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Full text of "Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson"

IN MEMORIAM 
FLORIAN CAJO 





GEOMETRY: 

TEE ELEMENTS OF EUCLID AND LEGENDRE 

SIMPLIFIED AND ARRANGED 
TO EXCLUDE FROM GEOMETRICAL REASONING 

WITH THE 

ELEMEx\TS OF PLAXE AND SPHEPJCAL TRIGONOMETRY, 

ANT) 

EXERCISES Ii\ ELEMENTARY GEOMETRY AND TRIGONOMETRY. 

ADAPTED FOK SCHOOLS AND COLLEGES. 



• L-AWR^i^CE' S:' BENSON, 

n 

Author of "The Truth of the Bible Upheld,"— London, 18G4; "Geometrical Disquisi- 
tions," — London, 1864 ; " Scientific Disquisitions Concerning the 
Circle and Ellipse," 1862. 
Member of the New York Association for the Advancement of Science and Art ; Hon. 
Mem. Phi Eappa Society, University of Georgia ; Brothers' Society, 
Yale College, etc., etc. 



[all rights reserved.] 



NEW YORK: 

PUBLISHED FOR THE AUTHOR BY DAVTES & KENT, 

No. 183 WILLLAM STREET. 
186V. 



15f 



Rev. Thomas A. Boone, Professor in Carolina Female College, An- 
sonville, North Carolina, writes : 

" Your new work on the Elements op Geometry (Book First) haa 
been submitted to the President of Carolina Female College. He has 
examined it critically, and indorses it as an evident advancement of the 
science, in that it simplifies and meets the capacity of learners, retains 
all the essentials of the science, and is equally as competent for mental 
discipline as the old Eeductio Ad Abmrdum." 



STBREOTTPEli AND ElECTROTTPEIB,- 

.-, 183 Wjjaiaar ^treeln Nw Y. . • 



CAJORI 



Entered, according to Act of Congress, in the year 1867, by 

LAWRENCE S. BENSON, 

In the Cleric's Office of the District Court of the United States for the Southern 

District of New York. 



TO 

PROFESSOR GERARDUS BEEKMAN DOCHARTY, LL.D., 

COLLEGE OP THE CITY OF NEW YORK. 



Sm — In permitting me to inscribe to you this Treatise of Elementary 
Geometry, you do me great lionor. Your experience and success as a 
Teaclier and an Author will readily enable you to give a full scrutiny to 
the design and compass of this volume. Much originality can not be 
expected in a subject which has been, for more than two thousand yeare, 
enriched by a great number of eminent men ; but in these days of practi- 
cability, a modification of this science may be attempted, as you have 
yourself thought proper to do, with a view of utilizing the important 
principles of Geometry, and presenting them in such a manner that 
though " no royal road to Geometry" can be found, the path to a know- 
ledge of it may be rendered so clear that the impediments wiU be in tihe 
learner himself. And to remove much diflBculty in acquiring an easy 
acquaintance with its numerous theorems and problems, I have thought 
proper to exclude the inelegant Bedrictio ad absurdum from the methods 
of geometrical reasoning which you have expressed — " a consummation 
most devoutly to he wislied" and which accomplishment, resulting from my 
labors, I now present for the benefit and use of those whose education is 
in the future. 

I hav3 the honor to be, 

Very respectfully, yours, 

Lawrence S. Benson. 
Vvw York, AprU mh, 1867. 



TESTIMONIALS. 



The College op the City of New York, » 
Cor. Lexington Avenue and 23d Street. 1 

New York, January Zd, 1867. 

I have had several interviews with Mr. Lawrence S. Benson on 
scientific subjects, and from his conversation, together with tlie Essays 
which he has published, I esteem him an excellent scholar and fine 
mathematician. He has a desire to establish the Elements of Euclid in 
all canes, independently of the demonstration known as the Beductio ad 
dbsurdum, " a consummation devoutly to be wished." 

Whatever aid or advice you can render him in the furtherance of this 
object will tend to the advancement of true science. 

Yours truly, 

G. B. DOCHARTY. 



Rooms op the New York Association for the Advancement ) 
of Science and Art, February )i8th, 18G7. I 

Extract from the transactions of the Association for the Advancement 
of Science and the Arts : 

" At a meeting of the New York Association held February 25, 1867, 
a paper on a new method of demonstrating the propositions of Geometry, 
denominated the Direct Method, in place of the one now in use,;ijid called 
the Indirect Method, was read by Lawrence S. Benson, Esq., which 
method the writer proposes to introduce into Schools and Academies. 

" After the reading of the paper, and the discussion of its merits, the 
subject was referred to Professor Fox, Principal of the Department of 
Free Schools of Cooper Union, and to Professor Cleveland Abbe, for ex- 
amination and report. It was also moved and carried that the Report 
when received be referred to the Section on Physical Science for final 
disposition. 

" The Section, after reading the Report of Professor Fox, the letter of 
Professor Abbe, and the opinion of Professor Docharty, who had been 
invited to examine the work, feel justified in commending this work as 
worthy of patronage. Professor Fox in his Report says : ' The design 
of arranging the Definitions, Axioms, and Propositions of Geometry, so 
as to use only the Direct Method of demonstration, is a good one, and 
when arranged in the form of a neat elementary text-book, will doubtless 
do much good, as the Direct Method is much more easily understood thaa 
the Lidirect Method, by beginners.' 

" L. D. Gale, M.D., 
Oen. Sec. of the New York Association for the 
Advancement of Science and tlie Art»." 



PREFACE. 



By way of preface, I will state what I have done in this edition, and 
explain the reason why I have done so. I have used such propositions 
only which are required to substantiate the principal theorems and 
problems by which the principles of Geometry have practical applica- 
tions in Trigonometry, Surveying, Mechanics, Engineering, Navigation, 
and Astronomy. I have generalized the various propositions found in 
the school editions of Geometry, and where particular cases arise under 
such general propositions, I have given the demonstrations for them. I 
have arranged the propositions to give the Direct Method of demonstra- 
tion in place of the Beductio ad ahsurdum or Indirect Method. 

My reasons for the foregoing changes are obvious to the experienced 
mind ; considering the extent and variety of modem education, the time 
devoted for the pupils to acquire rudimental knowledge becomes en- 
croached upon in order to make them acquainted with its numerous 
modifications ; and for the pupils to obtain such knowledge of the rudi- 
ments as will enable them to see the practical applications throughout 
all their extent and variety, which are very great in these days of ad- 
vancement and civilization, the rudiments which were taught centuries 
ago must be so abbreviated as to contain the essentials only. When materials 
for instructing the mind were scant, there were no opportunities to make 
close selection of them ; but now, when those materials are plentiful, a 
judicious selection of tlie best becomes imperatively necessary. And 
when geometrical principles have become extended by the Algebraic 
Analysis, and have been made practicable by Trigonometry, Surveying, 
Mechanics, Engineering, Navigation, and Astronomy, the mere mental 
exercises, which were regarded so beneiicent by the ancients, are unsuited 
for this practical age, which is continually bent on progress, while the 
intellect is sufficiently exercised by utilizing modern acquisitions; for 
this reason, I have reduced the number of propositions substantiating the 
principles of Geometiy, and I have classified them in such a manner that 
particular cases are enunciated by general propositions, a change which 
is likely to impress on the pupils the accuracy of geometrical principles, 
as they will be shown that geometrical principles are the same in all 
cases and under every circumstance. 

Many of the best geometers have objected to the Beductio ad dbsurdum 
in Geometry, while all geometers prefer the Direct Method of demonstra- 



VI PREFACE. 

tion. Any true proposition is susceptible of being directly demonstrated. 
And without entering into the merits or demerits of the Reductio ad ab' 
surdum, I have exchided it from geometrical reasoning, and have used 
the Direct Method only, a change which agrees with tlie spirit of the age, 
and fulfills the requirements of progress. I have omitted the various 
diagrams usually put among the definitions of Geometry, because when 
a magnitude is properly defined, the learner has a better conception of 
it from the definition than any diagram can give him ; and the omission 
of the diagrams will assist the mental exercise and cultivate the under- 
standing of the learner, which is the great object of geometrical study; 
and if the learner be made to draw the diagrams from the definitions, he 
will be better instructed than if they be given by the author. The 
time is not far distant when geometrical science may be attempted with- 
out using diagrams in the demonstrations. The diagrams are auxiliaries 
to the mind in the ascertainment of truth ; they are not necessary to the 
existence of truth, and " Geometry considers all bodies in a state of ab- 
straction, very different from that in which they actually exist, and the 
truths it discovers and demonstrates are pure abstractions, hypothetical 
truths." Hence, diagrams are like the pebbles used by Indians in count- 
ing, or other means of computing before the principle of numeration was 
discovered ; that when the intellect of man becomes more highly ex- 
panded and cultivated, diagrams will be regarded necessary to the first 
conceptions of geometrical knowledge, but altogether unsuited to a high 
development of geometrical science. 

I am greatly indebted to Hon. S. S. Randall, City Superintendent of 
the Board of Education of New York, for many valuable suggestions in 
the demonstrations and present arrangement of this work ; and also under 
many obligations to Professor Docharty, of the College of the City of New 
York ; and to L. D. Gale, M.D., General Secretary of the New York Asso- 
ciation for the Advancement of Science and Art, Cooper Union. 

Lawrence S. Benson, 

61 MoETON Street, City or New Yobk, 
April mh, 1867. 



THE ELEMENTS OF EUCLID AND LEGENDRE. 



BOOK FIRST. 
ON THE STRAIGHT LINE AND TRIANGLE. 

DEFINITIOXS. 

1. A definition is the precise term by which one thing is dis- 
tinguished from all other things. 

2. Mathematics is that science which treats of those abstract 
quantities known as numbers, symbols, and magnitudes. 

3. Geometry is that branch of Mathematics where the ex- 
tensions of magnitudes are considered without regard to the 
actual existence of those magnitudes. 

4. A m^agnitude has one or more of three dimensions, viz., 
length, breadth, and thickness. 

5. Geometers define a point, position without magnitude; 
but to give a point position, would entitle it to the three dimen- 
sions of magnitude, whereas a point in Geometry expresses no 
dimension. 

6. A diagram, represents the abstractions of magnitudes, 
whereby their dimensions are determined, and geometrical 
reasoning conducted without regard to the actual properties of 
those magnitudes. 

'7. A line expresses length only, and is capable of two con- 
ditions — it can be straight or curved ; when its length is always 
in one direction, it is straight ; but when there is a continual 
variation in the direction of its length, it is curved, or in brevity 
called a curve. 

Scholiiim. A straight line can not be defined as having all 
its points in the same direction, because the points of a line are 
its extremities, and the extremities of a curved line can be 






8 THE ELEMENTS OP [boOK I. 

placed on a straight line, and in this case the definition would 
not distinguish a straight line from a curve. And if a line be 
regarded composed of points, this would infer that a point has 
dimension ; but the intersection of lines is a point, which, how- 
ever, does not give position to the point, because a line is au 
abstraction, and position implies actual existence. 

8. A surface expresses an inclosure by not less than three 
straight lines, or by one curved line, or by one straight line and 
one curved line ; consequently a surface has breadth and length, 
and the extremities of surfaces are lines, and the intersection of 
one surface with another is a line. 

Scho. A plane surface, or sometimes called a plane^ is one in 
which any line can be drawn wholly in the surface; and a 
c*rved surface is one in which a curve only can be drawn 
wholly in the surface in the direction of the curvature. 

9. A volume or solid expresses an inclosure made by sur- 
faces, and has breadth, length, and thickness ; the extremities 
of a volume are surfaces, and the intersection of one volume 
with another is a surface. 

10. An angle is formed by two straight lines meeting each 
other ; the point of intersection of the lines is called the vertex 
of the ano-le. When one straicjht line meets another straight 
line, so as to make two adjacent angles, these angles are right 
angles when they are equal ; and when one angle is greater than 
the other angle, the greater angle is an obtuse angle, and the 
less angle is an acute angle. The straight line which makes 
the two adjacent angles equal is the perpendicular to the other 
6trai<;ht line. 

11. When two straight lines on the same plane never meet 
each other on whichever side they be produced, they are called 
parallel lines. 

12. Rectilinear surfaces are contained by straight lines, and 
are called polygons ; when a polygon has three sides, it is a 
triangle ; when it has four sides, it is a quadrilateral ; when 
it has five sides, it is z, pentagon ; when it has six sides, it is a 
hexagon / when it has seven sides, it is a heptagon / when it 
has eight sides, it is an octagon ; when it has nine sides, it is an 
enneagon / when it has ten sides, it is a decagon ; and so on, 
being distinguished by particular names derived from the Greek 



BOOK I.] EUCLID AND LEGENDRE. 9 

language, denoting the number of angles formed by the sides. 
The straight line drawn through two remote angles of a poly- 
gon of four or more sides, is a diagonal. 

13. When the triangle has its three sides equal, it is equi' 
lateral; when two of its sides are equal, it is isosceles; and 
when its sides are unequal, it is scalene. When its angles are 
equal, it is eqidangular y when one of its angles is a right angle, 
it is right-angled ; when one of its angles is an obtuse angle, it 
is obtuse-angled ; and when all its angles are acute, it is acute- 
angled. 

14. When a quadrilateral has its opposite sides parallel, it is 
2l parallelogram ; when it has two sides only parallel, or none 
of its sides parallel, it is a trapezium. 

15. When a parallelogram has a right angle, it is a rectangle ; 
when it has two adjacent sides equal, but no right angle, it is a 
rhombus. When the rectangle has its sides equal, it is a 
square / and when its opposite sides only are equal, it is an ob- 
long. When the parallelogram has its opposite sides only 
equal, and no right angle, it is a rhomboid. 

16. A plane surface contained by one line is a circle when 
every part of the line is equally distant from a point in the sur- 
face ; the point is the center of the circle, and the line is the 
circumference. 

17. The straight line drawn from the center to the circum- 
ference is the radius ; the straight line drawn from one part of 
the circumference through the center to another part of the 
circumference is the diameter, which divides the circle and cir- 
cumference each into two equal parts. When the straight line 
does not pass through the center, it is a chord. 

18. That portion of the circle contained by the semicircnm- 
ference and diameter is a semicircle; and that portion con- 
tained by the chord and a part of the circumference is a seg- 
ment ; a part of the circumference is an arc. 

19. If the vertex of an angle be the center of a circle, that 
part of the circumference intercepted by the sides of the angle 
will give the value of the angle ; hence, the angle is measured 
by an arc when its vertex is the center of the circle, liut 
when the vertex is in the circumference, the angle is subtended 
by the arc intercepted by its sides ; hence, equal angles will be 



10 THE ELEMENTS OF [BOOK I. 

measured by equal arcs, and subtended by equal arcs ; therefore 
equal arcs measure or subtend equal angles. 

20. Two arcs are supplementary when both together are 
equivalent to the semicircumferenoe. And two angles are sup- 
plementary when both together are equivalent to two right 
angles, and complementary when equivalent to one right 
angle. 

21. Things are equal when they have equal magnitudes and 
when they coincide in all respects; and are equivalent when 
they have equal magnitudes, but do not coincide in all respects. 

22. The term, each to each, or sometimes respectively, is a 
limiting expression, and is used to denote the equality of lines 
or magnitudes taken in the same order; for without this quali- 
fication, two lines or magnitudes said to be equal to two other 
lines or magnitudes, would imply that their sums are equal, 
when it would be desirous of meaning that they are equal in 
the same order in which they are expressed — a difference very 
important in the demonstration of a proposition. 

23. A proposition is demonstrated by superpositio7i when 
one figure is supposed applied to another, which is done in the 
first case of the third proposition of this book. 

24. One proposition is the converse of another when, in the 
language of logic, the subject of the latter is the predicate of 
the former, and the predicate of the latter is the subject of the 
former. 

METHOD OF REASONING. 

1. From the foregoing definitions, it is shown that the 
straight line and curve have certain relations, uses, and prop- 
erties which are important to be known. And in order that 
these relations, uses, and properties may be satisfactorily inter- 
preted, there are certain terms, expressive of certain facts or 
states of knowledge, by means of which the mind intuitively 
perceives a connection between the things known and those for 
elucidation, such as axioms^ hypotheses^ and postulates/ as 
demonstrations, theorems, problems, and lemmas,' as corollaries 
and scholiums. With the assistance of these, the mind is 
carried step by step in all its investigation of extension, and is 
able to discover by such investigation the properties, uses, and 
relations of geometrical magnitudes. They are the data by 



BOOK I.] EUCLID AND LEGENDRE. H 

which the hidden truths are revealed. Upon them a system of 
logic or argumentation is conducted, and by the conformity of 
the arguments and conclusions with the accepted truths, we 
have the science of Geometry. 

2. Proposition in Geometry is a general term, expressing the 
subjects to be considered, and is either a problem or theorem. 
When it is the first, there is something required to be per- 
formed, such as drawing a line or constructing a figure ; and 
whatever points, lines, angles, or other magnitudes are given to 
efiect the purpose, they are the data of the problem ; and when 
it is the latter, a truth is proposed for demonstration, and 
whatever is assumed or admitted to be true, and from which 
the proof is to be derived, is the hypothesis. 

3. Demonstration consists in evident deductions from clear 
premises, whereby the conclusion corroborates the premises and 
shows the argumentativeness of the deductions. In the course 
of demonstration, reference is often made to some previous 
proposition or definition. 

4. Sometimes inferences arise involving another principle, 
but do not require any long process of reasoning to establish 
their truth — these are corollaries. Any remark made from the 
demonstration of a proposition is a scholium. A proposition 
which is preparatory to one or more propositions, and is of no 
other use, is a lemma. 

5. And for the establishment of a proposition, there are four 
things required, viz. : the general enunciation, the particular 
enunciation, the construction, and the demonstration. 

6. The hypotheses of demonstration are known as axiom, and 
postulate; the former is assumed to prove the truth of a theorem, 
and the latter is granted to pei'form the requisites of a 
problem. 

7. An axiom is so evidently clear, that no process of reason- 
ing can make it more clear ; its truth is so easily recognized by 
the human mind, that so soon as the terms by which it is ex- 
pressed are understood, it is admitted; for instance, it is as- 
sumed as 

AXIOMS. 

1. Things which are equal to the same, or to equals, are equal 
to one another. 



12 THE ELEMENTS OF [bOOK I. 

2. If equals or the same be added to equals, the wholes are 
equal, 

3. If equals or the same be taken from equals, the remainders 
are equal. 

4. If equals or the same be added to unequals, the wholes 
are unequal. 

5. If equals or the same be taken from unequals, the re- 
mainders are unequal. 

6. Things which are doubles of the same, or of equals, are 
equal to one another. 

I. Things which are halves of the same, or of equals, are 
equal to one another. 

8. Magnitudes which exactly coincide with one another are 
equal. 

9. The whole is greater than its part. 

10. The whole is equal to all its parts taken together. 

II. All right angles are equal to one another. 

12. If a straight line meet two other straight lines which are 
in the same plane, so as to make the two interior angles on the 
same side of it, taken together, less than two right angles, these 
straight lines shall at length meet upon that side, if they be 
continually produced. 

These are the self-evident truths used by Euclid for geomet- 
rical demonstration ; but if the first eleven be considered for 
awhile, it will be seen that they can be reduced to two general 
axioms, viz., things which are equal to the same are equal, and 
things which are not equal to the same are unequal ; because 
when we add, subtract, multiply, or divide equals, the equality 
in each case is not destroyed; hence in each case equal to 
one another. And when we add unequals to or subtract un- 
equals from equals, the sums or remainders are not equal to the 
same, hence unequal to one another. And magnitudes which 
exactly coincide with one another are equal to the same, hence 
equal ; a whole and a part are not equal to the same, hence are 
unequal ; Avhile a whole and all its parts are equal to the same, 
hence are equal. From the definition of right angles, it is 
seen that when a straight line meets another straight line, so as 
to make the two adjacent angles formed by them equal to one 



BOOK I.] EUCLID AND LEGENDRE. 13 

another, the two adjacent angles are right angles; then these 
two right angles are equal ; and since all right angles agree with 
the definition, they are equal to the same thing, hence equal to 
one another. But the twelfth axiom is not self-evident, be- 
cause the converse has been demonstrated, viz.: that two 
straisrht lines which meet one another make with any third line 
the interior angles less than two right angles. Geometers perceiv- 
ing this blemish in the Elements of Euclid, have endeavored in 
many ways to remove it, but without complete success. They 
employed three methods for this purpose : 1, By adopting a 
new definition of parallel lines. 2. By introducing a new 
axiom. 3. By reasoning from the definition of parallel lines, 
and the properties of lines already demonstrated.* The diffi- 
culty with parallel lines is, that geometers have confounded a 
definition with a proposition. Definition 11 is perfectly legiti- 
mate, as it simply defines what kind of lines are parallel ; but 
when it is inferred from it that these lines are equally distant 
from each other, this is no axiomic inference, because the curve 
and its asymptote are two lines which never meet, however far 
they be produced on the same plane, but they are not equally 
distant from each other ; hence the inference that parallel lines 
are equally distant, embodies a question which requires a dem- 
onstration to establish ; and to establish this question has given 
perplexity to geometers, for though they have proven the lines 
equally distant at particular points, they have not proven them 
so at every point; and here consists the incompleteness of their 
demonstrations, and here is required some general demonstra- 
tion which will embrace every part of the lines, however so far 
they be produced on the same plane.f 

8. A postulate is a problem so easy to perform that it does 
not require any explanation of the manner of doing it, so that 
the geometer reasonably expects the method to be known ; for 
instance, it is granted as — 

* See notes to Playfair's Euclid, Legendre's Geometry, Leslie's Geom- 
etry, the ea-cursus to the tirst book of Camerer's Euclid, Berlin, 1825 ; 
Col. P. Thomson's Oeometry xoithout Axioim, Professors Thomson's and 
Simson's editions of Euclid — London, Glasgow, and Belfast. 

f See fifteenth and nineteenth propositions cf this book. 



14 THE ELEMENTS OF [bOOK I. 

POSTULATES, 

1. That a straight line can be drawn from any one point to 
any other point. 

2. That a terminated straight line may be extended to any 
length in a straight line. 

3. That a circle may be described from any center, at any 
distance from that center. 

EXPLAIS'ATION OF SIGSTS. 

In Algebra, the sign +, called Plus {more Jy), placed between 
the names of two magnitudes, is used to denote that these mag- 
nitudes are added together ; and the sign — , called Minus 
{less hy)^ placed between them, to signify that the latter is taken 
from the former. The sign z=, which is read equal to, signifies 
that the quantities between which it stands are equal to one 
another. The sign =o=, signifies that the quantities between 
which it stands are equivalent to one another. 

In the references, the Roman numerals denote the book, and 
the others, when no word is annexed to them, indicate the 
proposition ; otherwise the latter denote a definition, postulate, 
or axiom, as specified. Thus, III. 16 means the sixteenth 
proposition of the third book ; and I. ax. 2, the second axiom 
of the first book. So also hyp. denotes hypothesis, and const, 
construction. 



PROPOSITIONS. 



Prop. I. — Problem. — To describe an isosceles triangle on a 
finite straight line given in position. 

Let AX be the given straight line; it is required to describe 
an isosceles triangle having its base on AX. 

From a point C, without the line AX as 
a center, and a radius CA (I. post. 3), de- 
scribe a circle ABED, cutting the line AX 
in two points A and B; draw from these 
points the straight lines AE and BD (I. 
post. 1) passing through the center of the 
circle ; the triangle ACB is the one required. 
Because C is the center of the circle ABED (I. def 16), CA 




BOOK I.J EUCLID AND LEGENDRE. 15 

is equal to CB, therefore the triangle ACB has two sides equal ; 
hence (I. def. 13) it is isosceles, and is described on AX, which 
was required to be done. 

Corollary 1. But the angle EAB is subtended by the arc 
EB (I. def. 19), and the angle DBA is subtended by the arc 
DA ; since AE and DB pass through the center of the circle 
(const,), they are both diameters of the circle (I. def 17) ; hence 
the arcs DEB and ADE are each a semicii-cumference, and (I. 
ax. 1) are equal ; therefore the sum of tlie arcs BE and ED 
is equivalent to the sum of the arcs ED and DA ; the arc ED 
is common ; hence (I. ax. 3) we have the arc EB equal to the 
arc DA ; therefore the angles EAB and DBA are subtended 
by equal arcs, consequently (I. def 19) the angles are equal. 
Hence, in an isosceles triangle, the angles opposite the equal 
sides are equal. 

Cor. 2. The line AE, which forms with AB the angle EAB, 
intercepts the line DB at C, which forms with AB the angle 
DBA, and the line DB intercepts AE at C also. C being the 
center of the circle ABED, CB, that portion of BD intercepted 
hy AE, is equal to CA, that portion of AE intercepted by BD 
(I. def 16) ; but CB and CA are the sides of the triangle ACB 
(I. 1) ; hence, when two angles of a triangle are equal, the 
opposite sides to them are also equal, and the triangle is isos- 
celes (I. def 13). 

Pkop. II. — Problem. — To describe an equilateral triangle 
on a finite straight line given in magnitude. 

Let AB be the given straight line ; it is required to describe 
an equilateral triangle having AB for its base. 

From A as a center, and a radius 
AB (I. post. 3), describe the circle 
FCD ; and from B as a center, and 
a radius BA, describe the circle 
HCE. The circles having equal 
radii (I. ax. 1) are equal ; draw from 
C through the center B, CH ; and 
from C through the center A, CF (I. post 1) ; the triangle ACB 
is the one required. 

Because the circles have equal radii (const.), AC is equal to 




16 THE ELEMENTS OF [bOOK I. 

AB, and CB is equal to AB ; hence (I. ax. 1) the three sides 
of the triangle ACB are equal; the triangle (I. def. 13) is equi- 
lateral, and is described on AB, which was required to be done. 
Corollary 1. If AB be produced both ways (I. post. 2) to D 
and E, the angle DBC is subtended by the arc CD, and the 
angle FCB is subtended by the arc FB; the arcs OB and BF 
are together equivalent to the arcs BC and CD (I. ax. 1) ; 
hence (I. 1, cor. 1) the arc BF is equal to the arc CD, therefore 
(I. def 19) the angle FCB is equal to the angle DCB. Again: 
the angle ACH is subtended by the arc AH, and the angle 
CAE is subtended by the arc CE. But (I. ax. 1) the arcs AC 
and CE are equivalent to the arcs CA and AH, and (I. ax. 3) 
the arcs CE and AH are equal; therefore (I. def 19) the angles 
ACH and CAE are equal, but the angle ACH is the same as 
the angle FCB ; hence (I. ax. 1) the three angles of the triangle 
are equal ; therefore in an equilateral triangle the angles are 
equal. And in a manner similar to Cor. 2 of the first proi)osi- 
tion,it can be shown, conversely^ that when a triangle has three 
equal angles, the sides opposite them are also equal ; hence an 
equilateral triangle is also equiangular, and, conversely^ an equi- 
angular triangle is also equilateral. 

Prop. IH. — Theorem. — If two triangles have txoo sides of 
the one equal to two sides of the other, each to each, and have 
also an angle in one equal to an angle in the other simiiiarly 
situated with respect to those sides, the triangles have their 
bases or remaining sides equal / their other angles equal, each 
to each, viz., those to which the equal sides a.e opposite, and 
the triangles are equal. 

This general proposition has four cases, viz. : first, when the 
equal angles are contained by the respectively equal sides; 
Becond, when the equal angles are opposite to one pair of the 
respectively equal sides ; third, when the equal angles are op- 
posite to the other pair of the respectively equal sides ; and 
fourth, the limitation that when the least sides respectively of 
the triangles be equal, and the angles opposite the least sides 
"be equal, the angles opposite the greater of the respectively 
equal sides must be of the same kind, either both acute, or not 
acute. 



BOOK I.] 



EUCLID AND LEGENDRE. 



17 




First case. Let ABC and DEF be the two triangles having 
any two sides equal, each to each, 
viz., AC and CB equal to EF and 
DF, and the contained angles ACB 
and EFD equal ; the remaining sides 
AB and DE are equal, tlie angle CBA 
opposite AC equal to the angle FDE 
opposite FE, the angle CAB opposite 
CB equal to the angle FED opposite 
DF, and tlie triangles ABC and DEF 
are equal. 

If the triangle ABC be placed on the triangle DEF so that 
the vertex of the angle ACB will fall on the vertex of the 
angle DFE, the angle ACB being equal to the angle DFE 
(hyp.), the side CB will fall on FD, and the side CA will fall on 
FE ; CB and FD being equal (hyp.), the extremity B will fall 
on the extremity D. CA and FE being equal (hyp.), the ex- 
tremity A will fall on the extremity E ; and since AB is a 
straight line, it will coincide with DE (I. def. V), a straight line 
drawn from D to E. Therefore the triangle ABC has its three 
sides coinciding with the three sides of the triangle DEF ;; 
hence the angle CAB will fall on the angle FED, and be equal 
to it ; the angle CBA will fall on the angle FDE, and be equal 
to it ; consequently the two triangles have their three sides and. 
three angles equal, each to each, and (I. ax. 8) are equal. 

Second case. When the triangles ABC and DEF have the 
sides CA and CB respectively equal to FE and FD, and the 
angles ABC and EDF equal, respectively opposite to CA and 
FE, the remaining sides are equal ; the angle CAB opposite CB 
is equal to the angle FED opposite to FD, the angle ACB op- 
posite to AB is equal to the angle EFD opposite to DE, and 
the triangles are equal. 

Let the side DE b 

be put on AB so 

that D will fall 

on B, and the 

eqnal angles ABC 

and EDF will be 

on different sides of AB ; join CF (I. post. 1 ). BC and BF (hyp.) are 
2 




18 THE ELEMENTS OF [bOOK I. 

equal, the triangle CBF (I. def. 13) is isosceles, and (I. 1, cor. l) 
the angle BCF eqiial to the angle BFC; and hecause CA and, 
AF are equal (hyp.), the triangle CAF (I. def. 13) is isosceles, 
and (1, 1, cor. 1) the angle ACF is equal to the angle AFC ; 
then (I. ax. 2) the angles BCF and ACF are equal to the angles 
BFC and AFC, or the angle BCA equal to the angle BFA 
(I. ax. 1 and ax. 10). Hence we have in the triangles ABC 
and ABF, two sides, and the contained angle in each equal, 
.-«ach to each, therefore hj first case the triangles can be shown 
.-.equal in all respects. 

Third case. It can be proven in a similar manner as the second 

* case. When the angles CAB and FED of the second case are 

, obtuse, and the angles CBA and FDE of the third case are 

. obtuse, the proofs are given by the third axiom of the first book. 

Fourth case. When the triangles ABC and DEF have 

their least sides in each equal — viz., AB to DE — and another 

side in each, equal, the angles ACB and EFD being equal, the 

angles opposite the second pair of equal sides must be both 

acute or both not acute ; otherwise, two triangles can be formed 

having two sides .and an augl'-^ in each equal, each to each, and 

? the triangles unequal. For, in the 

triano-lcs ABC and ACD, the side 
AC is common, the angle BAC 
equal to angle CAD, the sides BC 
and CD can be equal, and the tri- 
angles (I. ax. 9) unequal; hence in two triangles when the 
greatest and least sides are respectively equal, and the equal 
angles opposite to the least sides be given, the angles opposite 
the greatest sides must both be not acute to determine the tri- 
angles ; but when in two triangles the two less sides of each 
are respectively equal, and the equal angles opposite tlie least 
sides be given, the angles opposite the other equal sides must 
both be acute, to determine the triangles. 

The equality of the triangles can be proven by {ho second 
and third cases, using the second axiom when the angles are 
acute, and the third axiom when the angles are obtuse ; but 
when the angles are right-angled, the equality of the triangles 
is shown from tlie first corollary to the first proposition without 
those axioms. 




!BOOK I.] 



EUCLID AND LEGENDKE. 



19 




B 



Peop. IV. — Theok. — If the three, sides of one triangle be 
■equal to the three sides of another, each to each: (1) the ajigles 
■of one triayigle are equal to the angles of the other, each to each^ 
viz., those to which tJie equal sides are 02)posite, and (2) the 
triangles are equal. 

Let ABC and DEF be the two tri- c p 

angles having then- three sides equal, 
viz., AB to DE, CA to FE, and CB 
to FD, the angles are equal, viz., ACB 
to EFD, CAB to FED, and CBA 
to FDE ; and the triangles are equal. 

If the side DE be placed on the 
side AB so that the triangles Avill 
fall on different sides of AB, D will 
fall on B, and E on A, because DE is equal to AB, and the 
triangle DEF will 
take the position 
BFA, BF being 
the same as DF, 
and FA the same 
as FE. Join CF, a 

and because (hyp.) BC is equal to BF, the angles BCF and 
BFC are equal (I. 1, cor. 1). It would be shown in a similar 
manner that the angles FCA and CFA are equal, therefore (I. 
ax. 2) the angles BCA and BFA are equal — that is (I. ax. 1), the 
angles BCA and DFE are equal. But (hyp.) the sides CB and 
FD are equal, and the sides CA and FE, and it has been shown 
that the contained angles are equal, therefore (I. 3, first case) 
the other angles are equal — that is, CAB to FED and CBA to 
FDE, and the triangles are equal. Wherefore, if the three 
sides, etc. 

Prop. V. — Prob. — To bisect a given angle, that is, to divide 
it into tico equal angles. 

Let BAC be the given angle ; it is required to bisect it. 

From A as a center, and AD less than AB (I. post. 3), de- 
scribe the arc DE ; draw the chord DE, then upon DE, on the 
side remote from A, describe an equilateral triangle (I. 2), DFE, 
then join AF ; AF bisects the angle BAC. 




20 



THE ELEMENTS OF 



[book I. 




Because AD is equal to AE (L def. 16), 
and AF is comraoii to the two triangles 
DAF and EAF, the two sides DF and 
EF are equal (I. 2) ; therefore the two tri- 
angles DAF and EAF have their three 
sides equal, each to each, and the triangles 
are equal (I. 4) ; consequently the angle 
DAF opposite DF is equal to the angle 
EAF opposite EF, and the angle BAG 
is bisected by the line AF; which was to be 
done. 

OTHERWISE, 

Let BAG be the given angle ; in AB take any two points as 
B and D, and cut off AG and AE respectively equal to AB and 
AD, join BE and GD, and the straight line joining the inter- 
section of BE and CD with the vertex A bisects BAG. The 
proof is easy, and is omitted to exercise the ingenuity of the 
pupU. 

