


PLANE   GEOMETRY.  — BOOK  IV.  165 

Draw  ED. 

From  F  draw  a  line  parallel  to  ED  meeting  AB  in  G. 

Then,  DG  i^  the  proportional  required.  q.e.p. 

Proof.      By  the  student.      Suggestion.     Refer  to  §  289. 

Proposition  XX 

319,  Problem,    To  find  a  mean  proportional  between  two 
given  lines.  . 


m [^         I 

A  C  D  B 

Data :  Any  two  lines,  as  I  and  m. 
Required  a  mean  proportional  between  I  and  m. 
Solution.     Draw  any  line,  as  AB. 
On  AB  take  AG  =  1,  and  GD  =  m. 
On  AD  as  a  diameter  describe  a  semicircumference. 
At  G  erect  a  perpendicular  to  AD  meeting  the  semicircumfer- 
ence in  E. 

Then,  GE  is  the  required  proportional.  q.e.f. 

Proof.     By  the  student.     Suggestion.    Refer  to  §  314. 

Ex.  426.  Three  lines  are  10^%  12c'n,  and  16°™.  Construct  their  fourth 
proportional. 

Ex.  427.    Two  lines  are  11cm  and  9*5™.    Construct  their  third  proportional. 

Ex.  428.  Two  lines  are  6^™  and  2°™  Construct  their  mean  propor- 
tional. 

Ex.  429.  Tangents  are  drawn  through  a  point  6™  from  the  circumference 
of  a  circle  whose  radius  is  9">.     Find  the  length  of  the  tangents. 

Ex.  430.  If  one  side  of  a  polygon  is  2  ft.  6  in.  long,  what  is  the  length  of 
the  corresponding  side  of  a  similar  polygon,  if  their  perimeters  are  respectively 
15  ft.  and  25  ft.  ? 

Ex.  431.  The  shortest  distance  from  a  given  point  to  the  circumference 
of  a  given  circle  is  2  ft.  The  length  of  a  tangent  from  the  same  point  to  the 
circumference  is  3  ft.    Find  the  diameter  of  the  circle. 

Ex.  432.  Five  straight  lines  passing  through  the  same  point  intercept 
segments  on  one  of  two  parallel  lines,  of  12^"^^  20dm,  28^™,  and  36<*m,  ^he 
segment  of  the  other  parallel  corresponding  to  the  20'*™  segment  is  15"*™. 
Find  the  other  segments. 


166 


PLANE   GEOMETRY.  — BOOK  IV. 


320.  Problem, 

ratio. 


Proposition  XXI 
To  divide  a  line  in  extreme  and  mean 


-  .c 


•-/P      • 


A  F  B  O 

Datum:  Any  line,  as  ^5. 

Required  to  divide  AB  in  extreme  and  mean  ratio. 

Solution.    At  one  extremity  of  AB,  as  A,  draw  AC  perpendicular 
to  AB  and  equal  to  \AB. 

With  C  as  a  center  and  ^  C  as  a  radius  describe  a  circumference. 

Through  C  draw  BE  cutting  the  circumference  in  D  and  meeting 
it  in  E.     On  AB  take  BF  =  BB  and  on  AB  produced  take  BG=BE. 

Then,  AB  :  BF  =  BF  :  AF, 

and  AGiBG  =  BG  :  AB', 

that  is,  §  296,  AB  is  divided  at  F  internally,  and  at  G  externally^ 
in  extreme  and  mean  ratio.  q.e.f. 

Proof.      §  315,  BE:AB=AB:BD', 

.-.   §  277,  BE  ~  AB  :  AB  =  AB  —  BD  :  BD,  (1) 

and,  §  276,  AB  -f  BE  :  BE  =  BD  +  AB  :  AB.  (2) 

Const.,  DE  =  2  AC  =  AB', 

hence,  BE  —  AB  =  BE  —  be  =  BD  =  BF. 

Substituting  in  (1)  for  BE  —  AB  its  equal  BF,  for  AB  —  BD  its 
equal  AF,  and  for  BD  its  equal  BF, 

BF:  AB  =AF:  BF, 
or,  §  275,  AB:BF  =  BF:  AF. 

Const.,  AB  +  BE  =  AG,  and  BD-\-AB=  BE. 

Substituting  in  (2)  for  AB  -^BE  its  equal  AG,  and  for  BD  -\-  AB 
its  equal  BE,  AG  :  BE  =  BE  :  AB. 

Since,  const.,  BE  =  BG, 

AQ  :  BG  ==  BQ  :  AB, 


PLANE   GEOMETRY.  — BOOK  IV.  167 


Proposition  XXII 

321.  Broblem,     Upon  a  given  line  to  construct  a  polygon 
similar  to  a  given  polygon. 


>b: 


Data:  Any  polygon,  as  ABODE,  and  any  line,  as  FG. 

Required  to  construct  on  FG  a  polygon  similar  to  ABODE. 

Solution.     Draw  AO  and  AD. 

At  F  and  G  construct  Z  t  and  Z  v  equal  respectively  to  Z  r 
and  Z  s. 

Produce  the  sides  from  F  and  G  until  they  meet  at  H. 

In  like  manner  on  FH  construct  A  FHJ,  and  on  FJ,  A  FJK, 
similar  respectively  to  A  AOD  and  ADE. 

Then,  FGHJK  is  the  required  polygon.  q.e.f. 

Proof.      By  the  student.     Suggestion.     Refer  to  §§  301,  309. 

SUMMARY 
322.   Truths  established  in  Book  IV. 

1.  Two  lines  are  parallel, 

a.  If  one  divides  two  sides  of  a  triangle  proportionally  and  the  other  is 
the  third  side.  §  291 

2.  Lines  are  in  proportion, 

a.  If  they  are  segments  of  two  sides  of  a  triangle  made  by  a  line  parallel 
to  the  third  side.  '  §  289 

b.  If  they  are  two  sides  of  a  triangle  and  their  corresponding  segments 
made  by  a  line  parallel  to  the  third  side.  §  290 

c.  If  they  are  two  sides  of  a  triangle  and  the  segments  of  the  third  side 
made  by  the  bisector  of  the  angle  opposite  that  side.  §  292 

d.  If  they  are  two  sides  of  a  triangle  and  the  external  segments  of  the 
third  side  made  by  the  bisector  of  the  exterior  angle  at  the  vertex  opposite 
that  side.  §293 


168  PLANE   GEOMETRY.  — BOOK  IV. 

e.  If  they  are  the  internal  and  external  segments  of  a  side  of  a  triangle 
made  by  the  bisectors  of  an  interior  and  exterior  angle  at  the  vertex 
opposite  that*  side.  §  298 

/.     If  they  are  the  altitudes  and  homologous  sides  of  similar  triangles. 

§305 

g.  If  they  are  segments  of  parallel  lines  made  by  lines  which  meet  in  a 
point.  §  308 

h.    If  they  are  homologous  sides  of  similar  polygons.  §  299 

i.  If  they  are  perimeters  of  similar  polygons  and  any  two  homologous 
sides.  §311 

3.  A  line  is  a  mean  proportional  between  two  other  lines, 

a.  If  it  is  the  perpendicular  to  the  hypotenuse  of  a  right  triangle  from 
the  vertex  of  the  right  angle,  and  the  other  lines  are  the  segments  of  the 
hypotenuse.  §  312 

b.  If  it  is  either  side  about  the  right  angle  of  a  right  triangle,  and  the 
other  lines  are  the  hypotenuse  and  the  segment  of  it  adjacent  to  that  side 
made  by  the  perpendicular  from  the  vertex  of  the  right  angle.  §  318 

c.  If  it  is  the  perpendicular  to  the  diameter  of  a  circle  from  any  point  in 
the  circumference,  and  the  other  lines  are  the  segments  of  the  diameter.    §  314 

d.  If  it  is  a  tangent  to  a  circle  from  any  point  without,  and  the  other  lines 
are  a  secant  from  the  same  point  and  its  external  segment.  §  315 

4.  Lines  pass  through  the  same  point, 

a.  If  they  are  non-parallel  lines  that  intercept  proportional  segments 
upon  two  parallel  lines.  §  308 

6.   Two  angles  are  equal, 

a.    If  they  are  homologous  angles  of  similar  polygons.  §  299 

6.   Two  triangles  are  similar, 

a.  If  the  angles  of  one  are  respectively  equal  to  the  angles  of  the 
other.  §  300 

b.  If  two  angles  of  the  one  are  respectively  equal  to  two  angles  of  the 
other.  §  301 

c.  If  they  are  right  triangles  and  an  acute  angle  of  one  is  equal  to  an 
acute  angle  of  the  other.  §  302 

d.  If  the  sides  of  one  are  proportional  respectively  to  the  sides  of  the 
other.  §  303 

e.  If  an  angle  of  one  is  equal  to  an  angle  of  the  other  and  the  including 
sides  are  in  proportion.  §  30(» 

/.    If  their  sides  are  parallel,  each  to  each.  §  307 


PLANE   GEOMETRY.— BOOK  IV.  169 

g.    If  their  sides  are  perpendicular,  each  to  each.  §  307 

h.    If  they  are  the  corresponding  triangles  of  similar  polygons  divided  by 

homologous  diagonals.  §  310 

7.   Two  polygons  are  similar, 

a.  If  they  have  their  homologous  angles  equal  and  their  homologous  sides 
proportional.  §  299 

b.  If  each  is  composed  of  the  same  numfcer  of  triangles  similar  each  to 
each  and  similarly  placed.  §  309 

SUPPLEMENTARY  EXERCISES 

Ex.  433.  Construct  a  triangle  whose  sides  are  6,  8,  and  10 ;  then  con- 
struct a  similar  triangle  whose  side  homologous  to  8  is  5. 

Ex.  434.   Divide  a  line  lO'^™  long  internally  in  extreme  and  mean  ratio. 

Ex.  435.  The  median  from  the  vertex  of  a  triangle  bisects  every  line 
drawn  parallel  to  the  base  and  terminated  by  the  sides,  or  the  sides 
produced. 

Ex.  436.  Two  circles  intersect  at  A  and  JB,  and  at  A  tangents  are  drawn, 
one  to  each  circle,  to  meet  the  circumference  of  the  other  in  C  and  D  respec- 
tively ;  BC,  BB,  and  AB  are  drawn.  Prove  that  BD  is  a  third  proportional 
to  ^C  and  AB. 

Ex.  437.  The  diameter  AB  of  a  circle  whose  center  is  0  is  divided  at 
any  point  O,  and  CD  is  drawn  perpendicular  to  AB,  meeting  the  circum- 
ference in  D  ;  OD  is  drawn,  and  CE  perpendicular  to  OD.  Prove  that  DE 
is  a  third  proportional  to  ^0  and  DC. 

Ex.  438.  In  the  triangle  ABC,  AD  is  the  median  to  BC;  the  angles 
ADC  and  ADB  are  bisected  by  DE  and  DE,  meeting  AC  and  AB  in  E  and 
F  respectively.     Then,  FE  is  parallel  to  BC. 

Ex.  439.  A  secant  from  a  given  point  without  a  circle  is  1  ft.  6  in.  long, 
and  its  external  segment  is  8  in.  long.  Find  the  length  of  a  tangent  to  the 
circle  from  the  same  point. 

Ex.  440.  The  radius  of  a  circle  is  6  in.  What  is  the  length  of  the  tan- 
gents drawn  from  a  point  12  in.  from  the  center  ? 

Ex.  441.  If  the  tangent  to  a  circle  from  a  given  point  is  2^^  and  the 
radius  of  the  circle  is  16^"^,  find  the  distance  from  the  point  to  the  circum- 
ference. 

Ex.  442.  If  from  the  vertex  D  of  the  parallelogram  ABCD  a  straight 
line  is  drawn  cutting  AB  at  E  and  CB  produced  at  F,  prove  that  OF  is  a 
fourth  proportional  to  AE,  AD,  and  AB. 


170  PLANE   GEOMETRY.  — BOOK  IV, 

Ex.  443.  If  the  segments  of  the  hypotenuse  of  a  right  triangle  made  by 
the  perpendicular  from  the  vertex  of  the  right  angle  are  6  in.  and  4  ft. ,  find 
the  length  of  the  perpendicular  and  the  length  of  each  of  the  sides  about  the 
right  angle. 

Ex.  444.  Eind  the  length  of  the  longest  and  of  the  shortest  chord  that 
can  be  drawn  through  a  point  7^  in.  from  the  center  of  a  circle  whose  radius 
is  19^  in. 

Ex.  445.  If  the  greater  segment  of  a  line  divided  internally  in  extreme 
and  mean  ratio  is  36  in.,  what  is  the  length  of  the  line  ? 

Ex.  446.  The  shorter  segment  of  a  line  divided  externally  in  extreme 
and  mean  ratio  is  240^™,     Find  the  length  of  the  greater  segment  in  meters. 

Ex.  447.  Find  the  shorter  segment  of  a  line  12^™  long  when  it  is  divided 
internally  in  extreme  and  mean  ratio.  When  it  is  divided  externally  in 
extreme  and  mean  ratio. 

Ex.  448.  The  tangents  to  two  intersecting  circles  drawn  from  any  point 
in  their  common  chord  produced  are  equal. 

Ex.  449.  If  the  common  chord  of  two  intersecting  circles  is  produced,  it 
will  bisect  their  common  tangents. 

Ex.  450.  ABC  is  a  straight  line,  ABD  and  BCE  are  triangles  on  the 
same  side  of  it,  having  angle  ABD  equal  to  angle  CBE  and  AB:BC  = 
BE  :  BD.     If  AE  and  CD  intersect  in  i^,  triangle  AFC  is  isosceles. 

Ex.  451.  If  in  the  triangle  ABC^  CE  and  BD  are  drawn  perpendicular 
to  the  sides  AB  and  AC  respectively,  these  sides  are  reciprocally  propor- 
tional to  the  perpendiculars  upon  them  ;  that  is,  AB  :  AC  =  BD  :  CE. 

Ex.  452.  ABCD  is  a  parallelogram.  If  through  O,  any  point  in  the 
diagonal  AC,  EF and  GH  are  drawn,  terminating  in  AB  and  DC,  and  in 
AD  and  BC  respectively,  iE'//is  parallel  to  GF. 

Ex.  453.  Lines  are  drawn  from  a  point  P  to  the  vertices  of  the  triangle 
ABC;  through  i>,  any  point  in  PA,  a  line  is  drawn  parallel  to  AB,  meeting 
PB  at  E,  and  through  E  a  line  parallel  to  BC,  meeting  PC  at  F.  If  FD  is 
drawn,  triangle  DEF  is  similar  to  triangle  ABC. 

Ex.  454,  If  two  lines  are  tangent  to  a  circle  at  the  extremities  of  a 
diameter,  and  from  the  points  of  contact  secants  are  drawn  terminated 
respectively  by  the  opposite  tangent  and  intereecting  the  circumference  at  the 
same  point,  the  diameter  is  a  mean  proportional  between  the  tangents. 

Ex.  455.  AB  and  AC  are  secants  of  a  circle  from  the  common  point  A, 
cutting  the  circumference  in  D  and  E  respectively.  Then,  the  secants  are 
reciprocally  proportional  to  their  external  segments  ;  that  is,  AB  :  AC  = 
AE :  AD. 

Suggestion.     Draw  CD  and  BE,  and  refer  to  §  322,  6,  b. 


PLANE   GEOMETRY.--- BOOK  IV,  171 

Ex.  456.  AB  and  CD  are  two  chords  of  a  circle  intersecting  at  E.  Prove 
that  AE  :  DE  =  CE  :  BE. 

Ex.  457.  Two  secants  intersect  without  a  circle.  The  segments  of  one 
are  4  ft.  and  20  ft.,  and  the  external  segment  of  the  second  is  16  ft.  Find  the 
length  of  the  second  secant. 

Ex.  458.  From  a  point  without  a  circle  two  secants  are  drawn,  whose 
external  segments  are  respectively  7<^™  and  Q*^™,  the  internal  segment  of  the 
latter  being  \S^^\    What  is  the  length  of  the  •first  secant  ? 

Ex.  459.  The  segments  of  a  chord  intersected  by  another  chord  are  7  in. 
and  9  in.,  and  one  segment  of  the  latter  is  3  in.     What  is  the  other  segment  ? 

Ex.  460.  Two  secants  from  the  same  point  without  a  circle  are  24'^'"  and 
32dm  long.  If  the  external  segment  of  the  less  is  5<*™,  what  is  the  external 
segment  of  the  greater  ? 

Ex.  461.  Through  a  point  ?•"  from  the  circumference  of  a  circle  a  secant 
28'n  long  is  drawn.  If  the  internal  segment  of  this  secant  is  XT'",  what  is  the 
radius  of  the  circle  ? 

Ex.  462.  If  from  any  point  in  the  <iiameter  of  a  circle  produced  a  tan- 
gent is  drawn  and  a  perpendicular  from  the  point  of  contact  is  let  fall  on  the 
diameter,  the  distances  from  the  point  without  the  circle  to  the  foot  of  the 
perpendicular,  the  center  of  the  circle,  and  the  extremities  of  the  diameter  are 
in  proportion. 

Suggestion.     Draw  the  radius  to  the  point  of  contact. 

Ex.  463.  If  the  sides  of  a  triangle  are  respectively  1.6^^^  .12Hm^  and  10™ 
long,  what  are  the  segments  into  which  each  side  is  divided  by  the  bisector 
of  the  opposite  angle  ? 

Ex.  464.  If  an  angle  of  one  triangle  is  equal  to  an  angle  of  another,  and 
the  perpendiculars  from  the  vertices  of  the  remaining  angles  to  the  sides 
opposite  are  proportional,  the  triangles  are  similar. 

Suggestion.     Refer  to  §  322,  6,  c  and  e. 

Ex.  465.  If  two  circles  are  respectively  6  in.  and  3  in.  in  diameter  and 
their  centers  are  10  in.  apart,  find  the  distance  from  the  center  of  the  smaller 
one  to  the  point  of  intersection  of  their  common  exterior  tangent  with  their 
line  of  centers  produced. 

Ex.  466.  Two  intersecting  chords  of  a  circle  are  38  ft.  and  34  ft.  respec- 
tively ;  the  segments  of  the  first  are  8  ft.  and  30  ft.  Find  the  segments  of 
the  second. 

Ex.  467.  What  is  the  length  of  a  chord  joining  the  points  of  contact  of 
the  tangents  drawn  from  a  point  13  in.  from  the  center  of  a  circle  whose 
radius  is  5  in.  ? 

Ex.  468.  Chords  AB  and  CD  of  a  circle  are  produced  in  the  direction 
of  B  and  D  respectively  to  meet  in  the  point  Ej  and  through  E  the  line  EF 
is  drawn  parallel  to  AD  to  meet  GB  produced  in  F.  Prove  that  EF  is  a 
mean  proportional  between  FB  and  FC 


172  PLANS  QnOMBTRY,^BOOK  IP, 

Ex.  469.  AB  is  a  diameter  of  a  circle,  and  through  A  any  straight  line 
is  drawn  to  cut  the  circumference  in  O  and  the  tangent  at  ^  in  Z>.  Prove 
that  ^C  is  a  third  proportional  to  AD  and  AB. 

Ex.  470.  From  any  point  in  the  base  of  a  triangle  straight  lines  are 
drawn  parallel  to  the  sides.  Prove  that  the  intersection  of  the  diagonals 
of  every  parallelogram  so  formed  lies  in  a  line  parallel  to  the  base  of  the 
triangle. 

EXo  471.  If  E  is  the  middle  point  of  one  of  the  parallel  sides  DC  of  the 
trapezoid  ABGD^  and  AE  and  BE  produced  meet  BG  and  AD  produced  in 
F  and  G  respectively,  prove  that  QF  is  parallel  to  AB. 

Ex.  472.  If  a  line  tangent  to  two  circles  cuts  their  line  of  centers,  the 
segments  of  the  latter  are  to  each  other  as  the  diameters  of  the  circles. 

Ex.  473.  The  bisector  of  the  vertical  angle  C  of  the  inscribed  triangle 
ABG  cuts  the  base  at  D  and  meets  the  circumference  in  E.  Prove  that 
AG'.GD=GE'.BG. 

Ex.  474.  Through  any  point  A  of  the  circumference  of  a  circle  a  tangent 
Is  drawn,  and  from  A  two  chords,  AB  and  AG\  the  chord  FG  parallel  to  the 
tangent  cuts  AB  and  ^  C  in  2>  and  E  respectively.    Prove  AB ;  AE= A  G :  AD. 

Ex.  475.  The  greatest  distance  of  a  chord  8  ft.  in  length  from  its  arc  is 
4  in.    Find  the  diameter  of  the  circle. 

Ex.  476.  If  two  circles  are  tangent  externally,  their  common  exterior 
tangent  is  a  mean  proportional  between  the  diameters  of  the  circles. 

Suggestion.  Draw  radii  to  the  points  of  contact,  draw  the  common 
interior  tangent  to  intersect  the  common  exterior  tangent,  and  connect  the 
point  of  intersection  with  the  centers. 

Ex.  477.  The  perpendicular  from  any  point  of  a  circumference  upon  a 
chord  is  a  mean  proportional  between  the  perpendiculars  from  the  same 
point  upon  the  tangents  drawn  at  the  extremities  of  the  chord. 

Suggestion.  Draw  lines  from  the  given  point  to  the  extremities  of  the 
chord,  and  refer  to  §  322,  6,  c. 

Ex.  478.  From  a  point  A  tangents  AB  and  ^C  are  drawn  to  a  circle 
whose  center  is  0,  and  BD  is  drawn  perpendicular  to  GO  produced.  Prove 
that  BD  is  a  fourth  proportional  to  AO^  GD^  and  GO. 

Suggestion.     Draw  AO  and  BG. 

Ex.  479.  From  a  point  E  in  the  common  base  of  two  triangles  ABG  and 
ABD^  straight  lines  are  drawn  parallel  to  ^C  and  AD^  meeting  BG  and  BD 
at  F  and  G  respectively.    Prove  that  FG  is  parallel  to  GD. 

Ex.  480.  If  tangents  to  a  circle  are  drawn  at  the  extremities  of  a  diam- 
eter, the  radius  is  a  mean  proportional  between  the  segra«»nts  of  any  third 
tangent  intercepted  between  them  and  divided  at  its  point  of  tangency. 

Suggestion.  Draw  lines  to  form  a  right  triangle,  havmg  the  third  tangen* 
for  its  hypotenuse  and  a  vertex  at  the  center. 


BOOK   V 


AREA  AND  EQUIVALENCE 


M 


D 


323.   The  amount  of  surface  in  a  plane  figure  is  called  its  Area. 

A  surface  is  measured  by  finding  how  many  times  it  contains 
some  given  square  which  is  taken  as  a  unit  of  measure. 

The  ordinary  units  of  measure  for  surfaces  are  the  square  inch, 
the  square  foot,  the  square  centimeter,  the  square  decimeter,  etc. 

Suppose  that  the  square  M  is  the  unit  of  measure,  and  that 
ABCD  is  the  rectangle  to  be  meas- 
ured. 

By  applying  M  to  ABCD  it  is 
evident  that  the  rectangle  may 
be  divided  into  as  many  rows  of 
squares,  each  equal  to  if,  as  the 
side   of  M  is   contained  times  in 

the  altitude  of  ABCB  ;  that  in  each  row  there  are  as  many  squares 
as  the  side  of  M  is  contained  times  in  the  base  of  ABCD-,  and 
therefore,  that  the  product  of  the  numerical  measures  of  the  base 
and  altitude  of  ABCD  is  equal  to  the  number  of  times  that  M  is 
contained  m.  ABCD. 

In  this  case  the  side  of  M  is  contained  4  times  \n  AD  and  6 
times  in  AB  ;  consequently,  M  is  contained  24  times  in  ABCD-, 
that  is,  the  rectangle  contains  24  square  units. 

Therefore,  if  the  side  of  a  square  is  a  common  measure  of  the 
base  and  altitude  of  a  rectangle,  the  product  of  the  numerical 
measures  of  the  base  and  altitude  expresses  the  number  of  times 
that  the  rectangle  contains  the  square,  and  is  the  numerical 
measure  of  the  surface,  or  the  area  of  the  rectangle. 


324.  For  the  sake  of  brevity,  the  product  of  the  base  and  altitude 
is  used  instead  of  the  product  of  the  numerical  measures  of  the  base 
and  altitude. 

178 


174 


PLANE   GEOMETRY.  — BOOK    V. 


The  product  of  two  lines  is,  strictly  speaking,  an  absurdity,  but 
since  the  expression  is  used  to  denote  the  area  of  a  rectangle  it 
follows,  that  the  geometrical  concept  of  the  prodiict  of  two  lines 
is  the  rectangle  formed  by  them. 

Thus,  AB  X  CD  implies  a  product,  which  is  a  numerical  result, 
but  it  must  be  interpreted  geometrically  to  mean  rect.  AB  •  CD. 

For  similar  reasons,  if  AB  represents  a  line,  AB^  must  be  inter- 
preted to  mean  geometrically  the  square  described  upon  the  line 
AB,  and  conversely,  the  square  described  upon  a  line  may  be 
indicated  by  the  square  of  the  line. 

325.  It  has  been  stated  in  §  36  that  equal  figures  may  be  made 
to  coincide,  consequently  such  figures  have  equal  areas. 

Figures,  however,  which  cannot  be  made  to  coincide  may  have 
equal  areas,  and  they  are  called  equivalent  figures. 

All  equal  figures  are  equivalent,  but  not  all  equivalent  figures 
are  equal. 

If  a  square  and  a  triangle  each  contains  one  square  foot  of  surface,  they  are 
equivalent ;  but  since  they  cannot  be  made  to  coincide,  they  are  not  equal. 

The  symbol  of  equivalence  is  =c=. 

326.  Since  equivalent  means  equal  in  area,  or  in  volume  as  will 
be  shown,  then,  §  222  may  be  extended  to  apply  to  equivalent 
magnitudes;  consequently,  if,  while  approaching  their  respective 
limits,  two  variables  are  always  equivalent,  their  limits  are  equivalent. 

Proposition  I 

327.  1 .  If  a  rectangle  is  3"  long  and  another  of  the  same  altitude  is 
6"  long,  how  do  they  compare  in  area  ?  How,  then,  do  rectangles  having 
equal  altitudes  compare  in  area? 

2.  How  do  rectangles  that  have  equal  bases,  but  different  altitudes, 
compare  in  area? 

Theorem.  Rectangles  ivhich  have  equal  altitudes  are  to 
ea/}h  other  as  their  bases. 

D  c  H 


1 

E 


PLANE   GEOMETRY.— BOOK   V. 


175 


Data :  Any  two  rectangles,  as  ABCD  and  EFGH,  whose  altitudes, 
AD  and  EH,  are  equal. 

To  prove  ABCD  :  EFGH  =  AB  :  EF. 

Proof.     Case  I.     When  AB  and  EF  are  commensurable. 

Suppose  that  iJ/  is  a  common  unit  of  measure  for  AB  and  EF. 

Apply  M  to  each  base,  and  suppose  that  it  is  contained  in  AB  7 
times  and  in  EF  4  times. 

Then,  AB  :EF  =  1  iL 

Divide  AB  into  7  equal  parts  and  EF  into  4  equal  parts,  and  at 
each  point  of  division  erect  a  perpendicular. 

ABCD  is  thus  divided  into  7  rectangles,  and  EFGH  into  4  rec- 
tangles. 

Since,  §  156,  these  rectangles  are  all  equal, 
ABCD  :  EFGH  =  7  :  4:. 
ABCD  :  EFGH  =  AB  :  EF.  Why  ? 

Case  II."   When  AB  and  EF  are  incommensurable. 


KO 


H 


M 


JB 


Since  AB  and  EF  are  incommensurable,  suppose  that  iHf  is  a 
common  unit  of  measure  for  AJ  and  EF,  and  that  JB  is  less 
than  M.     Draw  JK  W  AD. 

Then,  Case  I,         AJKD .:  EFGH  =AJ:EF; 
and  JBCK  is  less  than  any  one  of  the  rectangles  whose  base  is 
equal  to  M. 

Now,  if  M  is  indefinitely  diminished,  the  ratios  AJKD  :  EFGH 
and  AJ :  EF  remain  equal  and  indefinitely  approach  the  limiting 
ratios  ABCD  :  EFGH  and  AB  :  EF  respectively. 

Hence,  §  222,         ABCD  :  EFGH  =  AB  :  EF. 

Therefore,  etc.  q.e.d. 

328.  Cor.  Rectangles  which  have  equal  bases  are  to  each  other 
as  their  altitudes. 


176 


PLANE   GEOMETRY.  — BOOK   V, 


Proposition  II 

329.  Draw  two  rectangles  whose  bases  are  respectively  5''  and  3"  and 
altitudes  2"  and  4",  or  any  other  dimensions ;  divide  them  into  squares 
having  a  side  of  1".  How  many  square  inches  are  there  in  the  first  rec- 
tangle? In  the  second?  How  does  the  ratio  of  the  areas  of  the  two 
rectangles  compare  with  the  ratio  of  the  products  of  their  bases  by  their 
altitudes  ? 

Theorem,  Rectangles  are  to  each  other  as  the  products 
of  their  bases  by  their  altitudes. 


Data :  Any  two  rectangles,  as  A  and  B,  of  which  d  and  m  are 
the  bases,  and  e  and  n  the  altitudes,  respectively. 

To  prove  A:B  =  dxe'.mxn. 

Proof.     Construct  a  rectangle  c,  having  the  base  m  and  the 
altitude  e. 

^  :  (7  =  c?  :  m, 
C:5  =  e: n; 
A:B  =zd  xeimxn, 

Q.E.D. 

Proposition  III 


Then,  §  327, 
and,  §  328, 
hence,  §  287, 

Therefore,  etc. 


330.  How  many  square  inches  of  surface  are  there  in  a  rectangle  that 
is  5"  long  and  1"  wide?  5"  long  and  2"  wide?  5"  long  and  C"  wide? 
8''  long  and  1"  wide  ?  How  may  the  amount  of  surface,  or  the  area  of 
any  rectangle,  be  found? 

Theorem,  The  area  of  a  rectangle  is  equal  to  the  prod- 
uct of  its  base  by  its  altitude. 


Data:    Any    rectangle,    as   A, 
whose  base  is  d  and  altitude  6. 


To  prove  area  oi  A  =  d  xe. 


PLANE   GEOMETRY.  — BOOK   V.  177 

Proof.  Assume  that  the  unit  of  measure  is  a  square  M,  whose 
side  is  the  linear  unit. 

§329,       A:M  =  dxe:lxl,or  ^  =  ^^  =  dxe. 

M      1x1 

But,  §  323,  the  surface  of  A  is  measured  by  the  number  of 
times  it  contains  the  unit  of  measure  M; 

A 

—  =  areaof^. 

M 

But  —  =  c2  X  e. 

M 

Hence,  area  oi  A  —  dxe, 

Therefore,  etc.  q.e.d. 

Proposition  IV 

331.  1.  Draw  an  oblique  parallelogram,  and  on  the  same  base  a  rec- 
tangle having  an  equal  altitude.  How  do  the  triangles  thus  formed  at 
the  ends  of  this  figure  compare  ?  How  does  the  area  of  the  parallelo- 
gram compare  with  the  area  of  the  rectangle  ? 

2.  What  ratio  do  two  rectangles  have  to  each  other  ?  (§329)  What, 
then,  is  the  ratio  of  two  parallelograms  to  each  other? 

Theorem,  A  parallelogram  is  equivalent  to  the  rectangle 
which  has  the  same  base  and  altitude. 

Data:    Any    parallelogram,   as   ABCD,  ^_d e  c 

whose  base  is  AB  and  altitude  BE.  \   /  / 

To  prove  ABCB  equivalent  to  the  rec-  \/  / 

tangle  whose  base  is  AB  and  altitude  BE.  I v 

Proof.     Draw  AF  II  BE  and  meeting  OD  produced  in  F. 

Const.,  ABEF  is  a  rectangle  which  has  the  same  base  and  alti- 
tude as  ABCD. 

In  rt.  Abce  and  ADF,     BE  =  AF,  Why  ? 

and  BC  =  AD',  Why? 

Abce  =  A  adf.  Why  ? 

Hence,  Abce-\-  abed  ^Aadf  +  abed  ; 

that  is,  ABCD  =o=  ABEF. 

Therefore,  etc.  q.e.d. 

milne's  geom. — 12 


178  PLANE   GEOMETRY.  — BOOK    V. 

332.  Cor.  I.  The  area  of  a  parallelogram  is  equal  to  the  product 
of  its  base  by  its  altitude. 

333.  Cor.  II.  Parallelograms  are  to  each  other  as  the  products 
of  their  bases  by  their  altitudes;  conseqnentlj,  parallelograms  which 
have  equal  altitudes  are  to  each  other  as  their  bases,  imrallelograms 
which  have  equal  bases  are  to  each  other  as  their  altitudes,  and 
parallelograms  which  have  equal  bases  and  equal  altitudes  are 
equivalent. 

Proposition  V 

334.  1.  Draw  any  triangle,  and  through  two  of  its  vertices  draw 
lines  parallel  to  the  opposite  sides,  producing  them  until  they  meet. 
What  part  of  the  parallelogram  thus  formed  is  the  original  triangle? 
How  does  this  parallelogram  compare  with  a  rectangle  having  the  same 
base  and  altitude?    What  part  of  such  a  rectangle  is  the  triangle? 

2.  What  ratio  do  two  rectangles  have  to  each  other?  (§329)  What, 
then,  is  the  ratio  of  two  triangles  to  each  other? 

Theorem,  A  triangle  is  equivalent  to  one  half  the  rec- 
tangle which  has  the  same  base  and  altitude. 

E O 

Data:  Any  triangle,  as  ABC,  whose  base        /               ^' 
is  AB  and  altitude  CD.                                            !         ^^ 
To  prove        A  ABC  =o=  \  rect.  AB  •  CB.  l^^^ / 

A  B    D 

Proof.     Draw  AE  W  BC  and  CE  il  BA. 

Then,  ABCE  is  a  parallelogram,  AC  \^  its  diagonal, 
and,  §  152,  Aabc  =  A  AEC, 

or  A  ABC  ^^  ABCE. 

But,  §  331,  ABCE  o=  rect.  AB  •  CD. 

Hence,  AABC  -o^\  rect.  AB  •  CD.  q.e.d. 

335.  Cor.  I.  The  area  of  a  triangle  is  equal  to  one  half  the 
product  of  its  base  by  its  altitude. 

336.  Cor.  II.  Triangles  are  to  each  other  as  the  produxits  of  their 
bases  by  their  altitudes;  consequently,  triangles  which  have  equal 
altitudes  are  to  each  other  as  their  bases,  triangles  ivhich  have  equal 
bases  are  to  each  other  as  their  altitudes,  and  triangles  which  have 
equal  bases  and  equal  altitudes  are  equivalent. 


PLANE   GEOMETRY.  — BOOK   V,  179 

Proposition  VI 

337.  Draw  a  trapezoid  and  one  of  its  diagonals.  How  does  the  area 
of  the  trapezoid  compare  with  the  combined  areas  of  the  triangles  thus 
formed?  Since  both  triangles  have  the  same  altitude,  how  does  the 
area  of  the  trapezoid  compare  with  the  area  of  the  rectangle  which  has 
the  same  altitude  and  a  base  equal  to  the  sum  of  the  parallel  sides  of 
the  trapezoid  ? 

Theorem,  A  trapezoid  is  equivalent  to  one  half  the  rec- 
tangle which  has  the  same  altitude  and  a  base  equal  to 
the  sum  of  its  parallel  sides. 

D  C 

Data:  Any  trapezoid,  as  ABCD,  whose 
altitude  is  CE  and  whose  parallel  sides 
are  AB  and  CD. 

To  prove  ABCD=o=^  rect.  CE  •  {AB  +  en),  a 'e~b 

Proof.     Draw  the  diagonal  AC. 

Then,  ABCD  <^ /\  ABC -\- J^  ADC. 

§  334,  /^ABC  ^\ rect.  CE  •  ^5, 

and  A  ^D  C  =0=  i  rect.  CE  -  CD  \ 

hence,  t.ABC-\-I^ADC^\  rect.  CE  -  AB  ■\-\  rect.  CE  *  CD -^ 

that  is,  ABCD  ^\  rect.  CE  •  {AB  -f  CD). 

Therefore,  etc.  q.e.d. 

338.  Cor.  Tlie  area  of  a  trapezoid  is  equal  to  one  half  the 
product  of  its  altitude  by  the  sum  of  its  parallel  sides. 

339.  Sch.  It  will  be  observed  that  the  corollaries  §  332,  §  335, 
and  §  338  are  arithmetical  rules  for  computing  areas. 

Such  rules  are  readily  formed  from  the  theorems  to  which  they 
are  attached  by  employing  the  terms  product  and  equal  instead  of 
rectangle  and  equivalent. 

Ex.  481.  Triangles  on  the  same  base  and  having  their  vertices  in  the 
same  line  which  is  parallel  to  the  base  are  equivalent. 

Ex.  482.  The  parallel  sides  of  a  trapezoid  are  12d«n  and  S^"^,  and  their 
distance  apart  is  5<i"'.     What  is  the  area  of  the  trapezoid  ? 

Ex.  483.  The  area  of  a  trapezoid  is  52  sq.  in.,  and  the  sum  of  the  two 
parallel  sides  is  13  in.     What  is  the  distance  between  the  parallel  sides  ? 

Ex.  484.  The  area  of  a  triangle  is  36  sq.  ft.  If  its  base  is  9  ft,  what  is 
its  altitude  ? 


180  PLANE   GEOMETRY.  — BOOK   V. 

Proposition  VII 

340.  Draw  a  triangle  one  of  whose  sides  is  5'',  base  6",  and  altitude 
3",  or  other  dimensions ;  draw  another  triangle  having  an  equal  side  and 
altitude  but  any  base  whatever,  as  10",  and  the  angle  between  the  base 
and  given  side  equal  to  the  corresponding  angle  of  the  first  triangle. 
How  does  the  ratio  of  the  areas  of  these  triangles  compare  with  the 
ratio  of  the  products  of  the  sides  that  include  their  equal  angles  ? 

Theorem.  Two  triangles  having  an  angle  of  one  equal 
to  an  angle  of  the  other  are  to  each  other  as  the  products 
of  the  sides  including  the  equal  angles. 

Data:  Any  two  triangles,  as  ABC  and 

DEC,  having  the  common  angle  C. 

To  prove 

AABC:ADEC  =  ACXBG:DCXEC, 

A 

Proof.     Draw  AE. 

Since  BC  and  EC  may  be  regarded  as  the  bases  of  the  A  ABC 
and  AEC,  their  bases  are  in  the  same  straight  line ;  and  since  they 
have  their  common  vertex  at  A,  they  have  the  same  altitude. 

.-.§336,  AaBC:AAEC  =  BC:EC. 

In  like  manner,     A  AEC  :  A  DEC  =  AC  :  DC. 

Hence,  §  287,         A  ABC  :  Adec  =  AC  x  BC  :  DC  X  EC. 
Therefore,  etc.  q.e.d. 

341.  Cor.  If  the  products  of  the  sides  including  the  equal  angles 
are  equal,  the  triangles  ore  equivalent. 

Ex.  485.  Two  triangles  have  an  angle  in  each  equal,  the  including  sides 
(n  one  being  8  ft.  and  12  ft.,  and  in  the  other  6  ft.  and  20  ft.  The  area  of  the 
smaller  triangle  is  27  sq.  ft.    Find  the  area  of  the  larger  triangle. 

Proposition  VIII 

342.  Find  the  area  of  a  triangle  whose  base  is  14",  another  side  8", 
and  the  altitude  6",  or  other  dimensions;  also  the  area  of  a  similar  tri- 
angle whose  base  is  7",  or  any  other  convenient  length.  How  does  tlie 
ratio  of  the  areas  of  the  two  triangles  compare  with  the  ratio  of  the  areas 
of  the  squares  described  upon  their  bases  ?  With  the  ratios  of  the  squares 
upon  other  homologous  sides  or  lines  ? 


PLANE   GEOMETRY.  — BOOK  V.  181 

Theorem.  Similar  triangles  are  to  each  other  as  the 
squares  upon  their  homologous  sides. 

Data :  Any  two  similar 
triangles,  as  ABC  and  DEF. 

To  prove  AabcA DEF 
=  Iff :  ¥i?  =  etc. 

Proof.     §299,  jiA=/!:D', 

.-.§340,  AABC:ADEF  =  ABXA0:DEXDF,  (1) 

But,  §  299,  AC '.  DF  =  AB  :  DE.  (2) 

Multiplying  the  antecedents  by  AB  and  the  consequents  by  DEy 
§284,  AB  X  ACiDE  X  DF  =  Aff:DE\ 

Substituting  in  (1), 

A  ABC:  A  DEF  =  Iff :  De\ 

In  like  manner  the  same  may  be  proved  for  any  homologous 
sides. 

Therefore,  etc.  q.e.d. 

343.  Cor.  Similar  triangles  are  to  each  other  as  the  squares 
upon  any  of  their  homologous  lines. 

Ex.  486.  The  homologous  sides  of  two  similar  triangular  fields  are  in  the 
ratio  of  5  :  3.  How  many  times  the  area  of  the  second  field  is  the  area  of 
the  first  ? 

Proposition  IX 

344.  Divide  two  similar  polygons  into  triangles  by  diagonals  drawn 
from  a  pair  of  homologous  vertices.  Since  the  homologous  triangles  are 
similar,  to  what  is  the  ratio  of  their  areas  equal  ?  (§  342)  Write  all  the 
ratios  of  the  areas  of  each  pair  of  triangles.  Discover  from  the  ratios 
whether  every  pair  can  be  shown  to  have  the  same  ratio.  How,  then, 
does  the  ratio  of  the  sum  of  the  triangles  of  one  polygon  —  that  is,  its 
area  —  to  the  sum  of  the  triangles  of  the  other  compare  with  the  ratio  of 
any  two  corresponding  triangles  ?  (§  279)  Then,  how  does  the  ratio  of 
the  polygons  compare  with  the  ratio  of  the  squares  described  upon  their 
homologous  sides  ? 


182  PLANE   GEOMETRY.  — BOOK   V, 

Theoretn.     Similar  polygons  are  to  each  other  as  the 
•squares  upon  their  homologous  sides, 

D 

J 


Data :  Any  two  similar  polygons,  as  ABODE  and  FGHJK. 
To  prove         ABODE  :  FGHJK  =  AB' :  F(?  =  etc. 

Proof.     Draw  the  homologous  diagonals  AO,  AD,  FH,  and  FJ. 

Then,  §  310,  the  corresponding  triangles  thus  formed  are 
similar, 

and,  §  299,    AB  :  FG  =  BO :  GH=  OD:  HJ=DE:  Jir=  etc.; 
.-.  §  288,  AB^  :  FGT  =  BO' :  GH'  =  cF  :  H?  =  DE^ :  JK^  =  etc., 
also,  §  342,        A  ABO :  A  FGH  =  Iff  :  F(?  =  etc., 
A  AOD  :  A  FHJ  =  off  :  U?  =  etc., 
and  A  ABE :  A  FJK  =  D^ :  J?  =  etc. 

Since  the  ratios  of  these  A  are  all  equal,  by  §  279, 
A  ABO+A  AOD -{-A  ADE :  A  FGH-\-  A  FHJ-\-  A  FJK=:  A  ABC :  A  FQH', 
that  is,  ABODE  :  FGHJK  =  A  ABO :  A  FGH. 

But  A  ABO:  A  FGH  =  Iff  :  pff  =  etc. 

Hence,  abode  :  FGHJK  =  Iff :  i^  =  etc. 

Therefore,  etc.  q.e.d. 

345.  Cor.  I.  Similar  polygons  are  to  each  other  as  the  sqvxires 
upon  any  of  their  homologous  lines. 

346.  Cor.  II.  The  homologous  sides  of  any  similar  polygons  are 
to  each  other  as  the  square  roots  of  the  area^  of  those  polygons. 

Ex.  487.  In  two  similar  polygons  two  homologous  sides  are  15  ft.  and 
25  ft.  The  area  of  the  smaller  polygon  is  450  sq.  ft.  Find  the  area  of  the 
larger  one. 


PLANE   GEOMETRY.  — BOOK   V.  183 

Proposition  X 

347.  Draw  two  lines  respectively  3"  and  5"  long,  or  any  other 
lengths;  construct  a  square  on  each  and  a  square  on  their  sum;  also 
construct  the  rectangle  of  these  lines.  How  does  the  area  of  the  square 
on  their  sum  compare  with  the  combined  areas  of  the  other  squares  and 
double  the  area  of  the  rectangle  ? 

Theorem.  The  square  upon  the  sum  of  two  lines  is 
equivalent  to  the  sum  of  the  squares  upon  the  lines  plus 
twice  the  rectangle  formed  hy  the  lines* 


Data:  Any  two  lines,  as  AB  and  BC,  and  f 

their  sum  AC. 

To  prove  AC^  =o=  JLB^  +  BC^  +  2  rect.  AB  •  BO. 


.^D 


r 


^c 


B 

Proof.     On  reconstruct  the  square  ACDE;  draw  BG  I!  CD;  tak^ 
DK  equal  to  BC;  and  draw  FK  II  AC,  cutting  BG  in  H. 

Since  the  sides  of  HKDG  are  respectively  parallel  to  the  sides 
of  the  square  ACDE,  its  angles  are  rt.  A. 

§151,  GD  =  HK=BCy 

and  GH=DK', 

but,  const,  DK—BC) 

GD  =  HK=  GH=  DK—  BC,  Why  ? 

and,  §  143,    HKDG  is  a  square  whose  side  equals  BQ\ 
that  is,  HKDG  =  BC^. 

Similarly,  ABHF  =  Iff. 

Since  the  sides  of  BCHK  are  respectively  parallel  to  the  sides 
of  the  square  ACDE,  its  angles  are  rt.  A; 
and  since,  HB  =  AB, 

BCKH=  rect.  AB  .  BC. 

Similarly,  FHGE  =  rect.  AB  -  BC. 

But      A  CDE  =o  ABHF  +  HKD G -\-  B CKH  -\-  FHQB ; 

that  is,  AC^  =0=  Iff  H-  BC^  •{■  2  rect.  AB  -  BC. 

Therefore,  etc.  q.e.i>. 


184 


PLANE   GEOMETRY.  — BOOK   V. 


Proposition  XI 

348.  Draw  two  lines  respectively  3"  and  5"  long,  or  any  other  lengths; 
construct  a  square  on  each,  and  a  square  on  their  difference;  also  con- 
struct the  rectangle  of  these  lines.  How  does  the  area  of  the  square  on 
their  difference  compare  with  the  combined  areas  of  the  other  squares 
minus  double  the  area  of  the  rectangle  ? 

Theorem,  The  square  upon  the  difference  of  two  lines 
is  equivalent  to  the  sum  of  the  squares  upon  the  lines 
minus  twice  the  rectangle  formed  hy  the  lines. 

Data:   Any  two  lines,  as  AG  and  BC,  and    ^'  ^ 

their  difference  AB. 

To  prove  A?  =o=  AC^  +  BC^  —  2  rect.  AC  -  BC. 

Proof.  On  AC  construct  the  square  ACDE, 
on  BC  the  square  BGJC,  and  on  AB  the  square 
ABHF.     Produce  FH  to  meet  CD  in  K. 


Z  GBC  is  a  rt.  Z; 
Z.ABG  is  a  rt.  Z; 
Z  ABH  is  a  rt.  Z  ; 
GBH  is  a  straight  line. 
JCK  is  a  straight  line. 
A  G  and  J  are  rt.  Ay 
A  GHK  and  HKJ  are  vt  A) 
HGJK  is  a  rectangle. 
HB  =  AB,  and  BG  =  BC'y 
EG  =  AC, 
GJ  =  BC', 
HGJK  =  rect.  AC  'BC, 
FKDE  =  rect.  AC'BC. 
Const.,      ABHF  —  Iff,  ACDE  =  A^,  and  BGJC  =  B^. 
But  ABHF  o  ACDE  +  BGJC  —  {HGJK  -\-  FKDE)  ; 

that  is,  Iff  =0=  ic^  +  BC^  —  2  rect.  AC  •  BC. 

Therefore,  etc. 


Then, 

but 

Similarly, 
Now, 
also 


hence, 
also 

Similarly, 


Why? 
Why? 

Why? 

Why? 
Why? 

Why? 
Why? 


Q.E.D. 


PLANE   GEOMETRY.  — BOOK  V. 


185 


Proposition  XII 

349.  Draw  a  right  triangle  whose  sides  are  3",  4",  and  5",  or  any- 
other  right  triangle ;  construct  a  square  on  each  side  and  find  the  area 
of  each  square.  How  does  the  square  on  the  hypotenuse  compare  in 
area  with  the  sum  of  the  squares  on  the  other  sides  ? 

Theorem,  The  square  upon  the  hypotenuse  of  a  right 
triangle  is  equivalent  to  the  sum  of  the  squares  upon  the 
other  two  sides. 

First  Method 


Data:  Any  right  triangle,  as 
ABC\  the  square  on  the  hypote- 
nuse, as  ABDE ;  and  the  squares  on 
the  other  two  sides,  as  BCGF  and 
ACHJ  respectively. 

To  prove     ABDE  ^  BCGF  +  ACHJ. 

Proof.  From  C  draw  CKWbd, 
cutting  AB  in  L  and  meeting  ED  in 
K,  and  draw  CE  and  BJ. 


A  ACB,  ACH,  and  BCG  are  rt.  A ; 
§  58,  ACG  and  BCH  are  straight  lines. 

In  A  AEC  and  AJB,       AE  =  ab,  ac=  A  J, 


and 

Zeac=Zjab^, 

.-.  §100, 

A  AEC  =  A  AJB ; 

but,  §  334, 

AEKL  ^  2  A  AEC, 

and 

ACHJ ^2  A  AJB; 

.-. 

AEKL  =0=  ACHJ. 

In  like  manner, 

BDKL  =o  BCGF. 

But 

ABDE  ^  AEKL  -f  BDKL. 

Hence, 

ABDE  =c=  BCGF  +  ACHJ, 

Therefore, 

etc. 

Prove  that  BDKL 

=o  BCQF, 

Why? 
Why? 


Q.S.D. 


186 


PLANE   GEOMETRY.  — BOOK   V. 


Second  Method 

Proof.     Draw  the  perpendicular  CD. 

Then,  §  313,    AB:AC=AC:  AD, 

:2 


or 

and 

or 

Ax.  2, 
or 

Therefore,  etc. 


AC  =ABX  AD, 
AB  :  BC  =  BC :  DB, 
BC?  =  ABX  DB. 


AC  ■\-BC  r=  AB  {AD  +  DB)  =  AB  X  AB  =  Ab\ 


AB^  =  AC^  +  BG^ 


Q.E.D. 


350.  Cor.  I.  Either  side  of  a  right  triangle  is  equal  to  the  square 
root  of  the  difference  between  the  squares  of  the  hypotenuse  and  the 
other  side. 

The  following  is  an  easy  method  of  determining  integral  numbers  which 
are  measures  of  the  sides  of  right  triangles : 

Write  in  a  column  the  squares  of  the  numbers  of  the  scale  as  far  as  de- 
sired ;  subtract  each  square  from  all  of  the  others  following  it.  When  the 
remainder  is  a  perfect  square  its  square  root  is  the  measure  of  one  side  of  a 
right  triangle,  the  square  root  of  the  minuend  of  this  subtraction  is  the  meas- 
ure of  the  hypotenuse,  and  the  square  root  of  the  subtrahend  is  the  measure 
of  the  other  side.  By  taking  equimultiples  of  these  numbers  the  measures  of 
the  sides  of  similar  right  triangles  may  be  found. 

The  following  sets  of  numbers  are  some  of  the  integral  measures  of  the 
sides  of  right  triangles : 

3  4  5 

6  12  13 

7  24  25 

8  16  17 

9  40  41 


351.   Cor.  II.     The  ratio  of  the  diagonal  of  a  square  to  a  side 

is  V2. 


For,  in  the  square  ABCD,     AG^  —  AB^  -f-  B(f ; 


but      BC  =  AB 


hence,     AC  =2ab. 


Dividing  by  ab\     ^^^  =  2 ;  whence,  ^  =  V2. 
AB  ^B 


PLANE   GEOMETRY.  — BOOK    V.  187 

In  the  figure  on  page  185  : 

Ex.  488.    Prove  CE  perpendicular  to  JB. 

Ex  489.    If  the  lines  AH  and  BG  are  drawn,  prove  that  they  are  parallel. 

Et:.  490.  Prove  that  the  sum  of  the  perpendiculars  from  F  and  J  to  AB 
produced  is  equal  to  AB. 

Ex.  491.    Prove  that  J,  (7,  and  F  are  in  the  same  straight  line. 

Ex.  492.  If  the  lines  EJ  and  DF  are  drawn,  prove  that  the  sum  of  the 
angles  AEJ^  AJE,  BDF,  and  BFD  is  equal  to  one  right  angle. 

Ex.  493.  If  EM  and  DK  are  drawn  perpendicular  respectively  to  J  A  and 
FB  produced,  prove  that  the  triangles  AEM  and  BBN  are  each  equal  to 
triangle  ABC. 

Ex.  494.  If  the  lines  JH,  KC,  and  FG  are  produced,  prove  that  they 
meet  in  a  common  point. 

Ex.  495.  If  B  is  the  middle  point  of  the  side  BC  of  the  right  triangle 
ABC,  and  BE  is  drawn  perpendicular  to  the  hypotenuse  AB,  prove  that 

AtP-AE'^  -BE^. 

Ex.  496.  The  area  of  a  rectangle  is  26.4081  ^m  and  its  altitude  is  4.8*™. 
Fird  the  length  of  its  diagonal. 

^x.  497.  The  perpendicular  distance  between  two  parallel  lines  is  20  in. 
aiAd  a  line  is  drawn  across  them  at  an  angle  of  45°.  What  is  the  length 
0^  the  part  intercepted  between  the  parallel  lines  ? 

Ex.  498.  Find  the  area  of  a  right  isosceles  triangle,  if  the  hypotenuse  is 
1^0  rd.  in  length. 

Ex.  499.  The  diameter  of  a  circle  is  12<^™  and  a  chord  of  the  circle  is 
jQcm^     What  is  the  length  of  a  perpendicular  from  the  center  to  this  chord  ? 

Ex.  500.  Two  parallel  chords  in  a  circle  are  each  8  ft.  in  length,  and  the 
distance  between  them  is  6  ft.     Find  the  radius  of  the  circle. 

Ex.  501.  Two  sides  of  a  triangle  are  IS^m  and  15<i'"  and  the  altitude  on 
the  third  side  is  12dm.     pjnd  the  third  side  and  also  the  area  of  the  triangle. 

Ex.  502.  Two  parallel  lines  are  12  ft.  apart,  and  from  a  point  on  one  of 
them  two  lines,  one  20  ft.  and  the  other  13  ft.  long,  are  drawn  to  the  other 
parallel.     What  is  the  area  of  the  triangle  thus  formed  ? 

352.  When  from  the  extremities  of  a  given  straight  line  per- 
pendiculars are  let  fall  upon  an  indefinite  straight  line,  the  portion 
of  the  indefinite  line  between  the  perpendiculars  is  called  the 
projection  of  the  given  line. 

MN  is  the  projection  of  the  C|-^,„^^ 

line  CB  upon  the  line  AB.    If  |  ^~^^^£) 

the  point  B  is  in  the  line  AB,  \ ^ ^ ^ 

then  ilfl>  is  the  projection  of  Ol).  M  N  M  D 


188 


PLANE   GEOMETRY.  — BOOK  V. 


Proposition  XIII 

353.  Draw  a  triangle  whose  sides  are  2",  3",  and  4",  or  any  other 
oblique  triangle ;  construct  a  square  on  the  side  opposite  an  acute  angle ; 
construct  squares  on  the  other  two  sides  and  also  the  rectangle  of  one  of 
those  sides  and  the  projection  of  the  other  upon  that  side ;  find  the  area 
of  each  figure  constructed.  How  does  the  area  of  the  first  square  com- 
pare with  the  combined  area  of  the  otner  squares  less  twice  the  area  of 
the  rectangle? 

Theorem.  In  any  oblique  triangle  the  square  upon  the 
side  opposite  an  acute  angle  is  equivalent  to  the  sum  of  tlw 
squares  upon  the  other  two  sides  minus  twice  the  rectangle 
formed  by  one  of  those  sides  and  the  projection  of  the  other 
upon  that  side, 

c 


Data:  Any  oblique  triangle,  as  ABC,  in  which  A  is  an  acute 
angle  and  AD  the  projection  of  AO  on  AB,  or  AB  produced. 


To  prove 


BC 


AB  -{-AC  —  2  rect.  AB  .  AD. 


BD  =o=AB  -\-AD  —  2  rect.  AB  -  AD. 


Proof.     When  CD  lies  within  A  ABC, 
BD  =  AB  —  AD\ 

when  CD  lies  without  A  ABC, 

BD  =  AD-AB; 

and  in  either  case, 

§  348, 

Adding  CD  to  both  members  of  this  equation, 

BD^  -{-cff^  IS'  4-  aS  +Gff  —  2  rect.  AB 
But,  §  349,  bS  -\-gS^  b^,    . 

and  Iff  ^'ci?^  Iff, 

Hence,  substituting  sff  and  Iff  for  their  equivalents, 
Bff  =0=  Iff  +  Iff  —  2  rect.  AB  -  AD. 


AD. 


Q.E.I>. 


PLANE   GEOMETRY.  — BOOK   V.  189 


Proposition  XIV 

354.  Draw  a  triangle  whose  sides  are  2",  3'',  and  4",  or  any  other 
obtuse  triangle ;  construct  a  square  on  the  side  opposite  the  obtuse  angle; 
construct  squares  on  the  other  two  sides  and  also  the  rectangle  of  one  of 
those  sides  and  the  projection  of  the  other  upon  that  side  produced ;  find 
the  area  of  each  figure  constructed.  How  does  the  area  of  the  first  square 
compare  with  the  combined  area  of  the  other  squares  and  twice  the  area 
of  the  rectangle  ? 

Theoretn,  In  any  obtuse  triangle  the  square  upon  the 
side  opposite  the  obtuse  angle  is  equivalent  to  the  sum  of 
the  squares  upon  the  other  two  sides  plus  twice  the  rec- 
tangle formed  by  one  of  those  sides  and  the  projection  of 
the  other  upon  that  side. 

Data :  Any  obtuse  triangle,  as  ABC,  in 
which  B  is  the  obtuse  angle  and  BD  the  ^^ 

projection  of  BC  upon  AB  produced.  ^^    / 

To  prove  ^^         / 

AC^  ^  Iff  +  BC^  +  2  rect.  AB  •  BD.  /— / 

Proof.      .  AD  =  AB-\-BD'^ 

then,  §  347,  Iff  ^  Iff  +  sff  +  2  rect.  AB  •  BD. 

Adding  CD  to  both  members  of  this  equation, 

Iff  -^cff=o=  Iff  +  sff  +cff  -\-2  rect.  AB  -  BD. 
But,  §  349,  Iff -\- off  ^  Iff, 

and  sff  -\-c3^^  Bff. 

Hence,  substituting  Aff  and  Bff  for  their  equivalents, 

Ic^  ^  Iff  4-  BO^  +  2  rect.  AB  •  BD. 
Therefore,  etc.  q.e.d. 

Ex.  503.  The  diagonals  of  a  rhombus  are  30  in.  and  16  in.  What  is  the 
length  of  the  sides  ? 

Ex.  504.  A  square  lawn  with  the  walk  around  it  contains  ^  of  an  acre. 
If  the  walk  contains  |f  of  the  eniire  area,  what  is  the  width  of  the  walk  ? 

Ex.  505.  Find  the  area  of  a  field  in  the  form  of  a  trapezoid  whose  bases 
are  45  rd.  and  27  rd. ,  and  each  of  whose  non-parallel  sides  is  15  rd. 


190 


PLANE   GEOMETRY.  — BOOK   V. 


Proposition  XV 

355.  1.  Draw  a  triangle  whose  sides  are  2",  3",  and  4",  or  any  other 
oblique  trianglo,;  construct  the  squares  on  any  two  sides;  construct  a 
square  on  one  half  of  the  third  side  and  also  a  square  on  the  median  to 
that  side;  find  the  area  of  each  of  these  squares.  How  does  the  com- 
bined area  of  the  first  two  compare  with  double  the  combined  area  of 
the  other  two? 

2.  Construct  and  find  the  area  of  the  rectangle  of  the  third  side  and 
the  projection  of  the  median  upon  that  side.  How  does  the  difference  in 
the  area  of  the  first  two  squares  compare  with  double  the  area  of  this 
rectangle  ? 

Theorem,  In  any  oblique  triangle  the  sum  of  the  squares 
upon  any  two  sides  is  equivalent  to  twice  the  square  upon 
one  half  the  third  side,  plus  twice  the  square  upon  the 
median  to  that  side. 


E    . 


Data :   Any  oblique  triangle,  as  ABC,  and  the  median  CD,  making 
with  AB  the  obtuse  angle  ADC  and  the  acute  angle  BDC. 

To  prove  AC^  -\-BC^  =o=  2  AD^  -\-2cff. 

Proof.     Draw  CEJlab,  or  AB  produced. 

Then,  in  the  A  ADC  and  DBC  respectively, 
§  354,  AC^  ^Iff  -{-CD^-{-2  rect.  AD  •  DE,  (1) 

and,  §  353,  BC^  =^Dff  +CD^  —  2  rect.  DB  ■  DE.  (2) 

But,  data,  AD  =  DB. 

Substituting  for  DB  in  the  second  equation  its  equal  AD  and 
adding  (1)  and  (2),         ic'  +  5c'  =g=  2  Id'  +  2  off. 


Therefore,  etc. 


Q.E.D. 


356.  Cor.  In  any  oblique  triangle  'the  difference  of  the  squares 
upon  any  two  sides  is  equivalent  to  twice  the  rectangle  formed  by  the 
third  side  and  the  projection  of  the  median  upon  that  side. 


PLANE   GEOMETRY.  — BOOK   V,  191 


Proposition  XVI 

357.  Draw  a  circle  and  two  intersecting  chords ;  draw  two  chords, 
which  do  not  meet,  to  connect  the  extremities  of  the  given  chords,  thus 
forming  two  triangles.  What  angles  of  the  figure  are  equal?  Are  the 
triangles  equal,  equivalent,  or  similar  ?  How  does  the  ratio  of  the  longer 
segments  of  the  given  chords  (sides  of  the  similar  triangles)  compare 
with  the  ratio  of  their  shorter  segments?  How  does  the  rectangle  formed 
by  the  segments  of  one  chord  compare  with  the  rectangle  formed  by  the 
segments  of  the  other  ? 

Theorem.  If  two  chords  of  a  circle  intersect,  the  rectangle 
formed  by  the  segments  of  one  chord  is  equivalent  to  the 
rectangle  forrvied  hy  the  segments  of  the  other. 


Data:   Any  two  chords  of  a  circle,  as  AB    a/ 
and  CDj  intersecting,  as  at  J&.  / 

To  prove    rect.  ae-be^  rect.  DE  •  CE.  \ 


Proof.     Draw  AC  and  bd. 

Then,  in  A  AEC  and  DEB, 
§  225,        Z.  A=  Z.D,  each  being  measured  by  ^  arc  CB, 
and  /.G  =  Z.  B,  each  being  measured  by  -J-  arc  AD\ 

,',  §  301,  A  AEG  and  DEB  are  similar. 

Hence,  AE\DE—  GE :  BE,  Why  ? 

2tnd,  §  269,  AE  X  BE  =  DE  X  GE; 

that  is,  §  324,  rect.  AE-  BE^  rect.  DE  -  GE. 

Therefore,  etc.  q.e.d. 

358.  Cor.  If  a  chord  passes  through  a  fixed  point,  the  area  of  the 
rectangle  formed  hy  its  segments  is  constant  in  whatever  direction 
the  chord  is  drawn. 

Ex.  506.  A  ladder  25"*  long,  with  its  foot  in  the  street,  will  reach  on  one 
side  to  a  window  20"»  high,  and  on  the  other  to  a  window  16™  high.  What 
is  the  distance  between  the  windows  ? 


192 


PLANE   GEOMETRY.  — BOOK    V. 


Proposition  XVII 

359.  From  a  point  without  a  circle  draw  two  secants ;  draw  two  inter- 
secting chords  to  connect  the  points  of  intersection  of  tlie  secants  and 
the  circumference ;  select  two  triangles  each  of  which  has  a  secant  for 
one  of  its  sides.  What  angles  of  these  triangles  are  equal?  Are  the 
triangles  equal,  equivalent,  or  similar  ?  How  does  the  ratio  of  the  secants 
(sides  of  the  similar  triangles)  compare  with  the  ratio  of  their  external 
segments?  How  does  the  rectangle  formed  by  one  secant  and  its  ex- 
ternal segment  compare  with  the  rectangle  formed  by  the  other  and  its 
external  segments? 