Prop. VI. — Prob. — To bisect a given finite straight line. 
Let AB be the given line ; it is required to bisect it. 
Describe (L 2) upon it an equilateral triangle ABG, and 
C bisect (I. 5) the angle AGB by the straight 

line GD ; AB is bisected in the point D. 

Because AG is equal to GB, and CD 
common to the two triangles AGD, BGD, 
the two sides AG, GD are equal to BG, 
B CD, each to each ; and the angle AGD is 
equal (const.) to the angle BGD; therefore the base AD is 
equal (L 3) to the base DB, and the line AB is bisected in the 
point D ; which was to be done. 

Sc/io. In practice, the construction is effected more easily by 
describing arcs on both sides of AB, from A as a center, and 
"with any radius greater than the half of AB ; and then, by de- 
gcribing arcs intersecting them, with an equal radius, from B as 
center, the line joining the two points of intersection will bi- 
sect AB. The proof is easy. 




Peop. VII. — Peob. — To draw a straight litie perpendieuiar 



BOOK I.] 



EUCLID AND LKGENDRE. 



21 



to a given straight line, from a giveyi point in that straight 
line. 

Let AB be the given straight line, and C a point given in it ; 
it is required to draw a perpendicular from the point C. 

From C as a center, and a radius 
CE (I. post 3), describe the semicircle 
EHF; then (I. def. 16) EC is equal to 
CF, and on EF (I. 2) describe the ^ 
equilateral triangle EDF ; then a line 
from C to the vez'tex D is the perpen- 
dicular required. 

Because EC is equal to CF (I. def. 
16), ED is equal to DF (I. 2), and the angle DEC equal to 
angle DFC (I. 2, cor. 1) ; hence (I. 3) the triangles ECD and 
FCD are equal ; but the angle ECD is equal to the angle FCD, 
and therefore (I, def 1 0) DC is perpendicular to AB from C. 





Pkop. Vin. — Prob. — To draw a straight line perpendicular 
to a given straight line of an unlimited length, from a given 
•point without it. 

Let AB be the given straight line, which may be produced 
any length both ways, and let C be a point without it. It is 
required to draw a straight line from C 
perpendicular to AB. 

Take any point D upon the other side 
of AB, and from the center C, at the dis- 
tance CD, describe (L post. 3) the circle 
ADB meeting AB in A and B ; bisect (I. 6) AB in G, and join 
CG ; the straight line CG is the perpendicular required. 

Join CA, CB. Then, because AG is equal to GB, and CG 
common to the triangles AGC, BGC, the two sides AG, GC 
are equal to the two, BG, GC, each to each ; and the base CA 
is equal (I. def 1 6) to the base CB ; therefore the angle CGA 
is equal (I. 4) to the angle CGB ; and they are adjacent angles ; 
therefore CG is perpendicular (I. def 10) to AB. Hence, from 
the given point C a perpendicular CG has been drawn to the 
•given line AB ; which was to be done. 

Scho. This proposition and the preceding contain the only 
.two distinct cases of drawing a perpendicular to a given straight 



22 THE ELEMENTS OF [BOOK B- 

line through a given point ; the first, when the point is in the 
line ; the second, when it is zcithout it. 

In practice, the construction will be made rather more simple 
by describing from A and B, when found, arcs on the remote 
side of AB from C, with any radius greater than the half of 
AB, and joining their point of intersection with C. 

Prop. IX. — T^eor. — "When one straight line meets another 
straight line ayid forms two unequal angles on the same side 
of that line, the two angles will be equivalent to two right 
angles. 

^ Let the straight line DC meet the 

straight line AB and form the two un- 
equal angles DCA and DCB on the same 
side of AB ; the two angles will be equiva- 
c lent to two right angles. 

At C, where the line DC meets AB, draw a perpendicular to 
AB from C (I. '7), then the angles ACE and ECB are two right 
angles (I. def. 10). But (I. ax. 10) the angle ECB is equivalent 
to the angles ECD and DCB both together ; likewise the angles 
ACE and ECB together are equivalent to the angles ACD and 
DCB both together; hence (I. ax. 1) the angles ACD and DCB 
together ai-e equivalent to two right angles. Wherefore, when 
one straight line meets, etc. 

Cor. Hence, if the straight line DC be pi-oduced on the other 
side of AB, the four angles made by DC produced and AB 
are together equivalent to four right angles. 

Hence, also, all the angles formed by any number of straight 
lines intersecting one another in a common point are together 
equivalent to four right angles. 

Prop. X. — Theor. — 7)^, at a point in a straight line, tico 
other straight lines on the ojyposite sides make the adjacent 
angles together equivalent to two right atigles, those two straight 
lines are in one and the same straight line. 

Let DC be the straight line which makes, at the point C, 
with AC and CB, two adjacent angles ACD and DCB togethei- 
equivalent to two right angles ; AC and CB are in one and the- 
same straight line. 



— B 



BOOK I.] EUCLID AND LEGENDEE. 23 

From C draw a perpendicular to AC (I. ^ 

7), then the angle ACE is a right angle 
(I. def. 10). But (hyp.) ACD and DCB 
are together equivalent to two right 
angles, so ACE and ECB are equivalent * 
to two right angles (I. ax, 1) ; hence ACE being a right angle 
(const, and I. def. 10), ECB must also be a right angle; then 
EC is perpendicular to CB (I. def. 10), and the angles ACE and 
ECB are equal (I. ax. 11) ; therefore (I. def 10) EC is a straight 
line which makes two equal angles with AB. But AC and CB 
make with EC the same equal angles ; hence AC and CB are 
the same straight line with AB. And DC makes with AC and 
CB (hyp.) two adjacent angles equivalent to two right angles, 
but DC makes with AB (I. 9) the same angles equivalent to 
two right angles ; hence AC and CB are the same straight line 
with AB (I. def 7). 



OTIIER^VISE, 



It is proven (I. 9) that the angles ACD and DCB are to- 
gether equivalent to two right angles ; then, as C is a point in 
AB, AC can be one line and CB another (I. def 7, scho.) ; 
hence AC and CB are in one and the same straight line, be- 
cause the two unequal angles ACD and DCB are equivalent to 
two right angles (I. 9). Wherefore if, at a point, etc. 

Prop. XI. — Theoe. — If two straight lines cut one another^ 
the vertical or opposite angles are equal. 

Let the two straight lines AB, CD cut one another in E ; the 
angle AEC is equal to the angle DEB, and CEB to AED. 

Because the straight line AE makes with CD the angles 
CEA, AED, these angles are together equivalent (I. 9) to two 
risrht angles. Again : because DE n 
makes with AB the angles AED, DEB 
these also are together equivalent to 
two right angles; and CEA, AED ^ 
have been demonstrated to be equivalent to two right angles ; 
wherefore (I. ax. 11 and 1) the angles CEA, AED are equal to 
the angles AED, DEB. Take away the common angle AED, 
and (I. ax. 3) the remaining angles CEA, DEB are equal 




24 THE ELEMENTS OF [bOOK I. 

lu the same manner it can be demonstrated that the angles 
CEB, AED are equal. Therefore, if two straight lines, etc. 

Cor. If at a point in a straight line two other straight lines 
meet on the opposite sides of it, and make equal angles with the 
parts of it on opposite sides of the point, the two straight lines 
are in one and the same straight line. 

Let AEB be a straight line, and let the angles AEC, BED be 
equal, CE, ED are in the same straight line. For, by adding 
the angle CEB to the equal angles AEC, BED, we have BED, 
BEC together equal to AEC, CEB, that is (I. 9), to two right 
angles ; and therefore, by this proposition, CE, ED are in the 
same straight line. 

Scho. In the proof here given, the common angle is AED ; 
and CEB might with equal propriety be made the common 
angle. In like manner, in proving the equality of CEB and 
AED, either AEC or BED may be made the common angle. 
It is also evident, that when AEC and BED have been proved 
to be equal, the equality of AED and BEC might be inferred 
from the ninth proposition, and the third axiom. 

Prop. XIL — Pkob. — To describe a triangle of which the sides 
shall be equal to three given straight lines; but any txoo of 
these must be greater than the third. 

Let A, B, C be three given straight lines, of which any 
two are greater than the third ; it is required to make a tri- 
angle of which the sides shall be equal to A, B, C, each to each. 
Take an unlimited straight line DE, and let F be a point in 
it, and make FG equal to A, FH to B, and HK to C. From 

the center F, at the distance FG, 
describe (I. post. 3) the circle 
GLM, and from the center H, at 
-E the distance HK, describe the 
circle KLM. Now, because (hyp.) 
g FK is greater than FG, the cir- 
" ^ cumference of the circle GLM 

will cut FE between F and K, and therefore the circle KLM 
can not lie wholly within the circle GLM. In like manner, be- , 
cause (hyp.) GH is gi-onter than HK, the circle GLM can not 
lie wholly witluu the ciicle KLM. Neither can the circles be 




BOOK I.] ETTCLID AND LEGENDRE. 25 

wholly without each other, since (hyp.) GF and HK are to- 
o-ether jrreater than FH, The circles must therefore intersect 
each other ; let them intersect in the point L, and join LF, LH ; 
the triangle LFII has its sides equal respectively to the three 
lines A, B, C. 

Because F is the center of the circle GLM, FL is equal (I. 
def. 16) to FG; but (const.) FG is equal to A; therefore (I. ax. 
1) FL is equal to A, In like manner it may be shown that 
HL is equal to C, and (const.) FH is equal to B ; therefore the 
three straight lines LF, FH, HL are respectively equal to the 
three lines A, B, C ; and therefore the triangle LFH has been 
constructed, having its three sides equal to the three given lines, 
A, B, C ; which was to be done. 

Scho. It is evident that if MF, MH were joined, another tri- 
angle would be formed, having its sides equal to A, B, C, It 
is also obvious that in the practical construction of this problem, 
it is only necessary to take with the compasses FH equal to B, 
and then, the compasses being opened successively to the 
lenorths of A and C, to describe circles or arcs from F and H as 
centers, intersecting in L ; and lastly to join LF, LH. 

The construction in the proposition is made somewhat differ- 
ent from that given in Simson's Euclid, with a vicAv to obviate 
objections arising from the application of this proposition in the 
one that follows it. 

Prop. XIII. — Prob. — At a given point in a given straight 
line, to make a rectilineal angle equal to a given one. 

Let AB be the given straight line, A the given point in it, 
and C the given angle ; it is required to make an angle at A, 
in the straight line AB, that shall be equal to C. 

In the lines containing the angle C, 
take any points D, E, and join them, 
and make (L 12) the triangle AFG, 
the sides of which, AF, AG, FG, shall 
be equal to the three straight lines 

CD, CE, DE, each to each. Then, 
because FA, AG are equal to DC, 

CE, each to each, and the base FG to 
the base DE, the angle A is equal (I. 




26 



THE ELEMENTS OF 



[book 



4) to the angle C. Therefore, at the given jsoint A in the given 
straight line AB, the angle A is made equal to the given angle 
C ; which was to be done. 

Scho. The construction is easy, by making the triangles isos- 
celes. In doing this, arcs are described with equal radii from C 
and A as centers, and their chords are mad« equal. 

It is evident that another angle might be made at A, on the 
other side of AB, equal to C. 

Prop. XIV. — Theor. — If two angles of one triangle he equal 
to two angles of another^ each to each, and if a side of the one 
he equal to a side of the other siniiliarly situated with respect 
to those angles ; (l) the remaining sides are equal, each to each ; 
(2) the remaining angles are equal ; and (3) the triangles are 

equcd. 

This proposition is the converse of the third proposition, and 
is susceptible of three cases, viz. : first, when the equal sides are 
between the equal angles ; secondly and thirdly, when the equal 
sides are opposite to the equal angles similarly situated. 

Q J, Let ABC and DEF be two triangles 

M-hich have the angles ACB and EFD 
equal; the angles BAG and DEF 
equal, and the sides CA and FE equal ; 
then the sides CB and FD are equal, 
also the sides AB and DE ; the angles 
CBA and FDE are equal, and the tri- 
angles are equal to one another. 

If the triangles be placed so as to 
have their sides CB and FD in the same straight line, but the 
triangles be on opposite sides of that line, and the vertex C on 

the vertex F ; then because the angles 
AFD and EFD are equal (hyp.), the angle 
AFE is bisected by FD ; and because AF 
and FE are equal (hyp.), the triangle AFE 
is isosceles (I. def 13), and the line AE 
(T. 6) is also bisected by FD; hence (I. 1, 
cor. 1) the angles FAE and FEA are 
equal ; therefore (I. 3) the triangles FAH 
and FEII are equal. But the angles FAD 





BOOK I.] EUCLID AND LEGENDKE. 27 

and FED are eq^^al (byp.) ; tlien taking from each the equal 
angles FAII and FEH, there will remain (I. ax, 3) the angle 
HAD equal to the angle HED; hence (I. 1, cor. 2) the triangle 
AED is isosceles, and (I. def. 13 and I. cor. 2) the side AD is 
equal to the side ED, and AE being bisected by FD (I. 6), we 
have the triangles AHD and EHD (I. 3) equal ; therefore the 
angles HDA and HDE are equal; hence the triangles AFD 
and EFD have the sides AF and EF (hyp.) equal, the angles 
FAD and FED equal (hyp.), and the sides AD and DE equal 
(I. def 13 and I, cor. 2) ; therefore (I. 3) the angles FDA and 
FDE are equal ; the side FD is common, and the two triangles 
are equal. 

In a similar manner, the second and third cases can be dem- 
onstrated. Wherefore, if two angles of one triangle be, etc. 

Cor. Hence, from this proposition, it can also be shown that 
the second corollary to the first proposition is true. Let the 
triangle ABC have the angle CAB equal c 

to the angle CBA, then will AC be equal 
toCB. 

From the vertex C (I. S) draw CD per- 
pendicular to AB, then the angles CDA 
and CDB are both right angles (I. def 10), ^ 
the angles CAD and CBD are equal (hyp.), and the side CD coni- 
mon (const.); therefore (I. 14) the sides AC and CB are equal. 

8cho. It will be seen from propositions third, fourth, and 
fourteenth of this book, that two triangles are in every respect 
equal when the three sides of the one are respectively equal to 
the three sides of the other, when two triangles have two angles 
and a side in each equal, each to each, and when one angle and 
two sides of one are equal to one angle and two sides of the 
other, each to each, and the equal angles in each triangle simi- 
larly situated with respect to those sides, but with this limita- 
tion, that when two equal sides respectively of the triangles are 
the least sides, that the angles opposite the greater sides respec- 
tively of the triangles must be of the same kind, either both 
acute, both right-angled, or both obtuse. And from these prop- 
ositions it is shown that of the sides and angles of a triangle, 
three must be given to determine the triangle, and these three 
can not all be angles. Were only three angles given, the sides, 





28 THE ELEMENTS OF [bOOK I. 

as will appear from the twentieth prosposltion of this book, 
might be of any magnitude whatever. The pupil may occupy 
himself in proving these propositions by superposition or 
some other way, as by pursuing a course of demonstration 
different from what is given in the text, he will more readily 
familiarize himself with the process of geometrical reasoning. 

Prop. XV. — Theor. — Parallel straight liries are equally/ dis- 
tant from each other, however so far they be produced on the 
same plane. 

Let the straischt line AB 
be bisected (I. 6) at C, and 
let the perpendicular CD be 
drawn (I. 7) ; join AD and 
A c ' ^ BD, and if the triangle ADC 

be applied to the triangle BDC so that they will fall on difler- 
€nt sides of BD and have D and BD common, their sides DE 
and CB will be equally distant from each other, and so they 
will never meet, however so far they be produced on the same 
plane, and consequently (I. def 11) are parallel straight lines. 

Because DC is perpendicular to AB (const.), the angles ACD 
and BCD are both right angles (I, def 10), and are equal (I. ax. 
11) ; hence the triangles ADC and BDC have the side DC 
common, the sides AC and CB equal (const.), and the angles 
ACD and BCD equal (I. def 10 and ax. 11); therefore (I. 3) 
the triangles are equal, having the sides AD and BD equal ; 
the angle ADC equal to the angle BDC, and the angle DAC 
equal to the angle DBC. Now, when ADC is applied to BDC 
so that they will fall on difterent sides of BD, and have D and 
BD common (hyp.), since AD is equal to BD, the point A will 
fall on B ; hence the angle EBD will be the same as ADC, and 
equal to BDC, the angle BED the same as ACD, and equal to 
DCB, the angle EDB the same as DAC, and equal to DBC, the 
side DE the same as AC, and equal to AC ; the side EB equal 
to DC; therefore the triangles BDC and BDE are equal (I. 
ax. 8), having their sides and angles equal, each to each. 

Since DE and CB are straight lines, they have no variation 
in the direction of their lengths from D to E or from C to B 
(I. def 7) ; and because DC is equal to EB, DE and CB are 



BOOK I.] EUCLID AND LEGENDEE. 29 

equally distant from each other at their extremities, and having 
no variation in the direction of their lengths from D to E and 
from C to B (I. def. 7), they are also equally distant from each 
other at every part between D and E and C and B, each to 
each ; therefore DE and CB (I. post. 2) on being produced to 
any length are still the same straight lines, and will have no 
variation in the direction of their lengths (I. def. 7), conse- 
quently they will always be the distant DC or EB from each 
other, at every part, each to each ; and being always the same 
distant DC or EB from each other, will never meet, and are 
parallel straight lines (I. def 11). Wherefore, parallel straight 
lines, etc. 

Cor. 1. In like manner, it can be shown that DC and EB 
are parallel lines; hence DCBE is a parallelogram (I. def 14) ; 
and since the angles DCB and DEB are equal, and the angles 
EDB and BDC equal to the angles EBD and DBC (I. ax. 2), 
and the sides DE and CB equal, also the sides CD and BE 
equal, a parallelogram has its opposite sides and opposite angles 
equal. 

Cor. 2. Hence parallel lines, DE and CB, intercepted by a 
straight line, DB, make the alternate angles EDB and CBD 
equal; and, co?it»er5eZy, when the alternate angles EDB and CBD 
are equal, the lines DE and CB are parallel. 

Cor. 3. In the parallelogram, the opposite sides being equal, 
the straight lines which join the extremities of two equal and 
parallel straight lines toward the same parts — that is, the near- 
est extremities together — are themselves equal and parallel ; 
hence a quadrilateral which has two sides equal and parallel is 
(I. def. 14) a parallelogram. 

Cor. 4. Because the triangles DCB and DEB are equal, a 
diagonal, DB, bisects the parallelogram ; and if two parallelo- 
grams have an angle of the one equal to an angle of the other, 
and the sides containing those equal angles respectively equal, 
the parallelograms are equal, as the parallelograms can be bi- 
sected by diagonals subtended by the equal angles, and the tri- 
angles thus formed are equal (I. 3) ; hence (I. ax. 6) the par- 
allelograms are equal; hence, also, if a pai'allelograra and a tri- 
angle be upon the same or equal bases, and between the same 
parallels, the parallelogram is double the triangle. 



30 THE ELEMENTS OF [bOOK I. 

Cor. 5. Hence, also, parallelograms upon the same or equal 
bases, and between the same parallels, are equal ; and triangles 
upon the same or equal bases and between the same parallels, 
are equal. 

Cor. 6. Hence, from the preceding corollary, it is plain that, 
triangles or parallelograms between the same parallels, but 
upon unequal bases, are unequal. 

Cor. T. And a straight line drawn from the vertex of a tri- 
angle to the point of bisection of the base, bisects the triangle ; 
and if two triangles have two sides of the one respectively equal 
to two sides of the other, and the contained angles supple- 
mental (I. def 20), the triangles are equivalent; the converse is 
also true. 

Cor. 8. If through any point in either diagonal of a parallelo- 
gram straight lines be drawn parallel to the sides of the four 
parallelograms thus formed, those through which the diagonal 
does not pass, and which are called the com2)lements of the 
other two, are equivalent. 

Peop. XVI. — Theoe. — If a straight linefallupon tioo parallel 
straight Ihies, (l) it makes the alternate angles equal to one an- 
other ; (2) the exterior angle equal to the interior and remote 
upon the same side, and (3) the tico interior angles upon the 
same side together, equivalent to two right angles. 

Let the straight lines AB, CD be parallel, and let EF fall upon 
them; then (l) the alternate angles AGH, GHD are equal to 
one another ; (2) the exterior angle EGB is equal to the interior 
and remote upon the same side, GHD; and (3) the two in- 
terior angles BGH, GHD ujDon the same side are together 
equivalent to two right angles. 

Since AB, CD are parallel (hyp.), theper- 

B pendiculars (I. 7) MG, LH make the angles 

MGH, GIIL equal to one another (I. 15, 

cor. 2), and AG3I, LHD are two riglit 

.jj angles (I. def 10); if Ave add AGM to 

MGH they will be equal to AGH (I. ax. 

^' 10), and DHL added to LIIG are likewise 

equal to GIID ; hence (I. ax. 2) AGH and GHD are equal to 

one another. 




BOOK I.] EUCLID AND LEGENDRE. 31 

Second. AGII is equal to EGB (I. 11), therefore (I. ax, 1) 
EGB is equal to GHD. 

Third. Add to EGB and GHD, each, the angle BGII ; there- 
fore (I. ax. 2 and ax. 10) EGB and BGH are equivalent to the 
angles GHD and BGH, but EGB and BGH are equivalent to 
two right angles (I. 9) ; therefore, also, BGH and GHD are to- 
gether equivalent to two right angles. Wherefore, if a straight 
line, etc. 

Cor. 1. Hence, conversely, tAvo straight lines are parallel to 
one another, if another straight line falling on them (1) makes 
the alternate angles equal ; (2) the exterior angle equal to the 
interior and remote upon the same side of that line ; and (3) 
the two interior angles upon the same side together equivalent 
to two right angles. 

Let EF fall on AB and CD, the perpendiculars (I. 1) GM and 
HL make two riglit angles (I. def. 10), AGM and DHL. But 
AGH and DHG are equal (hyp.) ; hence (L ax. 3) MGH is 
■equal to LHG. But MGL and LHM are both right angles 
(const, and L def. 10) ; hence (I. ax. 3) HGL is equal to GHM; 
•therefore in the triangles HMG and GLH we have two angles 
in one equal to two angles in the other, each to each, and the 
side GH common, and (I. 14) the triangles are equal; hence 
GM and HL are equal, and (L 15) AB and CD are equally dis- 
tant from each other, and will never meet on being produced 
{L post. 2), and are parallel (L def' 11). 

Because EGB is equal to DHG (hyp.), and EGB equal to 
AGH (I. 11), the angle AGH is equal to DHG (L ax. 1), but 
they are alternate angles; therefore (L 16, cor. 1, part l) AB is 
parallel to CD. Again : because BGH and GHD are together 
equivalent to two right angles (hyp.), and AGH and BGH are 
also equivalent to two right angles (I. 9), the angles AGH and 
BGH are together equal to BGH and GHD (I. 7 and ax. 1) ; 
then (L ax. 3) AGH is equal to GHD, but they are alternate 
angles, therefore (I. 16, cor. 1, part l) AB and ED are parallel. 
Wherefore, two straight lines are parallel, etc. 

Cor. 2. When one angle of a parallelogram is a right angle, 
all the other angles are right angles; for since (I. 10) BGM and 
GMH are together equivalent to two right angles, if one of 
them be a right angle, the other must also be a right angle, and 



32 THE ELEMENTS OF [boOK I. 

(I. 15, cor. 1) the opposite angles are equal. A rectangle, then 
(L def. 15), has all its angles right angles. 

Cor. 3. If two straight lines make an angle, two others par- 
allel to them contain an equal or supplemental angle ; thus LGM, 
and the vertical angle produced by AB and the continuance of 
MG through G, are each equal to LHM, while the angle AGM, 
and its vertical angle contained by GB and the continuance of 
MG, are each equal to the supplement of LHM ; hence we can 
divide a given straight line AB into any proposed number of 
equal parts. 

Prop. XVH. — Theor. — Two straight lines which are not 
in the same straight line, and which nre parallel to a third 
straight line, are parallel to one another. 

Let the straight lines AB, CD be each of them parallel to 
the straight line EF ; AB is also parallel to CD. 

Let the straight line LH cut AB, CD, EF ; and because LH 
L cuts the parallel straight lines AB, EF, the 
angle LGB is equal (L 16, part 2) to the 
. J} angle LHF. Again : because the straight 
line LH cuts the parallel straight lines CD, 
" ^ EF, the angle LKD is equal (L 1 6, part 2) to 

^ ■ " — J. the angle LHF ; and it has been shown 

that the angle LGB is equal to LHF; 

wherefore, also, LGB is equal (L ax. 1) to 

LKD, the interior and remote angle on the same side of LH ; 

therefore AB is parallel (L 10, part 1) to CD. Wherefore, two 

straight lines, etc. 

Prop. XVIIL — Prob. — To draw a straight line parallel to a 
given straight line through a given poi^it without it. 

Let AB be the given straight line, and C the given point ; it 
is required to draw a straight line through C, parallel to AB. 

In AB take any point D, and join CD ; at the point C, in the 

c straight line CD, make (I. 13) the angle 

E — -^ F p^j, ^^^^^^ ^^ ^j^j, ^ ^^^ produce the 

A ^ B Straight line EC to any point F. 

Because (const.) the straight line CD, 
which meets the two straight lines AB, EF, makes the alternate 





BOOK I.] EUCLID AND LEGENDEE. 33 

angles ECD, CDB equal to one another, EF is parallel (I. 
15, eor, 2) to AB. Therefore the straight line EOF is drawn 
through the given point C parallel to the given straight linev 
AB ; which was to be done. • 

Prop. XIX, — Theok. — If a straight line meet two otho'^ 
straight lines xohich are in the same pkoie, so as to make the 
two interior angles on the same side of it, taken together, less 
than two right angles, these straight lines shall at length meet 
upon that side, if they he continually i^i'oduced. Axiom 
twelfth. Elements of Euclid. 

Let EF be the straight line meeting AB, ^ 
CD on the same plane, so tliat BLO, DOL 
are less than two right angles, the lines 
AB, CD will meet if continually produced. 

At the point O, draw GH (I. 1 8) parallel g 
to AB, then BLO, HOL are equivalent to 
two right angles (L IG), but DOL is less c 

than HOL (I. ax. 9) ; therefore BLO, DOL are less than BLO, 
HOL, and less than two right angles. But GH, which fornij^ 
with EF the angle HOL, is parallel (const.) with AB, which! 
forms with EF the angle BLO, and (L def 11) GH and AB- 
never meet each other, because (L 15) they are equally distant 
from each other, and (L 16, cor. l) the interior angles HOL^ 
BLO together equivalent to two right angles ; therefore.^ 
then, CD, which forms Avith EF the angle DOL less than HOL, 
can not be parallel with AB (L ax. 9) ; and not being parallel 
wath AB, CD and AB can not preserve the condition of par- 
allel lines (L def 11), and will meet. And the line CD making 
with EF the angle DOL less than HOL, on the side of EF, 
where the interior angles BLO, DOL are less than two ricrht 
angles, the line OD must therefore be between GH and AB, 
and is consequently nearer to AB than GH is to AB (L ax. 9) ; 
but CD making with EF the angle EOC greater than the angle 
GOE, which GH makes with EF, therefore CO must be with- 
out AB and GH, and consequently is farther from AB than GH 
is from AB ; hence the straight line CD, Avhich is made of CO 
and OD, has parts unequally distant from AB ; therefore since 
CD approaches AB on the side of EF where OD is, CD must 
3 



S^ THE ELEMENTS OF [bOOK I. 

meet AB on that side wliea continuallj^ produced, but OD 
makes the angle DOL less than HOL, therefore CD will meet 
AB on the side of EF where the angles BLO, DOL are less 
than BLO, HOL. Wherefore, if a straight line, etc. 

Cor. 1. Hence a straight line which intercepts one of two or 
more parallel straight lines will intercept the others if continu- 
ally produced ; hence, also, two straight lines which intercept 
each other are not both parallel to the same straight line. 

Prop. XX. — Theor. — If a side of any triangle be produced, 
(1) the exterior angle is equivalent to the tico interior and re- 
mote angles ; and (2) the three interior angles of every triangle 
are together equivalent to two right angles. 

Let ABC be a triangle, and let one of its sides BC be pro- 
educed to D; (1) the exterior angle ACD is equivalent to the 
two interior and remote angles CAB, ABC; and (2) the three 
interior angles, ABC, BCA, CAB, are together equivalent to 
two right angles. 

Through tlae point C draw (L 18) CE parallel to the straight 

line AB. Then, because AB is parallel to EC, and AC falls 

'>iipon them, the alternate angles BAC, ACE are (L 16, part 1) 

equal Again: because AB is ijarallel to 
EC, and BD falls upon them, the exterior 
angle ECD is equivalent (L I G, part 2) 
to the interior and remote ano-le ^VBC : 

'B^ ^ — D but the angle ACE has been sliown to 

F be equal to the angle BAC ; there- 
fore the whole exterior angle ACD is equivalent (I. ax. 2) to 
the two interior and remote angles CAB, ABC. To these 
equals add the angle ACB, and the angles ACD, ACB are 
equivalent (L ax. 2) to the three angles CBA, BAC, ACB ; but 
the angles ACD, ACB are equivalent (I. 9) to two right angles; 
therefore, also, the angles CBA, BAC, ACB are equivalent to 
two right angles. Wherefore, if a side, etc. 

Another proof of this important proposition can be given by 
producing the side AC through C to F. !N"ow (I. 11), the angle 
DCF is equal to the angle ACB ; EC being parallel (const.) to 
BA, the exterior angle ECD is equal to the intei'ior and remote 
angle ABC (L 10), and for the same reason the alternate angles 



r 




BOOK I.J EUCLID AND LEGENDRE. 35 

EGA and CAB are equal; hence we have the three angles of 
the triangle, ACB, CBA, and CAB, equal to the three angles 
DCF, DCE, and EGA, each to each ; but (I. 9) tlie angles DCF, 
DCE, and ECA are equivalent to two right angles ; therefore 
(I. ax. 1) the three angles ACB, CBA, and CAB of the triangle 
are likewise equivalent to two right angles. And when a par- 
allelogram (I. def 14) is formed by drawing (I. 18) pai-allels to 
BA and AC respectively, it can be shown by the sixteenth 
proposition tliat two adjacent angles of the parallelogram are 
equivalent to two right angles, and the four angles together 
equivalent to four right angles; since (I. 15, cor. 4) a diagonal 
bisects the parallelogram and forms two equal triangles, the 
angles are also equally divided, hence each triangle has its 
three angles equivalent to two right angles. 

Cor. 1. All tlie interior angles of any rectilineal figure, to- 
gether Avith four right angles, are equivalent to twice as many 
right angles as the figure has sides. 

For any rectilineal figure can be divided into as many tri- 
angles as the figure has sides, by draAving straight lines from a 
point within the figure to each of its angles; and by the prop- 
osition, all the angles of these triangles are equivalent to twice 
as many right angles as there are triangles — that is, as there are 
rsides of the figure ; and the same angles are equivalent to the 
angles of the figure, together with the angles at the point Avhich is 
the common vertex of the triangles — that is (I. 9, cor.), together 
with four right angles. Tlierefore all the angles of the figure, 
together Avitli four i-ight angles, are equivalent (I. ax. 1) to 
twice as many riglit angles as the figure has sides. 

Scho. 1. Another proof of this corollary may be obtained by 
■dividing the figure into triangles by lines drawn from any 
.ano-les to all the remote angles. Then each of the tAVO ex- 
treme triangles has tAVO sides of the polygon for tAvo of its sides, 
while each of the other triangles has only one side of the figure 
for one of its sides ; and hence the number of triangles is less 
by tAVO than the number of the sides of the figure. Biit the in- 
terior angles of the figure are evidently equivalent to all the 
interior angles of all the triangles — that is, to tAvice as many 
right angles as there are triangles, or tAvice as many right 
angles as the figure has sides, less the angles of two triangles — 



36 THE ELEMENTS OP [bOOK I. 

tlfet is, four riglit angles. Hence, iu any equiangular figure, 
the number of the sides being known, the magnitude of each 
angle compared with a right angle can be determined. Thus, 
in a regular pentagon, the amount of all the angles being twice 
five *right angles less four— that is, six right angles, each angle 
will be one fifth j^art of six right angles, or one right angle and 
one fifth. In a similar manner it would appear that in the 
regular hexagon, each angle is a sixth part of eight right angles, 
or a right angle and a third ; that in the regular heptagon, each 
is a right angle and three sevenths ; in the regular octagon, a 
rio-ht ano;le and a half, etc. 

Cor. 2. All the exterior angles of any rectilineal figure are 
together equivalent to four right angles. 

Because each interior angle ABC, and the adjacent exterior 
ABD, are together equivalent (I. 9) to 
two right angles, therefore all the in- 
terior, together with all the exterior 
angles of the figure, are equivalent to 
twice as many right angles as there are 
B sides of the figure — that is, hj the fore- 

going corollary, they are equivalent to all the interior angles 
of the figure, together with four right angles; therefore all the 
exterior angles are equivalent (I. ax. .3) to four right angles. 

Sc/to. 2. It is to be observed, that if angles be taken in the 
ordinary meaning, as understood by Euclid, this corollary and 
the foregoing are not applicable when the figures have re-entrant 

angles — that is, such as open outward. 
The second corollary will hold, how- 
ever, if the difference between each 
re-entrant angle and tAvo riiiht ansrles 
be taken from the sum of the other exterior anccles: and the 
former will be applicable, if, instead of the angle which opens 
externally, the difterence between it and four right angles be 
used. Both corollaries, indeed, Avill hold without change, if the 
re-entrant angle be regarded as internal and greater than two 
right angles; and if, to find the exterior angles, the interior be 
taken, in the algeljraic sense, from two right angles, as in this 
case, the re-entering angles will give negative or subtractive 
results. 





BOOK I.] EUCLID AND LEGENDEE. 87 

Cor. 3. If a triangle has a right angle, the remaining angles 
are together eqiiivaleut to a right angle ; and if one angle of a 
triangle be equivalent to the other two, it is a right angle. 

Cor. 4. The angles at the base of a right-angled isosctles tri- 
angle are each half a right angle. 

Cor. 5. If two angles of one triangle be equal to two angles 
of another, their remaining angles are equal. 

Cor. 6. Each angle of an equilateral triangle is one third of 
two right angles, or two thirds of one right angle. 

Cor. 1. Hence, a right angle may be trisected by describing 
an equilateral triangle on one of the lines containing the right 
angle. 

ScJio. 3. By this principle also, in connection with the fifth 
proposition, we may trisect any angle, Avhich is obtained by the 
successive bisection of a right angle, such as the half, the 
fourth, the eighth, of a right angle, and so on. 

Cor. 8. Any two angles of a triangle are less than two right 
angles. 

Cor. 9. Hence every triangle must have at least two acute 
angles. 

Peop. XXI. — Tiieor. — Iftxco sides of a triangle be unequal, 
(1) the greater side has the greater angle opposite to it ; and (2) 
conversely, if tioo angles of a triangle be unequal, the greater 
angle has the greater side opjiosite to it. 

D Let ABC be a triangle of which the side 

AB is greater than the side AC ; the angle 
ACB opposite AB is greater than the angle 
ABC opposite AC. 

Because AB is greater than AC, pro- 
duce AC (I. post. 2), and with A as a center, and a radius AB, 
describe a circle (I. post. 3) intercepting AC produced in D, 
and join BD ; the triangle ADB is isosceles (I. defs. 13 and 16) ; 
therefore the angle ADB is equal to the angle ABD (I. 1, cor. 1). 
But (I. 20) the exterior angle ACB is equivalent to the sum of 
the two remote interior angles CDB and DBC. And CDB is 
equal to DBA (I. 1, cor. 1), but ABC is less than ABD (I. ax. 
9) ; therefore ACB is greater than CBA. 

The proof can also be given by laying oflf on the greater side 





38 THE ELEMENTS OF [bOOK I. 

AB, tlie less side AC ; joining the vertex of the opposite angle 
with the point where AC terminates on AB ; and the demon- 
stration conducted similarly to the preceding. Again : by 
cutting oflf on AC a part equal to CB, hisect the angle BCA, 
and join the extremity of the part on AC equal to CB with the 
foot of the line bisecting the angle BCA ; this proof is given by 
means of (I. 3 and 20). 

Conversely: when the angle ACB is greater than the angle 
ABC, the side AB is greater than AC. At the vertex of BCA 

on AC make an angle ACD equal 
to the angle ABC (I. 13). Now, 
the angle ACD, the equal to the 
angle ABC, is subtended by AD, 
~^ ^ ^ and the angle ACB is subtended by 

AB, but AB is greater than AD ; hence the greater angle is 
subtended by the greater line ; therefore, in the triangle ACB, 
the greater angle ACB is subtended by a greater side than the 
less angle ABC. And (L def. 19) if the angles ACD and ABC 
be subtended by arcs, the arc subtending ACB is greater than 
the arc subtending ABC ; but the side AB is the chord of the 
arc subtendins: ACB, and AC is the chord of the arc subtending 
ABC ; therefore AB is greater than AC. Wherefore, if two 
sides of a triangle, etc. 

Cor. Hence any two sides of a triangle are together greater 
than the remaining side. 

Scho. The truth of this corollary is so manifest, that it is 
given as a corollary to avoid increasing the number of axioms. 
Archimedes defined the straight line the shortest distance be- 
tween two points ; hence two straight lines connecting three 
points not in the same direction, are together greater than one 
straight line connecting any two of those jDoiuts. 

Prop. XXII, — Tiieor. — If two triangles have tico sides of 
the one equal to two sides of the other, each to each, hut the 
angles contained hg those sides loiequal, the base or remaining 
side of the one ichich has the greater angle is greater than the 
base or reraaining side of the other. 

Let DEG and DEF be two triangles which have the sides 
DE common, the sides DG and DF equal, but the angle EDG 




HOOK I.] EUCLID AND LEGENDEE. 39 

greater tlian the angle EDF ; the side EG is also greater than 
the side EF. 

In the triang-les DEF and DEG Ave have 
(hyp.) DE commoii, DG equal to DF, and 
the an'He EDG crreater than the anHe 
EDF. Now, because DG and DF are equal, 
the angles DGF and DFG are equal (I. 1, 
cor. 1) ; but the angle DGF is greater than 
the angle EGF (I. ax. 9) ; therefore the angle 
DFG is greater than the angle EGF ; and much more is the 
angle EFG greater than the angle EGF. Then (I. 21), EG op- 
posite EFG is greater than EF opposite EGF. Wherefore, if 
two triangles have, etc. 

There are other cases of this proposition ; if a line equal to 
DE, the less side, be drawn through D, making with DF, on the 
same side of it with DE, an angle equal to EDG, the extremity of 
that line might fall on FE produced, or above, or helo^o it. Or 
the ano-le DFE could be in the triangle DEG, or on the other 
side of DE. And a very easy proof can be given by bisecting 
the angle FDG by a straight line cutting EG in a point, which 
call K, and joining DK and KF, for KG would be equal to 
KF (I. 3) ; adding EK, we would have EG equivalent to EK 
and KF; therefore (I. 21, cor.) EG greater than EF. 

Cor. Hence, conversely, if two triangles have two sides of 
the one equal to two sides of the other, each to each, but their 
bases unequal, the angles contained by the respectively equal 
sides of those triangles arc also unequal, the greater angle 
being in the triangle which has the greater base. For with a 
radius, DG, or its ecpial, DF, describe a circle (I. post. 3) from a 
center, D, and draw on the same side of DE with the angle 
EDF a line equal to EG from the other extremity of DE to the 
circumference, then in the triangles DEG and DEF we have 
(const.) DG equal to DF (I. def 16). But (hyp.) the side EF 
is less than EG ; hence the angle EDF is less than the angle 
EDG (I. ax. 9). 

Pkop. XXin. — Peob. — To describe a parallelogram upon a 
given straight line. 

Let DB be a given straight line ; it is required to describe a 



40 



THE ELEMENTS OF 



[book I. 



paralleloGjram. U])Ou it. From a point, D, on DB draw DF (I. 



F 



H 




B 



post. 1), then from the point F on 
DF draw FL parallel to DB (I. 18) ; 
and if through B, a point on DB, a 
parallel to DF (I. 18), he drawn, 
DBFG (I. def. 15) is a parallelogram. 
If from D a perpendicular, DII (I. 7), be drawn, then the 
angle HDB is a riglit angle (I. def 10) ; and from H a parallel 
to DB as HL be drawn (I. 18), and from B a parallel to HD 
as LB be drawn (I. 18), then DBIIL (I. def 15) is a rectangle. 
And if from D as a center, and a radius DB (I. post. 3), an arc 
be described intercepting DH, or DH produced, and a rectangle 
be described from that point where DH is intercepted, that 
rectangle will be a square (I. def 15). 

Co?: 1. Hence (const, and I. 15, cor. l) a square has all its 
sides equal, and (I. 15, cor. l) all its angles are right angles. 

Cor. 2. Hence the squares described on equal straight lines 
(I. 15, cor. 4) are equal. 

Cor. 3. If two squares be equal, their sides are equal (I. 
ax. 1). 

Cor. 4. If AB and AD, two adjacent sides of a rectangle 
BD, be divided into parts which are all equal, straight lines 
drawn through the points of section, parallel to the sides, divide 
the rectangle into squares which are all equal, and the number 
A B of which is equal to the product of the 

number of parts in AB, one of the sides, 
multiplied by the number of parts in 
AD, the other. For (const.) these 



figures are all parallelograms; and (I. 
c def. 15 and const.) the sides being equal, 
and the angles being (I. 16, part 2) equal to A, and there- 
fore right angles, hence (I. 15, cor. 4) they are all squares. Of 
these squares, also, there are evidently as many columns as there 
are parts in AB ; while in each column there are as many squares 
as there are parts in AD. The number of such squares con- 
tained in a figure is called, in the language of mensuration, the 
area of that figure. 

Cor. 5. Hence, since any parallelogram is equivalent (I. 15, 
cor. 5) to a rectangle on the same base and between the same 



BOOK I.] EUCLID AND LEGENDKE. 41 

parallels, it follows that the area of any 2^'^^'>'(Melogram is 
equivale7it to the product of its base and its "perpendicidar 
height / and the area of a square is computed by multiplying 
a side by itself 

Cor. 6. Hence, also (I. 15, cor. 4), the area of a triangle is 
computed by m.idtiplying any of its sides by the perpendicidar 
draion to that side from the opposite angle., and taking half 
the product ; and the area of a trapezium is found by multi- 
P'lying either diagonal by the sum of the perpendicidar s draion 
to it from the ayigles which it subtends, and taking half the 
product. When, in consequence of one of the angles being re- 
entrant, the perpendiculars lie on the same side of the diagonal, 
the difference of the perpendiculars must evidently be used in- 
stead of their sum. 

Cor. 7. Every polygon may be divided into triangles or tra- 
peziums by drawing diagonals ; and therefore the area of any 
polygon whatever can be computed by finding the areas of 
those component figures by the last corollary, and adding them 
together. 

Scho. This corollary and the two foregoing contain the ele- 
mentary principles of the mensuration of rectilineal figures, and 
they form a connection between arithmetic or algebra and ge- 
ometry. They also explain the origin of the expressions, " the 
square of a number," " the rectangle of two numbers," and " the 
product of two lines." 

Prop. XXIV. — Theor. — If parallelograms be described on 
two sides of any triangle, and their sides which are parallel to the 
sides of the triangle be produced until they meet, the sum of 
the parallelograms will be equivalent to the parallelogram de- 
scribed on the base of the triangle having its adjacent sides to 
the base parallel to the straight line joining the vertex of the 
triangle with the point of intersection of the sides of the other 
parallelograms produced, and terminated by the latter sides or 
those sides produced. 

Let BAG be a triangle ; the parallelograms MB AD and CEFA, 
described on the two sides BA and CA, respectively, are to- 
gether equivalent to the parallelogram BCHK, described on 
the base BC ; the parallelograms MBAD, CEFA, and BCIIK 



43 



THE ELEMENTS OF 



[book I. 




beino; described ao;reeablv to the 
proposition. 

Describe on BA and CA (L 
23) tlie parallelograms MB AD 
and CEFA ; and produce the 
M B L c E titles MD and EF until they 

meet in G ; draw GA, and produce it to L on the base BC (I. 
post. 2) ; describe the parallelogram BCIIK (I. 23) on BC. 
Then, since (const, and I. 15, cor. 1) BH and CK are parallel 
and equal to AG, they are parallel and equal to one another (I. 
ax. 1) ; also (1. 15, cor 3) HK is parallel and equal to BC ; hence (I. 
def. 15) BCHK is a parallelogram, and BLIIW and LCWK are 
also parallelograms. Xow, the parallelograms BLHW and 
HBAG (I. 15, cor. 5) are equal, and for similar reason the parallel- 
ograms HBAG and MBDA ; hence (I. ax. 1) MBDA is equal to 
BLHW. In similar manner, it can be shown tlnit CEFA is 
equiA^alent to LCKW; therefore the whole parallelogram BCHK 
(I, axs. 2 and 10) is equivalent to the sum of the two parallelo- 
grams MB AD and CEFA. Wherefore, if on any two sides of 
a triangle, etc. 

Cor. 1. A paiticular case of this proposition is, lohen the tri- 
angle is right-angled., then the sqiuires described on the legs — 
that is, the sides containing the right angle, are together equiva- 
lent to the square on the liypotlieniise — that is, the side opposite 
the right angle. 

Let ABC be a rio-ht-anoled 
triangle, having the right angle 
BAC ; the square described on 
tlie hypothenuse BC is equiva- 
lent to the sum of the squares 
on BA and AC. On BC, BA, 
Q and AC (L 23) desciibe the 
squares BCHK, BADE, and 
ACFG ; through A draw AL 
parallel to BH (L 18), and 
draw AH and DC (L post. 1). 
Tlien, because the angles 
BAC and BAE are both right 
angles (L def 10), the two 




BOOK I.] 



EUCLID AND LEGENDEE. 



43 



straight lines CA and AE are in the same straight line (I, 10). 
For like reason, BA and AF are in the same straight line. 
Again : because the angle IIBC is equal to the angle DBA (I, 
ax. 1), if the angle ABC be added to each, we have (I. ax. 2) 
HBA equal to DBC ; and because AB is equal to DB (const.), 
and BIT equal to BC, therefore (I. 3, case 1) the triangle ABH 
is equal to the triangle DBC. But (I. 15, cor. 4) the parallelo- 
gram BPIIL is double the ti-iangle ABH ; for like reason the 
square BADE is double the triangle DBC ; hence (I. ax. 6) 
BPIIL is equivalent to BADE. In like manner, PCLK can be 
shown equivalent to ACFE. Xow (I. ax. 10), BCHK is equiva- 
lent to BPHL and PCLK together; hence (L ax. 1) BCHK is 
equivalent to BADE and ACFG together. Wherefore, if the 
triangle is rio-ht-angled, etc. 




OTHERWISE, 

Let the squares on AB and 
BC fall on the same side of BC. 
Describe the square BAED on 
the side BA (L 23), and th.e 
square BCMN" on the side BC 
(L 23), and produce AE to F 
(I. post. 2) ; then tlirough D draw 
PF parallel to BC (I. 18). 

Because AF is parallel to BD 
(L def. 14, and 15, cor. 1), BC is 
equal to DF, and BA is equal to '^^ 
DE, and the angles BAC and DEF are both right angles (1. 
def 10), and equal (I. ax, 11); therefore the triangle BAC is 
equal to the triangle DEF (I. 3) ; and because BCED is com- 
mon to the square BAED and the parallelogram (I. def 14) 
BCDF, and the triangles BAC and DEF equal, the square 
BAED is equivalent to the parallelogram BCDF. And PF 
being parallel to BC (const.), the parallelograms (I. def 14) 
BCDF and BCPL have a common base, BC, and equal altitudes; 
hence (I. 15, cor. 5) they are equivalent, and (I. ax, 1) the 
square BAED is equivalent to the parallelogram BCPL. From 
A draw (I. 18) AK parallel to BM, and produce DE (I. post. 
2) to BM; then AK and BM being parallel (const.), ED and 



44 



THE ELEMENTS OF 



[book I. 



BA, being opposite sides of the same square, are also parallel 
(I. def. 15) ; hence MR is equal to BA, and equal also to DE 
(I. ax. 1). But BM is equal to BC (const.) ; therefore the par- 
allelogram BAjMR is equal to the parallelogram BCDF ; and 
BAMR having the same base and equal altitude with the par- 
allelogram BGMK, is equivalent to it (I. 15, cor. 5); hence 
BGMK is equivalent to BCDF, equivalent to BCPL, aud 
equivalent to the square BADE (I. ax. 1). 

Or, the square described upon BA is equivalent to the rect- 
angle of the hypothenuse B C and the part BG of the hypjoth- 
enuse nearest to BA intercepted by the p)erpe7idicular drawn 
from the vertex of the right angle to the liypothenuse. In a 
similar manner, it can be shown that the square described on 
A G is equivalent to the rectangle of the hypothenuse B (7, and 
the remaining part G G of the hypothenuse intercepted by the 
same p)erpendicxdar. But the two rectangles are equivalent to 
the square of the hyi^othenuse (I. ax. 10) ; hence the tico squares 
described on the sides AB and AG (I. ax. 1) are equivalent to 
the square described on the hypothenuse. 

E F And if we make the tiiangle 

an isosceles rio;ht-an(2;led tri- 
angle as ABC, the square de- 
sciibed on AB will contain 
four equal triangles, ACB,BCF, 
FCE, and ECA, while each of 
the squares described on AC 
and CB will contain two such triangles, and both together will 
be equivalent to the four equal triangles, or equivalent to the 
square on AB. The demonstration is very simple, and it would 
be well for the pupil to undertake it. 

Hence, conversely, if the square described upon one side of a 
triangle be equivalent to the sum of the square described upon 
the two other sides of the triangle, the angle contained by those 
two sides is a right angle ; and when those two sides form two 
equal squares, the triangle is a right-angled isosceles triangle. 

Scho. 1. The proof of the corollary can be shown, also, either 
by describing the square of the hypothenuse on the other side 
of BC; and the other squares sometimes on one side and some- 
times on the other; and since drawing a perpendicular from the 




BOOK I.] EUCLID AND LEGEXDKE. 45 

vertex of tlie riglit angle to the hypothenuse makes two ri(iht 
angles, so a line can be drawn from the same vertex to the 
point of bisection of the hypothenuse and make two supple- 
niental avgles and two equivalent triangles, and the demonsti'a- 
tion conducted by supplemental angles (I. 15, cor. V) instead of 
right angles. Proportion also gives neat and easy solutions to 
this corollary. (See V. 8, scho.) 

Cor. 2. If two right-angled triangles have their hypothenuses 
equal, and a side similarly situated in each also equal, the two 
triangles are equal by the third proposition of this book ; and, 
conversely, if the legs of a right-angled triangle be equal to the 
legs of another right-angled triangle, each to each, their hypoth- 
enuses can be in a similar manner shown equal. 

Cor. 3. Hence, also, we can find a square equivalent to the 
sum of more than two squares ; thus, let AB be the side of one 
square, and AC, perpendicular to it, the side of another squai'e ; 
join CB ; the square on CB (I, 24, cor. l) 
is equivalent to the sum of the squares on 
CA and AB. In like manner, if CD be 
drawn perpendicular to CB, and DB be 
drawn, the square on DB is equivalent to 
the squares on DC, CA, and AB, and by 
drawing a perpendicular to DB, a square can be found equiva- 
lent to the sum of four squares ; hence a square can be found 
equivalent to the sum of any number of squares. 

Cur. 4. Since (CB)- o (AB)= + (CA)=, we have (AB)=o 
(CB)" — (CA)-; hence a square can be found equivalent to the 
difference of two squares. 

Cor. 5. If a perpendicular be drawn from the vertex of the 
angle A, in the triangle BAC (diagram to cor. l), to P on the 
hypothenuse BC, cutting BC into two segments, BP and PC, 
the difference of the squares on the sides AB and AC is equiva- 
lent to the difference of the squares on the segments BP and 
PC. For the square on AB is equivalent to the squares on BP 
and PA, and the square on AC is equivalent to the squares on 
PC and PA; therefore (AC)^ — (AB)^ =o- (PC)' — (BP)^ 

Cor. 6. Hence the squares on txoo sides of a triangle are to- 
gether equivalent to txoice the square of half the remaining side^ 
and twice the square of the straight line from its point of M- 




46 



THE ELEMENTS OF 



[book I. 



section to the opposite anrjle. Suppose P in tlic triangle to te 
the point of bisection of the side BC ; tlion, when AP is per- 
pendicuhir to BC, we have (AB)- + (AC)= o (BP)- + (AP)' 
+ (PC)' + (AP)^ .0= 2 (BP)= + 2 (AP)\ And when AP is 
not perpendicular to BC, the equivalence of the squares on two 
sides to the same will be shown in tlic next book. 

Scho. 2. In proof of coj-. 5, the obvious principle is employed, 
that the difference of two magnitudes is the same as the differ- 
ence obtained after adding to each the same third magnitude. 
Thus the difference of the squares on BP and PC is the same as 
the difference between the sum of the squares BP and PA and 
of PC and PA. 



Pkop. XXy. — Theok. — Tlie side and diagonal of a square 
are incommeiisuraUe to one another — that is, there is no Ivie 
which is a measure of both. 

Let ABCD be a square, and BD one of its diagonals ; AB, 
BD are incommensurable. 

Cut off DE equal to DA, and join AE. Then, since (I. 1, 

cor. 1) the angle DEA is equal to the acute angle DAE, AEB 

A B i^ obtuse, and therefore (I. 22, cor.) in 

the triangle ABE, BE is less than AB, 
or than AD ; wherefore AD is not a 
measure of BD. Draw EF ])erpendicular 
to BD. Then tiie angles FAE, FEA, 
being the complements of the equal 
angles DAE, DEA, are equal, and 
therefore AF, FE are equal. But (I. 
20, cor. 4) ABD is half a riglit angle ; 
as is also BFE, since BEF is a right angle ; wherefore BE is 
equal to FE, and therefore to AF. From FB, which is evi- 
dently the diagonal of a square of which FE or EB is the side, 
cut off FG equal to FE, and join GE. Then it would be shown, 
as before, that BG is less than BE ; and therefore BE, the dif- 
ference between the side and diagonal of tlie square .VC, is con- 
tained twice in the side AB, with the remainder GB, wliich is 
itself the difference between the side FE or EB, and the diag- 
onal FB of another squai*e. By repeating the process, we 
Bhould find, in exactly the same manner, that BG would be 




BOOK I.] EUCLID A:SD LEGENDEE. 47 

contained twice in BE, with a remainder, which would be the 
difierence between the side and diagonal of a square described 
on I3G ; and it is evident that a like process might be repeated 
continually, as no excess of a diagonal above a side would 
be contained in the side witliout remainder; and as this pro- 
cess has no termination, theiv is no line, however small, which 
will be contained without remainder in both AB and BD ; they 
are, therefore, incommensurable. 

Scho. Tills proposition can be illustrated by numbers. Let 
10 be the side of tlie square ; then (I. 24, cor. 1) the diagonal 
will be expressed by the square root of 200, or 14'142 + ; there 
being no common multiple of 10 and 14-142 +, these numbers 
are incommensurable with each otlier. Or, wlicn any two lines 
are taken which by division and subdivision no common meas- 
ure can be found which can be contained in each v.'ithout a re- 
mainder, the two lines are said to be ijicommensicrable with 
each other, and such lines are tlie side and diagonal of a square ; 
and any two magnitudes wiiatever which have no common 
unit of measure are incommensurable with one another. 



E^TD or BOOK FIRST. 



BOOK SECOND.* 
ON" THE llECTANGLE AND SQUARE. 

DEFINITIONS. 

1. A rectangle is said to be contained by the two straight 
lines which are about any of the right angles. For the sake of 
brevity, the rectangle contained, by AB and CD is often ex- 
pressed simply by AB. CD, a jooint being placed between the 
letters denoting the sides of tlie rectangle ; and the square of a 
line AB is often written simply AB". 

2. A gno7non is the part of a parallelogram which remains 
when either of the parallelograms about one of the diagonals 
is taken away. 

PROPOSITIONS. 

Pkop. I. — TiiEOR. — If there be tioo straight lines, one of 
which is divided into any number of parts, the rectangle con- 
tained by the two lines is equivalent to the rectangles contained 
by the undivided line, and the several j^arts of the divided line. 

Let A and BC be two straight lines; and let BC be divided 
into any parts in tlie points D, E ; tlie rectangle contained by 
A and BC is equivalent to the rectangles contained by A and 
BD, A and DE, and A and EC. 

From B draAV (I. V) BG perpendicular to BC, and make it 
equal to A ; through G draw (I. 1 8) GH par- 
^ DEC ^^^^^ ^^ J3Q . ^^^^ through D, E, C draw DK, 

EL, CH parallel to BG. Then BH, BK, DL, 
and EH are evidently rectangles ; and BH is 
equivalent (I. ax. 10) to BK, DL, EH. But 
BH is contained by A and BC, for (H. dcf. 1) 
■^ it is contained by GB and BC, and GB is 

* The second and third books are arrantred very similarly to those 
books in the edition of Euclid i)y Professor Thomson of the University 
of Glasgow, Scotland. 



G K L n 



BOOK 11.^ EUCLID AND LEGENDEE. 49 

equal (const.) to A; and BK is contained by A and BD, for it 
is contained by GB and BD, of whicli GB is equal to A. Also 
DL is contained by A and DE, because DK, that is (I. 15, con 
4) BG, is equal to A ; and in like manner it is shown that EH 
is contained by A and EC. Therefore the rectangle contained 
by A and BC is eqidvalent to the several rectangles contained 
by A and BD, by A and DE, and by A and EC. Wherefore, 
if there be two straight lines, etc. 

Prop. II. — Theoe. — If a straight line he divided into any 
ttoo parts, the rectangles contained by thexchole and each of the 
parts are together equivalent to the square of the whole line. 

Let the straight line AB be divided into any two parts in the 
point C ; the rectangles AB.AC and AB.BC are equivalent tc» 
the square of AB, 

Upon AB describe (I. 23) the square AE, and through C 
draw (I. 18) CF parallel to AD or BE. Then 
AE is equal (I. ax. 10) to the rectangles AF and 
CE. But iVE is the square of AB, and AF is 
the rectangle contained by BA, AC ; for (II, 
def 1) it is contained by DA, AC, of which DA 
is equal to AB ; and CE is contained by AB, 
BC, for BE is equivalent to AB ; therefore the 
rectangles under AB, AC, and AB, BC are equivalent to the- 
square of AB. If, therefore, etc. 

8cho. This proposition may also be demonstrated in the fol- 
lowing manner : 

Take a straight line D equal to AB, Then (II, 1) the rect- 
angles AC.D and BCD are together equiva- 
lent to AB.D. But since D is equal to AB, ^ ^ ^ 

the rectangle AB.D is equivalent (I. def 15) 

to the square of AB, and the rectangles 

AC.D and BCD are respectively equivalent ° 

to ACAB and BCAB ; wherefoi-e the rect- 
angles ACAB and BCAB are together equivalent to the 
square of AB. 

In a manner similar to this, several of the following proposi- 
tions may be demonstrated. Such proofs, thougli perhaps not 
60 easily understood at first by the learner, are shorter than 
4 





so THE ELEMENTS OF [bOOK H. 

those given by Euclid; aud they have tlie advantage of being 
derived from those preceding them, instead of being estab- 
lished by continual appeals to original principles. 

Pkop. III. — Theor. — If a straight line he divided into any 
two 2ici7'ts, the rectangle contained hy the lohole and one of. the 
parts is equivalent to the square of that part^ together loith the 
rectangle contained hy the tioo parts. 

Let the straight line AB be divided into two parts in the 
point C ; the rectangle AB.BC is equivalent to the square of 
-BC, together with the rectangle AC.CB 

Upon BC describe (I. 23) the square CE ; produce ED toF; 

and through A draw (I. 18) AF parallel 
to CD or BE. Then the rectangle AE 
is equivalent (I. ax. 10) to the rectangles 
CE, AD. But AE is the rectangle con- 
tained by AB, BC, for it is contained 
by x^B, BE, of which BE is equal to 
BC ; and AD is contained by AC, CB, 
for CD is equal to CB ; also DB is the square of BC. Thei-e- 
fore the rectangle AB.BC is equivalent to the square of BC, 
together with the rectangle AC.CB. If, therefore, etc. 

8cho. Otherwise : Take a line D equal to CB. Then (II. 1) 
the rectangle AB.D is equivalent to the 

-^ ^ ^ rectangles BCD and AC.D ; that is (const. 

and L def 15), the rectangle AB.BC is 
^ equivalent to the square of BC togctlier with 

the rectangle AC.CB. 

Prop. IV.— Tiieor.— 7/" a straight line he divided into 
any tico j^^-^^ts, the square of the v-hole line is equivalent to 
the squares of the tioo i)arts\ together with twice their red 

angle. 

Let the straight line AB be divided into any two parts in C ; 
the square of AB is equivalent to the squares of AC and CB, 
together with twice the rectangle under AC and CB. 

On AB describe (L 23) the square of AE, and join BD; 
through C draw (I. 18) CGF parallel to AD or BE; and 
through G draw UK parallel to AB or DE. Then, l)ecause 



G 




/ 



BOOK II.] EUCLID AND LEGENDEE. 51 

CF is parallel to AD, aucl BD falls upon them, the exterior 
angle CGB is equivalent (I. 16) to the inte- 
rior and remote angle ADB ; but ADB is ^ ^ ? 

•equal (I. l) to ABD, because BA and AD 
are equal, being sides of a square; wherefore 
(I. ax. 1) the angle CGB is equal to GBC; 
iiad therefore the side BC is equal (I. 1, cor.) 
to the side CG. But (const.) the figure CK » ^ B 

is a parallelogram ; and since CBK is a right 
angle, and BC equal to CG, CK (I. def 15) is a square, and it 
is upon the side CB. For the like reason HF also is a square, 
and it is upon the side HG, which is equal (I. 15, cor. 1) to AC; 
therefore HF, CK are the squares of AC, CB. And because 
(I. 15, cor. 8) the complements AG, GEare equivalent, and that 
AG is the rectangle contained by AC, CB, for CG has been 
proved to be equal to CB ; therefore GE is also equivalent to 
the rectangle AC.CB; wherefore AG, GE are equivalent to 
twice the rectangle AC.CB. The four figures, therefore, HF, 
-CK, AG, GE are equivalent to the squares of AC, CB, and 
twice the rectangle AC.CB. But HF, CK, AG, GE make up 
the Avhole figure AE, which is the square of AB ; therefore the 
square of AB is equivalent to the squares of AC and CB, and 
twice the rectangle AC.CB. Wherefore, etc. 

Otherwise: AB^ = AB.AC + AB.CB (H. 2). But (H. 3) 
A.B. AC =AC=+ AC.CB, and AB.CB=zCB-+AC.BC. Hence 
<I. ax. 2) AB.AC + AB.BC, or AB-=AC^+CB-+2AC.CB. 

Cor. 1. It follows from this demonstration, that the parallel- 
ograms about the diagonal of a square are likewise squares. 

Cor. 2. Hence the square of a straight line • is equivalent to 
four times the square of its half; for the straight line being 
iDisected, the rectangle of the parts is equivalent to the square 
of one of them. 

Pkop. V. — Theor. — If a straight line be divided into two 
' equal parts, and also into tioo unequal parts ; the rectangle con- 
tained by the unequal parts., together with the square of the line 
hettoeen the points of section, is equivalent to the square of half 
the line. 

Let the straio;ht line AB be bisected in C, and divided un- 





H 


/ 


K L 


/ 



52 THE ELEMENTS OF [bOOK II. 

equally in D ; the rectangle AD.DB, together with the square 
of CD, is equivalent to the square of CB. 

On CB describe (I. 23) the square CF, join BE, and through 
D draw (I. 18) DHG parallel to CE or BF ; also through H 

draw KLM parallel to CB or EF ; and 
■^ c D B through A draw AK parallel to CL or 

BM. Then (I. 15, cor. 5) AL and CM 
^^ are equal, because AC is equal to CB ; 

and (I. 15, cor. 8) the complements CH 
E G F and HF are equivalent. Therefore (I. 

ax. 2) AL and CH together are equal to 
CM and HF together ; that is, AH is equivalent to the gnomon 
CMC To each of those add LG, and (I. ax. 2) the gnomon 
CMG, together with LG, is equivalent to AH together with 
LG. But the gnomon CMG, and LG make up the figure 
CEFB, which is the square of CB ; also AH is the rectangle 
imder AD and DB, because DB is equal (IL 4, cor. 1) to DH ; 
and LG is the square of CD. Thei-efore the rectangle AD.DB 
and the square of CD are equivalent to the square of CB. 
Wherefore, if a straight line, etc. 

Otherwise : Since, as is easily seen from the proof in the 
above, DF is equal to AL, take these separately from the entire 
figure, and there remain AH and LG equivalent to the square 
CF, as before. The proof may also be as follows: 

AD.DB = CD.DB + AC.DB (H. 1) or AD.DB =CD.DB+ 
CB.DB, because CB=:AC. Hence (H. 3) AD.DB =CD.DB + 
CD.DB+DB^=i:2CD.DB + DBl To each of these add CD=; 
then AD.DB + CD^=2CD.DB+DB^+CD^ or (H. 4) AD.DB 
+ CD'=CB\ 

Cor. 1. Hence the ditference of the squares of CB and CD is 
equivalent to the rectangle under AD and DB. But since AC 
is equal to CB, AD is equivalent to the sum of CB and CD, and 
DB is the difterence of these lines. Hence the difference of the 
squares of two straight lines is equivalent to the rectangle under 
their sum and difference. 

Cor. 2. Since the square of CB, or, which is the same, the 
rectangle AC.CB, is greater than the rectangle AD.DB by the 
equare of CD, it follows, that to divide a straight line into two 



BOOK n.] EUCLID AND LEGENDKE. 53 

parts, the rectangle of wliich may be the greatest possible, or, 
as is termed, a tnaxlmiim, tlie line is to be bisected. 

Cor. 3. Hence also tlie sum of the squares of tlie two parts 
into which a straight line is divided, is the least possible, or is, 
as it is termed, a ininimuni, Avlien the line is bisected. For (II. 
4) the square of tlie line is equivalent to the squares of the 
parts and twice their rectangle ; and therefore the greater the 
rectangle is, the less are the squares of the j^arts; but, by the 
foregoins; corollarv, the rectangle is a maximum when the line 
is bisected. 

C<yr. 4. Since (I. 24, cor. 5) the difference of the squares of 
the sides of a triangle is equivalent to the difference of the 
squares of the segments of the base, it follows, from the first 
corollary above, that the rectangle under the sum and differ- 
ence of the sides of a triangle is equivalent to the rectano-le 
under the sum and difference of the segments, intercepted be- 
tween the extremities of the base and the jjoint in which the 
perpendicular cuts the base, or tlie base produced. 

Cor. 5. Hence, also, if a straight line be drawn from the ver- 
tex of an isosceles triangle to any point in the base, or its con- 
tinuation, the difference of the squares of that line and either of 
the equal sides is equivalent to the rectangle under the seoments 
intercepted between the extremities of the base and the jDoint. 

Cor. 6. Since (I. 24, cor. 4) the square of one of the legs of a 
right-angled triangle is equivalent to the difference of the 
squares of the hypothenuse and the other leg, it follows (IT. 5, 
cor. 1) that the square of one leg of a right-angled triangle is 
equivalent to the rectangle under the sum and difference of the 
hypothenuse and the other. 

^cho. And a parallelogram can be constructed equivalent 
to a ffiven triangle, or anv given rectilinear fiorure havinor an 
angle equal to a given angle, or applied to a given straight line 
— that is, having that straight line for one of its sides, when 
the jjarallelogram shall be equivalent to a given triangle or 
given rectilinear figure, and have one of its angles equal to a 
given angle, by applying a parallelogram equivalent to the 
given triangle with an equal angle, to a given straight line, and 
then constructing an equal triangle to the given triangle (I. 15 
and 15, cor. 4). 



A C 


B D 


K 




H 


/ 


L 


/ 





54 THE ELEMENTS OF [bOOK II, 

Prop. VI. — Theok. — If a straight line he bisected, and be 
produced to any point, the rectangle contained by the whole 
line thus produced, and the part of it produced, together with 
the square of half the line bisected, is equivalent to the square 
of the straight line which is made %ip of the half and the p>art 
produced. 