Theorein,  If  from  a  point  without  a  circle  two  secants 
are  drawn,  the  rectangle  formed  hy  one  secant  and  its  ex- 
ternal segment  is  equivalent  to  the  rectangle  formed  by  the 
other  secant  and  its  external  segment. 

Data :  Any  point  without  a  circle,  as  A, 
and  any  two  secants  from  A,  as  AB  and  AC, 
cutting  the  circumference  in  E  and  D  re- 
spectively. 

To  prove    rect.  AB  •  AE  =0=  rect.  AC  •  AD. 

Proof.     Draw  BD  and  CE. 

Then,  in  A  ABD  and  ACE, 

Z^  is  common, 
and  Z.B  =  Z.C\ 

A  ABD  and  ACE  are  similar. 

Hence,  ab\AC=  ad\  ae, 

and  rect.  AB-  AE^ rect.  AC-  AD. 


Why? 
Why? 
Why? 


Therefore,  etc. 


Q.E.D. 


Ex.  507.   Find  the  area  of  a  triangle  each  of  whose  sides  is  12  ft. 

Ex.  508.  The  side  of  a  rhombus  is  29<='"  and  one  of  its  diagonals  is  40"°. 
What  is  the  length  of  the  other  diagonal  ? 

Ex.  509.  The  area  of  a  rhombus  is  1176  sq.  in,  and  one  of  its  diagonals  is 
42  in.     What  are  its  sides  and  the  other  diagonal  ? 

Ex.  510.  The  radius  of  a  circle  is  S^m  and  a  tangent  to  the  circle  is  IS**™. 
What  is  the  length  of  a  secant  drawn  from  the  same  point  as  the  tangent,  if 
the  secant  is  6*™  from  the  center  ? 


PLANE   GEOMETRY,  — BOOK    V.  193 


Proposition  XVIII 

360.  Draw  a  triangle  and  its  circumscribing  circle ;  bisect  the  vertical 
angle  and  produce  the  bisector  to  meet  the  circumference ;  connect  this 
point  of  meeting  with  the  point  of  intersection  of  the  base  and  shortest 
side.  What  angles  of  the  figure  are  equal?  What  triangles  are  similar? 
From  the  ratios  of  the  sides  of  similar  triangles  and  of  the  segments  of 
intersecting  chords  discover  how  the  rectangle  formed  by  the  sides  of  the 
given  triangle  compares  with  the  rectangle  formed  by  the  segments  of 
the  base  plus  the  square  upon  the  bisector  of  the  vertical  angle? 

Theorem,  If  the  bisector  of  the  vertical  angle  of  a  tri- 
angle intersects  the  base,  the  rectangle  formed  by  the  two 
sides  is  equivalent  to  the  rectangle  formed  by  the  seg- 
ments of  the  base  plus  the  square  upon  the  bisector. 

Data:  Any  triangle,  as  ABC,  and  the  bi-       /''^-^"^^ /    \,\ 

sector  of  its  vertical  angle,  as  CD,  intersect-   ^[^^ 4 — -t^^b 

ing  the  base  in  D.  \  j      /  \ 

To  prove  \  /   /''     / 

rect.  AC .  -BC7=c=  rect.  AD  -  BD -^  cff.  \^        //    y 

"'-- — ic-'-'' 

E 

Proof.  Circumscribe  a  circle  about  A  ABC,  produce  CD  to  meet 
the  circumference  in  E,  and  draw  EB. 

Then,  in  A  ADC  and  EBC, 
data,  Zacd=z/.  ecb, 

and  /.A  =  ZE;  Why? 

A  ADC  and  EBC  are  similar.  Why  ? 

Hence,  '  AC :  EC  =  CD  :  BC, 

and  rect.  AC  -  BC  ^  rect.  EC  •  CD, 

or    rect.  AC  •  BC  ^  rect.  (DE  +  CD)  •  CD  ^  rect.  DE  -  CD  -\-  cff. 

But,  §  357,         rect.  DE  -  CD  ^  rect.  AD  •  BD. 

Hence,  rect.  AC  >  BC  =o=  rect.  AD  *  BD  -{-  cff. 

Therefore,  etc.  q.e.d. 

Ex.  511.  A  pole  standing  on  level  ground  was  broken  75  ft.  from  the  top 
and  fell  so  that  the  end  struck  60  ft.  from  the  foot.  Find  the  length  of  the 
pole. 

milne's  geom-  —  33 


J  94 


PLANE   GEOMETRY.  — BOOK   V, 


Proposition  XIX 

361.  Draw  a  triangle,  a  line  representing  its  altitude,  the  circumscrib 
ing  circle,  and  a  diameter  from  the  vertex ;  connect  the  other  extremity 
of  this  diameter  with  the  point  of  intersection  of  the  base  and  shortest 
side.  What  right  triangles  are  similar?  How  does  the  ratio  of  their 
longest  sides  compare  with  the  ratio  of  their  shortest"  sides  ?  How  does 
the  rectangle  formed  by  the  sides  of  the  given  triangle  compare  with  the 
rectangle  formed  by  its  altitude  and  the  diameter  of  the  circumscribing 
circle  ? 

Theorem.  The  rectangle  formed  hy  any  two  sides  of  a 
triangle  is  equivalent  to  the  rectangle  formed  hy  the  alti- 
tude upon  the  third  side  and  the  diameter  of  the  circum- 
scribing circle. 


Data :  Any  triangle,  as  ABC;  a  diameter 
of  tbe  circumscribing  circle,  as  CD ;  and  the 
altitude  upon  AB,  as  CE. 

To  prove  rect.  AC  -  BC  ^  rect.  CD  •  CE. 

Proof.     Draw  DB. 

Data,  §  94,  Z  AEC  is  a  rt;  Z, 

§  227,  Z  DBC  is  a  rt.  Z. 

Then,  in  rt.  A  AEC  and  DBC, 

Za  =  ZD; 
.'.  §  302,  A  AEC  and  DBC  are  similar. 


Why? 


Hence, 
and 

Therefore,  etc. 


AC:  CD  =  CE:  BC, 
rect.  AC '  BC^  rect.  CD  •  CE. 


Q.E.D. 

Ex.  512.  Upon  the  diagonal  of  a  rectangle  28™  by  21™  a  triangle  equiva- 
lent to  the  rectangle  is  constructed.     What  is  the  altitude  of  the  triangle  ^ 

Ex,  513.  The  base  and  altitude  of  a  right  triangle  are  6  ft.  and  8  ft. 
respectively.  What  is  the  length  of  the  perpendicular  drawn  from  the 
vertex  of  the  right  angle  to  the  hypotenuse  ? 

Ex.  514.  The  parallel  sides  of  a  trapezoid  are  12  in.  and  16  in.  and  the 
non-parallel  sides  are  10  in.  What  is  the  area  of  the  triangle  formed  by 
joining  the  middle  point  of  the  shorter  base  with  the  extremities  of  the 
longer  ? 


PLANE   GEOMETRY.— BOOK    V, 


195 


Proposition  XX 

362.   Problem.    To  construct  a  square  equivalent  to  the 
sum  of  two  given  squares. 


E 

1 
i 

c 

"     1 

1 

A 

1 

I 

B 

'\ 

1 

F  D 

Data :    Any  two  squares,  as  A  and  B. 
Required  to  construct  a  square  equivalent  to  A-^B. 
Solution.     Draw  FB  equal  to  a  side  of  A.     At  one  extremity,  as 
Al  F,  draw  FF  _L  FB  and  equal  to  a  side  of  B.     Draw  ED.  . 
Construct  a  square   C,  having  each  of  its  sides  equal  to  EB. 
Then,  C  is  the  required  square.  q.e.f. 

^roof .     By  the  student.      Suggestion.     Refer  to  §  349. 


Proposition  XXI 

363.  JProblem.    To  construct  a  square  equivalent  to  the 
difference  of  two  given  squares. 


Data :  Any  two  squares,,  as  A  and  B. 
Required  to  construct  a  square  equivalent  to  A  —  B. 
Solution.     Draw  an  indefinite  line,  as  GB. 
At  G,  erect  a  perpendicular  to  GB,  as  GE,  equal  to  a  side  of  B. 
With  J5^  as  a  center  and  a  radius  equal  to  a  side  of  A,  describe 
an  arc  intersecting  GB  at  F.     Draw  EF. 

Construct  a  square   C,  having  each  of  its  sides  equal  to  GF. 
Then,  C  is  the  required  square.  q.e.f. 

Proof.     By  the  student.     Suggestion.     Refer  to  §  349o 


196 


PLANE   GEOMETRY.  — BOOK   V. 


Propositioa  XXII 

364.  Problem,    To  construct  a  square  equivalent  to  tha 
sum  of  any  number  of  given  squares. 


^       "    re 


-Fs 


H 


D  E 

Data :  Any  squares,  as  A,  B,  and  C. 

Required  to  construct  a  square  equivalent  to  A  +  B  -\-  c. 

Solution.     Draw  DE  equal  to  a  side  of  A. 

At  D  erect  a  perpendicular  to  DE,  as  DF,  equal  to  a  side  of  B. 

Draw  FE. 

At  F  erect  a  perpendicular  to  FE,  as  FG,  equal  to  a  side  of  C. 

Draw  GE. 

Construct  a  square  H,  having  its  sides  each  equal  to  GE. 

Then,  H  is  the  required  square.  q.e.d. 

Proof.     By  the  student.     Suggestion.    Eefer  to  §  349. 

Ex.  515.  Divide  a  triangle  into  two  equivalent  triangles  by  a  line  drawn 
through  any  vertex. 

Ex.  516.  Construct  a  triangle  equivalent  to  a  given  triangle  and  having 
the  same  base. 

Ex.  517.  Construct  an  isosceles  triangle  equivalent  to  a  given  triangle 
and  having  the  same  base. 

Ex.  518.    Construct  a  right  triangle  equivalent  to  a  given  triangle. 

Ex.  519.  Construct  a  triangle  equivalent  to  a  given  triangle,  having 
the  same  base  and  an  angle  at  the  base  equal  to  a  given  angle. 

Ex.  520.  Construct  a  triangle  similar  to  a  given  triangle  and  four  times 
the  given  triangle. 

Ex.  521.  Divide  a  parallelogram  into  two  equivalent  parts  by  a  line 
through  any  point  in  its  perimeter. 

Ex.  522.  Divide  a  rectangle  into  four  equivalent  parts  by  lines  through 
any  vertex. 

Ex.  523.  Construct  a  square  equivalent  to  a  triangle  whose  base  is  18cm 
and  altitude  4"". 

Ex.  524.  Construct  a  square  equivalent  to  a  rectangle  whose  dimensions 
are  16<^'"  and  4cm. 

Ex.  525.  Construct  a  square  equivalent  to  the  difference  between  two 
squares  whose  areas  are  25'<i «'"  and  IQn  cm. 


PLANE   GEOMETRY.  — BOOK   V, 


197 


Proposition  XXIII 

365.   JProblem,    To  construct   a  polygon  similar  to  two 
given  similar  polygons  and  equivalent  to  their  sum. 


<       c 


Data :  Any  two  similar  polygons,  as  A  and  B. 

Required  to  construct  a  polygon  similar  to  A  and  B  and  equiva- 
lent to  ^  +  ^• 

Solution.     Draw  a  line,  as  FD,  equal  to  m,  a  side  of  A. 

At  one  extremity,  as  F,  erect  a  perpendicular  FF,  equal  to  n, 
the  homologous  side  of  B.     Draw  FD. 

Taking  I,  a  line  equal  to  FD,  as  homologous  to  m  and  n,  construct 
a  polygon   C,  similar  to  A  and  B. 

Then,  C  is  the  required  polygon.  q.e.f. 

~"  Why? 


Proof. 

FD^  -{- FF^  =  FD^ ; 

.*. 

m^  +  n^  =  E 

Now,  §  344, 

A:  C  =  m^:^, 

and 

B:  C=n^:f'j 

.',  §  280, 

A  +  B:  C  =  m^-{-n^:P. 

But 

m'  +  n'  =  E 

Hence,  §  271, 

A  +  B^C. 

Ex.  526.   Construct  a  right  triangle  equivalent  to  a  given  square. 

Ex.  527.    Construct  a  right  triangle  equivalent  to  a  given  rectangle. 

Ex.  528.    Construct  a  right  triangle  equivalent  to  a  given  parallelogram. 

Ex.  529.    Construct  an  isosceles  triangle  equivalent  to  a  given  square. 

Ex.  530.  Construct  a  square  equivalent  to  the  sum  of  two  squares  whose 
sides  are  5  in.  and  10  in. 

Ex.  531.  Construct  a  square  equivalent  to  the  difference  of  two  squares 
whose  sides  are  IS''™  and  17<=™. 

Ex.  532.  Construct  a  polygon  similar  to  two  given  similar  polygons,  and 
equivalent  to  their  difference. 


198 


PLANE    GEOMETRY.  — BOOK   V. 


Proposition  XXIV 

366.   Problem,     To  construct  a  square  having  a  given 
ratio  to  a  given  square. 

K 


/F 


/H 


Data :  Any  square,  as  A,  and  any  ratio,  as  m :  n. 

Required  to  construct  a  square  B,  such  that  B  -.  A  =  m  :  n. 

Solution.     Draw  EF  equal  to  a  side  of  A,  and  draw  ED  making 
any  acute  angle  with  EF. 

On  ED  take  EG  equal  to  w,  and  GJ  equal  to  m. 

Draw  GF\  also  draw  JH II  GF,  meeting  EF  produced  at  H. 

On  EH  describe  a  semicircumference,  and  at  F  erect  FK  per- 
pendicular to  EH  and  meeting  the  semicircumference  in  K. 

On  a  line  equal  to  FK  as  a  side,  construct  the  square  B. 

Then,  B :  A  =  m  -.  n.  ,  q.e.f. 

Proof.     §  314, 
hence,  FK^  =  EF  x  FK  Why  ? 

Now, 


EF:FK=FK:FH', 
FK^  =  EF  X  FK 
FK^ :  eP  =  FK^  :  eP  ; 
FK^  :EF^  =  EF  X  FH:  EF  , 
or,  dividing  each  term  of  the  second  ratio  by  EF, 


FK   :  EF  : 

FH.EF 

FK^  :  EF^  ■ 


FH'.EF. 
GJ:EG-- 
m:n: 


m:  n 


B:  A  =  m 


§283, 
Const., 

that  is, 

Ex.  533. 
Ex.  534. 

gram. 

Ex.  535.  Construct  a  square  which  shall  be  equivalent  to  a  right  isos- 
celes triangle,  having  given  the  perpendicular  from  the  vertex  of  the  right 
angle  upon  the  hypotenuse. 


Construct  an  isosceles  triangle  equivalent  to  a  given  rectangle. 
Construct  an  isosceles  triangle  equivalent  to  a  given  parallelo- 


PLANE   GROMETRY.  —  BOOK    V.  199 


Proposition  XXV 

367.  Problem,  To  construct  a  triangle  equivalent  to  a 
given  polygon.  ^  j) 

Datum :  Any  polygon,  as  ABCDEF.     -  // 1  y^^/ 

Required  to  construct  a  triangle  /  ^\/    [/ 

equivalent  to  ABCDEF.  /y^^  i  ,\''1 

J  HA  B        G 

Solution.  Draw  DB,  and  from  C  tc  AB  produced  draw  CG  il  DB ; 
also  draw  DG. 

Draw  FA,  and  from  F  to  BA  produced  draw  FH  II  EA ;  also 
draw  EH. 

Draw  DH,  and  from  E  to  BA  produced  draw  EJ II  BH-,  also 
draw  DJ. 

Then,  JGD  is  the  required  triangle.  q.e.f. 

Proof.     In  the  polygons  AGDEF  and  ABCDEF^ 
ABDEF  is  common, 
and,  §  336,  A  DBG  ^  A  BBC ; 

AGDEF  ^  ABCDEF. 
In  the  polygons  HGDE  and  AGDEF, 

AGDE  is  common, 
and  Aj^^^IT^A^^J?';  Why? 

HGDE  =^  AGDEF. 
In  the  polygons  JGD  and  ir(?Z)^, 

ZTGD  is  common, 
and  Ahdj<^AHDE.  Why? 

Hence,  A  JGD  =0  ITG^D^  ^  AGDEF  ^  ABCDEF. 

Ex.  536.  Construct  a  parallelogram  equivalent  to  a  given  parallelogram 
and  having  an  angle  equal  to  a  given  angle. 

Ex.  537.  Bisect  a  given  parallelogram  (1)  by  a  line  passing  through  a 
given  point  within ;  (2)  by  a  line  perpendicular  to  a  side ;  (3)  by  a  line 
parallel  to  a  side. 


200 


PLANE   GEOMETRY.  — BOOK   V, 


Proposition  XXVI 

368.   Problem*     To   construct   a  square   equivalent   to   a 
given  parallelogram. 


J 


E    G 


H 


A  B  F  a  K 

Datum :  Any  parallelogram,  as  ABCD. 
Required  to  construct  a  square  equivalent  to  ABCD. 
Solution.     Draw  the  altitude  BE-^  also  draw  FG  equal  to  BE. 
Produce  FG  to  K,  making  GK  equal  to  AB.    On  FK  as  a  diameter 
describe  a  semicircumf erence,  draw  GJ  meeting  it  in  J  and  X  FK. 
On  a  line  equal  to  GJ  as  a  side,  construct  the  square  H. 
Then,  H  is  the  required  square.  q.e.f. 

Proof.     By  the  student.     Suggestion.     Refer  to  §  314. 

'  369.  Sch.  A  square  may  be  constructed  equivalent  to  a  given 
triangle  by  taking  for  its  side  a  mean  proportional  between  the 
base  and  one  half  the  altitude  of  the  triangle. 

To  construct  a  square  equivalent  to  any  given  polygon,  first  re- 
duce the  polygon  to  an  equivalent  triangle,  and  then  construct  a 
square  equivalent  to  this  triangle. 

Proposition  XXVII 
370.    Problem,    To  construct  a  rectangle  equivalent  to  a 
given  square,  and  having  the  sum  of  its  base  and  altitude 
^qual  to  a  given  line. 


i 


H 


Data :  Any  square,  as  A,  and  the  line  BC. 

Required  to  construct  a  rectangle  equivalent  to  A,  and  having 
the  sum  of  its  base  and  altitude  equal  to  BC. 


PLANE   GEOMETRY.  — BOOK   V. 


201 


Solution.     Upon  J5C  as  a  diameter  describe  a  semicircumference. 

At  one  extremity  of  BC,  as  B,  erect  a  perpendicular  to  BC,  as 
DBj  equal  to  a  side  of  A.  Draw  DE  II  5(7  meeting  the  semicir- 
cumference in  E.     Draw  EF  II  Z)J5  meeting  BC  m.  F. 

With  base  GF  and  altitude  -Bi^  construct  rectangle  H. 

Th-en,  H  is  the  required  rectangle.  q.e.f. 

Proof.  DB  =  EfI  Why? 

nff  =  EF^  <>  A. 

But,  §  314,  BF:EF  =  EF:CF) 

BF  X  OF  =  E?; 
that  is,  H^A. 


Proposition  XXVIII 

371.  Problem,  To  construct  a  rectangle  equivalent  to  a 
given  square,  and  having  the  difference  of  its  base  and 
altitude  equal  to  a  given  line. 


D 

N 

l\ 
1    \ 
1      \ 

A 

V     v 

\c 


H 


Data :  Any  square,  as  A,  and  the  line  BC. 

Required  to  construct  a  rectangle  equivalent  to  A,  and  having 
the  difference  of  its  base  and  altitude  equal  to  BC. 

Solution.     On  BC  as  a  diameter  describe  a  circumference. 

At  one  extremity  of  BC,  as  B,  erect  a  perpendicular  to  BC,  as 
BD,  equal  to  a  side  of  A. 

Through  0,  the  center  of  the  circle,  draw  DF  intersecting  the 
circumference  in  E  and  meeting  it  in  F. 

Then,  FD  —  ED  =  EF,  or  BC. 

With  base  FD  and  altitude  ED  construct  rectangle  H. 

Then,  H  is  the  required  rectangle.  q.e.f. 

Proof.     By  the  student.     Suggestion.     Refer  to  §§  316,  269, 


i02 


PLANE   GEOMETRY.  — BOOK   V. 


Proposition  XXIX 
372.   Problem,    To  construct  a  polygon  similar  to  a  given 
Tiolygon  and  equivalent  to  any  other  given  polygon. 


H 


I 


Data :  Any  two  polygons,  as  A  and  B. 

Required  to  construct  a  polygon  similar  to  A  and  equivalent 
to  B. 

Solution.     Find  c,  the  side  of  a  square  equivalent  to  A,  and 
d,  the  side  of  a  square  equivalent  to  B,  and  let  e  be  a  side  of  A. 
Find  a  fourth  proportional  to  c,  d,  and  e,  as  /. 
Upon  /  homologous  to  e  construct  H  similar  to  A. 
Then,  H  is  the  required  polygon.  q.e.f. 


Proof.     Const., 
Also 

But,  const.. 

But,  §  344, 

Hence,  §  272, 


H  is  similar  to  A. 

c  :d  =e  'J; 

(?:d^  =  e^\f\ 
4=o=c^,  and  £=od^; 

A'.B  =  e^:f. 

A'.H=e^'.p', 

A:H=A:B. 
H^B. 


Why 


SUMMARY 
"373.    Truths  established  in  Book  V. 
1.   A  rectangle  is  equivalent, 

a.  If  it  is  the  rectangle  formed  by  the  segments  of  one  of  two  intersecting 
chords,  to  the  reetangle  formed  by  the  segments  of  the  other.  §  357 

b.  If  it  is  formed  by  a  secant  and  its  external  segment,  to  a  rectangle 
formed  by  another  secant  from  the  same  point  and  its  external  segment.    §  359 

c.  If  it  is  formed  by  the  two  sides  of  a  triangle,  to  the  rectangle  formed 
by  the  segments  of  the  base,  made  by  the  bisector  of  the  vertical  angle,  plus 
the  square  upon  the  bisector.  §  360 


PLANE   GEOMETRY.  — BOOK   V.  203 

d.  If  it  is  formed  by  two  sides  of  a  triangle,  to  the  rectangle  formed 
by  the  altitude  upon  the  third  side  and  the  diameter  of  the  circumscribing 
circle.  "  §  361 

2.  Rectangles  are  in  proportion, 

a.  If  they  have  equal  altitudes,  to  their  bases.  §  327 

b.  If  they  have  equal  bases,  to  their  altitudes.  §  328 

c.  To  the  products  of  their  bases  by  their,  altitudes.  §  329 

3.  A  parallelogram  is  equivalent, 

a.  To  the  rectangle  which  has  the  same  base  and  altitude.  §  331 

b.  To  another  parallelogram  which  has  an  equal  base  and  an  equal  alti- 
tude. .  §  333 

4.  Parallelograms  are  in  proportion, 

a.  If  they  have  equal  altitudes,  to  their  bases.  §  333 

b.  If  they  have  equal  bases,  to  their  altitudes.  §  333 

c.  To  the  products  of  their  bases  by  their  altitudes.  §  333 

5.  A  triangle  is  equivalent, 

a.  To  one  half  the  rectangle  which  has  the  same  base  and  altitude.    §  334 

b.  To  another  triangle  which  has  an  equal  base  and  an  equal  altitude.   §  336 

c.  To  another  triangle  which  has  an  angle  equal  to  an  angle  of  the  first, 
and  the  products  of  the  sides,  including  the  equal  angles,  equal.  §  341 

6.  Triangles  are  in  proportion, 

a.  If  they  have  equal  altitudes,  to  their  bases.  §  336 

b.  If  they  have  equal  bases,  to  their  altitudes.  §  336 

c.  To  the  products  of  their  bases  by  their  altitudes.  §  336 

d.  If  they  have  an  angle  of  one  equal  to  an  angle  of  the  other,  to  the  prod- 
ucts of  the  sides  including  the  equal  angles.  §  340 

e.  If  they  are  similar  triangles,  to  the  squares  upon  their  homologous 
sides.  §  342 

/.  If  they  are  similar  triangles,  to  the  squares  upon  any  of  their  homolo- 
gous lines.  §  343 

7.  A  trapezoid  is  equivalent, 

a.  To  one  half  the  rectangle  which  has  the  same  altitude  and  a  base  equal 
to  the  sum  of  the  parallel  sides.  ~  §  337 

8.  The  square  upon  a  line  is  equivalent, 

a.  If  the  line  is  the  sum  of  two  lines,  to  the  sum  of  the  squares  upon  the 
lines  plus  twice  the  rectangle  formed  by  them.  §  347 

b.  If  the  line  is  the  difference  of  two  lines,  to  the  sum  of  the  squares  upon 
the  lines  minus  twice  the  rectangle  formed  by  them.  §  348 

c.  If  the  line  is  the  hypotenuse  of  a  right  triangle,  to  the  sum  of  the  squares 
upon  the  other  two  sides.  §  349 


204  PLANE  GEOMETRY.  — BOOK   V. 

d.  If  the  line  is  the  side  of  an  oblique  triangle,  opposite  an  acute  angle, 
to  the  sum  of  the  squares  upon  the  other  two  sides  minus  twice  the  rectangle 
formed  by  one  of  those  sides  and  the  projection  of  the  other  upon  that  side. 

§353 

e.  If  the  line  is  the  side  opposite  an  obtuse  angle  of  a  triangle,  to  the  sum 
of  the  squares  upon  the  other  two  sides  plus  twice  the  rectangle  formed  by 
one  of  those  sides  and  the  projection  of  the  other  upon  that  side.  §  354 

9.  The  sum  of  two  squares  is  equivalent, 

a.  If  they  are  the  squares  upon  any  two  sides  of  an  oblique  triangle,  to 
twice  the  square  upon  one  half  the  third  side  plus  twice  the  square  upon  the 
median  to  that  side.  §  356 

10.  The  difference  of  two  squares  is  equivalent, 

a.  If  they  are  the  squares  upon  any  two  sides  of  an  oblique  triangle,  to 
twice  the  rectangle  formed  by  the  third  side  and  the  projection  of  the  median 
upon  that  side.  §  356 

11.  Similar  polygons  are  in  proportion, 

a.    To  the  squares  upon  their  homologous  sides.  §  344 

6.   To  the  squares  upon  any  of  their  homologous  lines.  §  345 

12.  The  area  of  a  figure  is  equal, 

a.   If  it  is  a  rectangle,  to  the  product  of  its  base  by  its  altitude.  §  330 

6.   If  it  is  a  parallelogram,  to  the  product  of  its  base  by  its  altitude.    §  332 

c.  If  it  is  a  triangle,  to  half  the  product  of  its  base  by  its  altitude.     §  335 

d.  If  it  is  a  trapezoid,  to  half  the  product  of  its  altitude  by  the  sum  of  its 
parallel  sides.  §  338 

SUPPLEMENTARY  EXERCISES 

Ex.  538.  The  straight  line  joining  the  middle  points  of  the  parallel  sides 
of  a  trapezoid  bisects  the  trapezoid. 

Ex.  539.  The  lines  joining  the  middle  point  of  either  diagonal  of  a  quad- 
rilateral to  the  opposite  vertices  divide  the  quadrilateral  into  two  eq-uivalent 
parts. 

Ex.  540.  Two  triangles  are  equivalent,  if  they  have  two  sides  of  one 
respectively  equal  to  two  sides  of  the  other,  and  if  the  included  angles  are 
supplementary. 

Ex.  541.  0  is  any  point  on  the  diagonal  ^C  of  the  parallelogram  ABCD. 
If  the  lines  DO  and  BO  are  drawn,  prove  that  the  triangles  AOB  and  AOD 
are  equivalent. 

Ex.  542.  A  rhombus  and  a  rectangle  have  equal  bases  and  equal  areas. 
One  side  of  the  rhombus  is  15™  and  the  altitude  of  the  rectangle  is  12™. 
What  are  their  perimeters  ? 


PLANE  GEOMETRY.^ BOOK  V.  205 

Ex.  543.  The  area  of  a  rhombus  is  equal  to  one  half  the  product  of  its 
diagonals. 

Ex.  544.  The  diagonals  of  a  rhombus  are  64  rd.  and  37  rd.  What  is  the 
area  of  the  rhombus  ? 

Ex.  545.  The  base  of  a  triangle  is  75™,  and  its  altitude  is  60™.  Find  the 
perimeter  of  an  equivalent  rhombus,  if  its  altitude  is  45™. 

Ex.  546.  Find  the  area  of  a  rhombus,  if  the  sum  of  its  diagonals  is  12  in. 
and  their  ratio  is  3  :  5. 

Ex.  547.  A  man  travels  25  miles  east  from  a  certain  town,  and  another 
man  travels  36  miles  north  from  the  same  town.     How  far  apart  are  the  men  ? 

Ex.  548.  The  shortest  side  of  a  triangle  acute-angled  at  the  base  is  45  ft. 
long,  and  the  segments  of  the  base  made  by  a  perpendicular  from  the  vertex 
are  27  ft.  and  77  ft.    How  long  is  the  other  side  ? 

Ex.  549.  The  sides  of  a  triangle  are  25™  and  17™,  and  the  lesser  segment 
of  the  base  made  by  a  perpendicular  from  the  vertex  is  8™.  What  is  the 
length  of  the  base  ? 

Ex.  550.  In  a  right  triangle  the  base  is  3<i™,  and  the  difference  between 
the  hypotenuse  and  perpendicular  is  1^™.  What  are  the  hypotenuse  and 
perpendicular  ? 

Ex.  551.  In  a  right  triangle  the  hypotenuse  is  5<i™,  and  the  difference 
between  the  base  and  perpendicular  is  l^^™.     Find  the  base  and  perpendicular. 

Ex.  552.  The  sides  of  a  right  triangle  are  in  the  ratio  of  3,  4,  and  5,  and 
the  perpendicular  upon  the  hypotenuse  from  the  vertex  of  the  right  angle  is 
20  yd.     What  is  the  area  of  the  triangle  ? 

Ex.  553.  If  in  any  triangle  a  perpendicular  is  drawn  from  the  vertex  to 
the  base,  the  difference  of  the  squares  upon  the  sides  is  equivalent  to  the 
difference  of  the  squares  upon  the  segments  of  the  base. 

Ex.  554.  In  a  right  triangle  the  square  on  either  side  containing  the  right 
angle  is  equivalent  to  the  rectangle  contained  by  the  sum  and  the  difference 
of  the  other  sides. 

Ex.  555.  If  the  diagonals  of  a  quadrilateral  intersect  at  right  angles, 
prove  that  the  sum  of  the  squares  upon  one  pair  of  opposite  sides  is  equiva- 
lent to  the  sum  of  the  squares  upon  the  other  pair. 

Ex.  556.  The  altitude  of  an  equilateral  triangle  is  60  in.  How  long  are 
its  sides  ? 

Ex.  557.  Through  J)  and  E,  the  middle  points  of  the  sides  AC  and  BC 
of  the  triangle  ^J5 Cany  two  parallel  straight  lines  are  drawn  meeting  ^5  or 
AB  produced  in  the  points  F  and  G.  Prove  that  the  parallelogram  DFQE 
is  equivalent  to  half  the  triangle  ABC. 

Ex.  558.  The  four  triangles  into  which  a  parallelogram  is  divided  by  its 
diagonals  are  equivalent. 


206  PLANE   GEOMETRY.  — BOOK  V. 

Ex.  559.  The  diagonals  of  a  trapezoid  divide  it  into  four  triangles,  two 
of  which  are  si^nilar,  while  the  other  two  are  equivalent. 

Ex.  560.  .kU  any  trapezoid  the  triangle,  having  for  its  base  one  of  the  non- 
parallel  sides  and  for  its  vertex  the  middle  point  of  the  opposite  side,  is 
equivalent  to  one  half  of  the  trapezoid. 

Ex.  561.  The  triangle,  formed  by  drawing  a  line  from  any  vertex  of  a 
parallelogram  to  the  middle  point  of  one  of  the  opposite  sides,  is  equivalent 
to  one  fourth  of  the  parallelogram. 

Ex.  562.  A  triangle  is  equivalent  to  one  half  the  rectangle  of  its  perim- 
eter and  the  radius  of  the  inscribed  circle. 

Suggestion.  Draw  radii  to  the  points  of  contact  and  lines  from  the  ver- 
tices of  the  triangle  to  the  center  of  the  circle. 

Ex.  563.  If  the  perimeter  of  a  quadrilateral  circumscribed  about  a  circle 
is  400  ft.  and  the  radius  of  the  circle  is  25  ft. ,  what  is  the  area  of  the  quadri- 
lateral ? 

Ex.  564.  The  area  of  a  triangle  is  875  sq.  yd.  Find  its  base  and  altitude, 
if  they  are  in  the  ratio  of  14  to  5. 

Ex.  565.  The  homologous  sides  of  two  similar  fields  are  in  the  ratio  of  3 
to  5,  and  the  sum  of  their  areas  is  416Ha.     What  is  the  area  of  each  field  ? 

Ex.  566.  A  board  12  ft.  long  is  10  in.  wide  at  one  end  and  6  in.  at  the 
other.  What  length  must  be  cut  from  the  narrower  end  to  contain  a  square 
foot? 

Ex.  567.  The  side  of  one  equilateral  triangle  is  equal  to  the  altitude  of 
another.     What  is  the  ratio  of  their  areas  ? 

Ex.  568,  The  perimeter  of  an  isosceles  triangle  whose  base  is  its  shortest 
side  is  lOO^^""  ;  the  difference  between  the  base  and  an  adjacent  side  is  23'^'". , 
What  is  the  altitude  of  the  triangle  ?    What  is  its  area  ? 

Ex.  569.  Two  chords  on  opposite  sides  of  the  center  of  a  circle  are 
parallel ;  one  is  16  ft.  long  and  the  other  is  12  ft.  If  the  distance  between 
them  is  14  ft,,  what  is  the  diameter  of  the  circle  ? 

Ex.  570.  If  from  the  vertex  of  an  acute  angle  of  a  right  triangle  a  straight 
line  is  drawn  bisecting  the  opposite  side,  the  square  upon  that  line  is  less  than 
the  square  upon  the  hypotenuse  by  three  times  the  square  upon  half  the  line 
bisected. 

Ex.  571.  In  the  right  triangle  ABC,  BG^  =  3  AC\  If  CD  is  drawn 
from  the  vertex  of  the  right  angle  to  the  middle  point  of  AB,  angle  ACD 
equals  60°. 

Ex.  572.   If  ACB  and  ADB  are  two  angles  inscribed  in  a  semicircle,  and 
AE  and  BF  are  drawn  perpendicular  to  CD  produced,  prove  that 
C^^  -j-  CF^  =  DE^  +  DF^. 


PLANE  GEOMETRY.  — BOOK   V.  207 

Ex.  573.  If  lines  are  drawn  perpendicular  to  the  diagonals  of  a  square  at 
their  extremities,  a  second  square  is  formed  equivalent  to  twice  the  original 
square. 

Ex.  574.  The  square  upon  the  base  of  an  isosceles  triangle  whose  vertical 
angle  is  a  right  angle  is  equivalent  to  four  times  the  triangle. 

Ex.  575.  The  sum  of  the  squares  on  the  lines  joining  any  point  in  the  cir- 
cumference of  a  circle  with  the  vertices  of  an  inscribed  square  is  equivalent 
to  twice  the  square  of  the  diameter  of  the  circle. 

Ex.  576.  If  AF  and  BE  are  the  medians  drawn  from  the  extremities  of 
the  hypotenuse  of  the  right  A  ABC,  prove  that  4  AF^  +  4  BE^  =  5  AB^. 

Ex.  577.  If  perpendiculars  PF,  PD,  and  PE  are  drawn  from  any  point 
P  to  the  sides  AB,  BC,  and  ^C  of  a  triangle,  prove  that 

AF^  +  BD^  +  CE^  =  AE^  +  BF^  +  CL^. 

Ex.  578.  If  any  point  P  within  the  rectangle  ABCD  is  joined  to  the 
vertices,  prove  that  PA^  +  PG'^  =  PB^  +  PL^. 

Ex.  579.  If  CD  and  AE  are  the  perpendiculars  from  the  vertices  C  and 
A  of  the  acute  triangle  ABC  to  the  opposite  sides,  prove  that 

AC'^  =  BCxCE-\-ABx  AD. 

Suggestion.     Refer  to  §  373,  8,  d. 

Ex.  580.  If  ^C  and  BG  are  the  equal  sides  of  an  isosceles  triangle,  and 
AD  is  drawn  perpendicular  to  BC,  prove  that  AB^  =  2  BC  x  BD. 

Ex.  581.  The  sum  of  the  squares  on  the  diagonals  of  a  parallelogram  is 
equivalent  to  the  sum  of  the  squares  on  its  four  sides.- 

Ex.  582.  Three  times  the  sum  of  the  squares  on  the  sides  of  a  triangle  is 
equivalent  to  four  times  the  sum  of  the  squares  on  the  medians  of  the  triangle. 

Ex.  583.  Two  sides  of  an  oblique  triangle  are  137  and  111  respectively, 
and  the  difference  of  the  segments  of  the  third  side  made  by  a  perpendicular 
from  the  opposite  vertex  is  52.     What  is  the  third  side  ? 

Ex.  584.  The  chord  of  an  arc  is  80  in. ;  the  chord  of  half  the  arc  is  41  in. 
What  is  the  diameter  of  the  circle  ? 

Ex.  585.  Froni  a  point  without  a  circle  two  tangents  are  drawn  which 
with  the  chord  of  contact  form  an  equilateral  triangle  whose  side  is  18  in. 
Find  the  diameter  of  the  circle. 

Ex.  586.  If  the  center  of  each  of  two  equal  circles  is  on  the  circum- 
ference of  the  other,  the  square  on  the  common  chord  is  equivalent  to  three 
times  the  square  on  the  radius. 

Ex.  587.  A  tangent  and  a  secant  meet  without  a  circle,  forming  an  angle 
of  45°  ;  the  tangent  is  2  ft.  long  and  the  diameter  of  the  circle  is  4  ft  Find 
the  length  of  the  secant. 


208  PLANE   GEOMETRY.  — BOOK   V. 

PROBLEMS  OF   CONSTRUCTION 

Ex.  588.  Divide  a  given  parallelogram  into  two  equivalent  parts  by  a  line 
drawn  parallel  to  a  given  line. 

Ex.  589.  Divide  a  given  triangle  into  two  parts,  whose  ratio  is  3 : 4,  by 
a  line  drawn  from  one  vertex. 

Ex.  590.  Divide  a  given  parallelogram  into  two  parts,  whose  ratio  is  2  :  3, 
by  a  line  parallel  to  a  side. 

Ex.  591.  Construct  a  parallelogram  equivalent  to  the  sum  of  two  given 
parallelograms  of  equal  altitude. 

Ex.  592.  Construct  a  parallelogram  equnralent  to  the  difference  of  two 
given  parallelograms  of  equal  bases. 

Ex.  593.   Construct  a  square  equivalent  to  five  times  a  given  square. 

Ex.  594.  Transform  a  given  trapezoid  into  an  equivalent  isosceles  trape- 
zoid. 

Ex.  595.  Construct  a  rhombus  equivalent  to  a  given  parallelogram,  and 
having  one  side  of  the  parallelogram  for  a  diagonal. 

Ex.  596.    Construct  a  square  equivalent  to  a  given  rectangle. 

Ex.  597.  Construct  a  square  equivalent  to  four  sevenths  of  a  given 
square. 

Ex.  598.  Construct  a  rectangle  equivalent  to  a  given  square,  and  having 
a  given  side. 

Ex.  599.  Construct  a  rectangle  equivalent  to  a  given  rectangle,  and  hav- 
ing a  given  side. 

Ex.  600.    Construct  a  square  equivalent  to  a  given  rhombus. 

Ex.  601.  Construct  a  parallelogram  having  a  given  altitude,  and  equiva- 
lent to  a  given  parallelogram. 

Ex.  602.  Construct  a  rectangle  having  a  given  altitude  and  equivalent  to 
a  given  parallelogram. 

Ex.  603.  Construct  a  rhombus  having  a  given  side,  and  equivalent  to  a 
given  parallelogram. 

Ex.  604.  Construct  a  rhombus  having  a  given  altitude  and  equivalent  to 
a  given  parallelogram. 

Ex.  605.  Transform  a  triangle  into  an  equivalent  parallelogram  whose 
base  shall  be  the  base  of  the  triangle  and  one  of  whose  base  angles  shall  be 
equal  to  a  base  an^le  of  the  triangle. 

Ex.  606.  Construct  a  triangle  having  a  given  angle,  and  equivalent  to  a 
given  parallelogram. 

Ex.  607.   Construct  a  triangle  equivalent  to  a  given  trapezium. 

Ex.  608.   Construct  a  parallelogram  equivalent  to  a  given  trapezium. 


PLANE   GEOMETRY.  — BOOK   V.  209 

Ex.  609.    Construct  an  isosceles  triangle  on  a  given  base  and  equivalent 
to  a  given  trapezium. 

Ex.  610.    Construct  a  right  triangle  equivalent  to  a  given  triangle,  having 
given  one  of  the  sides  about  the  right  angle. 

Ex.  611.   Construct  a  right  triangle  equivalent  to  a  given  triangle,  having 
given  the  hypotenuse. 

Ex.  612.    Construct  a  triangle  equivalent-  to  a  given  triangle  and  having 
its  base  and  altitude  equal. 

Ex.  613.   Construct  an  equilateral  triangle  equivalent  to  a  given  triangle. 

Ex.  614.    Construct  an  equilateral  triangle  equivalent  to  a  given  square. 

Ex.  615.    Construct  a  rectangle  having  a  given  diagonal  and  equivalent 
to  a  given  rectangle. 

Ex.  616.    Construct  a  rectangle  having  a  given  diagonal  and  equivalent 
to  a  given  square. 

Ex.  617.    Construct  a  square  equivalent  to  a  given  trapezoid. 

Ex.  618.    Construct  a  triangle  equivalent  to  a  given  trapezoid. 

Ex.  619.    Construct  a  parallelogram  equivalent  to  a  given  trapezoid  and 
having  for  its  base  the  longer  base  of  the  trapezoid. 

Ex.  620.    Construct  a  triangle  equivalent  to  a  given  triangle  and  similar 
tto  another  given  triangle. 

Ex.  621.   Construct  a  parallelogram  equivalent  to  the  sum  of  two  given 
parallelograms. 

Ex.  622.    The  area  of  a  square  is  16.     Construct  a  square  that  shall  be  to 
it  in  the  ratio  of  5  to  3. 

Ex.  623.   Construct  a  hexagon  similar  to  a  given  hexagon,  having  its  ratio 
to  the  given  hexagon  as  5  is  to  3. 

Ex.  624.    Construct  a  square  equivalent  to  two  thirds  of  a  given  hexagon. 

Ex.  625.    Construct  a  square  equivalent  to  the  sum  of  a  given  pentagon 
and  a  given  parallelogram. 

Ex.  626.   Divide  a  given  triangle  into  two  equivalent  parts  by  a  line  per- 
pendicular to  one  side. 

Ex.  627.   Divide  a  given  triangle  into  two  equivalent  parts  by  a  line  paral- 
lel to  one  side. 

Ex.  628.   Bisect  a  given  trapezoid  by  a  line  parallel  to  the  bases. 

Ex.  629.   Bisect  a  given  quadrilateral  by  a  line  drawn  from  one  of  the 
-vertices. 

Ex.  630.   Bisect  a  given  quadrilateral  by  a  line  drawn  from  any  point 
an  its  perimeter. 

milne's  geom.  — 14 


210 


PLANE   GEOMETRY.  — BOOK   V. 


ALGEBRAIC   SOLUTIONS 


Ex.  631.    Given  the  sides  of  any 
triangle,  to  compute  the  altitude. 

Solution.    In  /lABC,  suppose  that 

angle  A  is  acute. 

A         c    B 

Then,  §  353,  a"^  =  b^  +  d^ -2c  x  AD; 

AD  =  ^'  +  ''-^\ 
2c 

In  A  ADC,  h^  =  b^-A&. 

Substituting  for  AD^  its  value, 

\        2c         I  4c2 

_  (2  6c  +  b^+  c2-  gg)  (2  bc-b^-  cH  g^) 
~  4c2 

_  {(5  +  c)2  -  g2}{g2  -  (6  -  c)^} 
4c2 

_  (6  +  c  +  g)  (6  +  c  -  g)(g  +  b  -  c)(a  -b  +  c) 
~  4c2 

g  +  6  +  c  =  2s, 

6  +  c  —  g  =  2(s  —  g), 

g  +  6  —  c  =  2(s  —  c), 

g  -  &  +  c  =  2(s-6). 

2s  X  2(s-g)x2(s-  6)x2(s-c). 


ADC 


Let 

then, 

and 
Hence, 


h^ 


4c2 


;i  =  *  Vs(s  -  g)  (s  -  &)  (s  -  c). 
c 


Ex.  632.    G^ivew  «^e  sides  of  any  tri- 
angle, to  compute  its  area. 

Denote  the  area  by  ^. 


a  \h 


Solution.     Ex.  631,  h=-  V8(«  -  g) (s  -  b) (s  -  c), 
c 


then, 


c  ..2 


A  =  -x^  Vs(s  -  g)(s  -  6)(«  -  c) 
^      c 


=  V8(«-a)(s-  6)(«-  c). 


PLANE   GEOMETRY.  — BOOK    V. 


211 


Ex.  633.  Given  the  sides  of  a  triangle,  to  compute  the 
medians. 

Solution.  §355,        a2  +  52  =  2m2  +  2(|y. 

Hence,  4  m2  =  2(a2  +  62)  -  c2, 

and  m  =  ^V2(a2  + 6?)- c2. 


Ex.  634.  Given  the  sides  of  a  tnangle,  to  compute 
the  bisectors  of  the  angles. 

Solution.  Circumscribe  a  circle  about  A  ABC; 
produce  CD  to  meet  the  circumference  in  E;  and 
draw  BE, 


§360, 

ah  =  ADxBD  +  k^; 

.-. 

k^  =  ah-ADy.BD, 

§292, 

AD:BD  =  b:a; 

.*. 

AD  +  BD :  AD  =  a  +  b  :bj 

and 

AD  +  BD:a  +  b  =  AD  :  6  =  BD :  a ; 

that  is, 

c         AD     BD 

a-\-b       6         a 

whence, 

AD=    ^\,  and  BD  =    «\; 

a+b                     a+b 

hence, 

(a  +  6)2 

^abd          ^'      "i 

""T      (a  +  6)2J 

a6{(a  +  6)2-c2} 

(a  +  6)2 

a6(a  +  6  +  c)(a  +  6-c) 

• 

(a +  6)2 

06  X  2  s  X  2(s  —  c) 

(a +  6)2 
0 

Hence, 

k  =  -^y/abs(is-c). 

1/  / 


E 


a  +  b 


212 


PLANE   GEOMETRY.  — BOOK    V, 


Ex.  635.  Given  the  sides  of  a  triangle  and  the  radius 
of  the  circumscribing  circle,  to  compute  the  area  of  the 
triangle. 

Solution.     Denote  the  radius  of  the  circle  by  r. 

Then,  §  361,  ab  =  2rh; 

abc  =  2  rch. 
But  ch  =  2A\ 

abc  =  4:Ar. 
abc 


Hence, 


A  = 


4r 


Ex.  636.   Given  the  sides  of  a  triangle,  to  compute  the 
radius  of  the  circumscribed  circle. 


Solution.     §  361, 
But,  Ex.  631, 

whence, 


ab  =  2  rh. 


h-- 

a)is- 

b){s- 

c); 

abz 

=  'i^Ks 

-«)(« 

-b){s 

-c), 

r  = 

_ 

abc 

4  Vs(s  —  a)  (s  —  b)  (s  —  c) 


Ex.  637.  The  three  sides  of  a  triangle  are  58  ft.,  51  ft.,  and  41  ft.  in 
length.     What  is  the  area  of  the  triangle  ? 

Ex.  638.  Find  the  altitude  on  each  of  the  sides  of  a  triangle  whose  sides 
are  respectively  7  in.,  9  in.,  and  11  in. 

Ex.  639.  If  the  sides  of  a  triangle  are  respectively  4™,  6™,  and  S^  long, 
what  are  its  three  medians  ? 

Ex.  640.  What  is  the  area  of  a  triangle,  if  the  radius  of  the  circumscribing 
circle  is  6.196™  and  the  sides  of  the  triangle  are  respectively  9'",  6™,  and  12" 
in  length  ? 

Ex.  641.  The  sides  of  a  triangle  are  respectively  12  in.,  11  in.,  and  9  in. 
in  length.    Find  the  radius  of  the  circumscribing  circle. 

Ex.  642.  The  sides  of  a  triangle  are  respectively  30^"",  50^",  and  70^". 
Find  the  lengths  of  the  three  angle  bisectors. 

Ex.  643.  If  two  sides  and  one  of  the  diagonals  of  a  parallelogram  are 
respectively  12  in.,  15  in.,  and  18  in.,  what  is  length  of  the  other  diagonal  ? 
What  is  the  area  of  the  parallelogram  ? 


BOOK  VI 

KEGULAR  POLYGONS  AND  MEASUREMENT  OF  THE  CIRCLE 

374.  A  polygon  which  is  equilateral  and  equiangular  is  called 
a  Regular  Polygon. 

An  equilateral  triangle  and  a  square  are  regular  polygons. 

Proposition  I 

375.  Draw  a  circle  and  inscribe  in  it  any  equilateral  polygon.  How 
do  the  arcs  subtended  by  the  sides  of  the  polygon  compare  ?  How  do 
the  arcs  intercepted  by  the  sides  of  the  angles  of  the  polygon  compare  ? 
How  do  the  angles  themselves  compare?  What  may  any  equilateral 
polygon  that  is  inscribed  in  a  circle  be  called  ? 

Theoretn.  Any  equilateral  polygon  inscribed  in  a  circle 
is  a  regular  polygon. 


Data:      Any    equilateral     polygon,    as    ^ 
ABODE,  inscribed  in  a  circle. 

To  prove  ABODE  a  regular  polygon. 


Proof.     §  196,     arc  AB  =  arc  BO  —  arc  OD  =  etc. ; 

arc  BODE  =  arc  ODEA  =  arc  DEAD  —  etc., 
and  Za  =  Zb  =  Zo=  etc. ;  Why  ? 

that  is,  ABODE  is  equiangular. 

But,  data,  abode  is  equilateral ; 

hence,  §  374,  ABODE  is  a  regular  polygon. 

Therefore,  etc.  q.e.d. 

213 


214  PLANE   GEOMETRY.  — BOOK  VL 

Proposition  II 

376.  1.  Divide  the  circumference  of  a  circle  into  any  number  of  equal 
arcs;  draw  the  chords  of  these  arcs  in  succession.  What  kind  of  an 
inscribed  polygon  is  thus  formed? 

2.  Draw  tangents  to  the  circle  at  the  extremities  of  the  chords  and 
produce  them  until  they  intersect.  How  do  they  compare  in  length? 
How  do  the  angles  formed  by  each  pair  of  tangents  compare  in  size? 
What  kind  of  a  circumscribed  polygon  has  been  formed? 

Theorem.  If  the  circumference  of  a  circle  is  divided  into 
any  number  of  equal  arcs, 

1.  The  chords  joining  the  extremities  of  the  arcs  in  sue- 
cession  form  a  regular  inscribed  polygon. 

2.  The  tangents  drawn  at  the  extremities  of  the  arcs 
form  a  regular  circumscribed  polygon. 

Data :  Any  circumference  divided  into 
equal  arcs  at  A,  B,  C,  etc. ;  the  chords  AB, 
BC,  CD,  etc. ;  and  the  tangents  GBH, 
HCJ,  etc. 

To  prove 

ABODE  and  FGHJK  regular   polygons. 

Proof.     1.  AB  —  BC=CD  =  etc., 

and  Za  =  Zb  =  Zc  =  etc. ; 

.*.  §  374,  ABODE  is  a  regular  polygon. 

2.  Z  BAG  =  Z  ABG  =  Z  OBH=z  etc. ;  Why  ? 

Aabg,  boh,  ODJ,  etc.,  are  equal  isosceles  A.  Why  ? 

Hence,  Zg  =  Zh=Zj=  etc., 

and  GB  =  BH=HO=  etc. ;  Why  ? 

tangents  GBH.  HO  J,  etc.,  are  equal. 

Hence,  fghjk  is  a  regular  polygon. 

Therefore,  etc.  .  q.e.d. 

377.  The  radius  of  a  circle  inscribed  in  a  regular  polygon  is 
called  the  apothem  of  the  polygon. 


PLANE   GEOMETRY.  — BOOK    VL  215 

378.  The  radius  of  a  circle  circumscribed  about  a  regular  poly- 
gon is  called  the  radius  of  the  polygon. 

379.  The  common  center*  of  the  circles  inscribed  in  and  circum- 
scribed about  a  regular  polygon  is  called  the  center  of  the  polygon. 

380.  The  angle  between  the  radii  drawn  to  the  extremities 
of  any  side  of  a  regular  polygon  is  called  the  angle  at  the  center 
of  the  polygon. 

Proposition  III 

381.  1.  Draw  any  regular  polygon;  pass  a  circumference  through 
three  of  its  vertices.     Does  it  pass  through  the  other  vertices?    Why? 

2.  With  the  same  center  and  a  radius  equal  to  the  apothem  describe 
a  circle.     How  many  sides  of  the  polygon  does  this  circle  touch?    Why? 

Theorem.  A  circle  may  he  circumscribed  about,  and  a 
circle  may  be  inscribed  in,  any  regular  polygon. 

D 


Datum :  Any  regular  polygon,  as  ABODE. 

To  prove  1.  That  a  circle  may  be  circum- 
iribed  about  ABODE. 
2.  That  a  circle  may  be  inscribed  in  ABODE. 


Proof.     1.   Describe  a  circle  passing  through  A,  B,  and  (7,  and 
from  the  center  0,  draw  OA,  OB,  00,  OD,  and  OE. 

In  the  A  BOO,  OB  =00;  Why  ? 

Z  OBO  =  Z  OOB.  Why  ? 

But,  §  374,  Z  ABO  =  Z  BOD ; 

.-.  Ax.  3,  Z  OB  A  =  Z  OOD. 

Also  AB  =  OD,  and  0B  =  00; 

hence,  §  100,  Aabo  =  AODO, 

and  OA  =  OD. 

Consequently,  the  circle  passing  through  A,  B,  and  O  passes 
through  D. 

In  like  manner  it  may  be  shown  that  this  circle  passes  through  E. 

Therefore,  a  circle  described  with  the  center  0  and  a  radius 
equal  to  OA  will  be  circumscribed  about  the  polygon. 


216  PLANE   GEOMETRY.  — BOOK    VI. 

2.  Since  the  sides  AB,  BC,  CD,  etc.,  are  equal  chords  of  the 
circumscribed  circle,  they  are  equally  distant  from  the  center 
and  the  perpendiculars  drawn  from  the  center  to  the  chords  are 
all  equal. 

Hence,  a  circle  described  with  the  center  0  and  a  radius  equal 
to  one  of  these  perpendiculars,  as  OH,  will  be  inscribed  in  the 
polygon,  for  each  of  the  sides  of  the  polygon  will  be  perpendicular 
to  a  radius  at  its  extremity  and  tangent  to  the  circle.  §  205 

Therefore,  etc.  q.e.d. 

382.  Cor.  I.  The  radius  drawn  to  the  vertex  of  any  interior 
angle  of  a  regular  polygon  bisects  the  angle, 

383.  Cor.  II.  Each  angle  at  the  center  of  a  regular  polygon  is 
equal  to  four  right  angles  divided  by  the  number  of  sides  of  the 
polygon. 

Proposition  IV 

384.  Draw  a  circle  and  a  regular  inscribed  polygon;  at  the  middle 
points  of  the  arcs  subtended  by  its  sides  draw  tangents  and  produce 
them  until  they  intersect.  How  do  they  compare  in  length  ?  How  do 
the  angles  formed  by  each  pair  of  tangents  compare  in  size?  What 
kind  of  a  circumscribed  polygon  has  been  formed? 

Theorem,  Tangents  to  a  circle  at  the  middle  points  of 
the  arcs  subtended  hy  the  sides  of  a  regular  inscribed  poly- 
gon form  a  regular  circumscribed  polygon. 

Data :  A  circle  whose  center  is  0 ;  any 
regular  inscribed  polygon,  as  ABODE-, 
and  the  tangents  FG,  GH,  HJ,  etc.,  at  the 
middle  points  L,  M,  P,  etc.,  of  the  arcs  AB, 
BO,  CD,  etc. 

To  prove  FGHJK  a  regular  circum- 
scribed polygon 

Proof. 


\^^^        ^Bl 

F 

L            G 

arc  AB  =  arc  BC  =  arc  CD  =  etc.. 

Why? 

arc  AL  =  arc  LB  =  arc  BM  =  etc. ; 

Why? 

arc  LM  =  arc  MP  =  etc. 

PLANE   GEOMETRY.  — BOOK   VL 


217 


But  the  sides  of  FGHJK  are  tangents  at  the  extremities  of  these 
arcs. 

Hence,  §  376,  FGHJK  is  a  regular  circumscribed  polygon. 
Therefore,  etc.  q.e.d. 

385.  Cor.  Regular  inscribed  and  circumscribed  polygons  of  the 
same  number  of  sides  may  be  so  placed  that  their  sides  are  parallel 
and  their  vertices  will  then  lie  upon  the  radii  (prolonged)  of  the 
inscribed  polygon. 

Proposition  V 

386.  Draw  two  regular  polygons  of  the  same  number  of  sides.  How 
do  the  homologous  angles  compare  in  size  ?  How  do  the  ratios  of  any 
two  pairs  of  homologous  sides  compare?  What  name  is  given  to  poly- 
gons that  have  such  relations  to  each  other? 

Theorem,  Regular  polygons  which  have  the  same  num- 
ber of  sides  are  similar, 

D 


Data :    Any  two  regular  polygons,  as  ABODE  and  FGHJK,  which 
have  the  same  number  of  sides. 

To  prove  ABCDE  and  FGHJK  similar. 

Proof.     By  §  166,  the  sum  of  the  angles  of  each  polygon  is  equal 
to  twice  as  many  right  angles  as  the  polygon  has  sides  less  two. 

Since,  §  374,  each  polygon  is  equiangular,  and  since  each  con- 
tains the  same  number  of  angles ; 

all  the  angles  of  both  polygons  are  equal. 

§  374,     AB  =  BC  =  CD  =  etc.,  and  FG  =  GH  =  HJ  =  etc. ; 
AB:FG=  BC:GH=etG. 

Hence,  §  299,     ABODE  and  FGHJK  are  similar. 


Therefore,  etc 


Q.E.D. 


218 


PLANE   GEOMETRY,  — BOOK   VL 


Proposition  VI 

387.  Draw  two  regular  polygons  of  the  same  number  of  sides ;  draw 
radii  to  the  extremities  of  a  pair  of  homologous  sides.  What  kind  of 
triangles  are  formed  ?  How  does  the  ratio  of  their  bases  compare  with 
the  ratio  of  the  radii?  With  the  ratio  of  the  apothems?  Since  the 
polygons  are  similar,  how  does  the  ratio  of  their  perimeters  compare  with 
the  ratio  of  any  two  homologous  sides  ?  With  the  ratio  of  their  radii  ? 
Of  their  apothems  ? 

Theorem,  The  perimeters  of  regular  polygons  of  the  same 
number  of  sides  are  to  eaxih  other  as  their  radii  and  also 
as  their  apothems. 

D 


Data :  Any  two  regular  polygons  of  the  same  number  of  sides, 
as  ABODE  and  FGHJK,  whose  radii  are  OA  and  PF,  and  apothems 
OL  and  PM,  respectively. 

Denote  their  perimeters  by  Q  and  S  respectively. 

To  prove  Q:S  =  OA:PF=  OL:  PM. 

Proof.     Draw  OB  and  PG. 

In  the  A  AOB  and  FPG, 
§383,  Zo  =  Zp, 

and  OA:PF=OB:PG', 

.-.  §  306,  A  AOB  and  FPG  are  similar. 

Hence,  AB:FG=  OA:  PF, 

and  AB:FG=  OL:  PM. 

But,  §  §  386,  311,  Q:S  =  AB:FG. 

Hence,  Q:S=  OA:PF=  OL:  PM. 

Therefore,  etc.  q.e.d. 

Ex.  644.  -il'he  apothem  of  a  regular  pentagon  is  41<=™  and  a  side  is  G^m. 
Find  the  perimeter  of  a  regular  pentagon  whose  apothem  is  82^". 


Why? 


Why? 


PLANE  GEOMETRY.  — BOOK   VI.  219 


Proposition  VII 

388.  Draw  a  regular  polygon  and  draw  the  radii  to  the  vertices  of  its 
angles.  How  does  each  triangle  thus  formed  compare  with  the  rectangle 
of  its  base  and  altitude  ?  How  does  the  sum  of  the  bases  of  the  triangles 
compare  with  the  perimeter  of  the  polygon  ?  Since  the  triangles  are  of 
equal  altitude,  how  does  the  polygon  compare  with  the  rectangle  of  its 
perimeter  and  apothem? 

Theorem,  A  regular  polygon  is  equivalent  to  one  half 
the  rectangle  formed  hy  its  perimeter  and  apothem. 


Data:   Any  regular  polygon,  as  ABODE, 

and  its  apothem  OF.  ■^^- 

Denote  its  perimeter  by  P.  \ 

To  prove    ABODE  ^  h  rect.  P-OF,  ^ 


Proof.     Draw  the  radii  OA,  OB,  00,  OD,  and  OE. 

These  radii  divide  the  polygon  into  triangles  whose  altitudes- 
are  each  equal  to  the  apothem  and  the  sum  of  whose  bases  i^ 
equal  to  the  perimeter. 

Now,  §  334,  A  ABO  =c=  I  rect.  AB  -  OF, 

A  BOO  =0=  J  rect.  BO  •  OF,  etc. 

Hence,    A  ABO  +  A  BOO  +  etc.  =o=  I  rect.  (AB  -\-BC+  etc.)  •  OF ; 
that  is,  ABODE  o  ^  rect.  P  •  OF. 

Therefore,  etc.  q.e.d. 

389.  Cor.  I.  The  area  of  a  regular  polygon  is  equal  to  one  half 
the  product  of  its  perimeter  by  its  apothem. 

390.  Cor.  II.  Regular  polygons  of  the  same  number  of  sides  are 
to  each  other  as  the  squares  upon  their  radii  and  also  as  the  squares 
upon  their  apothems.  §§  386,  345 

Ex.  645.  The  sides  of  a  regular  circumscribed  polygon  are  bisected  at 
the  points  of  tangency. 

Ex.  646.  The  angle  at  the  center  of  a  regular  polygon  is  the  supplement 
of  the  angle  of  the  polygon. 


220  PLANE   GEOMETRY.  — BOOK   VI. 


Proposition  VIII 

391.  Draw  a  circle  and  circumscribe  a  polj^gon  about  it.  How  does 
the  circumference  of  the  circle  compare  in  length  with  the  perimeter  of 
the  polygon?  How  does  the  circumference  compare  with  any  enveloping 
line? 

Theorem,  The  circumference  of  a  circle  is  less  than  the 
perimeter  of  a  circumscribed  polygon  or  any  enveloping 
line. 

Data:   Any  circumference,  as  ABC,  and  any- 
enveloping  line,  as  FGHJK. 

To  prove  ABC  <  fghjk. 

Proof.     Of  all  the  lines  enveloping  the  area 
ABC  there  must  be  a  least  line. 

Draw  DE  tangent  to  ABC,  and  cutting  FGHJK 
in  D  and  E. 

Then,  Ax.  10,  ED  <  EHJD\ 

FGEDK  <  FGHJK; 
hence,  FGHJK  is  not  the  least  enveloping  line. 

Similarly,  it  may  be  shown  that  no  other  line  than  ABC  can  be 
the  least  line  enveloping  the  area  ABC. 

Hence,  ABC  <  FGHJK 

Therefore,  etc.  q.e.d. 

Ex.  647.  Find  the  angle  and  the  angle  at  the  center  of  a  regular 
dodecagon.  ' 

Ex.  648.  If  the  radius  of  a  regular  inscribed  hexagon  is  r,  prove  that  its 
rfde  =  r,  and  its  apothem  =  ^  rVS. 

Ex.  649.  If  the  radius  of  an  inscribed  equilateral  triangle  is  r,  prove  that 
its  side  =  rVS,  and  its  apothem  =  ^  r. 

Ex.  650.  If  the  radius  of  an  inscribed  square  is  r,  prove  that  its 
side  =  rV2,  and  its  apothem  =  I  r V2. 

Ex.  651.  The  radius  of  a  circle  being  r,  find  the  area  of  an  inscribed 
equilateral  triangle. 

Ex.  652.  The  radius  of  a  circle  being  r,  find  the  area  of  an  inscribed 
square. 

Ex.  653.   Find  the  area  of  a  regular  hexagon  whose  side  is  10  ft. 


PLANE  GEOMETRY.  — BOOK   VL 


221 


Proposition  IX 

392.  Draw  a  circle  and  inscribe  in  it  a  regular  polygon ;  circumscribe 
about  it  a  similar  polygon  whose  sides  are  parallel  to  the  sides  of  the 
inscribed  polygon.  If  the  number  of  sides  of  each  polygon  is  increased 
indefinitely,  what  line  will  their  perimeters  approach  as  a  limit  ?  What 
will  their  areas  approach  as  a  limit  ? 

Theorem,  If  a  regular  polygon  is  circumscribed  about  a 
circle  and  a  similar  polygon  is  inscribed  in  the  circle,  and 
if  the  number  of  their  sides  is  indefinitely  increased, 

1.  Their  perimeters  approach  the  circumference  as  a 
common  limit. 

2.  The  polygons  approach  the  circle  as  a  common  limit. 

Data :  Any  two  regular  polygons  of  the 
same  number  of  sides,  as  A  and  -B,  respec- 
tively circumscribed  about  and  inscribed 
in  a  circle  whose  center  is  O.     " 

Denote  the  circle  by  S,  its  circumference 
by  C,  and  the  perimeters  of  the  polygons 
by  P  and  Q  respectively. 

To  prove  that,  if  the  number  of  sides  of 
the  polygons  is  indefinitely  increased, 

1.  P  and  Q  approach  C  as  their  common  limit. 

2.  A  and  B  approach  8  as  their  common  limit. 

Proof.  1.  Place  the  polygons  so  that  their  sides  are  parallel ; 
draw  a  line  from  0  to  G^  the  point  of  contact  of  the  side  BE  of 
the  circumscribed  polygon  and  draw  OD  which  by  §  385  will  pass 
through  the  extremity  H  of  the  side  HJ  of  the  inscribed  polygon^ 

Since  the  polygons  have  the  same  number  of  sides, 


§387, 

P\Q=OJ)\OG\ 

.-.  §  277, 

P—Q:Q=OD~OG:OG, 

and,  §  269, 

0GX(P-Q)=QX{0D-  OG); 

whence, 

P-Q  =  —X(0D-0G). 
OG      ^                 ^ 

(1) 

Kow,  Ax. 

10, 

OD<OG-{-  DG,  or   OD  —  0G<  BG. 

(2) 

But,  if  the  number  of  sides  of  each  polygon  is  increased  indefi- 


222 


PLANE  GEOMETRY.  — BOOK   VL 


nitely,  the  two  polygons  continuing  to  have  the  same  number  of 

sides,   the   length   of   each   side  decreases 

indefinitely   and  approaches   the   limit  0; 

therefore  DG^  which  is  half  the   side  DE, 

approaches  the  limit  0,  and  in  (2),  OD  —  OG 

approaches  the  limit  0. 

Hence,  from  (1),  P  —Q  approaches  the 
limit  0. 

Since,  §  391,    P  is  always  greater  than  C, 
and.  Ax.  10,         Q  is  always  less  than  C, 

the  difference  between  G  and  either  P  or  Q  is  less  than  P  —  Q] 
therefore,  the  difference  approaches  the  limit  0. 

Consequently,  P  and  Q  approach  G  as  their  common  limit. 

2.   Since  the  polygons  are  regular  and  similar, 
§  390,  A'.B=ob^:OC^', 

A  —  B'.B=off—off'.0^ 
Now  0b^—^^  =  D^; 

A  —  B:B  =  D^ :  0^, 


Why? 
Why? 


wrhence. 


£=^v5?. 


OG 


(3) 


But  DG,  which  is  half  of  the  side  BE,  approaches  the  limit  0. 

Hence,  from  (3),  A  —  B  approaches  the  limit  0. 

Since.  Ax.  8,        ^  is  always  greater  than  S, 
cLud  B  is  always  less  than  S, 

the  difference  between  S  and  either  ^  or  ^  is  less  than  A  —  B', 
therefore,  the  difference  approaches  the  limit  0. 

Consequently,  A  and  B  approach  S  as  their  common  limit. 

Therefore,  etc.  q.e.d. 

393.  Sch.  It  is  evident  that  this  is  a  special  case  of  the 
theorem,  which  may  be  proved,  that  the  perimeters  of  polygons 
inscribed  in  and  circumscribed  about  a  closed  convex  curve,*  when 
the  number  of  their  sides  is  indefinitely  increased,  approach  the 
curve  as  a  limit,  and  the  polygons  approach  the  figure  bounded  by 
the  curve  as  a  limit 

*A  convex  curve  is  a  curve  which  a  straight  line  can  cut  in  only  two 
points. 


PLANE   GEOMETRY,  — BOOK   VI. 


223 


Proposition  X 

394.  Draw  two  circles  and  inscribe  in  them  regular  polygons  of  the 
same  number  of  sides.  How  does  the  ratio  of  their  perimeters  compare 
with  the  ratio  of  their  radii  ?  How  does  the  ratio  of  the  circumferences 
compare  with  the  ratio  of  their  radii  ?     Of  their  diameters  ? 

Theoi^em.  Circumferences  are  to  each  other  as  their 
radii.  ^-r^^'^^r-^c 


B  E 

Data :  Any  two  circumferences,  as  ABC  and  DEF,  whose  radii 
are  OB  and  P^,  respectively. 

To  prove  ABC :  DEF  —  OB  :  PE. 

Proof.  In  ABC  and  DEF  inscribe  regular  polygons  of  the  sanw 
number  of  sides,  and  denote  their  perimeters  by  Q  and  S,  respeo 
tively. 

Then,  §  387,  Q:S=OB:  PE. 

If  the  number  of  sides  of  the  polygons  is  indefinitely  in- 
creased, the  polygons  still  remaining  regular  and  similar,  §  392, 
Q  and  s  approach  ABC  and  DEF,  respectively,  as  their  limits. 

Hence,  ABC :  DEF  =  OB  :  PE.  q.e.d. 

395.  Cor.  The  ratio  of  the  circumference  of  a  circle  to  its  diame- 
ter is  constant. 

The  ratio  of  the  circumference  of  a  circle  to  its  diameter 
is  represented  by  the  Greek  letter  tt  whose  approximate  value,  as 
shown  in  Ex.  698,  is  3.1416. 

If  the  circumference  of  a  circle  is  denoted  by  C,  its  diameter  by 

D,  and  its  radius  by  i?,  q 

7r=—} 
D 

.'.    C=7rD,  OT  2  7rR; 

that  is,  the  circumference  of  a  circle  is  equal  to  ir  times  the  diam  . 

eter  or2ir  times  the  radius. 


224  PLANE   GEOMETRY.  — BOOR    Vl. 

396.  Similar  arcs,  similar  sectors,  and  similar  segments  are  those 
which,  correspond  to  equal  central  angles. 

Proposition  XI 

397.  Draw  a  circle  and  circumscribe  about  it  any  regular  polygon. 
How  does  the  apothem  of  the  polygon  compare  with  the  radius  of  the 
circle  ?  If  the  number  of  sides  of  the  polygon  is  indefinitely  increased, 
how  does  the  limit  of  its  perimeter  compare  with  the  circumference  of 
the  circle?  How,  then,  does  the  polygon  at  its  limit  compare  with  the 
circle  ?  Since  the  polygon  is  equivalent  to  one  half  the  rectangle  of  its 
perimeter  and  apothem,  how  does  the  circle  compare  with  the  rectangle 
of  its  circumference  and  radius? 

Theorem,  A  circle  is  equivalent  to  one  half  the  rectangle 
formed  hy  Us  circumference  and  radius. 


Data:  Any  circle,  as  ABG,  whose  center 
is  O.  ^ 

Denote  its  circumference  by  C  and  its 
radius  by  R. 

To  prove  ABG^\  rect.  C  •  R. 


Proof.     Circumscribe  about  the  circle  any  regular  polygon,  as 
DEF,  and  denote  its  perimeter  by  P. 

Then,  the  apothem  of  the  polygon  is  equal  to  R, 

and,  §  388,  DEF  =0=  i  rect.  P  -  i?. 

Now,  if  the  number  of  sides  of  the  polygon  is  indefinitely  in- 
creased, 

§  392,  P  indefinitely  approaches  C  as  its  limit, 

and  BEF  indefinitely  approaches  ABQ  as  its  limit. 

But, however  great  the  number  of  sides, 
DEF  ^  ^  rect.  P  •  R. 

Hence,  §  326,  ABG  0=  ^  rect.  C  -  R. 

Therefore,  etc.  q.e.d^.. 

Arithmetical  Rule:  To  be  framed  by  the  student.  §  339" 


eLANE  GEOMETRY.  — BOOK   VL  226 

398.  Cor.  I.  The  area  of  a  circle  is  equal  to  it  times  the  square 
of  its  radius. 

§395,  C=2  7rR; 

.'.  Area  =  |(2 Tri?  x  R),  or  vl^. 

399.  Cor.  II.  The  areas  of  circles  are  to  each  other  as  the 
squares  of  their  radii. 

400.  Cor.  III.  The  area  of  a  sector  is  equal  to  one  half  the  prod- 
uct of  its  arc  hy  its  radius. 

401.  Cor.  IV.     Similar  sectors  are  to  each  other  as  the  squares  of 

their  radii.  _ 

Ex.  654.  If  the  circumferences  of  two  circles  are  314.16cm  and  157.08cm 
respectively,  and  the  radius  of  the  first  is  60^^,  what  is  the  radius  of  the 
second  ?    What  is  the  area  of  the  first  ? 

Ex.  655.  Find  the  circumference  and  area  of  a  circle  whose  radius 
is  2.5^"'. 

Ex.  656.  What  is  the  ratio  of  the  radii  of  two  circles,  if  the  area  of  one 
circle  is  twice  that  of  the  other  ? 

Ex.  657.  What  is  the  area  of  a  sector  whose  arc  is  |  of  the  circumference, 
if  the  radius  of  the  circle  is  18^™  ? 

Ex.  658.  If  the  radius  of  a  circle  is  63  in.,  how  long  is  the  arc  of  a 
sector  whose  angle  is  45°  ? 

Ex.  659.  Calling  the  equatorial  radius  of  the  earth  3962.8  miles,  what  ia 
the  length  of  a  degree  on  the  equator  ? 

Ex.  660.    Find  the  radius  of  a  circle  whose  area  is  6'^  ™. 

Ex.  661.    Find  the  circumference  of  a  circle  whose  area  is  100  sq.  in. 

Ex.  662.    What  is  the  area  of  a  circle  whose  circumference  is  100  ft.  ? 

Ex.  663.  What  is  the  area  of  a  circle  circumscribed  about  a  square  whose 
side  is  a  ? 

Ex.  664.  The  diagonals  drawn  from  a  vertex  of  a  regular  pentagon  to 
the  opposite  vertices  trisect  the  angle  at  that  vertex. 

Ex.  665.  If  the  chord  of  an  arc  is  72dm  and  the  chord  of  half  that  arc  is 
86.9dm,  what  is  the  diameter  of  the  circle  ? 

Ex.  666.  The  chord  of  an  arc  is  24  in.  and  its  altitude  is  9  in.  What  is 
the  diameter  of  the  circle  ? 

Ex.  667.  The  chord  of  half  an  arc  is  12m  and  the  radius  of  the  circle  is 
18m.     What  is  the  altitude  of  the  arc  ? 

Ex.  668.  The  altitude  of  an  equilateral  triangle  is  equal  to  one  and  a 
half  times  the  radius  of  the  circumscribed  circle. 

Ex.  669.  Find  the  central  angle  subtended  by  an  arc  whose  length  is 
equal  to  the  radius  of  the  circle. 


226 


PLANE   GEOMETRY.-^ BOOK   VI. 


Proposition  XII 

402.  Draw  two  similar  segments  and  their  radii.  What  kind  of 
sectors  and  what  kind  of  triangles  are  thus  formed?  How  does  the 
difference  in  the  area  of  each  sector  and  the  corresponding  triangle  com- 
pare with  the  area  of  the  corresponding  segment?  Since  the  ratio  of  the 
sectors  and  also  the  ratio  of  the  triangles  equals  the  ratio  of  the  squares 
of  the  radii,  how  does  the  ratio  of  the  segments  compare  with  the  ratio 
of  the  same  squares? 

Theorem,  Similar  segments  are  to  eaxih  other  as  the 
squares  upon  their  radii. 

Data :  Any  two  similar 
segments,  as  ^5Cand  DEF, 
whose  radii  are  AO  and 
DP  respectively. 

To  prove 

seg.  ABC ;  seg.  DBF 

— 2      2 

=  A0:  DP  . 
Proof.     Draw  OC  and  PF. 
In  the  A  ACQ  and  DFP,    AO  —  /.P, 
and  AO'.DP=CO\FP\ 

A  ACQ  and  dfp  are  similar; 
hence,  §  342,        AACOiA  DFP  =  AO^ :  Dp. 
But,  §  401,  sect.  ABCO :  sect.  DFFP  =  Iff :  D?; 

sect.  ABCO  :  sect.  DFFP  =  A  ACQ  :  A  DFP, 
and  sect.  ABCO :  A  ^CO=sect.  DFFP  :  A  DFP. 

Hence,  §  277, 
sect.  ABCO  —AAC0:AAC0  =  sect.  DFFP  —  A  DFP  :  A  DFP ; 
that  is,  seg.  ABC :  A  ACO  =  seg.  DEF :  A  DFP, 

or  seg.  ABC  :  seg.  DEF  =  A  ACO:  A  DFP. 


Why? 
Why? 
Why? 


Why? 


AAC0:ADFP  =  A0  :DP. 


But 

Hence,       seg.  ABC :  seg.  DEF 

Therefore,  etc. 


AO  .DP 


Why? 
Why? 

Q.E.D. 


PLANE   GEOMETRY.  —  BOOK   VL  227 

Proposition  XIII 
403.  Problem,    To  inscribe  a  square  in  a  circle. 

Datum :  Any  circle.  / 

Required  to  inscribe  a  square  in  the  circle.       I 


Solution.     Draw  any  two   diameters   at  right  angles  to  each 
other,  as  AB  and  CD. 

Draw  AD,  DB,  BC,  and  CA. 

Then,  ADBC  is  the  required  square.  q.e.f. 

Proof.     By  the  student. 

Proposition  XIV 
404.  Problem,    To  inscribe  a  regular  hexagon  in  a  circle^ 

Datum :  Any  circle.  //'  \\ 

Required  to  inscribe  a  regular  hexagon     ^C^  "K 

in  the  circle.  \\  // 


Solution.  From  A,  any  point  in  the  circumference,  as  a  center, 
and  with  a  radius  equal  to  the  radius  of  the  circle,  describe  an  arc 
intersecting  the  circumference  as  at  B. 

From  ^  as  a  center  with  the  same  radius,  describe  another  arc 
intersecting  the  circumference  as  at  C 

In  like  manner  determine  the  points  D,  E,  and  F. 

Draw  chords  connecting  these  points  in  succession. 

Then,  ABCDEF  is  the  required  hexagon.  q.e.f 

Proof.     By  the  student. 

Ex.  670.   To  circumscribe  an  equilateral  triangle  about  a  circle. 
Ex.  671.  To  circumscribe  a  square  about  a  circle. 


228  PLANE   GEOMETRY.  — BOOK   VI. 

Proposition  XV 
405.  Problem,    To  inscribe  a  regular  decagon  in  a  circle. 


Datum :  Any  circle.  jf'  \^ 

Required  to  inscribe  a  regular  decagon       ;  ^p  \ 

m  the  circle.  A         /  \  ]d 

B 

Solution.     Draw  any  radius,  as  OA,  and  divide  it  in  extreme 
and  mean  ratio  as  at  P ;  that  is,  so  that  AG :  PC  =  PO :  AP. 

From  ^  as  a  center  and  with  PO  as  a  radius,  describe  an  arc 
intersecting  the  circumference  as  at  B.     Draw  AB. 

From  5  as  a  center  with  the  same  radius,  or  AB,  describe  an  arc 
intersecting  the  circumference  as  at  C. 

In  like  manner  determine  the  points  D,  E,  F,  G,  H,  J,  and  K. 

Draw  chords  connecting  these  points  in  succession. 

Then,  ABCD-K  is  the  required  decagon.  q.e.f. 

Proof.     Draw  BP  and  BO. 

Const.,  AO  :  PO  =  PO  :  AP, 

and  AB  =  PO', 

.-.  in  A  ABO  and  ABP,     A0:AB  =  AB:  AP, 
and  Zais  common ; 

A  ABP  and  ABO  are  similar.  Why  ? 

Now,  A  ABO  is  isosceles ; 

A  ABP  is  isosceles, 
and  AB  =  BP  =  PO;  Why? 

hence,  A  BPO  is  isosceles. 

But  Z  APB  =  Zo  +  Z  PBO  =  2  ZO.  Why  ? 

Then,  Za-^Z  ABP  +  Z  APB  =  5  Z  O, 

and  Z  0  =  I  of  2  rt.  A,  or  j\  of  4  rt.  A; 

arc  AB  is  ^-^  of  the  circumference, 

and  the  chord  AB,  which  subtends  the  arc  AB,  is  a  side  of  the 
regular  inscribed  decagon  ABCD-K. 


PLANE   GEOMETRY,  — BOOK   VL  229 


Proposition  XVI 

406.  Problem,    To  inscribe  a  regular  pentadecagon  in  a 
circle.  k  ^ 


Datum :  Any  circle.  .  /  W 

Required  to  inscribe  a  regular  pentadeca-      I  m 

gon  in  the  circle.  V  / 

c 

Solution.  Draw  the  chord  AB  equal  to  a  side  of  the  regular 
inscribed  hexagon,  and  from  A  draw  the  chord  AC  equal  to  a  side 
of  the  regular  inscribed  decagon.     Draw  CB. 

From  5  as  a  center  with  CB  as  a  radius,  describe  an  arc  inter* 
secting  the  circumference  as  at  B. 

In  like  manner  determine  the  points  E,  F,  G,  H,  etc. 

Draw  chords  connecting  these  points  in  succession. 

Then,  CBBEF  etc.,  is  the  required  pentadecagon.  q.e.f. 

Proof.  Arc  ^^  =  i  of  the  circum., 

and  arc  AG  =  ^  oi  the  circum. ; 

.*.    arc  BG  =  arc  AB  —  arc  ^C  =  i  —  J^,  or  -^^  of  the  circum., 

and  the  chord  CB,  which  subtends  the  arc  CB,  is  a  side  of  the 
regular  inscribed  pentadecagon  GBDEF  etc. 

407.  Cor.  I.  By  joining  the  alternate  vertices  of  a,ny  regular 
inscribed  polygon  of  an  even  number  of  sides,  a  regular  polygon  oj 
half  the  number  of  sides  is  inscribed. 

408.  Cor.  II.  By  joining  to  the  vertices  of  any  regular  inscribed 
polygon  the  middle  points  of  the  arcs  subtended  by  its  sides,  a  regular 
polygon  of  double  the  number  of  sides  is  inscribed. 

Ex.  672.  To  inscribe  a  regular  octagon  in  a  circle. 

Ex.  673.  To  inscribe  a  regular  dodecagon  in  a  circle. 

Ex.  674.  To  circumscribe  a  regular  hexagon  about  a  circle. 

Ex.  675.  To  circumscribe  a  regular  octagon  about  a  circle. 

Ex.  676.  To  inscribe  a  regular  hexagon  in  an  equilateral  triangle. 


230 


PLANE   GEOMETRY.  — BOOK    VL 


Ex.  677.   To  divide  an  angle  of  an  equilateral  triangle  into  five  equal  parts. 

Ex.  678.  The  segment  of  a  circle  is  equal  to  f  of  a  similar  segment. 
What  is  the  ratio  of  their  radii? 

Ex.  679.  .  How  many  degrees  are  there  in  an  arc  18  in.  long  on  a  circum- 
ference whose  radius  is  5  ft.  ? 

Ex.  680.  The  radii  of  two  similar  segments  are  as  3:5.  What  is  the 
ratio  of  their  areas  ? 

Ex.  681.  In  a  circle  3  ft.  in  diameter  an  equilateral  triangle  is  inscribed. 
What  is  the  area  of  a  segment  without  the  triangle  ? 

Ex.  682.  Two  chords  drawn  from  the  same  point  in  a  circumference  to 
the  extremities  of  a  diameter  of  a  circle  are  6  in.  and  8  in.  respectively. 
What  is  the  area  of  the  circle  ? 


MAXIMA  AND   MINIMA 

409.  Of  any  number  of  magnitudes  of  the  same  kind  tlie 
greatest  is  called,  the  Maximum,  and  the  least  is  called  the 
Minimum. 

Of  all  chords  of  any  circle  the  diameter  is  the  maximum ;  and  of  all  lines 
from  any  point  to  a  given  line  the  perpendicular  is  the  minimum. 

410.  Figures  which  have  equal  perimeters  are  called  Isoperi- 
metric. 

Proposition  XVII 


411.  Theorem,  Of  all  triangles  having  two  given 
that  in  which  these  sides  are  perpendicular  to  each 
is  the  Tnaxijnujn. 

Data:  Any  two  triangles,  as  ABC 
and  ABD,  such  that  AC  =  AD,  AB  is 
common,  and  AC  and  AB  are  perpen- 
dicular to  each  other. 

To  prove      A  ABC  the  maximum. 

Proof.     Draw  EDl^AB. 

Area  of  A  ABC  =  -^  ^c  x  AB, 
and  area  of  A  ABD  =  ^ED  x  AB. 


sides, 
other 


But 

and 
Hence, 
Therefore,  etc. 


AD  >  ED,  and  AC=  AD) 

AC  >  ED, 
\ACX  AB>^ED  X  AB. 
A  ABC  is  the  maximum. 


Why? 
Why? 

Why? 

Q.E.D. 


PLANE   GEOMETRY.  — BOOK   VI.  231 


Proposition  XVIII 

412.   Theorem.    Of  all  isoperimetric  triangles  which  have 
the  same  hase,  the  isosceles  triangle  is  the  maximum. 

Data:    Any  two  isoperimetric  triangles                    c.-''  _      i^ 
upon  the  same  base  AB,  as  ABC  and  ABD^          ^^'^^^^^    \ 
of  which  ABC  is  isosceles.         -  ^<^^^    j. ^^-^ 

To  prove    A  ABC  the  maximum.  ^^^--^^T^^'i-^ 

Proof.     Produce  ^C  to  ^,  making  CE  equal  to  AC.     Draw  EB. 
Since  C  is  equidistant  from  A,  B,  and  E,  Z  ABE  may  be  inscribed 
in  a  semicircumf erence ; 

Z  ^5^  is  a  right  angle. 

Draw  DF  equal  to  i>5  meeting  EB  produced  in  i^ ;  CG  and  DH 
parallel  to  AB ;  CJ  and  DiT  perpendicular  to  AB ;  and  draw  ^J^. 
Then,  AE  =  AC  +  BC  =  AD  -[-  BD  =  AD  -{-  FD.  Why  ? 

But  AD-\-  FD>  AF]  Why? 

hence,  §  133,  EB  >  FB. 

But  G-B  =  1^5,  and  HB  =  ^i^5;  Why  ? 

GB>HB. 

Also,  (?s  =  CJ,  and  flS  =  D-ff,  the  altitudes  of  A  ^5C  and  ABDj 
/espectively ; 

CJ>DK. 

Now,  area  of  A  iBC  =  \AB  x  CJ, 

and  area  of  A^5Z)  =  ^^5  X  D^;  Why? 

area  of  A  ABC  >  area  of  A  ABD ; 
that  is,  A  ABC  is  the  maximuin. 

Therefore,  etc.  q.e.d. 

413.   Cor.     Of  all  isoperimetric  tHangles,  the  equilateral  triangle 
is  the  maximum. 

Ex.  683.   Of  all  equivalent  parallelograms  having  equal  bases,  the  rec- 
tangle  has  the  least  perimeter. 


232 


PLANE   GEOMETRY.  — BOOK   VI. 


Proposition  XIX 

414.   Theorem,    Of  isoperimetric  polygons  which  have  the 
same  nUmber  of  sides,  the  Tnaximum  is  equilateral. 

Data :  The  maximum  of  isoperimetric  poly-  - 

gons  of  a  given  mimber  of  sides,  as  ABCDEF. 


To  prove 


ABCDEF  equilateral. 


Proof.     Draw  AE.  b 

Then,  A  AEF  must  be  the  maximum  of  all 
the  A  that  can  be  formed  upon  AE  with  a 
perimeter  equal  to  that  of  A  AEF,  for  if  not,  a  greater  A,  as  AEG, 
could  be  substituted  for  A  ^^i^  without  changing  the  perimeter 
of  ABCDEF. 

But  it  would  be  impossible  to  enlarge  ABCDEF,  for,  data,  it  is 
the  maximum. 

Hence,  §  412,  A  AEF  is  isosceles, 

and  AF  =  EF. 

Similarly  any  two  consecutive  sides  may  be  shown  equal. 

Hence,  ABCDEF  is  equilateral.  q.e.d. 

Proposition  XX 

415.  Theorem,     Of  isoperimetric  regular  polygons,  that 
which  has  the  greatest  number  of  sides  is  the  m^ajcimum. 


Data :  Any  two  isoperimetric  regular  polygons,  as  ABC  and  D, 
of  which  D  has  one  side  more  than  ABC. 

To  prove  D  the  maximum. 

Proof.  To  E,  any  point  in  AB,  draw  CE  and  construct  the  A  CEF 
ec^ual  to  the  A  ACE. 


PLANE   GEOMETRY.  — BOOK   VL 


233 


Q.E.D. 


Then,  EBCF  ^  ABC, 

and  EBCF  and  D  are  isoperimetric. 

But,  §414,  D>EBCF', 

D>ABG\ 
that  is,  D  is  the  maximum. 

Therefore,  etc. 

416.  Cor.     The  area  of  a  circle  is  greater  than  the  area  of  any  ^ 

isoperimetric  polygon. 

Proposition  XXI 

417.  Theorem.  Of  regular  polygons  which  have  equal 
areas,  that  which  has  the  greatest  number  of  sides  has 
the  least  perimeter. 


Data :  Any  regular  polygons  which  have  equal  areas,  as  A  and 
Bj  of  which  A  has  a  greater  number  of  sides  than  B. 

To  prove  the  perimeter  of  A  less  than  the  perimeter  of  B. 

Proof.  Construct  the  regular  polygon  C,  having  its  perimeter 
equal  to  that  of  A  and  having  the  same  number  of  sides  as  B. 

Then,  §  415,  A>G.    _ 

But,  data,  A^^B-, 

B>C, 
the  perimeter  of  B  is  greater  than  that  of  C. 
the  perimeter  of  C  is  equal  to  that  of  A ; 
the  perimeter  of  B  is  greater  than  that  of  A ; 
the  perimeter  of  A  is  less  than  the  perimeter  of  B. 
Therefore,  etc.  q.e.d. 

418.  Cor.  The  circumference  of  a  circle  is  less  than  the  perimeter 
of  any  polygon  which  has  an  equal  area. 

Ex.  684.  Of  all  rectangles  of  a  given  area,  the  square  has  the  least 
perimeter. 


and,  §  346, 
But 

that  is, 


234 


PLANE   GEOMETRY.  — BOOK   VI. 


M- 


SYMMETRY 

419.  If  a  point  bisects  the  straight  line  joining  two  other 
points,  the  two  points  are  said  to  be  symmetrical  with  respect  to  a 
point,  and  this  point  is  called  the  center  of  symmetry. 

M  and  N  are  symmetrical  with  respect  to  the 
center  A,  if  A  bisects  the  straight  line  MN. 

420.  If  a  straight  line  is  the  perpen- 
dicular bisector  of  the  straight  line  joining 
two  points,  the  points  are  said  to  be  symr 
metrical  with  respect  to  a  straight  line,  and 
this  line  is  called  the  axis  of  symmetry. 

M  and  N  are  symmetrical  with  respect  to  the 
axis  XX',  if  XX  is  the  perpendicular  bisector  of 
the  straight  line  MN. 

421.  If  every  point  of  one  figure  has 
a  corresponding  symmetrical  point  in 
another,  the  two  figures  are  said  to  be 
symmetrical  with  respect  to  a  center  or  an 
axis. 


'V'o 


If  every  point  in  the  figure  ABC  has  a  sym- 
metrical point  in  A'B'  C  with  respect  to  O  as 
a  center,  then,  the  figures  ABC  and  A'B'C 
are  symmetrical  with  respect  to  the  center  O. 

If  every  point  in  the  figure  DEF  has  a  sym- 
metrical point  in  D'E'F'  with  respect  to  XX 
as  an  axis,  then,  the  figures  DEF  and  D'E'F' 
are  symmetrical  with  respect  to  the  axis  XX'. 

Two  plane  figures  that  are  symmetrical 
with  respect  to  an  axis  can  be  applied 
one  to  the  other  by  revolving  either  one 
about  the  axis ;  consequently  they  are 
equal,  and  if  two  figures  can  be  made  to 
coincide  by  revolving  one  of  them  about  an  axis  through  180®, 
they  are  symmetrical  with  respect  to  the  axis. 

422.  If  a  point  bisects  every  straight  line  drawn  through  it  and 
terminated  in  the  boundary  of  a  figure,  the  figure  is  said  to  be 
symmetrical  with  respect  to  a  point. 


PLANE   GEOMETRY,-~BOOK    VI. 


235 


If  0  bisects  every  straight  line  drawn 
through  it  and  terminated  by  the  boun- 
dary of  ABCDEF,  then,  ABCDEF  is 
symmetrical  with  respect  to  the  point  O. 


423.  If  a  straight  line  divides  a  plane  figure  into  two  parts 
which  are  symmetrical  with  respect  to  the  line,  the  figure  is  said 
to  be  symmetrical  with  respect  to  a  straight  line. 


If  the  parts  ABCD  and  AFED  are 
symmetrical  with  respect  to  XX',  then, 
the  figure  ABCDEF  is  symmetrical 
with  respect  t(*  the  straight  line  XX'. 


Proposition  XXII 

424.  Theorem,  A  quadrilateral  which  has  two  adja/^ent 
sides  equal  and  the  other  two  sides  equal,  is  symmetrical 
with  respect  to  the  diagonal  joining  the  vertices  of  the 
angles  formed  hy  the  equal  sides. 

Data:  A  quadrilateral,  as  ABCD, 
having  AB  =  AD,  CB  =  CD,  and  the 
diagonal  AG, 

To  prove  ABCD  symmetrical  with 
respect  to  ^C. 

Proof.     In  the  A  ABC  and  ADC, 
data,  AB  =  AD,  CB  =  CDy 

and  AC  is  common ; 

A  ABC  =  A  ADC,  Why  ? 

Z  BAC  =  Z  DAC,  and  Z  BCA  =  Z  DC  A.  Why  ? 

Hence,  if  ADC  is  turned  on  AC  as  an  axis,  it  may  be  made  to 
coincide  with  ABC. 

.'.  §  421,  ADC  and  ABC  are  symmetrical  with  respect  to  AC-, 
that  is,  §  423,      ABCD  is  symmetrical  with  respect  to  AC. 

Therefore,  etc.  q-E-d. 


236 


PLANE   GEOMETRY.  — BOOK   VL 


G 

Y 

F 

y_ 

^-         \9 

H 

^''      1 

E 

A 

M; 

X"" 

0 

D 

B 

C 
Y' 

Proposition  XXIII 

425.  Theorem.  If  a  figure  is  syjnmetrical  with  respect 
to  two  ojces  perpendicular  to  each  other,  it  is  syminetricaZ 
with  respect  to  their  intersection  as  a  center. 

Data:  A  figure,  as  ABCD-Hy 
symmetrical  with  respect  to 
the  two  perpendicular  axes, 
XX'  and  yy\  which  intersect 
at  o. 

To  prove  ABCD-H  symmet- 
rical with  respect  to  O  as  a 
center. 

Proof.  From  any  point  in  the  perimeter,  as  P,  draw  PMP'  _L  xx* 
and  PNQ  A.  YY^. 

Draw  MN,  P'O,  and  OQ. 

Now,  §420,  PM=P'M, 

and  ^M=ON',  Why? 

P'M=  ON, 
and,  §71,  P'M  W  0N\ 

consequently,  §  150,    MP' ON  is  a  parallelogram ; 

P'O  is  equal  and  parallel  to  MN. 

Similarly,  OQ  is  equal  and  parallel  to  MN. 

Hence,  points  P',  O,  Q  are  in  the  same  straight  line  P'OQ,  whict* 
is  bisected  at  0.  Why  5 

But  since  P  is  any  point  in  the  perimeter,  P'OQ  is  any  straigh* 
line  drawn  through  0. 

Hence,  §  422,  ABCD-H  is  symmetrical  with  respect  to  o  as  a 
center. 

Therefore,  etc.  q.e.d. 

Ex.  685.  A  segment  of  a  circle  is  symmetrical  with  respect  to  the  per^ 
pendicular  bisector  of  its  chord  as  an  axis. 

Ex.  686.  A  circle  is  symmetrical  with  respect  to  its  center  or  with  re- 
spect to  any  diameter  as  an  axis. 

Ex.  687.  A  parallelogram  is  symmetrical  with  respect  to  the  point  oi 
Intersection  of  its  diagonals  as  a  center. 


PLANE   GEOMETRY.  — BOOK   VL  237 

SUMMARY 
426.    Truths  established  in  Book  VL 

1.  Two  lines  are  equal, 

a.   If  they  are  sides  of  a  regular  polygon.  §  374 

2.  Lines  are  in  proportion, 

a.  If  they  are  the  perimeters  of  regular  polygons  of  the  same  number 
of  sides,  and  their  radii.  §  387 

b.  If  they  are  the  perimeters  of  regular  polygons  of  the  same  number 
of  sides,  and  their  apothems.  §  387 

c.  If  they  are  circumferences  and  their  radii.  §  394 

3.  Two  angles  are  equal, 

a.  If  they  are  angles  of  a  regular  polygon.  §  374 

4.  An  angle  is  bisected, 

a.  If  it  is  an  interior  angle  of  a  regular  polygon,  by  the  radius  drawn 
to  its  vertex.  §  382 

5.  A  polygon  is  regular, 

a.  If  it  is  equilateral  and  equiangular.  §  374 

b.  If  it  is  equilateral  and  inscribed  in  a  circle.  §  375 

c.  If  it  is  formed  by  chords  joining  the  extremities  of  arcs  which  are 
equal  divisions  of  the  circumference  of  a  circle.  §  376 

d.  If  it  is  formed  by  tangents  drawn  at  the  extremities  of  arcs  which 
are  equal  divisions  of  the  circumference  of  a  circle.  §  376 

e.  If  it  is  formed  by  tangents  to  a  circle  at  the  middle  points  of  the  arcs 
subtended  by  the  sides  of  a  regular  inscribed  polygon^  §  384 

6.  Polygons  are  similar, 

a.   If  they  are  regular  and  have  the  same  number  of  sides.  §  386 

7.  A  regular  polygon  is  equivalent, 

a.  To  half  the  rectangle  formed  by  its  perimeter  and  apothem.  §  388 

8.  A  circle  is  equivalent, 

a.   To  half  the  rectangle  formed  by  its  circumference  and  radius.     §  397 

9.  A  circumference  is  the  limit, 

a.  Of  the  perimeter  of  a  regular  inscribed  polygon  when  the  number  of 
its  sides  is  indefinitely  increased.  §  392 

b.  Of  the  perimeter  of  a  regular  circumscribed  polygon  when  the  num- 
ber of  its  sides  is  indefinitely  increased.  §  892 


238  PLANE   GEOMETRY.  — BOOK   VI. 

10.  A  circle  is  the  limit, 

a.  Of  a  regular  inscribed  polygon  when  the  number  of  its  sides  is  indefi- 
nitely increased.  §  392 

b.  Of  a  regular  circumscribed  polygon  when  the  number  of  its  sides  is 
indefinitely  increased.  §  392 

11.  Figures  are  in  proportion, 

a.  If  they  are  regular  polygons  of  the  same  number  of  sides,  to  the 
squares  upon  their  radii.  §  390 

b.  If  they  are  regular  polygons  of  the  same  number  of  sides,  to  the 
squares  upon  their  apothems.  §  390 

c.  If  they  are  circles,  to  the  squares  of  their  radii.  §  399 

d.  If  they  are  similar  sectors,  to  the  squares  of  their  radii.  §  401 

e.  If  they  are  similar  segments,  to  the  squares  of  their  radii.  §  402 

12.  The  area  of  a  figure  is  equal, 

a.  If  it  is  a  regular  polygon,  to  one  half  the  product  of  its  perimeter  by 
its  apothem.  §  389 

b.  If  it  is  a  circle,  to  one  half  the  product  of  its  circumference  by  its 
radius.  §  397 

c.  If  it  is  a  circle,  to  w  times  the  square  of  its  radius.  §  398 

d.  If  it  is  a  sector,  to  one  half  the  product  of  its  arc  by  its  radius.  §  400 


SUPPLEMENTARY  EXERCISES 

Ex.  688.  If  the  perimeter  of  each  of  the  figures,  equilateral  triangle, 
square,  and  circle  is  396  ft.,  what  is  the  area  of  each  figure  ? 

Ex.  689.  If  the  inscribed  and  circumscribed  circles  of  a  triangle  are  con- 
centric, the  triangle  is  equilateral. 

Ex.  690.  If  an  equilateral  triangle  is  inscribed  in  a  circle,  any  side  will 
cut  oflE  one  fourth  of  the  diameter  from  the  opposite  vertex, 

Ex.  691.  The  square  inscribed  in  a  circle  is  equivalent  to  one  half  the 
square  circumscribed  about  that  circle. 

Ex.  692.  A  circle  is  inscribed  in  a  square  whose  side  is  4  in.  How  much 
of  the  area  of  the  square  is  without  the  circle  ? 

Ex.  693.  What  is  the  width  of  the  ring  between  the  circumferences  of  two 
concentric  circles  whose  circumferences  are  48  ft.  and  36  ft.  respectively  ? 

Ex.  694.  Of  all  squares  that  can  be  inscribed  in  a  given  square,  the  mini- 
mum has  its  vertices  at  the  middle  points  of  the  sides. 

Ex.  695.  Every  equiangular  polygon  circumscribed  about  a  circle  is 
regular. 

Ex.  696.  In  any  regular  polygon  of  an  even  number  of  sides,  the  lines 
joining  opposite  vertices  are  diameters  of  the  circumscribing  circle. 


PLANE   GEOMETRY.— BOOK    VI. 


239 


Ex.  697.  Given  the  side  of  a  regular  inscribed  polygon  and  the  side  of  a 
similar  circumscribed  polygon^  to  compute  the  perimeters  of  the  regular  in- 
scribed and  circumscribed  polygons  of  double  the  number  of  sides. 

Data :  AB,  the  side  of  a  regular  inscribed 
polygon,  and  C*Z>,  the  side  of  a  similar  circum- 
scribed polygon,  tangent  to  the  arc  AB  at  its 
middle  point  E. 

Denote  the  perimeters  of  these  polygons  tty  P 
and  Q  respectively,  and  the  number  of  sides  in 
each  by  n  ;  denote  the  perimeters  of  the  inscribed 
and  circumscribed  polygons  which  have  2  n  sides 
by  S  and  T  respectively. 

Required  to^compute  the  value  of  S  and  of  T. 

Solution.  Through  A  and  B  draw  tangents  to  meet  CD  in  F  and  Cr 
respectively ;  also  draw  AE  and  BE. 

Then,  §  376,  AE  and  FO  are  sides  of  the  polygons  whose  perimeters  are 
iSand  r  respectively. 
P 


AB  = 


n 


AE 


8  T 

-^,  and  FG=^' 

2  w  2  n 


Draw  the  radii  00,  PO,  EG,  and  BO. 
Since,  §  385,  A  lies  in  CO, 


by  Ex.  221, 
.-.  §292, 
but 

and 

But 
.'.  substituting, 
or 
whence, 

Again,  in  the 

.-.  §301, 

and 

hence, 

and  substituting 


whence, 
and 


FO  bisects  Z^  OP,  or  Z  COE ; 
EF:CF=:EO:CO; 

P:Q  =  EO:CO; 

P:Q  =  EF:CF, 
P+  Q  :  P  =  EF  +  CF  :  EF  =  CE  : 

CE  =  iCD  =  ^,  andPP=iPO  =  ^ 

2n    4n 
P-\-Q:P  =  2Q:T; 

Q  +  P 
isosceles  ^ABE  and  AEF, 

ZABE  =  ZAEF', 
A  ABE  and  AEF  are  similar, 
AE:AB  =  EF:AE; 
AE^  =  ABx  EF, 
for  AE,  AB,  and  EF  their  values, 

4  n^      n     4  w  * 

/8'2  =  Pxr, 


EF. 


JS=  VPxT, 


240 


PLANE   GEOMETRY.  — BOOK    VL 


Ex.  698.  To  compute  the  approximate  ratio  of  a  circumference  to  its 
diameter. 

Solution.  If  the  diameter  of  a  circle  is  1,  the  side  of  a  circumscribed 
square  is  1,  and  its  perimeter  is  4 ;  the  side  of  an  inscribed  square  is  \  \/2, 
and  its  perimeter  is  2  V2,  or  2.82843. 

Thus,  §  =  4,  and  P  =  2  V2  for  computing  the  octagon. 

Substituting  these  values  in  the  formulae,    7"  =  ^  ^    ,     S  —  y/P  x  T 

Q-\-P 
(Ex.  697),  and  solving,  the  results  tabulated  below  are  found,  showing  the 

perimeters  to  five  decimal  places. 


No.  OF 


Computation  of  T 


Computation  of  S 


8 

16 

32 

64 

128 

256 

512 

1024 


y^2QxP 
Q+P 

2  X  4  X  2.82843 
4  +  2.82843 

2  X  3.31371  X  3.06147 
3.31371  +  3.06147 

2  X  3.18260  X  3.12145 


3.18260  + 
2  X  3.15172 


3.12145 
X  3.13655 


3.15172  +  3.13655 
2  X  3.14412x3.14033 


3.14412  + 
2  X  3.14222 


3.14033 
X  3.14128 


3.14222  + 
2x3.14175 


3.14128 
X  3.14151 


3.14175  + 
2  X  3.14163 


3.14151 
X  3.14157 


3.14163  +  3.14157 


3.31371 
3.18260 
3.15172 
3.14412 
3.14222 
3.14175 
3.14163 
3.14159 


s=Vp 

xT 

V2.82843  X 

3.31371 

V3.06147  X 

3.18260 

V3. 12145  X 

3.15172 

V3. 13655  X 

3.14412 

V3. 14033  X 

3.14222 

V3. 14128  X 

3.14175 

V3. 14151  X 

3.14163 

V3.14157X  3.14159 


3.06147 
3.12145 
3.13655 
3.14033 
3.14128 
3.14151 
3.14157 
3.14159 


The  results  of  the  last  two  computations  show  that  the  circumference  of  a 
circle  whose  diameter  is  1  is  approximately  3.1416  ;  that  is,  the  ratio  of  the 
diameter  of  a  circle  to  its  circumference  is  equal  to  the  ratio  of  1  to  3.1416, 
approximately. 

Ex.  699.  The  sides  of  an  inscribed  rectangle  are  30cm  and  40°™.  What 
is  the  area  of  the  part  of  the  circle  without  the  rectangle  ? 

Ex.  700.  What  is  the  area  of  a  figure  bounded  by  four  semicircumfer- 
ences  described  on  the  sides  of  a  three  foot  square  ? 

Ex.  701.  A  square  piece  of  land  and  a  circular  piece  of  land  each  con- 
tain one  acre.     Which  perimeter  is  the  greater,  and  how  much  ? 


PLANE  GEOMETRY. ^BOOK  VL  241 

Ex.  702.  The  area  of  an  inscribed  equilateral  triangle  is  one  half  the  area 
of  a  regular  hexagon  inscribed  in  the  same  circle. 

Ex.  703.  Of  all  triangles  that  have  the  same  vertical  angle  and  whose 
bases  pass  through  a  given  point,  the  minimum  is  the  one  whose  base  is 
bisected  at  that  point. 

Ex.  704.  An  arc  of  a  circle  whose  radius  is  6  ft.  subtends  a  central  angle 
of  20° ;  an  equal  arc  of  another  circle  subtends  a  central  angle  of  30°.  What 
is  the  radius  of  the  second  circle  ? 

Ex.  705.  Two  tangents  make  with  each  other  an  angle  of  60°,  and  the 
radius  of  the  circle  is  7  in.  What  are  the  lengths  of  the  arcs  between  the 
points  of  contact  ? 

Ex.  706.  If  the  apothem  of  a  regular  hexagon  is  10™,  what  is  the  area  of 
the  ring  between  the  circumferences  of  its  inscribed  and  circumscribed 
circles  ? 

Ex.  707.  If  a  circle  18<»»  in  diameter  is  divided  into  three  equivalent 
parts  by  two  concentric  circumferences,  what  are  their  radii  ? 

Ex.  708.  The  square  upon  the  side  of  a  regular  inscribed  pentagon  is 
equivalent  to  the  sum  of  the  squares  upon  the  radius  of  the  circle  and  the 
side  of  a  regular  inscribed  decagon. 

Ex.  709.  The  radius  of  a  regular  inscribed  polygon  is  a  mean  proportional 
between  its  apothem  and  the  radius  of  the  similar  circumscribed  polygon. 

Ex.  710.   If  the  radius  of  a  regular  inscribed  octagon  is  r,  prove  that  its 

side  =  r  V2  -  V2,  and  its  apothem  =  -  V2  +  y/2. 

2 

Ex.  711.   If  the  radius  of  a  regular  inscribed  decagon  is  r,  prove  that  its 

side  =  ~  (  VS  -  1),  and  its  apothem  =  ^  VlO  +  2  V6. 
2  4 

Ex.  712.  If  the  radius  of  a  regular  inscribed  dodecagon  is  y,  prove  that 
its  side  =  r  V2-V3,  and  its  apothem  =  -  V2  +  V3. 

Ex.  713.  If  the  radius  of  a  regular  inscribed  pentagon  is  r,  prove  that  its 

side  =  -  VlO  -  2  V5,  and  its  apothem  =  \  Ve  +  2  V6. 
2  4 

Ex.  714.  The  square  upon  a  side  of  an  inscribed  equilateral  triangle  is 
equivalent  to  three  times  the  square  upon  the  side  of  a  regular  inscribed 
hexagon. 

Ex.  715.  The  area  of  an  inscribed  square  is  16"9i».  pind  the  length  of  a 
side  of  a  regular  inscribed  octagon. 

Ex.  716.  If  the  radius  of  a  circle  is  r,  prove  that  a  side  of  a  regular 
circumscribed  hexagon  is  -^  Vs. 

Ex.  717.   The  area  of  a  regular  inscribed  dodecagon  is  equal  to  three 
•limes  the  square  of  the  radius. 
milne's  oeom.  — 16 


242  PLANE   GEOMETRY.  — BOOK   VI. 

Ex.  718,  Find  the  side  of  a  regular  hexagon  circumscribed  about  a  circle 
whose  diameter  is  1. 

Ex.  719.  The  apothem  of  an  inscribed  regular  hexagon  is  equal  to  one 
half  the  side  of  the  inscribed  equilateral  triangle. 

Ex.  720.  The  area  of  a  ring  bounded  by  two  concentric  circumferences  is 
equal  to  the  area  of  a  circle  whose  diameter  is  a  chord  of  the  outer  circumfer- 
ence and  is  tangent  to  the  inner  circumference. 

Ex.  721.  If  the  radius  of  a  circle  is  r,  find  the  area  of  a  segment  whose 
chord  is  one  side  of  a  regular  inscribed  hexagon. 

Ex.  722.  Three  equal  circles  with  a  radius  of  12  ft.  are  drawn  tangent  to 
each  other.     What  is  the  area  between  them  ? 

Ex.  723.  The  area  of  an  inscribed  regular  hexagon  is  equal  to  three 
fourths  that  of  a  regular  hexagon  circumscribed  about  the  same  circle. 

Ex.  724.  The  altitude  of  an  equilateral  triangle  is  equal  to  the  side  of  an 
equilateral  triangle  inscribed  in  a  circle  whose  diameter  is  the  base  of  the 
first  triangle. 

Ex.  725.   If  the  radius  of  a  circle  is  r  and  the  side  of  a  regular  inscribed 

polygon  is  a,  prove  that  the  side  of  a  similar  circumscribed  polygon  is    , 

v4r2— a"^ 

Ex.  726.  If  the  alternate  vertices  of  a  regular  hexagon  are  joined  by 
straight  lines,  another  regular  hexagon  is  formed  which  is  one  third  as  large 
as  the  original  hexagon. 

Ex.  727.  The  diagonals  of  a  regular  pentagon  divide  each  other  in 
extreme  and  mean  ratio. 


PROBLEMS  OF  CONSTRUCTION 

Ex.  728.   Construct  x,  ii  x  =  Vab. 

Ex.  729.    Inscribe  a  circle  in  a  given  sector. 

Ex.  730.  In  a  given  circle  describe  three  equal  circles  tangent  to  each 
other  and  to  the  given  circle. 

Ex.  731.  Divide  a  circle  into  two  segments  such  that  an  angle  inscribed 
in  one  shall  be  three  times  an  angle  inscribed  in  the  other. 

Ex.  732.  Construct  a  circumference  equal  to  the  sum  of  two  given 
circumferences. 

Ex.  733.    Inscribe  a  square  in  a  given  quadrant. 

Ex.  734.   Inscribe  a  square  in  a  given  segment  of  a  circle. 

Ex.  735.  Through  a  given  point  draw  a  line  so  that  it  shall  divide  a. given 
circumference  into  two  parts  having  the  ratio  3:7. 

Ex.  736.    Construct  a  circle  equivalent  to  twice  a  given  circle. 

Ex.  737.   Construct  a  circle  equivalent  to  three  times  a  given  circle. 


SOLID   GEOMETRY 


BOOK   YII 

PLANES  AND   SOLID  ANGLES 

427.  A  plane  is  a  surface  such  that  a  straight  line  joining  any 
two  of  its  points  lies  wholly  in  the  surface.     §  14. 

A  plane  is  considered  to  be  indefinite  in  extent,  but  in  a  diagram 
it  is  usually  represented  by  a  quadrilateral  segment. 

428.  The  student  will  be  aided  in  obtaining  correct  concepts  of 
the  truths  presented  in  the  geometry  of  planes  by  using  pieces  of 
cardboard  or  paper  to  represent  planes,  and  drawing  such  lines 
upon  them  as  are  required.  Pins  may  be  used  to  represent  the 
lines  which  are  perpendicular  or  oblique  to  the  planes. 

429.  1.  By  using  cardboard  to  represent  a  plane  and  the  point 
of  a  pin  or  pencil  to  represent  a  point  in  space,  discover  in  how 
many  directions  the  plane  may  be  passed  through  the  point. 

2.  By  using  a  card  as  before  and  the  points  of  a  pair  of  dividers 
to  represent  two  fixed  points  in  space,  discover  whether  the  num- 
ber of  directions  that  the  plane  may  take  is  greater  or  less  than 
when  it  was  passed  through  one  fixed  point. 

3.  Suppose  a  plane  is  passed  through  three  fixed  points  not  in 
the  same  straight  line,  how  many  directions  may  it  take  ?  How 
many  points,  then,  determine  the  position  of  a  plane  ? 

4.  Since  two  of  the  points  must  be  in  a  straight  line,  what  else 
besides  three  points  determine  the  position  of  a  plane  ? 

5.  Since  a  straight  line  through  the  other  point  may  intersect 
the  straight  line  joining  the  two  points,  what  else  will  determine 
the  position  of  a  plane  ? 

243 


244 


SOLID   GEOMETRY.  — BOOK    VII. 


6.  Since  a  straight  line  may  join  two  of  the  points  and  a 
straight  line  parallel  to  that  may  be  drawn  through  the  other 
point,  how  else  may  the  position  of  a  plane  be  determined  ? 

In  what  ways,  then,  may  the  position  of  a  plane  be  determined  ? 

430.  A  plane  is  determined  by  certain  points  or  lines,  when  it 
is  the  only  plane  which  contains  those  points  or  lines. 

A  plane  is  determined  by 

1.  Three  points  not  in  the  same  straight  line. 

2.  A  straight  line  and  a  point  ivithout  that  line, 

3.  Two  intersecting  straight  lines. 

4.  Two  parallel  straight  lines. 

431.  The  point  at  which  a  line  meets  a  plane  is  called  the 
Foot  of  the  line. 

432.  A  straight  line  that  is  perpendicular  to  every  straight 
line  in  a  plane  drawn  through  its  foot  is  perpendicular  to  the 
plane. 

In  this  case  the  plane  is  perpendicular  to  the  line. 

433.  A  straight,  line  that  is  not  perpendicular  to  every  line 
in  a  plane  drawn  through  its  foot  is  oblique  to  the  plane. 

434.  A  straight  line  and  a  plane  which  cannot  meet,  however 
far  they  may  be  produced,  are  parallel  to  each  other. 

435.  Two  planes  which  cannot  meet,  however  far  they  may  be 
produced,  are  parallel  to  each  other. 

436.  The  locus  of  the  points  common  to  two  non-parallel  planes 
is  the  Intersection  of  the  planes. 

437.  The  foot  of  the  perpendicular,  let  fall  from  a  point  to  a 
plane,  is  called  the  Projection  of  the  point  on  the  plane. 

438.  The  locus  of  the  projections  on  a  plane  of  all  points  in  a 
line  is  called  the  Projection  of  the  line. 

The  point  D  is  the  projection 
of  the  point  A  upon  the  plane 
MN,  and  DEF  is  the  projection 
of  the  line  ABC  on  the  plane 
Mir. 


SOLID   GEOMETRY.  — BOOK   VIL 


245 


439.  The  angle  which  a  straight  line  makes  with  a  plane  is  the 
acute  angle  between  the   line   and  its 
projection  on  the  plane,  and  is  called 
the  inclination  of  the  line  to  the  plane. 

MN  is  a  plane  ;  AB  a  straight  line  meeting 
MN\  and  AD  the  projection  of  AB  on  MN. 
Then,  angle  BAD  is  the  angle  which  AB 
makes  with  the  plane  MN. 

440.  The  distance  from  a  point  to  a  plane  is  understood  to  be 
the  perpendicular  distance  from  that  point  to  the  plane. 


Proposition  I 

441.  Place  two  planes  *  so  that  they  intersect.  What  kind  of  a  line 
is  the  line  of  their  intersection  ? 

Theorem.  The  intersection  of  two  planes  is  a  straight 
line. 

Data:  Any  two  intersecting  planes, 
as  MN  and  PQ. 

To  prove  the  intersection  of  MN  and 
PQ  a  straight  line. 

Proof.    Suppose  that  E  and  F  are  any 

two  of  the  points  in  which  MN  and  PQ 

intersect.     Draw  the  straight  line  EF.  '*2 

Since  E  and  F  are  points  in  the  plane  MN,  §  427,  the  straight 
line  joining  them  must  lie  in  MN;  and  since  they  are  also  points 
in  PQ,  the  straight  line  joining  them  must  lie  in  PQ. 

Hence,  EF  is  common  to  MN  and  PQ ; 

and  since,  §  430,  only  one  plane  can  contain  a  line  and  a  point  with- 
out that  line,  no  point  without  EF  can  be  common  to  MN  and  PQ; 
.*.  §  436,  EF  is  the  intersection  of  MN  and  PQ. 

But,  const.,  EF  is  a  straight  line  ; 

hence,  the  intersection  of  MN  and  PQ  is  a  straight  line.  q.e.d. 

♦  The  student  may  represent  planes  and  lines  as  suggested  in  §  428. 


246 


SOLID   GEOMETRY.  — BOOK   VII. 


Proposition  II 

442.  In  a  plane  draw  two  intersecting  straight  lines.  If  a  third 
straight  line  is  perpendicular  to  each  of  these  at  their  point  of  intersec- 
tion, what  is  its  direction  with  reference  to  the  plane? 

Theorem.  If  a  straight  line  is  perpervdicular  to  each  of 
two  other  straight  lines  at  their  point  of  intersection,  it  is 
perpendicular  to  the  plane  of  the  two  lines. 


Data :  Any  two  straight  lines,  as 
SB  and  CD,  intersecting  at  ^;  MN, 
the  plane  of  these  lines ;  and  HE,  a 
perpendicular  to  AB  and  CD  at  E. 

To  prove  HE  perpendicular  to  MN. 


Proof.  Through  E,  in  the  plane  MN^  draw  any  other  straight 
line,  as  JK;  also  draw  AC  intersecting  JK  in  L. 

Produce  HE  through  MN  to  F,  making  EF  =  HE,  and  draw  HAy 
HL,  HC,  FA,  FL,  and  FC. 

In  Aach  and  ACF,       AC  is  common, 
§  103,  HA  =  FA,  and  HC  =  FC; 

.'.  §  107,  A  ACH  =  A  ACF, 

and  /.HAC=AfaG\ 

.-.  §  100,  A  ALH  =  A  ALF, 

and  HL  =  FL; 

.'.  §106,  LE±HE; 

that  is,  HE±JK. 

Consequently,  HE  is  perpendicular  to  every  straight  line  drawn 
in  MN  through  E. 

Hence,  §  432,       HE  is  perpendicular  to  MN. 

Therefore,  etc.  q.e.d. 

443.  Cor.  A  straight  line,  which  is  perpendicular  to  a  plane  at 
any  point,  is  perpendicular  to  every  straight  line  which  can  be  drawn 
in  that  plane  through  that  point. 


SOLID   GEOMETRY.  — BOOK   VII. 


247 


Proposition  III 

444.  1.  At  any  point  in  a  given  straight  line  erect  two*  perpendicu- 
lars to  the  line,  and-  through  them  pass  a  plane.  What  is  the  direction 
of  the  plane  with  reference  to  the  given  line?  Can  any  perpendiculars 
be  drawn  to  the  given  line,  at  this  point,  which  do  not  lie  in  this  plane  ? 

2.  Can  any  other  plane  be  passed  through  this  point  perpendicular  to 
the  given  line  V 

3.  Through  a  point  without  a  straight  line  pass  as  many  planes  as 
possible  perpendicular  to  the  line.  How  many  such  planes  can  oe 
passed  through  the  point? 

Theorem,  Every  perpendicular  to  a  straight  line  at  a 
given  point  lies  in  a  plane  which  is  perpendicular  to  the 
line  at  that  point. 


Data:  Any  straight  line,  as  AB-,  and  a 
plane,  as  MN,  perpendicular  to  AB  at  E\ 
also  any  line,  as  EF,  perpendicular  to  AB 
at  E. 

To  prove  that  EF  lies  in  MN. 


Proof.  Suppose  that  the  plane  of  AB  and  EF  intersects  MN  in 
the  line  EH. 

Then,  §  443,  ABA. EH. 

Since,  §  51,  in  the  plane  of  AB  and  EF  only  one  perpendicular 
can  be  drawn  to  AB  at  E,  EF  and  EH  coincide,  and  EF  lies  in  MK. 

Hence,  every  perpendicular  to  AB  at  E  lies  in  the  plane  MN. 

Therefore,  etc.  q.e.d. 

445.  Cor.  I.  At  a  given  point  in  a  straight  line  one  plane  per- 
pendicular to  the  line  can  he  passed,  and  only  one. 

446.  Cor.  II.  Through  a  given  point  without  a  straight  line  one 
2^la7ie  perpendicular  to  the  line  can  be  passed,  and  only  one. 

Ex.  738.  What  is  the  locus  of  the  perpendiculars  to  a  given  straight  line 
at  a  given  point  in  the  line  ? 


248 


SOLID   GEOMETRY.  — BOOK   VIL 


Proposition  IV 

447.  1.  Erect  a  perpendicular  to  a  plane ;  connect  a  point  in  the  per- 
pendicular with  points  in  the  plane  which  are  equally  distant  from  the 
foot  of  the  perpendicular.  How  do  these  oblique  lines  compare  in 
length? 

2.  Connect  the  same  point  in  the  perpendicular  with  points  in  the 
plane  unequally  distant  from  the  foot  of  the  perpendicular.  How  do 
these  oblique  lines  compare  in  length  ? 

3.  Represent  a  perpendicular  and  several  other  lines  from  a  point  to 
a  plane.     Which  line  is  the  shortest? 

Theorem.  If  from  a  point  in  a  perpendicular  to  a 
plane  oblique  lines  are  drawn  to  the  plane, 

1.  Those  lines  which  meet  the  plane  at  equal  distances 
from  the  foot  of  the  perpendicular  are  equal. 

2.  Of  two  lines  which  meet  the  plane  at  unequal  dis- 
tances from  the  foot  of  the  perpendicular,  that  which 
meets  it  at  the  greater  distance  is  the  greater. 


Data:  A  perpendicular  to 
the  plane  MN,  as  CD,  and  the 
oblique  lines  CE,  CF,  and  CG, 
which  meet  Mlfso  that  DE=DF, 
and  DG  >  DE. 


To  prove 

1. 

CE  =  CF. 

2. 

CG  >  CE. 

Proof. 

1. 

Data,  CD±MN-', 

CD±  DE,  and  CD  ±  DF, 
In  rt.  A  EDC  and  FDC,       DE  =  DF, 

CD  is  common, 
and,  §  52,  Z  EDC  =  Z  fdG\ 

.'.  §  100,  A  EDC  =  A  FDCy 

and  CE  =  CF. 

2.   On  DG  take  DH=  DE,  and  draw  CH. 

Then,  CH=  CE. 

Hence,  §  132,  CG  >  CH,  or  CE. 


Why? 


Why? 

Q.B.D. 


SOLID   GEOMETBY.  —  BOOK  VIL 


•249 


448.  Cor.  I.  A  perpendicular  is  the  shortest  line  that  can  be 
drawn  from  a  point  to  a  plane. 

449.  Cor.  II.  Equal  oblique  lines  from  a  point  in  a  perpendicii' 
lar  to  a  plane  meet  the  plane  at  equal  distances  from  the  foot  of  the 
perpendicular;  and  of  two  unequal  oblique  lines  the  greater  meets 
the  plane  at  the  greater  distance  from  the.  foot  of  the  perpendicular, 

450.  Cor.  III.  Tlie  locus  of  a  point  in  space  equidistant  from 
all  points  in  the  circumference  of  a  circle  is  a  straight  line  passing 
through  the  center  and  perpendicular  to  the  plane  of  the  circle. 


Proposition  V 

451.  Erect  a  perpendicular  to  a  plane;  from  the  foot  of  the  perpen- 
dicular draw  a  straight  line  at  right  angles  to  any  other  straight  line  of 
the  plane ;  join  the  point  of  intersection  of  these  two  lines  with  any 
point  in  the  perpendicular.  What  is  the  direction  of  this  joining  line 
with  reference  to  the  line  in  the  plane  that  does  not  pass  through  the 
foot  of  the  perpendicular  ? 

Theorem,  If  from  the  foot  of  a  perpendicular  to  a 
plane  a  straight  line  is  drawn  at  right  angles  to  any 
straight  line  in  the  plane,  the  line  drawn  from  the  point 
of  meeting  to  any  point  in  the  perpendicular  is  perpen- 
dicular to  the  line  of  the  plane. 

Data:    A   perpendicular  .  %tf 

to  the  plane  MN,  as  CD; 
any  straight  line  in  MN,  as 
EF'j  DG  perpendicular  to 
EF;  and  CG  joining  any 
point  in  CD  with  G.  /  i>  /  ^ 

To  prove  CG  perpendicu- 
lar to  EF. 

Proof.  From  C  and  B  draw  lines  to  H  and  J,  two  points  in  EV^ 
equally  distant  from  Q. 

Then,  §  103,  HD  =  JD', 

.-.§447,  HC^JG\ 

hence,  §  106,  CGLhJ:, 

that  is,  CG  is  perpendicular  to  EF, 

Therefore,  etc.  Q.e.d 


M^ 


250 


SOLID   GEOMETRY.  —  BOOK    VII. 


452.  Cor.  The  locus  of  a  point  in  space  equidistant  from  the 
extremities  of  a  straight  line  is  the  plane  perpendicular  to  the  line 
at  its  middle  point. 

Proposition  VI 

453.  At  any  two  points  in  a  plane  erect  perpendiculars  to  the  plane. 
What  is  the  direction  of  the  perpendiculars  with  reference  to  each  other? 