Let the straight line AB be bisected in C, and produced to 
D ; the rectangle AD.DB, and the square of CB, are equiva- 
lent to the square of CD. 

Upon CD describe (I. 23) the square CF, and join DE ; 
through B draw (I. 18) BHG parallel to CE or DF ; through 

H draw KLM parallel to AD or EF, and 
through A draw AK parallel to CL or 
jj DM. Then because AC is equal to CB, 
the rectangle AL is equal (I. 15, cor. 5) to 
CH ; but (1.15, cor. 8) the complements CH, 
J, ^j J, HF are equivalent ; therefore, also, AL is 

equal to HF, To each of these add CM 
and LG ; therefore AM and LG are equivalent to the whole 
square CEFD, But AM is the rectangle under AD and DB, 
because (11. 4, cor. 1) DB is equal to DM; also, LG is the 
square of CB, and CEFD the square of CD. Therefore the 
rectangle AD.DB and the square of CB are equivalent to the 
square of CD. Wherefore, if a straight line, etc. 

Otherwise : Produce CA to N, and make CN equal to CD. 

To these add the equals CB and CA ; 
^ ^ ^ ^ ° therefore NB is equal (L ax. 2) to AD. 

But (IL 5) the rectangle NB.BD, oi- 
AD.BD, together with the square of CB, is equivalent to the 
square of CD ; which was to be proved. 

Prop. VIL — Theor. — If a straight line be divided into a?2y 
two parts, the squares of the whole line and of one of the p>arts 
are equivalent to ticice the rectangle contained by the whole and 
that part, together tcith the square of the other part. 

Let the straight line AB be divided into any two parts in the 
point C ; the squares of AB, BC are equivalent to twice the 
rectangle AB.BC, together with the square of AC. 



BOOK II.] EUCLID AND LEGENDRE. 65 

Upon AB describe (I. 23) the square AE, and construct the 
figure as in the preceding propositions. Then, 
because (I. 15, cor. 8) the complements CII, ^ c B 

FK are equivalent, add to each of thorn CK; 
the whole AK is therefore equal to the whole ^ 
CE ; therefore AK, CE are together double 
of AK. But AK, CE are the gnomon AKF, 



K 



together with the square CK ; therefore the i> F E 

gnomon AKF and the square CK are double 
of AK, or double of the rectangle AB.BC, because BC is equal 
(II. 4, cor. 1) to BK. To each of these equals add HF, which 
is equal (II. 4, cor. 1 ) to the square of AC ; therefore the gno- 
mon AKF, and the squares CK, HF are equivalent to twice the 
rectangle AB.BC and the square of AC. But the gnomon 
AKF, and the squares CK, HF make up the whole figure AE, 
together with CK; and these are the squares of AB and BC ; 
therefore the squares of AB and BC are equivalent to twice the 
rectangle AB.BC, together with the square of AC ; wherefore, 
if a straight line, etc. 

Otherwise: Because (II. 4) AB^ = AC=+BC-+2AC.BC, add 
BC= to both; then AB=+BC==AC= + 2BC=+2AC.BC. But 
(II. 3) BC=+AC.BC= AB.BC, and therefore 2BC=+2AC.BC 
=2AB.BC; wherefore AB-+BC^=AC=+2AB.BC. 

Cor. 1. Since AC is the diflierence of AB and BC, it follows 
that the square of the difterence of two straight lines is equiva- 
lent to the sum of their squares, wanting twice their rectangle. 

Cor. 2. Since (II. 4) the square of the sum of two lines ex- 
ceeds the sum of their squares by twice their rectangle, and 
since, by the foregoing corollary, the square of their difference 
is less than the sum of their squares by twice their rectangle, it 
follows that the square of the sum of two lines is equivalent to 
the square of their difference, together with four times their 
rectangle. 

Pkop. VIII. — TiiEOR. — If a straight line he divided into two 
equal, and also into tvw imequal parts, the squares of the un- 
equal parts are together double of the square of half the linCj 
and of the square of the line hetioeen the points of section. 

Let the straight line AB be divided equally in C, and un- 



56 



THE ELICMENTS OF 



[book n. 



E 



equally in D ; the square? of AD, DB are together double of 
the squares of AC, CD. 

From C draAV (I. 7) CE j)ei-|ie!Klicular to AB, and make it 
equal to AC or CB, and join EA, EB ; through D draw (I. 18) 
DF jjarallel to CE, and tlirougli V draw FG parallel to AB ; 
and join AF. Then because (const.) the triangles ACE, BCE 
are right-angled and isosceles, tlie angles CAE, AEC, CEB, 

O O 7 O 7 3 7 

EBC are (I. 20, cor. 4) each half a riglit angle. So also (I. 16, 
part 2) are EEC, BED, because FG is parallel to AB, and ED 
to EC ; and for the same reason EGF, ADF are right angles. 
The angle AEB is also a right angle, its parts being each half 
a right angle. Hence (I. 1) EG is equal to GF or CD, and ED 

to DB. Again (I. 24, cor. 1) : the square 
of AE is equivalent to the squares of AC, 
CE, or to tAvice the square of AC, since 
AC and CE are equal. In like manner, 
the square of EF is equivalent to twice 
the square of GF or CD. Now (I. 24, 
^ cor. 1), the squares of AD and DF, or 
of AD and DB, are equivalent to the square of AF ; and the 
squares of AE, EF, that is, twice the square of AC and twice 
the square of CD, are also equivalent to the square of AF; 
therefore (I. ax. 1) the squares of AD, DB are equivalent to 
twice the square of AC and twice the square of CD. If, there- 
fore, a straight line, etc. 

Otherwise: DB^+2BC.CD=BC=+CD^ (II. 7), or DB=+ 
2AC.CD=:AC= + CD^-; also (11. 4) AD==:AC^+CD=+2AC. 
CD. Add these equals togethei-, and from the sums take 
2AC.CD; then AD-+DB^=2AC^+2CD^ 




Prop. IX. — ^Theor. — If a straight line he bisected, and pro- 
duced to any point, the squares of the whole line thus produced, 
and of the part of it produced, are together double of the square 
of half the lifie bisected, and of the square of the line made up 
of the half and the part produced. 

Let the straight line AB be bisected in C, and produced to 
D ; the squares of AD, DB arc double of the squares of AC, 
CD. 

From C draw (I. 7) CE perpendicular to AB ; and make it 



BOOK II.] 



EUCLID AND LEGENDRE. 



57 



A 




equal to AC or CB ; join AE, EB, and tln-ough E and D draw 
(I. 1 8) EF parallel to AB, and 
DF parallel to CE. Then be- 
cause the straiu'lit line EF 
meets the parallels EC, FD, 
the ano-les CEF, EFD are 
equivalent (I. 9) to two right 
anixles : and therefore the an- 
gles BEF, EFD are less than 

two right angles; therefore (I. 19) EB, FD will meet, if pro- 
duced toward B, D ; let them meet in G, and join AG. Then 
it would be proved, as in the last proposition, that the angles 
CAE, xVEC, CEB, EBC are each half a right angle, and AEB 
a right angle. BDG is also a right angle, being equal (I. 16) 
to ECB, since (const.) EC, EG are parallel ; DGB, DBG are 
each half a right angle, being equal (I. 10 and I. 11) to CEB, 
CBE, each to each ; and FEG is half a right angle, being (I. 
16) equal to CBE. It would also be proved, as in the last 
proposition, that the squai-e of AE is twice the square of AC, 
and the square of EG twice the square of EF or CD. Now 
(I. 24, cor. l),the squares of AD, DG, or of AD, DB, are equiv- 
alent to tlie sipiare of AG; and the squares of AE, EG, or twice 
the square of AC and twice the square of CD, are also equiva- 
lent to the square of AG. Therefore (I. ax. 1) the squares of 
AD, DB are equivalent to twice the square of AC and twice 
the square of CD. If, therefore, a straight line, etc. 

Otherwise: Produce CA, making CH equal to CD. To 
these add CB, CA ; therefore IIB, AD are 
equal. Then (II. 8) IIB--+BD^ or AD=+ 
BD^ = 2CD=-f2AC-. 

&eho. The nine foregoing propositions may all be proved 
very easily by means of algebra, in connection with the princi- 
ples of mensuration, already established in the corollaries to 
the 23d proposition of the first book. Thus, to pi'ove the fourth 
proposition, let AC=:a, CB = 5, and, consequently, AB=a+5. 
Now, the area of the square described on AB will be found (I. 
23, cor. 4) by multiplying a-\-h by itself. This product is 
found, by performing tlie actual operation, to be «'+2aJ+^'; 
an expression, the first and third parts of which are, by the 



n A 



B D 



58 THE ELEMENTS OF [bOOK H. 

same corollary, the areas of the squares of AC and CB, and the 
second is twice the rectangle of those lines. 

In like manner, to prove the eighth, adopting the same nota- 
tion, Ave have the line which is made up of the whole and CB 
= a + 2$y and, multiplying this by itself, we get for the area ot 
the square of that line, cr+iab+ib", or a'-\-4{a-]-b)b, the 
first part of which is the area of the square of AC, and the 
second four times the area of the rectangle under AB and CB. 

It will be a useful exercise for the student to prove the other 
propositions in a similar manner. He will also find it easy to 
investigate various other relations of lines and their parts by 
means of algebra. 

All the properties delivered in these propositions hold also 
respecting numbers, if products be substituted for rectangles. 
Thus, 7 being equal to the sum of 5 and 2, the square, or sec- 
ond power of 7, is equal to the squares of 5 and 2 and twice 
their product ; that is, 49 = 25 + 4+20. 

Pkop. X. — Peob. — To divide a given straight line into txoo 
parts, so that the rectangle contained by the \chole and one of 
the parts may be equivalent to the square of the other part. 

Let AB be the given straight line ; it is required to divide it 
into two parts, so that the rectangle under the whole and one 
of the parts may be equivalent to the square of the other. 

Upon AB describe (I. 23) the square AD ; bisect (I. 6) AC 
in E, and join E with B, the remote extremity of AB ; produce 
CA to F, making EF equal to EB, and cut off AH equal to AF ; 
AB is divided in H, so that the rectangle AB.BH is equivalent 
to the square of AH. 

Complete the parallelogram AG, and produce GH to K. 
Then, since BAC is a right angle, FAH is also (I. 9) a right 
angle; and (I. def 15) AG is a square, AF, AH being equal 
by construction. Because the straight line AC is bisected in 
E, and produced to F, the rectangle CF.FA and the square of 
AE are together equivalent (II. 6) to the square of EF or of EB, 
since (const.) EB, EF are equal. But the squares of BA, AE 
are equivalent (I. 24, cor. 1) to the square of EB, because the 
angle EAB is a right angle; therefore the rectangle CF.FA 
and the square of AE are equivalent (I. ax. 1) to the squares of 



BOOK n.] 



EUCLID AND LEGENDRE. 



59 



BA, AE. Take away the square of AE, -u-hich is common to 
both; therefore the remainino- rectano-le CF.FA 



F 



G 



E 





H 


-" 





B 



D 



is equivalent (I. ax. 3) to the square of AB. 

But the figure FK is the rectangle contained 

by CF, FA, for AF is equal to FG ; and AD 

is the square of AB ; therefore FK is equal 

to AD. Take away the common part AK, 

and (I. ax. 3) the remainders FH and HD are 

equivalent. But HD is the rectangle AB.BH, 

for AB is equal to BD ; and FH is the square 

of AH. Therefore the rectangle AB.BH is equivalent to the 

square of AH ; Avherefore the straight line AB is divided in H, 

so that the rectangle AB.BH is equal to the square of AH ;, 

which was to be done. 

Sc/w. 1. In the practical construction in this proposition, and 
in the 2d cor. to 9 of the fifth book, which is virtually the same, it 
is sufficient to draw AE perpendicular to AB, making it equal 
to the half of AB, and producing it through A ; then, to make 
EF equal to the distance from E to B, and AH equal to AF. 
It is plain that BD might be bisected instead of AC, and that 
in this way another point of section would be obtained. 

While the enunciation in the text serves for ordinary pur- 
poses, it is too limited in a geometrical sense, as it comprehends 
only one case, excluding another. The following include& 
both : 

In a given straight line^ or its continuatio7i, to find ai^ointy 
such that the rectangle contained hy the given line^ and the seg- 
ment between one of its extremities and the required point, may 
be equal to the square of the segment between its other extremity 
and the same point. 

The point in the continuation of BA will be found by cutting 
ofiT a line on EC and its continuation, equal to EB, and describ- 
ing on the line composed of that line and AE a square lying on 
the opposite side of AC from AD ; as the angular j^oint of that 
square in the continuation of BA is the point required. The 
proof is the same as that given above, except that a rectangle 
corresponding to AK is to be added instead of being sub- 
tracted. 

Scho. 2. The line CF is equivalent to BA and AH ; and since 



60 THE ELEMENTS OF [bOOK H, 

it has been shown that the rectangle CF.FA is eqiiivalent to 
the square of BA or CA, it follows that if any straight line AB 
(see the next diagram) be divided according to this proposition 
in C, AC being the greater part, and if AD be made equal to 
AB, DC is similarly divided in A. So also if DE be made 



F 



E D A C B 



equal to DC, and EF to EA, EA is divided similarly in D, and 
FD in E ; and the like additions may be continued as far as we 
please. 

Conversely, if any straight line FD be divided according to 
this proposition in E, and if EA be made equal to EF, DC to 
DE, etc., EA is similarly divided in D, DC in A, etc. It fol- 
lows also, that the greater segment of a line so divided will be 
itself similarly divided, if a part be cut oif from it equal to the 
less ; and that by adding to the whole line its greater segment, 
another line will be obtained, which is similarly divided. 

Prop. XI. — Theok. — In an oUuse-angkd triangle^ the 
square of the greatest side exceeds the squares of the other tioo^ 
hy tioice the rectangle contained by either of the last-mentioned 
sides, and its continuation to meet a perpendicular drawn to it 
from the opposite angle. 

Let ABC be a triangle, having the angle ACB obtuse ; and 
let AD be perpendicular to BC produced ; the square of AB is 
equivalent to the squares of AC and CB, and twice the rectan- 
gle BC.CD. 

Because the straight line BD is divided into two parts in the 

point C, the square of BD is equivalent (II. 4) to the squares 

of BC and CD, and tAvice the rectangle 

BC.CD. To each of these equivalents add 

the square of DA ; and the squares of DB, 

DA are equivalent to the squares of BC, 

CD, DA, and twice the rectangle BC.CD. 

But, because the angle D is a right angle, 

the square of BA is equivalent (I. 24, cor. 1) to the squares of 

BD, DA, and tlie square of CA is equivalent to the squares of 




BOOK II.] EUCLID AND LEGENDRE. 61 

CD, DA ; therefore the square of BA is equivalent to the 
squares of BC, CA, and twice the rectangle BC.CD. Therefore, 
in an obtuse-angled triangle, etc. 

Pkop. XII. — TiiEOR. — I/i any triangle^ the square of a side 
subtending an acute a?igle is less than the squares of the other 
sides, by twice the rectangle contained by either of those sides, 
and the straight line hitercej^ted between the acute angle and 
the perpendicular drawn to that side from tJie opposite angle. 

Let ABC (see this figure and that of the foregoing proposi- 
tion) be any triangle, having the angle B acute ; and let AD 
be perpendicular to BC, one of the sides containing that angle; 
the square of AC is less than the squares of AB, BC, by twice 
the rectangle CB.BD. 

The squares of CB, BD are equivalent (II. 1) to twice the 
rectangle contained by CB, BD, and the square of DC. To 
each of these equals add the square of 
AD ; therefore the squares of CB, BD, 
DA are equivalent to twice the rect- 
angle CB.BD, and the squares of AD, 
DC. But, because AD is perpendicu- 
lar to BC, the square of AB is equiva- 
lent (I. 24, cor. 1) to the squares of 

BD, DA, and the square of x\.C to the squares of .VD, DC ; 
therefore the squares of CB, BA are equivalent to the square of 
AC, and twice the rectangle CB.BD ; tliat is, tlie square of AC 
alone is less than the squares of CB, BA, by twice the rectangle 
CB.BD. 

If the side AC be perpendicular to BC, then BC is the 
straight line between the perpendicular and the acute angle at 
B ; and it is manifest that the squares of AB, BC are equiva- 
lent (I. 24, cor. 1) to the square of AC and twice the square of 
BC. Therefore, in any triangle, etc. 

Scho. By means of this or the foregoing proposition, the area 
of a triangle may be computed, if the sides be given in num- 
bers. Thus, let AB = 17, BC = 28, and AC = 25. From AB=+ 
BC take AC; that is, from l7'4-2SUake 25=; the remainder 
448 is twice CB.BD. Dividing this by 56, twice BC, the quo- 
tient 8 is BD, Hence, from either of the triangles ABD, 




62 



THE ELEMENTS OF [bOOK II. 



ACD, we find the perpendicular AD to be 15 ; and thence the 
area is found, by taking half tlie product of BC and AD, to be 

210. 

The segments of the base are more easily found by means -of 
tJie 4th corollary to the fifth proposition of this book, in connec- 
tion with the pi-inciple, that if half the difference of two mag- 
nitudes le added to half their sum ^ the resvlt is the greater; and 
if half the difference he taken from half the sum, the remainder 
is the less. Thus, if 42, the sum of AB and AC, be multiplied 
Tjy 8, their difference, and if the product, 336, be divided by 28, 
the sum of the segments of the base, the quotient 12 is their 
difference. The half of this being added to the half of 28, the 
sum 20 is the greater segment CD ; and being subtracted from 
it, the remainder 8 is BD. 

To prove the principle mentioned alxjve, let AB be the 

greater, and BC the less of two 
^ ^ ^ ^ ^ magnitudes. Bisect AC in D, and 

' make AE equal to BC. Then AD 

or DC is half the sum, and ED or 
DB half the difference of AB and BC ; and AB the greater is 
equivalent to the sum of AD and DB, and BC the less is equiv- 
alent to the difference of DC and DB. 

Cor. Hence, when AP (I. 24, cor. 6) is not perpendicular to 
BC — the truth of the corollary can be shown by this and pre- 
-ceding propositions. 

Hence, if the sides of a ti-iangle be given in numbers, the line 
AD can be computed. Thus, if AC^ll, AB = 14, and BC = 7, 
we have AC-f BC= = 121 + 49 = 170, and 2AD-=:98. Then 170 
— 98 = 72, the half of Avhich is 36 ; and 6, the square root of 
this, is CD. 

Prop. XHI. — Vrov,.— To describe a square that shall he 
•equivalent to a given rectilineal figure. 

Let A be the given rectilineal figure ; it is required to de- 
scribe a square that shall be equal to it. 

Describe (H. 5 scho.) the rectangle BD equal to A. Then, if the 
sides of it, BE, ED, be equal to one another, it is a square, and 
what was required is done. But if they be not equal, produce 
one of them BE to F, and make EF equal to ED; bisect BF ia 



BOOK II.] 



EUCLID AND LEGENDEE. 



63 




O, and from the center G, at the distance GB, or GF, describe 
(I. post. 3) the semicircle BIIF; 
produce DE to H, and join GH. 
Therefore, because the straight 
line BF is divided equally in G, 
and unequally in E, the rectangle 
BE.EF, and the square of EG, 
are equivalent (II. 5) to the 

square of GF, or of GH, because GH is equal to GF. But the 
squares of HE, EG are equal (I. 24, cor, l) to the square of GH ; 
therefore the rectangle BE.EF and the square of EG are equiv- 
alent to the squares of HE, EG. Take away the square of EG, 
which is common, nvA the remaining rectangle BE.EF is equiv- 
alent to the square of EH. But the rectangle contained by 
BE, EF is the parallelogram BD, because EF is equal to ED ; 
therefore BD is equivalent to the square of EH ; but BD is 
equivalent to the figure A; therefore the square of EH is 
<^quivalent to A. The square described on EH, therefore, is the 
required square. 

Scho. Tliis is a particular case of the sixteenth proposition 
of the fifth book, and an easy solution can be effected by means 
of Proportion. 

Pkop. XIY. — TiiEOR. — T/ie siwi of the squares of the sides 
of a trapezium is equivalent to the sum of the squares of the 
diagonals, tor/ether icith four times the square of the straight 
line joining the points of bisection of the diagonals. 

Let ABCD be a trapezium, having its diagonals AC, BD 
bisected in E and F, aud let EF be joined ; the squares of AB, 
BC, CD, DA are together equivalent to the 
squares of AC, BD, together with four 
times the squai-e of EF. 

Join AF, EC. The squares of AB, AD 
are together equivalent (II. 12, cor.) to 
twice the sum of the squares of DF and 
AF; and the squares of BC, CD are equiv- 
alent to twice the sum of the squares of DF and CF. Add 
these equivalents together, and the sum of the squares of AB, 
BC, CD, DA is equivalent to four times the square of DF, to- 



B 




64 THE ELEMEXT3 OF [BOOK H. 

getlier with twice the sum of the squares of AF and CF. But 
tv/ice the squares of AF and CF are equivalent (11. 12, cor.) to four 
times the squares of AE and EF; and (II. 4, cor. 2) four times 
the square of DF is equivalent to the square of BD, and four 
times the square of AE to the square of AC. Hence the squares 
of AB, BC, CD, DA are equivalent to the squares of AC and 
BD, together with four times the square of EF. Therefore, the 
sum, etc. 

Cor. Hence the squares on the diagonals of a parallelogram 
are together equivalent to the sum of the squares on its sides — 
for in the case of a parallelogram the line EF vanishes, as the 
diagonals of a parallelogram bisect each other. 

/S'cAo. Hence, if we have the sides and one of the diagonals 
of a parallelogram in numbers, we can compute the remaining 
diagonal. Thus, if AB, DC be each =9, AD, BC each = 7, and 
AC =8, we have AB^+BC'+CD= + DA= = 81 + 49+81 + 49 = 
260, and AC-r=64. Taking the latter from the fonner, and ex- 
tracting the square root, we find BD = 14. 



END OF BOOK SECOND. 



BOOK THIRD. 

ON THE CIRCLE, AND LINES AND ANGLES DE- 
PENDING ON IT, AND RECTILINEAL FIGURES 
DESCRIBED ABOUT THE CIRCLE. 

DEFIXinOIfS. 

1. A STRAIGHT line is said to touch a circle, or to be a tangent 
to it, when it meets the circle, and being produced does not 
cut it. 

2. Circles are said to touch one another, which meet, but do 
not cut one another. 

3. In a circle, chords are said to be equally distant from the 
center, when the perpendiculars drawn to them from the center 
are equal. 

4. And the chord on which the greater perpendicular falls, 
is said to \)e farther from the center. 

5. An angle in a segment of a circle is an angle contained 
by two straight lines drawn from any point in the arc of the 
segment to the extremities of its chord ; 

6. And an angle is said to stand upoti the arc intercepted 
between the straight lines that contain the angle. 

7. A sector of a circle is a figure contained by any arc of the 
circle, and two radii drawn through its extremities. 

8. A quadrant is a sector whose radii are perpendicular to 
each other. It is easy to show by superposition, that a quad- 
rant is half of a semicircle, and therefore a fourth part of the 
entire circle. 

9. Similar segments of circles are those which contain equal 
angles. 

10. Concentric circles are those which have the same center. 

11. A regular polygon is equilateral. 

12. When the sides of one rectilineal figure pass through the 

5 



66 THE ELEMENTS OF [bOOK HI. 

angular points of another, the figures not coinciding with one 
another, the interior figure is said to be inscribed in the exte- 
rior, and the exterior to be circutnscribed, or described, about 
the interior one. 

13, When all the angular points of a rectilineal figure are 
upon the circumference of a circle, the rectilineal figure is said 
to be inscribed in the circle, and the circle to be circumscribed, 
or described, about the rectilineal figure. 

14. When each side of a rectilineal figure touches a circle, 
the rectilineal figure is said to be circumscribed, or described, 
about the circle, and the circle to be inscribed in the rectilineal 
figure. 

PROPOSITIONS. 

Prop. I. — Prob. — To find the center of a given circle,. 

Let ABC be the given circle, and draw any chord AB. 

Bisect AB (I. 6) by the perpendicular 
EDO drawn to the circumference on 
both sides AB. 

Since EDC bisects AB (const.) and is 

perpendicular to AB (L 6), the angles 

AEB and ACB are also bisected by 

EDC, and the subtended arcs ACB and 

AEB (I. 5) are likewise bisected by 

EDC; therefore the arcs CA+AEo 

the arcs CB+BE; hence CA + AE or 

CE is the semicircumference, and the perpendicular EDC is a 

diameter — consequently it passes through the center of the 

circle. 

Then taking any other chord FIT, and in same manner it can 
be shown that the perpendicular MN which bisects the chord 
is also a diameter, and since all diameters pass through the 
center of the circle, the point at which they intersect each 
other, being the only point they have in common, that point is 
the center of the circle. 

Cor. 1. Hence, to find center of any regular polygon (L 
6), bisect the sides by perpendiculars drawn from the points 
of bisection (I. 7), and the point where the perpendiculars inter- 
eect each other is the center of the polygon. 




BOOK m,] EUCLID AND LEGENDRB. 67 

Cor. 2. In a triangle, straight lines drawn from the points 
of bisection of the three sides to the opposite angles all pass 
throue'h the same point. 

8cho. From the preceding coroUarj'^ it can be shown that 
each of the straight lines is divided into two segments at the 
common point of bisection, of which the segment nearest the 
angle is double the other. 

Prop. II. — Theor. — If a straight line drav^nfrom the center 
of a circle bisect a chord which does not pass through the ce?i- 
ter, it cicts it at right angles/ and {2) if it cut it at right angles, 
it bisects it. 

Let ABC be a circle ; and let ED, a straight line drawn 
from the center E, bisect any chord AB, which does not pass 
through the center, in the point D ; ED cuts AB at right 
angles. 

Join EA, EB. Then in the triangles ADE BDE, AD is 
equal to DB, DE common, and (I. def 16) 
the base EA is equal to the base EB ; 
therefore (I. 4) the angles ADE, BDE are 
equal; and consequently (I. def 10) each 
of them is a right angle; wherefore ED 
cuts AB at right angles. 

Next, let ED cut AB at right angles ; ED also bisects it. 

The same construction being made, because the radii EA, 
EB are (I. def 16) equal, the angle EAD is equal (I. 1, cor.) to 
EBD ; and the right angles ADE, BDE are equal ; therefore 
in the two triangles EAD, EBD there are two angles in one 
equal to two angles in the other, each to each, and the side 
ED, which is opposite to one of the equal angles in each, is 
common; therefore (L 14) AD is equal to DB. If a straight 
line, therefore, etc. > 

Cor. 1. Hence, in an isosceles triangle, a straight line drawn 
from the vertex bisecting the base is perpendicular to it ; and 
a straight line drawn from the vertex perpendicular to the base, 
bisects it. 

Cor. 2. Let the straight line AB cut the concentric circles 
ABC, DEF in the points A, D, E, B ; AD is equal to EB, and 
AE to DB. From the common center G, draw GH perpen- 




68 



THE ELEMEi^TS OF 



[book ni. 



dicular to AB. 




Then (III. 2) AH is equal to HB, and DH to 
HE. From AH take DH, and from HB 
take HE, and the remainders, AD, EB, 
are equal. To these equals add DE, and 
the sums AE, DB are equal. 

Cor. 3. Any numljer of parallel chords 
in a circle are all bisected by a diameter 
perpendicular to them. 



Pkop. hi. — Theok. — Tico chords of a circle which are not 
both diameters, can not bisect each other. 

From the definition of the circle, the center is the only point 

in the circle which is equally distant 
D from all parts of the circumference ; then 

any chord which passes through the 
center is bisected at the center ; the 
diameter is the only chord (HI, 1) 
which passes through the center, there- 
fore any other chord AB can not bisect 
a diameter CD. And when neither chord 
AB nor EF is a diameter, their point of 
intersection not being- the center of the 
circle, is unequally distant from the circumference ; therefore 
the chords AB and EF are not bisected by each other. 




Prop. IV. — Theor. — Iftico circles cut 09ie another, they have 
not the same center. 

Let ABC and DBE be two circles which cut one another in 

B; they will not have the 
same center. For AC is the 
diameter of ABC (III. 1), and 
DE is the diameter of DBE. 
But the intersection of the 
two diameters AC and FH 
is the center of the circle 
ABC (in. 1), and the inter- 
section of the two diameters 
LM and DE is the center of the circle DBE. Now the points 




BOOK m.] EUCLID AND LEGENDRE. 69 

of intersection of these diameters are different, therefore ABC 
has not the same center with DBE. Wherefore, if two cir- 
cles, etc. 

Cor. 1. Hence, if one circle touches another internally, they 
have not the same center. 

Cor. 2. One circle can not cut another in more than two 
points, nor touch another in more than one point. 

Prop. V. — Theoe. — If from any jjoint icithin a circle, 
which is not the center, straight lines be drawn to the circum- 
ference ; (1) the greatest is that tchich passes through the cen- 
ter, and (2) the continuation of that line to the circumference, 
in the opposite direction, is the least y (3) of others, one nearer 
to the line passing through the center is greater than one more 
remote ; and (4) from, the same point there can be drawn ordy 
two equal straight lines, one upon each side of either the longest 
or shortest line, and making equal angles vnth that line. 

Let ABCD be a circle, E its center, and AD a diameter, in 
which let any point F be taken, which is not the center ; of all 
the straight lines FA, FB, FC, etc., that can be drawn from F 
to the circumference, FA is the greatest, and FD the least ; and 
of the others, FB is greater than FC. 

1. Join BE,CE. Then (I. 21, cor.) BE, EF are greater than 
BF ; but AE is equal to EB ; therefore AF, that is, AE, EF, is 
greater than BF. 

2. Because CF, FE are greater (I. 21, cor.) than EC, and EC 
is equal to ED ; CF, FE are greater than ED. Take away the 
common part FE, and (I. ax. 5) the re- 
mainder CF is greater than the remain- 
der FD. 

3. Again : because BE is equal to CE, 
and FE common to the triangles BEF, 
CEF ; but the angle BEF is greater than 
CEF ; therefore (L 22) the base BF is 
greater than the base CF. 

4. Make (I. 13) the angle FEH equal 

to FEC, and join FH. Then, because CE is equal to HE, EF 
common to the two triangles CEF, HEF, and the angle CEF 
equal to the angle HEF ; therefore (I. 3) the base FC is equal 




TO THE ELEMENTS OF [bOOK III. 

to the base FH, and the angle EFC to the angle EFH. 
But, besides FH, no other straight line can be drawn from F 
to the circumference equal to FC (L ax, 9). 

Prop, VI, — Theok, — If from any point without a circle 
straight lines he drawn to the circrimference; (l) of those which 
fall npon the concave part of the circumference, the greatest is 
that which passes through the center ; unci (2) of the rest, one 
nearer to the greatest is greater than one more remote. (3) 
Hut of those which fall upon the convex part, the least is that 
which when produced passes through the center' and (4) of the 
rest, one nearer to the least is less than one more remote. And 
(5) only two equal straight lines can be drawn from the point 
to either part of the circumference, one upon each side of the 
line passing through the center, and making equal angles 
with it. 

Let ABF be a circle, M its center, and D any point without 
it, from which let the straight lines DA, DE, DF be drawn to 
the circumference. Of those which fall upon the concave pait 
of the circumference AEF, the greatest is DlNLiV, which passes 
through the center ; and a line DE nearer to it is greater than 
DF, one more remote. But of those which fall upon the con- 
vex circumference LKG, the least is DG, the external part of 
DMA ; and a line DK nearer to it is less than DL, one more 
remote, 

1. Join ME, MF, ML, MK ; and because MA is equal to ME, 
add MD to each ; therefore AD is equal to EM, MD ; but (I. 
21, cor,) EM, MD are together greater than ED; therefore, also, 

AD is greater than ED. 

2. Because ME is equal to MF, and 
MD common to the triangles EMD, 
FMD, but the angle EMD is greater than 
FMD ; therefore (L 22) the base ED is 
greater than the base FD. 

3, Because (I, 21, cor.) MK, KD are 
greater than MD, and MK is equal to 
MG, the remainder KG is greater (I, ax. 
5) than the remainder GD ; that ie, GD L* 
less than KD. 




BOOK m.] 



EUCLID AND LEGENDKE. 



71 



4, Because MK is equal to ML, and MD common to the 
triangles KMD, LMD, but the angle DMK less than DML; 
therefore the base DK is less (I. 22) than the base DL. 

5. Make (I. 13) the angle DMB equal to DMK, and join 
DB. Then, because MK is equal to MB, MD common to the 
triangles KMD, BMD, and the angle KMD equal to BMD ; 
therefore (I. 3) the base DK is equal to the base DB, and the 
angle MDK to the angle MDB, But, besides DB, there can 
be no straight line drawn from D to the circumference equal to 
DK (I. ax. 9). 



Prop. VII. — Theoe. — Equal chords in a circle are equally 
distant from the center ; and (2) chords which are equally dis- 
tant from the center are equal to one another. 

Let the chords AB, CD, in the circle ABDC, be equal to one 
another; they are equally distant from the center. 

Take (TIL 1) E the center of the circle, and draw (L 7) EF, 
EG perpendiculars to AB, CD ; join also EA, EC. Then, be- 
cause the straight line EF, passing through the center, cuts the 
chord AB, which does not pass through the center, at right 
angles, it also (III. 2) bisects it ; where- 
fore AF is equal to FB, and AB is dou- 
ble of AF. For the same reason, CD is 
double of CG ; but AB is equal to CD ; 
therefore AF is equal (I. ax. V) to CG. 
Then, in the right-angled triangles EF A, 
EGC, the sides EA, AF are equal to the 
sides EC, CG, each to each, therefore 
(L 3) the sides EF, EG are equal. But 

chords in a circle are said (IIL def, 3) to be equally distant 
from the center, when the perpendiculars drawn to them from 
the center are equal; therefore AB, CD are equally distant 
from the center. 

Next, if the chords AB, CD be equally distant from the cen- 
ter, that is, if FE be equal to EG, AB is equal to CD. For, 
the same construction being made, it may, as before, be demon- 
strated that AB is double of AF, and CD of CG ; and because 
tlie right-angled triangles EFA, EGC have the sides AE, EF 
equal to CE, EG, each to each, the sides AF, CG are also (L 3) 




72 THE ELEMENTS OF [bOOK m. 

equal to one another. But AB is double of AF, and CD of 
CG ; wherefore AB is equal (I. ax. 6) to CD. Therefore equal 
chords, etc. 

Cor. Hence, the diameter of a circle is the greatest chord ; 
(2) of others, one nearer to the center is greater than one more 
remote ; and (3) the greater is nearer to the center than the 
less. 

Prop. VIII. — Thkoe. — The straight line drawn perpendicu- 
lar to a diameter of a circle^ through its extremity.^ falls icith- 
out the circle ; hut any other straight line drawn through that 
point cuts the circle. 

Let ABC be a circle, of which D is the center, and AB a 
diameter ; if AE be drawn through A perpendicular to AB, it 
falls without the circle. 

In AE take any i:)oint F ; and draw 
DF, meeting the circumference in C. 
Because DAF is a right angle, it is 
greater (I. 20) than DFA; and there- 
fore (I. 21) DF is greater than DA. 
But (I. def. 16) DA is equal to DC; 
therefore DF is greater than DC, and 
the point F is therefore without the 
circle ; and in the same manner it may 
be shown, that any other point in AF, 
except the point A, is without the circle. 

Again : any other straight line drawn through A cuts the 
circle. 

Let AG be drawn in the angle DAF, and draw (I. 8) DH 
perpendicular to AG, and meeting the circumference in C. 
Then, because DHA is a right angle, and DAH less than a 
right angle, being a part of DAE, the side DH is less (I. 20) 
than the side DA, But (L def 16) DK is equal to DA; there- 
fore DH is less than DK ; the point H is therefore within the 
circle ; and AG cuts the circle, since its continuation through 
A must fall on the opposite side of EAL, and must therefore be 
without the circle. Therefore the straight line, etc. 

Cor. From this it is manifest that the straight line which is 
drawn at right angles to a diameter of a circle from its extrem- 




BOOK in.] 



EUCLID AND LEGENDRE. 



73 



ity touches (III. def. 1) the ch'cle; and that it touches it only 
in one poiut, because at every point except A, it falls without 
the circle. It is also evident, that there can be but one tan- 
gent at the same point of a circle. 

Pkop. IX. — Pbob. — From a given pointy either -without a 
given circle^ or in its circumference^ to draw a straight line 
touchi^ig the circle. 

First : let A be a given point without the given circle BCD ; 
it is required to draw from A a straight line touching the 
circle. 

Find (III. 1) E the center of the circle, and draw AE cutting 
the circumference in D ; from the center E, at the distance EA, 
describe (I. post. 3) the circle AFG; 
from D draw (I. 7) DF at right an- 
gles to EA; and draw EBF, and 
join AB. AB touches the circle 
BCD. 

Because E is the center of the cir- q 
cles, EA is equal to EF, and ED to 
EB ; therefore the two sides AE, 
EB are equal to the two FE, ED, 
each to each, and they contain the 
angle AEF common to the two tri- 
angles AEB, FED ; therefore the angle EBA is equal (I. 3) 
to EDF, and is, therefore, a right angle, because (const.) 
EDF is a right angle. Now, since EB is drawn from the cen- 
ter, it is part of a diameter of which B is one extremity ; but a 
straight line drawn from the extremity of a diameter at right 
angles to it touches (III. 8, cor.) the circle; therefore AB 
touches the circle ; and it is drawn from the given point A ; 
which was to be done. 

Secondlj : if the given point be in the circumference of the 
circle, as the point D, draw DE to the center E, and DF at 
right angles to DE ; DF touches (III 8, cor.) the circle. 

Cor, 1. If AB be produced to H, AH is bisected (III. 2) in 
B. Hence a chord in a circle touching a concentric one is 
bisected at the point of contact. 

Scho. 1. It is evident that from any point A without the cir- 





74 THE ELEMENTS OF [BOOK HI. 

cle, two tangents may be drawn to the circle, and that these 
are equal to one another, being equal respectively to the equal 
lines DF and DF'. 

Scho. 2. The construction of the first case of this problem is 
as easily efiected in practice, by describing a circle on AE as 
diameter, as its circumference will cut that of the given circle 
in the points B and B^ The reason of this will be evident 
from the twelfth proposition of this book. 

Cor. 2. Hence in a given straight line AB, a point may be 
found such that the difference of its distances from two given 

points C, D, may be equal to a given 
straight line. Join CD, and from D as 
center, with a radius, DE equal to the 
given difference, describe a circle ; draw 
CP"G perpendicular to AB, and make FG 
D equal to FC ; through C, G describe a cir- 
cle touching the other circle ; join B, D, 
the centers of the two circles, and draw BC ; BC, BD are evi- 
dently the required lines. 

Cor. 3. In the same manner, if a circle were described fi-om 
D as center, with the sum of two given lines as radius, and a 
circle were described through C and G touching that circle, 
straight lines drawn from the center of that circle to C and D 
would be equal to the given sum. 

Pkop. X. — Theor. — The angle at the center of a circle is 
double of the angle at the circuntference., xipon the same base^ 
that is, upon the same part of the circumference. 

In the circle ABC, let BEC be an angle at the center, and 
BAC an angle at the circumference, which 
have the same arc BC for their base ; BEC 
is double of BAC. 

Draw AE, and produce it to F ; and 
first, let E, the center of the circle, be 
wiiliin the angle BAC. Because EA is 
equal to EB, the angle EAB is equal (I. 1, 
"p ^ cor.) to EI>A ; therefore the angles EAB, 

EBA are together double of EAB ; but (I. 20) the angle BEP 
is equal to EAB, EBA; therefore also BEF is double of EAB. 





BOOK in.] EUCLID AND LKGENDRE. 75 

For the same reason, the angle FEC is double of the angle 
EAC ; therefore the whole angle BEC is double of the whole 
BAG. 

Again : let E the center of the circle be without the an^le 

CD O 

BAG ; it may be demonstrated, as in the 
first case, that the angle FEG is double of 
FAG, and that FEB, a part of the first, is 
double of FAB a part of the other; there- 
fore the remaining angle BEC is double of 
the remaininsj ansjcle BAG. The angle at 
the center, therefore, etc. 

Scho. That if two magnitudes be double 
of two others, each of each, the sum and diff'erence of the first 
two are respectively double of the sum and difference of the 
other two. It is thus proved by Play fair: "Let A and B, G 
and D be four magnitudes, such that A = 2G, and B = 2D; 
then A+B = 2(G + D). For since A = G + G, and B=D-f-D, 
adding equals to equals, A4-B = (G + D) + G + D = 2(G + D). 
So, also, if A be greater than B, and therefore G greater than 
D, since A=:G4-G, and B=D4-D, taking equals from equals, 
A— B = (G— D) + (G— D), that is A— B = 2(G— D)." 

The following is an outline of another proof of the second 
case: from the triangle BGE we have (I. 20) BGG=BEG-!-B 
=:BEG4-EAG (I. 1). We have also from the triangle AGG, 
in a similar manner, BGG = BAG + G = BAG + EAG = 2BAG + 
EAG. Hence (I. ax. 1) BEG + EAG = 2BAG + EAG. Take 
away EAG, etc. If in the first diagram, GE were produced to 
meet AB, the first case might be proved in a similar manner. 
The second case might also be proved by drawing from G to 
AB a straight line meeting it in a point H, and making Avith 
AC an angle equal to BAG. Then, by taking the difference 
between the equal angles EAC, EGA, and the equal ones HAG, 
lie A, we have EAB and EGH equal, and therefore EBA= 
EGH. But EGB=HGG; and therefore (I. 20, cor. 5) BEG = 
BHG=:2BAG. Tl e same proof, with some obvious variations, 
would be applicable in the first case. 

There is evidently a third case, viz., when AB or AG passes 
through the center; but though this case is not given in a sep- 
arate form, its proof is contained in that of either of the others. 



76 



THE ELEMENTS OF 



[book nic 



PpwOp. XL — Theor. — In a circle, (1) the angle in a semicir- 
cle is a right angle ; (2) the angle in a seginent greater than a 
semicircle is acute ; and (3) the angle in a segment less than a 
semicircle is obtuse. 

Let ABC be a circle, of which F is the center, BC a diame- 
ter, and consequently BAG a semicircle ; and let the segment 
BAD be greater, and BAE less than a semicircle ; the angle 
BAG in the semicircle is a right angle ; but the angle BAD 

in the segment greater than a semicircle 
is acute ; and the angle BAE in the seg- 
ment less than a semicircle is obtuse. 

Draw AF and produce it to G. Then 
(III. 10) the angle BAG at the circum- 
ference, is half of BFG at the center, 
both standing on the same arc BG ; and 
for the same reason, GAG is half of 
GFC. Therefore the whole angle BAG is half of the angles 
BFG, GFG ; and (I. 9) these are together equivalent to two 
right angles ; therefore DAG is a riglit angle, and it is an angle 
in a semicircle. But (I. ax. 9) the angle BAD is less, and BAE 
greater than the right angle BAG ; therefore an angle in a seg- 
ment greater than a semicircle is acute, and an angle in a seg- 
ment less than a semicircle is obtuse. 




Pbop. XII. — Theoe. — If a straight line touch a circle, the 
straight line drawn from the center to the point of contact is 
'perpendicular to the line touching the circle. 
Let the straight line FH touch the circle ABGD at the point 

C ; the straight line GA drawn from 
that point to the center of the circle is 
perpendicular to the line touching the 
circle. Draw a diameter BD parallel to 
FH (I. 18). and with the diameter 
BD as a base, and the point C as a ver- 
tex, make the triangle BGD. But BGD 
is a right angle (III. 11). FGB and 
HCD are equivalent to a right angle 
(L 9), and because BD is parallel to FH (const.), FGB is 
equal to GBD, and HGD is equal to CDB (L 15, cor. 2) ; and 




BOOK in.] 



EUCLID AND LEGENDKE. 



77 



at the point C draw a perpendicular CA to FH (I. 7), it will 
also be perpendicular to BD (I. 17), then FCB and BCE are 
equivalent to BCD (I. ax. 1) ; taking a^vay the common angle 
BCE, we have FCB equal to ECD ; but FCB is equal to 
CBD, hence ECD is equal to CBD ; and in same manner it can 
be shown that ECB is equal to CDB. The triangles EBC and 
EDC having EC common, the angle EBC equal to the angle 
ECD, and the angles BEC and CED both right angles, are 
equal (I. 3), therefore EB is equal to ED, and the diameter BD 
is bisected by CA, then CA passes through the center of the 
circle (III. 1). Wherefore, if a straight line, etc. 

Cor. Hence, conversely, if a straight line touch a circle, a 
straight line drawn from the point of contact, perpendicular to 
the tangent, passes through the center. 



Prop. XIII. — Theoe. — If one circle touch another inter- 
nally or externally in any point, the straiyht line rohich joins 
their centers being jyroducecl, passes through that point. 

Let al)Q and DEC be two circles which touch one another 
internally, their centers will be in the same straight line with 
their point of contact. 

At the point of contact C draw the tangent FH, and at this 
point erect a perpen- 
dicular CaE (I. 7). B p D 
The diameter of aJC 
is perpendicular to the 
tangent at the point of 
contact (HI. 12), hence 
Ca is the diameter of 
a6C, and passes 
through the center of 
a5C(IH. 1). NowFH 

is also tangent to the circle DEC (const.) at the point C ; 
hence, the perpendicular CE (HI. 12) is the diameter of DEC, 
and passes (HI. 1) through the center of DEC. But Ca and 
CE are in the same straight line, hence the centers of the cir- 
cles ahQi and DEC, and the point of contact C, are in the 
straight line CaE. Wherefore, if one circle touch, etc. 

Or, let ABC and CDE be two circles which touch one an- 




78 THE ELEMENTS OF [bOOK IH. 

Other externally in the point C ; their centers will be in the 
same straight line with the point of contact. 

At the point C draw a tangent FH, which will be perpen- 
dicular to the diameter of ABC (III. 12). FH being tangent 
at the point of contact of the circles, is also pei-pendicular to 
the diameter of CDE (IIL 12). ACF is a right angle (I. def. 
10), and FCE for the same reason is a right angle, and both 
equal to one another (I. def, 10, and ax. 11), hence are two 
right angles ; then (I. 10) AC and CE form the same straight 
line; but AC passes through the center of ABC, and CE passes 
through the center of CDE, being their respective diameters 
(IIL 1), therefore the centers of the circles are in the same 
straight line with the point C. Wherefore, if two circles touch 
each other, etc. 

Prop. XIV. — Tueor. — Similar segmevU of circles upon 
equal bases are equal to one another, and have equal arcs. 

Let AEB, CFD be similar segments of circles upon the equal 
straight lines or bases, AB, CD ; the segments are equal ; and 
likewise the arcs AEB, CFD are equal. 

For if the segment AEB be applied to the segment CFD, so 

that the point A may 
E y_ be on C, and the 

straight line AB on 





CD, the point B will 
D coincide with D, be- 
cause AB is equal to 
CD ; therefore the straight line AB coinciding with CD, the 
segment AEB must coincide (I. def 16) with the segment 
CFD, and is therefore equal (I. ax. 8) to it ; and the arcs AEB, 
CFD are equal, because they coincide. Therefore similar seg- 
ments, etc. 

Prop. XV. — Prob, — A segment of a circle being given, to 
complete the circle of lohich it is a segment. 

Assume three points in the arc of the segment, and find (IIL 
1) the center of the circle. From that center, at the distance 
between it and any point in the arc describe a circle, and it will 
evidently be the one required- 



BOOK ni.] 



EUCLID AND LEGENDEE. 



79 



Prop. XV L — Treor. — Ifi equal circles, or in the same circle, 
equal a?igles stand upon equal arcs, whether they are at the 
centers or the circumferences. 

Let ARC, DEF be equal circles, having the equal angles 
BGC, EHF at their centers, and BAG, EDF at their circum- 
ferences ; the arc BKC is equal to the arc ELF. 

Join BG, EF ; and because the circles ABG, DEF are equal, 
their radii are equal ; therefore the two sides BG, GG are equal 





to tlic two, EH, HF ; and (hyp.) the angles G and H are equal ; 
therefore (I. 3) the base BG is equal to the base EF. Then, 
because the angles A and D are equal, the segment BAG is 
similar (III, def 8) to the segment EDF ; and they are upon 
equal straight lines BG, EF ; but (III. 14) similar segments of 
circles upon equal straight lines have equal arcs ; therefore the 
arc BAG is equal to the arc EDF. But the whole circumfer- 
ence ABG is equal to the whole DEF, because the circles are 
equal ; therefore the remaining arc BKG is equal (I. ax. 3) to 
the remaining arc ELF. Wherefore, in equal circles, etc. 

Cor. 1. Gonversely, in equal circles, or in the same circle, the 
angles which stand upon equal arcs are equal to one another, 
•whether they are at the centers or the circumferences. (I. 
def 19.) 

Cor. 2. Hence, in a circle, the arcs intercepted between par- 
allel chords are equal. For if a straight line be drawn trans- 
versely, joining two extremities of the chords, it will (I. 16) 
make equal angles with the chords ; and therefore the arcs on 
"which these stand are eqiial. 

Cor. 3. Hence, in equal circles, equal chords divide the cir- 
cumferences into parts which are equal, each to each. 



80 



THE ELEMENTS OF 



[book nr. 



D 




Pkop. XVII. — Peob. — To bisect a given arc of a circle. 
Let ADB be a giA-en arc ; it is required to bisect it. 
Draw AB, and (I. 6 and 7) bisect it in C, by the perpendicu- 
lar CD ; the arc ABD is bisected in the point D. 

Join AD, DB. Then, because AC is equal to CB, CD com- 
mon to the triangles ACD, BCD, and the angle ACD equal to 

BCD, each of them being a right angle; 
therefore (I. 3) AD is equal to BD. But 
(III. 16, cor. 3), in the same circle, equal 
lines cut off equal arcs, the greater equal 
AC B to the greater, and the less to the less; 

and AD, DB are each of them less than a 
semicircle, because (III. 1) DC^ or DC produced, passes througli 
the center; wherefore the arc AD is equal to the are DB; 
therefore the given arc is bisected in D ; which was to be done. 

Prop. X'N'III. — Theoe. — If a straight line touch a circle, 
and from thepoifit of contact a straight line be drawn dividing 
the circle into tico segments ^ the angUs made by this line with 
the tangent are equivaletit to the angles which are in the alter- 
nate segments. 

Let the straight line DE touch the circle BAG in the point 
B, and let the straight line BA be drawn dividing the circle 
into the segments AGB, AB ; the angle ABE is equal to any an- 
gle in the segment AGB, and the angle ABD to any angle in AB. 

If AB (fig. 1 ) be perpendicular to DE, it passes (III. 1 2, cor.) 
through the center, and the segments being therefore semicir- 
cles, the angles in them are (HI. 11) right angles, and conse- 
quently equal to those which AB makes with DE. 

But if BA (fig. 2) be not perpendicular to DE, draw BF per- 
■ A F 




D B E D B E 

pendicular to it ; join FA and produce it to E; join also CA, 



BOOK in.] 



EUCLID AND LFGKNDRE. 



81 



CB, C being any point in the arc ACB. Then (TIT. 12, cor.) 
BF is a diameter, and (III. 11, and I. 9) the angles BAF, BAE 
are riglit angles. Therefore, in the triangles J^AE, FBE, the 
angle E is common, and the angles BAE, FBE equal, being 




right angles; wherefore (I. 20, cor. 5) the remaining angle 
ABE is equal to the remaining angle F, which is an angle in 
the remote or alternate segment BFA. 

Again : the two angles ABD, ABE are equal (I. 9) to two 
right angles ; and because ACBF is a quadrilateral in the circle, 
the opposite angles C and F are also equivalent (I. 20, cor. 1) 
to two right angles ; therefore (I. ax. 1) the angles ABD, ABE 
are together equal to C and F*. From these equals take away 
the angles ABE and F, which have been proved to be equal ; 
then (I. ax. 3) the remaining angle ABD is equal to the re- 
maining angle C, which is an angle iu the remote segment 
ACB. If, therefore, a straight line, etc.. 

ScliO. 1. The first case is wanting in most editions of Euclid. 
In the second diagram, FA and DB will meet (I. 19) if pro- 
duced, the angles FBE, BFA together being evidently less 
than two right angles. 

Scho. 2. Let AB be a fixed chord, and ^ 

through B draw any other line DE cut- 
ting the circle in C ; and join AC. Now 
(III. 16, cor. 1), wherever C is taken in 
the arc ACB, the angle ACE is con- 
stantly of the same magnitude ; and so 
also (I. 9) is the exterior angle ACD. 
If C be now taken as coinciding with B, 
the straight line DE will become the 
tangent D'BE', AC will coincide with AB, axid the angles 
6 




82 THE ELEMENTS OF [bOOK HI. 

AGE, ACD will become ABE', ABD'. If again C take the 
position C^ the angles ACE, ACD will become AC'^'', 
AC"T>'\ Now, the equality of ABE', ACE, and of ABD', 
ACD'' is what is provecl in the eighteenth proposition, and 
from the equality of ACD and AC'D" cor. 2 follows by the 
addition of ACB. 

Cor. 1. Angles in the same segment of a circle are equal to 
one another. 

Cor. 2. The opposite angles of any quadrilateral figure de- 
scribed in a circle are together equivalent to two right angles; 
and conversely^ if two opposite angles of a quadrilateral be to- 
gether equal to two right angles, a circle may be described 
about it. * 

Cor. 3. If the circumference of a circle be cut by two 
straight lines which are perpendicular to one another, the 
squares of the four segments between the point of intersection 
of the two lines and the points in which they meet the circum- 
ference, are together equal to the square of the diameter. 

Prop. XIX. — Prob. — TJpon a given straight line, to describe 
a segment of a circle containing an angle equal to a given 
angle. 

Let AB be the given straight line, and C the given angle ; it 
is required to describe on AB a segment of a circle containing 
an angle equal to C. 

First : if C be a right angle, bisect (I. 6) AB in F, and from 

the center F, at the distance FB, de- 
scribe the semicircle AHB ; therefore 
(III. 11) any angle AHB in the semicir- 
cle is equal to the right angle C. 

But if C be not a right angle, make 
(I. 13) the angle BAD equal to C, and 
(I. 7) from A draw AE pei-pendicular to AD ; bisect (I. 5 and 
6) AB by the perpendicular FG, and join GB. Then, because 
AF is equal to FB, FG common to the triangles AFG, BFG, 
and the angle AFG equal to BFG, therefore (I. 3) AG is 
equal to GB ; and the circle described from the center G, at 
the distance GA, will pass through the point B ; let this be the 
circle AIIB, Then, because from the poiut A, the extremity 




[book III. 



EUCLID AND LEGENDRE. 



83 




of the diameter AE, AD is drawn at right angles to AE, AD 

(III. 8, cor.) touches the circle; and (III. 18) because AB, 

drawn from the point of contact A, cuts 

the circle, the angle DAB is equal to any 

angle in the alternate segment AHB ; but 

DAB is equal to C ; therefore also C is 

equal to any angle in the segment AHB ; 

wherefore upon the given straight line 

AB the segment AHB is described, which 

contains an angle equal to C j which was 

to be done. 

Scho. It is evident there may be two 
segments answering the conditions of the 
problem, one on each side of the given 
line. It is also plain, that when C is an 
acute angle, and the segment is to be above AB, G is above 
AB; but when obtuse, it is below it. It is likewise plain, that 
the angle BAE is the complement of the given angle C ; that 
is, the difference between it and a right angle. 

Cor. Hence from a given circle can be cut off a segment 
which shall contain an angle equal to a given angle. 



Prop. XX. — Theor. — If two chords of a circle cut one an- 
other, the rectangle contained by the segments of one of them is 
equal to the rectangle contained by the segments of the other. 

In the circle LBM, let the two chords LM, BD cut one an- 
other in the point F ; the rectangle ].<F.FM ^ 
is equal to the rectangle BF.FD. 

If LM, BD both pass through the center, 
80 that F is the center, it is evident that 
LF, FM, BF, FD, being (I. def 16) all 
equal, the rectangle LF.FM is equal to the 
rectangle BF.FD. 

But let one of them, BD, pass through 
the center and cut the other, AC, which does not pass through 
the center, at right angles, in the point E. Then, if BD be 
bisected in F, F is the center. Join AF; and because BD 
which passes through the center, is perpendicular to AC, AE, 
EC are (III. 2) equal to one another. Now, because BD is 




84 THE ELEMENTS OF [bOOK HI. 

divided equally in F, and unequally in E, the rectangle BE.ED, 
and the square of EF are equivalent (II. 5) to the square of FB ; 
that is (I. 23, cor. 2), to the square of FA. But (I. 24, cor. I) 
the squares of AE, EF are equivalent to the square of FA ; 
therefore the rectangle BE.ED and the square of EF are 
equivalent to the squares of AE, EF. Take away the common 
square of EF, and the remaining rectangle BE.ED is equivalent 
to the remaining square of AE; that is, to the rectangle 
AE.EC. 

Next, let BD pass through the center, and cut AC, which 
does not pass through the center, in E, but not at right angles. 
Then, as before, if BD be bisected in F, F is the center of the 

circle. Join AF, and (I. 8) draw FG per- 
pendicular to AC ; therefore (III. 2) AG is 
equal to GC ; wherefore (II. 5) the rectan- 
gle AE.EC and the square of EG are equiv- 
alent to tlie square of AG. To each of 
these equivaleuts add the square of GF; 
therefoi'e the rectangle AE.EC and the 
squares of EG, GF are equivalent to the 
squares of AG, GF ; but (I. 24, coi-. l) the squares of EG, GF are 
equivalent to the square of EF; and the squares of AG, GF are 
equivalent to the square of AF ; therefore the rectangle AE.EC 
and the square of EF are equivalent to the square of AF ; that 
is (I. 28, cor. 2), to the square of FB. But (II. 5) the square 
of FB is equivalent to the rectangle BE.ED, together with the 
square of EF ; therefore the rectangle AE.EC and the square 
of EF are equivalent to the rectangle BE.ED and the square 
of EF; take away the common square of EF, and the remain- 
ing rectangle AE.EC is equivalent to the remaining rectangle 
BE.ED. 

Lastly : let neither of the lines pass through the center, and 
through E, the point of intersection, draw a diameter. Then, 
the rectangle AE.EC is equivalent, as has been shown, to the 
rectangle DE.EB ; and, for the same reason, the rectangle of 
the other chord is equivalent to the same rectangle DPIEB ; 
therefore (I. ax. 1) the rectangle AE.EC is equivalent to the 
rectangle of the other chord. If, therefore, two chords of a 
circle, etc. 




BOOK III.] EUCLID AND LEGENDI E. 85 

Scho. The second and third cases may he thus demonstrated 
in one: 

Join FC. Then the rectangle AE.EC is equivalent (IT. 5, 
cor. 5) to the difference of the squares of AF and FE, or ot DF 
and FE,or (II. 5, cor. l)to the rectangle DE.EB. 

Proportion, however, affords much the easiest method of 
demonstrating both this proposition and the following. 

Peop. XXI. — TiiEOR. — If from any point without a c'rcle 
two straight li?ies be draim, one of which cuts the'circle, and 
the other touches it ; the rectangle contained hy the whole line 
which cuts the circle^ and the />«?•« of it without the circle, is 
equivalent to the square of (he line which touches it. 

Let D he any point without the circle ABC, and DCA, DB 
two straight lines drawn from it, of which DCA cuts the circle 
in C and A, and DB touches it in B ; the rectangle AD.DC is 
equivalent to the square of DB. 

There are two cases — first : let DCA pass through the center 
E, and join EB. Therefore (III. 12) the 
angle EBD is a right angle; and (II. 6) 
"because the straight line AC is bisected in 
E, and produced to D, the rectangle AD.DC 
and the square of EC are together equiva- 
lent to the square of ED ; and CE is equal 
to EB ; therefore the rectangle AD.DC and 
the square of EB are equivalent to the 
square of ED. But (I. 24, cor. l) the 
square of ED is equivalent to the squares a 

of EB, BD, because EBD is a right angle ; 
therefore the rectangle AD.DC and the square of EB are 
equivalent to the squares of EB, BD ; take away the common 
square of EB, and the remaining rectangle AD.DC is equiva- 
lent to the square of the tangent DB. 

Second : if DCA do not pass through the center, take (III. 1) 
the center E, and draw (I. 8) EF perpendicular to AC, and 
join EB, EC, ED. Then, because the straight line EF, which 
passes through the center, is perpendicular to the chord AC, 
AF is equal (III. 2) to FC. And (II. 6) because AC is bisected 
iu F, and produced to D, the rectangle AD.DC and the square 




86 



THE ELEMENTS OF 




[book ni. 

of FC are equivalent to the square of FD. To each of these 

equals add the square of FE ; therefore the rectangle AD.DC, 

and the squares of CF, FE are equivalent to 

the squares of DF, FE ; but (I. 24, cor. 1) 

the square of ED is equivalent to the squares 

of DF, FE, because EFD is a right angle ; 

and the square of EC, or (I, 23, cor. 2) of 

EB, is equivalent to the squares of CF, FE ; 

therefore the rectangle AD.DC and the 

square of EB are equivalent to the square 

of ED. But (I. 24, cor. 1) the squares of 

EB, BD are equivalent to the square of ED 

because EBD is a right angle; therefore the 

rectangle AD.DC and the square of EB 

are equivalent to the squares of EB.BD. Take away the 

common square of EB ; therefore the remaining rectangle 

AD.DC is equivalent to the square of DB ; wherefore, if from 

any point, etc. 

Scho. The second case may be demonstrated more briefly 
thus: Join AE. Then the rectangle AD.DC is equal (II. 5, 
cor. 5) to the difference of the squares of ED and EC, or of ED 
and EB, or (I. 24, cor. 1) to the square of DB. 

Cor. 1. If from any point without a circle there be drawn 
two straight lines cutting it, as AB, AC, the 
rectangles contained by the whole lines and 
the parts of them without the circle are 
equivalent to one another, viz., the rectangle 
BA.AE to the rectangle CA.AF; for each 
rectangle is equivalent to the square of the 
tangent AD. 

Cor. 2. If two straight lines intersect each 
other, so that the rectangle under the seg- 
ments of one of them is equal to the rectangle 
under the segments of the other, their ex- 
tremities lie in the circumference of a circle. 

Scho. By means of the first corollary, it would be shown in 
a similar manner, that if two straight lines meet in a point, and 
if they be so divided that the rectangle under one of them and 
its segment next the common point is equal to the rectangle 




BOOK. III.] 



EUCLID AND LEGENDRE. 



87 



under the otlier and its corresponding segment, the points of 
section and the extremities remote from the common point lie 
in the circumference of the same circle. 



D 



Pkop. XXII. — TiiEOR. — If frotn a point without a circle 
there be draw:i two straight lines, one of ichich cuts the ctrcle^ 
and the other meets it ; and if the rectangle contained by the 
whole line which cuts the circle, and the part of it without the 
circle, be equivalent to the square of the line which meets it, the 
line which meets the circle touches it. 

If from a point without the circle ABC two straight lines 
DCA and DB be drawn, of which DCA cuts the circle, and DB 
meets it ; and if tlie rectangle AD.DC be equivalent to the 
square of DB ; DB touches the circle. 

Draw (III. 9) the straight line DE touching the circle ABC ; 
find (III. 1) the center F; and join FE, FB, 
FD. Then (III. 12) FED is a right angle; 
and (fll. 21) because DE touches the circle 
ABC, and DCA cuts it, the rectangle AD.DC 
is equivalent to the square of DE. But (hyp.) 
the rectangle AD.DC is equivalent to the 
square of DB ; therefore the square of DE is 
equivalent to the square of DB ; and the 
straight line DE equal (I. 23, cor. 3) to the 
straight line DB. But FE is equal to FB, 
and the base FD is common to the two tri- 
angles DEF, DBF ; therefore (I. 4) the angle DEF is equal to 
DBF; but DEF is a right angle; therefore, also, DBF is a 
right angle ; and FB is a part of a diameter, and the straight 
line which is drawn at riiiht angles to a diameter, from its ex- 
tremity, touches (III. 8, cor.) the circle ; therefore DB touches 
the circle ABC. Wherefore, if from a point, etc. 

Prop. XXIII. — Prob, — To divide a given straight line into 
two j)arts, such that the sjuare of one of them may be equivalent 
to the rectangle contained by the other, and a given straight line. 

Let AB, AC be two given straight lines ; it is required to 
divide AB into two parts, such that the square of one of them 
may be equivalent to the rectangle under AC and the other. 




88 



THE ELEMENTS OF 



[book m. 



On CB, the sum of the given lines, describe the semicircle 
CDB, and draw AD jterpendicular to CB ; bisect CA in E, and 
join DE ; and make EF equal to ED ; then the square of AF 
is equivalent to the rectangle AC.FB. 

Describe the semicircle CGA, cutting ED in G, and join GF. 
Then, because FE, EG are respectively 
equal to DE, EA, and the angle FED 
common, GF is equal to AD, and the an- 
gle EGF to EAD, whicli is a right angle, 
and therefore GF touches the circle CGA. 
Hence (III. 20) the rectangle CA.AB is 
equivalent to the square of AD, or of FG, 

or (HI. 21) to the rectangle CF.FA. But (11. 1) the rectanglo 
CA.AB is equivalent to the two CA.AF, CA.FB, and (II. 3) 
the rectangle CF.FA is equivalent to CA.AF and the square of 
AF. From these equivalents take away the rectangle CA.AF, 
and there remains the rectangle CA.FB equivalent to the 
square of AF. • 

Scho. The tenth proposition of the second book is the par- 
ticular case of this problem in which the given lines are equal. 




Prop. XXIV. — Prob. — To draw a common tangent to two 
given circles. 

Let BDC, FUG be given circles ; it is required to di-aw a 
common tangent to them. 

Join their centers A, E, and make BK equal to FE ; from A 




as center, with AK as radius, describe a circle cutting another, 
described on AE as diameter, in M ; draw AM, meeting tlie 
circle BCD in D ; and draw Eli parallel to AM; join Dil; it 
touches both the given circles. 



BOOK III.] 



EUCLID AND LEGENDRE. 



89 



For MD, EII, which (const.) are parallel, are equal to one 
another, because each of them is equal to KB; therefore (I. 15, 
cor. B) Dll is parallel to ]\1E. Now (III. 11), the angle AME 
in a semicircle is a right angle; and therefore (I. 16) the angles 
ADII, DUE are right angles, and (III. 8, cor.) DH touches 
both the ciicles, since it is perpendicular to the radii AD, EH. 

ScJio. In the figure the circles lie on the same side of the 
tangent, which is therefore exterior to them ; but the tangent 
can be transverse, or lie between the circles. It is plain also, 
tliat in these figures, by using the point L instead of M, an- 
other exterior and another transverse tangent would be ob- 
tained ; and this will always be so, when each of the circles 
lies wholly without the other, and does not touch it. But if 
the circles touch one another externally, the two transverse 
tangents coalesce into a single line passing through their point 
of contact ; if they cut one another, there will be two exterior 
tangents, but no transverse one ; if one of the circles touch the 
other internally, they carf have only one common tangent, and 
this passes through their point of contact ; and, lastly, if one of 
them lie wholly within the other without touching it, they can 
have no common tangent. 

If the circles be equal, the points of contact of the exterior 
common tangents are the extremities of the diameters perpen- 
dicular to the line joining the centers ; for (I. 15, cor. 3, and 1 6) 
the lines connecting the extremities of these toward the same 
parts are perpendicular to the diameters, and therefore (III. 8, 
cor.) they touch the circles. 

Prop. XXV.— Prob.— 7b insa-ibe a7iy regular polygon in a 
given circle. 

Let ABDC be the given circle ; it is 
required to inscribe any polyg<m in it. 