Theorem.  Two  straight  lines  perpendicular  to  the  same 
plane  are  parallel. 


Data :  Any  two  straight  lines 
perpendicular  to  plane  MN,  as 
CD  and  EF. 

To  prove   CD  and  EF  parallel. 


M 


Proof.     Draw  CF  and  DF,  and  through^ J^  draw  GHl.  DF. 

Then,  §443,  EF±GH, 

const.,  DF±  GHy 

and,  §  451,  CF±GH; 

.'.  §  444,  EF,  DF,  and  CF  lie  in  the  same  plane. 

Hence,  CD  and  EF  lie  in  the  same  plane. 

But,  §  443,  CD  ±  DF,  and  EF ±  DF; 

hence,  §  71,  CD  and  EF  are  parallel. 

Therefore,  etc. 

454.  Cor.  I.  If  one  of  two  parallel  straight 
lines  is  perpendicular  to  a  plane,  the  other  is  also 
perpendicular  to  the  plane. 


455.  Cor.  II.  Two  straight  lines  that  are 
parallel  to  a  third  straight  line  in  another  plane 
are  parallel  to  each  other. 


Q.E.D. 


N 


Ex.  739.  The  length  of  a  perpendicular  from  a  given  point  to  a  plane  is 
5dm,  What  is  the  diameter  of  the  circle  which  is  the  locus  of  the  foot  of  an 
oblique  line  drawn  from  the  same  point  to  the  plane,  if  the  oblique  line  is 
13<»™  long  ? 


SOLID   GEOMETRY.  — BOOK   VIL  251 


Proposition  VII 

456.  1.  At  any  point  in  a  plane  erect  as  many  perpendiculars  to  the 
plane  as  possible.     How  many  can  be  erected  ? 

2.  Choose  a  point  above  or  below  the  plane,  and  from  that  point  draw 
as  many  perpendiculars  to  the  plane  as  possible.  How  many  such  per- 
pendiculars can  be  drawn  ? 

Theorem,  From  a  given  point  only  one  perpendicular 
to  a  given  plane  can  be  drawn. 

B\      jC 

Data:   Any  plane,  as  MN,  and  any  I   / 

point,  as  A.  I  '^ 

To  prove  that  from  A  only  one  per-  /  ^ 

pendicular  to  MN  can  be  drawn. 


M 


Proof.     Case  I.      When  the  given  point  is  in  the  given  plane. 

Draw  AB  X  MN,  and  from  A  draw  any  other  line,  as  AC. 

li  AC  is  perpendicular  to  MN, 
§  453,  AB  and  AC  are  parallel  to  every  line  that  is  perpendicular 
to  MN; 
but,  §  70,  this  is  impossible ; 

AC  is  not  perpendicular  to  MN. 

Hence,  only  one  perpendicular  to  MN  can  be  drawn  from  A. 

Case  II.      When  the  given  point  is  without  the  given  plane. 

Draw  AB  _L  MN,  and  from  A  draw  l\ 

any  other  line  to  MN,  as  AC. 

If  ^C  is  perpendicular  to  MN,  §  453, 
AB  and  AC  are  parallel  to  every  line 
that  is  perpendicular  to  MN; 
but,  §  70,  this  is  impossible ;  ^^ 


C 


AC  is  not  perpendicular  to  MN. 

Hence,  only  one  perpendicular  to  MN  can  be  drawn  from  A. 
Therefore,  etc.  q.e.d. 


252  SOLID   GEOMETRY.  — BOOK    VII. 


Proposition  VIII 

457.  1.  Represent  two  parallel  straight  lines,  only  one  of  which  is  in 
a  given  plane.  What  is  the  direction  of  the  other  line  with  reference  to 
the  plane  ? 

2.  Represent  a  plane  and  a  straight  line  parallel  to  it ;  pass  any  plane 
through  the  line  so  that  it  intersects  the  given  plane.  What  is  the 
direction  of  the  intersection  with  reference  to  the  given  line  ? 

3.  Represent  two  intersecting  planes  and  a  straight  line  parallel  to 
their  intersection.  What  is  the  direction  of  this  line  with  reference  to 
each  of  the  planes  ? 

4.  Represent  any  two  straight  lines  in  space ;  through  one  of  them 
pass  any  plane.  Can  this  plane  be  turned  on  the  line  as  an  axis  into  a 
position  parallel  to  the  other  given  line  ? 

5.  Represent  any  two  straight  lines  in  space  and  a  point  without 
them ;  represent  a  line  passing  through  this  point  and  parallel  to  one  of 
the  given  lines;  represent  a  second  line  through  the  same  point  and 
parallel  to  the  other  given  line ;  represent  the  plane  of  these  lines  that 
intersect  at  the  given  point.  What  is  the  direction  of  this  plane  with 
reference  to  each  of  the  given  lines  ? 

Theorem.  If  a  straight  line  without  a  plane  is  parallel 
to  any  straight  line  in  the  plane,  it  is  parallel  to  the  plane. 


Data:    Any  straight  line  in 

plane    MN,    as    EF,    and    any  / — j -i -jN 

straight  line  without  MN  and 
parallel  to  EF,  as  CD. 

To  prove     CD  parallel  to  MN. 

Proof.     Through  CD  and  EF  pass  the  plane  ED. 

Now,  if  CD  is  not  parallel  to  MN,  it  must  meet  MN  in  the  inter 
section  of  MN  and  ED,  that  is,  in  EF. 

But,  data,  CD  cannot  meet  EF-, 

hence,  -        CD  cannot  meet  MNi 

that  is,  CD  is  parallel  to  MN.  Q.e.d. 

458.  Cor.  I.  If  a  straight  line  is  parallel  to  a  plane,  the  intersec- 
tion of  the  plane  with  any  plane  passed  through  the  line  is  parallel 
to  the  line. 


SOLID   GEOMETRY.  — BOOK  VIL  253 

459.  Cor.  II.  A  straight  line  parallel  to  the  intersection  of  two 
planes  is  parallel  to  each  of  the  planes. 

460.  Cor.  III.  Through  any  given  straight  line  a  plane  may  be 
passed  parallel  to  any  other  given  straight  line  not  intersecting  the 
first;  if  the  lines  are  not  parallel,  only  one  such  plane  can  he  passed. 

461.  Cor.  IV.  Through  a  given  poiiyt  a  plane  may  he  passed 
parallel  to  any  two  given  straight  lines  in  space;  and  if  the  lines 
are  not  parallel,  only  one  such  plane  can  he  passed. 

Proposition  IX 

462.  Represent  two  planes  each  perpendicular  to  the  same  straight 
line.     What  is  the  direction  of  the  planes  with  reference  to  each  other  ? 

Theorem,  Two  planes  perpendicular  to  the  same  straight 
line  are  parallel. 


Data  :  Any  two  planes  perpendicu- 
lar  to  EF,  as  MN  and  PQ. 

To  prove       MN  and  PQ  parallel. 


7 


F 

Proof.  If  MN  and  PQ  are  not  parallel,  they  will  meet,  and  thus 
there  will  be  two  planes  passing  through  the  same  point  and  per- 
pendicular to  the  same  line  EF. 

But,  §  446,  this  is  impossible. 

Hence,  MN  and  PQ  cannot  meet; 

that  is,  MN  and  PQ  are  parallel. 

Therefore,  etc.  Q-ed- 

Ex.  740.  If  a  straight  line  and  a  plane  are  perpendicular  to  another 
straight  line,  they  are  parallel. 

Ex.  741.  If  a  hne  is  equal  to  its  projection  on  a  plane,  it  is  parallel  to 
the  plane. 

Ex.  742.  If  a  line  makes  equal  angles  with  three  intersecting  lines  in  the 
same  plane,  it  is  perpendicular  to  that  plane. 

Ex.  743.  If  a  plane  bisects  a  straight  line  at  right  angles,  any  point  in 
the  plane  is  equidistant  from  the  extremities  of  the  line. 


264  SOLID   GEOMETRY.— BOOK    VIL 


Proposition  X 

463.  1.  Represent  two  parallel  planes  each  intersected  by  a  third 
plane.  In  what  direction,  with  reference  to  each  other,  do  the  lines  of 
intersection  extend  ? 

2.  Represent  two  parallel  straight  lines  included  between  two  parallel 
planes.     How  do  the  lines  compare  in  length  ? 

Theorem,  The  intersections  of  two  parallel  planes  hy  a 
third  plane  are  parallel. 


Data:  Any  two  parallel  planes,  as 
MN  and  PQ,  intersected  by  a  third   m^ 
plane,  as  RS,  in  GH  and  JK,  respec- 
tively. 

To  prove      GH  and  JK  parallel.  p^ 


Proof.     §  435,         MN  and  PQ  cannot  meet ; 
.*.  GH  and  JK,  which  are  lines  lying  in  MN  and  PQ  respectively, 
cannot  meet. 

But  GH  and  JK  lie  in  the  same  plane  i?S; 

hence,  GH  and  JK  are  parallel. 

Therefore,  etc.  q.e.d. 

464.  Cor.  I.  Parallel  straight  lines  included  between  parallel 
planes  are  equal. 

465.  Cor.  II.     Two  parallel  planes  are  everywhere  equally  distant. 

Ex.  744.  Draw  a  perpendicular  to  a  given  plane  from  any  point  without 
the  plane. 

Ex.  745.  Erect  a  perpendicular  to  a  given  plane  at  a  given  point  in  the 
plane. 

Ex.  746.  A  line  parallel  to  two  intersecting  planes  is  parallel  to  their 
intersection. 

Ex.  747.  If  two  lines  are  parallel,  the  intersections  of  any  planes  pass- 
ing through  them  are  parallel. 

Ex.  748.  If  a  plane  is  passed  through  a  diagonal  of  a  parallelogram,  the 
perpendiculars  to  it  from  the  extremities  of  the  other  diagonal  are  equal. 


SOLID   GEOMETRY.  — BOOK  VI I.  255 


Proposition  XI 

466.  1.  Represent  a  straight  line  perpendicular  to  one  of  two  parallel 
planes.  What  is  the  direction  of  the  line  with  reference  to  the  other 
plane  ? 

2.  Pass  as  many  planes  as  possible  through  a  point  and  parallel  to  a 
given  plane.     How  many  such  planes  can  be  passed  ? 

3.  Represent  two  intersecting  straight  lines  each  parallel  to  a  given 
plane.  What  is  the  direction  of  the  plane  of  these  lines  with  reference 
to  the  given  plane  ? 

Theorem,  A  straight  line  perpendicular  to  one  of  two 
parallel  planes  is  perpendicular  to  the  other. 


Data :  Any  two  parallel  planes,  as 
MN  and  PQ,   and  any   straight   line   M' 
perpendicular  to  MN,  as  EF. 

To  prove  EF  perpendicular  to  PQ. 


7 


H 


Proof.  Through  EF  pass  any  two  planes,  as  EH  and  EK,  inter- 
secting MN  in  EG  and  EJ,  and  PQ  in  FH  and  FK,  respectively. 

Then,  §  463,  EG  II  FH,  and  EJ  II  FK; 

and,  §  443,  EF±EG  smd  EJ-, 

EF ± FH  3iiid  FK.  Why? 

Hence,  §  442,       EF  is  perpendicular  to  PQ. 

» 
Therefore,  etc.  q.e.d. 

467.  Cor.  I.  Through  a  given  point  one  plane,  and  only  one, 
can  he  passed  parallel  to  a  given  plane. 

468.  Cor.  II.  If  two  intersecting  straight  lines  are  each  parallel 
to  a  given  plane,  the  plane  of  those  lines  is  parallel  to  the  given 
plane. 

Proposition  XII 

469.  Draw  an  angle  in  one  plane,  and  in  another,  an  angle  whose  sides 
are  respectively  parallel  to  the  sides  of  the  first  angle  and  extending  in 
the  same  direction.  How  do  the  angles  compare  in  size?  In  what  direc- 
tion do  their  planes  extend  with  reference  to  each  other? 


256 


SOLID   GEOMETRY.  — BOOK   VII. 


Theorem,  If  two  angles,  not  in  the  same  plane,  have  their 
sides  respectively  parallel  and  extending  in  the  same  direc- 
tion, they  are  equal  and  their  planes  are  parallel. 

Data ;  Any  two  angles,  as  F  and 
J,  in  the  planes  MN  and  PQ  respec- 
tively, having  the  sides  FE  and  FG 
parallel  to  and  extending  in  the 
same  direction  with  JH  and  JK 
respectively. 

To  prove  Zf=Z.J,  and  MN II  PQ. 

Proof.  1.  Take  FE=:.JH,  and  FG  =  JK\  and  draw  FJ^  EH,  GK, 
EG,  and  HK. 

Then,  §  150,     fehj  and  FGKJ  are  parallelograms ; 
EH=FJ=z  GK,  and  EH  II  FJ  II  GK', 
EQKH  is  a  parallelogram,  and  EG  =  HK. 
A  EFG  =  A  HJKf 

Zf  =  Zj. 

FEWpq,  and  FGWpQ; 

MN  11  PQ. 


Hence, 

and 

2.  §  457, 
hence,  §  468, 

Therefore,  etc. 


Why? 
Why? 
Why? 


Q.E.D. 


Proposition  XIII 


470.  Represent  two  straight  lines  intersected  by  three  parallel  planes. 
If  one  line  is  divided  into  segments  which  are  in  the  ratio  of  2  :  3,  what 
is  the  ratio  of  the  segments  of  the  other  line?  If  one  line  is  divided  into 
segments  which  are  in  any  ratio  whatever,  how  does  the  ratio  of  the 
segments  of  the  other  line  compare  with  this  ratio  ? 

Theorem,  If  two  straight  lines  are  intersected  hy  three 
parallel  planes,  their  corresponding  segments  are  propor- 
tional. 


Data:  Any  two  lines,  as  AB  and 
CD,  intersected  by  any  three  paral- 
lel planes,  as  MN,  PQ,  and  RS,  in  the 
points  A,  L,  B,  and  C,  G,  D,  respec- 
tively. 

To  prove      AL:LB=CG:  GD. 


M^ 

L 

\       \ 

r/ 

L.\a?  / 

r 

\  \ 

/ 

'V 

rI- 

--Dj 

SOLID    GEOMETRY.  — BOOK    VIL  257 

Proof.  Draw  AD  intersecting  PQ  in  0;  and  draw  LO,  OG,  AC, 
and  JS-D. 

Then,  §§  430,  463,    LO  II  bd,  and  OG  II  AG; 
.-.  §  289, 
and 
hence,  AL :  LB  ==  CQ :  GD, 

Therefore^  etc.  '  q.e.d. 

DIHEDRAL    ANGLES 

471.  The  opening  between  two  intersecting  planes  is  called  a 
Dihedral  Angle,  or  simply  a  Dihedral. 

The  intersection  of  the  planes  is  called  the  edge  of  the  dihedral 
angle,  and  the  two  planes  are  called  its  faces. 

472.  A  dihedral  angle  may  be  desig-  y\ 
nated  by  letters  at  four  points,  two   in                     /^   \ 
its  edge  and  one  in  each  face,  the  two  i>^    a     \ 
letters  at  the  edge  being  written  between                   \    \    \ 
the  other  two.  /      ^— -\-— .^^ 

When  but  one  dihedral  angle  is  formed  /  e X    -\^ 

at  the  same  edge  it  is  designated  simply      ^/ \/ 

by  two  letters  at  this  edge. 

AB  is  the  edge,  and  BO  and  BD  are  the  faces  of  the  dihedral  angle  AB, 
or  G-AB-D. 

473.  The  angle  formed  at;  any  point  in  the  edge  of  a  dihedral 
angle  by  two  perpendiculars  to  the  edge,  one  in  each  face,  is  called 
the  Plane  Angle  of  the  dihedral  angle. 

EF  and  GF  m  BG  and  BD  respectively,  both  perpendicular  to  AB  at  F, 
form  the  plane  angle  EFQ  of  the  dihedral  angle  G-AB-D. 

The  plane  angle  is  of  the  same  size  at  whatever  point  in  the 
edge  it  is  constructed.     (§§  71,  469). 

The  size  of  a  dihedral  angle  does  not  depend  upon  the  extent 
of  its  faces,  but  upon  their  difference  in  direction. 

474.  Two  dihedral  angles  which  can  be  made  to  coincide  are 
equal. 

milne's  geom.  — 17    . 


268  SOLID   GEOMETRY.  — BOOK    VII. 

475.  Dihedral  angles  are  adjacent,  right,  acute,  obtuse,  comple- 
mentary, supplementary,  or  vertical,  according  as  their  plane  angles 
conform  to  the  definitions  of  those  terms  given  in  plane  geometry. 

Two  dihedral  angles  are  adjacent,  if  they  have  a  common  edge,  and  a 
common  face  between  them ;  they  are  rights  if  they  are  formed  by  two 
perpendicular  intersecting  planes ;  they  are  vertical^  if  the  faces  of  one  are 
prolongations  of  the  faces  of  the  other. 

EXERCISES 

476.  1.  Eepresent  a  plane  meeting  another  plane.  How  does 
the  sum  of  the  two  dihedral  angles  thus  formed  compare  with  two 
right  dihedral  angles  ? 

2.  Represent  two  adjacent  dihedral  angles  whose  sum  is  equal 
to  two  right  dihedral  angles.     How  do  their  exterior  faces  lie  ? 

3.  Eepresent  two  intersecting  planes.  How  do  the  vertical 
dihedral  angles  compare  in  size  ? 

4.  Eepresent  two  parallel  planes  intersected  by  a  third  plane. 
How  do  the  alternate  interior  dihedral  angles  compare  in  size  ? 
How  do  the  corresponding  dihedral  angles  compare  ?  To  how 
many  right  dihedral  angles  is  the  sum  of  the  two  interior  dihedral 
angles  on  the  same  side  of  the  intersecting  plane  equal  ? 

5.  Eepresent  two  planes  intersected  by  a  third  plane.  In 
what  direction  do  the  two  planes  extend  with  reference  to  each 
other,  if  the  alternate  interior  dihedral  angles  are  equal  ?  If  the 
corresponding  dihedral  angles  are  equal?  If  the  sum  of  the 
interior  dihedral  angles  on  the  same  side  of  the  intersecting  plane 
is  equal  to  two  right  dihedral  angles  ? 

6.  Eepresent  two  dihedral  angles  whose  corresponding  faces 
are  parallel.  How  do  these  dihedrals  compare  in  size,  if  both 
corresponding  pairs  of  faces  extend  in  the  same  direction  from 
their  edges  ?     If  both  extend  in  opposite  directions  ? 

Discover  whether  it  is  possible  for  the  dihedrals  to  have  their 
faces  parallel  and  yet  not  be  equal. 

7.  Eepresent  two  dihedral  angles  whose  corresponding  faces 
are  perpendicular  to  each  other.  How  do  the  dihedrals  compare 
in  size,  if  both  are  acute  9    If  both  are  obtuse  ? 

Discover  whether  it  is  possible  for  the  dihedrals  to  have  their 
faces  perpendicular  and  yet  not  be  equal. 


SOLID   GEOMETRY.  —  BOOK    VII. 


259 


Proposition  XIV 

477.    Represent  two  dihedral  angles  whose  plane  angles  are  equal. 
How  do  the  dihedral  angles  compare  in  size  ? 

Theorem,    Two  dihedral  angles  are  equal,  if  their  plane 
<ingles  are  equal. 


Data:  Any  two  dihedral  angles,  as  AB  and  EF,  whose  plane 
angles  CAB  and  GEH  are  equal. 

To  prove  dihedral  /.  AB  =  dihedral  Z  EF. 

Proof.  Suppose-  that  dihedral  Z  AB  is  applied  to  dihedral 
Z  EF  in  such  a  way  that  the  plane  Z  CAD  coincides  with  the 
equal  plane  Z  GEH. 

Then,  point  A  coincides  with  point  E, 

and,  §  430,        plane  CAD  coincides  with  plane  GEH-, 
.-.  §  456,  AB,  the  perpendicular  to  plane  CAD,  coincides  with  EF, 
the  perpendicular  to  plane  GEH. 

Since  AB  coincides  with  EF,  and  AC  with  EG, 
§  430,  plane  BC  coincides  with  plane  FG. 

In  like  manner  it  may  be  proved  that 

plane  BD  coincides  with  plane  FH. 

Hence,  §  474,  dihedral  Z  AB  =  dihedral  Z  EF. 

Therefore,  etc.  q.e.d. 

Ex.  749.  If  two  planes  intersect  each  other,  the  vertical  dihedral  angles 
are  equal. 

Ex.  750.  If  a  plane  intersects  two  parallel  planes,  the  alternate  interior 
dihedral  angles  are  equal. 

Ex.  751.  The  line  ABC  pierces  three  parallel  planes  in  A^  B,  and  C, 
respectively,  and  the  line  DEF  pierces  the  same  planes  in  2),  E,  and  F, 
respectively.  If  AB  is  6  in.,  BC  8  in.,  and  DF  12  in.,  what  is  the  length 
of  DE  and  of  EF? 


260 


SOLID   GEOMETRY,  — BOOK    VII. 


Proposition  XV 

478.  Represent  two  dihedral  angles  whose  plane  angles  are  in  the 
ratio  of  3  to  4,  or  any  other  ratio.  What  is  the  ratio  of  the  dihedial 
angles  ? 

Theorem,  Dihedral  angles  are  to  each  other  as  their 
plane  angles. 


H 


Data:  Any  two  dihedral  angles,  as  C-^^-D  and  G-EF-U^  whose 
plane  angles  are  CAD  and  G^fl"  respectively. 

To  prove  C-AB-D  :  G-EF-H=z  A  CAD  :  Z  GEH. 

Proof.  Suppose  that  A  CAD  and  GEH  have  a  common  unit  of 
measure,  as  Z  CAJ,  which  is  contained  in  Z  CAD  three  times  and 
in  Z  GEH  four  times. 

Then,  Z  CAD  :  Z  GEH  =3:4. 

Divide  the  plane  angles  CAD  and  GEH  into  parts  each  equal  to 
Z  CAJ,  and  through  the  several  lines  of  division  and  the  edges 
AB  and  EF  pass  planes.. 

By  §  477,  these  planes  divide  C-AB-D  into  three  and  G-EF-H 
into  four  equal  parts ; 

C-AB-D  :  G-EF-H  =  3  :  4. 

Hence,     C-AB-D  :  G-EF-H  =  Z  CAD  :  Z  GEH. 

By  the  method  of  limits,  exemplified  in  §  223,  the  same  may  be 
:  proved  when  the  dihedral  angles  are  incommensurable. 

Therefore,  etc.  q.e.d. 

479.  Sch.  It  is  evident  that  the  plane  angle  of  a  dihedral  may 
be  taken  as  the  measure  of  the  dihedral. 

Ex.  752.  If  the  sum  of  two  adjacent  dihedral  angles  is  equal  to  two 
right  dihedral  angles,  their  exterior  faces  are  in  the  same  plane. 


SOLID   GEOMETRY.^ BOOK    VIL  261 


Proposition  XVI 

480.  1.  Represent  two  planes  that  are  perpendicular  to  each  other; 
in  one  of  them  draw  a  straight  line  perpendicular  to  their  intersection. 
What  is  the  direction  of  the  line  with  reference  to  the  other  plane  ? 

2.  Represent  two  planes  perpendicular  to  each  other  and  a  straight 
line  perpendicular  to  one  of  them  at  any  ^point  of  their  intersection. 
How  does  this  line  lie  with  reference  to  the  other  plane? 

Theorem.  If  two  planes  are  perpendicular  to  each  other, 
a  straight  line  drawn  in  one  of  thein  perpendicular  to  their 
intersection  is  perpendicular  to  the  other. 

Data:  Any  two  planes  perpen- 
dicular to  each  other,  as  MN  and 
PF,  intersecting  in  EF,  and  any 
line  in  PF  perpendicular  to  EF,  as 
GH. 

To  prove 

GH  perpendicular  to  MN. 

Proof.     In  MN  draw  HJl.  EF. 


\L 


Then,  §  473,  the  angle  GHJ  is  the  plane  angle  of  the  right  di- 
hedral angle  P-EF-N 
.'.  §  475,  Z  GHJ  is  a  rt.  Z. 

Data,  Z  GHE  is  a  rt.  Z  ; 

hence,  GHl.  HJ  and  EF  at  their  point  of  intersection ; 
and,  §  442,  GH  is  perpendicular  to  MN. 

Therefore,  etc.  q.e.d. 

481.  Cor.  If  tivo  planes  are  perpendicular  to  each  other,  a  per- 
peyidicular  to  one  of  them  at  any  point  of  their  intersection  lies  in 
the  other. 

Ex.  753.  Find  the  locus  of  all  points  equidistant  from  two  parallel  planes. 

Ex.  754.  Parallel  lines  which  pierce  the  same  plane  make  equal  angles 
with  it. 

Ex.  755.  If  two  planes  intersect  each  other,  the  sum  of  the  two  adjacent 
dihedral  angles  on  the  same  side  of  either  plane  is  equal  to  two  right  dihedral 
angles. 

Ex.  756.  If  a  plane  intersects  two  parallel  planes,  the  interior  dihedral 
angles  on  the  same  side  of  the  intersecting  plane  are  supplementary. 


262  SOLID   GEOMETRY.  — BOOK    VIL 


Proposition  XVII 

482.  1.  Represent  a  straight  line  perpendicular  to  a  plane.  What 
is  the  direction,  with  reference  to  this  plane,  of  every  plane  passing 
through  the  perpendicular? 

2.  Represent  a  dihedral  angle  and  a  plane  perpendicular  to  its  edge. 
What  is  the  direction  of  this  plane  with  reference  to  each  of  the  faces  of 
the  dihedral  ? 

Theorem,  If  a  straight  line  is  perpendicular  to  a  plane, 
every  plane  passed  through  the  line  is  perpendicular  to 
that  plane. 

Data :  Any  straight  liue  perpen- 
dicular to  plane  MN,  as  CD,  and 
any  plane  passing  through  CD,  as 
EF. 

To  prove 

EF  perpendicular  to  MN. 


M' 


iN 


E 


Proof.     In  MN  draw  DH A.  EG,  the  intersection  of  EF  and  MN. 

§443,  CD  A.  EG', 

.-.  §  473,    Z  CDH  is  the  plane  angle  of  dihedral  Z  F-EG-N. 

But,  §  443,  Z  CDH  is  a  rt.  Z  ; 

hence,  F-EG-N  is  a  right  dihedral  angle ; 

that  is,  EF  is  perpendicular  to  MN. 

Therefore,  etc.  q.e.d. 

483.  Cor.  A  plane  perpendicular  to  the  edge  of  a  dihedral  angle 
is  perpendicular  to  each  of  its  faces. 

Ex.  757.  Find  the  locus  of  all  points  in  space  equidistant  from  two  given 
points. 

Ex,  758.  Find  the  locus  of  all  points  at  a  given  distance  from  a  given 
plane. 

Ex.  759.  A  line  and  its  projection  on  a  plane  determine  a  second  plane 
perpendicular  to  the  fii*st. 

Ex.  760.  If  a  line  is  parallel  to  one  plane  and  perpendicular  to  another, 
the  two  planes  are  perpendicular  to  each  other. 

Ex.  761.  D  is  any  point  in  the  perpendicular  AF  from  A  to  the  side  BC 
of  the  triangle  ABC.  If  DE  is  perpendicular  to  the  plane  ABC,  and  GB 
pas.ainff  through  E  is  parallel  to  BC,  then,  AE  is  perpendicular  to  GH. 


SOLID   GEOMETRY.  —  BOOK   VII. 


263 


Proposition  XV  III 

484.  1.  Represent  two  intersecting  planes  each  perpendicular  to  a 
third  plane.  What  is  the  direction  of  their  intersection  with  reference 
to  the  third  plane  ? 

2.  What  is  the  direction  of  the  third  plane  with  reference  to  the  inter- 
section of  the  other  two  planes  V 

3.  Represent  two  planes  perpendicular  to  each  other  and  another 
plane  perpendicular  to  each  of  them.  What  is  the  direction  of  the  inter- 
section of  any  two  of  these  planes  with  reference  to  the  third  plane? 
What  is  the  direction  of  each  intersection  with  reference  to  the  other 
two  intersections? 

Theorem,  If  two  intersecting  planes  are  each  perpendic- 
ular to  a  third  plane,  their  intersection  is  perpendicular  to 
the  third  plane. 

Data :  Any  two  planes,  as  PQ 
and  RS,  intersecting  in  EF,  and 
perpendicular  to  a  third  plane, 

as  MN. 

To  prove 

EF  perpendicular  to  MN. 

Proof.  At  F,  the  point  common  to  the  three  planes,  erect  a 
perpendicular  to  the  plane  MN. 

By  §  481,  thjs  perpendicular  lies  in  both  PQ  and  i?/S,  and  hence 
must  coincide  with  their  intersection  EF. 

Consequently,      EF  is  perpendicular  to  MN. 

Therefore,  etc.  q.e.d. 

485.  Cor.  I.  A  plane  perpendicular  to  each  of  two  intersecting 
planes  is  perpendicular  to  their  intersection. 

486.  Cor.  II.  If  a  plane  is  perpendicular  to  two  planes  which 
are  perpendicular  to  each  other,  the  intersection  of  any  two  of  these 
planes  is  perpendicular  to  the  third  plane,  and  ea/ih  of  the  three 
intersections  is  perpendicular  to  the  other  two. 

•  Ex.  762.  If  from  a  point  within  a  dihedral  angle  perpendiculars  are 
drawn  to  its  faces,  the  angle  contained  by  these  perpendiculars  is  equal  to 
the  plane  angle  of  the  adjacent  dihedral  angle  formed  by  producing  one  of 
the  planes. 


264 


SOLID   GEOMETRY.  — BOOK  VII. 


Proposition  XIX 

487.  Represent  a  straight  line  oblique  to  a  plane.  How  many  pUues 
can  be  passed  through  that  line  perpendicular  to  the  given  plane  ? 

Theorem.  Through  any  straight  line  not  perpendicular 
to  a  plane,  one  plane  perpendicular  to  that  plane  can  he 
passed,  and  only  one. 

Data:  Any  plane,  as  MN,  and  any 
straight  line  not  perpendicular  to  MN, 
as  CD. 

To  prove  that  through  CD  one  plane 
perpendicular  to  MN  can  be  passed, 
and  only  one. 

Proof.     From  any  point  of  CD,  as  E,  draw  EF  ±  MN. 

Through  CD  and  EF  pass  a  plane,  as  GD,  intersecting  MN. 

Then,  §  482,  GD  is  perpendicular  to  MN. 

Now,  if  any  other  plane  perpendicular  to  MN  could  be  passed 
through  CD, 
§  484,  CD  would  be  perpendicular  to  MN. 

But,  data,  CD  is  not  perpendicular  to  MN; 

hence,  only  one  plane  perpendicular  to  MN  can  be  passed  through  CD. 

Therefore,  etc.  q.e.d. 

Proposition  XX 

488.  Represent  two  intersecting  planes  forming  a  dihedral  angle,  and 
a  third  plane  bisecting  this  angle ;  select  any  point  in  the  bisecting  plane. 
How  do  the  distances  of  the  point  from  the  faces  of  the  angle  compare  V 

Theorem.  Every  point  in  the  plane  which  bisects  a 
dihedral  angle  is  equidistant  from  the  faces  of  the  angle. 


Data:  Any  dihedral  angle,  as 
C-AB-D ;  the  plane  bisecting  it,  as 
AK',  and  any  point  in  AK,  as  H.  ^ 

To  prove  U  equidistant  from  the 
faces  CB  and  AD. 


SOLID   GEOMETRY,  — BOOK   VII. 


265 


Proof.     Draw  HE  and  HF,  the  perpendiculars  from  H  to  CB  and 

AD  respectively,  and  through,  them  pass  a  plane  intersecting  the 
planes  CB,  AD,  and  AK  in  the  lines  EG,  FG,  and  HG  respectively. 

Then,  §  482,         plane  EGFHl.  CB  and  AD ; 
.-.   §  485,  EGFHl.  AB. 

Hence,  §  443,  EG,  FG,  and  HG  are  perpendicular  to  AB. 

Data,  C-AB-K=D-AB-K'j 

hence,  §  479,  Z  EGH  =  Z  FGH. 

In  rt.  A  EGH  and  FGH,  GH  is  common, 
and  ZEGH=/.FGH'y 

AEGH=AFGH,  Why? 

and  HE=HF', 

that  is,  §  440,  H  is  equidistant  from  the  faces  CB  and  AD. 

Therefore,  etc.  q.e.d. 

Proposition  XXI 

489.  Represent  a  line  oblique  to  a  given  plane  and  represent  its  pro- 
jection upon  that  plane.  How  does  the  acute  angle  formed  by  the  line 
with  its  projection  compare  in  size  with  the  angle  which  the  line  makes 
with  any  other  line  of  the  plane  ? 

Theorem,  The  acute  angle  formed  by  a  line  and  its  pro- 
jection upon  a  plane  is  the  least  angle  which  the  line  makes 
with  any  line  of  the  plane. 

Data:  Any  plane,  as  MN-, 
any  line,  as  CD,  meeting  MN  in 
C;  the  projection  of  CD  upon 
MN,  as  CE;  and  any  other  line 
drawn  in  MN  through  C,  as  CF. 

To  prove 
Z  DCE  less  than  Z  DGF. 

Proof.     Take  CG  =  GE,  and  draw  DG  and  DE. 

In  A  CED  and  CGD,       CD  is  common, 

CE  =  CG,  and  ED  <GD',  Why  ? 

hence,  §  130,  Z  DCE  is  less  than  Z  DCF. 

Therefore,  etc  q.e.d. 


266  SOLID  GEOMETRY.  — BOOK  VII. 

Proposition  ZXII 

490.  Represent  two  straight  lines  in  different  planes  and  a  common 
perpendicular  to  them.     How  many  such  perpendiculars  can  there  be  ? 

Theorem,  Between  two  straight  lines,  not  in  the  same 
plane,  one  common  perpendicular  can  be  drawn,  and  only 
one. 

c, -^ 

Data :  Any  two  straight  lines  | 

not  in  the  same  plane,  as  AB  ! 

and  CD.  /]        '  !  V"  I^r'T"^ 

To  prove  that  one  common  /   | 

perpendicular  can  be  drawn  to  /  ^__S/k     ' 

AB  and  CJ5,  and  only  one.  / [a_j 

Proof.  1.  Through  any  point,  as  K,  in  AB  draw  LK  II  CB^  and 
let  MN  be  the  plane  of  LK  and  AB ;  through  CD  pass  a  plane,  as 
QD,  perpendicular  to  MN,  intersecting  MN  in  GB.  and  AB  in  H. 

Then,  §  457,  CD  11  MN  and,  §  458,  GH II  CD. 

In  the  plane  GD  draw  ED  ±  GH. 

Then,  HdLcd,  Why? 

and,  §  480,  HD  A.  MN; 

/.  §443,  HDA.AB. 

2.  Now,  suppose  that  any  other  line,  as  JK,  is  perpendicular  to 
AB  and  CD. 

In  MN  draw  KL  II  CD,  and  in  GD  draw  JE±  QH. 

Then,  hyp.  and  §  72,         JK±KL; 
hence,  §  442,  JK±3IN. 

But,  §480,  JEA.MN. 

Hence  there  are  two  perpendiculars  from  J  to  MN', 
but,  §  456,  this  is  impossible. 

Hence,  JK  is  not  perpendicular  to  AB  and  CD,  and  HD  is  the 
only  perpendicular  common  to  those  lines. 

Therefore,  etc.  q.e.d. 

Ex.  763.  The  shortest  line  that  can  be  drawn  between  two  straight  lines 
not  in  the  same  plane  is  their  common  perpendicular. 


SOLID   GEOMETRY.^ BOOK   VIL 


26? 


POLYHEDRAL  ANGLES 

491.  The  angle  formed  by  three  or  more  planes  which  meet 
at  but  one  point  is  called  a  Polyhedral  Angle,  or  Polyhedral. 

The  point  in  which  the  planes  meet  is  called  the  vertex;  the 
intersections  of  the  planes  are  called  the 
edges;  the  portions  of  the  planes  incliided 
between  the  edges  are  called  the /aces;  and 
the  angles  formed  by  the  edges  are  called 
the  face  angles  of  the  polyhedral  angle. 

In  the  polyhedral  angle  Q-ABCD,  Q  is  the 
vertex;  QAB,  QBC,  etc.,  are  the  faces;  QA, 
QB,  etc.,  are  the  edges;  and  angles  AQB,  BQO^ 
etc. ,  are  the  face  angles. 

492.  The  faces  of  a  polyhedral  angle  are  of  indefinite  extent, 
but  for  convenience  they  are  represented  as  limited  by  an  inter- 
secting plane  called  the  Base. 

ABCD  is  the  base  of  Q-ABGD. 

493.  A  polyhedral  angle  whose  base  is  a  convex  polygon  is 
called  a  Convex  Polyhedral  Angle. 

494.  Polyhedral  angles  which  have  their  face  angles  and  their 
dihedral  angles  equal,  each  to  each,  and  arranged  in  the  same 
order,  are  equal,  for  they  can  be  made  to  coincide.  Polyhedral 
angles  which  have  their  face  angles  and  their  dihedral  angles 
equal,  each  to  each,  and  arranged  in  reverse  order,  are  symmetrical, 
but  not  generally  equal. 

The  trihedral  angles  Q-ABG 
and  Q'-A'B'C  are  symmetrical, 
when  the  face  angles  AQB,  BQC, 
CQA  are  equal  respectively  to 
the  face  angles  A'Q'B',  B'Q'G'^ 
G'Q'A',  and  the  dihedral  angles 
QA,  QB,  QG  are  equal  respec- 
tively to  the  dihedral  angles  Q'A', 
Q'B',Q'G>. 

Two  symmetrical  polyhedral  angles  cannot,  generally,  be  made 
to  coincide. 


268  SOLID   GEOMETRY.  — BOOK    VI I. 

495.  If  the  edges  of  one  of  two  polyhedral  angles  are  pro- 
longations of  oho  edges  of  the  other  through  their  common  ver- 
tex the  angles  are  called  Vertical  Polyhedral  Angles. 

496.  A  polyhedral  angle  having  three  faces  is  called  a  trihedral 
angle;  one  having  four  faces  is  called  a  tetrahedral  angle;  etc. 

Proposition  XXIII 

497.  Represent  two  vertical  polyhedral  angles.  How  do  the  face 
angles  of  one  compare  with  the  face  angles  of  the  other?  How  do  the 
dihedral  angles  of  one  compare  with  the  dihedral  angles  of  the  other? 
Are  they  arranged  in  the  same  or  in  a  reverse  order  in  the  two  polyhe- 
drals  ?     What  name  is  given  to  such  polyhedral  angles  ? 

Theorem.  Two  vertical  polyhedral  angles  are  symrnet' 
rical. 


B' 


Data:    Any  two  vertical  polyhedral  angles, 
as    Q-ABGD  and   Q-A'B'CfD'. 
To  prove  Q-ABCD  and  Q-A^B^CfD'  symmetrical. 


B 


c 


Proof.     §  59,       face  Z  AQB  =  face  Z  A^QB^, 
and  face  Z  BQC,  etc.  =  face  Z  b'qc',  etc.,  respectively. 

§§  473,  477,  dihedral  Zqa  =  dihedral  Z  QA', 
and  dihedral  Z  QB,  etc.  =  dihedral  Z  QB',  etc.,  respectively. 

But  the  face  and  dihedral  angles  of  Q-ABCD  are  arranged  in 
an  order  which  is  the  reverse  of  the  equal  face  and  dihedral  angles 
of  Q-A'B'C'D'. 

Hence,  §  494,  Q-ABCD  and  Q-A'b'&D'  are  symmetrical. 

Therefore,  etc.  q.e.d. 

Ex.  764.  A  plane  can  be  passed  perpendicular  to  only  one  edge  and  only 
two  faces  of  a  polyhedral  angle. 

Ex.  765.  Every  point  within  a  dihedral  and  equidistant  from  its  faces 
lies  in  the  plane  which  bisects  that  dihedral. 

Ex.  766.  The  sides  of  an  isosceles  triangle  are  equally  inclined  to  any 
plane  through  its  base. 


SOLID   GEOMETRY. ^ BOOK   VIL  269 

Proposition  XXIV 

498.  Represent  a  trihedral  angle.     How  does  the  sum  of  any  two  of 

its  face  angles  compare  with  the  third  face  angle  ? 

Theorem,  The  sum  of  any  two  face  angles  of  a  tri- 
hedral angle  is  greater  than  the  third  face  angle. 

The  theorem  requires  proof  only  when  the  q 

third  angle  is  greater  than  each  of  the  others.  /K 

Data:  Any  trihedral  angle,  as  Q-ABC,  hav-  /     y\ 

ing  one  face  angle,  as  AQC,  greater  than  either         /        \\\   ' 

of  the  other  face  angles.  /  \  \  \ 

To  prove  i     ^^^^^^  \  /\ 

Zaqb-^Zbqc greater  than  ZAQa  ^b\ 

Proof.  In  the  face  AQC  draw  QD,  making  Zaqd  =  ZAQB; 
through  any  point,  as  n,  of  QD  draw  ADO  in  the  plane  AQC; 
take  QB  =  QD,  and  through  line  AC  and  point  B  pass  a  plane. 

Then,  Aaqb  =  Aaqd,  and  AB  =  AD,  Why  ? 

In  A  ABC,  AB-^BC>AD-^  D0\  Why  ? 

but  AB=zAB', 

.'.  Ax.  5,  BC>DC. 

In  A  bqc  and  DQC,  qb  =  QD, 

QC  is  common, 
and  BC>DC; 

Z  BQC  is  greater  than  Z  DQC.  Why  ? 

Const.,  ZaQB  =  ZaqD; 

.-.  Ax.  4,    Z  AQB  +  Z  BQC  is  greater  than  Z  AQD  4-  Z  DQC, 
or  Z  AQB  +  Z  BQC  is  greater  than  Z  AQC. 

Therefore,  etc.  q.e.d. 

Proposition  XXV 

499.  Represent  any  convex  polyhedral  angle ;  around  some  point  in 
a  plane  as  a  common  vertex  construct  in  succession  angles  equal  to  the 
face  angles  of  this  polyhedral.  How  do^s  their  Sum  compare  with  four 
right  angles? 


270  SOLID   GEOMETRY.  — BOOK    VIL 

Theorem,     The  sum  of  the  face  angles  of  any  convex 
polyhedral  angle  is  less  than  four  right  angles. 

Q 

Datum :  Any  convex  polyhedral  angle,  /// 1  \ 

as  Q.                               ,  /    //  1  \ 

To  prove  that  the  sum  of  the  face  angles  /        'e\     \ 

of  Q  is  less  than  four  right  angles.  /---'/      '1  --  \ 


\c 


Proof.     Pass  a  plane  intersecting  the  edges  of  Q  in  A,  B,  c,  etc. 

Then,  ABODE  is  a  convex  polygon. 

From  0,  any  point  within  ABODE,  draw  OA,  OB,  00,  etc. 

The  number  of  triangles  whose  common  vertex  is  Q  equals  the 
number  whose  common  vertex  is  O. 

Hence,  the  sum  of  the  angles  of  the  triangles  whose  vertex  is 
Q  equals  the  sum  of  the  angles  of  the  triangles  whose  vertex  is  O. 

But  in  the  trihedral  angles  whose  vertices  are  A,  B,  O,  etc., 
§  498,   Z  QBA  +  Z  QBO  is  greater  than  Z  ABO,  or  Z  ABO  -f  Z  OBO, 
and        Z  QOB  +  Z  QCD  is  greater  than  Z  BOD,  or  Z  BOO  +  Z  i)(70. 

Hence,  reasoning  in  a  similar  manner  regarding  the  other  base 
angles  of  the  triangles,  the  sum  of  the  base  angles  of  all  the 
triangles  whose  vertex  is  Q  is  greater  than  the  sum  of  the  base 
angles  of  the  triangles  whose  vertex  is  O. 

Therefore,  the  sum  of  the  face  angles  at  Q  is  less  than  the  sum 
of  the  angles  at  0.  Why  ? 

But  the  sum  of  the  angles  at  0  equals  four  right  angles. 

Hence,  the  sum  of  the  face  angles  of  Q  is  less  than  four  right 
angles. 

Therefore,  etc.  q.e.d. 

Proposition  XXVI 

500.  Represent  two  trihedral  angles  having  the  three  face  angles  of 
one  equal  respectively  to  the  three  face  angles  of  the  other.  How  do 
the  trihedrals  compare?  Can  there  be  two  trihedrals  which  fulfill  the 
same  conditions  and  yet  not  be  equal?  What  name  is  given  to  such 
trihedrals? 


SOLID   GEOMETRY.— BOOK    VTL 


271 


Theorem.  Two  trihedral  angles  are  either  equal  or  sym- 
metrical, if  the  three  face  angles  of  one  are  equal  to  th^ 
three  face  angles  of  the  other,  each  to  each. 


C  C 


Data :  Any  two  trihedral  angles,  as  Q  and  Q',  having  the  face 
angles  AQB,  BQC,  AQC  equal  to  the  face  angles  A'Q'b',  b'q'O',  A'Q'&y 
each  to  each. 

To  prove         Q  either  equal  or  symmetrical  to  Q'. 

Proof.  On  the  edges  of  Q  and  Q'  take  the  equal  distances  QAj 
QB,  QC,  Q'A',  Q'B',  Q'C',  and  draw  AB,  BC,  AC,  A'B',  B'C',  A'C'. 

Then,  A  QAB,  QBC,  QAC  are  equal  to  A  Q'a'b',  Q'b'c',  Q'A'&y 
each  to  each.  Why  ? 

Hence,      AABO  =  Aa'b'c',  and  Z BAC=  Z b'a'c'.  Why? 

On  the  edge  QA  take  AD  and  on  Q'A'  take  A'n'  =  AD.  At  D  and 
D'  construct  the  plane  A  HDK  and  H'd'k'  of  the  dihedrals  QA  and 
Q'A'  respectively,  the  sides  meeting  AB,  AC,  A'b',  and  A'c'  as  at  H, 
K,  h'  and  E^  respectively,  inasmuch  as  A  QAB,  QAC,  etc.,  are  acute, 
being  angles  of  isosceles  A  QAB,  etc.     Draw  HE  and  H'K', 

Then,  const.,  AD  =  A'd', 

and  Z  DAH  =ZD'A'n'] 

rt.  AADH=  rt.  AA'd'h',  and  AH=  A'H* ; 
similarly,  AK=  A'k', 

and,  since  Z  BAC  =  Z  b'a'&  ; 

A  AHK  =  A  A'H'K',  and  HK  =  H'K' ; 
but  DH  =  D'H",  and  DK^d'e*', 

A  HDK  =  A  H'D'K*,  and  Z  HDK  =  Z  H'D'K*. 
Hence,  §  477,   dihedral  ZQA  =  dihedral  Z  Q'A', 


Why  . 
Why? 


272  SOLID   GEOMETRY.  — BOOK  VIL 

In  like  manner  it  may  be  shown  that  the  dihedral  angles  QB 
and  QC  are  equal  to  the  dihedral  angles  Q'^'  and  Q^&  respectively. 

Hence,  §  494,  if  the  equal  angles  are  arranged  in  the  same  order, 
as  in  the  first  two  figures,  the  two  trihedral  angles  are  equal; 
but  if  they  are  arranged  in  the  reverse  order,  as  in  the  first  and 
third  figures,  the  two  trihedral  angles  are  symmetrical. 

Therefore,  etc.  q.e.d. 

501.  Cor.  If  two  trihedral  angles  have  three  face  angles  of  the 
one  equal  to  three  face  angles  of  the  other,  then  the  dihedral  angles 
of  the  one  are  respectively  equal  to  the  dihedral  angles  of  the  other. 

SUPPLEMENTARY   EXERCISES 

Ex.  767.  If  a  straight  line  is  parallel  to  a  plane,  any  plane  perpendicular 
to  the  line  is  perpendicular  to  the  plane. 

Ex.  768.  If  a  straight  line  intersects  two  parallel  planes  it  makes  equal 
angles  with  them. 

Ex.  769.  If  a  line  is  parallel  to  each  of  two  planes,  the  intersections 
which  any  plane  passing  through  it  makes  with  the  planes  are  parallel. 

Ex.  770.  The  projections  of  parallel  straight  lines  on  any  plane  are  either 
parallel  or  coincident. 

Ex.  771.  Find  the  locus  of  points  which  are  equidistant  from  three  given 
points  not  in  the  same  straight  line. 

Ex.  772.  From  any  point  within  the  dihedral  angle  A-BC-D,  EF  and 
EG  are  drawn  perpendicular  to  the  faces  AC  and  BD,  respectively,  and  GH 
perpendicular  to  ^O  at  H.    Prove  that  FH  is  perpendicular  to  BG. 

Ex.  773.  If  a  plane  is  passed  through  the  middle  point  of  the  common 
perpendicular  to  two  straight  lines  in  space,  and  parallel  to  both  lines,  it 
bisects  every  straight  line  drawn  from  any  point  in  one  line  to  any  point  in 
ihe  other  line. 

Ex.  774.  If  the  intersections  of  several  planes  are  parallel,  the  perpen- 
diculars drawn  to  them  from  any  point  lie  in  one  plane. 

Ex.  775.  If  two  face  angles  of  a  trihedral  are  equal,  the  dihedral  angles 
opposite  them  are  also  equal. 

Ex.  776.  A  trihedral  angle,  having  two  of  its  dihedral  angles  equal,  may 
be  made  to  coincide  with  its  symmetrical  trihedral  angle. 

Ex.  777.  In  any  trihedral  the  three  planes  bisecting  the  three  dihedrals 
intersect  in  the  same  straight  line. 

Ex.  778.  In  any  trihedral  the  planes  which  bisect  the  three  face  angles, 
and  are  perpendicular  to  those  faces,  respectively,  intersect  in  the  same 
straight  line. 


BOOK   VIII 

POLYHEDRONS 

502.  A  solid  bounded  by  planes  is  called  a  Polyhedron. 

The  intersections  of  the  planes  which  bound  a  polyhedron  are 
called  its  edges;  the  intersections  of  the  edges  are  called  its 
vertices;  and  the  portions  of  the  planes  included  by  its  edges  are 
called  its  faces. 

The  line  joining  any  two  vertices  of  a  polyhedron,  not  in  the 
same  face,  is  called  a  diagonal  of  the  polyhedron. 

503.  A  polyhedron  having  four  faces  is  called  a  tetrahedron; 
one  having  six  faces  is  called  a  hexahedron;  one  having  eight 
faces  is  called  an  octahedron;  one  having  twelve  faces  is  called  a 
dodecahedron;  one  having  twenty  faces  is  called  an  icosahedron. 

504.  If  the  section  made  by  any  plane  cutting  a  polyhedron  is 
a  convex  polygon,  the  solid  is  called  a  Convex  Polyhedron. 

Only  convex  polyhedrons  are  considered  in  this  work. 

PRISMS 

505.  A  polyhedron  two  of  whose  faces  are  equal 
polygons,  which  lie  in  parallel  planes  and  have 
their  homologous  sides  parallel,  and  whose  other 
faces  are  parallelograms,  is  called  a  Prism. 

The  two  equal  and  parallel  faces  of  the  prism 
are  called  its  bases;  the  other  faces  are  called  lat- 
eral faces;  the  intersections  of  the  lateral  faces  are 
called  lateral  edges;  the  sum  of  the  lateral  faces  is  called  the  lat- 
eral, or  convex  surface ;  and  the  sum  of  the  areas  of  the  lateral 
faces  is  called  the  lateral  area  of  the  prism. 

The  lateral  edges  of  a  prism  are  parallel  and  equal.     §  153. 

The  perpendicular  distance  between  the  bases  of  a  prism  is  its 
altitude. 

milnb's  geom.  — 18  273 


274 


SOLID   GEOMETRY.  — BOOK    VIII. 


506.  A  prism  is  called  triangular,  quadrangular, 
hexagonal,  etc.,  according  as  its  bases  are  triangles, 
quadrilaterals,  hexagons,  etc. 

507.  A  prism  whose  lateral  edges  are  perpendicu- 
lar to  its  bases  is  called  a  Right  Prism. 


508.  A  prism  whose  lateral  edges  are  not  perpendicular  to  its 
bases  is  called  an  Oblique  Prism. 

509.  A  right  prism  whose  bases  are  regular  polygons  is  called 
a  Regular  Prism. 

510.  A  section  of  a  prism  made  by  a  plane  perpendicular  to  its 
lateral  edges  is  called  a  Right  Section. 


511.  The  part  of  a  prism  included  between 
one  base  and  a  section  made  by  a  plane  oblique 
to  that  base,  and  cutting  all  the  lateral  edges,  is 
called  a  Truncated  Prism. 


512.  A  prism  whose  bases  are  parallelograms 
is  called  a  Parallelepiped. 

513.  A  parallelopiped  whose  lateral  edges  are 
perpendicular  to  its  bases  is  called  a  Right  Paral- 
lelopiped. 

514.  A  parallelopiped  all  six  of  whose  faces  are 
rectangles  is  called  a  Rectangular  Parallelopiped. 

515.  A  parallelopiped  whose  six  faces  are  all  squares  is  called 
a  Cube. 

516.  The  quantity  of  space  inclosed  by  the  surfaces  which 
bound  a  solid  is  called  the  Volume  of  the  solid. 

A  solid  is  measured  by  finding  how  many  times  it  contains 
some  other  solid  adopted  as  the  unit  of  measure. 

The  units  of  measure  for  volume  are  the  cubic  inch,  the  cubic 
foot,  the  cubic  yard,  the  cubic  centimeter,  the  cubic  decimeter,  etc. 


SOLID   GEOMETRY.  — BOOK    VIII. 


275 


Suppose  that  the  cube  M  is  the  unit 
of  measure  and  that  AB  is  the  rectan- 
gular parallelopiped  to  be  measured. 

Apply  an  edge  qf  M  to  each  edge  of 
AB  and  at  the  points  of  division  pass  - 
planes  respectively  perpendicular  to 
those  edges.     These  planes  divide  AB 
into  cubes,  each  equal  to  the  unit  M. 

It  is  evident  that  there  will  be  as  many  layers  of  these  cubes 
as  the  edge  of  M  is  contained  times  in  the  altitude  of  AB,  that 
each  layer  will  contain  as  many  rows  of  cubes  as  the  edge  ot  M  is 
contained  times  in  the  width  of  AB,  and  that  each  row  will  con- 
tain as  many  cubes  as  the  edge  of  M  is  contained  times  in  the 
length  of  ^5;  and,  therefore,  that  the  product  of  the  numerical 
measures  of  the  three  dimensions  of  AB  is  equal  to  the  number  of 
times  that  M  is  contained  in  AB. 

In  this  case  the  edge  of  *M  is  contained  4  times  in  BE,  3  times 
in  DA,  and  5  times  in  DC;  consequently,  there  are  5  cubes  in  each 
cow,  3  rows  in  each  layer,  and  4  layers  in  the  parallelopiped ;  that 
ts,  M  is  contained  in  AB  5  x  3  x  4  =  60  times,  or  the  rectangular 
parallelopiped  contains  60  cubic  units. 

Therefore,  if  the  edge  of  3f  is  a  common  unit  of  measure  of  the 
three  dimensions  of  a  rectangular  parallelopiped,  the  product  of 
the  numerical  measures  of  the  three  dimensions  expresses  the  num- 
ber of  times  that  the  rectangular  parallelopiped  contains  the  cube, 
and  is  the  numerical  measure  of  the  volume  of  the  rectangular 
parallelopiped. 

517.  For  the  sake  of  brevity,  the  product  of  the  three  dimensions 
is  used  instead  of  the  product  of  the  mimerical  measures  of  the 
three  dimensioris. 

The  product  of  three  lines  is,  strictly  speaking,  an  absurdity, 
but  since  the  expression  is  used  to  denote  the  volume  of  a  rectan- 
gular parallelopiped,  it  follows  that  the  geometrical  concept  of  the 
product  of  three  lines  is  the  rectangular  parallelopiped  whose  edges 
they  are. 

Thus,  DC  X  DA  X  DE  implies  a  product,  which  is  a  numerical 
result,  but  it  must  be  interpreted  geometrically  to  mean  the  rectan- 
gular parallelopiped  whose  edges  are  DC,  DA,  and  DE. 


276  SOLID   GEOMETRY.  — BOOK   VIII. 

For  similar  reasons,  the  cube  of  a  line  must  be  interpreted  geo- 
metrically to  mean  the  cube  constructed  upon  the  line  as  an  edge, 
and  conversely,  the  cube  constructed  upon  a  line  may  be  indicated 
by  the  cube  of  the  line. 

518.  Solids  which  have  the  same  form  are  similar;  those  which 
have  the  same  volume  are  equivalent;  and  those  which  have  the 
same  form  and  volume  are  equal. 

Proposition  I 

519.  1.  Form  *  a  prism.  Since  the  faces  are  parallelograms,  how  does 
each  face  coi'npare  with  a  rectangle  having  the  same  base  and  altitude  ? 
Considering  a  lateral  edge  as  the  base  of  each,  how  does  the  sum  of  the  alti- 
tudes compare  with  the  perimeter  of  the  right  section  ?  Since  the  lateral 
edges  are  equal,  how  does  the  lateral  surface  of  a  prism  compare  with 
the  rectangle  of  its  lateral  edge  and  the  perimeter  of  a  right  section? 

2.   To  what  rectangle  is  the  lateral  surface  of  a  right  prism  equivalent? 

Theorem,  The  lateral  surface  of  a  prism  is  equivalent 
to  the  rectangle  formed  hy  a  lateral  edge  and  the  perime- 
ter of  a  right  section. 


Data :  Any  prism,  as  AD\  of  which  AA^  is  a 
lateral  edge,  and  FGHJK  any  right  section. 
To  prove  lateral  surface  of 

AD'  ^  rect  AA' .  (FG  -i-GH-}-  etc.). 


Proof.    §  505,   AB',  BCf,  cd\  etc.,  are  parallelograms, 
data,  FGHJK  is  a  right  section  ; 

.-.  §  443,  FG  A.  AA',  GH±  BE',  HJ  A.  CC',  etc. 

Now  the  lateral  surface  oi  AD'  ^  AB'  +  BC'  +  etc. ; 

*  Objective  representations  of  the  solids  referred  to  in  this  and  the  follow- 
ing books  will  aid  the  student  very  greatly  in  acquiring  the  correct  geomet- 
rical concepts.  Solids  made  from  wood  or  glass  may  be  procured,  but  it  will 
be  far  better  for  the  student  to  form  them  for  himself  from  some  plastic  mate- 
rial, like  clay  or  putty.  He  can  then  cut  them  readily  in  any  desired  manner 
with  a  thin-bladed  knife. 


SOLID   GEOMETRY.— BOOK   VIII.  277 

but,  §  331,  AB'  =0=  rect.  AA'  •  FO^ 

BC'  ^  rect.  BB'  -  GH,  etc.; 
AB'  -\-B(f  +  etc.  =0=  rect.  AA'  -  FG-\-  rect.  BB'  •  GH  +  etc., 
or  lat.  surf,  of  AD'  ^  rect.  aa'  •  FG  +  rect.  BB'  •  GH+  etc. 

But,  §  505,  AA'  =  £^'  =  cc'  =  etc. 

Hence,  lat.  surf,  of  AD'  =c=  rect.  AA'  -  (FG  -{-  GH-\-  etc.).        q.e.d. 

520.  Cor.  The  lateral  surface  of  a  right  prism  is  equivalent  to 
the  rectangle  formed  by  its  altitude  and  the  perimeter  of  its  base. 

Arithmetical  Rules :  To  be  framed  by  the  student.  §  339 

Proposition  II 

521.  1.  Form  a  prism;  cut  it  by  parallel  planes.  What  figures  are 
the  sections  made  by  these  planes?    How  do  they  compare? 

2.  How  does  any  section  of  a  prism  parallel  to  the  base  compare  with 
the  base  ? 

3.  How  do  all  right  sections  of  the  same  prism  compare  ? 

Theorem,  The  sections  of  a  prism  made  hy  parallel 
planes  are  equal  polygons. 

Data:  Any  prism,  as  PR,  cut  by  any  par- 
allel planes,  as  AD  and  FJ,  making  the  sec- 
tions ABODE  and  FGHJK. 

To  prove  ABODE  =  FGHJK. 

Proof.         §  463,  AB  II  FG,  BO  II  GH,  etc.; 

.-.  §  469,  Z  ABO  =  Z  FGH,  Z  BOD  =  Z  GHJj  etc. 

Also,  §  151,  AB  =  FG,  BO  =  GH,  etc. 

Then,  ABODE  and  FGHJK  are  mutually  equiangular  and  equi- 
lateral, and  one  can  be  applied  to  the  other  so  that  they  will  ex- 
actly coincide. 

Hence,  §  36,  ABODE  =  FGHJK.  q.e.d. 

522.  Cor.  I.  Any  section  of  a  prism  parallel  to  the  base  is  equal 
to  the  base. 

523.  Cor.  II.     All  right  sections  of  the  same  prism  are  equals 


278 


SOLID   GEOMETRY.  — BOOK    VIII. 


Proposition  III 

524.  1.  Form  two  prisms  such  that  three  faces  including  a  trihedral 
angle  of  one  are  equal  to  the  corresponding  faces  of  the  other  and  simi- 
larly placed  in  each  prism.     How  do  the  prisms  compare  ? 

2.  Form  two  truncated  prisms  such  that  three  faces  including  a  tri- 
hedral angle  of  one  are  equal  to  the  corresponding  faces  of  the  other 
and  similarly  placed  in  each  prism.     How  do  the  prisms  compare  ? 

3.  Form  two  right  prisms  having  equal  bases  and  equal  altitudes. 
How  do  they  compare? 

Theorem,  Two  prisms  are  equal,  if  three  faces  includ- 
ing a  trihedral  angle  of  one  are  equal  to  three  faces  in- 
cluding a  trihedral  angle  of  the  other,  each  to  each,  and 
these  faxes  are  similarly  placed. 

Data :  Any  two  prisms,  as  AJ 
and  A' J',  having  the  faces  AG, 
AD,  and  AK  equal  to  the  faces 
A'G\  A'n',  and  A'K',  each  to 
each,  and  similarly  placed. 

To  prove  AJ=A'J'.  ^* 

B  C  B'         C 

Proof.  Data,  the  face  angles  BAE,  BAF,  and  EAF  are  equal  to 
the  face  angles  B'A'e',  B'a'f',  and  E'a'f',  respectively; 

.-.  §  500,     trihedral  angle  ^-5^J^  =  trihedral  angle  A'-B'e'f'. 

Apply  prism  A'J^  to  AJ  so  that  the  faces  of  trihedral  Z  A'  shall 
coincide  with  the  equal  faces  of  the  trihedral  Z  A. 

Then,  the  points  O'  and  D'  fall  upon  C  and  B,  respectively,  and, 
§  505,  C'h'  and  n'j^  take  the  direction  of  CH  and  DJ,  respectively. 

Since  the  points  F',  G',  K*  coincide  with  F,  G,  K,  respectively, 
§  430,  the  planes  of  the  upper  bases  must  coincide. 

Then,  H^  coincides  with  H,  and  J'  with  J. 

Hence,  the  prisms  AJ  and  A^J^  coincide  in  all  their  parts ; 
that  is,  AJ=A'j'.  Q.E.i>. 

525.  Cor.  I.  Tico  tmncated  prisms  are  equal,  if  three  faces  in- 
cluding a  trihedral  angle  of  one  are  equal  to  three  faces  including  a 
trihedral  angle  of  the  other,  each  to  each,  and  these  faces  are  simi- 
larly placed. 


SOLID   GEOMETRY.  — BOOK    VIII.  279 

526.  Cor.  II.  Two  right  prisms  are  equal,  if  they  have  equal 
bases  aiid  equal  altitudes. 

Proposition  IV 

527.  Form  any  oblique  prism ;  on  a  base  equal  to  a  right  section  of 
the  oblique  prism  form  a  right  prism  whose  altitude  is  equal  to  a  lateral 
edge  of  the  oblique  prism.     How  do  these  prisms  compare  in  volume  ? 

Theorem,  An  oblique  prism  is  equivalent  to  a  right 
prism  which  has  its  base  equal  to  a  right  section  of  the 
oblique  prism,  and  its  altitude  equal  to  a  lateral  edge  of 
the  oblique  prisma. 

Data:  Any  oblique  prism,  as  AD^ ^  a  right  sec- 
tion of  it,  as  FGHJK]  and' a  lateral  edge,  as  AA'. 

To  prove  AD^  equivalent  to  a  right  prism  whose 
base  is  FGHJK  and  altitude  equal  to  AA\ 

Proof.  Produce  AA^  to  F^  making  FF^  =  AA', 
and  at  F'  pass  a  plane  perpendicular  to  FF'  cut- 
ting all  the  faces  of  AD'  produced  and  forming 
the  right  section  f'G'h'j'k'  parallel  to  FGHJK.         b       c 

Then,  §  521,     section  F'G'h'j'k'  =  section  FGHJK, 
and  FJ'  is  a  right  prism  whose  base  is  FGHJK  and  altitude  equal 
to  AA'. 

In  the  truncated  prisms  AJ  and  A'j*, 
§  505,  the  bases  AD  and  a'd'  are  equal. 

Const.,  AA'  =  FF',  and  BB'  =  GG' -, 

.-.  Ax.  3,  AF=A'F',  and  BG  =  B'g'. 

AB  and  FG  are  equal  and  parallel  to  A'b'  and  F'g',  respectively; 
and  A  FAB,  ABG,  etc.,  of  the  face  AG  are  equal  respectively  to 
Af'a'b',  A'b'G',  etc.,  of  the  face  A'g'-,  ^        Why? 

.-.  AG  and  A'g'  are  mutually  equiangular  and  equilateral,  and  one 
can  be  applied  to  the  other  so  that  they  will  exactly  coincide. 

Hence,  §  36,  AG  =  A'G'. 

In  like  manner  AK  may  be  proved  equal  to  A'K*. 

Hence,  §  525,  prism  AJ  =  prism  A'J^. 

Adding  to  each  the  prism  FD', 
then^  AD'=o=FJ'.  9.E.D. 


280 


SOLID   GEOMETRY.  — BOOK   VIII. 


Proposition  V 

528.    Form  any  parallelepiped.     How  do  the  opposite  faces  compare  ? 
In  what  direction  do  they  extend  with  reference  to  each  other? 

Theoretn.     The   opposite  fdces  of  a  parallelopiped   are 
equal  and  parallel. 

Data :   Any  parallelopiped,  as  AG,  and 
any  opposite  faces  of  AG,  2i^  AF  and  DG. 

To  prove  AF  and  DG  equal  and  parallel. 

Proof.    AB  il  DC,  and  BF  II  CG-,    Why  ? 
.-.  §  469,      Zabf=Zdcg. 

Also,  AB  =  DC,  and  BF=CG',   Why  ? 

AF=zDG, 
and,  §  469,  AF  II  DG. 


Why? 

Q.E.D. 


Proposition  VI 

629.  1.  Form  any  parallelopiped;  pass  planes  through  any  three 
pairs  of  diagonally  opposite  edges.  What  plane  figures  are  formed  by 
these  edges  and  the  intersections  of  these  planes  with  the  faces  of  the 
parallelopiped?  How  do  the  diagonals  of  these  parallelograms  corre- 
spond with  the  diagonals  of  the  parallelopiped?  How  do  the  segments  of 
each  diagonal  compare  in  length  ? 

Theorem.  The  diagonals  of  a,  parallelopiped  bisect  each 
other. 

Data :  Any  parallelopiped,  as  AG, 
whose  diagonals  are  AG,  BH,  CE,  and 
DF. 

To  prove  that  AG,  BH,  CE,  and  DF 
bisect  each  other. 

Proof.    Through  the  opposite  edges 
AE  and  CG  pass  a  plane. 
§  505,  AE  and  CG  are  equal  and  parallel ; 

ACGE  is  a  parallelogram ; 
and,  §  154,  diagonals  AG  and  CE  bisect  each  other  at  0. 

In  like  manner,  AG,  BH,  and  AG,  DF  also  bisect  each  other  at  0. 

Hence,  AG,  BH,  CE,  and  DF  bisect  each  other.  q.e.d. 

530.  Cor.  The  diagonals  of  a  rectangtdar  parallelopiped  are 
epial. 


SOLID   GEOMETRY.— BOOK    VIIL 


281 


Proposition  VII 

531.  1.  Form  two  rectangular  parallelopipeds  whose  bases  are  equal 
and  whose  altitudes  are  in  the  ratio  of  2:3,  or  any  other  ratio.  How 
does  the  ratio  of  their  volumes  compare  with  the  ratio  of  their  altitudes  ? 

2.  How  does  the  ratio  of  two  rectangular  parallelopipeds  having  two 
dimensions  in  common  compare  with  the  ratio  of  their  third  dimensions? 

Theorem.  Rectangular  parallelopipeds  which  have  equal 
bases  are  to  each  other  as  their  altitudes. 


T>A 


A  "  B 

Data:  Any  two  rectangular  parallelopipeds,  as  ^  and  B,  whose 
bases  are  equal  and  whose  altitudes  are  CD  and  EF  respectively. 

To  prove  A:B=CD:BF. 

Proof.  Suppose  the  altitudes  CD  and  BF  have  a  common 
measure  which  is  contained  in  CD  3  times  and  in  BF  5  times. 


Then, 


CD:BF=3:5. 


I 


Divide  CD  into  three  and  BF  into  five  equal  parts  by  applying 
this  common  measure  to  them,  and  through  the  several  points  of 
division  pass  planes  perpendicular  to  these  lines. 

§  462,  these  planes  are  parallel  to  each  other  and  to  the  bases 
of  A  and  B ; 

.-.  §§  523,  526,  A  is  divided  by  these  parallel  planes  into  three, 
and  B  into  five  equal  rectangular  parallelopipeds ; 
^:^  =  3:5. 

Hence,  A:  B=  CDiBF. 

By  the  method  of  limits  exemplified  in  §  327  the  same  may  be 
proved  when  the  altitudes  are  incommensurable. 

Therefore,  etc.  q.e.d. 

532.  Cor.  Rectangular  parallelopipeds  which  have  two  dimensions 
in  common  are  to  each  other  as  their  third  dimensions. 


282 


SOLID   GEOMETRY.  — BOOK    VIII. 


Proposition  VIII 

533.  1.  Form  two  rectangular  parallelepipeds  whose  altitudes  are 
equal  and  the  areas  of  whose  bases  are  in  the  ratio  of  2:3,  or  any  other 
ratio.  How  does  the  ratio  of  their  volumes  compare  with  the  ratio  of 
their  bases  ? 

2.  If  two  rectangular  parallelepipeds  have  one  dimension  in  common, 
how  does  the  ratio  of  their  volumes  compare  with  the  ratio  of  the  prod- 
ucts of  their  other  two  dimensions? 

Theorem,  Rectangular  parallelopipeds  which  have  equal 
altitudes  are  to  each  other  as  their  bases. 


h 

/  \    ^ 

/ 

h 

A""""A 
/  ^/ 

'_ i.../ 

/ 
/ 

A 

/ 

/ 
/ 

/ 

/ 

! 

/ 

/' 

L V 

Data :  Any  two  rectangular  parallelepipeds,  as  A  and  5,  which 
have  a  common  altitude,  as  ^,  and  the  dimensions  of  whose  bases 
are  d,  e,  and  m,  n,  respectively. 


To  prove 


A:Bz=d  X  e:m  X 


Proof.     Construct  a  third  rectangular  parallelopiped  C,  having 
the  altitude  ^,  and  the  dimensions  of  its  base  d  and  n. 


Then,  §  532, 
and 
hence,  §  287, 

Therefore,  etc. 


A'.C=e:ny 
C:B  =  d:m; 
A:B=:d  xe:m  xn. 


Q.B.J). 


534.  Cor.  Rectangular  parallelopipeds  which  have  one  dimension 
in  common  are  to  each  other  as  the  products  of  their  other  two  dimen- 
sions. 


SOLID   GEOMETRY.  — BOOK   VIII. 


283 


Proposition  IX 

535.  Form  any  two  rectangular  parallelopipeds,  and  also  a  third  one 
whose  base  is  equal  to  the  base  of  the  first  and  whose  altitude  is  equal 
to  that  of  the  second.  By  comparing  each  of  the  first  two  with  the 
third,  discover  how  the  ratio  of  their  volumes  compares  with  the  ratio  of 
the  products  of  their  three  dimensions. 

Theorem,  Rectangular  parallelopipeds  are  to  each  other 
as  the  products  of  their  three  dimensions. 


n 

/l^ 

^ 

Ac/ 

y\^ 

X 

n\       ! 

\y 

y' 

ym 

/ 

Data :  Any  two  rectangular  parallelopipeds,  as  A  and  B,  whose 
dimensions  are  d,  e,  /,  and  I,  m,  n,  respectively. 

To  prove  A:B  =  dxexf:lxmxn. 

Proof.     Construct  a  third  rectangular  parallelopiped  C,  having 
the  dimensions  d,  e,  and  n. 

Then,  §  532,  A'.C  =  f:n, 

and,  §  534,  C :  B  =  d  x  e:l  x  m-^ 

hence,  §  287,  A.B  =  dxexf'.lxmxn. 

Therefore,  etc.  q.e.d. 


Proposition  X 

536.  Form  any  rectangular  parallelopiped,  and  a  cube,  whose  edge  is 
some  linear  unit,  as  a  unit  of  measure.  Since  the  ratio  of  these  solids  is 
equal  to  the  ratio  of  the  products  of  their  three  dimensions,  find  the 
measure  of  the  volume  of  the  parallelopiped  in  terras  of  its  three  dimen- 
sions. 

Theorem.  The  volume  of  a  rectangular  parallelopiped 
is  equal  to  the  product  of  its  three  dimensions. 


284 


SOLID   GEOMETRY.— -BOOK  VIII, 


Data:  Any  rectangular  parallelo- 
piped,  as  A,  whose  dimensions  are 
d,  e,  and  /. 

To  prove 

volume  oi  A  =  dxe  xf. 


^ 

A 

M 

A  1^ 

y^ 

1 

} 

-} 

i 

1 

1 

Proof.     Assume  that  the  unit  of  volume  is  a  cube,  if,  whose 
edge  is  the  linear  unit. 

Then,  §  535,        ^  :3f  =  (Z  x  e  x/:  1  X  1  X  1, 


or 


JJf     1  xl  xl  "^ 


But,  §  516,  the  volume  of  A  is  measured  by  the  number  of 
times  it  contains  the  unit  of  measure,  M\ 


M 


=  volume  of  A, 


But 


~=dxex/ 


Hence,     volume  ot^  =  dxex/ 
Therefore,  etc. 


Q.E.D. 


537.   Cor.     Tlie  volume  of  a  rectangular  paraUelopiped  is  equal 
to  the  product  of  its  base  by  its  altitude. 


Proposition  XI 

538.  1.  Form  any  oblique  parallelepiped.  How  does  it  compare  in 
volume  with  a  rectangular  parallelepiped  having  an  equivalent  base  and 
the  same  altitude  ? 

2.  What,  then,  is  the  measure  of  the  volume  of  any  parallelepiped  in 
terras  of  its  base  and  altitude  ? 

Theorem.  Any  parallelepiped  is  equivalent  to  a  rec- 
tangular parallelepiped  having  the  same  altitude  and  an 
equivalent  base. 


SOLID   GEOMETRY.  — BOOK  VIII. 


285 


Data :  Any  parallelepiped,  as  A&,  whose  base  is  ABCD. 
To  prove  AC^  equivalent  to  a  rectangular  parallelopiped  having 
the  same  altitude  and  a  base  equivalent  to  ABCD. 


Proof.  On  AB  produced  take  EF  equal  to  AB,  and  through  E 
and  F  pass  planes  _L  EF,  as  EH^  and  FG'.  Produce  the  faces  AC, 
a'g',  AB',  and  DC'  to  intersect  the  planes  eh'  and  FG',  forming  the 
right  parallelopiped  EG'. 

Then,  §  527,  AC'  =0=  EG'. 

On  HE  produced  take  MJ  equal  to  HE,  and  through  M  and  J 
pass  planes  ^MJ,  as  ML'  and  Jit.  Produce  the  faces  EG,  E'g', 
eh',  and  FG'  to  intersect  the  planes  ML'  and  JK',  forming*  the 
right  parallelopiped  JL'. 

Then,  §  527,  EG'=(>JL'; 

AC'oJL'. 

Const.,  EF  =  AB,  and  AF  II  i)G! ; 

.-.§333,  EFGHo^ABCD. 

Also,  const.,  ifJ"  =  HE,  and  ^J  II  GS'; 

JELM  o  ^i^^G^Zf  =0=  ^5C2).  Why  ? 

Const.,  plane  ^'g'j'  II  plane  AGJ,  and  hence  the  three  solids 
have  the  same  altitude. 

Const.,  §  482,  faces  EH^  and  FG'  are  perpendicular  to  AGJ-, 
hence,  faces  JM^  and  ElJ  are  perpendicular  to  AGJ. 

Also,  const.,  §  482,  faces  ML'  and  Jit  are  perpendicular  to  AGJ-, 
the  faces  of  JlJ  are  rectangles; 
hence,  §  514,      JL'  is  a  rectangular  para-llelopiped. 

But  AC'  =0=  JL',  and  JKLM  =c=  ^5CZ>. 


286  SOLID   GEOMETRY.— BOOK  VIII. 

Hence,  ACf  is  equivalent  to  a  rectangular  parallelopiped  having 
the  same  altitude  and  a  base  equivalent  to  ABCD. 

Therefore,  etc.  q.e.d. 

539.  Cor.  The  volume  of  any  parallelopiped  is  equal  to  the 
product  of  its  base  by  its  altitude. 

Proposition  XII 

540.  Form  a  parallelopiped  ;  divide  it  into  two  triangular  prisms  by 
a  plane  passing  through  two  diagonally  opposite  edges.  How  do  these 
prisms  compare  in  volume?  What,  then,  is  the  volume  of  a  triangular 
prism  in  terms  of  its  base  and  altitude  ? 

Form  any  prism ;  divide  it  into  triangular  prisms  by  planes  through 
a  lateral  edgCo  What  is  the  volume  of  each  triangular  prism?  What, 
then,  is  the  volume  of  any  prism  ? 

Theorem.  The  plane  passed  through  two  diagonally  op- 
posite edges  of  a  parallelopiped  divides  it  into  two  equiva- 
lent triangular  prisms. 

H — ^ 

Data:  Any  parallelopiped,  as  AG,  and       ^(^i^ ^ P'/ 

a  plane,  as  ACGE,  passing  through  two      AuzA- ---riiiJr.Ji 

diagonally  opposite  edges,  as  AE  and  GG.       I    j  ~^^d/^ 

To  prove  prism  ABC-F  =o  prism  ADC-H.      /  dI -L-Jo 

A  B 

Proof.  Through  the  parallelopiped  AG  pass  a  plane  forming 
the  right  section  JKLM  and  intersecting  ACGE  in  JL. 

§  528,  AF  II  DG,  and  AH  II  5G ; 

hence,  §  463,  JK  II  ML,  and  JM  II  KL ; 

JKLM  is  a  parallelogram,  and  JL  is  its  diagonal. 

Hence,  A  JKL  =  A  JML.  Why  ? 

Now,  §  527,  prism  ABC-F  is  equivalent  to  a  right  prism  whose 
base  is  JKL,  and  whose  altitude  is  AE; 

also  prism  ADC-H  is  equivalent  to  a  right  prism  whose  base  is 
JML,  and  whose  altitude  is  AE. 

But,  §  526,  these  two  right  prisms  are  equal ; 
hence,  ABC-F  =0=  ADC-H. 

Therefore,  etc.  q.b.d. 


SOLID   GEOMETRY.  — BOOK  VIII.  287 

541.  Cor.  I.     A  triangular  prism  is  equivalent  to  one  half  of  a 
paralldopiped  having  the  same  altitude  and  a  base  twice  as  great. 

542.  Cor.  II.     The  volume  of  a  triangular  prism  is  equal  to  the 
product  of  its  base  by  its  altitude.  ^^<^^ 

543.  Cor.  III.     The  volume  of  any  prism  is  equal        /  //'    j 
to  the  product  of  its  base  by  its  altitude.  j  j  /     I 

544.  Cor.  IV.     Prisms  are  to  each  other  as  the    1 1-^-  J 
products  of  their  bases  by  their  altitudes;   conse-  \f  '~~'v 
quently,  prisms  which  have  equivalent  bases  are  to  each  other  as 
their  altitudes;  prisms  which  have  equal  altitudes  are  to  each  other 
as  their  bases;  and  prisms  which  have  equivalent  bases  and  equal 
altitudes  are  equivalent. 

PYRAMIDS 

545.  A, polyhedron  whose  base  is  a  polygon,  and  whose  lateral 
faces  are  triangles  which  have  a  common  vertex,  is 

called  a  Pyramid.  ^ 

.  The  common  vertex  of  the  triangles  which  form 
the  lateral  faces  of  a  pyramid  is  called  the  vertex     / 
of   the  pyramid;   the  lines   in  which  the  lateral    / 
face?^  intersect  are  called  the  lateral  edges;  the  sum  V 

of  the  lateral  faces  is  called  the  lateral,  or  convex       

surface;  and  the  sum  of  the  areas  of  the  lateral  faces  is  called 
the  lateral  area  of  the  pyramid. 

The  perpendicular  distance  from  the  vertex  of  a  pyramid  to 
the  plane  of  its  base  is  its  altitude. 

546.  A  pyramid  is  called  triangular,  quadrangular,  etc.,  accord- 
ing as  its  base  is  a  triangle,  quadrilateral,  etc.  ^ 

547.  A  pyramid  whose  base  is  a  regular  polygon, 
and  whose  vertex  lies  in  the  perpendicular  to  the 
base  erected  at  its  center,  is  called  a  Regular  Pyra- 
mid. I 

Since  a  regular  polygon  may  be  inscribed  in  a    ^ ^^ 

circle,  it  is  evident,  from  §  450,  that  the  vertex  of  a  regular  pyrar 
mid  is  equidistant  from  the  vertices  of  the  polygon  forming  its 


288  SOLID    GEOMETRY.— BOOK   VIII. 

base ;  and  hence  the  lateral  edges  of  a  regular  pyramid  are  equal, 
and  its  lateral  faces  are  equal  isosceles  triangles. 

548.  The  perpendicular  distance  from  the  vertex  of  a  regular 
pyramid  to  the  base  of  any  one  of  its  lateral  faces  is  called  the 
Slant  Height  of  the  pyramid. 

The  slant  height  is  therefore  the  altitude  of  the  equal  isosceles 
triangles  which  form  the  lateral  faces  of  the  pyramid. 

549.  The  part  of  a  pyramid  included  between  its  base  and 
a  plane  which  cuts  all  its  lateral  edges  is  called  a  Truncated 
Pyramid.  

550.  A  truncated  pyramid  whose  bases  are  par-     /T^^f^  \ 
allel  is  called  a  Frustum  of  a  pyramid.  /    ;.  .  I   A  \ 

The  perpendicular  distance  between  the  bases  of  //'  I  \ 
a  frustum  is  its  altitude.  ^^>^^   1  ^/^ 

The  lateral  faces  of  the  frustum  of  a  regular  ^^1^ 

pyramid  are  equal  isosceles  trapezoids,  and  the  common  altitude 
of  these  trapezoids  is  the  slant  height  of  the  frustum. 

Proposition  XIII 

551.  1.  Form  a  regular  pyramid.  Since  the  lateral  faces  are  triangles, 
how  does  each  compare  with  the  rectangle  having  the  same  base  and 
altitude  ?  How  does  the  altitude  of  each  compare  with  the  slant  height 
of  the  pyramid  ?  How  does  the  sum  of  the  bases  of  the  lateral  faces  com- 
pare with  the  perimeter  of  the  base  of  the  pyramid?  How,  then,  does 
the  lateral  surface  of  a  regular  pyramid  compare  with  the  rectangle  of 
the  perimeter  of  its  base  and  its  slant  height  ? 

2.  Form  the  frustum  of  a  regular  pyramid.  What  plane  figures  are 
its  faces?  To  what,  then,  is  the  surface  of  each  equivalent?  How 
does  the  sum  of  the  upper  bases  of  the  faces  compare  with  the  perim- 
eter of  the  upper  base  of  the  frustum?  The  sum  of  the  lower  bases  of 
the  faces  with  the  perimeter  of  the  lower  base  of  the  frustum  ?  How, 
then,  does  the  lateral  surface  of  the  frustum  of  a  regular  pyramid  com- 
pare with  the  rectangle  of  its  slant  height  and  the  sum  of  the  perimeters 
of  its  bases  ? 

Theorem,  The  lateral  surface  of  a  regular  pyramid  is 
equivalent  to  one  half  the  rectangle  formed  hy  the  perim- 
eter of  its  hose  and  its  slant  height. 


SOLID   GEOMETRY.  — BOOK  VIII.  289 


Data:    Any  regular  pyramid,  as   Q-ABCDE,  //  '\\ 

whose  slant  height  is  QH.  //  j  \  \ 

To  prove  lateral  surface  of  /  / 1  [e\   \ 

Q-ABCDE  =G=  i  rect.  (AB  +  BC  +  etc.)  .  QH,  ^/'  / 7         \    \d 

i^ ^ 

Proof.     §  545,  lat.  surf,  of  Q-ABCDE  ^  A  QAB  +  A  QBC  -\-  etc., 
and,  since,  §  548,  the  altitudes  of  these  triangles  =  QH, 

A  QAB  ^  i  rect.  AB  •  QH,  Why  ? 

A  QBC^^  rect.  BC  •  Qfl,  etc. ; 
.-.  A  QAB  +  A  QBC-\-  etc.  =0=  i  rect.  AB  -  QH+^  rect.  5C  •  Qfl-^-  etc., 
or        lat.  surf,  of  Q-ABCDE  ^  ^  rect.  {AB  +  BC  +  etc.)  •  QH. 
Therefore,  etc.  q.e.d. 

552.  Cor.  The  lateral  surface  of  a  frustum  of  a  regular  pyra- 
mid is  equivalent  to  one  half  the  rectangle  formed  by  its  slant  height 
and  the  sum  of  the  perimeters  of  its  bases. 

Arithmetical  Rules :  To  be  framed  by  the  student.  §  339 

Ex.  779.  The  perimeter  of  the  base  of  a  regular  pyramid  is  14  ft.  and 
its  slant  height  is  6  ft.    What  is  its  lateral  area  ? 

Proposition  XIV 

553.  1.  Form  a  pyramid ;  cut  it  by  a  plane  parallel  to  its  base.  How 
do  the  ratios  of  the  segments  of  the  lateral  edges  compare  with  each 
other,  and  with  the  ratio  of  the  segments  of  the  altitude? 

2.    Is  the  section  equal,  equivalent,  or  similar  to  the  base  ? 

Theorem,  If  a  pyramid  is  cut  hy  a  plane  parallel  to  its 
base, 

1.  The  lateral  edges  and  the  altitude  are  divided  propor- 
tionally. 

2.  The  section  is  a  polygon  similar  to  the  base, 

milnb's  geom.  —  la 


290 


SOLID   GEOMETRY.  — BOOK  VIIL 


Data:  Any  pyramid,  as  Q-ABCDE,  wtose  altitude  is  QO,  and 
any  plane  parallel  to  the  base,  as  MJSf,  which  cuts  the  pyramid  in 
the  section  FGHJK,  and  the  altitude  in  P. 


To  prove    1.    QF .  qa=  QG:  QB  =  QP:  Q0  =  etc. 
2.  FGHJK  and  ABCDE  similar. 

Proof.     1.   Through  Q  pass  a  plane  parallel  to  ABCDE. 

Then,  the  lateral  edges  and  the  altitude  are  intersected  by  three 
parallel  planes ; 
.-.  §  470,  QF :  QA  =  QG:  QB=  QP:  QO,  etc. 

2.    §  463,  FG  II  AB,  GH  11  BC,  HJ  II  CD,  etc. ; 

.  §  469,  Z  FGH  =  Z  ABC,  Z  GHJ  =  Z  BCD,  etc. 

Also,  §  306,  A  QFG,  QGH,  etc.,  are  similar  to  A  Q^5,  QBC,  etc., 
each  to  each ; 

FG:AB=QG:  QB,  and  (?^:  BC=  QG:  QB] 
hence,  FG:  AB  =  GH:  BC. 

In  like  manner,  GH:  BC  —  HJ:  CD,  etc. 

Hence,  FGHJK  and  ABCDE  are  mutually  equiangular  and  have 
their  homologous  sides  proportional ; 
that  is,  §  299,    FGHJK  and  ABCDE  are  similar. 

Therefore,  etc.  q.e.d. 

554.    Cor.  I.     Parallel  sections  of  a  pyramid  are  to  each  other  as 
the  squares  of  their  distances  from  the  vertex. 

For,  §  344,  FGHJK :  ABCDE  =  F^  :  i? ; 

but,  §  553,  FG  :  AB  =  QG:  QB  =  QP  :  QO; 

bence,  FGHJK :  ABCDE  =  QF' :  off. 


SOLID   GEOMETRY.  — BOOK  VIII.  291 

655.  Cor.  II.  If  two  pyramids  have  equal  altitudes,  sections 
'parallel  to  their  bases  and  equally  distant  from  their  vertices  are  to 
each  other  as  the  bases. 

For,  §  554,         FGHJK :  ABODE  =  Qp  :  Q(?, 
and  F'G'H' :  A^B^C'  =  Q^F^  :  Q^l 

But  QP  =  Q'P',  and  QO  =  Q'O' ; 

FGHJK :  AB ODE  =  F'G'H'  :  A'B'&j 
or  FGHJK :  F' G'H'  =  ABODE  :  A'B'C'.  Why  ? 

556.  Cor.  III.  If  two  pyramids  have  equal  altitudes  and  equiva- 
lent bases,  p^.ctions  parallel  to  their  bases  and  equally  distant  from 
their  vertices  are  equivalent. 

Proposition  XV 

657.  Form  two  triangular  pyramids  which  have  equivalent  bases  and 
equal  altitudes.     How  do  they  compare  in  volume? 

Theorern,  Triangular  pyramids  which  have  equivalent 
bases  and  equal  altitudes  are  equivalent. 


Data:  Any  two  triangular  pyramids,  as  Q-ABO  and  T^DEF, 
with  equivalent  bases  ABC  and  DEF,  and  equal  altitudes. 

To  prove  Q-ABO  ^  t-def. 

Proof.  Divide  the  equal  altitudes  into  equal  parts,  each  n  units 
long,  and  through  the  points  of  division  pass  planes  parallel  to 
ABO  and  DEF  respectively. 

By  §  556,  the  corresponding  sections  of  the  two  pyramids, 
formed  by  these  planes,  are  equivalent. 


292  SOLID   GEOMETRY.— BOOK  VIII. 

If  the  pyramids  are  not  equivalent,  suppose  Q-ABC  is  the 
greater.  On  its  base,  and  on  each  section  as  a  lower  base,  construct 
a  prism  with  lateral  edges  parallel  to  ^  Q  and  altitude  equal  to  n. 


B  E 

On  each  section  of  T-def,  as  an  upper  base,  construct  a  prism 
with  lateral  edges  parallel  to  DT  and  altitude  equal  to  n. 

Then,  §  544,  each  prism  in  T-DEF  is  equivalent  to  the  prism 
next  above  it  in  Q-ABC ;  consequently,  the  difference  between  the 
two  sets  of  prisms  is  the  lowest  prism  of  the  first  set. 

Now,  if  ?i  is  decreased  indefinitely,  the  lowest  prism  is  decreased 
indefinitely,  and  the  difference  between  the  two  sets  of  prisms 
may  be  made  less  than  any  assigned  volume,  however  small. 

But  the  sum  of  all  the  prisms  of  the  first  set  is  greater  than 
Q-ABC  and  the  sum  of  all  the  prisms  of  the  second  set  is  less 
than  T-DEF',  therefore,  the  difference  between  Q-ABC  smd  T-DEF 
is  less  than  the  difference  between  the  two  sets  of  prisms,  and 
consequently,  less  than  any  assigned  volume,  however  small. 

Hence,  Q-ABC  =o  T-DEF. 

Therefore,  etc.  q.e.d. 

Proposition  XVI 

558.  1.  Form  any  triangular  prism  and  divide  it  into  three  triangular 
pyramids.  How  do  the  pyramids  compare  with  each  other?  To  what 
part,  then,  of  the  prism  is  the  pyramid,  which  has  the  same  base  and 
altitude,  equivalent  ? 

2,  Form  any  pyramid ;  divide  it  into  triangular  pyramids  by  planes 
through  a  lateral  edge.  What  is  the  volume  of  each  triangular  pyramid? 
What,  then,  is  the  vohime  of  any  pyramid? 

Theorem..  A  triangular  pyramid  is  equivalent  to  one 
third  of  a  triangular  prism  which  has  tJie  same  base  and 
altitude. 


SOLID   GEOMETRY.-^  BOOK  VIII. 


293 


Data:   Any  triangular  pyramid,  as   Q-ABC,  and   a  triangular 
prism,  as  DQE~ABC,  which  has  the  same  base  and  altitude. 


To  prove  Q-ABC  =0=  |  D QE-ABC. 

Proof.  Through  QC  and  QD  pass  a  plane  intersecting  the 
parallelogram  ACED  in  CD.  Then,  D QE-ABC  is  composed  of  the 
three  triangular  pyramids  Q-ABC,  Q-ACD,  and  Q-CDE. 

AACD  =  ACDEy  Why? 

that  is,  Q-ACD  and  Q-CDE  have  equal  bases,  and  the  same  altitude ; 

.-.   §  557,  Q-ACD  ^  Q-CDE. 

Regarding  C  as  the  vertex  and  DQE  as  the  base  of  Q-CDE, 
§  505,  Q-ABC  and  C-DQE  have  equal  bases  and  equal  altitudes; 

.-.   §  557,  Q-ABC  =0  C-DQE,  or  Q-CDE', 

consequently,  Q-ABC  =0=  Q-ACD  =0=  Q-CDE. 


Hence, 


Q-ABC  =0=  i  D  QE-ABC. 


Q.E.D. 


559.  Cor.  I.     The  volume  of  a  triangular  pyramid  is  equal  to 
one  third  the  product  of  its  base  by  its  altitude. 

560.  Cor.  II.  The  volume  of  any  pyramid  is 
equal  to  one  third  the  product  of  its  base  by  its 
altitude. 

561.  Cor.  III.  Pyramids  are  to  each  other  as 
the  products  of  their  bases  by  their  altitudes;  con- 
sequently, pyramids  which  have  equivalent  bases 

are  to  each  other  as  their  altitudes;  pyramids  which  have  equal 
altitudes  are  to  each  other  as  their  bases;  and  pyramids  which  have 
equivalent  bases  and  equal  altitudes  are  equivalent. 


294 


SOLID   GEOMETRY,  — BOOK   VIIL 


Proposition  XVII 

562.  Form  the  frustum  of  a  triangular  pyramid ;  divide  it  into  three 
triangular  pyramids  one  of  which  shall  have  for  its  base  the  lower  base 
of  the  frustum,  another  the  upper  base,  and  both  the  altitude  of  the 
frustum  for  their  altitude.  It  will  be  shown  that  the  third  pyramid  is 
equivalent  to  a  pyramid  whose  altitude  is  the  altitude  of  the  frustum 
and  whose  base  is  a  mean  proportional  between  the  bases  of  the  frustum. 

To  the  sum  of  what  triangular  pyramids,  then,  is  the  frustum  of  a 
triangular  pyramid  equivalent? 

Theorem,  A  frustum  of  a  triangular  pyramid  is  equiv- 
alent to  the  sum  of  three  pyramids  of  the  same  altitude  as 
the  frustum,  and  whose  bases  are  those  of  the  frustum  and 
a  mean  proportional  between  them. 


Data :  A  frustum  of  any  triangular  pyramid,  as  ABC-DEF,  whose 
bases  are  ABC  and  DEF. 
Denote  its  altitude  by  H. 

To  prove  ABC-DEF  equivalent  to  the  sum  of  three  pyramidc* 
whose  common  altitude  is  H  and  whose  bases  are  ABC,  VEF,  ana 
a  mean  proportional  between  them.  * 

Proof.  Through  the  points  A,  E,  C,  and  B,  E,  C  pass  planes, 
thus  dividing  the  frustum  into  the  three  pyramids  E-ABC,  C-BEF, 
and  E-ADC. 

Then,  E-ABC  and  C-DEF  have  the  bases  ABC  and  DEF,  respec- 
tively, and  the  common  altitude  H. 

It  remains  to  prove  E-ADC  equivalent  to  a  pyramid  whose  alti- 
tude is  H  and  whose  base  is  a  mean  proportional  between  ABC  and 
DEF, 


SOLID   GEOMETRY.  — BOOK  VIII.  295 

In  the  face  DB,  draw  EJ II  DA,  and  pass  tlie  plane  EJC-,  also  draw  JD. 

§  457,  EJ  II  plane  ACFD ; 

hence,       iJ  and  J  are  equally  distant  from  plane  ACFD-, 

E-AD  C  ^  J- AD  C.  Why  ? 

Kow,  D  may  be  regarded  as  the  vertex,  and  AJC  as  the  base  of 
J- ADC. 

Then,  E-ADC  is  equivalent  to  D-AJC,  whose  altitude  is  H. 

Draw  JK  II  ^F. 

Then,  §  469,    Z  ^J^  =  Z  DEF,  Z  e/^ZT  =  Z  EDF, 
and,  §  151,  AJ=DE; 

AAJK  =  Adef,  Why? 

and  AK  =  DF.  Why  ? 

Now  the  altitudes  of  A  ^5C  and  ^JC  on  AB  are  equal, 
and  thte  altitudes  of  A  ^J'C  and  AJK  on  ^c  are  equal; 
•.  §  336,  A  ABC :  A  AJC  =  AB:AJ=AB:  DE, 

and  AAJC:AAJK=AC:AK=AC:DF, 

But,  §  553,  A  ABC  and  D-E^i^  are  similar ; 

hence,  AB:DE  =  AC :  DF ;  Why  ? 

AABC:AAJC  =  AAJC:  AAJK. 

But  A  AJK  =  A  DEF; 

AABG:AAJC  =  AAJC:  ADEF; 
that  is,    A  ^JC  is  a  mean  proportional  between  A  ABC  and  DEF. 

Hence,  ABC-DEF  is  equivalent  to  the  sum  of  three  pyramids, 
whose  common  altitude  is  H  and  whose  bases  are  ABC,  DEF,  and 
a  mean  proportional  between  them. 

Therefore,  etc.  q.e.d 

563.  Cor.  I.     A  frustum  of  any  pyramid  /^^v^^^^^-^ 
is  equivalent  to  the  sum  of  three  pyramids  of  I  j  ^^\\ 
the  same  altitude  as  the  frustum  and  whose  I           \   \ 
bases  are  those  of  the  frustum  and  a  mean  \- {-----     \--^\ 
proportional  between  them.  \j      ^^^-^\^y^ 

564.  Cor.  II.  The  volume  of  a  frustum  of  any  pyramid  is  equal 
to  one  third  the  product  of  its  altitude  by  the  sum  of  its  bases  and  a 
mean  proportional  between  them. 


296  SOLID   GEOMETRY,  — BOOK   VIIL 


Proposition  XVIII 

565.  Form  a  truncated  triangular  prism  and  through  one  of  its  upper 
vertices  pass  planes  dividing  it  into  three  triangular  pyramids.  Since 
any  face  of  a  pyramid  may  be  considered  as  its  base,  discover  whether 
each  one  of  these  pyramids  is  equivalent  to  some  one  of  the  three  pyra- 
mids whose  common  base  is  the  base  of  the  prism  and  whose  vertices  are 
the  three  vertices  of  the  inclined  section. 

Theorem,  A  truncated  triangular  prism  is  equivalent  to 
the  sum  of  three  pyramids  whose  comm^on  base  is  the  base 
of  the  prism,  and  whose  vertices  are  the  three  vertices  of  the 
inclined  section. 


Dit 


Data :  Any  truncated  triangular  prism, 
as  ABC-DEF,  and  the  three  pyramids 
E-ABC,   D-ABC,  and  F-ABC. 

To  prove  ABC-DEF  ^  E-ABC  -f  D-ABC 

+  F-ABC.  ^^ 


Proof.  Pass  planes  through  E,  A,  C,  and  E,  D,  C,  dividing  the 
prism  into  the  pyramids  E-ABC,  E-ACD,  and  E-DCF. 

D-ABC  may  be  regarded  as  having  ACD  for  its  base  and  B  for 
its  vertex. 

Now,  §  505,  AD  II  BE  II  CF',  and  hence,  §  457,  BE  II  plane  ACD-, 

B-ACD  and  E-ACD  have  equal  altitudes; 
hence,  §  557,  D-ABC  =o  B-ACD  o  E-ACD. 

F-ABC  may  be  regarded  as  having  ACF  for  its  base  and  B  for  its 
vertex ; 

but  A  ACF  <>  A  DCF,  and  BE  II  plane  DCF ;  Why  ? 

hence,  F-ABC  =o=  B-A  CF  <)=  E-D CF ;  Why  ? 

E-ABC  4-  D-ABC  +  F-ABC  ^  E-ABC  +  E-ACD  +  E-DCF. 

But  ABC-DEF  ^  E-ABC  +  E-ACD  +  E-DCF ; 

hence,  ABC-DEF  o=  E-ABC  +  D-ABC  +  F-ABC. 

Therefore,  etQ.  (^.e.v. 


SOLID   GEOMETRY.  — BOOK  VIII. 


297 


566.  Cor.  I.  The  volume  of  a  truncated  right  triangular  prism 
is  equal  to  the  product  of  its  base  by  one  third  the  sum  of  its 
lateral  edges. 

1.  What  is  the  direction  of  the  lateral  edges 
AD,  BE,  CF  with  reference  to  the  base  ABC? 

2.  How,  then,  do  AD,  BE,  and  CF  (compare 
with  the  altitudes  of  the  three  pyramids  whose 
sum  is  equivalent  to  ABC-BEF  ? 

3.  To  what  is  the  volume  of  each  of  these 
pyramids  equal  ? 

4.  To  what,  then,  is  the  volume  of  ABC-DEF  equal  ? 

567.  Cor.  II.     The  volume  of  any  truncated  triangular  prism  is 
equal  to  the  product  of  its  right  section 
by  one  third  the  sum  of  its  lateral  edges. 

1.  If  GHK  is  a  right  section,  to  what 
is  the  volume  of  GHK-DEF  equal  ? 

2.  To  what  is  the  volume  of  GHK- 
ABC  equal  ? 

3.  To  what,  then,  is  the  volume  of 
ABC-DEF  equal? 

Ex.  780.  What  is  the  lateral  area  of  a  right  prism  whose  altitude  is  12  in. 
and  the  perimeter  of  whose  base  is  20  in.? 

Ex.  781.  Find  the  ratio  of  two  rectangular  parallelopipeds,  if  their  alti- 
tudes are  each  7™  and  their  bases  3'"  by  4"'  and  7™  by  9™,  respectively. 

Ex.  782.  Find  the  ratio  of  two  rectangular  parallelopipeds,  if  their  di- 
mensions are  2dn\  4dm,  3dm,  and  6^™,  7'^",  S*!*",  respectively. 

Ex.  783.  What  is  the  volume  of  a  rectangular  parallelepiped  whose  edges 
are  20.5'",  12.75™,  and  8.6™,  respectively  ? 

Ex.  784.  The  altitude  of  a  regular  hexagonal  prism  is  12  ft.,  and  each 
side  of  its  base  is  10  ft.     What  is  its  volume  ? 

Ex.  785.  What  is  the  volume  of  a  pyramid  whose  altitude  is  18<i™  and 
whose  base  is  a  rectangle  10^™  by  6<i™  ? 

Ex.  786.  What  is  the  volume  of  a  truncated  right  triangular  prism,  if 
each  side  of  its  base  is  3  ft.  and  its  edges  are  3  ft.,  4  ft.,  and  6  ft.,  respectively  ? 

Ex.  787.  What  is  the  lateral  area  of  the  frustum  of  a  square  pyramid 
whose  slant  height  is  13™,  each  side  of  the  lower  base  being  3.5™,  and  of  the 
upper  base  2™  ? 


298 


SOLID  GEOMETRY.^BOOK   VIIL 


Proposition  XIX 

568.  Form  two  tetrahedrons  which  have  a  trihedral  angle  of  one 
equal  to  a  trihedral  angle  of  the  other.  Considering  homologous  faces 
of  these  trihedrals  as  bases  of  the  tetrahedrons,  how  does  the  ratio  of 
their  volumes  compare  with  the  ratio  of  the  products  of  their  bases  by 
their  altitudes?  (§  561)  How  does  the  ratio  of  their  bases  compare 
with  the  product  of  the  ratios  of  the  homologous  edges  which  include 
the  basal  face  angles  of  the  equal  trihedrals?  (§  340)  How  does  the 
ratio  of  the  altitudes  of  the  tetrahedrons  compare  with  the  ratio  of  the 
third  edges  of  the  equal  trihedrals?  (§  299)  From  these  equal  ratios 
discover  how  the  ratio  of  the  volumes  of  the  tetrahedrons  compares  with 
the  ratio  of  the  products  of  the  three  edges  of  the  equal  trihedral  angles. 

Theorem,  Tetrahedrons  which  have  a  trihedral  angle  of 
one  equal  to  a  trihedral  angle  of  the  other  are  to  each  other 
as  the  products  of  the  edges  of  the  equal  trihedral  angle^^ 


Data :  Any  two  tetrahedrons,  as  Q-ABC  and  T-DEF,  which  have 
the  trihedral  angles  Q  and  T  equal. 

To  prove     Q-ABC :  T-DEF  =  QA  X  QB  X  QC :  TD  X  TE  X  TF. 

Proof.      Apply   T-DEF  to   Q-ABC  SO  that  the  equal  trihedral 
angles  T  and  Q  coincide. 

Draw  CO  and  FP  perpendicular  to  the  plane  QAB. 

Then,  their  plane  intersects  QAB  in  QPO. 

Now,  CO  and  FP  are  the  altitudes  of  the  triangular  pyramids 
G-QAB  and  F-QDE ; 
.-.    §  561,         C-QAB  :  F-QDE  =  QAB  X  CO  :  QDE  X  FP.  (1) 

But,  §  340,  QAB  :  QDE  =  QA  X  QB  :  QD  X  QE. 

Substituting  in  (1), 

C-QAB :  F-QDE  =  QA  X  QB  X  CO  :  QD  X  QE  X  FP.    (2) 


SOLID   GEOMETRY.  —  BOOK   VIII.  299 

Now,  rt.  A  QOC  and  QPF  are  similar ;  Why  ? 

.-.   §  299,  CO  :  FP  =  QG :  QF. 

Substituting  in  (2), 

C-QAB  :  F-QDE  =  QA  X  QB  X  QC :  QD  X  QE  X  QF] 
that  is,  Q-ABO :  T-BFF  =  QA  X  QB  X  QC :  TD  X  TE  X  TF. 

Therefore,  etc.  q.e.d, 

SIMILAR    AND   REGULAR   POLYHEDRONS 

569.  Polyhedrons  which  have  their  corresponding  polyhedral 
angles  equal,  and  have  the  same  number  of  faces  similar  each  to 
each,  and  similarly  placed,  are  called  Similar  Polyhedrons. 

Faces,  edges,  angles,  etc.,  which  are  similarly  placed  in  similar 
polyhedrons  are  called  homologous  faces,  edges,  angles,  etc. 

570.  Since  the  homologous  sides  of  similar  polygons  are  pro- 
portional, the  homologous  edges  of  similar  polyhedrons  are 
proportional. 

571.  Since  similar  polygons  are  proportional  to  the  squares 
upon  any  of  their  homologous  lines,  the  homologous  faces  of 
similar  polyhedrons  are  proportional  to  the  squares  upon  any  of 
their  homologous  edges. 

572.  From  §  571  it  is  evident  that  the  entire  surfaces  of  simi- 
lar polyhedrons  are  proportional  to  the  squares  upon  any  of  their 
homologous  edges. 

Proposition  XX 

573.  1.  Form  two  similar  polyhedrons,  and  if  possible  divide  them 
into  the  same  number  of  tetrahedrons,  similar  each  to  each. 

2.  How  does  the  ratio  of  any  two  homologous  lines  in  two  similar 
polyhedrons  compare  with  the  ratio  of  any  two  homologous  edges  ? 

Theorem.  Similar  polyhedrons  may  he  divided  into  the 
same  number  of  tetrahedrons,  similar  each  to  each,  and 
similarly  placed. 


mo 


SOLID  GEOMETRY.  — BOOK  VIII. 


Data :  Any  two  similar  polyhedrons,  as  AJ  and  A' J'. 
To  prove  that  AJ  and  A'j'  may  be  divided  into  the  same  num- 
ber of  tetrahedrons,  similar  each  to  each,  and  similarly  placed. 

K 


Proof.  Select  any  trihedral  angle  in  AJ,  as  B,  and  through  the 
extremities  of  its  edges,  as  A,  G,  C,  pass  a  plane.  Also  through 
the  homologous  points,  A',  G',  C',  pass  a  plane. 

Then,  §  310,  in  the  tetrahedrons  B-AGC  and  B'-A'g'C',  the 
faces  BACj  BAG,  BGC  are  similar  to  the  faces  bWc',  B'A'G'y 
b'g'C',  each  to  each. 

Hence,  AG :  A'G'  =  BG :  B'g'  =  CG :  C'G', 

and  AG:A'G'  =  AB:A'B'  =  AO:A'Cf; 

AG:A'G'=  CGiC'G' =AC:A'C'. 


Hence, 


face  ACG  is  similar  to  face  A'c'G' 


Why? 


.'.  the  homologous  faces  of  these  tetrahedrons  are  similar. 

Also,  §  500,  the  homologous  trihedral  angles  of  these  tetra- 
hedrons are  equal. 

Hence,  §  569,  tetrahedrons  B-AGC  and  b'-A'g'C'  are  similar. 

Now,  if  these  similar  tetrahedrons  are  removed  from  the  simi- 
lar polyhedrons  of  which  they  are  a  part,  the  polyhedrons  which 
remain  will  continue  to  be 'similar;  for  the  faces  and  polyhedral 
angles  of  the  original  polyhedrons  will  be  similarly  modified. 

By  continuing  to  remove  similar  tetrahedrons  from  them,  the 
original  polyhedrons  may  be  reduced  to  similar  tetrahedrons,  and 
they  will  then  have  been  divided  into  the  same  number  of  tetra- 
hedrons, similar  each  to  each,  and  similarly  placed. 

Therefore,  etc.  q.e.d. 

674.  Cor.  Homologous  lines  in  similar  polyhedrons  are  propor- 
tional to  tJieir  homologous  edges. 


SOLID   GEOMETRY.  — BOOK  VJII.  301 


Proposition  XXI 

575.  Form  two  similar  tetrahedrons.  How  do  the  trihedral  angles 
at  the  vertices  compare  ?  Then,  how  does  the  ratio  of  the  tetrahedrons 
compare  with  the  product  of  the  ratios  of  the  homologous  edges  of  the 
corresponding  trihedral  angles?  (§  568)  How  do  the  ratios  of  these 
edges  compare  with  each  other  ?  Then,  how  does  the  ratio  of  the  tetra- 
liedrons  compare  with  the  ratio  of  the  cubes  of  any  two  homologous 


Theorem,    Similar  tetrahedrons  are  to  each  other  as  the 
tubes  of  their  homologous  edges.  . 


B 

Data :  Any  two  similar  tetrahedrons,  as  Q-ABC  and  T-DEF. 

To  prove  Q-ABC :  T-DEF  =  O?  :  ^  =  etc. 

Proof.     §  569,     trihedral  Z  Q  =  trihedral  Z  T; 

/.  §  568,    Q-ABC :  T-DEF  =  QA  X  QB  X  QC :  TD  X  TE  X  TF, 

Q-ABC       QAX  QB  X  QC       QA       QB       QC 

or  = =  —  X  —  X    —  • 

T-DEF       TD  X  TE  X  TF       TD       TE       TF 

But         QA:TD=  QB:TE=  QC:TF,OV  —  = =  — -• 

TD       TE       TF 

-u-                            Q-ABC      QA  ^QA  ^  QA      QA^ . 
Hence, =  —  x  —  x  — -  =  ^=r.  j 

T-DEF       TD       TD       TD       tD 
that  is,  Q-ABC  :  T-DEF  =QA^:T^. 

In  like  manner,  the  same  may  be  proved  for  any  two  homolo- 
gous edges. 

Therefore,  etc.  Q-e.i>. 


302  SOLID   GEOMETRY.  — BOOK  VIIL 

576.  Cor.  Similar  polyhedrons  are  to  each  other  as  the  cubes  of 
their  homologous  edges. 

1.  Into  what  may  two  similar  polyhedrons  be  divided  ?     §  573 

2.  How  do  the  ratios  of  these  portions  compare  with  the  ratios 
of  the  cubes  of  their  homologous  edges  ?  §  575 

3.  How  do  these  ratios  compare  with  the  ratios  of  the  cubes  of 
any  two  homologous  edges  of  the  polyhedrons  ?  §  574 

4.  How,  then,  does  the  ratio  of  the  sums  of  these  portions 
compare  with  the  ratio  of  the  cubes  of  any  two  homologous  edges 
of  the  polyhedrons  ? 

577.  A  polyhedron  whose  faces  are  equal  regular  polygons, 
and  whose  polyhedral  angles  are  equal,  is  called  a  Regular 
Polyhedron. 

578.  1.  What  is  the  least  number  of  faces  that  a  convex  poly- 
hedral angle  may  have  ?  How  does  the  sum  of  the  face  angles  of 
any  convex  polyhedral  angle  compare  with  360°  ?  Since  each 
angle  of  an  equilateral  triangle  is  60°,  may  a  convex  polyhedral 
angle  be  formed  by  combining  three  equilateral  triangles  ? 
Four  ?  Five  ?  Six  ?  Why  ?  Then,  how  many  regular  convex 
polyhedrons  are  possible  with  equilateral  triangles  for  faces  ? 

2.  How  many  degrees  are  there  in  the  angle  of  a  square  ?  May 
a  convex  polyhedral  angle  be  formed  by  combining  three  squares  ? 
By  combining  four  ?  Why  ?  Then,  how  many  regular  convex 
polyhedrons  are  possible  with  squares  for  faces  ? 

3.  Since  each  angle  of  a  regular  pentagon  is  108°,  may  a  convex 
polyhedral  angle  be  formed  by  combining  three  regular  penta- 
gons ?  By  combining  four  ?  Why  ?  Then,  how  many  regular 
convex  polyhedrons  are  possible  with  regular  pentagons  1 3r  faces  ? 

4.  Since  each  angle  of  a  regular  hexagon  is  120°,  may  a  convex 
polyhedral  angle  be  formed  by  combining  three  regular  hexagons  ? 
Why  ?  By  combining  three  regular  heptagons  ?  Why  ?  What 
is  the  limit  of  the  number  of  sides  of  a  regular  polygon  that  may 
be  used  in  forming  a  convex  polyhedral  angle,  and  therefore  in 
forming  a  regular  convex  polyhedron  ? 

What,  then,  is  the  greatest  number  of  regular  convex  poly- 
hedrons possible  ? 


SOLID   GEOMETRY.  — BOOK  VIII  303 

o79.  There  are  only  five  regular  convex  polyhedrons  possible, 
called  from  the  number  of  their  faces  the  tetrahedron,  the  hexahe- 
dron, the  octahedron,  the  dodecahedron,  and  the  icosahedron. 

The  tetrahedron,  octahedron,  and  icosahedron  are  bounded 
by  equilateral  triangles;  the  hexahedron  by  squares;  and  the 
dodecahedron  by  pentagons. 

580.  The  point  within  a  regular  polyhedron  that  is  equidistant 
from  all  the  faces  of  the  polyhedron  is  called  the  center  of  the 
polyhedron. 

The  center  is  also  equidistant  from  the  vertices  of  all  the  poly- 
hedral angles  of  the  polyhedron. 

Therefore,  a  sphere  may  he  inscribed  in,,  and  a  sphere  may  he 
circumscribed  about,  any  regular  polyhedron,  §§  640,  641 

Proposition  XXII 

581.  Problem,  Upon  a  given  edge  to  construct  the  regular 
polyhedrons. 

Datum :  An  edge,  as  AB.  jv 

Required  to  construct  the  regular  polyhe- 
drons on  AB. 

Solutions.     1.    The  regular  tetrahedron. 

Upon  AB  construct  an  equilateral  triangle,         Ni^-'^ 

HS  ABC.  ^ 

At  the  center  of  A  ABC  erect  a  perpendicular,  and  take  a  point 
D  in  this  perpendicular  such  that  DA  =  AB. 
Draw  lines  from  D  to  the  vertices  of  A  ABC. 
Then,  the  polyhedron  D-ABC  is  a  regular  tetrahedron.        q.e.f. 
Proof.      By  the  student.     Suggestion.     Refer  to  §§  450,  500. 

rr 

2.    The  regidar  hexahedron.  /r  y^ 

Upon  AB  construct  the  square  ABCD,  and    e 
upon  its  sides  construct  the  squares  AF,  BG, 
CH,  and  BE  perpendicular  to  ABCD. 

Then,  the  polyhedron  ^G  is  a  regular  hexa- 
hedron. Q.E.F. 


DI 


Proof.     By  the  student.     Suggestion.     Refer  to  §  600. 


304 


SOLID  GEOMETRY.-- BOOK  VUL 


3.  The  regular  octahedron. 

Upon  AB  construct  the  square  ABCD,  and 
through  its  center  0  pass  a  line  perpendicular 
to  its  plane. 

On  this  perpendicular  take  the  points  E 
and  F  such  that  AE  and  AF  are  each  equal 
to  AB. 

Draw  lines  from  E  and  F  to  the  vertices 
of  ABCD. 

Then,  the  polyhedron  E-ABCD-F  is  a  regular  octahedron,  q.e.f. 

Proof.     By  the  student.     Suggestion.    Refer  to  §  460, 

4.  The  regular  dodecahedron. 

Upon  AB  construct  a  regular  pentagon  ABODE,  and  to  each  side 
of  it  apply  an  equal  pen- 
tagon so  inclined  to  the 
plane  of  ABODE  as  to 
form  trihedral  angles  at 
A,  B,  C,  D,  E. 

Then,  a  convex  sur- 
face FHKMP,  composed 
of  six  regular  pentagons, 
has  been  constructed. 

Construct  a  convex  surface  f'h'k'm'p'  equal  to  FHKMP,  and 
apply  one  to  the  other  so  as  to  form  a  single  convex  surface. 

The  surface  thus  formed  is  that  of  a  regular  dodecahedron,   q.e.f. 

Proof.     By  the  student.     Suggestion.     Refer  to  §  600. 

5.  The  regular  icosahedron. 

Upon  AB  construct  a  regular  pentagon  ABODE;  at  its  center 
erect  a  perpendicular ;  and  take  a  point  Q  in 
this  perpendicular  such  that  QA  =  AB. 

Draw  lines  from  Q  to  the  vertices  of  the 
pentagon  forming  a  regular  pentagonal  pyra- 
mid Q- ABODE. 

Complete  the  pentahedral  angles  at  A,  B,  0, 
D,  E  by  adding  to  each  three  equilateral  tri- 
angles, each  equal  to  A  QAB, 


SOLID   GEOMETRY.  —  BOOK   VIII. 


Construct  a  regular  pentagonal  pyramid  Q'-A'b'c'b'e'  equal  to 
Q-ABCDE,  and  join  it  to  the  convex  surface  already  formed  so  as 
to  form  a  single  convex  surface. 

The  surface  thus  formed  is  that  of  a  regular  icosahedron.     q.e.f. 

Proof.     By  the  student.     Scggestion.    Refer  to  §  460. 

582.  Sch.  The  five  regular  polyhedrons  may  be  made  from 
cardboard  in  patterns  as  given  below,  by  cutting  half  through 
along  the  dotted  lines,  folding,  and  pasting  strips  of  paper  along 
the  edges. 


TETRAHEDRON 


HEXAHEDRON 


OOTABBDBON 


DODKOAHEDRON 


I006AHBDB0N 


583. 


FORMULA 

Notation 


B    =  base. 

5    =  ipper  base. 

p    =  perimeter  of  base. 

p'  =  perimeter  of  upper  base. 

P"  =  perimeter  of  right  section. 

H   =  altitude. 

milne's  geom. — 20 


L  =  slant  height. 

E  =  lateral  edge. 

d] 

e  =  dimensions   of   a   paral 

/  lelopiped. 

A  =  lateral  area. 

r  =  volume. 


306  SOLID   GEOMETRY,  — BOOK   VIIL 

Prism. 

^  =  ^  X  P"    .    .    . .     §  519 

A=HXP  (E-iglit  Prism) §  520 

V  =  BxH §§542,543 

Rectangular  Parallelepiped. 

V  =  dxexf §536 

r  =  B  XH  (True  also  for  any  parallelopiped)   §§  537,  539 

Pyramid. 

A  =  ^P  XL  (Regular  Pyramid) §  551 

V  =  lBxH §§559,560 

Frustum  of  a  Pyramid. 

A  =  ^L(p-{-P')  (Regular  Pyramid) §  552 

v  =  ^H(B-\-b-\-^Wxl>) §564 

SUPPLEMENTAKY  EXERCISES 

Ex.  788.  The  edge  of  a  cube  is  5^  in.    What  are  its  volume  and  surface  ? 

Ex.  789.  What  are  the  entire  area  and  volume  of  a  right  prism  4.5  ft.  in 
altitude,  if  the  bases  are  equilateral  triangles  13  in.  on  a  side  ? 

Ex.  790.  What  is  the  total  area  of  a  regular  triangular  pyramid  whose 
slant  height  is  IS*^™  and  each  side  of  whose  base  is  9^™  ? 

Ex.  791.  What  is  the  volume  of  a  triangular  pyramid  whose  altitude  is 
11  ft.  and  the  sides  of  whose  base  are  3  ft.,  4  ft.,  and  5  ft.? 

Ex.  792.  What  is  the  edge  of  a  cube  whose  volume  equals  that  of  a  rec- 
tangular parallelopiped  whose  edges  are  9  in.,  12  in.,  and  16  in.? 

Ex.  793.  The  altitude  of  a  prism  is  6^"^  and  the  area  of  its  base  is  2.5'^  ^™. 
What  is  the  altitude  of  a  prism  of  the  same  volume,  if  the  area  of  its  base  is 

5.75sq  dm  ? 

Ex.  794.  The  homologous  edges  of  two  similar  tetrahedrons  are  as  4 :  6. 
What  is  the  ratio  of  their  surfaces  ?    What  is  the  ratio  of  their  volumes  ? 

Ex.  795.  What  is  the  altitude  of  a  pyramid  whose  volume  is  36^"  m  and 
the  sides  of  whose  triangular  base  are  6™,  5™,  and  4™  ? 

Ex.  796.  The  area  of  the  upper  base  of  the  frustum  of  a  pyramid  is 
48  sq.  ft.  and  that  of  the  lower  base  is  72  sq.  ft.  If  the  altitude  of  the  frus- 
tum is  60  ft.,  what  is  ita  volume  ? 


SOLID   GEOMETRY— BOOK    Vltl.  307 

Ex.  797.  What  is  the  altitude  of  the  frastum  of  a  regular  hexagonal 
pyramid,  if  its  volume  is  IG^um  and  the  sides  of  its  bases  are  respectively 
1.5™  and  2.5'"? 

Ex.  798.  A  pyramid  20  ft.  high  has  100  sq.  ft.  in  its  base ;  a  section 
parallel  to  the  base  has  an  area  of  55  sq.  ft.  How  far  is  the  section  from  the 
base? 

Ex.  799.  What  is  the  volume  of  an  oblique  truncated  triangular  prism 
whose  edges  are  5"\  7'",  and  9'",  and  the  area  of  whose  right  section  is  IQ^™  ? 

Ex.  800.    What  is  the  edge  of  a  cube  whose  entire  area  is  1*^™  ? 

Ex.  801.  The  base  of  a  pyramid  contains  121  sq.  ft. ;  a  section  parallel 
to  the  base  and  3  ft.  from  the  vertex  contains  49  sq.  ft.  What  is  the  altitude 
of  the  pyramid  ? 

Ex.  802.  What  is  the  lateral  area  of  a  regular  hexagonal  pyramid  whose 
base  is  inscribed  in  a  circle  whose  diameter  is  15  ft.,  the  altitude  of  the  pyra- 
mid being  8  ft.  ?     What  is  the  volume  of  the  pyramid  ? 

Ex.  803.   Any  lateral  edge  of  a  right  prism  is  equal  to  the  altitude. 

Ex.  804.  The  square  of  a  diagonal  of  a  rectangular  parallelepiped  is 
equal  to  the  sum  of  the  squares  of  its  three  edges. 

Ex.  805.  If  the  edges  of  a  tetrahedron  are  all  equal,  the  sum  of  the 
angles  at  any  corner  is  equal  to  two  right  angles. 

Ex.  806.  The  section  of  a  triangular  pyramid  made  by  a  plane  parallel 
to  two  opposite  edges  is  a  parallelogram. 

Ex.  807.    The  lateral  faces  of  right  prisms  are  rectangles. 

Ex.  808.  The  section  of  a  prism  made  by  a  plane  parallel  to  a  lateral 
edge  is  a  parallelogram. 

Ex.  809.  Thife  diagonal  of  a  cube  is  equal  to  the  product  of  its  edge  and 
V8. 

Ex.  810.  The  volume  of  a  regular  prism  is  equal  to  the  product  of  its 
lateral  area  and  one  half  the  apothem  of  the  base. 

Ex.  811.  Any  straight  line  passing  through  the  center*  of  a  parallelo- 
piped  and  terminated  by  two  faces  is  bisected  at  the  center. 

Ex.  812.  If  any  two  non-parallel  diagonal  planes  of  a  prism  are  perpen- 
dicular to  the  base,  the  prism  is  a  right  prism. 

Ex.  813.  The  base  of  a  pyramid  is  16  sq.  ft.  and  its  altitude  is  7  ft. 
What  is  the  area  of  a  section  parallel  to  the  base,  if  it  is  2  ft.  6  in.  from  the 
apex  ? 

Ex.  814.  The  edges  of  a  rectangular  parallelepiped  are  3  in.,  4  in.,  and 
6  in.  What  is  the  area  of  its  diagonal  planes  and  the  length  of  its  diagonal 
line? 

*  The  center  of  a  parallelepiped  is  the  intersection  of  its  diagonals. 


308  SOLID   GEOMETRY— BOOK    VIIL 

Ex.  815.  A  portion  of  a  railway  embankment  is  18  ft.  by  380  ft.  at  the 
top,  and  40  ft.  by  380  ft.  at  the  bottom.  If  its  height  is  12  ft.,  how  many 
cubic  yards,  or  loads,  of  earth  does  it  contain  ? 

Ex.  816.  If  the  four  diagonals  of  a  four-sided  prism  pass  through  a  com- 
mon point,  the  prism  is  a  parallelopiped. 

Ex.  817.  If  a  pyramid  is  cut  by  a  plane  parallel  to  its  base,  the  pyramid 
.  cut  off  is  similar  to  the  given  pyramid. 

Ex.  818.  The  lateral  area  of  a  right  prism  is  less  than  the  lateral  area  of 
any  oblique  prism  having  the  same  base  and  altitude. 

Ex.  819.  If  a  section  of  a  pyramid  made  by  a  plane  parallel  to  the  base 
bisects  the  altitude,  the  area  of  the  section  is  one  fourth  the  area  of  the  base, 
and  the  pyramid  cut  off  is  one  eighth  of  the  original  pyramid. 

Ex.  820.  The  volume  of  a  right  triangular  prism  is  equal  to  one  half  the 
product  of  any  lateral  face  by  its  distance  from  the  opposite  edge. 

Ex.  821.  If  the  diagonals  of  three  unequal  faces  of  a  rectangular  paral- 
lelopiped are  given,  compute  the  edges. 

Ex.  822.  What  is  the  lateral  area  of  a  regular  pyramid  whose  slant 
height  is  10  ft.,  the  base  being  a  pentagon  inscribed  in  a  circle  whose  radius 
is  6  ft.  ?    What  is  the  volume  ? 

Ex.  823.  The  volume  of  a  rectangular  parallelopiped  is  336c" '",  its  total 
area  is  320sqm^  and  its  altitude  is  4™.     What  are  the  dimensions  of  its  base  ? 

Ex.  824.  A  pyramid  weighs  30^8,  and  its  altitude  is  12^"'.  A  plane 
parallel  to  the  base  cuts  off  a  frustum  which  weighs  15^^.  What  is  the  alti- 
tude of  the  frustum  ? 

Ex.  825.  Each  side  of  the  base  of  a  regular  triangular  pyramid  is  3  in., 
and  its  altitude  is  8  in.     What  are  its  lateral  edge  and  lateral  area  ? 

Ex.  826.  The  volume  of  a  regular  tetrahedron  is  equal  to  the  product  of 
the  cube  of  its  edge  and  ^^  \/2. 

Ex.  827.  The  volume  of  a  regular  octahedron  is  equal  to  the  product  of 
the  cube  of  its  edge  and  \  V2. 