Since (I. 9, cor.) all the angles formed 
by any number of straight lines inter- 
secting each other in a common point 
are equivalent to four right angles, and 
(I, 20, cor. 1) any rectilinear figure can 
"be divided into as many triangles as the 
figure has sides, by straight lines joining the extremities of the 




90 THE ELEMENTS OF [BOOK m. 

sides with a poiut within the figure — a regular polygon being 
equilateral — the straight lines connecting the extremities of 
those sides with the center of the regular polygon will divide 
the regular polygon into equal triangles ; for if ABC be a regu- 
lar polygon of three sides, AB is equal BC, and also equal to 
CA, and draw from E (III. 1, cor. 1), the center of the polygon, 
EA, EB, and EC. Now the triangles AEB, BEC, and CEA, 
having their bases equal (hyp.), their other sides common, 
(const.) are equal; therefore (I. 4) the angles AEB, BEC, and 
CEA are equal ; but those angles are equivalent to form right 
angles (I. 20, cor. 1) ; hence, each is one third of four right 
angles, or two thirds of two right angles ; consequently, a reg- 
ular polygon can be inscribed in a circle by making an angle 
on the diameter (I. 13) with the center of the circle the vertex 
of the angle, equal to that part of four right angles that the 
regular polygon has sides, viz. : if the regular polygon has four 
sides, each angle at the center will be one fourth of four right 
angles ; if five sides, one fifth of fourS-ight angles; and so on. 
Or, by bisecting (I. 5) the angle or the arc (III. 11), a regular 
polygon of double the number of sides can be inscribed in the 
circle. Having by these means the angle at the center of a 
regular polygon, the chord of the arc intercepted by the sides 
of the angles will be the side of the regular polygon required, 
from which (I. 12 and 13) the regular polygon can be inscribed 
in the given circle, which was to be done. 

Cor. 1. If the sides of the angles be produced beyond the 
circumference (I. post. 2), and parallels to the exterior sides of 
the polygon (I. 18) be drawn touching the circle (III. 9), then 
a similar and regular polygon can be described about the given 
circle. 

Cor, 2. Find the center of a given regular polygon (III. 1, 
cor. 1), and a circle can be inscribed in the polygon, or circum- 
scribed about it, by taking the straight line drawn from the 
center of the polygon to an extremity of the side of the poly- 
gon i'or a radius lor the circumscribed circle, and a line drawn 
from the center to the point of bisection of the side for a radius 
of the inscribed circle. 

Cor. 3. Since the triangles AEB, BEC, and AEC are equiv- 
alent respectively to the rectangles (I. 23, cor. 6) under the ra- 



BOOK in.] EUCLID AND LEGENDRE. 01 

dius of an inscribed circle and the halves of AB, BC, and CA, 
it follows (IL 1) that the area of ABC is equivalent to tlie rect- 
angle under the radius and half the perimeter. Hence, if the 
sides be given in numbers, the length of the radius may be 
computed by calculating (II. 12, scho.) the area, and dividing 
it by half the sum of the sides, or its double by the sum of the 
sides. 

Cor. 4. Since the angles formed about the center of a circle 
are together equivalent to four right angles (I. 9, cor.), and 
since the angles of a triangle are together equivalent to two 
right angles (I. 20), it follows that when an equiangular trian- 
gle is formed (I. 2) having a vertex at the center and the radius 
for a side, each angle of the triangle is one tliird of two riglit 
angles, or one sixth of four right angles ; hence, six equal 
angles can be formed (I. 1.3) about the center of a circle, and 
since the sides about tliese angles are intercepted by the cir- 
cumference (const.), they are equal (I. def 16) ; and (I. 2, cor.) 
an equiangular triangle being also equilateral, the side of the 
triangle opposite the angle at the center of the circle is equal 
to the radius of the circle (T. ax. 1) ; therefore the radius can 
be made to subtend six equal arcs of the circumference. 

Cor. 5. When in the case of a general proposition to descrV^e 
a circle touching three given straight Imes ichich do vot pass 
through the same point, and which are not all parallel to one 
another. If two of the lines be parallel, there may evideiidy 
be two equal circles, one on each side of the line falling on the 
parallels, each of which will touch the three given lines, and 
their centers will be the intersections of the lines bisectimr the 
angles made by the parallels with the third line. But if the 
lines form a triangle by their intersections, there will be four 
circles touching them; one, inscribed, and the others each 
touching one side externally and the other two produced. The 
centers of the external circles will be the intersections of the 
lines bisecting two exterior angles; and the line bisecting the 
remote interior angle will pass through the same point. The 
method of proof is tl.e same as that given in the proposition. 

Cor. 6. If straight lines be drawn from the center of one of 
the external circles to the vertices of the triangle, the three tri- 
angles formed by the sides of the triangle and the straight lines 



92 THE ELEMENTS OF [bOOK HI. 

are respectively equivalent to the rectangles (I. 15, cor. 4) 
under the radius of that circle and halves of tlie sides of the 
triangle. And if the triangle formed by the b.ide of the original 
triangle nearest the center of the circle, and the lines drawn to 
the vertices at the extremities of that side, be taken from the 
sum of the two other triangles, there remains the original tri- 
angle equivalent to the rectangle under the radius, and the ex- 
cess of half the sum of the two other sides of the orin-inal trian- 
gle above half the side nearest the center of the circle, or, which 
is the same thing, to the rectangle under the radius, and the 
excess of half the perimeter of the original triangle above the 
Bide nearest the center of the circle. The radius, therefore, of 
any of the external circles may be computed by dividing the 
area of the original triangle by the excess of half its perimeter 
above the side which the circle touches externally. 

Scho. The polygons considered in this proposition, and those 
which may be derived from them by the process of bisecting 
the angles or arcs subtending the sides of the polygons, are the 
only ones till lately which geometers have been able to de- 
scribe by elementary geometry, that is, by means of the 
straight line and circle. M. Gauss, of Gottingen, in \n?, Disqui- 
sitioiies Arithmeticoe, has shown that by elementary geometry, 
every regular poly. on may be inscribed in a circle, the number 
of whose sides is a power of 2 increased by unity, and is a 
prime number, or a number which can not be produced by the 
multiplication of two whole numbers, such as 17 — the fourth 
power of 2 increased by unity — and polygons of 257 and C5537 
sides. But the investigation is too complex and difficult for an 
ordinary school text-book. 



END OP b60K third. 



BOOK FOURTH.* 

THE GENERAL THEORY OF PROPORTION". 

DEFINITIONS. 

1. A LESS number or magnitude is said to measure a greater, 
or to be a measure^ a part^ or a suhmuUxple of tlie greater, when 
the less is contained a certain number of times, exactly, in the 
greater; and 

2. The greater is said to be a multiple of the less. 

3. Magnitudes which can be compared in respect of quantity, 
that is, which are either equal to one another, or unequal, are 
Baid to be of the sam,e kind, or homogeneous. 

Scho. 1. Thus, lines, whether straight or curved, are magni- 
tudes of the same kind, or are homogeneous, since they may be 
equal or unequal. In like manner, surfaces, solids, and angles 
form three other classes of homogeneous magnitudes. On the 
contrary, lines and surfaces, lines and angles, surfaces and 
solids, etc., are heterogeneous. Thus, it is obviously improper 
to say, that the side and area of a square are equal to one an- 
other, or are unequal. So likewise we cannot say that the area 
and one of the angles of a triangle are equal to each other, or 
are unequal ; and they are therefore heterogeneous. 

4. If there be two magnitudes of the same kind, the relation 
which one of them bears to the other in respect of quantity, is 
called its ratio to the other. 

The first term, or magnitude, is called the antecedent of the 
ratio, and the second the consequent. 

5. If there be four magnitudes, and if any like multiples 
whatever be taken of the first and third, and any whatever of 

* The accurate but prolix method of Euclid is substituted by the fol- 
lowing more concise method, by employing the notations and simple 
principles of Algebra. See Sup. to Book V. Thombon's Euclid. 



^^ THE ELEMENTS OF [bOOK IV. 

the soconrl anfl fourth ; and if, according as the multiple of the 
first is gi-eater than the multiple of the second, equal to it, or 
less, the multiple of the third is also greater than the multiple 
of the fourth, equal to it, or less ; then the first of the magni- 
tudes is said to have to the second the same ratio that the third 
has to the fourth. 

6. Magnitudes which have the same ratio are called propor- 
tionals ; and equality or identity of ratios constitutes /irojoor- 
tion or analogy. 

When magnitudes are proportionals, the relation is expressed 
briefly by saying, that the first is to the second, a« the third to 
the fourth, the fifth to the sixth, and so on. 

1. When of the multiples of four magnitudes, taken as in the 
fifth definition, the multiple of the first is greater than that of 
the second, but the multiple of the third is not greater than 
that of the fourth ; then the first is said to have to the second 
a greater ratio than the third has to the fourth ; and, on the 
conti-ary, the third is said to have to the iomth a. less ratio than 
the first has to the second. 

8. When there are three or more magnitudes of the same 
kind, such that the ratios of the first to the second, of the sec- 
ond to the third, and so on, whatever may be their number, are 
all equal; the magnitudes are said to be continual propor- 
tionals. 

9. The second of three continual proportionals is said to be a 
mean proportional between the other two. 

10. When there is any number of magnitudes of the same 
kind, greater than two, the first is said to have to the last the 
ratio compounded of the ratio which the first has to the second, 
and of the latio which the second has to the third, and of the 
ratio which the third has to the fourth, and so on to the last 
magnitude. 

For exam].]e, if A, B, C, D be four magnitudes of the same 
kind, the first, .>, is said to have to the last, D, the ratio com- 
pounded of tlie ratios of A to B, B to C, and C to D. 

11. When iln-ee magnitudes are continual proportionals, the 
ratio of tlie first to the third is said to be duplicate of the ratio 
of the first Xo the second, or of the second to the third. 

12. When iour magnitudes are continual proportionals, the 



BOOK rv.] EUCLID AND LEGENDEE. 95 

ratio of the first to the fourth is said to be triplicate of the ratio 
of the first to the second, of the ratio of the second to the third, 
or of tlie ratio of the third to the fourth. 

Scho. 2. In continual proportionals, by their own nature, and 
that of compound ratio, the ratio of the first to the third is 
compounded of two equal ratios ; and the ratio of the first to 
the fourth, of three equal ratios ; and hence we see the reason 
and the propriety of calling the first duplicate ratio, and the 
second triplicate. It is plain, that on similar principles, the 
ratio of the first to the fifth would be said to be quadruplicate 
of the ratio of the first to the second, of the second to the third, 
etc., and thus we might form other similar terms at pleasure. 

The terras subduplicate, subtriplicate^ and sesquiplicate^ 
which are sometimes employed by mathematical wiiters, are 
easily understood after the explanations given above. In con- 
tinual proportionals, the ratio of the first terra to the second is 
said to be subcJuplicate of the ratio of the first to the third, and 
subtriplicate of that of the first to the fourth. Again : if there 
be four continual proportionals, the ratio of the first to the 
fourth is said to be sesquiplicate of the ratio of the first to the 
third ; or, which amounts to the same, the ratio which is com- 
pounded of another ratio and its subduplicate, is sesquiplicate 
of that ratio. 

1 3. In proportionals, the antecedent terms are called homolo- 
gous to one another, as also the consequents to one another. 

Geometers make iise of the following technical words to 
denote diiferent modes of deriving one proportion from an- 
other, by changing either the order or the magnitudes of the 
terms. 

14. Alternately : this word is used when there are four pro- 
portionals of the same kind ; and it is inferred that the first has 
the same ratio to the third which the second has to the fourth ; 
or that the first is to the third as the second to the fourth ; as 
is shown in the fourth proposition of this book. 

15. By inversion: when there are four proportionals, and it 
is inferred that the second is to the first as the fourth to the 
third. Prop. 3, Book IV. 

16. By composition: when there are four proportionals, and 
it is inferred, that the first, together with the second, is to the 



96 THE ELEMENTS OF [BOOK IV. 

second as the tliii-fl, together with the fourth is to the fourth. 
Tenth Prop., Book IV. 

17. By division : when there are four proportionals, avd it 
is inferred that the exjcess of the first above the second is to the 
second, as the excess of the third above the fourth is to the 
foui-th. Tenth Prop., Book IV. 

18. My coyiveraion : when there are four proportionals, and 
it is inferred that the first is to its excess above tlie second, as 
the third to its excess above the fourth. Eleventh Prop., Book 
IV. 

Sc'ho. The substance of the five preceding definitions may be 
exhibited briefly in the following manner, the signs + and — 
denoting addition and subtraction, as has been explained 
already at the beginning of the second book: 

Let A: B:: C:D; 
Alternately, A : C : : B : D ; 

By inversion, B : A : : D : C ^ 

By composition, A + B : B : : C + D : D ; 

By division. A— B : B : : C— D : D ; 

By conversion, A : A — B : : C : C — D. 

19. Ex mquo^ or ex equali (scil. distantid)^ from equality of 
distance: when there is any number of magnitudes more than 
two, and as many others, which, taken two in the one rank, and 
two in the other, in direct order, have the same ratio; and it is 
inferred that the first has to the last of the first rank the same 
ratio which the first of the other rank has to the last. This is 
demonstrated in the thirteenth proposition of this book. 

20. Ex cequo^ inversely : when there are three or more mag- 
nitudes, and as many others, which taken two and two in a 
cross order, have the same ratio ; that is, when the first magni- 
tude is to the second in the first rank, as tlie last but one is to 
the last in the second rank; and the second to the third of the 
first rank, as the last but two is to the last but one of the second 
rank, and so on ; and it is inferred, as in the preceding defini- 
tion, that the first is to the last of the first rank, as the first to 
the last of the other rank. This is proved in the fourteenth 
proposition of this book. 



BOOK IV.] EUCLID AND LEGENDRE. 97 

AXIOMS. 

1. Like multiples of the same, or of equal magnitudes, are 
equal to one another. 

2. Those magnitudes of which the same, or equal magni- 
tudes, are like multiples, are equal to one another. 

3. A multiple of a greater magnitude is greater than the 
same multiple of a less. 

4. That magnitude of which a multiple is greater than the 
same multiple of another, is greater than that other magnitude. 

EXPLANATION OF SIGNS. 

1. The product arising from multiplying one number by an- 
other is expressed by writing the letters representing them, one 
after the other, without any sign between them ; and some- 
times by placing between them a point, or the sign x. 

2. A product is called a power ^ when tlie factors are all the 
same. Thus, AA, or as it is generally written, A^, is called the 
second power, or the square of A ; AAA, or A', the third 
power, or cube of A; AAA A, or K\ its fourth power, etc. 

In relation to these powers, A is called their root. Thus, A 
is the second or square root of A'^, the third or cube root of A% 
the fourth root of A^, etc. In like manner, the second or square 
root of A is a number Avhich, when multiplied by itself, pro- 
duces A; the th rd or cube ro ^ of A is such a number, that if 
it be multiplied by itself, and the product by the same root 
again, the final product will be A. The square root of A is 
denoted by -/A or A*, its cube root by -^A, or A*, its fourth 
root by Ai, etc. 

3. The quotient arising from dividing one number by an- 
other is denoted by writing the dividend as the numerator of a 
fraction, and the divisor as its denominator. 

4. The signs =, =0, >, <, signify respectively equal to^ 
equivalent to, greater than, less than. 

PROPOSITIONS. 

Prop. I. — Treor, — If there be four numbers such that the 
quotients obtained by dividing the first by the second, and the 

1 



98 THE ELEMENTS OF [bOOK IV. 

third hy the fourth, are equal ; the first has to the second the 
same ratio that the third has to the fourth. 

Let A, B, C, D be four magnitudes, such that o— fj > ^^^^ 

A: B:: C :D. 

For, let m and n be any whole numbers, and multiply the 

AC 

fractions — and ^ by m, and divide the product by n y then 

— =- = — ^-. Now, if mA be srreater than riB, mC will also be 
nB riD 

greater than nD ; for, if this were not so, — rr would not be 

equal to -y:- In like manner it might be shown, that if mA 

be equal to 7iB, mC will be equal to iiQ ; and that if raK be 

less than nB, rriG will be less than wD. But wA, raQ are any 

like multiples whatever of A, C ; and ?iB, 7^D any whatever of 

B, D ; and therefore (IV. def 5) A : B : : C : D. Therefore, if 

there be four numbers, etc. 

AC 1 

Scho. This proposition is the same, when ^ or Y\—P ^^ ^iP 

being a whole number. 

A C 
Cor, If AD=BC, by dividing by B and D, we get ^=t^> 

and therefore, by this proposition, A : B : : C : D. Hence, if 
the product of two numbers be equal to that of two others, the 
one pair may be taken as the extremes and the other as the 
means of an analogy. 

Prop. II. — Theor. — If any four numbers he proportional^ 

and if the first be divided by the second, and the third by the 

fourth, the quotients are equal. 

A C 
Let A : B : : C : D ; then ^= jt 

If A and B be whole numbers, let the first and third terms 
be multiplied by B, and the second and fourth by A, and the 
products are AB, AB, BC, AD. Now, since the first and sec- 
ond of these are the same, the third and fourth are (IV. def. 6) 
equal; that is, AD=BC; and by dividing these by B and D, 

A C 

we find (IV. ax, 2) =t = ^. 



BOOK IV.] EUCLID AND LEGENDEE. 99 

If A and B be fractions, let A=— , and B=-, so that niA— 

' in n 

E, and nB=F; the numerators and denominators E, F, m, n 

E F 

beinii whole numbers. Then (hyp.) — : - : : C : D. Multiply 

the first and third of these by mF, and the second and fourth 
l)y wE, and the products are EF, EF, mYQ, and nED. Now, 
the first and second of these being the same, the third and 
fourth (IV. def 5) are equal ; that is, nED=mFC, or mnP^ — 
wiftBC, since E=imA, and F=wB. Hence, by dividing these 
by m and w, we get (IV. ax. 2) ADi=BC ; and the rest of the 
proof is the same as in the first case. Therefore, if any four 
numbers, etc. 

Scho. 1. If either A or B be a whole number, the proof is in- 
cluded in the second part of the demonstration given above. 
Thus, if A be a whole number, we have simply E = A and m=: 
1, and everything will proceed as above. The proof would also 
be readily obtained by substituting for B as before, but retain- 
inc: A unchanged. 

If A and B be incommensurable, such as the numbers ex- 
pressing the lengths of the diagonal and side of a square, the 
lengths of the diameter and circumference of a circle, etc., their 
ratios may be approximated as nearly as we please. Thus the 
diagonal of a square is to its side, as |f : 1, nearly ; as \%\ : 1, 
more nearly; as \%\\ : 1, still more nearly, etc. Hence, in 
such cases we can have no hesitation in admitting the truth of 
the proposition, as we see that it holds with respect to numbers 
the ratio of which differs from that of the proposed numbers by 
a quantity which may be rendered as small as we please — 

smaller, in fact, than anything that can be assigned. 

A 1 

Scho. 2. This proposition is the same, when T>=i? or -, p 

being a whole number. 

Scho. 3. From this proposition and the foregoing, it appears, 
that if two fractions be equal, the numerator of the one is to its 
denominator as the numerator of the other to its denominator ; 
and that if the first and second of four proportional numbers be 
made the numerator and denominator of one fraction, and the 
third and fourth those of another, the two fractions are equal. 



100 THE ELEMENTS OF [bOOK TV. 

This is the same in substance as that the two expressions, A : 

A C 
B : : C : D, and |T=t-o are equivalent, and may be used for 

one another. 

Cor. 1. It appears in the demonstration of this proposition, 
that AD = BC; that is, if four numbers be pT'oportionals, the 
product of the extremes is equal to the product of the means. 
Hence, if the product of the means be di\ided by one of the ex- 
tremes, the quotient is the other; and thus we have a proof of 
the ordinary ai'ithmetical rule for finding a fourth proportional 
to three uiven numV)ers. 

Cor. 2. It is evident, that if A be greater than B, C must be 
greater than D ; if equal, equal ; and if less, less ; as otherwise 

=5" and =^ could not be equal. 

A C 
Cor. 3. If A : B : : C : D, and consequently —=y^, by multi- 

plying these fractions by—, we get — 7^= — :Fr, or mA : wB : : 

mC : nT>. 

A 

Cor. 4. If A be greater than B, the fraction — is evidently 

T> C C 

greater than -^, and the fraction -^ less than ry ; that is, of two 

unequal numbers, the greater has a greater ratio to a third than 
the less has ; and a thii'd number has a greater ratio to the 
less than it has to the greater. 

A B 

Cor. 5. Conversely, if -^ be greater than -r^, A is greater than 

— be less than =, 
A B' 



B ; and, if — be less than z^, A is also greater than B. 



Prop. III. — Theor. — If four numbers he proportionals, they 
are proportionals also when taken inversely. 

If A : B : : C : D ; then, inversely, B : A : : D : C. 

For (IV. 2, cor. 1) BC = AD ; and hence by dividing by A 

and C, we obtain -r=p5 or (I^- 2, scho. 2) B : A : : D : C. 

Therefore, if four numbers, etc. 



/ 






BOOK IV,] EUCLID AND LEGENDRE. 101 

Prop. IY. — Theor. — If four numbers be proportionals, they 
are also proportionals when taken alternately. 

If A : B : : C : D ; tlien, alternately, A : C : : B : D. 

For (IV, 2, cor. 1) AD=;:BC ; whence, by dividing by C and 

A B 

D we get 7s =yt; or (IV. 2, scho. 2) A : C : : B : D, There- 

fore, if four numbers, etc, 

Scho. When the first and second terms are not of the same 
kind as the third and fourth, the terms can not be taken altern- 
ately, as ratios would thus be instituted between heterogene- 
ous macrnitudes. 



o 



Prop, V. — Theor. — Ejual numbers have the same ratio to 
the same number ^ and the same has the same ratio to equal 
numbers. 

Let A and B be equal numbers, and C a third ; then A : C : : 
B : C, and C : A : : C : B. 

A B 

For, A and B being equal, the fractions -^ and y, are also 

equal, or, which is the same, A : C : : B : C ; and, by inversion, 
(IV. 3) C : A : : C : B, Therefore, equal numbers, etc. 

Prop. VI. — Theor. — JVmnbers which have the same ratio 
to the same nwnber are equal ; and those to which the same 
has the same ratio are equal. 

If A : C : : B : C, or if C : A : : C : B, A is equal to B. 

A B 

For, since —7=., ,by multiplying by C we get A = B. 

The proof of the second part is the same as this, since, by in- 
version (IV. 3), the second analogy becomes the same as the 
first. Therefore, numbers, etc. 

Prop. VII. — Theor. — Ratios that are equal to the sameratio 

are equal to one another. 

If A : B : : C : D, and E : F : : C : D ; then A : B : : E : F. 

-. . AC , E C . „ ,^ . A E . ^ 

ror, smce p=-r., and = - ,therciore (1. ax. 1) p=p, ; that , 

is, (IV. 2, scho. 2) A : B : : E : F. Therefore, ratios, etc. 



102 THE ELEMENTS OF [bOOK 17. 

Pkop. Vin. — Theok. — Ofnumhers which are proportionals^ 
as any one of the antecedents is to its consequent^ so are all the 
antecedents taken together to all the conseqzients. 

If A : B : : C : D : : E : F ; then A : B : : A+C+E : B+D 
+F. 

ACE 

Since :^==: ==-7, put each fraction equal to q, and multiply 

hj the denominators ; then A=zBq, C=:Dq, and E^Fg-. 
Hence, by addition, A+C-}-E = (B + D-f F)^'/ and by dividing 

by B+D+F, we get ^=WTWZw' ^^^ ?— g 5 ^^^ there- 

foi'e 4"= t"!"^tS > or A : B : : A+C + E : B+D + F. There- 
a ±>+jj+i^ 

fore, etc. 

Prop. IX. — Theor. — 3fagmtudes have the same ratio to one 
another that their like multiples have. 

Let A and B be two magnitudes ; then, n being a whole 
number, A : B : : nA : nB. 

For :jj=— ^, or A : B : : nA : wB. Therefore, magnitudes, 
etc. 

Prop. X. — ^Theor. — If four mimbers be prop>ortionals ; then 
(1) hy composition, the sum of the first and seco7\d is to the 
second, as the sum of the third and fourth to the fourth ; and 
(2), 5y division, the excess of the first above the second is to the 
second, as the excess of the third above the fourth is to the 
fourth. 

If A : B : : C : D ; then, by composition, A+B : B : : C+D: 
D; and by division, A— B : B : : C — D : D. 

1. Since (hyp.) vi = i-:i and since -rr = ^: add the latter frac- 

^ "^ ^ ' B D B D 

tions to the former, each to each, and there results — rr~ = 

5^, or A+B : B : : C +D : D. 

2. By subtracting the latter pair of the same fractions from 

the former, each from each, we obtain — =—- = —=-— ; or A — 

B i) 

B : B : : C— D : D. If, therefore, etc. 



BOOK rV.] EUCLID AND LEGENDRE. 103 

Cor. By dividing the fractions which were found above by- 
addition, by those which were found by subtraction, we get 

:^^=^^; or (IV. 2, scho. 2) A+B : A-B : : C+D : C- 

D; that is, if four numbers be proportional, the sura of the fii-st 
and second terms is to their difference, as the sum of the third 
and fourth terms is to their difference. It is evident, that if B 
be greater than A, the analogy would become B+A : B— A : : 
D + C : D— C. 

Prop. XI. — Theoe. — If foumumhers he proportional ; then,, 
by conversion, the first is to its excess abuve the second^ as the 
third to its excess above the fourth. 

If A : B : : C : D ; then, by conversion, A : A — B : : C : 
C— D. 

x^ • n ;,- .BD -. AC. 

Jbor, smce (hyp. and mver.) —=^^ and smce — = -; take 

the former fractions from the latter, each from each, and there 

remains — -r — = — ^ — , or (by inver.) A:A — B:: C:C — D. 

Therefore, if four numbers, etc. 

Prop. XII. — Theor. — If there be members forming two or 
more analogies which have common consequents^ the sum of 
all the first antecedents is to their common consequent, as the 
sum of all the other antecedents is to their common consequent. 

If A : B : : C : D, and E : B : : F : D ; then A + E : .B : : 
C+F : D. 

For (hyp.) -j5=— ,and p = Tx; and hence, by addition, — -^— 
=5^, or A+E : B : : C + F : D. If, therefore, etc. 

Prop. XIII. — Theor. — If there be three or more niiyn^ers, 
and OS many others, which, taken two and two in order, have 
the sa7ne ratio ; then, ex aequo, the first has to the last of the 
first rank the sayne ratio that the first has to the last of the 
second rank. 

If the two ranks of numbers. A, B, C, D, and E, F, G, II, be 



104 THE ELKMENT8 OF [bOOK IV. 

Buch that A : B : : E : F, B : C : : F : G, and C : D : : G : 11 ; 
then A : D : : E : H. 

A E B F C C 

For, since (hyp.) -- =-, - = - and ^ = - ; by multiplying 

together the first, third, and fifth fractions, and the second, 
fourth, and sixth, we obtain jTnn^KnxT 5 ^r, by dividing the 

terras of the first of these fractions by BC, and those of the 

A E 

second by FG, j)=g, or A : D : : E : II. Therefore, if there 

be three, etc. 

This proposition might also be enunciated tlius : If there be 
numbers forming two or more analogies, such that the conse- 
quents in each are the antecedents in the one immediately fol- 
lowing it, an analogy will be obtained by taking the antece- 
dents of the first analogy and the consequents to the last for 
its antecedents and consequents. 

Prop. XIV. — Theok. — If there be three or more numbers, 
and as many others, which, taken two and two in a cross order, 
have the same ratio ; then, ex tequo inversely, the first has to 
the last of the first rank the same ratio uhlch the first has to 
the last of the second ranTc. 

If the two ranks of numbers. A, B, C, D, and E, F, G, H,be 
such that A : B : : G : H, B : C : : F : G, and C : D : : E : F ; 
then, ex mqrco inversely, A : D : : E : H. 

A C B F P F 

For, since (hyp.) -^=^, ^=^,and ^ =^, by multiplying to- 
gether the fractions as in the preceding proposition, we get 

ABC GFE , ,,..,., . , . ^ 

^pY)=TT7TT^j whence, by dividing the terms of the first of 

these fractions by BC, and those of the second by GF, we ob- 

A E 
tain jx = fT, or A : D : : E : H. If, therefore, etc. 

This proposition may also be enunciated thus: If there be 
numbers forming two or more analogies, such that the means 
of each are the extremes of the one immediatelj' following it, 
another analogy may be obtained by taking the extremes of 
the first analogy and the means of the last for its extremes and 
means. 



BOOK IV.] EUCLID AND LEGENDKE. 105 

Prop. XV. — Theor. — If there he numbers forming tico or 
more analogies^ the products of their corresponding terms are 
proportionals. 

If A : B : : C : D, E : F : : G : II, and K : L : : M : N ; then 
AEK : BFL : : CGM : DUN. 

^ ,, - A C E G TC M ^ . . 

For (hyp.) j^ =0, ^=|j>' aucl jy = ^; and taking the pro- 
ducts of the corresponduig terms of these fractions, we obtain 

A FTC OC \f 

^^=^^., or AEK : BFL :: CGM : DHN. Therefore, if 

Bi'L DHIS' 

there be numbers, etc. 

Cor. 1. Hence, if there be two analoj^ies consistinsr of the 
sane terms, A, B, C, D, we have A" : B' : : C^ ; D^ ; if there be 
three, we have A' : B' : : C* : D', etc. ; and it thus appears, 
that like powers of proportional numbers are themselves pro- 
portional. 

Cor. 2. Like roots of proportional numbers are proportional. 
Thus, if A : B : : C : D, let 4/ A : VB : : VC : VE. Then, by 
the preceding corollary, A : B : : C : E. But (hyp.) A : B : : 
C : D ; and therefore (IV. 7) C : E : : C : D, and (IV. 6) E = 
D, and consequently V : A -/ B : : VC : VE, or VD. 

Prob. XVI. — Theor. — The sum of the greatest and least of 
four p. oportional members is greater than the sum of the other 
two. 

If A : B : : C : D, and if A be the greatest, and therefore 
(IV. 2, cor. 2) D the least ; A and D are together greater than 
B and C. 

For (by conversion) A : A — B :: C : C — D, and, altern- 
ately, A : C : : A— B : C— D. But (hyp.) A>C, and there- 
fore' (IV. 2, cor. 2) A— B > C — D. To each of these add B ; 
then A>B + C— D. Add again, D; then, A-hD>B4-C. 
Therefore, etc. 

Cor. Hence the mean of three proportional numbers is less 
than half the sum of the extremes. 

Prob. XVII. — Theor. — In numbers which are continual 
proportiojials, the first is to the third as the second power of 
the first to the second power of the second y the first to the 



106 THE ELEMENTS OF [bOOK IV. 

fourth as the third power of the first to the third power of the 
second ; the first to the fifth as the fourth power of the first to 
the fourth power of the second ; and so on. 
1 If A, Vy, C, D, E, etc., be continual proportionals; A : C : : 
A^ B^ ; A : D : : A^ : B'; A : E : : A^ : B', etc. 

For, since (IV. def. 8) A : B : : B : C, and since A : B : : 
A : B, we have (IV. 15) A^ : B^ : : AB : BC, or, dividing the 
third and fourth terms by B, A* : B' : : A : C. 
Again : since A* : B' : : A : C, 

and A : B : : C : D we have (IV. 15) 
A' : B' : : AC : CD, or dividing the third and fourth terms 
by C, A^ : B^ : : A : D ; and so on, as far as we please. There- 
foi'e, etc. 

Cor. Hence (IV. defs. 11 and 12) the ratio which is duplicate 
of that of any two numbers, is the same as the ratio of their 
squares ; that which is triplicate of their ratio, the same as the 
ratio of their cubes, etc. 

Prop, XVIII. — ^Theor. — A ratio xcMch is compounded of 
other ratios, is the same as the ratio of the products of their 
homologous terms. 

Let the ratio of A to D be compounded of the ratios of A to 
B, B to C, and C to D ; the ratio of A to D is tlie same as that 
of ABC, the product of the antecedents, to BCD, the product 
of the consequents. 

For, since A : D : : A : D, multiply the terms of the sec- 
ond ratio by BC ; then (IV. 9) A : D : : ABC : BCD. There- 
fore, etc. 

Prop. XIX. — Theor. — In numbers which are continual 
pro2)ortionalsj the difference of the first and second is to the 
first, as the difference of the first and last is to the sum of all 
t/ie terms excejjt the last. 

If A, B, C, D, E be continual proportionals, A — B : A : : 
A— E: A + B + C + D. 

For, since (hyp.) A : B : : B : C : : C : D : : D : E, we 
have (IV. 8) A : B 



(conv.) A : A — B : 
A— B : A :: A— E 



A+B+C+D : B + C + D + E. Hence 
A + B + C + D : A— E; and (inver.) 
A+B + C + I). 



BOOK IV.] EUCLID AND LEGENDRE. 107 

It is evident that if A were the least term, and E the great- 
est, we should get in a similar manner, B^ — A : A : : E — A : 
A-f-B-f-C+D. Therefore, in numbers, etc. 

Cot. If the series be an infinite decreasing one, the last term 
will vanish, and if S be put to denote the sum of the series, the 
analogy will become A — B : A : : A : S ; and this, if rA be 
put instead of B, and the first and second terms be divided by 
A, will be changed into 1 — r : 1 : : A : S. The number r is 
called the common ratio, or common m,ultiplier, of the series, 
as by multiplying any term by it, the succeeding one is ob- 
tained. 



END OP BOOK FOURTH. 



BOOK FIFTH. 

DEFINITIONS. 

1. Similar rectilineal figures are those which have their 
several angles equal, each to each, and the sides about the 
equal angles proportionals. 

2. Two magnitudes are said to be reciprocally proportional 
to two others, when one of the first pair is to one of the second, 
as the remainincc one of the second is to the remaining one of 
the first. 

3. A straight line is said to be cut in extreme andmean ratio^ 
■when the whole is to one of the segments as that segment is to 
the other. 

4. The altitude of any figure is the straight line drawn from 
its vertex perpendicular to its base. 

5. A sti-aight line is said to be cut harmonically, Avhen it is 
divided into three segments, such that the whole line is to one 
of the extreme segments as the other extreme segment is to the 
middle one. 

PROPOSITIONS. 

Prop. I. — Theor. — Triangles and parallelograms of the 
same altitude are one to another as their bases. ^ 

Let the triangles ABC, ACD, and the parallelograms EC, 
CF have the same altitude, viz., the perpendicular drawn from 
the point A to BD ; then, as the base BC is to the base CD, so 
is the triangle ARC to the triangle ACD, and the parallelo- 
gram EC to the parallelogram CF. 

Produce BD both ways, and take any number of straight 
lines BG, Gil, each equal to BC ; and any number DK, KL, 
each equal to CD; and join AG, AH, AK, AL. Then, because 
CB, BG, Gil are all equal, the triangles ABC, AGB, AUG are 
(L 15, cor.) all equal. Therefore, whatever multiple the base 



BOOK v.] 