Ex.  828.  Any  plane  passing  through  the  center  of  a  parallelopiped 
divides  it  into  two  equal  solids. 

Ex.  829.   The  lateral  area  of  a  regular  pyramid  is  greater  than  its  base. 

Ex.  830.  The  lateral  edge  of  the  frustum  of  a  regular  triangular  pyramid 
is  4^  ft.,  a  side  of  one  base  is  5  ft.,  and  of  the  other  4  ft.  What  is  the 
volume  ? 

Ex.  831.  The  sum  of  the  perpendiculars  from  any  point  within  a  regular 
tetrahedron  to  each  of  its  four  faces  is  equal  to  its  altitude. 

Ex.  832.  In  a  regular  tetrahedron  an  altitude  is  equal  t;Q  tiUfee  times  the 
perpendicular  from  its  foot  on  a  face. 


BOOK   IX 


CYLINDERS  AND  CONES 


584.  A  surface,  generated   by  a  moving  straight  line  which 
always  remains  parallel  to  its  original  posi- 
tion and  continually  touches  a  given  curved 
line,  is  called  a  Cylindrical  Surface. 

The  moving  straight  line  is  called  the  gen- 
eratrix, and  the  given  curved  line  is  called 
the  directrix. 

The  generatrix  in  any  position  is  called 
an  element  of  the  surface. 

585.  A  solid  bounded  by  a  cylindrical  surface  and  two  parallel 
planes  which  cut  all  its  elements  is  called  a  Cylinder. 

The  plane  surfaces  are  called  the  bases  and  the  cylindrical  sur- 
face is  called  the  lateral  surface  of  the  cylinder. 

All  elements  of  a  cylinder  are  equal.     §  464. 

The  perpendicular  distance  between  its  bases  is  the  altitude  of 
the  cylinder. 

586.  A  section  of  a  cylinder  made  by  a  plane  perpendicular  to 
its  elements  is  called  a  Right  Section. 

587.  A  cylinder  whose  elements  are  perpendicular  to  its  base  is 
called  a  Right  Cylinder. 

588.  A  cylinder  whose  elements  are  not  perpendicular  to  its 
base  is  called  an  Oblique  Cylinder. 

589.  A  cylinder  whose  bases  are  circles  is  a  Circular  Cylinder. 
The  straight  line  joining  the  centers  of  the  bases  of  a  circular 

cylinder  is  called  the  axis  of  the  cylinder. 

590.  A  right  circular  cylinder  is  called  a  Cylinder  of 
Revolution,  because  it  may  be  generated  by  the  revolu- 
tion of  a  rectangle  about  one  of  its  sides. 

Cylinders  of  revolution  generated  by  similar  rectan- 
gles revolving  about  homologous  sides  are  similar. 

300 


310  SOLID   GEOMETRY.  — BOOK  IX. 

591.    A  plane  which  contains  an  element  of  a  cylinder  and  does 
not  cut  the  surface  is  a  Tangent  Plane  to  the  cylinder. 
The  element  is  called  the  element  of  contact. 

692.  Any  straight  line  that  lies  in  a  tangent  plane  and  cuts  the 
element  of  contact  is  a  Tangent  Line  to  the  cylinder. 

593.  When  the  bases  of  a  prism  are  inscribed  in  the  bases  of  a 
cylinder  and  its  lateral  edges  are  elements  of  the  cylinder,  the 
prism  is  said  to  be  inscribed  in  the  cylinder. 

594.  When  the  bases  of  a  prism  are  circumscribed  about  the 
bases  of  a  cylinder  and  its  lateral  edges  are  parallel  to  the  ele- 
ments of  the  cylinder,  the  prism  is  said  to  be  drcumscnhed  about 
the  cylinder. 

Proposition  I 

595.  1.  Form  a  cylinder  and  cut  it  by  any  plane  through  an  element 
of  its  surface  (§  519  n.).  What  plane  figure  is  the  section  made  by  the 
cutting  plane? 

2.  If  the  cylinder  is  a  right  cylinder,  what  plane  figure  does  such  a 
plane  make? 

Theorem,  Any  section  of  a  cylinder  made  hy  a  plane 
passing  through  an  element  is  a  parallelogram. 

Data :  Any  section  of  the  cylinder  EF,  as 
ABCD,  made  by  a  plane  passing  through  AB, 
an  element  of  the  surface. 

To  prove    ABCD  a  parallelogram. 

Proof.  The  plane  passing  through  the  ele- 
ment AB  cuts  the  circumference  of  the  base 
in  a  second  point,  as  D.     Draw  BO  II  AB.  ^{ 

Then,  §  63,  DC  is  in  the  plane  BAD ;  ^ 

and,  §  584,  DC  is  an  element  of  the  cylinder. 

Hence,  DC,  being  common  to  the  plane  and  the  lateral  surface 
of  the  cylinder,  is  their  intersection. 

Also,  §  463,  AD  II  BC; 

hence,  §  140,  ABCD  is  a  parallelogram. 

Therefore,  etc.  q.e.d 


SOLID   GEOMETRY.  — BOOK  IX. 


311 


596.    Cor.     Any  section  of  a  right  cylinder  made  by  a  plane  pass- 
ing through  an  element  is  a  rectangle. 


Proposition  II 

597.    1.   Form  a  cylinder.     How  do  its  bases  compare? 

2.  Cut  the  cylinder  by  parallel  planes. which  cut  all  its  elements. 
How  do  the  sections  thus  made  compare  with  each  other  ? 

3.  How  does  a  section  made  by  a  plane  parallel  to  the  base  compare 
with  the  base? 

Theorem,    The  hases  of  a  cylinder  are  equal. 

Data :  Any  cylinder,  as  MG,  whose  bases  are 
HG  and  MN. 

To  prove  HG  =  MN. 

Proof.  Take  any  three  points  in  the  perim- 
eter of  the  upper  base,  as  D,  E,  F,  and  from 
them  draw  the  elements  of  the  surface  DA,  EB, 
FC,  respectively. 

Draw  AB,  BC,  AC,  BE,  EF,  and  DF. 

§§  585,  584,  AB,  BE,  and  CF  are  equal  and 
parallel ; 

.-.    §  150,  AE,  AF,  and  BF  are  parallelograms; 
and  AB  =  BE,  AC  =  DF,  BC=EF'j 

hence,  A  ABC  =  A  BEF. 

Apply  the  upper  base  to  the  lower  base  so  that  BE  shall  fall 
upon  AB. 

Then,  F  will  fall  upon  C. 

But  F  is  any  point  in  the  perimeter  of  the  upper  base,  there- 
fore, every  point  in  the  perimeter  of  the  upper  base  will  fall  upon 
the  perimeter  of  the  lower  base. 

Hence,  §  36,  HG  =  MN. 


Why? 
Why? 


Therefore,  etc. 


Q.E.D. 


598.  Cor.  I.     The  sections  of  a  cylinder  made  by  parallel  planes 

cutting  all  its  elements  are  equal. 

599.  Cor.  II.     Hie  axis  of  a  circular  cylinder  passes  through  the 
centers  of  all  the  sections  parallel  to  the  bases. 


812  SOLID   GEOMETRY.  — BOOK  IX, 

Proposition  III 

600.  To  what  is  the  lateral  surface  of  any  prism  equivalent?  (§  519) 
If  the  number  of  its  lateral  faces  is  indefinitely  increased,  what  solid 
does  the  prism  approach  as  its  limit  ?  How,  then,  does  the  lateral  sur- 
face of  any  cylinder  compare  with  the  rectangle  formed  by  an  element 
and  the  perimeter  of  a  right  section  ? 

Theorem,  The  lateral  swrfax^e  of  a  cylinder  is  equivo/- 
lent  to  the  rectangle  formed  by  an  element  of  the  surface 
and  the  perimeter  of  a  right  section. 

Data:  Any  cylinder,  as  FK\  any  right 
section  of  it,  as  ABODE ;  and  any  element 
of  its  surface,  as  FG. 

Denote  the  lateral  surface  of  FK  by  S, 
and  the  perimeter  of  its  right  section 
by  P. 

To  prove       S  ^  rect.  FG  •  P. 

Proof.  Inscribe  in  the  cylinder  a  prism ; 
denote  its  lateral  surface  by  s'  and  the 
perimeter  of  its  right  section  by  P'. 

Then,  §  593,  each  lateral  edge  is  an  element  of  the  cylinder, 
and,  §  585,  the  elements  are  all  equal ; 
.-.  §  519,  S'  0=  rect.  FG  •  P'. 

Now,  if  the  number  of  lateral  faces  of  the  inscribed  prism  is 
indefinitely  increased, 

§  393,  P'  approaches  P  as  its  limit ; 

S'  approaches  S  as  its  limit. 

But,  however  great  the  number  of  faces, 
S'  =0=  rect.  FG  .  P'. 

Hence,  §  326,  S  =0=  rect.  FG  •  P. 

Therefore,  etc.  q.e.d. 

601.  Cor.  The  lateral  surface  of  a  cylinder  of  revolution  is 
equivalent  to  the  rectangle  formed  by  its  altitude  and  the  circum- 
ference of  its  base. 

Arithmetical  Rules :  To  be  framed  by  the  student.  §  339 


SOLID   GEOMETRY.  — BOOK  IX. 


313 


6U2.  formulae :  Let  A  denote  the  lateral  area,  T  the  total  area 
H  the  altitude,  and  R  the  radius  of  the  base  of  a  cylinder  f 
revolution. 

Then,  §  395,  ^  =  2  Tri?  x  fl", 

and,  §  398,  7=  2 Tri?  x  H-\-'2  ttR'  =  2 iri?(fl--f-  R), 

Proposition  IV 

603.  Compute  the  areas  of  any  two  similar  cylinders  of  revolution, 
as  those  whose  altitudes  are  4"  and  2"  and  whose  radii  are  2"  and  1", 
respectively.  How  does  the  ratio  of  their  lateral  areas,  or  of  their  total 
areas,  compnre  with  the  ratio  of  the  squares  of  their  altitudes,  or  with 
the  ratio  ol  the  squares  of  their  radii  ? 

Theoretn,  The  lateral  areas,  or  the  total  areas,  of  simU 
lar  eylinders  of  revolution  are  to  each  other  as  the  square& 
of  their  altitudes,  or  as  the  squares  of  their  radii. 

Data:  Any  two  similar  cylinders  of 
revolution,  whose  altitudes  are  H  and  S^, 
and  radii  R  and  i?',  respectively. 

Denote  their  lateral  areas  by  A  and  A\ 
and  their  total  areas  by  T  and  T\  respec- 
tively. 

To  prove   1.    A\A^  =  H'^ :  B.^  =  i?' :  i?'^. 
2.    r;r'  =  ^2.^/2^^.^f2 

Proof.     1.   Since  the  generating  rectangles  are  similar, 
R'~      ~ 


H 


§§  299,  279, 
.-.  §  602, 


H'     R'  +  B 
^x^ 


7» 


H' 


or 


J___  2  7rRH 

A'~2'7rR'H'       R'      IT      IT 
A:A'  =  H^:H'^=R':R". 


9     8^09     r^  2'kR{R^-H)  ^R[  R  +  H\^ie  __H\ 

or  T:T'  =  H^:H"'=R^:R"'. 

Therefore,  etc.  Q.e.b 

604.  Cor.  The  lateral  areas,  or  the  total  areas,  of  similar  cylin- 
ders of  revolution  are  to  each  other  as  the  squares  of  any  of  their 
like  dimensions. 


314  SOLID   GEOMETRY.  — BOOK  IX. 

Proposition  V 

605.  To  what  is  the  volume  of  any  prism  equal?  (§  543)  If  tha 
number  of  its  lateral  faces  is  indefinitely  increased,  what  solid  does 
the  prism  approach  as  its  limit  ?  To  what,  then,  is  the  volume  of  anj 
cylinder  equal  ? 

Theorem.  The  volume  of  any  cylinder  is  equal  to  the 
product  of  its  base  by  its  altitude. 

Data :  Any  cylinder,  as  A,  whose  base  is  B 
and  altitude  H. 

Denote  its  volume  by  F. 

To  prove  V=  B  x  H. 

Proof.  Inscribe  in  the  cylinder  a  prism, 
and  denote  its  volume  by  F'  and  its  base 
bys'. 

Then,  the  altitude  of  the  prism  is  H, 

and,  §  543,  F'  =  5'  X  H. 

Now,  if  the  number  of  lateral  faces  of  the  inscribed  prism  is 
indefinitely  increased, 

§  393,  5'  approaches  B  as  its  limit ; 

F'  approaches  F  as  its  limit. 
But,  however  great  the  number  of  faces, 

F'  =  5'  X  H. 
Hence,  §  222,  V=BxH. 

Therefore,  etc.  q.e.d. 

606.  Formula :  Let  B  denote  the  radius  of  the  base  of  a  cylinder 
of  revolution. 

Then,  §398,  B  =  7rlf', 

V=TTiexH. 

Proposition  VI 

607.  Compute  the  volumes  of  any  two  similar  cylinders  of  revolution, 
as  those  whose  altitudes  are  4"  and  2"  and  whose  radii  are  2"  and  1" 
respectively.  How  does  the  ratio  of  their  volumes  compare  with  the 
ratio  of  the  cubes  of  their  altitudes,  or  with  the  ratio  of  the  cubes  of  their 
radii? 


SOLID   GEOMETRY.  — BOOK  IX. 


315   . 


Theorem,  The  volumes  of  similar  cylinders  of  revolution 
are  to  each  other  as  the  cubes  of  their  altitudes,  or  as  the 
cubes  of  their  radii. 

Data :  Any  two  similar  cylinders  of  rev- 
olution, whose  altitudes  are  H  and  H',  and 
radii  R  and  i2'  respectively. 

Denote  their  volumes  by  V  and  r'  re- 
spectively. 

To  prove     V-.V^  =  H^\  H^  =  :^:  r'^ 

Proof.     Since  the  generating  rectangles  are  similar, 


H 


299, 
.  §  606, 


R       H 
R'      H*' 

V 

ttR^H        R^       H 
-kR^'H'       R"  ^  H' 

V:  V'  —H^i  H'^  -. 

ir'3     R" 

or  V:  V  =  H'' :  H""  =  I^  :  R^. 

Therefore,  etc.  q.e.d. 

608.    Cor.     The  volumes  of  similar  cylinders  of  revolution  are  to 
each  other  as  the  cubes  of  any  of  their  like  dimensions. 


CONES 

A  surface,  generated  by  a  moving  straight  line  which 
passes  through  a  fixed  point  and  continually  touches  a  given 
curved  line,  is  called  a  Conical  Surface. 

The  moving  straight  line  is  called  the  generatrix,  the  fixed 
point  the  vertex,  and  the  given  curved  line  the  directrix. 

The  generatrix  in  any  position  is  called 
an  element  of  the  surface. 

If  the  generatrix  extends  on  both  sides 
of  the  vertex,  the  whole  surface  consists  of 
two  portions  which  are  called  the  lower 
and  upper  nappes  respectively. 

Q-ABCD  and  Q-A'B'C'D'  are  the  lower  and 
upper  nappes  respectively  of  a  conical  surface  of 
which  A  A'  is  the  generatrix,  Q  the  vertex,  ABGD 
the  directrix,  and  AA' ,  BB'^  CC,  DD',  etc.,  are 
elements. 


.  316  SOLID   GEOMETRY.  — BOOK  IX. 

610.  A  solid  bounded  by  a  conical  surface  and  a  plane  which 
cuts  all  its  elements  is  called  a  Cone. 

The  plane  surface  is  called  the  base  and  the  conical  surface  is 
called  the  lateral  surface  of  the  cone. 

The  perpendicular  distance  from  its  vertex  to  the  plane  of  its 
base  is  the  altitude  of  the  cone. 

611.  A  cone  whose  base  is  a  circle  is  called  a  Circular  Cone. 
The  straight  line  joining  the  vertex  and  the  center  of  the  base 

of  a  cone  is  called  the  axis  of  the  cone. 

612.  A  cone  whose  axis  is  perpendicular  to  its  base  is  called  a 
Right  Cone. 

613.  A  cone  whose  axis  is  not  perpendicular  to  its  base  is 
called  an  Oblique  Cone. 

614.  A  right  circular  cone  is  called  a  Cone  of  Revolution,  be- 
cause it  may  be  generated  by  the  revolution  of  a  right  triangle 
about  one  of  its  perpendicular  sides. 

All  the  elements  of  a  cone  of  revolution  are 
equal,  and  any  one  of  them  is  called  the  slant 
height  of  the  cone. 

Cones   of    revolution,   which   are   generated   by 
similar  right  triangles  revolving  about  homologous     K~"l}~^       J 
perpendicular  sides,  are  similar. 

616.  A  plane  which  contains  an  element  of  a  cone  and  does 
not  cut  the  surface  is  a  Tangent  Plane  to  the  cone. 

The  element  is  called  the  element  of  contact. 

616.  Any  straight  line  that  lies  in  a  tangent  plane  and  cuts  the 
element  of  contact  is  a  Tangent  Line  to  the  cone. 

617.  The  portion  of  a  cone  included  between  its 
base  and  a  section  parallel  to  the  base  and  cutting 
all  the  elements  is  called  a  Frustum  of  a  cone. 

The  base  of  the  cone  is  called  the  lower  base  of 
the  frustum,  and  the  parallel  section  the  upper  base. 

The  perpendicular  distance  between  its  bases  is  the  altitude  of 
the  frustum. 

The  portion  of  an  element  of  a  cone  of  revolution  included 
between  the  parallel  bases  of  a  frustum  is  the  slant  height  of  the 
frustum. 


SOLID  GEOMETRY,— BOOK  IX. 


317 


618.  When  the  base  of  a  pyramid  is  inscribed  in  the  base  of  a 
cone  and  its  lateral  edges  are  elements  of  the  cone,  the  pyramid 
is  said  to  be  inscribed  in  the  cone. 

619.  When  the  base  of  a  pyramid  is  circumscribed  about  the 
base  of  the  cone  and  its  vertex  coincides  with  the  vertex  of  the 
cone,  the  pyramid  is  said  to  be  circumscribed  about  the  cone. 


Proposition  VII 

620.    Form  a  cone  and  cut  it  by  any  plane  through  its  vertex, 
plane  figure  is  the  section  made  by  the  cutting  plane  ? 


What 


Theorem,    Any  section  of  a  cone  made  hy  a  plane  pass- 
ing through  its  vertex  is  a  triangle. 


Data :  Any  cone,  as  Q-ABD,  and  any  section 
of  it,  as  CQD,  made  by  a  plane  passing  through 
the  vertex  Q. 

To  prove  CQD  a  triangle. 


Proof.     Draw  the  straight  lines  QC  and  QD. 

Then,  §  609,  QC  and  QD  are  elements  of  the  cone, 
and,  §  427,  since  QC  and  QD  each  have  two  points  in  common 
with  the  plane  CQD,  they  lie  in  that  plane ; 

.*.  QC  and  QD  are  the  intersections  of  the  cutting  plane  and  the 
lateral  surface. 

Also,  §  441,  CD  is  a  straight  line. 

Hence,  §  85,  CQD  is  a  triangle. 

Therefore,  etc.  q.e.d. 

Ex.  833.  What  is  the  lateral  area  of  a  cylinder  of  revolution  whose  alti- 
tude is  18  ft.  and  the  diameter  of  whose  base  is  6  ft.  ? 

Ex.  834.  What  is  the  volume  of  a  cylinder  of  revolution  whose  altitude 
is  7  ft.  and  the  circumference  of  whose  base  is  5  ft.  ? 

Ex.  835.  How  many  cubic  feet  are  there  in  a  cylindrical  log  14  ft.  long 
and  2.5  ft.  in  diameter  ? 

Ex.  836.  The  altitude  of  a  cylinder  of  revolution  is  16^""  and  the  diameter 
of  its  base  is  11dm.     What  is  its  total  area  ?    What  is  its  volume  ? 


318 


SOLID   GEOMETRY.— BOOK  IX, 


Proposition  VIII 

621.  !•  Form  a  circular  cone  and  cut  it  by  any  plane  parallel  to  its 
base.     What  plane  figure  is  the  section  made  by  the  cutting  plane? 

2.   In  what  points  will  the  axis  of  the  cone  pierce  all  the  sections  that 
are  parallel  to  the  base  ? 

Theorem.    Any  section  of  a  circular  cone  made  hy  a  plane 
parallel  to  the  base  is  a  circle. 

Data :  Any  circular  cone,  as  Q-ABC  and  any 
section  of  it,  as  DEF,  made  by  a  plane  parallel  to 
the  base. 

To  prove  DEF  a  circle. 

Proof.  Draw  QO,  the  axis  of  the  cone  piercing 
BEF  in  P. 

Through  QO  and  any  elements,  QA,  QB,  etc.,  pass 
planes  cutting  the  base  in  the  radii  OA,  OB,  etc.,   a\ 
and  the  parallel  section  in  the  straight  lines  P2>, 
PE,  etc. 

Data,  planes  DEF  and  ABC  are  parallel ; 

.-.  §  463,  PB  II  OA,  and  PE  II  OB. 

Consequently,  A  QPD  and  QOA  are  mutually  equiangular  and 
therefore  similar ;  likewise  A  QPE  and  QOB  are  similar. 

Hence,  §  299,  PB  :  OA  =  QP  :  QO, 

and  PE:OB=  QP:  QO; 

PB  :  OA  =  PE  :  OB. 

But  OA  =  OB',  Why? 

.-.  §  272,  PB  =  PE', 

that  is,  all  the  straight  lines  drawn  from  the  point  P  to  the  perim- 
eter of  the  section  BEF  are  equal. 

Hence,  §  173,  BEF  is  a  circle. 

Therefore,  etc.  q.e.d. 

622.  Cor.     The  axis  of  a  circular  cone  passes  through  the  centers 
of  all  the  sections  that  are  parallel  to  the  base. 


Ex.  837.    The  total  area  of  a  cylinder  of  revolution  is  659"<i' 
altitude  is  IS**'".     What  is  the  diameter  of  its  base  ? 


and  its 


SOLID   GEOMETRY.  — BOOK  IX. 


319 


Proposition  IX 

623.  To  what  is  the  lateral  surface  of  any  regular  pyramid  equivalent  ? 
If  the  number  of  its  lateral  faces  is  indefinitely  increased,  what  solid 
does  the  pyramid  approach  as  its  limit  ?  How,  then,  does  the  lateral  sur- 
face of  a  cone  of  revolution  compare  with  the  rectangle  formed  by  the 
circumference  of  its  base  and  its  slant  height? 

Theoretn,  The  lateral  surface  of  a  cone  of  revolution  is 
equivalent  to  one  half  the  rectangle  formed  by  the  circum- 
ference of  its  base  and  its  slant  height. 

Data:  Any  cone  of  revolution,  as  Q-ABFB, 
whose  slant  height  is  L,  and  the  circumference 
of  whose  base  is  C. 


Denote  its  lateral  surface  by  S. 

To  prove  S^^  rect.  C -  L. 

Proof.  Inscribe  in  the  cone  a  regular  pyramid 
of  any  number  of  faces  and  denote  its  lateral  sur- 
face by  S',  its  slant  height  by  L',  and  the  perimeter 
of  its  base  by  P. 

Then,  §  618,  each  lateral  edge  of  the  pyramid  is  an  element  of 
the  cone ; 
and,  §  551,  s'  o=  i  rect.  P  •  L'. 

Now,  if  the  number  of  lateral  faces  of  the  inscribed  pyramid  is 
indefinitely  increased, 

§  392,  p  approaches  C  as  its  limit ; 

L'  approaches  L  as  its  limit, 

and  s'  approaches  s  as  its  limit. 

But,  however  great  the  number  of  faces, 
S'  ^  1  rect.  P  '  L'. 

Hence,  §  326,  S^^  rect.  C  •  L. 

Therefore,  etc.  Q.E.D. 

Arithmetical  Rule  :  To  be  framed  by  the  student. 

624.   Formulae :  Let  R  denote  the  radius  of  the  base  of  a  cone 
of  revolution,  A  its  lateral  area,  and  T  its  total  area. 

Then,  §  395,  A  =  ^(2  ttR  x  L)  =  ttRL, 

and,  §  398,  T=  ttRL  +  irPi^  =  ttR^L  +  R), 


320  SOLID   GEOMETRY.  — BOOK  IX, 

Proposition  X 

625.  ^  Compute  the  areas  of  any  two  similar  cones  of  revolution,  as 
those  whose  altitudes  are  8"  and  4",  slant  heights  10"  and  5",  and  the 
radii  of  whose  bases  are  6"  and  3",  respectively.  How  does  the  ratio  of 
their  lateral  areas,  or  of  their  total  areas,  compare  with  the  ratio  of  the 
squares  of  their  altitudes,  or  with  the  ratio  of  the  squares  of  the  radii  of 
their  bases  ? 

Theorem,  The  lateral  areas,  or  the  total  areas,  of  similar 
cones  of  revolution  are  to  ea/ih  other  as  the  squares  of  their 
altitudes,  or  as  the  squares  of  the  radii  of  their  bases. 

Data :  Any  two  similar  cones  of  revo-  a 

lution,  whose   altitudes  are  H  and  ^,  /!\ 

slant  heights  L  and  i',  and  the  radii  of         /   i   \  /t\ 

whose  bases  are  R  and  R\  respectively.  /      P  \  /  i  '\ 

Denote  their  lateral  areas  by  A  and  /''p  t~~~""A  /"'^~~\ 
A\  and  their  total  areas  by  T  and  r',  re-  (;-— -^  1  ^.-.j^^ 
spectively. 

To  prove  1.    A-.A'^h'^:  H'^  =  i?- :  R'^ 

2.    T:Tf  =  H^:H'^  =  R'':R". 

Proof.     1.   Since  the  generating  triangles  are  similar, 

§§299,279,  E.=.L  =  ^=±±1.; 

'         '  H'      L'       R'      L'-\-R'' 

a  aoA  -^         ttRL        R       L        R^        H 

or  A:A'  =  H^:H'^=I^:R'K 

2     Sfi24.      T-  7rR(L-\-R)    ^R  (  L -{- R\_Iil'  _^ 
^  '    r'       TrR\L'  +  R')       R'\L'-^R'J       R"      H'^ 

or  T:T'  =  H^:H'^  =  R^:  R'^. 

Therefore,  etc.  q.e.d. 

626.  Cor.  TTie  lateral  areas,  or  the  total  areas,  of  similar  cones 
of  revolution  are  to  each  other  as  the  squares  of  their  like  dimensions. 

Ex.  838.  What  is  the  lateral  area  of  a  cone  of  revolution  whose  slant 
height  is  13  ft.  and  the  diameter  of  whose  base  is  6  ft.  ? 

Ex.  839  What  is  the  ratio  of  the  lateral  surface  of  a  right  circular 
cylinder  to  that  of  a  right  circular  cone  having  the  same  base  and  altitude, 
if  the  altitude  is  5  times  the  radius  of  the  base  t 


SOLID   GEOMETRY.— BOOK  IX.  321 

Proposition  XI 

627.  1.  To  what  is  the  lateral  surface  of  a  frustum  of  any  regular 
pyramid  equivalent?  If  the  number  of  its  lateral  faces  is  indefinitely 
increased,  what  solid  does  the  frustum  of  the  pyramid  approach  as  its 
limit?  How,  then,  does  the  lateral  surface  of  the  frustum  of  any  cone 
of  revolution  compare  with  the  rectangle  formed  by  its  slant  height  and 
the  sum  of  the  circumferences  of  its  bases  ? 

2.  How  does  the  circumference  of  a  section  equidistant  from  the  bases 
compare  with  half  the  sum  of  the  circumferences  of  the  bases  ? 

Theorem,  The  lateral  surface  of  a  frustujn  of  a  cone  of 
revolution  is  equivalent  to  one  half  the  rectangle  formed 
by  its  slant  height  and  the  sum  of  the  circumferences  of 
its  bases. 

Data:  Any  frustum  of  a  cone  of  revolution,  as  A, 
whose  slant  height  is  L,  and  the  circumferences  of 
whose  lower  and  upper  bases  are  C  and  C',  respec- 
tively. 

Denote  the  lateral  surface  of  the  frustum  hj  S. 

To  prove        S^^  rect.  L  -  (c -\- C'). 

Proof.  Inscribe  in  the  frustum  of  tlie  cone  the  frustum  of  a 
regular  pyramid,  and  denote  its  lateral  surface  by  s\  its  slant 
iieight  by  L\  and  the  perimeters  of  its  lower  and  upper  bases  by 
P  and  P',  respectively. 

Then,  §  552,  S'  ^  i-  rect.  L'  -  {P  ^  P'). 

Now,  if  the  number  of  lateral  faces  of  the  inscribed  frustum  is 
indefinitely  increased, 
§  392,    P  and  P'  approach  C  and  C',  respectively,  as  their  limits; 

L'  approaches  i  as  its  limit, 
and  S'  approaches  S  as  its  limit. 

But,  however  great  the  number  of  faces, 
S'  o=  i  rect.  z' .  (P  +  P'). 

Hence,  §  326,         S^^  rect.  i  •  (C  +  C'). 

Therefore,  etc.  q.e.d. 

628.  Cor.  The  lateral  surface  of  a  frustum  of  a  cone  of  revolu- 
tion is  equivalent  to  the  rectangle  formed  by  its  slant  height  and  the 
circumference  of  a  section  equidistant  from  its  bases. 

Arithmetical  Rules :  To  be  formed  by  the  student. 
milne's  geom.  ^21 


822  SOLID   GEOMETRY,  — BOOK  IX. 


Proposition  XII 

629.  To  what  is  the  volume  of  any  pyramid  equal?  If  the  number 
of  its  lateral  faces  is  indefinitely  increased,  what  solid  does  the  pyra- 
mid approach  as  its  limit?  To  what,  then,  is  the  volume  of  any  cone 
equal? 

Theorem,  The  volume  of  any  cone  is  equal  to  one  third 
the  product  of  its  base  hy  its  altitude. 


Data:  Any  cone,  as  Ay  whose  base  is  B  and 
altitude  H. 

I>enote  its  volume  by  V. 

To  prove  v=i:lbxh. 


Proof.  Inscribe  in  the  cone  a  pyramid,  and  denote  its  volume 
by  V'  and  its  base  by  B\ 

Then,  the  altitude  of  the  pyramid  is  H, 

and,  §  560,  r  =  ^B'xH. 

Now,  if  the  number  of  lateral  faces  of  the  inscribed  pyramid  is 
indefinitely  increased, 

§  393,  jB'  approaches  B  as  its  limit ; 

F'  approaches  V  as  its  limit. 

But,  however  great  the  number  of  faces, 

Hence,  §  222,  V=\BxH. 

Therefore,  etc.  q.e.d. 

630.   Formula :  Let  R  denote  the  radius  of  a  cone  of  revolution 
Then,  §398,  B  =  tt^', 

V=^TrI^  XB. 

Ex.  840.  The  slant  height  of  a  right  circular  cone  is  21  ft.  and  its  altitude 
is  16  ft.     What  is  its  total  area  ? 

Ex.  841.  The  slant  height  of  a  right  circular  cone  is  6™  and  the  radius 
of  its  base  is  6"™.     What  is  its  lateral  area  ?    What  is  its  volume  ? 


SOLID   GEOMETRY.  — BOOK  IX.  323 


Proposition  XIII 

631.  Compute  the  volumes  of  any  two  similar  cones  of  revolution,  as 
those  whose  altitudes  are  8"  and  4",  and  the  radii  of  whose  bases  a^e  6" 
and  3"  respectively.  How  does  the  ratio  of  their  volumes  compare  with 
the  ratio  of  the  cubes  of  their  altitudes,  or  with  the  ratio  of  the  cubes 
of  the  radii  of  their  bases  ? 

Theorem,  The  volumes  of  similar  cones  of  revolution  are 
to  each  other  as  the  cubes  of  their  altitudes,  or  as  the  cubes 
of  the  radii  of  their  bases. 

Data :  Any  two  similar  cones  of  revo-  A 

lution,   whose  altitudes  are  H  and  H',  /  \\ 

and  the  radii  of  whose  bases  are  R  and  /    j    \  A 

R\  respectively.  /        '      \  /  lA 

Denote  their  volumes  by  V  and  F*,  re-  .  /-''^  i  ""A  1''e'\~'\ 
spectiv^y.  V"  J       \^I_-^ 

To  prove      V:  r'  =  H^:  h'^  =  I^i  B'K 

Proof.     Since  the  generating  triangles  are  similar, 

^299,  |  =  |; 

or  V'.  r  =  H^:H'^  =  Ii^:Ii'\ 

Therefore,  etc.  q.e.d. 

632.  Cor.  The  volumes  of  similar  cones  of  revolution  are  to  each 
other  as  the  cubes  of  any  of  their  like  dimensions. 

Ex.  842.    What  is  the  volume  of  a  cone  whose  altitude  is  13  ft.  and  the 

circumference  of  whose  base  is  9  ft.  ? 

Ex.  843.  What  is  the  total  area  of  the  frustum  of  a  cone  of  revolution 
whose  slant  height  is  n^"^  and  the  radii  of  whose  bases  are  5^™  and  3^^™  ? 

Ex.  844.  How  far  from  the  base  must  a  cone  of  revolution  whose  altitude 
is  15  in.  be  cut  by  a  plane  parallel  to  the  base  so  that  the  volume  of  the  frus- 
tum shall  be  one  half  that  of  the  entire  cone  ? 

Ex.  845.  At  what  distances  from  the  vertex  must  a  cone  of  revolution  be 
cut  by  planes  parallel  to  the  base  to  divide  it  into  three  equivalent  solids  ? 

Ex.  846.  A  plane  parallel  to  the  base  of  a  cone  of  revolution  cuts  the 
altitude  at  a  point  |  of  the  distance  from  the  vertex.  What  is  the  ratio  of 
the  volume  of  the  cone  cut  off  to  that  of  the  original  cone  ? 


324  SOLID  GEOMETRY.  — BOOK  IX. 

Proposition  XIV 

633.  To  what  is  the  volume  of  a  frustum  of  any  pyramid  equal?  If 
the  number  of  its  faces  is  indefinitely  increased,  what  solid  does  the 
frustum  of  the  pyramid  approach  as  its  limit?  To  what,  then,  is  the 
volume  of  the  frustum  of  any  cone  equal  ? 

Theorem.  The  volume  of  a  frustum  of  any  cone  is  equal 
to  one  third  the  product  of  its  altitude  hy  the  sum  of  its 
bases  and  a  mean  proportional  between  them. 

Data :  Any  frustum  of  any  cone,  as  A,  whose 
altitude  is  H,  and  whose  lower  and  upper  bases 
are\B  and  b  respectively. 

Denote  the  volume  of  the  frustum  by  V. 


To  prove       V=^\H(b  +  b  +V-B  x  h). 


• 


Proof.  Inscribe  in  the  frustum  of  the  cone  the  frustum. of  a 
pyramid,  and  denote  its  volume  by  F'  and  its  lower  and  upper 
bases  by  B'  and  &',  respectively. 

Then,  the  altitude  of  the  frustum  of  the  pyramid  is  Hy 
and,  §  564,  r  =  \  h(b'  +  b' +  ^W^Cb^). 

Now,  if  the  number  of  lateral  faces  of  the  inscribed  frustum  is 
indefinitely  increased, 

§  393,  B^  and  6'  approach  B  and  h  respectively  as  their  limits ; 
F'  approaches  V  as  its  limit. 

But,  however  great  the  number  of  faces, 

r  =  iH(B'  +  &'  +  V^'  X  6'). 

Hence,  §  222,      F  =  ^  h(b  +  &  +  V^  X  b). 

Therefore,  etc.  _  q.e.d. 

634  Formula :  Let  R  and  R'  denote  the  radii  of  the  baees  of  a 
frustum  of  a  cone  of  revolution. 

Then,  §  398,        B^ttI^,  b  =  irR^\ 
and  -VBxb  =  vRR' ; 

Ex.  847.  What  is  the  volume  of  the  frustum  of  a  cone  whose  altitude  is 
21  ft.  and  the  circumferences  of  whose  bases  are  17  ft.  and  13  ft.  respectively  ? 


SOLID   GEOMETRY.  — BOOK  IX. 


325 


635.                                    FORMULA 

Notation 

B   =  base. 

R 

=  radius  of  base. 

h    z=z  upper  base. 

R' 

=  radius  of  upper  base. 

C   =  circumference  of  base. 

H 

=  altitude. 

C'  =  circumference  of  upper 

L 

=  slant  height. 

base. 

A 

=  lateral  area. 

C"  =  circumference    of    mid- 

T 

=  total  area. 

section. 

V 

=  volume. 

Cylinder  of  Revolution. 
A=C  X  H. 
A  =  2  ttRH. 

T  =  2  ttR  (H -^  R). 

V—  B  X  H.    (True  also  for  any  cylinder.) 
V  =  ttR^H.  % 


Cone  of  Revolution. 

A  =  ^C  X  L. 

A  =  ttRL. 

T=zttR{L-\-  R). 

V=\B  X  H.     (True  also  for  any  cone.)' 

F  =  i  TT^H. 


Frustum  of  a  Cone  of  Revolution. 

^  =  |Z((7+C'). 
A  =  LX  C". 


V=^H(B  +  b  H-V5  X  b). 

r=l7rH(R'+R'^  +  RR'). 


(True  also  for  any  frustum.) 


SUPPLEMENTARY   EXERCISES 

Ex.  848.    What  is  the  lateral  area,  total  area,  and  volume  of  a  cylinder 
of  revolution  whose  diameter  is  8  in.  and  altitude  12  in.  ? 

Ex.  849.   What  is  the  lateral  area,  total  area,  and  volume  of  a  cone  of 
revolution  whose  base  is  10^™  in  diameter  and  whose  altitude  is  12^™  ? 
•   Ex.  850.    What  is  the  lateral  area,  total  area,  and  volume  of  a  frustum 
of  a  cone  of  ravolution,  the  radii  of  whose  bases  are  6  in.  and  4  in.,  respec- 
tively, and  whose  altitude  is  9  in.  ? 


326  SOLID  GEOMETRY.^BOOK  IX. 

Ex.  851.  On  a  cylindrical  surface  only  one  straight  line  can  be  drawn 
through  a  given  point. 

Ex.  852.  The  intersection  of,  two  planes  tangent  to  a  cylinder  is  parallel 
to  an  element. 

Ex.  833.  How  many  square  yards  of  canvas  are  required  for  a  conical 
tent  18  ft.  high  and  10  ft.  in  diameter? 

Ex.  854.  How  many  cubic  feet  are  there  in  a  piece  of  round  timber  40  ft. 
long,  whose  ends  are  respectively  3  ft.  and  1  ft.  in  diameter  ? 

Ex.  855.  A  cylindrical  vessel  is  40<im  long  and  20^™  in  diameter.  What 
is  the  weight,  in  grams,  of  the  water  that  it  will  hold  ? 

Ex.  856.  What  is  the  weight,  in  grams,  of  a  piece  of  lead,  if  when  put 
Under  water  in  a  cylindrical  tank  24<^i"  in  diameter  it  causes  the  level  of  the 
water  to  rise  S^m,  the  specific  gravity  of  lead  being  11.4  ? 

Ex.  857.  A  cylindrical  cistern  is  6.4™  deep  and  5.2™  in  diameter.  How 
long  will  it  take  to  fill  it,  if  2^1  flow  into  it  per  minute  ? 

Ex.  858.  A  plane  through  a  tangent  to  the  base  of  a  circular  cylinder 
and  the  element  drawn  to  the  point  of  contact,  is  tangent  to  the  cylinder, 

Ex.  859.  A  plane  through  a  tangent  to  the  base  of  a  circular  cone  and 
the  element  drawn  to  the  point  of  contact,  is  tangent  to  the  cone. 

Ex.  860.  The  volumes  of  two  similar  cylinders  of  revolution  are  as  27  :  64. 
If  the  diameter  of  the  first  is  3  ft.,  what  is  the  diameter  of  the  second  ? 

Ex.  861.  A  cylindrical  vessel  holds  1728  grams  of  water.  What  are  the 
dimensions  of  the  vessel,  if  the  diameter  is  one  third  of  the  altitude  ? 

Ex.  862.  Show  that  any  lateral  face  of  a  pyramid  circumscribed  about  a 
circular  cone  is  tangent  to  the  cone. 

Ex.  863.  What  is  the  height  of  a  cylinder  4.8<^™  in  diameter,  if  it  is 
equivalent  to  a  cone  of  revolution  6.6<i™  in  diameter  and  6.4^™  high  ? 

Ex.  864.  A  cylindrical  vessel  is  12*=™  in  diameter  and  20"™  high.  How 
many  grams  of  mercury  would  it  hold,  the  specific  gravity  of  mercury 
being  13.6  ? 

Ex.  865.  The  volumes  of  two  similar  cones  of  revolution  are  to  each  other 
as  512  :  729.     What  is  the  ratio  of  their  lateral  areas  ? 

Ex.  866.  How  many  centigrams  of  alcohol  will  a  cylindrical  bottle  hold, 
if  the  bottle  is  8<=™  in  diameter  and  24^™  high,  the  specific  gravity  of  alcohol 
being  .79  ? 

Ex.  867.  The  specific  gravity  of  marble  is  2.8.  What  is  the  weight,  in 
kilograms,  of  a  conical  piece  of  marble,  if  the  radius  of  its  base  is  20^™  and  its 
height  50c™  ? 

Ex.  868.  The  slant  height  of  a  cone  of  revolution  is  3™.  How  far  from 
the  vertex  must  the  elements  be  cut  by  a  plane  parallel  to  the  base  in  order 
that  the  lateral  surface  may  be  divided  into  two  equivalent  parts  ? 

Ex.  869.  If  the  altitude  of  a  cylinder  of  revolution  is  equal  to  the 
diameter  of  its  base,  the  volume  is  equal  to  the  product  of  its  total  area  by 
one  third  of  its  radius. 


BOOK  X 

SPHERES 

^36.    A  solid  bcOiiided  by  a  surface,  every  point  of  which  is 
equally  distant  from  a  point  within,  is  called  a  Sphere.''^ 

The  point  within  is  called  the  center. 

A  sphere  may  be  generated  by  the  revolu- 
tion of  a  semicircle  about  its  diameter  as  an 
axis. 


637.  A  straight  line  drawn  from  the  center 
to  any  point  of  the  surface  of  a  sphere  is 
called  a  radius. 

A  straight  line  which  passes  through  the  center  of  a  sphere, 
and  whose  extremities  are  in  the  surface,  is  called  a  diameter. 

638.  A  line  or  plane  which  has  one,  and  only  one,  point  in 
common  with  the  surface  of  a  sphere  is  tangent  to  the  sphere. 

The  sphere  is  then  said  to  be  tangent  to  the  line  or  plane. 

639.  Two  spheres  whose  surfaces  have  one,  and  only  one,  point 
in  common  are  tangent  to  each  other. 

640.  When  all  the  faces  of  a  polyhedron  are  tangent  to  a 
sphere,  the  sphere  is  said  to  be  inscribed  in  the  polyhedron. 

641.  When  all  the  vertices  of  a  polyhedron  lie  in  the  surface 
of  a  sphere,  the  sphere  is  said  to  be  circumscribed  about  the  poly- 
hedron. 

*  In  teaching  Spherical  Geometry,  the  class-room  should  be  furnished  with 
a  spherical  blackboard,  on  which  the  student  should  draw  the  diagrams 
required.  It  is  also  advised  that  each  student  be  provided  with  some  sort  of  a 
blackened  or  slated  sphere  for  use  in  the  preparation  of  lessons  in  cases 
where  figures  are  to  be  drawn  on  its  surface.  A  hemispherical  cup  to  fit  the 
sphere  will  enable  him  to  draw  great  circles  on  the  sphere. 

327 


328  SOLID   GEOMETRY.— BOOK  X. 

642.    A-x.  18.    All  radii  of  the  same  sphere,  or  of  equal  spheres, 
are  equal. 

19.   All  diameters  of  the  same  sphere,  or  of  equal  spheres,  are 


20.    Two  spheres  are  equal,  if  their  radii  or  diameters  are  equal. 

Proposition  I 

643.  1.  Form  a  sphere  and  cut  it  by  any  plane  (§  519  n.).  What 
plane  figure  is  the  section  thus  formed? 

2.  If  a  line  joins  the  center  of  the  sphere  with  the  center  of  a  circle 
of  the  sphere,  what  is  its  direction  with  reference  to  the  plane  of  the 
circle  ? 

3.  Cut  a  sphere  by  planes  which  are  equally  distant  from  its  center. 
How  do  the  sections  thus  formed  compare  ? 

4.  If  the  cutting  planes  are  unequally  distant  from  the  center,  which 
circle  is  the  larger  ? 

Theorem,  Any  section  of  a  sphere  -made  hy  a  plane  is 
a  circle. 


Data :  A  sphere ;  its  center  0 ;  and  any  section,  as  ABD. 
To  prove  ABD  a  circle. 

Proof.  Draw  OC  perpendicular  to  the  plane  ABD ;  draw  the 
radii  OA  and  OD  to  any  two  points  in  J;he  perimeter  of  the  sec- 
tion ;  and  draw  CA  and  CD. 

Since  0  is  a  point  in  the  perpendicular  0(7, 

and,  Ax.  18,  OA  =  OD, 

§  449,  CA  =  CD. 

But  A  and  D  are  any  two  points  in  the  perimeter  of  section 
ABD. 

Hence,  §  173,     ABD  is  a  circle  whose  center  is  C.  q.e.d. 


SOLID  GEOMETRY.  — BOOK  X.  329 

644.  Cor.  I.  77ie  line  Joining  the  center  of  a  sphere  to  the  center 
of  a  circle  of  the  sphere  is  perpendicular  to  the  plane  of  the  circle. 

645.  Cor.  II.  Circles  of  a  sphere  made  by  planes  equally  distant 
from  the  center  are  equal. 

646.  Cor.  III.  Of  two  circles  of  a  sphere  made  by  planes  un- 
equally distant  from  the  center,  the  nearer  is  the  larger. 

647.  A  section  of  a  sphere  made  by  a  plane  which  passes 
through  the  center  is  called  a  Great  Circle  of  the  sphere. 

648.  A  section  of  a  sphere  made  by  a  plane  which  does  not 
pass  through  the  center  is  called  a  Small  Circle  of  the  sphere. 

649.  The  diameter,  which  is  perpendicular  to  the  plane  of  a 
circle  of  a  sphere,  is  called  the  Axis  of  the  circle. 

650.  The  ends  of  the  axis  of  a  circle  of  a  sphere  are  called 
the  Poles  of  the  circle. 

651.  1.  Form  a  sphere  and  cut  it  by  any  plane,  thus  forming 
a  circle  of  the  sphere.  Through  what  point  of  the  circle  does 
its  axis  pass  ? 

2.  Form  a  sphere  and  cut  it  by  two  parallel  planes,  thus  form 
ing  two  parallel  circles.  How  are  their  axes  situated  with  refer- 
ence to  each  other?  How,  then,  are  the  poles  of  one  of  these 
circles  situated  with  reference  to  the  poles  of  the  other  ? 

3.  By  passing  planes  form  any  two  great  circles  of  the  same, 
or  of  equal  spheres.     How  do  they  compare  with  each  other  ? 

4.  Form  a  sphere  and  divide  it  into  two  parts  by  a  great  circle. 
How  do  these  parts  compare  with  each  other  ? 

5.  By  passing  planes  form  any  two  great  circles  of  a  sphere. 
Since  their  intersection  passes  through  the  center  and  is  a  diame- 
ter of  each  circle,  how  do  two  great  circles  divide  each  other  ? 

6.  Form  two  great  circles  of  a  sphere  by  passing  two  planes 
through  it  perpendicular  to  each  other.  Where  do  these  circles 
pass  with  reference  to  each  other's  poles?  If  two  great  circles 
pass  through  each  other's  poles,  what  is  the  direction  of  their 
planes  with  reference  to  each  other  ? 

7.  Form  a  sphere  and  pass  a  plane  through  its  center  and  any 
two  points  on  its  surface.  What  kind  of  a  circle  is  the  section 
thus  formed  ?  What  kind  of  an  arc,  then,  may  be  drawn  through 
any  two  points  on  the  surface  of  a  sphere  ? 


330  SOLID   GEOMETRY.  — BOOK  X. 

8.  Form  a  sphere  and  pass  a  plane  through  any  three  points 
on  its  surface.  What  plane  figure  is  the  section  thus  formed? 
How  many  planes  may  be  passed  through  the  three  points? 
How  many  circles,  then,  may  be  drawn  through  any  three  points 
on  the  surface  of  a  sphere  ? 

652.  Tlie  axis  of  a  circle  passes  through  the  center  of  that  circle. 

653.  Parallel  circles  have  the  same  axis  and  the  sayne  poles. 

654.  Cfreat  circles  of  the  same  sphere,  or  of  equal  spheres,  are 
equal. 

655.  Any  great  circle  of  a  sphere  bisects  the  sphere. 

656.  Ttvo  great  circles  of  the  same  sphere  bisect  each  other. 

657.  Two  great  circles  whose  planes  are  perpendicular  pass 
through  each  other^s  poles;  and  conversely. 

658.  Through  tico  given  points  on  the  surface  of  a  sphere  an  arc 
of  a  great  circle  may  be  draivn. 

659.  Through  three  given  points  on  the  surface  of  a  sphere  one 
circle  may  be  drawn,  and  only  one. 

Proposition  II 

660.  Form  a  sphere  and  select  any  two  points  on  its  surface ;  through 
these  points  and  the  center  of  the  sphere  pass  a  plane.  What  kind  of  a 
circle  is  this  section?    Then,  what  kind  of  an  arc  joins  the  given  points? 

Join  them  by  any  other  line  on  the  surface.  Which  line  represents 
the  shortest  distance  between  the  given  points  ? 

Theorem,  The  shortest  distance  on  the  surface  of  a  sphere 
hetween  any  two  -points  on  that  surface  is  the  arc,  not  greater 
than  a  semicircujnference,  of  the  great  circle  which  joins 
them.  ,  ^-^-^ — ^ 


Data :  Any  two  points  on  the  surface  of  a  sphere,  as  A  and  B^ 
joined  by  the  arc  of  a  great  circle,  as  AB,  not  greater  than  a  semi- 


SOLID   GEOMETRY.— BOOK  X.  331 

circumference  ;  also  any  other  line  on  the  surface  joining  A  and  B, 
as  AECB. 

To  prove  AB  less  than  AECB. 

Proof.  Take  any  point  in  AECB,  as  D,  and  pass  arcs  of  great 
circles  through  A  and  D,  and  B  and  D.  Draw  OA,  OB,  and  OD  from 
0,  the  center  of  the  sphere. 

■  Then,  Aaob,  ADD,  and  BOD  are  the  face  A  of  the  trihedral  Z 
whose  vertex  is  at  0 ; 

.-.  §  498,         Z  AOD  -f  Z  jBOJD  is  greater  than  Z  ^05. 

But,  §  224,  arcs  AD,  BD,  and  AB  are  the  measures  of  A  AOD, 
BOD,  and  ^05  respectively ; 

arc  AD  +  arc  BD  >  arc  ^5. 

In  like  manner,  joining  any  point  in  AED  with  A  and  D,  and  any 
point  in  DCB  with  Z>  and  B  by  arcs  of  great  circles,  the  sum  of 
these  arcs  will  be  greater  than  arc  AD  -\-  arc  BD,  and  therefore 
greater  than  arc  AB. 

If  this  process  is  indefinitely  repeated,  the  points  common  to 
AECB  and  the  path  from  ^  to  ^  on  the  great  circle  arcs  will 
approach  as  near  each  other  as  we  please,  and  the  sum  of  these 
arcs  will  continually  increase  and  approach  AECB  as  a  limit. 

But  the  sum  of  the  great  circle  arcs  is  always  greater  than  AB. 

Therefore,  AB  is  less  than  AECB.  q.e.d. 

661.  By  the  distance  between  two  points  on  the  surface  of  a 
sphere  is  meant  the  shortest  distance;  that  is,  the  arc  of  a  great 
circle  joining  them. 

66^.  The  distance  from  the  nearer  pole  of  a  circle  to  any  point 
in  its  circumference  is  called  the  Polar  Distance  of  the  circle. 

Proposition  III 

663.  1.  Form  a  sphere  and  cut  it  by  ^any  plane;  pass  planes  through 
the  axis  of  the  circle  thus  formed  and  any  points  in  its  circumference. 
What  kind  of  arcs,  then,  connect  the  pole  of  the  circle  and  the  points  of 
its  circumference?  How  do  these  arcs  compare?  Then,  how  do  the 
distances  from  the  pole  of  a  circle  of  a  sphere  to  all  the  points  in  its 
circumference  compare? 


382 


SOLID   GEOMETRY.  —  BOOK  X. 


2.  If  the  circle  is  a  great  circle,  what  part  of  a  circumference  is  its 
polar  distance  ? 

3.  By  passing  planes  form  two  equal  circles  of  the  same  or  of  equal 
spheres.     How  do  their  polar  distances  compare  ? 

4.  Select  a  point  on  the  surface  of  a  sphere  which  is  at  a  quadrant's 
distance  from  each  of  two  other  points.  Where  is  this  point  situated 
with  reference  to  a  pole  of  a  great  circle  that  passes  through  the  other 
two  points  ? 

Theorem,  All  points  in  the  circumference  of  a  circle  of 
a  sphere  are  equally  distant  from  a  pole  of  the  circle. 


■p' 

Data :  Any  circle  of  a  sphere,  as  ABC,  and  its  poles,  P  and  P'. 

To  prove  all  points  in  the  circumference  of  ABC  equally  distant" 
from  P  and  also  from  P\ 

Proof.     Draw  great  circle  arcs  from  P  to  any  points  in  the 
circumference  of  ABC,  as  A,  B,  and  C. 

§§  649,  652,  PP'  A.  ABC  at  its  center ; 

.*.  §  450,  chords  PA,  PB,  and  PC  are  equal ; 

hence,  §  196,  arcs  PA,  PB,  and  PC  are  equal. 

In  like  manner,  arcs  P^A,  P'B,  and  P'C  may  be  proved  equal. 

But  A,  B,  and  C  are  any  points  in  the  circumference  of  ABC. 

Hence,  §  661,  all  points  in  the  circumference  of  ABC  are  equally 
distant  from  P  and  als^  from  P'. 

Therefore,  etc.  q.e.d. 

664.  Cor.  I.     TJie  polar  distance  of  a  great  circle  is  a  quadrant* 

665.  Cor.  II.     The  polar  distances  of  equal  circles  on  the  same^ 
or  on  equal  spheres,  are  equal. 


*  In  Spherical  Geometry  the  term  quadrant  generally  means  the  quadrant 
of  a  great  circle. 


SOLID   GEOMETRY.  — BOOK  X.  333 

666.  Cor.  III.  A  point,  which  is  at  the  distance  of  a  quadrant 
from  each  of  iwo  other  points  on  the  surface  of  a  sphere^  is  a  pole 
of  the  great  circle  passing  through  those  points. 

667.  Sch.  I.  By  using  the  facts  demonstrated  in  §  663  and  in 
§  664  we  may  draw  the  circumferences  of 

small  and  great  circles  on  the  surface  of  a  .-Jl 

material  sphere.  ,  IV'^ 

To  draw  the  circumference  of  a  circle, 
take  a  cord  equal  to  its  polar  distance,  and, 
placing  one  end  of  it  at  the  pole,  cause  a 
pencil  held  at  the  other  to  trace  the  cir- 
cumference, as  in  the  fignire. 

To  describe  the  circumference  of  a  great 
circle,  a  quadrant  must  be  used  as  the  arc. 

668.  Sch.  II.  By  means  of  §  QQQ  we  are  enabled  to  pass  the 
circumference  of  a  great  circle  through  any 

two  points,  as  A  and  B,  on  the  surface  of  a 
material  sphere  in  the  following  manner : 

From  each  of  the  given  points,  as  poles, 
and  with  a  quadrant  arc,  draw  arcs  to  inter-       \^ 
sect,  as  at  0.     The  circumference  described        \ 
from  this  intersection  with  a  quadrant  arc 
will  be  the  circumference  required. 

Proposition  IV 

669.  1.  If  a  plane  is  perpendicular  to  a  radius  of  a  sphere  at  its  ex- 
tremity, how  many  points  do  the  sphere  and  the  plane  have  in  common  ? 
What  name  is  given  to  such  a  plane  ? 

2.  How  many  points  do  a  straight  line  and  a  sphere  have  in  common, 
if  the  line  is  perpendicular  to  a  radius  of  the  sphere  at  its  extremity? 
What  name  is  given  to  such  a  line? 

3.  What  is  the  direction  of  every  plane  or  line,  that  is  tangent  to  a 
sphere,  with  reference  to  the  radius  drawn  to  the  point  of  contact? 

4.  If  a  straight  line  is  tangent  to  any  circle  of  a  sphere,  how  does  it 
lie  with  reference  to  a  plane  tangent  to  the  sphere  at  the  point  of  contact? 

5.  If  a  plane  is  tangent  to  a  sphere,  what  is  the  relation  to  the  sphere 
of  any  line  drawn  in  that  plane  and  through  the  point  of  contact? 

6.  If  two  straight  lines  are  tangent  to  a  sphere  at  the  same  point, 
what  is  the  relation  of  the  plane  of  those  lin^s  tQ  the  sphere  ? 


334 


SOLID   GEOMETRY.  — BOOK  X. 


Theorem,    A  plane  perpendicular  to  a  radius  of  a  sphere 
at  its  extremity  is  tangent  to  the  sphere. 


'ml 


J 


Data :   Any  sphere,  and  a  plane,  as  MN,  perpendicular  to  a 
radius,  as  OP,  at  its  extremity  P. 

To  prove  MN  tangent  to  the  sphere. 

Proof.     Take  any  point  except  P  in  MN,  as  A,  and  draw  OA. 

Then,  §448,  OP<OA', 

point  A  is  without  the  sphere. 
But  A  is  any  point  in  MN  except  P ; 

every  point  in  MN  except  P  is  without  the  sphere. 
Hence,  §  638,     MN  is  tangent  to  the  sphere  at  P.  q.e.d. 

670.  Cor.  I.     Any  straight  line  perpendicular  to  a  radius  of  a 
sphere  at  its  extremity  is  tangent  to  the  sphere. 

671.  Cor.  II.     A.iy  plane  or  line  tangent  to  a  sphere  is  perpen- 
dicular to  the  radius  drawn  to  the  point  of  contact. 

672.  Cor.  III.     A  straight  line  tangent  to  any  circle  of  a  sphere 
lies  in  the  plane  tangent  to  the  sphere  at  the  point  of  contact. 

673.  Cor.  IV.     Any  straight  line  drawn  in  a  tangent  plane  and 
through  the  point  of  contact  is  tangent  to  the  sphere  at  that  point. 

674.  Cor.  V.     Any  two  straight  lines  tangent  to  a  sphere  at  the 
same  point  determine  the  tangent  plane  at  that  point. 


Proposition  V 

675.  Select  any  four  points  not  in  the  same  plane;  form  a  tetrahe- 
dron of  which  these  points  are  the  vertices ;  and  at  the  centers  of  the 
circles  circumscribed  about  any  two  of  its  faces  erect  perpendiculars. 
How  do  the  distances  from  any  point  in  either  perpendicular  to  the 
vertices  of  the  face  to  which  it  is  perpendicular  compare  ?    If  these  per- 


SOLID   GEOMETRY.  — BOOK  X.  335 

pendiculars  intersect,  how,  then,  do  the  distances  from  their  intersection 
to  the  four  given  points  compare  ?  Is  there  any  other  point  that  is  equi- 
distant from  the  four  given  points  ?  What  surface,  then,  may  be  passed 
through  these  four  points?  How  many  such  surfaces  may  be  passed 
through  them? 

Theorem,  Through  any  four  points  not  in  the  same 
plane  one  spherical  surface  may  be  passed,  and  only  one. 

Data :  Any  four  points  not  in  the  same 
plane,  as  A,  B,  c,  D. 

To  prove  that  one  spherical  surface  may  be 
passed  through  A,  B,  C,  B,  and  only  one. 

Proof.  Suppose  H  and  G  to  be  the  centers 
of  circles  circumscribed  about  the  triangles 
BOB  and  ACB,  respectively. 

Draw  HK±  plane  BOB,  and  GE  ±  plane  ACB. 

§  450,  every  point  in  HK  is  equidistant  from  points  B,  C,  B,  and 
every  point  in  GE  is  equidistant  from  points  A,  C,  B. 

Erom  H  and  G  draw  lines  to  L,  the  middle  point  of  CB. 

Then,  §  106,  HL  ±  CB,  and  GL±CB', 

.*.  §  444,  the  plane  through  HL  and  GL  is  perpendicular  to  CB, 
and,  §  483,  this  plane  is  perpendicular  to  planes  BCB  and  ACB. 

Const.,  GE  J.  Tpl^ne  ACB  Sit  G', 

.-.  §  481,  '     GE  lies  in  the  plane  HLG. 

In  like  manner  it  may  be  shown  that  HK  lies  in  plane  HLG. 

Hence,  tlie  perpendiculars  HK  and  GE  lie  in  the  same  plane, 
and,  being  perpendicular  to  planes  which  are  not  parallel,  they 
must  intersect  at  some  point,  as  at  O. 

Since  O  is  an  the  perpendiculars  HK  and  GE,  it  is  equidistant 
from  B,  C,  B,  and  from  A,  C,  B. 

Hence,  O  is  equidistant  from  A,  B,  C,  B,  and  the  surface  of  the 
sphere,  whose  center  is  O  and  radius  OB,  will  pass  through  the 
points  A,  B,  C,  B. 

Now,  §  450,  the  center  of  any  sphere  whose  surface  passes 
through  the  four  points  A,  B,  C,  B  must  be  in  the  perpendiculars 
HK  and  GE. 

Hence,  0,  the  intersection  of  HK  and  GE,  must  be  the  center  of 
the  07ily  sphere  whose  surface  can  pass  through  Aj  B,  G,  B. 

Therefore,  etc.  q.e.d. 


336  SOLID   GEOMETRY.  — BOOK  X. 

676.  Cor.  I.     A  sphere  may  he  circumscribed  about  any  tetra- 
hedron. 

677.  Cor.  II.     The  four  perperidiculars  to  the  faces  of  a  tetrahe- 
dron through  their  centers  meet  at  the  same  point. 


Proposition  VI 

678.  Form  any  tetrahedron  and  pass  planes  bisecting  any  three  of  its 
dihedral  angles  which  have  one  face  in  common.  How  do  the  distances 
from  the  point  of  intersection  of  these  bisecting  planes  to  the  faces  of  the 
tetrahedron  compare  ?  What  figure,  then,  may  be  inscribed  in  any  tetra- 
hedron ? 

Theorem,    A  sphere  may  be  inscribed  in  any  tetrahedron. 
Ad  aP 


Datum :   Any  tetrahedron,  as  D-ABC. 

To  prove  that  a  sphere  may  be  inscribed  in  D-ABC. 

Proof.  Bisect  any  three  of  the  dihedral  angles  which  have  one 
face  common,  as  AB,  BC,  AC,  by  the  planes  GAB,  OBC,  OAC,  respec- 
tively. 

By  §  488,  every  point  in  the  plane  OAB  is  equidistant  from  the 
faces  ABC  and  ABD. 

Also  every  point  in  plane  OBC  is  equidistant  from  the  faces 
ABC  and  BCD;  and  every  point  in  the  plane  OAC  is  equidistant 
from  the  faces  ABC  and  ACD. 

Therefore,  point  0,  the  intersection  of  these  three  planes,  is 
equidistant  from  the  four  faces  of  the  tetrahedron. 


SOLID   GEOMETRY.  — BOOK  X,  337 

Hence,  §  638,  a  sphere  with  0  as  a  center  and  with  a  radius  equal 
to  the  distance  from  0  to  any  face  will  be  tangent  to  each  face, 
and,  §  640,  it  will  be  inscribed  in  the  tetrahedron. 

Therefore,  etc.  q.e.d. 

679.  Cor.  The  six  planes  which  bisect  the  six  dihedral  angles  of 
a  tetrahedron  intersect  in  the  same  point.     . 

Proposition  VII 

680.  Problem,    To  find  the  radius  of  a  material  spliere. 


p 


p 


/ 


Datum:  Any  sphere,  as  APP'. 

Required  to  find  the  radius  of  APP^. 

Solution.  From  any  point  P  as  a  pole  describe  any  circumfer- 
ence on  the  surface.  §  667.  From  any  three  points  in  this  cir- 
cumference, as  A,  B,  C,  measure  the  chord  distances  AB,  BC,  AC. 
(Use  compasses  with  curved  branches.) 

Construct  the  Aa'b'c'  having  its  sides  equal  respectively  to 
AB,  BC,  AC,  and  circumscribe  about  it  a  circle. 

With  the  radius  OB'  as  a  side,  and  PB'  equal  to  the  chord  of  the 
arc  joining  P  and  B,  as  hypotenuse,  construct  the  rt.  Apb'O. 

Draw  a  line  from  B'  perpendicular  to  B'P,  and  produce  it  to 
meet  PO  produced  in  P'. 

Then,  PP'  thus  determined  is  equal  to  the  diameter  of  the  sphere, 
and  its  half,  PG,  is  the  required  radius.  q.e.f. 

Proof.     By  the  student. 

Ex.  870.  Equal  straight  lines  whose  extremities  are  in  the  surface  of  a 
sphere  are  equally  distant  from  the  center  of  the  sphere. 

Ex.  871.    The  six  planes  which  bisect  at  right  angles  the  six  edges  of  a 
tetrahedron  all  intersect  at  the  same  point. 
milne's  geom. — 22 


338  SOLID   GEOMETRY.  —  BOOK  X. 

SPHERICAL   ANGLES  AND   POLYGONS 

681.  The  angle  betiveen  two  intersecting  curves  is  the  angle  con- 
tained by  the  two  tangents  to  the  curves  at  their  intersection. 

682.  The  angle  between  two  intersecting  arcs  of  great  ciicles 
is  called  a  Spherical  Angle. 

683.  A  portion  of  the  surface  of  a  sphere  bounded  by  three  or 
more  arcs  of  great  circles  is  called  a  Spherical 
Polygon. 

The  bounding  arcs  are  the  sides  of  the  poly- 
gon; the  angles  which  they  form  are  the 
angles  of  the  polygon ;  and  the  points  of  inter- 
section are  the  vertices  of  the  polygon. 

An  arc  of  a  great  circle  joining  any  two 
non-adjacent  vertices  of  a  spherical  polygon  is  a  diagonal. 

684.  The  planes  of  the  sides  of  a  spherical  polygon  form  a 
polyhedral  angle  whose  vertex  is  the  center 
of  the  sphere,  and  whose   face   angles   are 
measured  by  the  sides  of  the  polygon. 

O- ABODE  is  a  polyhedral  angle  whose  vertex  O 
is  the  center  of  the  sphere  and  whose  face  angles 
JE'OD,  DOC,  etc.,  are  measured  by  the  sides  ED, 
DC,  etc.,  of  the  spherical  polygon  ABODE. 

685.  A  spherical  polygon  whose  corresponding  polyhedral  angle 
is  convex  is  called  a  Convex  Spherical  Polygon. 

Spherical  polygons  will  be  regarded  as  convex  unless  otherwise 
specified. 

686.  A  spherical  polygon  of  three  sides  is  called  a  Spherical 
Triangle. 

A  spherical  triangle  is  right,  oblique,  equilateral,  isosceles,  etc., 
under  the  same  conditions  that  plane  triangles  are  right,  oblique,  etc. 

687.  The  sides  of  a  spherical  polygon,  being  arcs,  are  usually 
measured  in  degrees,  minutes,  and  seconds. 

688.  Any  two  points  on  the  surface  of  a  sphere  may  be  joined 
by  two  arcs  of  a  great  circle,  one  of  which  will  usually  be  greater 
and  the  other  usually  less  than  a  semicircumference. 

Unless  otherwise  stated  the  less  arc  is  always  meant. 


SOLID   GEOMETRY.  —  BOOK  X  839 


Proposition  VIII 

689.  1.  Form  a  sphere  and  construct  on  it  a  spherical  angle  (§  667) ; 
pass  planes  through  the  sides  of  the  angle  and  the  center  of  the  sphere ; 
on  the  part  thus  cut  out  draw  a  great  circle  arc  with  the  vertex  of  the 
angle  as  a  pole,  and  draw  the  radii  to  the  extremities  of  this  arc.  How 
does  the  angle  between  the  radii  compare  with  the  angle  contained  by 
the  tangents  to  the  sides  of  the  spherical  angle  at  their  intersection? 
What  arc  measures  the  angle  between  the  radii?  Then,  what  arc 
measures  the  given  spherical  angle? 

2.  How  does  a  spherical  angle  compare  with  the  dihedral  angle 
formed  by  the  planes  of  its  sides? 

Theorem,  A  spherical  angle  is  measured  hy  the  are  of 
a  great  circle  described  from  its  vertex  as  a  pole  and  irv- 
eluded  hy  its  sides,  produced  if  necessary. 


Data:  Any  spherical  angle,  as  APB, 
and  the  arc  of  a  great  circle,  as  AB,  de- 
scribed from  the  vertex  P,  and  included 
between  the  sides  AP  and  BP. 

To  prove   Z  APB  measured  by  arc  AB. 


Proof.     Draw  PT  and  PT'  tangent  to  AP  and  J5P,  respectively 
at  P  J  also  draw  the  radii  OA,  OB,  and  the  diameter  PPK 

Const.,  §  205,  PT±PP'  in  plane  PAP*, 

and  since,  data,  PA  is  a  quadrant, 

O^  ± PP'  in  plane  P^IP*;  Why  ? 

..  §  71,  PTWOA,  ' 

and  similarly,  PT'  II  OB ; 

.-.§469,  Ztpt'  =  Zaob. 

But,  §  224,        Z^O^'is  measured  by  arc  AB;  * 

Z  TPT'  is  measured  by  arc  AB; 
that  is,  §  681,       Z  APB  is  measured  by  arc  AB. 

Therefore,  etc.  ^  QJLD, 


840  SOLID   GEOMETRY,  — BOOK  X. 

690.  Cor.  I.  A  spherical  angle  has  the  same  measure  as  the  di- 
hedral angle  formed  by  the  planes  of  its  sides. 

691.  Cor.  II.  If  two  sides  of  a  spherical  triangle  are  quadrants, 
the  third  side  measures  the  angle  opposite. 

692.  Cor.  III.  If  each  side  of  a  spherical  triangle  is  a  quadrant, 
each  angle  is  a  right  angle. 

693.  Cor.  IV.  If  two  arcs  of  great  circles  cut  each  other,  their 
vertical  angles  are  equal. 

694.  Cor.  V.  The  angles  of  a  spherical  polygon  are  equal  to  the 
dihedral  angles  between  the  planes  of  the  sides  of  the  polygon. 

695.  Sch.  Since,  §§  684,  694,  the  sides  and  angles  of  a  spheri- 
cal polygon  have  respectively  the  same  measures  as  the  face  and 
dihedral  angles  of  the  corresponding  polyhedral  angle,  we  may, 
from  any  property  of  polyhedral  angles,  infer  an  analogous  property 
of  spherical  polygons  ;  and  conversely. 

Proposition  IX 

696.  1.  Forin  a  sphere  and  draw  on  it  a  spherical  triangle;  pass 
planes  through  the  sides  of  the  triangle  and  the  center  of  the  sphere, 
and  thus  cut  out  the  corresponding  trihedral  angle.  How  does  the  sum 
of  any  two  face  angles  of  the  trihedral  compare  with  the  third  face 
angle?  How,  then,  does  the  sum  of  any  two  sides  of  the  spherical  tri- 
angle compare  with  the  third  side  ? 

2.  How  does  any  side  compare  with  the  difference  of  the  other  two 
sides  ? 

Theoretn.  The  sum  of  any  two  sides  of  a  spherical  tri- 
angle is  greater  than  the  third  side. 


Data:   Any  spherical  triangle,  as  ABC,  on  the  sphere  whose 
center  is  0.  ^ 


SOLID  GEOMETRY.  — BOOK  X.  341 

To  prove  the  sum  of  any  two  sides,  as  AC -\- BCy  greater  than 
the  third  side  AB. 

Proof.     In  the  corresponding  trihedral  angle  0-ABC, 
§  498,  Z  AOC  +  Z  BOG  is  greater  than  Z  AOB ; 

.-.  §  695,  AO  -{-  BC  is  greater  than  AB. 

Therefore,  etc.  -  q.e.d. 

697.  Cor.  I.  A7iy  side  of  a  spherical  triangle  is  greater  than  the 
difference  of  the  other  tivo  sides. 

Proposition  X 

698.  Form  a  sphere  and  draw  on  it  a  spherical  polygon ;  pass  planes 
through  the  sides  of  the  polygon  and  the  center  of  the  sphere,  and  thus 
cut  out  the  corresponding  polyhedral  angle.  How  does  the  sum  of  the 
face  angles  of  the  polyhedral  compare  with  four  right  angles?  How, 
then,  does  the  sum  of  the  sides  of  the  spherical  polygon  compare  with 
the  circumference  of  a  great  circle  ? 

Theorem,  The  sutu  of  the  sides  of  a  spherical  polygon  is 
less  than  the  circumference  of  a  great  circle. 


Data:  Any  spherical  polygon,  as  ABCD,  on  the  sphere  whose 
center  is  0. 

To  prove  AB  +  BC  -{■  CD  -{-  DA  <  the  circumference  of  a  great 
circle. 

Proof.     In  the  corresponding  polyhedral  angle  0-ABCD, 
§  499,  Z  AOB  +  Z  BOC  +  Z  COD  +  Z  DOA  <  4  rt.  ^ ; 

•.  §  695,  AB  +  BC+CD-\-DA<  the  circum.  of  a  great  circle. . 

Therefore,  etc.  q.e.d. 


342  SOLID   GEOMETRY.  — BOOK  X. 

699.  If  from  the  vertices  of  a  spherical  triangle  as  poles  arcs 
of  great  circles  are  described,  these  arcs  form  by  their  intersections 
a  second  triangle  which  is  called  the  Polar  Triangle  of  the  first. 

If  ^ ,  ^,  and  C  are  the  poles  of  the  great  circle  arcs 
B'C,  A'C,  and^'i?',  respectively,  then,  ^'5' C  is  the 
polar  triangle  of  ABC. 

If  from  A,  B,  and  C  as  poles  entire  great 
circles  instead  of  arcs  are  described,  these 
circles  will  divide  the  surface  of  the  sphere 
iuto  eight  spherical  triangles. 

Of  these  eight  triangles,  that  one  is  the  polar  of  ABC  whose 
vertex  A'  corresponding  to  A  lies  on  the  same  side  of  BC  as  the 
vertex  A ;  and  in  the  same  way  the  other  corresponding  vertices 
may  be  determined. 

Proposition  XI 

700.  On  the  surface  of  a  sphere  draw  a  spherical  triangle ;  draw  also 
its  polar  triangle.  Test  the  first  triangle  to  see  if  it  is  the  polar  triangle 
of  the  second. 

Theorem,  If  one  of  two  spherical  triangles  is  the  polar 
triangle  of  the  other,  then  the  other  is  the  polar  triangle  of 
that  one. 

Data :  Any  spherical  triangle,  as  ABC,  and  • 

its  polar  triangle,  a'b'C'. 

To  prove  ABC  the  polar  triangle  of  A'b'C'. 

Proof.    §  699,  A  is  the  pole  of  arc  B'c'-, 

. .  §  664,  5'  is  at  a  quadrant's  distance  from  A. 

Also,  C  is  the  pole  of  arc  A'b'-, 

B'  is  at  a  quadrant's  distance  from  C. 

Hence,  §  666,         B'  is  the  pole  of  arc  AC. 

In  like  manner  it  may  be  shown  that, 

a'  is  the  pole  of  arc  BC,  and  C'  the  pole  of  arc  AB. 

Hence,  §  699,  ABC  is  the  polar  triangle  of  A^B'c*. 

Therefore,  etc.  q.e.d. 


SOLID   GEOMETRY.  — BOOK  X.  343 

Proposition  XII 

701.  On  the  surface  of  a  sphere,  draw  a  spherical  triangle  and  its 
polar  triangle ;  select  an  angle  of  one  of  these  and  extend  its  sides,  if 
necessary,  to  meet  the  opposite  side  of  the  other  triangle.  What  part  of 
a  circumference  is  the  distance  from  each  point  of  meeting  to  the  farthest 
extremity  of  that  opposite  side  ?  Then,  how^does  the  arc  intercepted  on 
that  side  by  the  sides  of  the  given  angle  compare  with  two  quadrants  less 
the  whole  side,  or  with  180°  less  the  whole  side?  How,  then,  does  the 
measure  of  the  given  angle  compare  with  the  supplement  of  the  opposite 
side  of  the  polar  triangle  ? 

Theorem.  In  two  polar  triangles  any  angle  of  the  one  is 
measured  by  the  supplement  of  the  opposite  side  of  the  other. 

Data :  Any  two  polar  triangles,  as  ABC  and 
a'b'c',  and  any  angle  of  either  triangle,  as  A. 

To  prove  Z  A  measured  by  180°  —  B'c'. 

Proof.  Produce  the  arcs  AB  and  AC  until 
they  meet  B'c'  at  the  points  G  and  H,  respec- 
tively. 

Since,  data,  A  is  the  pole  of  arc  GH, 

§  689,  Z  ^  is  measured  by  GH. 

Since,  data,  B'  and  C'  are  the  poles  of  arcs  AH  and  AG,  respec- 
tively, 

arcs  b'h  and  C'G  are  quadrants. 

Jlence,         B'H+  c'G  =  ?i  semicircumference ; 
that  is,  B'c'  -^  GH=a,  semicircumference  =  180° ; 

GH=1S0°  -B'C'. 
But  Z  ^  is  measured  by  GH. 

Hence,  Z  ^  is  measured  by  180°  —  B'C'. 

Therefore,  etc.  q.e.d. 

702.  Two  polar  triangles  are  also  called  Supplemental  Triangles. 
For,  if  we  denote  the  number  of  angle  degrees  in  each  angle  by 

the  letter  at  the  vertex,  and  the  number  of  arc  degrees  in  each 
opposite  side  by  the  corresponding  small  letter,  we  have  th(' 
following  relations : 

,Za  =  180°  -a',  Zb  =  180°  -b',  Zc  =  180°  -  c', 
'z^'  =  180°-a,    Z^'  =  180°-6,    Z  C' =  180°  -  c 


344  SOLID   GEOMETRY.  — BOOK  X, 


Proposition  XIII 

703.  On  the  surface  of  a  sphere  draw  a  spherical  triangle ;  draw  also 
its  polar  triangle.  What  is  the  measure  of  each  angle  of  the  given  tri- 
angle ?  What,  then,  is  the  sum  of  the  measures  of  its  three  angles  ?  Since 
the  sum  of  the  arcs  which  form  the  sides  of  the  polar  triangle  is  greater 
than  an  arc  of  0°  and  (§  698)  less  than  an  arc  of  360°,  what  is  the  greatest 
number  and  also  the  least  number  of  degrees  that  there  can  be  in  the 
sum  of  the  angles  of  the  given  triangle  or  any  spherical  triangle? 
Express  your  conclusions  in  terms  of  right  angles. 

Theorem,  The  sum  of  the  angles  of  a  spherical  triangle 
is  greater  than  two  and  less  than  sijc  right  angles. 

Data:    Any    spherical    triangle,    as    ABC,  j^ 

whose  angles  are  A,  B,  and  C. 

To  prove  1.   Z  ^  +  Z  if  +  Z  C>  2  rt.  zi. 

2.   Z  ^  +  Z  5  +  Z  C  <  6  rt.  Z. 

Proof.  1.  Construct  A'B'&y  the  polar  trian- 
gle oiABC,  and  denote  the  number  of  degrees  in 
B'C',  A'c',  and  A'B'hy  a',  h\  and  c',  respectively. 

Then,  §  701,  Z  J  =  180°-a',  Z  5=180°-5',  and  Z  (7=180' 
.-.  Ax.  2,       Z^  +  Z^  +  Zc=540°-(a'4-&'  +  c'). 

But,  §  698,  5'C'  +  ^'C'  +  A^B^  <  the  circum.  of  a  great  circle ; 
that  is,  a/  +  6'  +  c'<360°; 

Z^-fZ^  +  Zo  180°,  or  2  rt.  A. 

2.  a'-f-6'-hc'>0°; 

hence,  Z^  +  Z^  +  Z(7< 540°,  or  6  rt.  A 

Therefore,  etc.  q.e.Dc 

704.  Cor.  A  spherical  triangle  may  have  tivo,  or  even  three, 
right  angles;   or  it  may  have  two,  or  even  three,  obtuse  angles. 

Ex.  872.  The  sides  of  a  spherical  triangle  are  65°,  86°,  and  98°.  What 
are  the  angles  of  its  polar  triangle  ? 

Ex.  873.  The  angles  of  a  spherical  triangle  are  53°,  77°,  and  92°.  What 
are  the  sides  of  its  polar  triangle  ? 

Ex.  874.  The  angles  of  a  spherical  triangle  are  65°,  80°,  and  110°.-  What 
are  the  sides  of  its  polar  triangle  ? 


kO 


SOLID   GEOMETRY.  — BOOK  X. 


345 


705.  A  spherical  triangle  having  two  right  angles  is  said  to  be 
hirectangidar ;  and  one  having  three  right  angles  is  said  to  be 
trirectangular. 

706.  The  excess  of  the  sum  of  the  angles  of  a  spherical  tri- 
angle over  two  right  angles  is  called  tKe  Spherical  Excess  of  the 
•triangle. 

707.  The  excess  of  the  sum  of  the  angles  of  a  spherical  polygon 
over  two  right  angles,  taken  as  many  times  as  the  polygon  has 
sides  less  two,  is  called  the  Spherical  Excess  of  the  polygon. 

If  a  polygon  has  n  sides,  its  spherical  excess  is  equal  to  the 
sum  of  the  spherical  excesses  of  the  n  —  2  spherical  triangles  into 
which  the  polygon  may  be  divided  by  diagonals  from  any  vertex. 

708.  Spherical  triangles  in  which  the  sides  and  angles  of  the 
one  are  equal  respectively  to  the  sides  and  angles  of  the  other,  but 
arranged  in  the  reverse  order,  are  called  Symmetrical  Spherical 
Triangles. 

Two  spherical  triangles  are  symmetrical, 
when  the  vertices  of  one  are  at  the  ends 
of  the  diameters  from  the  vertices  of  the 
other. 

Triangles  ABC  and  A'B'C  are  symmetrical 
spherical  triangles. 

Symmetrical    spherical    triangles    are 
mutually  equilateral  and  equiangular,  and  the  equal  sides  are  oppo- 
site the  equal  angles,  yet  they  cannot  generally  be  made  to  coincide. 

To  make  the  symmetrical  triangles  ABC  and 
A'B'C  coincide,  any  arc  BC  must  be  made  to 
coincide  with  its  equal  B'C.  This  can  be  done 
in  only  two  ways  —  with  B  either  on  B'  or 
on  C.  When  superposed  with  B  on  C,  unless 
the  triangles  are  isosceles,  angles  B  and  C  are 
unequal  and  the  triangles  will  not  coincide ;  with 
B  on  B',  A  and  A'  fall  on  opposite  sides  of  B'C 
and  the  triangles  will  not  coincide. 

Symmetrical  spherical  triangles  which  are  isosceles  can  b^ 
made  to  coincide. 


346  SOLID   GEOMETRY.  — BOOK  X. 

Proposition  XIV 

709.  On  the  surface  of  a  sphere  draw  two  symmetrical  triangles. 
How  do  they  compare  in  area?     Are  the  triangles  equal  or  equivalent? 

Theorem.  Two  symmetrical  spherical  triangles  are 
equivalent. 

Data:  Any  two  symmetrical  spherical  tri- 
angles, as  ABC  and  A'b'c\ 

To  prove       A  ABC  ^  A  a'b'C'. 

Proof.     Case  I.     When  they  are  isosceles. 

If  isosceles,  they  may  be  made  to  coincide ; 
2iYesi  ABC  =  Siresi  a' b'C'. 

Case  II.     When  the  triangles  are  not  isosceles. 

Suppose  P  and  P'  to  be  the  poles  of  the  small  circles  passing 
lihrough  the  points  A^  B,  C,  aud  A',  B',  C',  respectively. 

Data,  arcs  AB,  AC,  5C=  arcs  a'b',  a'c',  b'c',  respectively; 
•.  §  196,  chords  of  arcs  AB,  AC,  BC  =  chords  of  arcs  A'B',  A'c',  B'C', 
■respectively ;   hence,  §  107,  the  plane  triangles  formed  by  these 
chords  are  equal ; 
.-.  §  208,  the  small  circles  through  A,  B,  C,  and  A',  b',  C'  are  equal. 

Draw  the  great  circle  arcs  PA,  PB,  PC,  p'a',  P'b',  p'C'. 

Then,  §  665,  these  arcs  are  equal. 

Now,  §§  695,  501,  the  angles  of  APAB  are  equal  to  the  angles 
of  A  P'a'b',  respectively,  and  the  equal  parts  of  the  triangles  are 
in  reverse  order ; 

.-.  §§  708,  686,  A  PAB  and  P'a'b'  are  symmetrical  and  isosceles, 
ind.  Case  I,  area  PAB  =  area  p'a'b'. 

In  like  manner,  area  PBC= area  P'B'c',  and  area  PAC=  area  P'A'c' ; 
area  PAB  +  PBC  +  PAC  =  area  p'a'b'  +  p'b'C'  +  P'A'c', 
IT       area  ABC  =  area  A' B'c' ;  that  is,  A  ABC  <^  A  a'b'c'. 

If  the  pole  P  should  be  without  A  ABC,  then  P'  would  be  with- 
out A  a'b'c',  and  each  triangle  would  be  equivalent  to  the  sum 
of  two  isosceles  triangles  diminished  by  the  third ;  consequently, 
the  result  would  be  the  same  as  before. 

Therefore^  etc.  9.E.D. 


SOLID   GEOMETRY.  — BOOK  X,  347 

Proposition  XV 

710.  1.  On  the  same  sphere,  or  on  equal  spheres,  draw  two  spherical 
triangles  having  two  sides  and  the  included  angle  of  one  equal  to  the 
corresponding  parts  of  the  other,  and  arranged  in  the  same  order.  Can 
the  triangles  be  made  to  coincide?     Then,  how  do  they  compare? 

2.  Draw  two  spherical  triangles  as  befoi'e,  but.  with  the  given  equal 
parts  arranged  in  the  reverse  order ;  draw  another  triangle  symmetrical 
to  one  of  these.  How  does  it  compare  with  the  other?  Then,  are  the 
given  triangles  equal  or  equivalent? 

Theorem.  Two  triangles  on  the  same,  or  on  equal  spheres, 
having  two  sides  and  the  included  angle  of  one  equal  to 
two  sides  and  the  included  angle  of  the  other,  each  to  each, 
are  either  equal  or  equivalent. 

*Data:  Two  spherical  triangles,  as  ABC  and 
DBF,  in  which  AB  =  DE,  AC  —  DF,  and  angle  A 
=  angle  D. 

Case  I.  When  the  given  equal  parts  of  the 
two  triangles  are  arranged  in  the  same  order. 

To  prove  A  ABC  =  A  DEF. 

Proof.  The  A  ABC  can  be  applied  to  the  A  DFF,  as  in  the  cor- 
responding case  of  plane  triangles,  and  they  will  coincide. 

Hence,  §  36,  A  ABC  =  A  DFF. 

Case  II.  When  the  given  equal  parts  of  the  two  triangles  are 
arranged  in  reverse  order,  as  in  triangles  ABC  and  A'b'c'  in  which 
AB  =  A'b',  AC  =  A'C',  and  ZA=zZa'. 

To  prove  A  ABC  o  A  a'b'c'. 

Proof.  Suppose  the  A  DFF  to  be  symmetrical  with  respect  to 
the  A  A'b'c'. 

Then,  §  708,  the  sides  and  angles  of  A  DFF  are  equal  respec- 
tively to  those  of  A  A'B'C' ; 

.-.  in  the  A  ABC  and  DFF,  Z  A  =  Z  D,  AB  =  DF,  and  AC  =  DF, 
and  the  equal  parts  are  arranged  in  the  same  order ; 
.-.  Case  I,  A  ABC  =  A  DFF. 

But,  §  709,  A  A'B'C'  =0=  A  DFF. 

Hence,  A  ABC  =7- A  a'b'c'.  q.e.d. 


348  SOLID   GEOMETRY.  — BOOK  X. 


Proposition  XVI 

711.  1.  On  the  same  sphere,  or  on  equal  spheres,  draw  two  spherical 
triangles  having  a  side  and  two  adjacent  angles  of  one  equal  to  the  cor- 
responding parts  of  the  other,  and  arranged  in  the  same  order.  Can 
these  triangles  be  applied  to  each  other  so  that  they  will  coincide? 
Then,  how  do  they  compare  ? 

2.  Draw  two  spherical  triangles  as  before,  but  with  the  given  equal 
parts  arranged  in  the  reverse  order ;  draw  another  triangle  symmetrical 
to  one  of  these.  How  does  it  compare  with  the  other?  Then,  are  the 
given  triangles  equal  or  equivalent  ? 

Theorem,  Two  triangles  on  the  same  sphere,  or  on  equal 
spheres,  having  a  side  and  two  adjacent  angles  of  one  equal 
to  a  side  and  two  adjacent  angles  of  the  other,  each  to  each, 
are  either  equal  or  equivalent. 

Proof.  One  of  the  triangles  may  be  applied 
to  the  other,  or  to  its  symmetrical  triangle,  as 
in  the  corresponding  case  of  plane  triangles. 


Therefore,  etc.  q.e.d. 

Proposition  XVII 

712.  On  the  same  sphere,  or  on  equal  spheres,  draw  two  mutually 
equilateral  spherical  triangles.  How  do  the  angles  of  one  compare  with 
the  angles  of  the  other?  If  the  equal  parts  are  arranged  in  the  same 
order  in  each,  how  do  the  triangles  compare  ?  If  the  equal  parts  are  in 
reverse  order,  are  the  triangles  equal  or  equivalent?  > 

Theorem,  Two  mutually  equilateral  triangles  on  the 
same  sphere,  or  on  equal  spheres,  are  mutually  equiangu- 
lar, and  are  either  equal  or  equivalent. 

Proof.     By  §§  695,  501,  the  triangles  are  mutually  equiangular; 
they  are  equal  or  symmetrical.  Why? 

If  they  are  symmetrical, 
then,  §  709,  they  are  equivalent. 

Hence,  they  are  either  equal  or  equivalent. 

Therefore,  etc.  '  Q.e.d. 


SOLID   GEOMETRY.  —  BOOK  X.     '  349 


Proposition  XVIII 

713.  1.  On  the  surface  of  a  sphere,  draw  an  isosceles  spherical  trian- 
gle ;  draw  the  arc  of  a  great  circle  from  its  vertex  to  the  middle  of  the 
opposite  side.  In  the  two  triangles  thus  formed,  how  do  the  sides  of  one 
compare  with  the  sides  of  the  other?  Then,  how  do  the  angles  of  one 
compare  with  the  angles  of  the  other  ?  In  the  original  triangle,  how  do 
the  angles  opposite  the  equal  sides  compare  with  each  other  ? 

2.  How  does  the  great  circle  arc  from  the  vertex  to  the  middle  of  the 
base  of  an  isosceles  spherical  triangle  divide  the  vertical  angle?  What 
is  its  direction  with  reference  to  the  base  ?  Into  what  kind  of  triangles 
does  it  divide  the  given  triangle  ? 

Theorem,  In  an  isosceles  spherical  triangle,  the  angles 
opposite  the  equal  sides  are  equal. 


Data:    An  isosceles  spherical  triangle,  as 
ABC,  in  which  AB  =  AC. 

To  prove  /.B  =  /.c. 


Proof.  Draw  the  arc  of  a  great  circle,  as  AB,  from  the  vertex  A, 
bisecting  the  side  BC. 

Then,  in  A  ABB  and  ACB, 

AD  is  common, 
AB  =  AC,BB  =  BC;  Why? 

that  is,  the  triangles  are  mutually  equilateral ; 
.-.  §  712,      A  ABB  and  ACB  are  mutually  equiangular. 
Hence,  Zb  =  ZC. 

Therefore,  etc.  q.e.d. 

714.  Cor.  The  arc  of  a  great  circle  drawn  from  the  vertex  of  an 
isosceles  spherical  triangle  to  the  middle  of  the  base  bisects  the  vertical, 
angle,  is  perpendicular  to  the  base,  and  divides  the  triangle  into  two 
symmetrical  triangles. 

Ex.  875.  ■  If  the  sides  of  a  spherical  triangle  are  50°,  75°,  and  110°.  what 
are  the  angles  of  its  polar  triangle  ? 

Ex.  876.  If  the  sides  of  a  spherical  triangle  are  54°,  89°,  and  103°,  what 
is  the  spherical  excess  of  its  polar  triangle  ? 


350  SOLID   GEOMETRY.  — BOOK  X 


Proposition  XIX 

715.  On  the  surface  of  a  spliere,  draw  two  mutually  equiangular  trian- 
gles ;  draw  also  their  polars.  How  do  the  sides  of  their  polars  compare, 
each  to  each  ?  Then,  how  do  the  angles  of  the  polars  compare,  each  to 
each?  How,  then,  do  the  sides  of  the  given  triangles  compare,  each 
to  each  ?  If  the  equal  parts  in  the  given  triangles  are  arranged  in  the 
same  order  in  each,  how  do  the  triangles  compare?  If  the  equal  parts 
are  arranged  in  reverse  order,  are  the  triangles  equal  or  equivalent  ? 

Theorem,  Two  mutually  equiangular  triangles  on  the 
same  sphere,  or  on  equal  spheres,  are  mutually  equilateral, 
and  are  either  'equal  or  equivalent. 


Data :  Two  spherical  triangles,  as  A  and  B,  that  are  mutually- 
equiangular. 

To  prove    A  A  and  B  mutually  equilateral,  and  either  equal  or 
equivalent. 

Proof.     Suppose  A  ^'  to  be  the  polar  of  A  A,  and  A  5'  the  polar 
^iAB. 

Data,  A  A  and  B  are  mutually  equiangular ; 

•  •'.  §  701,    their  polar  A,  A'  and  B\  are  mutually  equilateral; 
hence,  §  712,  A  A'  and  B^  are  mutually  equiangular ; 
.-.  §  701,  A  A  and  B  are  mutually  equilateral. 

Hence,  §  712,  A  A  and  B  are  either  equal  or  equivalent. 

Therefore,  etc.  ,  q.e.d. 

716.  Cor.  I.     If  two  angles  of  a  spherical  triangle  are  equal,  the 
sides  opposite  these  angles  are  equal,  and  the  triangle  is  isosceles. 

717.  Cor.  II.     If  three  planes  are  passed  through  /^VrPX 
the  center  of  a  sphere,  each  perpendicular  to  the  other  /:-"'h'i~~  "A 
two,  they  divide  the  surface  of  the  sphere  into  eight  \~tJ^ — / 
equal  tri rectangular  triangles.                                §  695  x^ilX^ 


SOLID   GEOMETRY,  — BOOK  X.  351 

Proposition  XX 

718.  1.  On  the  surface  of  a  sphere  draw  a  spherical  triangle,  two  of 
whose  angles  are  unequal.  How  do  the  sides  opposite  these  angles  com- 
pare?    Which  one  is  the  greater? 

2.  Draw  a  spherical  triangle,  two  of  whose  sides  are  unequal.  How 
do  the  angles  opposite  these  sides  compare?    ^Which  one  is  the  greater? 

Theorem,  If  two  angles  of  a  spherical  triangle  are  un- 
equal, the  sides  opposite  are  unequal,  and  the  greater  side 
is  opposite  the  greater  angle;  conversely,  if  two  sides  are 
unequal,  the  angles  opposite  are  unequal,  and  the  greater 
angle  is  opposite  the  greater  side. 

Data:  A  spherical  triangle,  as  ABC,  in 
which  the  angle  ACB  is  greater  than  the 
angle  ABC. 

To  prove  AB  >  AC. 

Proof.     Draw  CD,  the  arc  of  a  great  circle,  making  Z  BCD  —  /.h- 

Then,  §  716,  DB  =  CD. 

Now,  §  696,  AD^CD>  AC, 

AD-\-DB>  AC, 

or  ■  AB>AC. 

Conversely:  Data:  A  spherical  triangle,  as  ABC,  in  which  the 
side  AB  is  greater  than  the  side  AC. 

To  prove  Z.ACB  greater  than  Z  B, 

Proof.     If  /.acb  =  Zb, 

then,  §  716,  AB  =  AC, 

which  is  contrary  to  data. 

If  Z.  ACB  is  less  than  Z  B, 

then,  Z  £  is  greater  than  Z  ACB, 

and  AC>AB, 

which  is  also  contrary  to  data. 

Therefore,  both  hypotheses,  namely,  that  ^ACB  =  Ab  and 
that  /.ACB  is  less  than  /.B,  are  untenable. 

Consequently,       Z  ACB  is  greater  than  Z  B. 

Therefore,  etc.  q.b.j>. 


352  SOLID   GEOMETRY.  — BOOK  X, 

SPHERICAL  MEASUREMENTS 

719.  The  portion  of  the  surface  of  a  sphere  included  between 
two  parallel  planes  is  called  a  Zone. 

The  perpendicular  distance  between  the 
planes  is  the  altitude  of  the  zone,  and  the 
circumferences  of  the  sections  made  by 
the  planes  are  called  the  bases  of  the  zone. 

If  one  of  the  parallel  planes  is  tangent 
to  the  sphere,  the  zone  is  called  a  zone  of 
one  base. 

ABCD  is  a  zone  of  the  sphere. 

720.  The  portion  of  the  surface  of  a  sphere  bounded  by  two 
semicircumferences  of  great  circles  is  called  a  Lune. 

The  angle  between  the  semicircumferences  jj^ 

which  form  its  boundaries,  is  called  the  angle 
of  the  lune. 

ABCD  is  a  lune  of  which  BAD  is  the  angle. 

721.  Lunes  on  the  same  sphere,  or  on  equal 
spheres,  having  equal  angles  may  be  made  to 
coincide,  and  are  equal. 

722.  A  convenient  unit  of  measure  for  the  surfaces  of  spherical 
figures  is  the  spherical  degree^  which  is  equal  to  -^^  of  the  surface 
of  a  hemisphere. 

Like  the  unit  of  arcs,  it  is  not  a  unit  of  fixed  magnitude,  but 
depends  upon  the  size  of  the  sphere  upon  which  the  figure  is 
drawn. 

It  may  be  conceived  of  as  a  birectangular  spherical  triangle 
whose  third  angle  is  an  angle  of  one  degree. 

The  distinction  between  the  three  different  uses  of  the  term 
degree  should  be  kept  clearly  in  mind;  an  angular  degree  is  a 
difference  of  direction  between  two  lines,  and  it  is  the  360th  part 
of  the  total  angular  magnitude  about  a  point  in  a  plane  (§  35) ; 
an  arc  degree  is  a  line,  which  is  the  360th  part  of  the  circumfer- 
ence of  a  circle  (§  224) ;  a  spherical  degree  is  a  surface,  which  is 
the  360  th  part  of  the  surface  of  a  hemisphere,  or  the  720th  part 
of  the  surface  of  a  sphere. 


SOLID   GEOMETRY.  — BOOK  X.  353 

Proposition  XXI 

723.  Represent  an  axis  and  a  li  iie  oblique  to  it,  but  not  meeting  it ;  draw 
lines  from  the  extremities  and  middle  point  of  this  line  perpendicular  to 
the  axis ;  from  the  nearer  extremity  draw  a  line  parallel  to  the  axis ; 
also  a  line  perpendicular  to  the  given  line  at  its  middle  point  and 
terminating  in  the  axis.  If  the  given  line  revolves  about  the  axis,  what 
kind  of  a  surface  will  it  generate  ?  To  what  is  this  surface  equivalent  ? 
(§  628)  By  means  of  the  proportion  of  lines  from  similar  right  tri- 
angles, express  the  surface  in  terms  of  the  projection  of  the  given  line  on 
the  axis  arid  the  circumference  of  a  circle  whose  radius  is  the  perpen- 
dicular from  the  middle  point  of  the  given  line.  Would  this  result  hold 
true,  if  the  line  should  meet  the  axis  or  be  parallel  to  it  ? 

Theorem,  The  surface  generated  hy  a  straight  line  re- 
volving about  an  axis  in  its  plane  is  equivalent  to  the 
rectangle  formed  hy  the  projection  of  the  line  on  the  axis 
and  the  circumference  whose  radius  is  a  perpendicular 
erected  at  the  middle  point  of  the  line  and  terminated  hy 
the  axis. 

Data:  Any  line,  as  AB,  revolving  about  an  u 

axis,  as  MN\   its  projection  upon  MN,  as   C7);  a    __ 

and  EO  perpendicular  to  AB  at  its  middle  point  /l 

and  terminating  in  the  axis.  "7^^ — 

To  prove  surface  AB  ^  rect.  CD  •  2  ttEO.  Z__.i__^ 

Proof.     Draw  EF  ±  MN  and  AK  II  MN.  ^      ^ 

If  AB  neither  meets  nor  is  parallel  to  'MN  it 
generates  the  lateral  surface  of  a  frustum  of  a 
cone  of  revolution  whose  slant  height  is  AB  and  axis  CZ)j 
.-.  §  628,  surface  AB  ^  rect.  AB  •  2  ttEF. 

§  307,  A  ABK  and  EOF  are  similar, 

and  AB:  AK=EO:  EF; 

rect.  AB'  EF^  rect.  AK  -  EO. 

But,  §  151,  AK=  CD; 

hence,  rect.  AB  •  EF  o=  rect.  CD  •  EO, 

and  rect.  AB  •  2  irEF  ^  rect.  CD  •  2  ttEO  ; 

that  is,  surface  AB  =o  rect.  CD  •  2  -n-EO. 

If  AB  meets  axis  MN,  ot  is  parallel  to  it,  a  conical  or  a  cylindri 
cal  surface  is  generated,  and  the  truth  of  the  theorem  follows. 

Therefore,  etc.  q.e.d. 

milne's  gbom. — 23 


C 


N 


354  SOLID   GEOMETRY.  —  BOOK  X, 


Proposition  XXII 

724.  Draw  a  semicircumference  and  inscribe  in  it  a  regular  semi- 
polygon.  How  does  the  sum  of  the  projections  of  the  sides  of  the  poly- 
gon on  the  diameter  of  the  semicircle  compare  with  the  diameter  ?  How 
do  the  perpendiculars  to  the  sides  of  the  polygon  at  their  middle  points 
compare  in  length,  if  they  terminate  in  the  diameter  ?  If  the  figure  is 
revolved  about  the  diameter  as  an  axis,  to  what  is  the  surface  generated 
by  the  perimeter  of  the  semipolygon  equivalent  ?  How  does  the  perim- 
eter of  the  semipolygon  at  its  limit  compare  with  the  semicircumference, 
if  the  number  of  its  sides  is  indefinitely  increased?  What  is  the  limit 
of  the  perpendicular  to  the  middle  point  of  a  side  of  the  semipolygon? 
How,  then,  does  the  surface  of  a  sphere  compare  with  the  rectangle 
formed  by  its  diameter  and  the  circumference  of  a  great  circle  ? 

Theoi'eiti,  The  surface  of  a  sphere  is  equivalent  to  the 
rectangle  formed  hy  its  diameter  and  the  circunfiference  of 
^  great  circle. 

Data:  A  sphere,  whose  center  is  0,  generated  by 

the   revolution  of   the   semicircle   ABCB  about  the  -^ 

diameter  AD.  / 

Denote  the  surface  of  the  sphere  by  s,  and  its  1 

radius  by  i?.  ^ 

To  prove  S  =0=  rect.  AD  -  2  irE. 

Proof.  Inscribe  in  the  semicircle  half  of  a  regular  polygon  of 
An  even  number  of  sides,  as  ABCD,  and  let  S'  denote  the  surface 
generated  by  its  sides. 

Draw  BE  and  CF  ±  AD,  and  the  perpendiculars  from  0  to  the 
chordsf  AB,  BC,  and  CD. 

§§  202,  200,  these  perpendiculars  are  equal,  and  bisect  the  chords. 

Then,  §  723,         surface  AB  =0  rect.  AE  -  2  ttOH, 

surface  BC  ^  rect.  EF  •  2  irOH, 

and  surface  CD  ^  rect.  FD  •  2.TT0H. 

But  the  sum  of  the  projections  AE,  EF,  and  FD  equals  the  diam- 
eter AD ; 

S' <>  rect.  ^z> .  2  ttO^. 

Now,  if  the  number  of  sides  of  the  inscribed  semipolygon  is 
indefinitely  increased, 


D 


SOLID   GEOMETRY.  — BOOK  X.  355 

§  392,  the  semiperimeter  will  approach  the  semicircumference  as 
its  limit ; 

OH  will  approach  R  as  its  limit, 

and  s'  will  approach  S  as  its  limit. 

But,  however  great  the  number  of  sides  of  the  semipolygon, 

S'  =0  rect.  AD'2  7rbH. 

Hence,  §  326,  s  ^  rect.  AD  -  2  irB. 

Therefore,  etc.  q.e.d. 

725.  Cor.  I.     The  area  of  the  surface  of  a  sphere  is  equal  to  the 
product  of  its  diameter  by  the  circumference  of  a  great  circle. 

726.  Cor.  II.     §  725,  area  =  AB  x  27rR  =  2  R  x  2  7rR  =  4:7rR^', 

that  is,  the  area  of  the  surface  of  a  sphere  is  equal  to  the  area  of 
four  great  circles. 

727.  Cor.  III.     Let  R  and  i?'  denote  the  radii,  D  and  D'  the 
diameters,  and  A  and  A'  the  areas  of  the  surfaces  of  two  spheres. 

Then,  §  726,       ^  =  4  ttR'',  and  ^'  =  4  irR'^  • 

A:  A'  =  4.7rR:'  lA-rrR"  =  R'  :  R"  =  B'-.  D"  ;  ' 

that  is,  the  areas  of  the  surfaces  of  two  spheres  are  to  each  other  as 
the  squares  of  their  radii,  or  as  the  squares  of  their  diameters. 

728.  Cor.  IV.     Area  of  a  zone,  as  BC  =:EF  x2tvR\ 

that  is,  the  area  of  a  zone  is  equal  to  the  product  of  its  altitude  by 
the  circumference  of  a  great  circle. 

729.  Cor.  V.     Zones  on  the  same  sphere,  or  on  equal  spheres^  aru 
to  each  other  as  their  altitudes. 

730.  Cor.  VI.     §  728,  area  of  a  zone  of  one  base,  as 

AB  =  AE  X  2  ttR  =  ttAE  X  AB. 
Draw  BB.     Then,  §  313,  AE  x  AB  =  Iff ; 

area  of  zone  AB  =  irAB  ; 
that  is,  the  area  of  a  zone  of  one  base  is  equal  to  the  area  of  a 
circle  whose  radius  is  the  chord  of  the  generating  arc. 


356  ^       SOLID   GEOMETRY.  — BOOK  X, 


Proposition  XXIII 

731.  Divide  the  surface  of  a  sphere  into  hemispheres  by  a  great 
circle ;  on  one  of  the  hemispheres  form  two  opposite  triangles  by  draw- 
ing two  great  circle  arcs  to  intersect;  complete  the  circumferences 
of  which  these  arcs  are  parts.  By  comparing  one  of  these  opposite 
triangles  with  a  triangle  on  the  other  hemisphere  that  completes  a  lune 
of  which  the  other  of  the  given  triangles  is  a  part,  discover  how  the  sum 
of  the  given  triangles  compares  with  a  lune  whose  angie  is  the  angle 
between  the  given  arcs. 

Theorein.  If  two  arcs  of  great,  circles  intersect  on  the 
surface  of  a  hemisphere,  the  sum  of  the  two  opposite  tri- 
angles thus  formed  is  equivalent  to  a  lune  whose  angle 
is  the  angle  between  the  given  arcs. 

Data:  Opposite  A,  as  AEB  and  DUCy 
formed  by  two  great  circle  arcs,  as  AED 
and  BEC,  on  the  hemisphere  E-ABDC. 

To  prove  A  AEB  -\-  A  DEC  ^  lune  AEBF. 

Proof.  Produce  arcs  AED  and  BEC  around 
the  sphere  intersecting  as  at  F. 

§  QbQj    arc  BE  =  arc  AE  (each  being  the  supplement  of  arc  AE), 
arc  CE  =  arc  BE  (each  being  the  supplement  of  arc  BE), 
and,  §  693,  Z  DEC  =  Z  AEB  =  Z  AEB ; 

.-.§710,  ADEC  =o=AAFB. 

Adding  A  AEB  to  each  side  of  this  expression  of  equivalence, 

A  AEB  +  A  DEC  ^  A  AEB  -f-  A  AFB'. 
Hence,  A  AEB  +  A  DEC  =o  lune  AEBF.  q.e.d. 

Proposition  XXIV 

732.  On  the  surface  of  a  sphere  draw  a  lune  whose  angle  is  to  four 
right  angles  as  3  :  12 ;  from  the  vertex  of  its  angle  as  a  pole  describe  the 
circumference  of  a  great  circle.  ^Vhat  is  the  ratio  of  the  arc  included 
between  the  sides  of  the  lune  to  the  whole  circumference?  Divide  the 
circumference  into  12  equal  parts  and  through  the  points  of  division  and 
the  poles  pass  great  circle  arcs.  Into  how  many  equal  lunes  do  these  arcs 
divide  the  surface  of  the  sphere  ?  The  given  lune  ?  How,  then,  does  the 
ratio  of  the  given  lune  to  the  surface  of  the  sphere  compare  with  the  ratio 
of  the  angle  of  the  lune  to  four  right  angles  ? 


SOLID   GEOMETRY.  — BOOK  X,  357 

Theorem  s  A  lune  is  to  the  surface  of  a  sphere  as  the 
angle  of  the  lune  is  to  four  right  angles. 

Data:  A  lime,  as  ACFD,  whose  angle  is  OAl),  on  the  sphere 
whose  center  is  O.  ^ 

Denote  the  lune  by  L  and  the  surface  /^''^^^^X 

of  the  sphere  by  S.^  /I 

To  prove      L:  S  =  Z  GAD  :  4  rt.  A.  J'"""""  J 

Proof.     With  ^  as  a  pole,  describe  the       \^  I 

circumference  of  a  great  circle  BCEH.  \  I 

Then,  §  689,  arc  CD  measures  Z  CAD,  and  ^---_j^5^ 

circumference  BCEH  measures  4  rt.  A.  ^ 

Suppose  arc  CD  and  BCEH  have  a  common  unit  of  measure,  as 
CJ,  contained  in  CD  m  times  and  in  BCEH  n  times. 

Then,  CD  :  BCEH  =  m  :  n, 

or  Z  CAD  :  4  rt.  A  =  m:n. 

Beginning  at  C,  divide  BCEH  into  parts,  each  equal  to  the 
unit  of  measure  CJ,  and  through  the  points  of  division  and  the 
poles,  A  and  F,  of  this  circumference  pass  great  circles. 

By  §§  689,  721,  these  circles  divide  the  whole  surface  of  the 
sphere  into  n  equal  lunes  of  which  the  given  lune  contains  m. 

Then,  L-.  S  =  m'.n. 

Hence,  L:  S  =  Z  CAD  :  4  rt.  A. 

By  the  method  of  limits  exemplified  in  §  223,  the  same  may 
be  proved,  when  arc  CD  and  BCEH  are  incommensurable. 

Therefore,  etc.  q.e.d. 

733.  Cor.  I.     Let  A  denote  the  degrees  in  the  angle  of  a  lune. 
Then,  L  :  S  =  A  :  360°. 

Since,  §  722,  S  contains  720  spherical  degrees,- 
i  :  720  =  ^  :  360 ; 
whence,  L  =  2  A-, 

that  is,  the  numerical  measure  of  a  lune  expressed  in  spherical 
degrees  is  twice  the  numerical  measure  of  its  angle  expressed  in 
angular  degrees. 

734.  Cor.  II.  Lunes  on  the  same  sphere,  or  on  equal  spheres,  are 
to  each  other  as  their  angles. 


358  SOLID   GEOMETRY.^ BOOK  X. 

Proposition  XXV 

735.  Oil  a  sphere  draw  a  spherical  triangle  and  complete  the  great 
circles  whose  arcs  are  its  sides.  How  many  triangles  having  a  common 
vertex  with  the  given  triangle  occupy  the  sm-face  of  a  hemisphere? 
Since  the  given  triangle  plus  any  one  of  the  others  is  equivalent  to  a 
lune  whose  angle  is  equal  to  one  of  the  angles  of  the  given  triangle,  or  to 
twice  as  many  spherical  degrees  as  that  angle  contains  angular  degrees, 
how  does  three  times  the  given  triangle  plus  the  other  three  compare 
with  twice  the  number  of  spherical  degrees  that  there  are  angular  de- 
grees in  the  angles  of  the  given  triangle?  How  many  spherical  degrees 
are  there  in  the  four  triangles  occupying  the  surface  of  the  hemisphere? 
Then,  discover  how  the  number  of  spherical  degrees  in  the  given  triangle 
compares  with  the  sum  of  the  angular  degrees  in  its  angles  less  180°, 
that  is,  w4th  its  spherical  excess. 

Theorem,  A  spherical  triangle  is  equivalent  to  as  many 
spherical  degrees  as  there  are  angular  degrees  in  its  spJieri- 
eal  excess. 

Data :  A  spherical  triangle,  as  ABC,  whose 
spherical  excess  is  E  degrees. 

To  prove   AABCoE  spherical  degrees. 

Proof.     Complete  the  great  circles  whose 
arcs  are  sides  of  A  ABC. 

These  circles  divide  the  surface  of  the 
sphere  into  eight  spherical  triangles,  any 
four  of  which  having  a  common  vertex,  as  A,  form  the  surface  of 
a  hemisphere,  whose  measure  is  360  spherical  degrees. 

§  731,  A  ABC  -\-  Aab'c'  ^2i  lune  whose  angle  equals  angle  A ; 
,-.  §  733,  A  ABC  +  A  AB'C'  <i=  2  A  spherical  degrees.      (1) 

In  like  manner,  A  ABC  +  A  AB'C  o=  2  5  spherical  degrees,  (2) 
and  A  ABC  +  A  ABC'  ^2c  spherical  degrees.      (3) 

Adding  (1),  (2),  and  (3), 
3  A  ABC  -f  A  AB'C'  4-  A  AB'C  -f-  A  ABC'  =o  2  (^  -f  J5  +  C7)  sph.  deg. 

But  A ^J5C+A^5'C''4-A^fi'a4-A^£C'=  360  spherical  degrees; 
.-.  2  A^5C7-|-360  spherical  degrees  =c=  2(^14- 5 4-C)  spherical  degrees ; 
hence,  A  ABC  o  (A  +  B  -\-  C  —  180)  spherical  degrees ; 

that  is,  §  706,  Aabc^e  spherical  degrees.  q.e.p. 


SOLID   GEOMETRY.^ BOOK  X.  359 

Proposition  XXVI 

736.  Draw  any  spherical  polygon-  and  divide  it  into  spherical  tri 
Angles  by  diagonals  from  any  vertex.  To  how  many  spherical  degrees 
is  each  triangle  equivalent  ?  How  does  the  number  of  spherical  degrees 
in  the  sum  of  the  triangles  compare  with  the  number  of  angular  de- 
grees in  the  sum  of  the  spherical  excesses* of  the  triangles?  To  how 
many  spherical  degrees,  then,  is  any  spherical  polygon  equivalent? 

Theorem^  Any  spherical  -polygon  is  equivalent  to  as 
many  spherical  degrees  as  there  are  angular  degrees  in  its 
spherical  excess. 


Data:    Any  spherical   polygon,  as  ABCDF,         / 
whose  spherical  excess  is  E  degrees.  ^^ 


To  prove    ABCDF  =c=  E  spherical  degrees. 


Proof.  Divide  the  polygon  into  spherical  triangles  by  diagonals 
/rom  any  vertex,  as  A. 

By  §  735,-  each  triangle  is  equivalent  to  as  many  spherical  de- 
grees as  there  are  angular  degrees  in  its  spherical  excess. 

Hence,  the  polygon  is  equivalent  to  as  many  spherical  degrees 
as  there  are  angular  degrees  in  the  sum  of  the  spherical  excesses 
of  the  triangles ;  that  is,  §  707,  in  the  spherical  excess  of  the 
polygon. 

Hence,  J^CDi^^-E  spherical  degrees. 

Therefore,  etc.  q.e.d. 

• 

737.  Cor.  The  area  of  any  spherical  polygon  is  to  the  area  of 
(he  surface  of  the  sphere  as  the  number  which  expresses  its  spherical 
excess  is  to  720. 

Ex.  877.  The  angle  of  a  lune  is  40°.  What  part  of  the  surface  of  the 
sphere  is  the  lune  ? 

Ex.  878.  What  is  the  area  of  a  spherical  triangle  whose  angles  are  86°, 
120°,  and  110°,  if  the  radius  of  the  sphere  is  lO^m  ? 

Ex.  879.  The  area  of  the  surface  of  a  sphere  is  160  sq.  in. ;  the  angles 
of  a  spherical  triangle  on  this  sphere  are  93°,  117°,  and  132°.  What  is  the 
area  of  the  triangle  ? 

Ex.  880.  Two  spherical  triangles  on  the  same  sphere,  or  on  equal  spheres, 
are  equivalent,  if  the  perimeters  of  their  polar  triangles  are  ec^ual. 


360 


SOLID   GEOMETRY.  — BOOK  X. 


738.   A  solid  bounded  by  a  spherical  polygon  and  the  planes  of 
its  sides  is  called  a  Spherical  Pyramid. 

The  center  of  the  sphere  is  the  vertex  of  the         ^       ^"^.-4 
pyramid,  and  the  spherical  polygon  is  its  base. 

0- ABODE  is  a  spherical  pyramid  whose  vertex  is 
0  and  base  ABODE. 


739.  The  portion  of  a  sphere  contained 
between  two  parallel  planes  is  called  a 
Spherical  Segment. 

The  sections  made  by  the  parallel  planes  are  the  bases  of  the 
spherical  segment ;  the  perpendicular  distance  between  its  bases 
is  the  altitude  of  the  segment. 

If  one  of  the  parallel  planes  is  tangent  to  the  sphere,  the  seg- 
ment is  called  a  segment  of  one  base. 


740.  The  portion  of  a  sphere  bounded  by  a  lune  and  the  planes 
of  its  sides  is  called  a  Spherical  Wedge,  or  Ungula. 

741.  The  portion  of  a  sphere  generated  by 
the  revolution  of  a  circular  sector  about  a 
diameter^  of  the  circle  is  called  a  Spherical 
Sector. 

The  zone  generated  by  the  arc  of  the  circu- 
lar sector  is  the  base  of  the  spherical  sector. 


a 


742.  MNEFAB  is  a  semicircle,  AD  and  BC  are  lines  from  the 
semicircumference  perpendicular  to  the  diameter  MN,  and  OE  and 
OF  are  radii.  Then,  if  the  semicircle  is  revolved 
about  MN  as  an  axis,'it  generates  a  sphere. 

The  arc  AB  generates  a  zone  whose  altitude  is  DC, 
and  whose  bases  are  the  circumferences  generated 
by  the  points  A  and  B. 

The  figure  ABCD  generates  a  spherical  segment 
whose  altitude  is  DC  and  whose  bases  are  the  circles 
generated  by  AD  and  BC. 

The  arc  BM  generates  a  zone  of  one  base,  and  the  figure  BCM  a 
spherical  segment  of  one  base. 

The  circular  sector  OEF  generates  a  spherical  sector  whose  bound- 
ing surfaces  are  its  base,  the  zone  generated  by  the  arc  EFj  and 
the  conical  surfaces  generated  by  tha  radii  OE  and  OF. 


SOLID   GEOMETRY.  — BOOK  X, 


361 


Proposition  XXVII 

743.  Represent  a  polyhedron  circumscribed  about  a  sphere.  If  pyra- 
mids are  formed  having  the  faces  of  the  polyhedron  as  bases  and  the 
center  of  the  sphere  as  a  common  vertex,  how  will  the  altitudes  of  these 
pyramids  compare  with  each  other  and  with  the  radius  of  the  sphere  ? 
What  is  the  volume  of  each  pyramid?  What,  then,  is  the  volume  of  the 
sum  of  the  pyramids?  If  the  number  of  faces  of  the  polyhedron  is 
indefinitely  increased,  how  will  its  volume  compare  with  the  volume  of 
the  sphere  ?    To  what,  then,  is  the  volume  of  a  sphere  equal  ? 

Theorem,  The  volume  of  a  sphere  is  equal  to  the  prod- 
uct of  its  surface  by  one  third  of  its  radius. 

D 


^^^m. 


/ 


I 


H  B 

Data :  A  sphere  whose  center  is  0,  surface  S,  and  radius  E. 

Denote  its  volume  by  F. 

To  prove  V=S  x\R. 

Proof.     Circumscribe    about   the    sphere    any   polyhedron,   as 
D-ABCj  and  denote  its  surface  by  s'  and  its  volume  by  v'. 

Form  pyramids,  as  0-ABC,,etG.,  having  the  faces  of  the  poly- 
hedron as  bases  and  the  center  of  the  sphere  as  a  common  vertex. 

Then  these  pyramids  will  have  a  common  altitude  equal  to  R, 
and,  §  560,  the  volume  of  each  pyramid  =  its  base  x  I-  R. 


V' 


S'  x^R. 


Now,  if  the  number  of  pyramids  is  indefinitely  increased  by 
passing  planes  tangent  to  the  sphere  at  the  points  where  the 
edges  of  the  pyramids  cut  the  surface  of  the  sphere, 

-s'  approaches  S  as  its  limit}  .-.  v'  approaches  V  as  its  limit. 


362  SOLID   GEOMETRY.-^ BOOK  X. 

But,  however  great  the  number  of  pyramids, 

r'  =  s'  x^R. 
Hence,  §  222,  V  =  s  x^B.  q.e.d. 

744.  Formulae :    F=sx^i?  =  | ttR^  =  | TrD\ 

745.  Cor.  I.  Let  R,  R'  denote  the  radii,  n,  d'  the  diameters, 
and  V,  r'  the  volumes,  respectively,  of  two  spheres. 

Then,  §  744,         F  =  |  ttR^  and  f'  =  |  ttR'^  ; 

F:  V'  =  ^ttR^  :  ^ttR'^  =  R^  :  R'^  =  D^  :  D'^; 

that  is,  the  volumes  of  tivo  spheres  are  to  each  other  as  the  cubes  of 
their  radii,  or  as  the  cubes  of  their  diameters. 

746.  Cor.  II.  The  volume  of  a  spherical  pyramid  is  equal  to  the 
product  of  its  base  by  one  third  of  the  radius  of  the  sphere. 

747.  Cor.  III.  The  volume  of  a  spherical  sector  is  equal  to  the 
product  of  the  zone  which  forms  its  base  by  one  third  of  the  radius 
of  the  sphere. 

748.  Formula:  Let  R  denote  the  radius  of  a  sphere,  G  the 
circumference  of  a  great  circle,  F  the  volume  of  a  spherical  sector, 
and  H  and  z  the  altitude  and  area,  respectively,  of  the  corre- 
sponding zone. 

Then,  since  G  =  2irR,  and  z  =  2  ttRH,  f  =  |  ttR^H, 


Proposition  XXVIII 

749.  Draw  a  semicircle ;  to  the  extremities  of  any  arc  of  the  semicir- 
cumference  draw  radii,  and  from  their  extremities  draw  lines  perpendicular 
to  the  diameter.  If  this  figure  is  revolved  about  the  diameter  as  an  axis, 
what  kind  of  a  solid  is  generated  by  the  part  included  between  the  per- 
pendiculars and  the  given  arc  ?  Between  the  radii  and  the  given  arc  ? 
Between  each  radius  and  the  perpendicular  from  its  extremity?  Find 
an  expression  for  the  volume  of  the  spherical  segment  in  terms  of  the 
spherical  sector  and  the  two  cones. 

Theorem,  The  volume  of  a  spherical  segment  is  equal  to 
one  half  the  product  of  its  altitude  hy  the  sum  of  its  hoses, 
plus  the  volume  of  a  sphere  of  which  that  altitude  is  a 
diameter. 


SOLID   GEOMETRY.  —  BOOK  X, 


363 


Data:  A  spherical  segment,  as  that  generated 
by  the  revolution  of  ABCD  about  MN  as  an  axis. 

Denote  the  volume  of  the  segment  by  F;  its 
altitude  CD,  by  H\  the  radii  of  its  bases  AD  and 
BC,  by  r  and  r',  respectively  ;  and  the  radius  of  the 
semicircle  by  R. 

To  prove  F  =  |  H(Tr7^  +  Trr'^)  +  i-  7rH\ 

Proof.     Draw  the  radii  OA  and  OB. 

The  volume  of  the  segment  generated  by  ABCD  equals  the 
volume  of  the  spherical  sector  generated  by  OAB  plus  the  volume 
of  the  cone  generated  by  OBC  minus  the  volume  of  the  cone 
generated  by  OAD-, 

.'    §§748,629,    V=^TrR''H^\irr^''OC-\7r7^0D. 


But 


H=  OC  —  OD, 


DC  ,  and  R^  —  7^=  OD 


Then,  F=  -l-7r)2  R\OC  -  0D)-\-{R'^  -  OC^)OG-(R^  -  OD^)OD\ 
=  ^Tr\2  I^(OC  -  0D)+  RXOC -  OD)-(OC^  -OD^)l 
=  I  7rH\3  R:'—(0a^  +  0(7  X  OD  -f  0D^)\. 

But,  §  348,    (OC ~  ODf=  0&  -  2  OC  y.0D^0D^=H^\ 

0C^+  OC  X  OD  -\-  OD^  =  f{0(f  -\-  OD^)--^ 

=  3i2«-f(r^^-r«)-^. 


Hence, 


Therefore,  etc. 


Q.E.D, 


750.   Cor.     Let  the  segment  be  a  segment  of  one  base,  as  that 
generated  by  MBAD. 

Then,  the  radius  /  =  0, 

and  V=\Trr^H+^TrH^] 

that  is,  the  volume  of  a  spherical  segment  of  one  base  is  equal  to 
one  half  the  volume  of  the  cylinder  having  the  same  base  and  the 
same  altitude  plus  the  volume  of  a  sphere  of  which  that  altitude  is 
the  diameter. 


364 


SOLID   GEOMETRY.  — BOOK  X, 


751. 

B  =  base. 

D  =  diameter. 

R  =  radius. 

r  =  radius  of  lower  base. 

r'  =  radius  of  upper  base: 

Sphere. 


FORMULA 


Notation 


A=4.7rR'   . 

r=sxiR 


Zone. 


A  =  2  ttRH 


H=  altitude. 

S  =  surface. 

A  —  area  (or  area  of  surface). 

V  =  volume. 


Spherical  Pyramid. 

V=B x^R 

Spherical  Sector. 

V=B x^R   

V=^7r^H      ..... 

Spherical  Segment. 


§726 
§743 
§744 
§744 

§728 

§746 

§747 
§748 

§749 


SUPPLEMENTARY  EXERCISES 

Ex.  881.    What  is  the  volume  of  a  sphere  whose  radius  is  9  in.  ? 

Ex.  882.  The  circumference  of  a  great  circle  of  a  sphere  is  36  ft.  What 
is  the  area  of  the  surface  of  the  sphere  ? 

Ex.  883.  The  diameter  of  a  sphere  is  13^™.  How  many  cubic  decimeters 
does  it  contain  ? 

Ex.  884.    The  volume  of  a  sphere  is  ISTO^"  ™.     What  is  its  radius  ? 

Ex.  885.  The  area  of  the  surface  of  a  sphere  is  69  sq.  ft.  What  is  its 
diameter  ? 


SOLID   GEOMETRY.  — BOOK  X.  366 

Ex.  886.  The  edge  of  a  cube  is  16cm.  what  is  the  volume  of  the  circum- 
scribed sphere  ? 

Ex.  887.  If  the  sides  of  a  spherical  triangle  are  75°,  93°,  and  110°,  what 
is  the  spherical  excess  of  its  polar  triangle  ? 

Ex.  888.  The  angles  of  a  spherical  triangle  are  98°,  110°,  and  160°.  What 
is  the  area  of  a  symmetrical  triangle  on  the  same  sphere,  the  radius  being 
12'»? 

Ex.  889.  Find  the  volume  of  a  triangular  spherical  pyramid,  the  angles 
of  the  base  being  58°,  116°,  and  145°,  and  the  diameter  of  the  sphere  being 
20  in. 

Ex.  890.  What  is  the  area  of  a  lune  whose  angle  is  60°  on  the  surface  of 
a  sphere  whose  radius  is  6  in.  ? 

Ex.  691.  What  is  the  area  of  a  zone  whose  altitude  is  3^™  on  the  surface 
of  a  sphere  whose  radius  is  %^^  ? 