EUCLID AND LEGENDRE. 



loa 




HC is of BC, the same multiple is the triangle AUG of ABC. 
For the same reason, wliatver multiple LC is of CD, the same 
multiple is the triangle ALC 
of ADC. Also, if the'base HC - 
be equal to CL, the triangle 
AHC is equal (I, 15, cor.) to 
ALC ; and if the base HC be 
greater than CL, likewise (L 
15, cor. 6) the ti-iangle AHC is 

greater than ALC; and if less, less. Therefore, since there are 
four magnitudes, viz., the two bases, BC, CD, and the two tri- 
angles ABC, ACD; and of the base BC, and the triangle ABC, 
the first and third, any like multiples whatever have been taken, 
viz., the base HC, and the triangle AHC ; and of the base CD, 
and the triangle ACD, the second and fourth, have been taken 
any like multiples whatever, viz., the base CL, and the triangle 
ALC ; and that it has been shown that, if the base HC be 
greater than CL, the triangle AHC is greater than ALC ; if 
equal, equal; and if less, less; therefore (IV. def. 5) as the 
base BC is to the base CD, so is the triangle ABC to the trian- 
gle ACD. 

Again : because (L 15, cor.) the parallelogram CE is double 
of the triangle ABC, and the parallelooi-ani CP"" of the triangle 
ACD, and that (IV. 9) magnitudes have the same ratio which 
their like multiples have ; as the triangle ABC is to the trian- 
gle ACD, so is the parallelogram EC to the parallelogram CF. 
But it has been shovvn, that BC is to CD, as the triangle ABC 
to the triangle ACD ; and as the tiiansjfle ABC is to the trian- 
gle ACD, so is the parallelogram EC to the parallelogram CF; 
therefore (TV. 7) as the base BC is to the base CD, so is the 
parallelogram EC to the parallelogram CF. Wheiefore, trian- 
gles, etc. 

Scho. This proposition may be briefly demonstiated thus: 

Let a perpendicular drawn from A to BD be called P. Then, 

■J^P.BC will be equivalent to the area of the triangle ABC, and 

Ap.CD that of ACD. Dividing, therefore, the former of these 

, t, , 1 il'-KC BC ABC ,„, „ 

equals by the latter, we get yp-^ or, qy)~A<Jd' °^" ^ ' 

scbo. 2) BC : CD : : ABC : ACD. In extending this method 



110 THE ELEMENTS OF [BOOK V. 

of proof to the parallelograms, we have merely to hbc P instead 
ofiP. 

Cor. 1. From this it is plain, that triangles and parallelo- 
grams which have equal altitudes, are one to another as their 
bases. 

Let the figures be placed so as to have their bases in the 
same straight line ; and perpendiculars being drawn from the 
vertices of the triangles to the bases, the straight line Avhich 
joins the vertices is parallel (I, 15, cor.) to that in Avhich their 
bases are, because the perpendiculars ai-e both equal and paral- 
lel to one another. Then, if the same construction be made as 
in the proposition, the demonstration will be the same. 

Cor. 2. Hence, if A, B, C be any three straight lines, we have 
A : B : : A.C : B.C. 

Cor. 3. So, likewise, if the straight lines A, B, C, D be pro- 
portional, and E and F be any other straight lines, we shall 
have, according to the preceding corollary, and the seventh 
proposition of the fourth book, A.E : BE : : C.F ; D.F. 



Prop. II. — Theor. — If a straight line he parallel to the base 
of a triangle, it cuts the other sides, or those produced, propor- 
tionally, and the segments between the parallel and the base are 
homol'jgous to one another ; and (2) if the sides of a triangle, 
or the sides produced, be cut proportionally, so that the seg- 
ments between the points of section and the base are homologous 
to one another, the straight line which jo. ns the points of sec- 
tion is parallel to the base. 

The enunciation of this proposition which is given by Dr. 
Simson and others, is defective, and might lead to error in 
its application, as it does not point out what lines are homolo- 
gous to one another in the analogies. 

It is plain that, instead of one proposition, this is in reality 
two, which are converses of one another. 

1. Let DE be parallel to BC, one of the sides of the triangle 
ABC ; BD : DA : : CE : EA. 

Join BE, CD. Then (L 15, cor.) the triangles BDE, CDE 
are equivalent, because they are on the same base DE, and be- 
tween the same parallels. DE, BC. Now ADE is another tri- 





BOOK v.] EUCLID AND LEGENDRK 111 

angle, and (IV. 5) equal magnitudes have to the same the same 
ratio; therefore, as the triangle 

BDE to ADE, so is the triangle ^ ^ » 

CDE to ADE. But, 

(V. ]) as the triangle BDE 
to ADE, so is BD to DA ; 

Because, having the same alti- 
tude, viz., the perpendicular 
drawn from E to AB, they are 
to one another as their bases; ^ c B O 

and for the same reason, 

as the triangle CDE to ADE, so is CE to EA. 

Therefore (IV. 1) as BD : DA : : CE : E A. 

2. Next, let the sides AB, AC of the triangle ABC, or those 
produced, be cut proportionally in the points D, E ; that is, so 
that BD : DA : : CE : EA, and join DE ; DE is parallel to BC. 

The same construction being made, because (hyp.) 

as BD : DA : : CE : EA ; and (V. 1) 

as BD to DA, so is the triangle BDE to the triangle ADE; and 
as CE to EA, so is the triangle CDE to ADE ; therefore (IV. 
7) the triangle BDE is to ADE, as the triangle CDE to ADE ; 
that is, the triangles BDE, CDE have the same ratio to ADE ; 
and therefore (IV. 6) the triangles BDE, CDE are equal ; and 
they are on the same base DE, and on the same side of it ; 
therefore (I. 15, cor.) DE is parallel to BC. Wherefore, if a 
straight line, etc. 

Cor. The triangles which two intersecting straight lines 
form with two parallel ones, have their sides which are on the 
intersecting lines proportional ; and those sides are homologous 
which are in the same straight line; and (2), conversely, if two 
straight lines form with two intersecting ones triangles which 
have their sides that are on the intersecting lines proportional, 
the sides which are in the same Rtrai<irht line with one another 
being homologous, those straight lines ai-e parallel. 

1. Let DE and BC (first and second figures) be the parallels, 
and let them be cut by the straight lines BD, CE, which inter- 
sect each other in A; then BA : AC : : DA : AE. For, since 
BD : DA : : CE : EA, we have, by composition in the first fig- 



112 THE ELEMENTS OF [BOOK V. 

tire, and by division in the second, BA : DA : : CA : EA, and, 
alternately, BA : AC : : DA : AE. 

2. But if BA : AC : : DA : AE, DE and BC are parallel. 
For, alternately, BA : DA :: CA : EA ; then, in the first figure 
by division, and in the second by composition, we have BD : 
DA : : CE : EA ; and therefore, by the second part of this 
proposition, DE is pai-allel to BC. 

Prop. III. — TheoPw — Tlie sides about the equal aiigles of 
equiangular triangles are proportionals ; and those ichich are 
ojyposite to the equal angles are homologous sides^ that is, are 
the antecedents or consequents of the nitios. 

Let ABC, DCE be equiangidar triangles, having the angle 
ABC equal to DCE, and ACB to DEC, and consequently (I. 
20, cor. 5) BAC equal to CDE ; the sides about the equal 
angles are proportionals ; and those are the homologous sides 
which are opposite to the equal angles. 

Let the triangles be placed on the same side of a straight 
line BE, so that sides BC, CE, which are opposite to equal 
ano-les, may be in that straiu^ht line and contii^nous to one an- 
other; and so that neither the equal angles ABC, DCE, nor 
ACB, DEC at the extremities of those sides may be adjacent. 

Then, because (I. 20) the angles 
ABC, ACB are together less than 
two right angles, ABC and DEC, 
which (hyp.) is equal to ACB, are 
also less than two right angles ; 
wherefore (I. 19) BA, ED will meet, 
if produced ; let them be produced 
and meet in F. Again : because 
the angle ABC is equal to DCE, BF is parallel (L 16, cor.) to 
CD ; and, because the angle ACB is equal to DEC, AC is par- 
allel to FE. Therefore, FACD is (I. def 15) a parallelogram; 
and consequently (L 15, cor.) AF is equal to CD, and AC to 
FD. Now (V. 2) because AC is parallel to FE, one of the 
sides of the triangle FBE, 

BA : AF : : BC : CE. 
But AF is equal to CD ; therefore (IV. 5) 

as BA : CD : : BC : CE, 




BOOK v.] EUCLID AND LI GKNDRE. 113 

and alternately (IV. 4) as AB : BC : : DC : CE. 
Ai^ain : (V. 2) because CD is parallel to 15F, as 

BC : CE : : FD : DE; but FD is equal to AC; therefore, 
as BC : CE : : AC : DE ; and, alternately, 
as BC : CA : : CE : ED. 
Therefore, because it has been proved that 

AB : BC : : DC : CE, and as BC : CA : : CE : ED; 
ex mquo (IV. 13), BA : AC : : CD : DE. Therefore, the 
Bides, etc. 

Scho. 1. Hence (V, def l) equianirnlar triangles are similar. 

Cor. If two angles of one triangle be respectively equal to 
two angles of another, their sides are proportional, and the 
sides opposite to equal angles are homologous. For (I. 20, cor. 
5) the remaining angles are equal, and therefore the triangles 
are equiangular. 

Scho. 2. In a similar manner we may produce a given 
straight line, so that the whole line so produced may have to 
the part produced the ratio of two given straight lines. Thus, 
if BA be the line to be produced, make at B an angle of any 
magnitude, and take BE and CE equal to the other given lines; 
join AC, and draw FE parallel to it. Then, since FE is paral- 
lel to AC, a side of the triangle ABC, we have (V. 2) BF : 
AF : : BE : CE, so that BF has to AF the given ratio. 

Prop. IV. — Theor. — The straight line which bisects an aru- 
gle of a triangle, divides the opposite side into segments ichich 
have the same ratio to one another as the adjacent sides of the 
triangle have ; and (2) if the segm.ents of the base have the 
same ratio as the adjacent sides, the straight line draicn from 
the vertex to the point of section, bisects the vertical angle. 

1. Let the angle BAC of the triangle ABC be bisected by 
the straight line AD ; then BD : DC : : BA : AC. 

Through C draw (I. 18) CE par- 
allel to DA; then (I. 16, cor. l) 
BA produced will meet CE; let 
them meet in E. Because AC 
meets the parallels AD, EC, the g D c 

angle ACE is equal (I. 16) to the 

alternate angle CAD ; and because BAE meets the same paral- 
8 




114 



THE ELEMENTS OF 



[book 



lels, the angle E is equal (I. ] 6, part 2) to BAD ; therefore (I. 
ax. 1) the angles ACE, AEC are equal, because they are re- 
spectively equal to the equal angles, DAC, DAB; and conse- 
quently AE is equal (I. 1, cor.) to AC. Now (V. 2) because 
AD is parallel to EC, one of the sides of the triangle BCE, 
BD : DC : : BA : AE ; but AE is equal to AC; therefore (IV. 
5) BD : DC : : BA : AC. 

2. Let now BD : DC : : BA : AC, and join AD ; the angle 
BAC is bisected by AD. 

The same construction being made, because 
(hyp.) BD : DC : : BA : AC; and 
(V. 2) BD : DC : : BA : AE, 
since AD is parallel to EC ; therefore (IV. 1) BA : AC : : BA : 
AE; consequently (IV. 6) AC is equal to AE ; and (I. 1) the 
angles AEC, ACE are therefore equal. But (I. 16) the angle 
BAD is equal to E, and DAC to ACE ; wherefore, also, BAD 
is equal (I. ax. 1) to DAC; and therefore the angle BAC is 
bisected by AD. The straight line, therefore, etc. 

And if an exterior angle of a triangle be bisected by a straight 
line which also cuts the base produced, the segments between 
the bisecting line and the extremities of the base have the 
same ratio to one another as the other sides of the triangle 
have ; and (2) if the segments of the base produced have the 
same ratio which the other sides of the triangle have, the 
straight line drawn fiom the vertex to the point of section 
bisects the exterior angle of the triangle. 

1. Let an exterior angle CAE of any triangle ABC be 
bisected by AD which meets the opposite side produced in D ; 
then BD: DC:: BA : AC. 

Through C draw (T. 18) CF parallel to AD; and because 
AC meets the parallels AD, FC, the angle ACF is equal (I. 16) 

to CAD ; and because the straight line 
FAE meets the parallels AD, FC, the 
angle CFA is equal to DAE ; therefore, 
also, ACF", CFA are (I. ax. 1 ) equal to 
one another, because they are respect- 
ively equal to the equal angles DAC, 
DAE ; and consequently (I. cor.) AF is 
equal to AC. Then (V. 2) because AD is parallel to FC, a side 




BOOK v.] EUCLID AND LEGENDEE. 115 

of the triangle BCF, BD : DC : : BA : AF ; but AF is equal to 
AC ; as therefore BD : DC : : BA : AC. 

2. Let now BD : DC : : BA : AC, and join AD ; the angle 
CAD is equal to DAE. 

The same construction being made, because 
BD : DC : : BA : AC; and that (V. 2) BD : DC : : BA : AF; 
therefore (IV. V) BA : AC : : BA : AF ; wherefore (IV. 6) AC 
is equal to AF, and (I. 1) the angle AFC to ACF. But (I. 16) 
the angle AFC is equal to EAD, and ACF to CAD ; therefore, 
also (I. ax. 1), EAD is equal to CAD. Wherefore, etc. 

Cor. If G be the point in which BC is cut by the straight 
line bisecting the angle BAC, 

we have (V. 4) BG : GC : : BA : AC ; 
and by this proposition, BD : DC : : BA : AC ; 
whence (IV. 7) BD : DC :: BG : GC, and therefore (V. def 5) 
BD is divided harmonically in G and C. 

8cho. If the triangles be isosceles, the line bisecting the ex- 
terior angle at the vertex is parallel to the base. In this case, 
the segments may be regarded as infinite, and therefore equal, 
their difference, the base, being infinitely small in comparison 
of them. 

Prop. V. — Theor. — If the sides of two triangles^ about each 
of their angles^ he proportionals, the triangles are equiangular, 
and have their equal angles opposite to the homologous sides. 

Let the triangles ABC, DEF have their sides proportionals, 
that AB : BC : I DE : EF ; and BG : CA : : EF : FD ; and 
consequently, ex aequo, BA : AC : : ED : DF; the triangles are 
equiangular, and the equal angles are opposite to the homolo- 
gous sides, viz., the angle ABC equal to DEF, BCA to EFD, 
and BAC to EDF. 

At the points E, F, in the straight 
line EF, make (L 13) the angle FEG 
equal to B, and EFG equal to C. 
Then (V. 3, cor.) the triangles ABC, 
GEF have their sides opposite to the 
equal angles proportionals ; wherefore, 

AB:BC::GF:EF; but (hyp.) 
AB : BC : : DE ; EF. 




116 THE ELEMENTS OF [bOOK V. 

Therefore (TV. 7) DE : EF : : GF : EF ; whence, since DE and 
GF have the same ratio to EF, they are (IV. 6) equal. It 
may be shown in a similar manner that DF is equal to EG; 
ani because, in the triangles DEF, GEF, DE is equal to FG, 
EF common, and DF equal to GE ; therefore (I. 4) the angle 
DEF is equal to GFE, DFE to GEF, and EDF to EGF. Then, 
because the angle DEF is equal to GFE, and (const.) GFE to 
ABC ; therefore the angle ABC is equal to DEF. For the 
same reason, ACB is equal to DFE, and A to D. Therefore 
the ti'iangles ABC, DEF are equiangular. Wherefore, if the 
sides, etc. 

Prop. VI.— Theor. — If two triangles have one angle of the 
one equal to one angle of the other^ and the sides about the 
equal angles proportionals ; the remaining angles are equal^ 
each to each^ viz.y those which are opposite to the homologous 
sides. 

Let the triangles ABC, GEF, of the previous diagrams, have 
the angles ABC, EFG equal, and the sides about those angles 
proportionals ; that is, BA : BC : : GF : EF ; the angle BAG 
is equal to EGF, and ACB to EFG. 

Make (I. 13) the angle FED equal to either of the angles 
ABC, GFE ; and the angle EFD equal to ACB. Then (V. 3, 
cor.), 

BA : BC : : DE : EF. But (hyp.) 
BA : BC : : GF : EF ; 

and therefore (IV. 7) GE : EF : : DE : EF; wherefore ED is 
equal (IV. 6) to FG. Now EF is common to the two trian- 
gles GEF, DEF; and the angle GFE is equal (const.) to 
DEF ; therefore the angle EFD is equal (I. 3) to FEG, and 
D to G. But (const.) the angle EFD is equal to ACB ; there- 
fore ACB is equal to FEG ; and (hyp.) the angle ABC is equal 
to GEF ; wherefore, also (I. 20, cor 5), the remaining angles A 
and G are equal. Therefore, if two triangles, etc. 

Prop. VII. — Theor. — If two triangles have two sides of the 
one proportional to tioo sides of the other, and if the angles 
opposite to one pair of the homologous sides be equal, and 
those opposite to the other pair be either both acute, or not 



BOOK v.] 



EUCLID AND LEGENDRE. 



lit 



G 



acute, the angles contained by the proportional sides are 
equal. 

Let ABC and EFG be two triangles which have tlie sides 
CB, CA proportional to GF, GE ; the angles CBA, GFE equal, 
and the ether angles CAB, GEF 
acute ; then the angles ACB, EGF, 
contained by the proportional sides, 
are equal. 

If the triangle ABC be applied to 
EFG, so that^ CB will foil on GF, 
and the vertex C on the vertex G, 
and make GP equal to CB ; then, 
because the angle CBA is equal to 
the angle GFE, AB will take the 
direction OP parallel to EF (I. 
16). Since OP is parallel to EF, 

GOP is equal to GEF (I. 16, cor.), and we have GF : GP :: 
GE : GO (V. 3) ; but by hypothesis GF : CB : : GE : CA, and 
GP is equal to CB. Hence, CA is equal to GO ; therefore OP, 
which joins the extremities of GP and GO, is equal to AB, 
which joins the extremities of CB and CA (I. ax. 1), and the 
triangles ABC, OPG are equal. Hence, the angles GOP, 
CAB are equal (I. 16, cor, 1), but the angle GOP is equal to 
the angle GEF (I. 16, cor. 1); consequently (I. 14), the re- 
maining angle of GOP is equal to the remaining angle of EFG, 
and the angles ACB, EGF are equal. In the same manner it 
can be shown that the angles ACB, EGF are equal when 
CAB, GEF are not acute. Wherefore, if two triangles have, 
etc. 




Prop. VIII. — Theor. — Tn a right-angled triangle, if a per- 
pendicular be drawn from the right angle to the hypothenuse, 
the triangles on each side of it are simdar to the whole triangle, 
and to one another. ' 

Let ABC be a right-angled triangle, having the riijht angle 
BAC; and from the point A let AD be drawn perpendicular to 
the hypothenuse BC ; the triangles ADB, ADC are sitnilai- to 
the whole triangle ABC, and to one another. 

Because the angle BAC is equal (I. ax. 11) to ADB, each of 




118 THE ELEMENTS OF [bOOK V. 

them beiiiff a right anejle, and that the ansjle B is common to 
the two triangles ABC, ABD ; the remaining angle C is equal 

(I. 20, cor. 5) to the remaining angle 
BAD. Therefore the triangles ABC, 
ABD are equiangular, and (V. 3) the 
sides about their equal angles are 
proportionals ; wherefore (V. def, l) 
the triancfles are similar. In the 
same manner it mioht be demon- 
strated, that the triangle ADC is equiangular and similar to 
ABC ; and the triangles ADB, ADC, being each equiangular 
to ABC, are (I. ax. 1) equiangular, and therefore (V. 3 and 
def. 1) similar to each other. Therefore, etc. 

Cor. From this it is manifest, that the perpendicular drawn 
from the right angle of a right-angled triangle to the hypothe- 
nuse, is a mean proportional (IV. def 9) between the segments 
of the hypothenuse; and also that each of the sides is a mean 
proportional between the hypothenuse and its segment adjacent 
to that side. For (V. 3) in the triangles BDA, ADC, 

as BD : DA : : DA : DC ; in the triangles ABC, DBA, 

as BC : BA : : BA : BD ; and, in the triangles ABC, ACD, 

as BC : CA : : CA : CD. 

Scho. This proposition affords an easy way of solving the 
first corollary to the twenty-fourth proposition of the first book, 
as follows: 

Let ABC be a triangle, right-angled at A ; the square of the 
hypothenuse BC is equivalent to the squares of the legs AB, 
AC. 

Draw AD perpendicular to BC. Then (V. 8, cor.) BC : 
BA : : BA : BD, and BC : CA : : CA : CD. Hence (IV. 2, 
cor. 1) the rectangle BC.BD is equivalent to the square of AB, 
and the rectangle BC.CD to the square of AC. Hence (I. 
ax. 2) BC.BD + BC.CD, or (II. 2) BC==^AB-+ACl 

Prop. IX. — Puob. — To find a third proportioyial to two 
given straight lines. 

Let A and B be two given straight lines ; it is required to 
find a third proportional to them. 



EUCLID AND LEGENDKE. 



119 



BA 




BOOK v.] 

Take two straight lines CF, CG, containing any angle C ; 
and upon these make CD equal to A, and DF, CE each equal 
to B. Join DE, and (I. 18) draw P'G parallel 
to it. EG is the third proportional required. 

For (V. 2) since DE is parallel to FG, CD : 
DF : : CE : EG. But (const.) CD is equal to 
A, and DF, CE each equal to B; therefore A : 
B :: B : EG; wherefore to A and B the third 
proportional EG is found, which was to be 
done. 

aS'c'/^o, Other modes of solving this problem 
may sometimes be employed with advantage. 
The following are among the most useful : 

1. Draw AD (fig. to prop. 8) perpendicular to the indefinite 
straight line BC, and make DB, DA equal to the given lines ; 
join AB, and draw AC perpendicular to it ; DC is the third 
proportional required. For (V. 8, cor.) BD : DA : : DA : DC. 

2. Draw BC perpendicular to AB, and having made BAand 
AC equal to the less and greater of the given lines, draw CD 
and BE perpendicular to AC ; AD will be a third proportional 
to AB and AC, and AE to AC and AB. This follows from the 
third proposition of this book, since the triangles ABC, ACD 
are equiangular, as are also ABC, AEB. 

The angles ABC, ACD, etc., are hex-e made right angles. 
They may be of any magnitude, 
however, provided they be equal. 
It is sufficient, therefore, to draw 
two straight lines, AB, AC, making 
any angle ; to cut off AB, AC equal 
to the given ))roportionals ; and 
then, BC being joined, to make the 

angle ACD equal to ABC, and to draw BE parallel to CD; or 
to make the angle ABE equal to AL'B, and to draw CD paral- 
lel to BE. 

This method affords an easy means of continuing a vseries of 
lines in contiimal proportion, both ways, when any two succes- 
sive terms are given. Thus, after CD and 1)E are drawn, it is 
only necessary to draw DF, EG, etc., parallel to BC, and F'H, 
GK, etc., parallel to CD ; as AD, x\F, iVH, etc., will be the siic- 



G B D H 




120 



THE ELEMENTS OF 



[cook V. 




A C B 



ceedinpj terms of the ascending scries, and AE, AG, AK, etc., 
those of the desceiidiiify one. 

Prop. X. — Prob. — To find a fourth proportional to three 
given straight lines. 

Let A, B, C be three given straight lines; it is required to 
find a fourth proportional to them. 

Take two straight lines DE, DF, contain- 
ing any angle EDF, and make DG equal to 
( A, GE equal to B, and DH equal to C ; and 
having joined GH, draw (I. 18) EF parallel 
to it through the point E; HF is the fourth 
proportional required. 

For (V. 2) since Gil is parallel to EF, as 

DG : GE : : DH : HF; but DG is equal to 

A, GK to B, and DH to C ; therefore as A : 

B : : C : HF; wherefore, to the three given 

straight lines. A, B, C, the fourth proportional HF is found ; 

which was to be done. 

^cho. 1. The solution of this problem may also be effected in 
several different ways; some of which may be meiely indicated 
to the student, as the proofs present no difficulty ; and it is evi- 
dent that, with slight modification, they are applicable in solv- 
ing the ninth proposition, which is only a particular case of the 
tenth. 

1. Let AE,EC (last fig. to HL 20) be the second and third 
terms, placed contiguous, and in the same straight line, and 
draw BE, making any angle with AC, and equal to the first 
term ; through the three jioints, A, B, C, describe a circle cut- 
ting BE produced in D; ED is the fourth proportional. If 
AB and CD be joined, the proof will be obtained by means of 
the triangles ABE, CDE, which are similar. 

2. Draw AB, AC (fig. to IH. 21, cor.) making any angle, and 
make AB, AE equal to the second and third tei'uis; then if AC 
be taken equal to the first term, and a circle be described pass- 
ing thiough B, C, and E, and meeting AC in F, AF will be 
the required line. If BF and EC be joined, the triangles ABF 
and ACE are similar, and hence the proof is immediately ob- 
tained. 



BOOK v.] EUCLID AND LEGENDE"S. 121 

3, Make BD and DA (fig. to V. 8) perpendicular to each 
otlicT, and equal to tlie first and second terms; join AB, and 
draw AC perpendicular to it; in DA, produced through A, if 
necessary, take a line equal to the third term, through the 
upper extremity of which draw a line parallel to AC ; the line 
intercepted on DC, produced if necessary, hetween D and tliis 
parallel, is the fourth proportional required. ^ 

Cor. 1. If four straight lines be proportionals, the rectangle 
contained by the extremes is equivalent to that conta,ined by 
the means; and (2) if the rectangle contained by the extremes 
be equivalent to that contained by the means, the four straight 
lines are proportionals (V. 2, cor.). 

iScko. 2. This corollary, of which the corollary immediately 
following is a case, affords the means of deriving the equality 
of rectangles, and the proportion of straight lines containing 
them, from one another. It evidently corresponds to IV. 
prop. 1, cor., and prop. 2, cor. 1, and it might be regarded as an 
immediate result of those corollaries, without any distinct 
proof, if the lines were expressed (T-. 2-3, cor. 4) by lineal units, 
and the rectangles by superficial ones. This corollary and the 
third proposition of this book, when employed in connection 
with one another, form one of the most powerful instruments 
in geometrical investigations, and they facilitate in a peculiar 
degree .the application of algebra to such inquiries. 

Cor. 2. If three straight lines be proportionals, the rectangle 
contained by the extremes is equivalent to the square of the 
mean ; and (2) if the rectangle contained by the extremes be 
equivalent to the square of the mean, the three straight lines 
are proportionals. 

Prop. XI. — Prob. — To find a mean proportional hetween two 
given straight lines. 

Let AB, BC be two given straight lines ; it is required to find 
a mean proportional between them. 

Place AB, BC in a straight line, and upon AC as diameter 
describe the semicircle ADC ; fi-om B (T. 7) draw BD at right 
angles to AC ; BD is the mean proportional between AB and 
BC. 

Join AD, DC. Then, because the angle ADC in a semicircle 




122 THE ELEMENTS OF [bOOK V. 

is (III. 11) a right angle, and because in the right-angled tri- 
angle ADC, DB is drawn from the right 
angle perpendicular to AC, DB is a mean 
proportional (V. 8, cor.) between AB, BC, 
the segments of the base. Therefore be- 
tween AB, BC, the mean proportional DB 
is found ; Avhich was to be done. 
Scho. Out of several additional ways of solving this prob- 
lem, the following may be mentioned : 

1. On the greater extreme as diameter describe a semicircle ; 
from the diameter cut oif a segment equal to the less extreme, 
through the extremity of which draw a perpendicular cutting 
the circumference ; and the chord drawn from that intersection 
to the extremity of the diameter common to the two extremes 
is the required mean. The proof is manifest from III. 11, and 
V. 8, cor. 

2. Make AD, DC (2d fig. to III. 21) equal to the given ex- 
tremes ; on AC as chord describe any circle, and a tangent 
drawn from D will be the required mean. The proof of this is 
obtained by joining AB and BC, as the triangles ADB, BDC 
are similar. 

When one mean is determined, others may be found between 
it and the given extremes, and thus three means will be insert- 
ed between the given lines ; and by finding means between 
each successive pair of the five terms of which the series then 
consists, the number of means will be increased to seven. By 
continuing the process Ave may find fifteen nieans, thirty-one 
means, or any number which is less by one than a power of 2. 
We cannot find, however, by elementary geometry, any other 
number of means, such as two, four, or five. 

Cor. A given straight line can be divided in extreme and 
mean ratio. 

Prop. XTI. — Theor. — J^quivalent pnrnllelograms inhich Imve 
an an.fjle of the one equal to an angle of the other, have their 
sides about those angles reciprocally i/roportioiHil ; atid (2) 
parallelograms lohich have an angle of the one equal to an an^ 
gle of the < titer, and the sides a,bout those angles reciprocally 
proportional., are equioale^it to one another. 



BOOK v.] EUCLID AND LEGENDKE. 123 

1. Let AB, BC be equivalent parallelograms, which have 
the angles at B equal ; the sides about those angles are recip- 
rocally proportional; that is, DB : BE : : GB : BF. 

Let the sides DB, BE be placed in the same straight line, 
and contiTUOus, and let the parallelograms be on opposite sides 
of DE ; then (I. 10, cor.) because the angles at B are equal, 
FB, BG are in one straight line. Complete the parallelogram 
FE, and (IV. 5) because AB is equal to BC, and that FE is 
another parallelogram, 

AB: FE::BC:FE. But (V. 1) 
as AB to FE, so is the base DB to BE ; and 
as BC to FE, so is the base GB to BF ; 
therefore (IV. 1) as DB : BE : : GB : BF. The sides, therefore, 
of the parallelograms AB, BC, about their 
equal angles, are (V. def 2) reciprocally 
proportional. 




2. But let the sides about the equal an- 
gles be reciprocally proportional, viz., DB : 
BE : : GB : BF ; the parallelograms AB, 
BC are equivalent. 

The same construction being made, because, 
as DB : BE : : GB : BF; and (V. 1) 
as DB : BE : : AB : FE ; and as GB : BF : : BC : FE ; 
therefore (IV. 1) as AB : FE : : BC : FE ; wherefore (IV. 6) 
the parallelogram AB is equivalent to the parallelogram BC. 
Therefore equivalent parallelograms, etc. 

Scho. 1. If AD, CG were produced to meet, it would be easy 
to show, that AB and BC would be the complements of the 
parallelograms about the diagonal of the whole parallelogram 
AC. 

In the demonstration, it should in strictness be proved that 
AF and CE meet when produced. This follows from I. 19. 

Cor. Hence, equivalent triangles which have an angle of the 
one equal to an angle of the other, have their sides about those 
angles reciprocally proportional ; and (2) triangles which have 
an angle of the one equal to an angle of the other, and the sides 
about those angles reciprocally proportional, are equivalent to 
one another. 

JScho. 2. It is evident from the seventh corollary to the fif- 



124 



THE ELEMENTS OF 



[book V. 



teenth proposition of the first book, that this proposition is 
true as well as when the angles are supplemental as when they 
are equal. 



Prop. XIIT. — Prob. — TIpon a given straight line to describe 
a figure similar to a. given rectilineal fi ure, and such that the 
given line shall be homologous to an assigned side of the given 
figure. 

Let AB be the given straight line, on which it is reqnirefi to 
describe a rectilineal figure similar to a given rectilineal figure, 
and such that AB may be homologous to CD, a side of the 
given figure. 

First, let the given rectilineal figure be the triangle CDE. 
Make the angles BAF, ABF respectively equal to DCP], CDE; 
and (V, 3, cor.) the triangle ABF is similar to CDE, and has 
AB homologous to CD. 

Again: let the given figure be the quadrilateral CDOE. 
Join DE, and, as in the first part, describe the triangle ABF 
having the angles BAF, ABF respectively equal to DCE, CDE, 

and also the triangle BFH having the 
angles FBH, BFH respectively equal 
to^EDG, DEC. Then (I. 20, cor. 5) 
the angles BHF, DGE are equal, and 
(const.) A and C are equal. Also, since 
(conet.) ABF, FBH are respectively 
equal to CDE, EDG, the whole ABH 
is equal to the whole CDG. For the 
same reason AFH is equal to CEG ; and therefore the quadri- 
lateral figui-es A BHF, CDGE are equiangular. But likewise 
these figures have their sides about the equal angles propor- 
tional. For the triangles ABF, CDE being equiangular, and 
also BFH, DEG; as BA : AF : : DC : CE; and^as FH. : 
nB::EG: GD. Also, in the same triangles, AF : FB :: 
CE : ED ; and as FB : FH : : ED : EG ; therefore, ex mquo, 
AF : FH : : CE : EG. In the same maimer it may be pi'oved, 
that AB : BH : : CD : DG; whcrefoi-e (V. def 1) the figures 
ABIIF, CDGE are similar to one another. 

Next : let the given figure be CDKGE. Join DG ; and, as in 
the second case, deiscribe the figure ABHF, similar to CDGE, 




BOOK v.] EUCLID AND LKGENDRE. 125 

and similarly situated; also, describe the triangle BHL having 
the angles BHL, IIBL respectively equal to DGK, GDK. Then 
(I. ax. 2) the whole angles FIIL, ABL are respectively equal 
to the whole angles EGK, CDK ; and (I. 20, cor. 5) the angles 
L and K are equal. Therefore the figures ABLHF, CUKGE 
are equiangular. Again : because the quadrilaterals ABHF, 
CDGE, and the triangles BLH, DKG are similar; as FH : 
HB :: EG : GD, and HB : HL : : GD : GK ; therefore, ex 
cequo, FH : HL : : EG : GK. In like manner it may be shown, 
that AB : BL : : CD : DK ; and because the quadrilateials 
ABHF, CDGE, and the triangles BLH, DKG are similar, the 
sides ?,bout the angles A and C, L and K, AFH and CEG are 
proportional. Therefore (V. def l) the five-sided figures 
ABLHF, CDKGE are similar, and the sides AB and CD are 
homologous. In the same manner, a rectilineal figure of six or 
more sides may be described on a given straight line, similar to 
one given ; which was to be done. 

Scho. In practice, if AB be parallel to CD, the cojistruction 
is most easily efll-cted by drawing AF and BF parallel to CE 
and DE ; then FH and BH parallel to EG and DG ; and lastly, 
HL and BL parallel to GK and DK. The doing of this is 
much facilitated by employing tlie useful instrument, the par- 
allel ruler. For the easiest methods, however, of performing 
this and many other problems, the student must have recourse 
to works that treat expressly on such subjects, particulai'ly 
treatises on practical geometry, surveying, and the use of 
mathematical instruments. 