Ex.  892.  What  is  the  volume  of  a  spherical  sector  whose  altitude  is  3.5", 
if  the  radius  of  the  sphere  is  11™  ? 

Ex.  893.  What  is  the  volume  of  a  spherical  wedge  whose  angle  is  72°,  the 
volume  of  the  sphere  being  1728  cu.  in.  ? 

Ex.  894.  What  is  the  area  of  a  zone  of  one  base,  if  the  chord  of  its 
generating  arc  is  13*^*^^  ? 

Ex.  895.  The  area  of  a  zone  of  a  sphere  20^™  in  diameter  is  1508<i*i™ 
What  is  the  altitude  of  the  zone  ? 

Ex.  896.  The  angles  of  the  base  of  a  triangular  spherical  pyramid  are 
90°,  121°,  and  135°.  What  is  the  volume  of  the  pyramid,  the  volume  of  the 
sphere  being  194  cu.  in.  ? 

Ex.  897.    Spherical  polygons  are  to  each  other  as  their  spherical  excesses. 

Ex.  898.  The  base  of  a  spherical  pyramid  is  a  trirectangular  triangle. 
What  part  of  the  sphere  is  the  pyramid  ? 

Ex.  899.  The  surface  of  a  sphere  is  equivalent  to  the  lateral  surface  of  the 
circumscribing  cylinder. 

Ex.  900.  What  is  the  spherical  excess  of  a  triangle  whose  area  is 
261 .8  sq.  in. ,  if  the  radius  of  the  sphere  is  10  in.  ? 

Ex.  901.  A  lune  and  a  trirectangular  spherical  triangle  on  the  same 
sphere  are  to  each  other  as  the  angle  of  the  lune  is  to  an  angle  of  45°. 

Ex.  902.    Trirectangular  triangles  on  equal  spheres  are  equal. 

Ex.  903.  The  diameters  of  two  spheres  are  12  in.  and  14  in.  respectively. 
What  is  the  ratio  of  their  surfaces  ?     What  is  the  ratio  of  their  volumes  ? 

Ex.  904.  The  areas  of  the  surfaces  of  two  spheres  are  as  144  to*  24.  What 
is  the  ratio  of  their  diameters  ?     What  is  the  ratio  of  their  volumes  ? 

Ex.  905.  The  diameters  of  the  sim  and  earth  are  in  the  ratio  of  109  : 1. 
What  is  the  ratio  of  their  volumes  ? 


366  SOLID   GEOMETRY.  — BOOK  X. 

Ex.  906.  How  many  quarts  of  water  will  a  hemispherical  kettle  hold,  if 
its  inside  diameter  is  12  in.  ? 

Ex.  907.  If  lines  are  drawn  from  any  point  in  the  surface  of  a  sphere  to 
the  ends  of  a  diameter,  they  are  perpendicular  to  each  other. 

Ex.  908.  What  is  the  circumference  of  a  small  circle  of  a  sphere  whose 
diameter  is  9^™,  if  the  circle  is  at  a  distance  of  S^m  from  the  center  ?  • 

Ex.  909.  The  'dihedral  angles  of  a  spherical  pyramid  are  40°,  80°,  and 
120°,  and  its  edge  is  9  ft.     What  is  the  volume  of  the  pyramid  ? 

Ex.  910.  Tlie  dihedrals  of  a  trihedral  angle  whose  vertex  is  at  the  centei 
of  a  sphere  are  75°,  90°,  and  130°.  What  is  the  volume  of  the  part  of  the 
sphere  included  by  the  faces  of  the  trihedral,  the  radius  of  the  sphere  being 

Ex.  911.  What  is  the  radius  of  a  sphere  which  is  equivalent  to  the  sum 
of  two  spheres  whose  radii  are  respectively  4  in.  and  7  in.  '? 

Ex.  912.  How  many  cubic  decimeters  does  a  segment  of  a  single  base 
contain,  if  it  is  cut  from  a  sphere  12<i«i  in  diameter,  the  altitude  of  the  seg- 
ment being  ^^^  ? 

Ex.  913.  In  a  sphere  whose  diameter  is  20  ft. ,  what  is  the  volume  of  a 
segment,  the  bases  of  which  are  on  the  same  side  of  the  center,  one  3  ft.  and 
the  other  5  ft.  from  it  ? 

Ex.  914.  Find  the  area  of  the  surface  of  a  sphere  inscribed  in  a  cube 
whose  surface  is  726  sq.  ft. 

Ex.  915.  A  trirectangular  triangle  and  a  lune  on  the  same  sphere  are  in 
the  ratio  of  14  :  9.  What  is  the  angle  of  the  Imie,  and  what  part  of  the  sur- 
face of  the  sphere  is  the  lun€  ? 

Ex.  916.  Find  the  area  of  a  spherical  quadrilateral  whose  angles  are  120°, 
130°,  140°,  and  150°,  the  volume  of  the  sphere  being  1000  cu.  ft. 

Ex.  917.  The  base  of  .a  cone  of  revolution  is  the  gi'eat  circle  of  a  sphere, 
and  its  altitude  is  the  radius  of  the  sphere.  What  is  the  ratio  of  the  surface 
of  the  sphere  to  the  lateral  surface  of  the  cone  ? 

Ex.  918.  The  base  of  a  cone  is  equal  to  a  great  circle  of  a  sphere,  and  its 
altitude  is  equal  to  the  diameter  of  the  sphere.  What  is  -the  ratio  of  their 
volumes  ? 

Ex.  919.  Find  the  altitude  of  a  zone  whose  area  is  equal  ti  that  of  a 
great  circle  of  the  sphere,  the  radius  of  the  sphere  being  8^"'. 

Ex.  920.  How  many  spherical  bullets^  in.  in  diameter  can  be  molded 
from  a  spherical  piece  of  lead  \  ft.  in  diameter? 

Ex.  921.  A  cannon  ball  put  into  a  cylindrical  tub  2  ft.  in  diameter  causes 
the  water  in  the  tub  to  rise  2  in.     What  is  the  diameter  of  the  cannon  ball  ? 

Ex.  922.  A  sectiion  parallel  to  the  base  of  a  hemisphere  bisects  its  altitude. 
What  is  the  ratio  of  the  volumes  of  the  spherical  segments  thus  formed  ? 

Ex.  923.  The  volume  of  a  sphere  is  to  that  of  the  circumscribed  cube  as 
ir  is  to  6. 


SOLID   GEOMETRY.  — BOOK  X.  367 

Ex.  924.   The  volume  of  a  sphere  is  to  that  of  the  inscribed  cube  as  ir 

is  to  — . 

V3 

Ex.  925.  The  surface  of  a  sphere  is  to  the  total  surface  of  the  circum- 
scrihing  cylinder  as  2  is  to  3. 

Ex.  926.  The  volume  of  a  sphere  is  to  the  volume  of  a  circumscribing 
cylinder  as  2  is  to  3. 

Ex.  927.  A  sphere  is  cut  by  five  parallel  planes  at  the  distance  of  2*'", 
3dm^  4dm,  and  5<im  from  each  other,  respectively.  What  are  the  relative 
areas  of  the  zones  included  between  the  planes  ? 

Ex.  928.  The  sides  opposite  the  equal  angles  of  a  birectangular  spherical 
triangle  are  quadrants. 

Ex.  929.  The  slant  height  of  a  cone  of  revolution  is  equal  to  the  diameter 
of  its  base.    What  is  the  ratio  of  its  total  area  to  that  of  the  inscribed  sphere  ? 

Ex.  930.  The  smallest  circle  whose  plane  passes  through  a  given  point 
within  a  sphere  is  that  one  whose  plane  is  perpendicular  to  the  radius  through 
the  given  point. 

Ex.  931.  The  intersection  of  the  surfaces  of  two  spheres  is  the  circum- 
ference of  a  circle  whose  plane  is  at  right  angles  to  the  line  joining  the 
centers  of  the  spheres,  and  whose  center  is  on  that  line. 

Ex.  932.  \Yhat  is  the  area  of  the  circle  of  intersection  of  two  spheres 
whose  radii  are  respectively  5^"!  and  8<i™,  if  their  centers  are  10^™  apart  ? 

Ex.  933.  What  is  the  weight  of  an  iron  ball,  the  area  of  whose  surface  is 
28q  m^  the  specific  gravity  of  iron  being  7.5  ? 

Ex.  934.  If  the  exterior  diameter  of  a  spherical  shell  is  12  in.,  what  should 
be  the  thickness  of  its  wall  in  order  that  it  may  contain  696. 9  cu.  in.  ? 

Ex.  935.  What  is  the  weight  of  a  hollow  iron  shell  whose  wall  is  2  in. 
thick,  if  it  will  hold  31|  pounds  of  water,  the  specific  gravity  of  iron  being  7.5  ? 

Ex.  936.  An  equilateral  triangle  revolves  about  its  altitude.  Compare 
the  volumes  of  the  solids  generated  by  the  triangle,  the  inscribed  circle,  and 
the  circumscribed  circle. 

Ex.  937.  From  a  sphere  whose  surface  is  69  sq.  ft.  a  segment  of  one  base 
is  cut,  which  has  an  altitude  of  3  ft.  What  is  the  convex  surface  of  the 
segment  ? 

Ex.  938.  What  is  the  radius  of  a  sphere  inscribed  in  a  regular  tetrahedron 
whose  entire  area  is  16  sq.  ft.  ? 

Ex.  939.  What  is  the  area  of  the  surface  of  a  sphere  inscribed  in  a  regu- 
lar tetrahedron  whose  edge  is  6  in.  ? 

Ex.  940.  How  much  of  the  surface  of  the  earth  could  a  man  see,  if  he 
were  at  the  distance  of  a  diameter  above  it  ? 

Ex.  941.  •  How  far  from  the  surface  of  the  earth  must  a  man  be  in  order 
that  he  may  see  one  fifth  of  it  ? 


368  SOLID   GEOMETRY.  — BOOK  X. 

Ex.  942.  All  arcs  of  great  circles  drawn  through  the  pole  of  a  given  great 
circle  are  perpendicular  to  its  circumference. 

Ex.  943.  The  sum  of  the  squares  of  three  chords  perpendicular  to  each 
other  at  any  point  in  the  surface  of  a  sphere  is  equal  to  the  square  of  the 
diameter. 

Ex.  944.  If  a  zone  of  one  base  is  a  mean  proportional  between  the 
remaining  surface  of  the  sphere  and  its  total  surface,  how  far  is  the  base 
of  the  zone  from  the  center  of  the  sphere  ? 

Ex.  945.  If  any  number  of  lines  in  space  meet  in  a  point,  the  feet  of  the 
perpendiculars  drawn  to  these  lines  from  another  point  lie  in  the  surface  of  a 
sphere. 

PROBLEMS   OF   CONSTRUCTION 

Ex.  946.    Bisect  an  arc  of  a  great  circle. 

Ex.  947.    Bisect  a  spherical  angle. 

Ex.  948.  At  a  given  point  in  a  given  arc  of  a  great  circle  construct  a 
spherical  angle  equal  to  a  given  spherical  angle. 

Ex.  949.  Construct  a  spherical  triangle,  the  poles  of  the  respective  sides 
being  given. 

Ex.  950.  Construct  a  spherical  triangle,  having  given  two  sides  and  the 
included  angle. 

Ex.  951.  Construct  a  spherical  triangle,  having  given  a  side  and  two  adja- 
cent angles. 

Ex.  952.*  Construct  a  spherical  triangle,  having  given  the  three  sides. 

Ex.  953.   Construct  a  spherical  triangle,  having  given  the  three  angles. 

Ex.  954.  Draw  an  arc  of  a  great  circle  perpendicular  to  a  given  spherical 
arc  from  a  point  without. 

Ex.  955.  Draw  an  arc  of  a  great  circle  perpendicular  to  a  given  spherical 
arc  at  a  point  in  it. 

Ex.  956.  Pass  a  plane  tangent  to  a  sphere  at  a  given  point  on  the  surface 
of  the  sphere. 

Ex.  957.  Pass  a  plane  tangent  to  a  sphere  through  a  given  straight  line 
without  the  sphere. 

Ex.  958.  Cut  a  given  sphere  by  a  plane  passing  through  a  given  straight 
line  so  that  the  section  shall  have  a  given  radius. 

Ex.  959.  Through  a  given  point  on  a  sphere  draw  a  great  circle  tangent 
to  a  given  small  circle. 

Ex.  960.  Through  a  given  point  on  a  sphere  draw  a  great  circle  tangent 
to  two  equal  small  circles  whose  planes  are  parallel. 

Ex.  961.  Describe  a  circle  to  pass  through  three'given  points  on  the  sur- 
face of  a  sphere. 

Ex.  962.   Circumscribe  a  circle  about  a  given  spherical  triangle. 


GEOME  TR  Y.  —  RE  VIE  W.  369 


EXERCISES  FOR   REVIEW 

Ex.  1.  The  perpendicular  erected  at  the  middle  point  of  one  side  of  a 
triangle  meets  the  longer  of  the  other  two  sides. 

Ex.  2.  Of  the  bisectors  of  two  unequal  angles  of  a  triangle,  produced  to 
the  point  of  intersection,  the  bisector  of  the  smaller  angle  is  the  longer. 

Ex.  3.  The  straight  lines  which  join  the  'middle  points  of  the  opposite 
sides  of  any  quadrilateral  bisect  each  other. 

Ex.  4.  If  a  line  is  drawn  from  the  middle  point  of  one  base  of  a  trapezoid 
to  pass  through  the  intersection  of  the  diagonals,  it  will  bisect  the  other  base. 

Ex.    5.    If  the  opposite  sides  of  a  pentagon  are  produced  to  intersect,  the 
um  of  the  angles  at  the  vertices  of  the  triangles  thus  formed  is  equal  to  two 
right  angles. 

Ex.  6.  The  sum  of  the  four  lines  drawn  from  the  vertices  of  any  quad- 
rilateral to  any  point  except  tlie  intersection  of  the  diagonals  is  greater  than 
the  sum  of  the  diagonals. 

Ex.  7.  If  the  internal  bisector  of  one  base  angle  of  a  triangle  and  the 
external  bisector  of  the  other  base  angle  are  produced  until  they  meet,  the 
angle  included  between  them  is  equal  to  half  the  vertical  angle  of  the  tri- 
angle. 

Ex.  8.  The  angle  contained  by  the  bisectors  of  two  exterior  angles  of  any 
triangle  is  equal  to  half  the  sum  of  the  two  adjacent  interior  angles. 

Ex.  9.  If  each  of  two  angles  of  a  quadrilateral  is  a  right  angle,  the  bisec- 
tors of  the  other  angles  are  either  perpendicular  or  parallel  to  each  other. 

Ex.  10.  If  the  side  CB  of  the  triangle  ABC  is  greater  than  the  side  GA, 
and  CA  is  produced  to  D  and  CB  to  E^  making  AD  and  BE  equal,  AE  is 
greater  than  DB. 

Ex.  11.  In  the  triangle  ABC  a  straight  line  AD  is  drawn  perpendicular 
to"  the  straight  line  BD  which  bisects  angle  B.  Prove  that  a  straight  line 
through  D  parallel  to  BC  bisects  AC. 

Ex.  12.  If  one  side  of  a  triangle  is  greater  than  the  other,  any  line  from 
the  vertex  of  the  included  angle  to  the  base  is  less  than  the  longer  side. 

Ex.  13.  Lines  drawn  from  one  vertex  of  a  parallelogram  to  the  middle 
points  of  the  opposite  sides  trisect  a  diagonal. 

Ex.  14.  No  two  straight  lines  drawn  from  two  vertices  of  a  triangle  and 
terminated  by  the  opposite  sides  can  bisect  each  other. 

Ex.  15.  The  base  of  a  triangle  whose  sides  are  unequal  is  divided  into 
two  parts  by  a  straight  line  bisecting  the  vertical  angle.  Prove  that  the 
greater  part  is  adjacent  to  the  greater  side. 

Ex.  16.    If  two  exterior  angles  of  a  triangle  are  bisected,  and  from  the 
point  of  intersection  of  the  bisectors  a  straight  line  is  drawn  to  the  vertex  of 
the  third  angle,  this  line  bisects  that  angle. 
milne's  geom. — 24 


370  GEOME  TRY.  — RE  VIE  W. 

Ex.  17.  ABC  is  a  triangle  ;  D  is  the  middle  point  of  50,  and  E  of  AD ; 
BE  produced  meets  AC  in  F.     Prove  that  ^40  is  trisected  in  F. 

Ex.  18.  In  the  triangle  ABC  the  sides  AB^  BC,  and  CA  are  trisected  at 
the  consecutive  points  D  and  E,  F  and  G^,  and  H  and  /i  respectively.  Prove 
that  the  lines  EF^  GH,  and  KD,  when  produced,  form  a  triangle  equal  to 
ABC. 

Ex.  19.  If  one  of  the  equal  sides  of  an  isosceles  triangle  is  produced 
below  the  base  to  a  oertain  length,  if  an  equal  length  is  cut  off  above  the 
base  from  the  other  equal  side,  and  if  the  two  points  are  joined  by  a  straight 
line,  this  line  is  bisected  by  the  base. 

Ex.  20.  ABC  is  a  triangle,  and  BE  and  CF  are  drawn  perpendicular  to 
AG,  a  line  through  ^  ;  2>  is  the  middle  point  of  BC.  Show  that  FD  equals 
ED. 

Ex.  21.  The  angle  contained  by  the  bisectors  of  the  base  angles  of  any 
triangle  is  equal  to  the  vertical  angle  of  the  triangle  plus  half  the  sum  of  the 
base  angles. 

Ex.  22.  The  bisectors  of  two  angles  of  an  equilateral  triangle  intersect, 
and  from  their  point  of  intersection  lines  are  drawn  parallel  to  any  two  sides. 
Prove  that  these  lines  trisect  the  third  side. 

Ex.  23.   The  opposite  sides  of  a  regular  hexagon  are  parallel. 

Ex.  24.  If  in  a  quadrilateral  the  diagonals  are  equal  and  two  sides 
are  parallel,  the  other  sides  are  equal. 

Ex.  25.  If  from  any  point  in  the  base  of  an  isosceles  triangle  perpendicu- 
lars are  drawn  to  the  equal  sides,  their  sum  is  equal  to  the  perpendicular 
drawn  from  either  extremity  of  the  base  to  the  opposite  side. 

Ex.  26.  The  sum  of  the  perpendiculars  from  any  point  within  an  equi- 
lateral triangle  to  its  sides  is  equal  to  the  altitude. 

Ex.  27.  If  from  the  vertex  of  any  triangle  two  lines  are  drawn,  one  of 
which  bisects  the  angle  at  the  vertex  and  the  other  is  perpendicular  to  the 
base,  the  angle  between  these  lines  is  half  the  difference  of  the  angles  at  the 
base  of  the  triangle. 

Ex,  28.  In  any  triangle,  the  sides  of  the  vertical  angle  being  unequal, 
the  median  drawn  from  the  vertical  angle  lies  between  the  bisector  of  that 
angle  and  the  longer  side. 

Ex.  29.  In  any  triangle,  the  sides  of  the  vertical  angle  being  unequal, 
the  bisector  of  that  angle  lies  between  the  median  and  the  perpendicular 
drawn  from  the  vertex  to  the  base. 

Ex.  30.  Lines  are  drawn  through  the  extremities  of  the  base  of  an 
isosceles  triangle,  making  angles  with  it,  on  the  side  opposite  the  vertex, 
each  equal  to  one  third  of  a  base  angle  of  the  triangle,  and  meeting  the  sides 
produced.     Prove  that  the  three  triangles  thus  formed  are  isosceles. 


GEOMETRY.  — REVIEW.  371 

Ex.  31.  If  two  circumferences  are  tangent  internally  and  the  radius  of 
the  larger  is  the  diameter  of  the  smaller,  any  chord  of  tiie  larger  drawn  from 
the  point  of  contact  is  bisected  by  the  circumference  of  the  smaller. 

Ex.  32.  If  perpendiculars  are  drawn  to  any  chord  at  its  extremities  and 
produced  to  intersect  a  diameter  of  the  circle,  the  points  of  intersection  are 
equally  distant  from  the  center. 

Ex.  33.  If  perpendiculars  are  drawn  from  the  ends  of  a  diameter  of  a 
circle  upon  any  secant,  their  feet  are  equally  distant  from  the  points  in  which 
the  secant  intersects  the  circumference. 

Ex.  34.  Given  an  arc  of  a  circumference,  the  chord  subtended  by  it,  and 
the  tangent  at  one  extremity.  Prove  that  the  perpendiculars  dropped  from 
the  middle  point  of  the  arc  upon  the  tangent  and  chord,  respectively,  are 
equal. 

Ex.  35.  The  bisectors  of  the  angles  contained  by  the  opposite  sides  (pro- 
duced) of  an  inscribed  quadrilateral  intersect  at  right  angles. 

Ex.  36.  If  two  opposite  sides  of  an  inscribed  quadrilateral  are  equal,  the 
other  two  sides  are  parallel. 

Ex.  37.  In  a  given  square,  inscribe  an  equilateral  triangle  having  its 
vertex  in  the  middle  of  a  side  of  the  square. 

Ex.  38.  Find,  in  a  side  of  a  triangle,  a  point  from  which  straight  lines, 
drawn  parallel  to  the  other  sides  of  the  triangle  and  terminated  by  them, 
are  equal. 

Ex.  39.  Construct  a  triangle,'  having  given  the  base,  one  of  the  angles 
at  the  base,  and  the  sum  of  the  other  two  sides. 

Ex.  40.  Construct  a  triangle,  having  given  the  base,  one  of  the  angles  at 
the  base,  and  the  difference  of  the  other  two  sides. 

Ex.  41.  Construct  a  triangle,  having  given  the  perpendicular  from  the 
vertex  to  the  base,  and  the  difference  between  each  side  and  the  adjacent 
segment  of  the  base. 

Ex.  42.  If  two  circles  are  each  tangent  to  two  parallel  lines  and  a  trans- 
versal crossing  them,  the  line  of  centers  is  equal  to  the  segment  of-  the 
transversal  intercepted  between  the  parallels. 

Ex.  43.  If  through  the  point  of  contact  of  two  circles  which  are  tangent 
to  each  other  externally  any  straight  line  is  drawn  terminated  by  the  circum- 
ferences, the  tangents  at  its  extremities  are  parallel  to  each  other. 

Ex.  44.  If  two  circles  are  tangent  to  each  other  externally  and  parallel 
diameters  are  drawn,  the  straight  line  joining  the  opposite  extremities  of 
these  diameters  will  pass  through  the  point  of  contact. 

Ex.  45.    Construct  the  three  escribed  *  circles  of  a  given  triangle. 

*  A  circle  tangent  to  one  side  of  a  triangle  and  to  the  other  two  sides  produced 
is  called  an  escribed  circle. 


372  GEOMETRH. ^REVIEW, 

.    Ex.  46.   Construct  an  isosceles  right  triangle,  having  given  the  sum  of 
the  hypotenuse  and  one  side. 

Ex.  47.  Construct  a  right  triangle,  having  given  the  hypotenuse  and  the 
sum  of  the  sides. 

Ex.  48.  Construct  a  right  triangle,  having  given  the  hypotenuse  and  the 
difference  of  the  sides. 

Ex.  49.  A  and  B  are  two  fixed  points  on  the  circumference  of  a  circle 
and  CD  is  any  diameter.     What  is  the  locus  of  the  intersection  of  CA  and 

Ex.  50.  Construct  a  triangle,  having  given  a  median  and  the  two  angles 
into  which  the  angle  is  divided  by  that  median. 

Ex.  51.  Construct  a  triangle,  having  given  the  base,  the  difference 
between  the  sides,  and  the  difference  between  the  angles  at  the  base. 

Ex.  52.  Construct  an  isosceles  triangle,  having  given  the  perimeter  and 
altitude. 

Ex.  53.  The  circles  described  on  two  sides  of  a  triangle  as  diameters 
intersect  on  the  third  side,  or  the  third  side  produced. 

Ex.  54.  ABC  is  a  triangle  having  AG  equal  to  BC',  D  is  any  point  in 
AB.  Prove  that  the  circles  circumscribed  about  triangles  ADC  and  DBC 
are  equal. 

Ex.  55.  Construct  a  triangle,  having  given  two  sides  and  the  median 
to  the  third  side. 

Ex.  56.  Construct  a  triangle,  having  given  its  perimeter,  and  having  its 
angles  equal  to  the  angles  of  a  given  triangle. 

Ex.  57.  Construct  a  triangle,  having  given  one  side  and  the  medians 
to  the  other  sides. 

Ex.  58.  Construct  a  circle  of  given  radius  to  touch  a  given  circle  and  a 
given  straight  line.     How  many  such  circles  may  there  be  ? 

Ex.  59.  Construct  a  circle  of  given  radius  which  shall  be  tangent  to  two 
given  circles.     How  many  solutions  may  there  be  ? 

Ex.  60.  If  an  equilateral  triangle  is  inscribed  in  a  circle  and  from  any 
point  in  the  circumference  lines  are  drawn  to  the  vertices,  the  longest  of 
these  lines  is  equal  to  tb^.  sum  of  the  other  two. 

Ex.  61.  If  two  circles  intersect  each  other,  two  parallel  lines  passing 
through  the  points  of  intersection  and  terminated  by  the  exterior  arcs  are 
equal. 

Ex.  62.  An  isosceles  triangle  has  its  vertical  angle  equal  to  an  exterior 
angle  of  an  equilateral  triangle.  Prove  that  the  radius  of  the  circumscribed 
circle  is  equal  to  one  of  the  equal  sides  of  the  isosceles  triangle. 

Ex.  63.  If  a  chord  of  a  circle  is  extended  by  a  length  equal  to  tlie  radius, 
and  from  the  extremity  a  secant  is  drawn  through  the  center  of  the  circle, 
the  length  of  the  greater  included  arc  is  three  times  the  length  9l  \\x^  less. 


GEOME  TR  Y.  —  RE  VIE  W.  373 

Ex.  64.  Construct  three  circles  having  equal  diameters  and  being  tangent 
to  each  other. 

Ex.  65.  Construct  two  circles  of  given  radii  to  touch  each  other  and  a 
given  straight  line  on  the  same  side  of  it, 

Ex.  66.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle, 
and  the  point  at  which  the  base  is  cut  by  the  bisector  of  the  vertical  angle. 

Ex.  67.  Construct  a  circle  to  touch  a  given  circle  and  also  to  touch  a 
given  straight  line  at  a  given  point. 

Ex.  68.  Construct  a  circle  to  touch  a  given  straight  line,  and  to  touch  a 
given  circle  at  a  given  point. 

Ex.  69.  Construct  a  circle  to  touch  a  given  circle,  have  its  center  in  a 
given  line,  and  pass  through  a  given  point  in  that  line. 

Ex.  70.    If  a  :h  =  c  :d,  prove  that 

a'^ -\- ab -{- b'^ :  a^  -  ab  +  b^  =  c^  +  cd -{■  d^  :  c^  -  cd  +  d^. 

Ex.  71.   Given  three  lines  a,  b,  and  c.     Construct  as  =  —• 

4     r  "' 

Ex.  72.    Construct  x,  having  given  -  =  -. 

Ex.  73.  The  diagonals  of  a  trapezoid  divide  each  other  into  segments 
which  are  proportional. 

Ex.  74.  If  one  side  of  a  right  triangle  is  double  the  other,  the  perpen- 
dicular from  the  vertex  upon  the  hypotenuse  divides  the  hypotenuse  into 
parts  which  are  in  the  ratio  of  1  to  4. 

Ex.  75.  ABCD  is  an  inscribed  quadrilateral.  The  sides  AB  and  DC  are 
produced  to  meet  at  E.     Prove  triangles  ACE  and  BDE  similar. 

Ex.  76.  If  AB  is  a  chord  of  a  circle,  and  CD  is  any  chord  drawn  from 
the  middle  point  C  of  the  arc  AB  cutting  the  chord  AB  at  E^  prove  that  the 
chord  ^C  is  a  mean  proportional  between  CE  and  CD. 

Ex.  77.  If  perpendiculars  are  drawn  from  two  vertices  of  a  triangle  to 
the  opposite  sides,  the  triangle  cut  off  by  the  line  jbining  the  feet  of  the  per- 
pendiculars is  similar  to  the  original  triangle. 

Ex.  78.  AB  is  the  hypotenuse  of  the  right  triangle  ABC.  If  perpen- 
diculars are  drawn  to  AB  at  A  and  B,  meeting  BC  produced  at  E,  and  AC 
produced  at  Z>,  the  triangles  ACE  and  BCD  are  similar. 

Ex.  79.  If  two  circles  intersect  in  the  points  A  and  J5,  and  a  secant 
through  B  cuts  the  circumferences  in  C  and  D  respectively,  the  straight  lines 
AC  and  AD  are  in  the  same  ratio  as  the  diameters  of  the  circles. 

Ex.  80.    Inscribe  a  square  in  a  given  right  isosceles  triangle. 

Ex.  81.  From  the  obtuse  angle  of  a  triangle  draw  a  line  to  the  base, 
which  shall  be  a  mean  proportional  between  the  external  segments  into 
wnich  It  divides  the  base. 


374  ,  GEOMETRY.— REVIEW, 

Ex.  82.  Through  a  given  point  draw  a  straight  line,  so  that  the  parts  of 
it  intercepted  between  that  point  and  perpendiculars  drawn  to  it  from  two 
other  given  points  may  have  a  given  ratio. 

Ex.  83.  Show  that  the  diagonals  of  a  trapezoid,  one  of  whose  bases  is 
double  the  other,  cut  each  other  at  a  point  of  trisection. 

Ex.  84.  A  tangent  to  a  circle  at  the  point  A  intersects  two  parallel  tan- 
gents whose  points  of^  contact  are  D  and  E,  in  B  and  C  respectively.  BE 
and  CD  intersect  at  F.  Prove  that  the  line  AF  is  parallel  to  the  tangents 
BD  and  CE. 

Ex.  85.  The  angle  C  of  the  triangle  ABC  is  bisected  by  CD,  which  cuts 
the  base  AB  at  Z) ;  0  is  the  middle  point  of  AB.  Prove  that  OD  has  the 
same  ratio  to  OA  that  the  difference  of  the  sides  has  to  their  sum. 

Ex.  86.  A  and  B  are  two  points  on  the  circumference  of  a  circle  of 
which  0  is  the  center  ;  tangents  at  A  and  B  meet  at  E ;  and  from  A  the  line 
AD  is  drawn  perpendicular  to  OB.     Prove  BE :  B0  =  BD  :  AD. 

Ex.  87.  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 
circumference  in  F  and  G  respectively.  Prove  that  CFDG  has  any  two  of 
its  adjacent  sides  in  the  same  ratio  as  the  remaining  two. 

Ex.  88.  Two  circles  whose  centers  are  O  and  P  intersect  in  A,  and  the 
tangent  to  each  at  A  meets  the  circumference  of  the  other  in  C  and  B  respec- 
tively.    Prove  that  AB:AC  =  OA:  PA. 

Ex.  89.  0  is  the  center  of  the  circle  inscribed  in  the  triangle  ABC ;  AG 
meets  BC  in  D.     Trove  AO  :  DO  =  AB -\- AC  :  BC. 

Ex.  90.  ABC  is  an  isosceles  triangle;  the  perpendicular  to  AC  bX  C 
meets  the  base  AB,  or  the  base  produced,  at  ^ ;  Z>  is  the  middle  point  of  AE. 
Prove  that  ^C  is  a  mean  proportional  between  AB  and  AD. 

Ex.  91.  AB  and  CD  are  two  parallel  straight  lines  ;  E  is  the  middle 
point  of  CD ;  AC  and  BE  meet  in  F,  and  AE  and  BD  meet  in  G.  Prove 
that  FG  is  parallel  to  AB. 

Ex.  92.  If  two  circles  are  tangent  to  each  other,  either  internally  or 
externally,  any  two  straight  lines  drawn  through  the  point  of  contact  will  be 
cut  proportionally  by  the  circumferences. 

Ex.  93.  From  one  of  the  points  of  intersection  of  two  intersecting  circles 
a  diameter  of  each  circle  is  drawn.  Prove  (1)  that  the  line  joining  the 
extremities  of  these  diameters  passes  through  the  other  point  of  intersection  ; 
and  (2)  that  this  line  is  parallel  to  the  line  of  centers  of  the  circles. 

Ex.  94.  If  in  a  right  triangle  a  perpendicular  is  drawn  from  the  vertex  of 
the  right  angle  to  the  hypotenuse,  and  circles  are  inscribed  in  the  triangles 
thus  formed,  the  diameters  are  proportional  to  the  sides  of  the  given  right 
angle. 

Ex.  95.  The  distance  from  the  center  of  a  circle  to  a  chord  8<*™  long  is 
4dm,     What  is  the  distance  from  the  center  to  a  chord  6<^™  long  ? 


GEOMETRY.  — REVIEW.  875 

Ex.  96.  If  a  chord  18^"^  long  is  bisected  by  another  chord  22^°^  long, 
what  are  the  segments  of  the  latter? 

Ex.  97.  If  two  intersecting  chords  divide  the  circumference  of  a  circle 
into  parts  whose  lengths  taken  in  order  are  as  1,  1,  2,  and  5,  what  angles  do 
the  chords  make  with  each  other  ? 

Ex.  98.  The  square  on  the  hypotenuse  of  a  right  isosceles  triangle  is 
equivalent  to  four  times  the  triangle. 

Ex.  99.  If  the  sides  of  a  triangular  field  are  respectively  llHm^  gnm^  and 
8^1"  long,  how  many  hektares  are  there  in  the  area  of  the  field  ? 

Ex.  100.  The  sides  of  a  triangle  are  respectively  39,  42,  and  45  inches  in 
length.     What  is  the  radius  of  the  inscribed  circle  ? 

Ex.  101.  The  sides  of  a  triangle  are  respectively  5  ft.,  5  ft.,  and  6  ft. 
What  is  the  diameter  of  the  circumscribed  circle  ? 

Ex.  102.  A  triangular  field  has  its  sides  respectively  16  rd.,  24  rd.,  and 
36  rd.  long.  What  is  the  length  of  a  line  from  the  middle  of  the  longest 
side  to  the  opposite  comer  ?    What  is  the  area  of  the  field  ? 

Ex.  103.  If  a  chord  IQcm  long  is  5"n  distant  from  the  center  of  a  circle, 
what  is  the  radius  of  the  circle,  and  the  distance  from  the  end  of  the  chord 
to  the  end  of  the  radius  that  is  perpendicular  to  the  chord  ? 

Ex.  104.  How  many  square  meters  are  there  in  the  area  of  the  quadri- 
lateral ABCD,  if  AB  =  6™,  BG  =  11"",  CD  =  4%  AD  =  13™,  and  the  diago- 
nal ^(7=15°^? 

Ex.  105.  If  two  equivalent  triangles  have  a  comm'on  base,  and  lie  on 
opposite  sides  of  it,  the  base,  or  the  base  produced,  will  bisect  the  line  join- 
ing their  vertices. 

Ex.  106.  ABC  is  a  given  triangle.  Construct  an  equivalent  triangle, 
having  its  vertex  at  a  given  point  in  BC,  and  its  base  in  the  same  straight 
Hue  as  AB. 

Ex.  107.  Through  the  vertex  A  of  the  parallelogram  ABCD  draw  a  line 
meeting  the  side  CB  produced  in  F^  and  the  side  CD  produced  in  E.  Prove 
that  the  rectangle  of  the  produced  parts  of  the  sides  is  equivalent  to  the 
rectangle  of  the  sides. 

Ex.  108.  ABC  is  a  right  triangle  having  its  right  angle  at  B.  At  A  and 
C  perpendiculars  to  ^C  are  erected  to  meet  CB  and  AB  produced  in  E  and 
F  respectively,  and  EF  is  drawn.  Prove  that  the  triangles  BEF  and  ABC 
are  equivalent. 

Ex.  109.  The  square  upon  the  altitude  of  an  equilateral  triangle  is 
equivalent  to  three  times  the  square  upon  half  of  one  of  the  sides  of  the 
triangle. 

Ex.  110.  If  from  a  point  D  in  the  base  AB  of  the  triangle  ABC  straight 
lines 'are  drawn  parallel  to  the  sides  AC  and  BC  respectively,  so  as  to  meet 
BC  m  F  and  ^O  in  ^,  triangle  EFG  is  a  mean  proportional  between  tri- 
angles  ADE  and  DBF. 


876  GEOMETRY.— REVIEW. 

Ex.  111.  Two  sides  Of  a  triangle  are  70™  and  65"^,  and  the  difference  of 
the  segments  of  the  other  side  made  by  a  perpendicular  from  the  opposite 
vertex  is  9*".    What  is  the  length  of  the  other  side  ? 

Ex.  112.  The  sum  of  two  sides  of  a  triangle  is  128  ft. ,  and  a  perpendicular 
from  the  vertex  opposite  the  other  side  divides  that  side  into  segments  of 
60  ft.  and  28  ft.     What  are  the  sides  of  the  triangle  ? 

Ex.  113.  Two  sides  of  a  triangle  are  in  proportion  to  each  other  as  6  is  to 
5,  and  the  adjacent  segments  of  the  other  side  made  by  a  perpendicular  from 
the  opposite  vertex  are  36  ft.  and  14  ft.  respectively.     What  are  the  sides  ? 

Ex.  114.  The  difference  of  the  two  sides  of  an  oblique  triangle,  obtuse- 
angled  at  the  base,  is  9"^,  and  the  segments  of  the  base  produced  made  by  a 
perpendicular  from  the  opposite  vertex  are  30™  and  9™.     What  are  the  sides  ? 

Ex.  115.  A  flag  pole  140  ft.  long,  standing  on  an  eminence  30  ft.  high, 
broke  so  that  the  top  struck  the  level  ground  at  a  distance  from  the  base  of 
the  pole  equal  to  the  length  of  the  part  standing.  What  was  the  length  of 
the  part  broken  off  ? 

Ex.  116.  If  from  the  extremities  of  the  base  of  any  triangle  lines  are 
drawn  bisecting  the  other  two  sides,  these  lines  intersect  within  the  triangle 
and  form  another  triangle  on  the  same  base  equivalent  to  one  third  of  the 
original  triangle. 

Ex.  117.  Upon  the  sides  of  a  right  triangle,  as  homologous  sides,  three 
similar  polygons  of  any  number  of  sides  are  constructed.  Prove  that  the 
polygon  upon  the  hypotenuse  is  equivalent  to  the  sum  of  the  polygons  upon 
the  other  two  sides. 

Ex.  118.  ABCD  is  a  rectangle,  and  BD  is  its  diagonal ;  a  circle  whose 
center  is  0  is  inscribed  in  the  triangle  DEC ;  EO  and  FO  are  drawn  perpen- 
dicular to  AB  and  AB  respectively.  Then,  the  rectangle  AFOE  is  equiva- 
lent to  one  half  the  rectangle  ABCD. 

Ex.  119.  If  squares  are  described  upon  the  three  sides  of  a  right  triangle, 
and  the  extremities  of  the  adjacent  sides  of  any  two  squares  are  joined,  the 
triangle  so  formed  is  equivalent  to  the  given  triangle. 

Ex.  120.    Inscribe  a  circle  in  a  given  rhombus. 

Ex.  121.  A  segment  whose  arc  is  60°  is  cut  off  from  a  circle  whose 
radius  is  15  ft.    What  is  the  area  of  the  segment  ? 

Ex.  122.  If  the  bisectors  of  all  the  angles  of  a  polygon  meet  in  a  point, 
a  circle  may  be  inscribed  in  the  polygon. 

Ex.  123.  If  the  area  of  a  certain  circle  is  154«<J™,  how  many  degrees  are 
there  in  an  angle  at  the  center,  if  it  is  subtended  by  an  arc  of  the  circumfer- 
ence 6. 5""  long  ? 

Ex.  124.  Prove  that  an  equiangular  polygon  inscribed  in  a  circle  is 
regular,  if  the  number  of  its  sides  is  odd. 


GEOMETRY.  — REVIEW,  3T7 

Ex.  125.  Two  parallel  chords*  in  a  circle  are  the  sides  of  regular  in- 
scribed polygons,  one  a  hexagon  and  the  other  a  dodecagon.  If  the  radius 
of  the  circle  is  11  in.,  how  far  apart  are  the  chords  ? 

Ex.  126.  What  is  the  length  of  the  side  of  a  square  equivalent  to  a 
circle  in  which  a  chord  of  30<i™  has  an  arc  whose  height  is  S^^™  ? 

Ex.  127.  If  a  4-inch  pipe  will  fill  a  cistern  in  2  hr.  30  min.,  how  long  will 
it  take  a  2-inch  pipe  to  fill  it  ? 

Ex.  128.  If  an  equilateral  triangle  and  a  regular  decagon  each  has  a 
perimeter  of  Q^^  what  is  the  difference  in  area  between  them  ? 

Ex.  129.  If  an  equilateral  triangle  is  inscribed  in  a  circle,  the  line  joining 
the  middle  points  of  the  arcs  cut  off  by  two  of  its  sides  will  be  trisected  by 
those  sides. 

Ex.  130.  If  the  area  between  three  equal  circles,  each  tangent  to  the 
other  two,  is  40sq  ™,  what  are  the  radii  of  the  circles  ? 

Ex.  131.   Construct  a  circle  equal  to  three  fourths  of  a  given  circle. 

Ex.  132.  If  a  circle  is  circumscribed  about  a  right  triangle,  and  on  each 
of  its  sides  as  a  diameter  a  semicircle  is  described  exterior  to  the  triangle, 
the  sum  of  the  areas  of  the  crescents  thus  formed  is  equal  to  the  area  of  the 
triangle. 

Ex.  133.  The  area  of  an  inscribed  regular  octagon  is  equal  to  that  of 
a  rectangle  whose  sides  are  respectively  equal  to  the  sides  of  the  inscribed 
and  circumscribed  squares. 

Ex.  134.  If  the  radius  of  a  circle  is  r,  prove  that  the  area  of  a  regular 
inscribed  octagon  is  2  r^\/2. 

Ex.  135.  The  area  of  a  circle  is  a  mean  proportional  between  the  areas 
of  any  two  similar  polygons,  one  of  which  is  circumscribed  about  the  circle 
and  the  other  is  isoperimetric  with  the  circle.     (^Oalileo^s  TJieorem.) 

Ex.  136.  Prove  that  the  sum  of  the  perpendiculars  drawn  to  the  sides 
of  a  regular  polygon  from  any  point  within  is  equal  to  the  apothem  of  the 
polygon  multiplied  by  the  number  of  its  sides. 

Ex.  137.  If  upon  the  sides  of  a  regular  hexagon  squares  are  constructed 
outwardly,  the  exterior  vertices  of  these  squares  are  the  vertices  of  a  regular 
dodecagon. 

Ex.  138.  A  horse  is  tethered  to  a  hook  on  the  inner  side  of  a  fence  which 
bounds  a  circular  grass  plot.  His  tether  is  so  long  that  he  can  just  reach  the 
center  of  the  plot.  The  area  of  so  much  of  the  plot  as  he  can  graze  over  is 
3/(4  IT  —  3\/3)  sq.  rd. ;  find  the  length  of  the  tether  and  the  circumference  of 
the  plot.     {Harvard.) 

Ex.  139.  If  equal  straight  lines  are  drawn  from  a  given  point  to  a  given 
plane  they  make  equal  angles  with  the  plane. 

Ex.  140.  Two  planes  which  are  each  perpendicular  to  a  third  plane  are 
parallel,  if  their  intersections  with  the  third  plane  are  parallel. 


378  '  GEOMETRY.  — REVIEW. 

Ex.  141.  The  line  joining  the  extremities  of  two  equal  lines  which  are 
perpendicular  to  a  plane,  on  the  same  side  of  it,  is  parallel  to  the  plane. 

Ex.  142.  Through  a  given  line  in  a  given  plane  pass  a  plane  to  make 
a  given  angle  with  the  given  plane. 

Ex.  143.  Through  a  given  line  parallel  to  a  given  plane  pass  a  plane  to 
make  a  given  angle  with  the  given  plane. 

Ex.  144.  Through  the  edge  of  a  given  dihedral  angle  pass  a  plane  to 
bisect  that  angle. 

Ex.  145.  Find  the  locus  of  the  points  in  space  which  are  equidistant  from 
two  parallel  lines. 

Ex.  146.  If  a  straight  line  is  perpendicular  to  a  plane,  its  projection  on 
any  other  plane  is  perpendicular  to  the  intersection  of  the  two  planes. 

Ex.  147.  If  two  planes  are  perpendicular,  a  straight  line  drawn  from  any 
point  of  one  plane  perpendicular  to  the  other  will  lie  in  the  first  plane. 

Ex.  148.  The  projection  of  a  straight  line  upon  a  plane  is  a  straight 
line. 

Ex.  149.  Find  the  locus  of  the  points  in  space  which  are  equidistant 
from  three  given  planes. 

Ex.  150.  Find  the  locus  of  the  points  in  space  equidistant  from  two  in- 
tersecting straight  lines. 

Ex.  151.  Two  trihedrals  are  equal  or  symmetrical,  if  two  face  angles, 
and  the  dihedral  between  their  faces,  in  one  are  equal,  each  to  each,  to  the 
corresponding  parts  in  the  other. 

Ex.  152.  Find  the  locus  of  the  points  in  space  which  are  equidistant 
from  three  given  straight  lines  in  the  same  plane. 

Ex.  153.  Find  the  locus  of  the  points  in  a  given  plane  which  are  equi- 
distant from  two  given  points  without  the  plane. 

Ex.  154.  The  angles  AOB  and  AOG  in  different  planes  are  equal.  Prove 
that  the  plane  bisecting  the  dihedral  angle  between  their  planes  is  perpen- 
dicular to  the  plane  BOG. 

Ex.  155.  The  planes  through  any  two  pairs  of  lines  that  pass  through 
a  point  intersect  in  a  line  which  passes  through  the  same  point. 

Ex.  156.  In  a  given  plane  find  a  point  equidistant  from  three  given 
points  without  the  plane. 

Ex.  157.  Through  a  given  point  in 'space,  draw  a  straight  line  which 
shall  cut  two  given  straight  lines  not  in  the  same  plane. 

Ex.  158.  Find  the  locus  of  the  points  in  space  which  are  equidistant 
from  two  given  planes  and  also  equidistant  from  two  given  points. 

Ex.  159.  Two  planes  are  perpendicular  respectively  to  two  non-parallel 
lines  which  are  not  in  the  same  plane.  Prove  that  their  intersection  is 
perpendicular  to  any  plane  that  is  parallel  to  both  lines. 


GEOMETRY.  — REVIEW.  379 

Ex.  160.  From  the  vertex  of  a  trihedral  angle  a  line  is  drawn  within  the 
angle.  Prove  that  the  sum  of  the  angles  which  this  line  makes  with  the  edges 
is  less  than  the  sum,  but  greater  than  half  the  sum,  of  the  face  angles. 

Ex.  161.  Two  trihedrals  are  equal  or  symmetrical,  if  two  dihedrals  and 
the  included  face  angle  of  one  are  equal,  each  to  each,  to  the  corresponding 
parts  of  the  other. 

Ex.  162.  A  plane  parallel  to  two  sides  of  a-  quadrilateral  in  space  (that 
is,  a  quadrilateral  whose  sides  do  not  all  lie  in  the  same  plane)  divides  the 
other  two  sides  proportionally. 

Ex.  163.  In  any  trihedral,  the  three  planes,  passed  through  the  edges 
perpendicular  to  the  opposite  faces  respectively,  intersect  in  the  same  straight 
line. 

Ex,  164.  In  any  trihedral,  the  three  planes,  passed  through  the  edges  and 
the  bisectors  of  the  opposite  face  angles  respectively,  intersect  in  the  same 
straight  line. 

Ex.  165.  What  is  the  edge  of  a  cubical  vessel  that  holds  one  half  ton  of 
water  ? 

Ex.  166.  Represent  the  base  edge  of  a  regular  four-sided  pyramid  by  e, 
its  altitude  by  h,  and  its  total  surface  by  T.  Compute  the  base  edge  in  terms 
of  h  and  T. 

Ex.  167.  What  is  the  difference  in  volume  between  the  frustum  of  a 
pyramid  and  a  prism,  each  12'i'"  high,  if  the  bases  of  the  frustum  are  squares 
whose  sides  are  lO'i'"  and  S^™  respectively,  and  the  base  of  the  prism  is  a 
eection  of  the  frustum  parallel  to  its  bases  and  midway  between  them  ? 

Ex.  168.   What  is  the  volume  of  a  regular  tetrahedron  whose  edge  is 

10dm? 

Ex.  169.  The  altitude  of  a  regular  hexagonal  pyramid  is  13  in.,  and  its 
slant  height  is  16  in.     What  is  its  lateral  edge  ? 

Ex.  170.  The  lateral  faces  of  a  regular  quadrangular  pyramid  are  equi- 
lateral triangles,  and  its  altitude  is  9™.     What  is  the  area  of  the  base  ? 

Ex.  171.  The  altitude  of  a  frustum  of  a  regular  quadrangular  pyramid  is 
10<=m,  and  the  sides  of  its  bases  are  respectively  IG"^"*  and  6'='".  What  is  the 
lateral  area  of  the  frustum  ? 

Ex.  172.  If  the  altitude  of  a  pyramid  is  h,  at  what  distance  from  the 
vertex  will  it  be  cut  by  a  plane  parallel  to  the  base  and  dividing  the  pyra- 
mid into  two  parts  which  are  in  the  ratio  of  3  :  4  ? 

Ex.  173.  At  what  distances  from  the  vertex  will  a  lateral  edge  of  a  pyra- 
mid be  cut  by  two  planes  parallel  to  the  base,  if  they  divide  the  pyramid  into 
three  equivalent  parts,  the  length  of  the  edge  being  m  ? 

Ex.  174.  If  the  base  edge  of  a  regular  square  pyramid  is  m,  and  its  total 
surface  is  7',  what  is  its  volume  ? 

Ex.  175.  The  perimeter  of  the  base  of  a  regular  quadrangular  pyramid 
is  p,  and  the  area  of  a  section  through  two  diagonally  opposite  edges  is  A. 
What  is  the  lateral  area  of  the  pyramid  ? 


380  GEOME  TRY.  — RE  VJE  W. 

Ex.  176.  If  two  tetrahedrons  have  the  faces  including  a  trihedral  of  one 
similar  to  the  faces  including  a  trihedral  of  the  other,  each  to  each,  and 
similarly  placed,  the  tetrahedrons  are  similar. 

Ex.  177.  If  two  tetrahedrons  have  a  dihedral  angle  of  one  equal  to  a 
dihedral  angle  of  the  other,  and  the  faces  including  these  dihedrals  similar, 
each  to  each,  and  similarly  placed,  the  tetrahedrons  are  similar. 

Ex.  178.  The  perpendicular  from  the  middle  point  of  the  diagonal  of  a 
rectangular  parallelopiped  upon  a  lateral  edge  bisects  the  edge,  and  is  equal 
to  one  half  the  projection  of  the  diagonal  upon  the  base. 

Ex.  179.  In  any  polyhedron  the  number  of  edges  increased  by  two  is 
equal  to  the  number  of  vertices  increased  by  the  number  of  faces.  {Eider's 
Theorem. ) 

Ex.  180.  The  sum  of  the  face  angles  of  any  polyhedron  is  equal  to  four 
right  angles  taken  as  many  times,  less  two,  as  the  polyhedron  has  vertices. 

Ex.  181.  If  a  plane  is  tangent  to  a  circular  cone,  its  intersection  with  the 
plane  of  the  base  is  tangent  to  the  base. 

Ex.  182.  If  a  plane  is  tangent  to  a  circular  cylinder,  its  intersection  with 
the  plane  of  the  base  is  tangent  to  the  base. 

Ex.  183.  What  are  the  dimensions  of  a  cylindrical  measure  whose  alti- 
tude is  half  its  diameter,  if  it  holds  a  half  bushel  ? 

Ex.  184.  Find  the  weight  of  the  water  that  will  be  contained  in  a  vertical 
pipe  40  ft.  high  and  1  ft.  in  diameter.  Also  find  the  pressure  per  square  inch 
on  the  base  of  the  pipe. 

Ex.  185.  A  rectangle  revolves  successively  about  two  adjacent  sides 
whose  lengths  are  m  and  n  respectively.  Compare  the  volumes  of  the 
cylinders  thus  generated. 

Ex.  186.  A  right  triangle  revolves  successively  about  the  perpendicular 
sides  whose  lengths  are  m  and  n  respectively.  Compare  the  volumes  of  the 
cones  thus  generated. 

Ex.  187.  If  the  sides  including  the  right  angle  of  a  right  triangle  are  m 
and  n,  what  is  the  area  of  the  surface  generated  by  revolving  the  triangle 
about-its  hypotenuse  as  an  axis  ? 

Ex.  188  Find  the  altitude  of  a  cylinder  of  revolution  of  radius  r,  if  the 
cylinder  is  equivalent  to  a  rectangular  parallelopiped  whose  dimensions  are 
I,  m,  and  n. 

Ex.  189.  Find  the  altitude  of  a  cone  of  revolution  of  radius  r,  equivalent 
to  a  rectangular  parallelopiped  whose  dimensions  are  I,  m,  and  n. 

Ex.  190.  The  altitudes  of  two  equivalent  cylinders  of  revolution  are  in 
the  ratio  a  :b.     If  the  radius  of  one  is  r,  what  is  the  radius  of  the  other  ? 

Ex.  191.  Find  the  altitude  of  a  regular  quadrangular  prism  whose  base 
edge  is  w,  the  prism  being  equivalent  to  a  cylinder  of  revolution  whose  alti- 
tude is  h  and  radius  r. 


GEOME  TR  Y.  —  RE  VIE  W.  381 

Ex.  192.  How  must  the  dimensions  of  a  cylinder  of  revolution  be  in- 
creased in  order  to  form  a  similar  cylinder  whose  total  surface  shall  be  n 
times  that  of  the  original  cylinder  ? 

Ex.  193.  How  must  the  dimensions  of  a  cylinder  of  revolution  be  increased 
in  order  to  form  a  similar  cylinder  whose  volume  shall  be  n  times  that  of  the 
original  cylinder  ? 

Ex.  194.  What  is  the  radius  of  the  base  of  a  circular  cone  whose  altitude 
is  A,  the  longest  and  the  shortest  elements  being  /  and  V  respectively  ? 

Ex.  195.  What  is  the  slant  height  of  a  frustum  of  a  cone  of  revolution 
whose  lateral  surface  is  8  and  whose  lower  and  upper  bases  are  B  and  h 
respectively  ? 

Ex.  196.  A  cone  of  revolution  whose  radius  is  r  and  altitude  h  is  divided 
into  two  equivalent  parts  by  a  plane  parallel  to  the  base.  What  is  the  total 
area  of  the  frustum  thus  formed  ? 

Ex.  197.  The  volume  of  a  cylinder  of  revolution  is  equal  to  the  area  of 
its  generating  rectangle  multiplied  by  the  circumference  generated  by  the 
point  of  intersection  of  the  diagonals  of  the  rectangle. 

Ex.  198.  On  each  base  of  the  frustum  of  a  cone  of  revolution  there  is  a 
cone  whose  vertex  is  in  the  center  of*  the  other  base.  If  the  radii  of  the  lower 
and  upper  bases  are  r  and  r'  respectively,  what  is  the  radius  of  the  circle  of 
intersection  of  the  two  cones  ? 

Ex.  199.  A  stone  bridge  20  ft.  wide  has  a  circular  arch  of  140  ft.  span  at 
the  water  level.  The  crown  of  the  arch  is  140  (1  —  ^  VS)  ft.  abave  the  sur- 
face of  the  water.  How  many  square  feet  of  surface  must  be  gone  over  in 
cleaning  so  much  of  the  under  side  of  the  arch  as  is  above  water  ?  {Harvard.) 

Ex.  200.  What  part  of  the  whole  surface  of  a  sphere  is  a  spherical  trian- 
gle whose  angles  are  57°  57',  75°  27',  and  100°  36'  ? 

Ex.  201.  What  is  the  volume  of  a  right  cone  whose  altitude  is  15  ft., 
inscribed  in  a  sphere  whose  radius  is  10  ft.  ? 

Ex.  202.  How  far  from  the  base  of  a  hemisphere  must  a  plane  be  passed 
to  divide  the  surface  into  two  equivalent  zones  ? 

Ex.  203.  The  volume  of  a  spherical  segment  of  one  base  is  V  and  its 
altitude  is  h.     What  is  the  radius  of  the  sphere  ? 

Ex.  204.  Find  an  expression  for  the  volume  of  a  cube  inscribed  in  a 
sphere  whose  radius  is  r. 

Ex.  205.  Two  equal  circles  intersect  in  a  diameter.  If  a  plane  is  passed 
perpendicular  to  that  diameter,  prove  that  the  fcfur  points  in  which  it  inter- 
sects the  circumferences  lie  in  the  circumference  of  a  circle. 

Ex.  206.  The  square  on  the  diameter  of  a  sphere  and  the  square  on  the 
edge  of  an  inscribed  cube  are  to  each  other  as  3  is  to  1. 

Ex.  207.  Find  an  expression  for  the  altitude  of  a  zone  of  a  sphere  whose 
radius  is  r,  the  area  of  the  zone  being  equal  to  that  of  a  great  circle  of  the 
sphere. 


382  GEOMETRY.  — REVIEW. 

Ex.  208.  Find  an  expression  for  tlie  altitude  of  a  zone  whose  area  is  A 
on  a  spliere  whose  volume  is  V. 

Ex.  209.  Assuming  the  atmosphere  to  extend  to  a  height  of  50  miles 
above  the  earth's  surface  and  the  earth  to  be  a  sphere  whose  radius  is  approxi- 
mately 4000  miles,  what  is  the  volume  of  the  atmosphere  ? 

Ex.  210.  Assuming  the  earth  to  be  a  sphere  whose  radius  is  approxi- 
mately 4000  miles,  how  far  at  sea  is  a  lighthouse  visible,  if  it  is  80  ft.  high  ? 

Ex.  211.  A  swimmer,  whose  eye  is  at  the  surface  of  the  water,  can  just 
see  the  top  of  a  buoy  a  mile  distant.  If  the  buoy  is  8  in.  out  of  the  water, 
what  is  the  radius  of  the  earth  ? 

Ex.  212.    How  high  above  the  surface  of  the  earth  must  a  man  be  in 

order  that  he  may  see  -  of  it  ? 
n 

Ex.  213.  What  is  the  area  of  the  zone  illuminated  by  a  taper  h  deci- 
meters from  the  surface  of  a  sphere  whose  radius  is  r  decimeters  ? 

Ex.  214.  In  a  cube  whose  edge  is  1  ft.  there  are  inscribed  a  cylinder, 
a  cone,  a  sphere,  and  a  square  pyramid.  What  is  the  volume  of  each  of 
these  solids  ? 

Ex.  215.  A  cylindrical  boiler  with  hemispherical  ends  has  a  total  length 
of  12  ft.  and  its  circumference  is  10  ft.  What  is  its  surface  ?  What  weight 
of  water  is  required  to  fill  it  ? 

Ex.  216.  Find  the  diameter  of  a  sphere  which  is  circumscribed  about 
a  regular  square  pyramid  whose  base  is  4  in.  square  and  altitude  8  in. 

Ex.  217.  A  sphere  8  in.  in  diameter  has  a  3-inch  hole  bored  through 
its  center.     What  is  the  remaining  volume  ? 

Ex.  218.  What  is  the  volume  of  the  portion  of  a  sphere  lying  outside 
of  an  inscribed  cylinder  of  revolution  whose  altitude  is  h  and  radius  r  ? 

Ex.  219.    Inscribe  a  circle  in  a  given  spherical  triangle. 

Ex.  220.  Find  the  locus  of  the  centers  of  the  sections  of  a  given  sphere 
made  by  planes  passing  through  a  given  straight  line. 

Ex.  221.  Find  the  locus  of  the  centers  of  the  sections  of  a  given  sphere 
made  by  planes  passing  through  a  given  point  without  the  sphere. 

Ex.  222.  Having  given  the  radius,  construct  a  spherical  surface  to  pass 
through  three  given  points. 

Ex.  223.  Having  given  the  radius,  construct  a  spherical  surface  to  pass 
through  two  given  points  and  be  tangent  to  a  given  plane  or  to  a  given  sphere. 

Ex.  224.  Having  given  the  radius,  construct  a  spherical  surface  to  pass 
through  a  given  point  and  be  tangent  to  two  given  planes. 

Ex.  225.  Having  given  the  radius,  construct  a  spherical  surface  to  be  tan- 
gent to  three  given  planes. 


GEOMETRY.  —  TABLES. 


383 


METRIC  TABLES 
Measures  of  Length 


10  Millimeters  (m™) 
10  Centimeters 
10  Decimeters 
10  Meters 
10  Dekameters 
10  Hektometers 


=  1  Centimeter  (c™) 
=  1  Decimeter  (dm) 
=  1  Meter  (") 
=  1  Dekameter  (^m) 
=  1  Hektometer  (Hm) 
=  1  Kilometer  (J^) 


Measures  of  Surface 
100  Sq.  Millimeters  («i  ""'n)  =  1  Sq.  Centimeter  (^^  =»"> 


100  Sq.  Centimeters 
100  Sq.  Decimeters 
100  Sq.  Meters 
100  Sq.  Dekameters 
100  Sq.  Hektometers 


=  1  Sq.  Decimeter  (^i  dm) 
=  1  Sq.  Meter  («q™) 
=  1  Sq.  Dekameter  ("Q  Dm) 
=  1  Sq.  Hektometer  (sqHm) 
=  1  Sq.  Kilometer*8'iKm) 


A  square  hektometer  is  also  called  a  hektare  (h*>. 


Measures  of  Volume 
1000  Cu.  Millimeters  (c"  mm)  =  1  Cu.  Centimeter  («"  cm) 
1000  Cu.  Centimeters  =  1  Cu.  Decimeter  (cu  dm) 

1000  Cu.  Decimeters  =  1  Cu.  Meter  (^um) 


Measures  of  Capacity 


10  Milliliters  (ml) 

10  Centiliters 

10  Deciliters 

10  Liters 

10  Dekaliters 

10  Hektoliters 


=  1  Centiliter  (•!) 
=  1  Deciliter  (di) 
=  1  Liter  d) 
r=  1  Dekaliter  (DD 
=  1  Hektoliter  (Hi) 
=  1  Kiloliter  (KD 


The  liter  contains  a  volume  equal  to  a  cube  whose  edge  is  a  decimeter. 


384 


GEOMETRY.-^  TABLES, 


Measures  of  Weight 
10  Milligrams  ("s)  =  1  Centigram  (««) 


10  Centigrams 
10  Decigrams 
10  Grams 
10  Dekagrams 
10  Hektograms 


=  1  Decigram  (<i8> 
=  1  Gram  (s) 
=  1  Dekagram  (I>k) 
=  1  Hektogram  (««) 
=  1  Kilogram  (Kg) 


The  weight  of  a  gram  is  the  weight  of  a  cubic  centimeter  of  distilled  water 
t  its  greatest  density. 

Metric  Equivalents 


1  Meter 

=  39.37  in.  =  1.0936  yd. 

1  Yard 

=  .9144°^ 

1  Kilometer 

=  .62138  Mile 

1  Mile 

=  1.6093Km 

1  Hektare 

=  2.471  Acres 

1  Acre 

=  .4047Ha 

1  Liter 

_  f  .908  qt.  dry 

~  11.0567  qt.  liquid 

1  qt.  dry 
1  qt.  liq. 

=  1.1011 
=  .94631 

1  Gram 

=  15.432  Grains 

1  Grain 

=  .06488 

1  Kilogram 

=  2.2046  lb. 

1  Pound 

=  .4536Kg 

^  Approximate  Metric  Equivalents 

1  Decimeter  =  4  in.  ^  ^^.^^^  ^  f  t%  qt-  dry 

1  Meter  =  40  in.  1 1|  qt.  liquid 

1  Kilometer  =  f  Mile  1  Hektoliter  =  2|  bu. 

1  Hektare      =  2|  Acres  1  Kilogram  =  2^  lb. 

Notes.  —  1.  The  specific  gravity  of  a  substance  (solid  or  liquid)  is  the 
ratio  between  the  weight  of  any  volume  of  the  substance  and  the  weight  of  a 
like  volume  of  distilled  water  at  its  greatest  density ;  consequently,  since  a 
cubic  centimeter  of  distilled  water  at  its  greatest  density  weighs  one  gram, 
the  weight  of  any  substance  may  be  found  if  its  specific  gravity  and  volume 
are  known. 

2.  A  cubic  foot  of  water  weighs  62^  lb.,  or  1000  oz. 

3.  A  bushel  contains  2150.42  cu.  in. 

4.  A  gallon  contains  231  cu.  in. 


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