Prop. XIV. — Theor. — Similar plane figures are to one an- 
other in the duplicate ratio of their homologous sides. 

Let ABC, DEF be sitnilar triangles, having the angles B and 
E equal, and AB : BC :: DE : EF, so that (IV. def. 13) the 
side BC is homologous to EF. A 

The ti'iangle ABC has to the trian- 
gle DEF the duplicate ratio of 
that which BC has to EF. 

Take (V. 9) BG a third propor- 
tional to BC, EF, so that BC : EF 
: : EF : BG, and join GA. Then, because 




126 THE ELEITENTS OF [bOOK V. 

AB : BC : : DE : EF ; alternately, 
AB : DE : : BC : EF ; but (const.) 
as BC : EF : : EF : BG ; 
therefore (IV. V) as AB : DE : : EF : BG. The sides, therefore, 
of the triangles ABG, DEF -which are about the equal angles, 
are reciprocally proportional, and therefore (V. 12, cor.) the 
triangles ABG, DEF' are equal. Again : because BC : EF : : 
EF : BG ; and that if three straight lines be proportionals, the 
first is said (IV, def 11) to have to the third the duplicate ratio 
of that which it has to the second ; BC therefore has to BG the 
duplicate ratio of that which BC has to EF. But (V. 1) asBC 
to BG, so is the triangle ABC to ABG. Therefore (IV. 7) the 
triangle ABC has to ABG the duplicate ratio of that which BC 
has to EF. But the triangle ABG is equal to DEF ; wherefore, 
also, the triangle ABC has to DEF the duplicate ratio of that 
which BC has to EF. Therefore, etc. 

Scho. 1. From this it is manifest, that if three straight lines 
be proportionals, as the first is to the third, so is any triangle 
upon the first to a similar triangle similarly described on the 
second. 

The third proportional might be taken to EF and BC, and 
placed from E along EF produced ; and then a triangle equal 
to ABC would be formed by joining D with the extremity of 
the produced line. 

Scho. 2. The above case might also be proved by making 
BH equal to EF, joining AH, and drawing through H a par- 
allel to AC. The triangle cut off by the parallel is equal (1. 14) 
to DEF. But (V. 1) the triangle ABC is to the triangle ABH 
as BC to BH; and, for the same reason, the triangle ABH is to 
the triangle cut off by the parallel, or to DEF, as BC to BH, or 
(const.) as BH to BG. Therefore, ex cequo^ the triangle ABC 
Las to the triangle DEF the same ratio that BC has to BG, or 
(IV. def. 11) the duplicate ratio of that which BC has to EF. 

Again : let ABCD and AEFG be two squares, then will 
they be to one another in the duplicate ratio of their sides — as 
it has been previously demonstrated that similar triangles are 
to one another in the duplicate ratio of their liomologous sides. 
Therefore : 

ABD : AEF in the duplicate ratio of AB to AE ; and ADC : 



BOOK v.] 



EUCLID AND LEGENDEE. 



127 



D 


/ 


/ 





F 



AFG in the duplicate ratio of AB to AE ; hence, by compo- 

pition, ABD+ADC : AEF+ AFG in the 

duplicate ratio of AB to AE ; but ABD 

+ ADC=ABDC, and AEF+AFG = 

AEFG ; wherefore the squares are to 

one another in the duplicate ratio of 

their sides. 

For like reason, the, triangles AIID, 
AFC are to one another in the duplicate 

ratio that PH is to DF, or that AD is to AC ; hence, similar 
triangles are one to another in the duplicate ratio of their alti- 
tudes or bases ; since the segments AHD, AFC have the same 
bases and altitudes as the triangles AHD and AFC, and being 
segments of quadrants, are similar ; 
hence, they are to one another in the du- 
plicate ratio of their bases and altitudes ; 
therefore, by composition, AED + seg. 
AHD : ABC + seg. AFC in the duplicate 
ratio of their homologous sides, or simi- 
lar polygons are to one another in the 
duplicate ratio of their homologous sides, 
and quadrants of circles are one to an- 
other in the duplicate ratio of their radii ; hence, semicircles 
are one to another in the duplicate ratio of their diameters, and. 
circles are one to another in the duplicate ratio of their diame- 
ters. Wherefore, similar surfaces are, etc. 

Cor. 1. Therefore, universally, if three lines be proportionals, 
the first is to the third as any plane figure upon the first to a 
similar and similarly described figure upon the second. 

Cor. 2. Because all squares are similar figures, and all circles 
are similar figures, the ratio of any two squares to one another 
is the same as the duplicate ratio of their sides ; and the ratio 
of any two circles to one another is the same as the duplicate 
ratio of their diameters ; hence, any two similar plane figures 
are to one another as the squares or circles (IV. 7) described on 
their homolocrous sides. 

Cor. 3. Because the sides of similar plane figures are propor- 
tional, therefore (IV. 8) their perimeters or peripheries are 
propoi'tional to the homologous sides ; hence, the perimeters of 




128 THE ELKMENTS OF [bOOK V. 

similar polyo^ons are proportional to their apotliems ; the cir- 
cumfei-enees of circles are proportional to their diameters. 

Cor. 4. Hence plane figures which are similar, to the same 
figure are similar to one another (I. ax. l). 

Cor. 5. If four straight lines be proportionals, the similar 
plane figures, similarly desci-ibed upon them, are also propor- 
tionals; and, conversely., if the similar plane figures, similarly- 
described upon four straight lines, be proportionals, those lines 
are proportionals. 

Cor. 6. Similar polygons inscribed in circles are to one an- 
other as the squares of the diameters. 

Cor. 1. When there are three parallelograms, AC, CH, CF, 
the first, AC (IV. def 10), has to the third, CF, the ratio which 
is compounded of the ratio of the first, AC, to the second, CIT, 
and of the ratio of CH to the third, CF; but AC is to CH as 
their bases ; and CH is to CF as their bases ; therefore AC has 
to CF the ratio which is compounded of ratios that are the 
same with the ratios of the sides. 

Scho. 3. Dr. Simson remarks in his Note on this corollaiy, 
that "nothing is usually reckoned more difficult in the elements 
of geometry by learners, than the doctrine of compound ratio." 
This distinguished geometer, however, has both freed the text 
of Euclid from the errors introduced by Theon or others, and 
has explained the subject in such a manner as to remove the 
difficulties that were formeily felt. According to liim, " every 
proposition in which compound ratio is made use of, may with- 
out it be both enunciated and demonstrated ;" and " the use of 
compound ratio consists wholly in this, that by means of it, cir- 
cumlocutions may be avoided, and thereby propositions may- 
be more briefly either enunciated or demonstrated, or both 
may be done. For instance, if this corollary were to be enun- 
ciated, without mentioning compound ratio, it might be done 
as follows: If two pai-allelograms be equiangular, and if as a 
side of the first to a side of the second, so any assumed straight 
line be made to a second straight line ; and as the other side 
of the first to the other side of the second, so the second 
straight line be made to a third ; the first parallelogram is to 
the second as the first straight line to the third ; and the 
demonstration would be exactly the same as we now have it. 



BOOK v.] EUCLTD AND LEGENDKE. 129 

But the ancient cceometers, when they observed this ennncia- 
ti(ni could l)e made nhortev, l)y giving a name to the ratio 
which the first straiglit line lias to the last, by which name the 
intermediate ratios miglA likewise be signified, of the first to 
the second, and of the second to the third, and so on, if there 
■were more of them, they called this ratio of the fiist to the last, 
the ratio compounded of the ratios of the first to the second, 
and of the second to the third straight line; that is, in the 
present example, of the ratios which are the same with the 
ratios of the sides." 

Scho. 4. The seventh corollary will be illustrated by the fol- 
lowing proposition, which exhibits the subject in a different, 
and, in some respects, a preferable light : 

Triangles which have an angle of the one equal to an angle 
of the othei\ are proportional to the rectangles covitcined by 
the sides about those angles ; and (2) equiangular parallelO' 
grams are proportional to the rectangles co7itained by their ad- 
jacent sides. 

1. Let ABC, DBE be two triangles, having the angles ABC, 
DBE equal; the first triangle is to the second as AB.BC is to 
DB.BE. 

Let the triangles be placed with their equal angles coinciding, 
and join CD. Then (V. 1) AB is to DB as the triangle ABC 
to DBC. But (V. 1, cor. 2) AB : DB :: AB.BC :DB.BC; 
theiefore (IV. 7) AB.BC is to DB.BC as the triangle ABC to 
DBC. Li the same maimer it would 
be shown that DB.BC is to DB.BE as 
the triangle DBC to DBE ; and, there- 
fore, ex ceq^io^ AB.BC is to DB.BE as 
the triangle ABC to DBE. 

2. If parallels to BC through A and 
D, and to AB through C and E were 

drawn, parallelograms would be formed which would be re- 
spectively double of the triangles ABC and DBE, and which 
(IV. 9) would have the same ratio as the triangles; that is, the 
ratio of AB.BC to DB.BE; and this proves the second part of 
the proposition. 

Comparing this proposition and the corollary, we see that the 
ratio which is compounded of ihe ratio of the sides, is the same 
9 




130 



THE ELEMENTS OF 



[book V. 



as the ratio of their rectangles, or the same (T. 23, cor. 5) as the 
ratio of their products, if they he e:q)ressed in numbers. This 
conclusion might also be derived from the proof given in the 
text. For (const.) DC : CE : : CG : K ; whence (V. 10, cor.) 
K.DC = CE.CG. But it was proved that BC : K : : AC : CF; 
or (V. 1) BC.DC : K.DC : : AC : CF; or BC.DC : CE.CG :: 
AC : CF; because K.DC=CE.CG. 

The twelfth proposition of this book is evidently a case of 
this proposition; and the fourteenth is also easily derived 
from it. 




Pijop. XV. — TriEOPw — The parallelograms ahout the diago- 
nal of any parallelogram are similar to the whole, and to one 
another. 

Let ABCD be a parallelogram, and EG, HK the parallelo- 
grams about the diagonal AC ; the parallelograms EG, HK are 
similar to the whole parallelogram, and to one another. 

Because DC, GF are parallels, the angles ADC, AGF are 

(I. 16) equal. For the same reason, be- 
cause BC, EF are parallels, the angles 
ABC, AEF are equal ; and (T. 15, cor. 1) 
each of the angles BCD, EFG is equal 
to the opposite angle DAB, and there- 
fore they are equal to one another ; 
Avherefore, in the parallelograms, the 
angle ABC is equal to AEF, and BAC common to the two tri- 
angles BAC, EAF ; therefore (V. 3, cor.) as AB ; BC : : AE : 
EF. And (IV. 5), because the opposite sides of parallelograms 
are equal to one another, AB : AD :: AE : AG; and DC : 
BC : : GF : EF; and also CD : DA : : FG : GA. Therefore 
the sides of the parallelograms BD, EG about the equal angles 
are proportionals; the parallelograms are, therefore (V. def l), 
similar to one another. In the same manner it would be shown 
that the parallelogram BD is similar to HK. Therefore each 
of the parallelograms EG, HK is similar to BD. But (V. 14, 
cor.) rectilineal figures which are similar to the same figure, are 
similar to one another; therefore the parallelogram EG is simi- 
lar to HK. Wherefore, etc. 

Scho. Hence, GF : FE : : FH : FK. Therefore the sides of 



BOOK v.] 



EUCLID AND LEGENDRE. 



131 



the paralleloojrains GK and EH, about the equal angles at F, 
are reciprocally proportional ; and (V. 12) these parallelograms 
are equivalent ; a conclusion which agrees with the eighth 
corollary to the fifteenth proposition of the first book. 

Prop. XVI. — Prob. — To describe a rectllmeal figure which 
shall be similar to one given rectilineal figure., and equivalent 
to one another. 

Let ABC and D be given rectilineal figures. It is required 
to describe a figure similar to ABC, and equal to D. 

Upon the straight line BC describe (II. 5, scho.) the parallelo- 
gram BE equivalent to ABC ; also upon CE describe the parallel- 
ogram CM equivalent to D, having the angle FCE equal to 
CBL. Therefore (I. 16 and 10) BC and CF are in a straight 
line, as also LE and EM. Between BC and CF find (V. 11) a 
mean proportional GH, and ^ 

on it describe (V. 13) the 
figure GHK similar, and 
similarly situated, to ABC; 
GHK is the figure required. 

Because BC : GH : : GH : 
CF, and if three straight 
lines be proportionals, as the 
first is to the third, so is (V. 

14, cor. 2) the figure upon the first to the similar and similarly 
described figure upon the second ; therefore, 

as BC to CF, so is ABC to KGH ; but (V. 1) 
as BC to CF, so is BE to EF ; 
therefore (IV. 7) as ABC is to KGH, so is BE to EF. But 
(const.) ABC is equivalent to BE ; therefore KGH is equivalent 
(IV. ax. 4) to EF; and (const.) EF is equivalent to D; where- 
fore, also, KGH is equivalent to D; and it is similar to ABC. 
Therefore the rectilineal figure KGH has been described, simi- 
lar to ABC and equivalent to D ; which was to be done. 




Prop. XVII. — Theor. — If two similar parallelograms have 
a comm,on angle, and be similarly situated, they are about the 
same diagonal. 

Let ABCD, AGEF be two similar parallelograms having a 



132 



THE ELEMENTS OF 



[book V. 



C 



E 



F 



B 



common angle CAB, tlicy will be about the same diagonal 

AD. Similar parallelograms have 
their sides about equal angles pro- 
portional (V. def. l). Draw the 
diagonals EG and CB ; hence, AB : 
AG : : AC : AE ; therefore EG ia 
parallel to CB (V. 3), and the angles 
AEG, ACB are equal (I. 16); like- 
wise the angles EGA, CBA. The 
triangles E AG, FG A are equal (I. 15, cor. 5); likewise the tri- 
angles CAB, DBA ; therefore AF is equal to EG, and AD is 
equal to CB ; but AF is the diagonal also of AGEF, and is in 
the same straight line with AD, the diagonal of ABCD. 
Wherefore, if two similar parallelograms, etc. 

Cor. Hence, equiangular parallelograms have to one another 
the ratio which is compounded of the ratio of their sides; 
hence, triangles which have one angle of the one equal, or sup- 
plemental, to one angle of the other, have to one another the 
ratio which is compounded of the ratio of the sides containing 
those angles. 



Pkop. XVIII. — ^Theok. — Of all the parallelof/rmns that can 
he inscribed in any triangle^ that which is described on the half 
of one of the sides as base is the greatest. 

Let ABC be a triangle, having BC, one of its sides, bisected 
in D; draw (I. 18) DE parallel to BA, and EF to BC ; let also 

G be any other point in BC, and 
describe the parallelogram GK; 
FD is greater than KG. 

If G be in DC, through C draw 
CL parallel to BA, and produce 
FE, KH, Gil as in the figure. 
Then (I. 15, cor. 8) the comj>le- 
ments LTI and IID are equivalent ; and since the bases CD, 
DB are equal, the parallelograms ND, DK (I. 15, cor. 5) are 
equivalent. To LH add ND, and to HD add DK ; then (I. 
ax. 2) the gnomort MND is equivalent to the parallelogram 
KG. But (L ax. 9) DL is greater than MXD ; and therefore 




BOOK V.J EUCLID AND LFGENDEE. 133 

FD, which (T. 15, cor. 5) is equal to DL, is greater than KG, 
which is equivalent to the gnomon JNIND. - 

If G were in BD, since BD is equal to DC, AE is equal (V. 
2) to EC, and AF to FB; and by drawing through A a jiaral- 
lel to BC, meeting DE produced, it would he proved in the 
same manner that FD is greater than the inscribed ])aralk'lo- 
gram applied to BG. Therefore, of all the parallelogi-ams, etc. 

Cor. Since (V. lo) all parallelograms having one angle coin- 
ciding with BCL, and their diagonals with CA, arc similar, it 
follows from this proposition that if, on the segments of a given 
straight line, BC, two parallelograms of the same altitude be 
described, one of them, DL, similar to a given parallelogram, 
the other, DF, will be the greatest possible when the segments 
of the line are equal. 

Scho. The parallelogram FD exceeds KG by the parallelo- 
gram OM similar to DL or DF, and described on OH, which 
is equal to DG, the difference of the bases BD and BG. Hence 
we can describe parallelograms equivalent and similar to given 
rectilineal figures. 

The enunciation of this proposition here given is much more 
simple and intelligible than that of Euclid, and the proof is 
considerably shortened, Euclid's enunciation, as given by Dr. 
Simson, is as follows: "Of all pai'allelograms applied to the 
same straight line, and deficient by parallelograms, similar and 
similarly situated to that which is described upon the half of 
the line ; that which is applied to the half, and is similar to its 
defect, is the greatest." It may be remaiked, that this piopo- 
sition, in its simplest case, is the same as the second corollary 
to the fifth proposition of the second book. 

Prop. XIX. — ffHEOR. — In equal circles^ or in the same cir- 
cle^ angles, whether at the centers or circumferences^ have the 
same ratio as the arcs on which they stand have to one another^ 
so also have the sectors. 

Let, in the equal circles ALB, DNE, the angles LGK, KGC, 
CGB, DHN, NHM, and IMITF be at the centers, and the angles 
BAG and EDF be at the circumferences, then will those angles 
have to each other the same ratio as the arcs KL, KC, CB, DN, 
NM, MF, and FE have to one another. 



134 



THE ELEMENTS OF 



[book V. 




Since (T. def. 19) all angles at the center of a circle are 
measured by the arcs intercepted by the sides of the angles, 

and all ancjles at the circum- 

ference are subtended by the 

arcs intercepted by the sides 

of the angles, and (III. 16) 

equal angles will have equal 

arcs whether they be at the 

center or the circumference; 

hence, the same ratio which 

the arcs have to each other, will the angles also have to one 

another — that is, when the arcs be greater, the angles will be 

greater; less, less ; and equal, equal. 

And since the sectors are contained (III. def V) by the sides 
of the anirles and the arcs, the same ratio between the sectors 
will evidently exist as there is between the arcs. Wherefore, 
in equal circles, etc. 

Cor. Hence, conversely^ arcs of the same or equal circles will 
have the same ratio as the angles or sectors which they measure 
or subtend — when greater, greater ; less, less ; or equal, equal. 

Pkop. XX. — ^Peob. — Tlie area of a regular inscribed poly ' 
gon^ and that of a regular circumscribed one of the same nvirv- 
her of sides being given y to find the areas of the regidar in- 
scribed and circumscribed polygons having double the number 
of sides. 

].et A be the center of the circle, BC a side of the inscribed 
polygon, and DE parallel to BC, a side of the circumscribed 
one. Draw the perpendicular AFG, and the tangents BH, 

CK, and join BG; then BG will be a side 
of the inscribed polygon of double the 
number of sides; and (111.^6, cor. l) IIK is 
a side of the similar cii-cumscribed one. 
/ \\ // \ Now, as a like construction would be 
/ x/^ \ Tinule at each of the remaining angles of 

MAN ^^Ijp iwlygon, it will be sufficient to con- 
sider the i)art here represented, as the tri- 
angles connected with it are evidently to each other as the poly- 
gons of which they are parts. For the sake of brevity, then. 



n 



G K 




BOOK v.] EUCLID AND LEGENDKS. 135 

let P denote the polygon whose side is BC, and P' that whose 
side is DE; and, in like manner, let Q and Q' represent those 
whose sides are BG and HK ; P and P/ therefore, are given ; 
Q and Q' required. 

Now (V. ]) the triangles ABF, ABG are proportional to 
their bases AF", AG, as they are also to the polygons P, Q ; 
therefore AF : AG : : P : Q. The triangles ABG, ADG are 
likewise as their bases AB, AD, or (V. 3) as AF, AG; and 
they are also as the polygons Q and P' ; therefore AF : AG : : 
Q : P' ; wherefore (IV. 7)" P : Q : : Q : P' ; so that Q is a mean 
proportional between the given polygons P, P'; and, re))re- 
senting them by numbers, we have Q-=PP'', so that the area 
of Q will be comjyuted by -tnaltlphjing P by P', and extracting 
the square root of the product. 

Again : because AH bisects the angle GAD, and because 
the triangles AHD, AHG are as their bases, Me have Gil : HD 
: : AG : AD, or AF : AB : : AHG : AHD. But we have 
already seen that AF : AG :: P : Q; and therefore (IV. 7) 
AHG : AHD : : P : Q. Hence (IV. 11) AUG : ADG : : P : 
P-|-Q; whence, by doubling the antece- 
dents, 2 AHG : ADG : : 2P : P+Q. But 

D IT P K" "R 

it is evident, that whatever part the tri- 



angle ADG is of P^, the same part of the 
polyg )n Q' is the triangle AHK, which is 
double of the trianojle AHG, Hence the 
last analogy becomes Q' : P^ : : 2P : P+Q. 
Now (IV. 2, cor. 1) the product of the ex- 
tremes is equal to the product of the means; and therefore Q' 
will be computed by dividing twice the product o/'P and V by 
P+Q; and the mode of finding Q has been pointed out 
already. 

Prop. XXI. — ^Theor. — Of regular polygons which have 
eqiial perimeters., that which has the greater mimber of sides is 
the greater. 

Let AB be half the sides of the polygon which has the less 
number of sides, and BC a perpendicular to it, which will evi- 
dently pass through the center of its inscribed or circumscribed 
circle; let C be that center, and join AC. Then, ACB will be 




136 



THE ELEMENTS OF 



[book V. 




the angle at tlie center subtended by the half side AB. ]\I;ike 
BCD equal to the angle subtended at the center of the other 
polygon by half its side, and from C as center, with CD as ra- 
dius, describe an arc cutting AC in E, and CB i)roduced in F. 
Then, it is plain, that the angle ACB i'* to four right angles as 
AB to the common peiimeter; and four right angles are to 
DCB, as the common pei-imeter to the half of 
a side of the other polygon, which, for brevi- 
ty, call S; then, ex mquo^ the angle ACB is 
to DC]} as AB to S. But (V. 19) the angle 
ACB is to DCB as the sector ECF to the 
sector DCF; and consequently (IV. 1) the 
sector ECF is to DCF as AB to S, and, by 
division, the sector ECD is to DCF as AB — 
S to S. Now the triangle ACD is greater 
than the sector CED, and DCB is less than DCF. But (V. 1) 
these triangles are as their bases AD, DB ; therefore AD has to 
DB a greater i-atio than AB — S to S. Hence AB, the sum of 
the first and second, has to DB, the second, a greater ratio than 
AB, the sum of the third and fourth, has to S, the fourth ; and 
therefore (IV. 2, cor. 5) 8 is greater than DB. Let then BG 
be equal to S, and draw GH parallel to DC, meeting FC pro- 
duced in H. Then, since the angles GHB, DCB are equal, BH 
is the perpendicular drawn from the center of the polygon hav- 
ing the greater number of sides to one of the sides ; and since 
this is greater than BC, the like perpendicular in the other 
polygon, while the perimeters are equal, it Ibllows that the area 
of that which has the gi-eater number of sides is greater than 
that of the other. 



Prop. XXIT. — Theor. — If the diameter of a circle 1 e divided 
into any two parts, AB, BC, and if semicircles, ADBjBECy be 
described on opposite sides of these, the circle is divided by 
tJieir arcs into ttoo figures, GDE, FED, the boundary of each <f 
tchich is equal to the circumference q/'FG ; and which are such 
that AC : BC : : FG : FED, and AC : AB :: ¥G : GDE. 

For (V. 14, cor. 3) the circumferences of circles, and conse- 
quently the halves of their circumferences, are to one another as 
their diameters ; thereiore AB is to AC as the arc ADB to 




BOOK v.] EUCLID AND LEGENDRE. 137 

AFC, and BC is to AC as the arc BEC to AFC. Hence (IV. 
1) AC is to AC as the compound arc 
ADEC to AFC; therefore ADEC is p 

equal to half the circumference ;and the 
entire boundaries of the fiirures GDE, 
FED are each equal to the circumfer- 
ence of FG. 

Again (V. 14, cor, 2): circles, and 
consequently semicircles, are to one an- 
other as the squares of their diameters ; 
therefore AC' is to AB^ as the semicircle 

AFC to the semicircle ADB. Hence, since (H. 4) AC' = AB» 
-f2AB.BC + BC^ we find by conversion that AC is to 2AB. 
BC + BC% as the semicircle AFC to the remaining space 
BDAFC; whence, by inversion, 2AB.BC + BC' is to AC' as 
BDAFC to the semicircle AFC. But BC' is to AC= as the 
Bemicircle BEC to the semicircle AFC; and therefore (IV. 1) 
2AB.BC + 2BC' is to AC as the compound figure FDE to the 
semicircle AFC. But (II. 3) 2AB.BC+2BC = 2AC.BC, and 
(V. 1) 2AC.BC : AC^ : : BC : ^AC. Hence the preceding 
analogy becomes BC to ^AC, as FED to the semicircle, or by 
doubling the consequents, and by inversion, AC to BC, as FG 
to FED ; and it would be proved, in the same manner, that 
AC : AB : : FG : GDE. 

Cor. Hence we can solve the curious problem, in which it is 
required to divide a circle into any proposed number of parts, 
equal in area and boundary; as it is only necessary to divide 
the diameter into the proposed number of equal parts, and to 
desci'ibe semicircles on opposite sides. Then, whatever part 
AB is of AC, the same part is AEG of the circle. Their bound- 
aries are also equal, the boundary of each being equal to the 
circumference of the circle. 

Scho. Another solution would be obtained, if the circumfer- 
ence were divided into the proposed number of equal parts, and 
radii drawn to the points of division. This division, hoM'ever, 
can be eftected only in some particular cases by means of ele- 
mentary geometry. 

Pkop. XXIII. — Prob» — To divide a given circle ABC into 



138 



THE ELEMENTS OF 



[book V. 




any proposed number of equal parts by means of concentric 
circles. 

Divide the radius AD into the proposed number of equal 
parts, suppose three, in the points E, F, and through these 
points draw perpendiculars to AD, meeting a semicircle de- 
scribed on it as diameter in G, H; 
from D as center, at the distances DG, 
DH, describe the circles GL, HK ; their 
circumferences divide the circle into 
equal parts. 

Join All, DH. Then (V. 17, cor. 2) 
AD, DH, DF being continual propor- 
tionals, AD is to DF as a square de- 
scribed on AD is to one described on 
DH. But (V. 14, cor. 2) circles are proportional to the squares 
of their diameters, and consequently to the squares of their 
radii. Hence (IV. 7) AD is to FD as the circle ABC to the 
circle HK ; and therefore, since FD is a third of AD, HK is a 
third of ABC. It would be proved in a similar manner that 
AD is to ED as ABC to GL. But ED is two thirds of AD, 
and therefore GL is two thirds of ABC ; wherefore the space 
between the circumferences of GL, HK is one third of ABC, 
as is also the remaining space between the circumferences of 
ABC and GL. 

Cor. Hence it is plain that the area of any annulus^ or nng, 
between the circumferences of two concentric circles, such as 
that between the circumferences of ABC and GL is to the cir- 
cle ABC as the difference of the squares of the radii DM, DL 
to the square of DM ; or (II. 5, cor. 1, and HI. 20) as the rect- 
angle AL.LM, or the square of the perpendicular LB to the 
square of DM; and it therefore follows (V. 14) that the ring is 
equivalent to a circle described with a radius equal to LB. 



Prop. XXIV. — Theor. — If on BC the hypothenuse of a 
ri^/it-angled triangle ABC, a semicircle, BAC, be described on 
the same side as the triangle, and if semicircles, ADB, AEC, ba 
described on the legs, falling without the triangle, the lunes or 
crescents ADB, AEC, bounded by the arcs of the semicircles^ 
are together equal to the right-angled triangle ABC. 



BOOK v.] 



EUCLID AND LEGENDRE. 



139 




For (V, 14 and lY. 9) the semicircle ADB is to the semicircle 
BAG as the square of AB to the square of BC, and AEC to 
BAG as the square of AC to the square of BC ; whence (IV. 1) 
the two semicircles ADB, AEG taken 
together, have to BAG the same ratio 
as the sum of the squares of AB, AG 
to the square of BC, that is, the ratio 
of equality. From these equals take 
the segments AFB, AGG, and there 
remain the luncs DF, EG equal to the triangle ABC. 

Cor. If the legs AB, AC be equal, the arcs AB'B, AGG are 
equal, and each of them an arc of a quadrant ; also the radius 
drawn from A is perpendicular to BC ; and since the halves of 
equals are equal, each lune is equal to half of the triangle 
ABC. 

-If, therefore, ABC be a qnadi'ant, and 
on its choi'd a semicircle be described, the 
lune comprehended between the circum- 
ferences is equal to the triangle ABC ; 
and since (II. 13) a square can be found 
equal to ABC, we can thus effect the 
quadrature of a space, ADC, bounded 
by arcs of circles. 




Prop. XXV. — Prob. — To find the area of a circle. 

Sc/io. 1. The approximate area of a circle can be found by 
means of the twentieth proposition of this book, by what is 
called the method of exha'ustions, giving an error in excess ; 
viz., the approximate area thus obtained is square of radius 
multiplied by .3.1415926, etc. 

Geometry being an exact science, and its conclusions being 
derived from accurate principles, the apj)roximate area for the 
circle is not consistent with the strictness of geometrical rea- 
soning, and the area of the circle must be established exactly 
before it can be regarded a geometrical truth. The reason why 
the method of e haustions gives the a]yjyroxim,ate result, is be« 
cause — by the twentieth proposition of this book — the circum- 
scribed and inscribed regular and similar polygons about the 
circle are supposed^ by continually doubling the number of their 



140 THE ELEMENTS OF [BOOK V. 

sides, to he made equivalent to the circle; but Carnot, in his 
Reflexions sur la Metaphysique da Calcul Infinitesimal^ 
states, "That the ancient geometers did 'not consider it con- 
sistent with the strictness of o-eometrical reasnTiing to re>2:ard 
curve lines as polygons of a great number of sides." Now, 
the area of any regular ])olygon is the rectangle of its a])othem 
and semi-perimeter; but this area is derived from the sixth cor- 
ollary of the twenty-third proposition of the first book — since 
it has been shown in the first corollary of the twentieth propo- 
sition of the same book, that any rectilineal figure can be 
divided into as many triangles as the figure has sides ; there- 
fore, in case of a regular polygon, when triangles are formed ia 
it by straight lines drawn from the center to the extremities of 
the several sides of the polygon, the area of the polygon be- 
comes by the tenth axiom of the firj^t book equivalent to the 
sum of these triangles ; hence (I. 23, cor. 6) each triangle is the 
rectangle of the apothem of the polygon and a semi-side of the 
polygon; therefore the area of the polygon is (I. ax. 10) the 
rectangle of its apothem and its semi-perimeter. Since (I. 23, 
C01-. 4) the aiea of a triangle is derived from the properties of 
parallel straight lives, and any polygon has its sides straight 
lines (I. def 12), the pre perties o^ parallel straight lines are 
applicable to all polygons; but the circle being formed by a 
curve li7ie, the properties of parallel straight lines are not ap- 
plicable to it ; hence the reason is evident why the ancient 
geometers objected to the curve line being regarded a polygon 
of a great number of sides. Euclid, in his ^/e??<e/i^<f, endeav- 
ored to sustain the proposition, that the circle is the I'cctangle of 
its radius and semicircumfereiice, by what is called the indirect^ 
apogogic, or Meductio ad ahsurdwm, method. Now, every 
true pro])osition can be directly demonstrated, and a fair test 
of the truth or falsity of this proposition can be in the success 
or failure of it being directly demonstrated. I have given the 
d reel demonstrations for every other ])roposition in geometry; 
but I can not do so in this case — therefore I believe the proj)0- 
sition fallacious. Archimedes has shown that the relation of 
diameter to the circumference of a circle expressed in numbers, 
to be as 7 to 22 — which is practically correct. Among isoperi- 
metrical figures, the circle contains the greatest area ; there- 



BOOK v.] EUCLID AND LEGENDRE. 141 

fore when 22 expresses the circumference of a circle, the perim- 
eter of its equivalent square must he greater than 22; and if a 
cube be inechauically constructed upon a base whose perimeter 
is 24.2487 + , it will be equivalent to a cylinder of same height, 
the diameter of whose base is 7. 

Now, when 24.2487+ expresses the perimeter of a square, 
each of its sides (I. 23, cor. 1) will be 6.0621 + ; and its area 
■will be 36.75, or three times square of the radius of the circle. 
Hence we get by mechanical construction less than what is ob- 
tained by the method of exhaustions. The geometrical con- 
firmation of the mechanical construction is given in the second 
corollary to the seventeenth proposition of the sixth book. 

Scho. 2. Euclid has endeavored to demonstrate that the cir- 
cle is the rectangle of circumference ajid semi-i'adius. Now, 
the square equal to circle is somewhere between the inscribed 
and circumscribed squares, and its area is equal to its perime- 
ter multiplied by less than semi-radius; consequently the rect- 
angle of circumference and semi-radius will produce more than 
area of circle. 

{^Or Thomson^ s Eficlid, Appendix, DooJc I., Prop. JTJTXZX) 

" The area of a circle is equal to the rectangle under its ra- 
dius, and a straight line equal to half its circumference. Let 
AB be the radius of the circle BC ; the area of BC is equal to 
the rectangle under AB and a straight line D equal to half the 
circumference. 

" For if the rectangle AB. D be not equal to the circle BC, it 
is equal to a circle either greater or less than BC. First, sup- 
pose, if possible, the rectangle AB. D to be the area of a circle 
EF, of which the radius AE is greater than AB." Here Euclid 
is inconsistent with his own proposition : at the very stait he 
bases his argument upon a contradiction. He premises that 
the area of a circle is equal to the rectangle under its radius, 
and a straight line equal to half its circumference ; then sujy- 
pose, i. 6., asks to be granted for the sake of argument, that 
that same rectangle is equal to a larger circle. Why does he 
resort to this subterfuge? It will be said to show the Heductio 
ad absurduni ; very well, let us follow his argument : "and let 
GHK be a regular polygon described about the circle BC, such 



142 THE ELKMENT8 OF [bOOK V. 

that its sides flo not meet the circumference of EF. Then, by 
dividing this polygon into triangles by radii drawn to G, H, 
K, etc., it would be seen that its area is equal to the rectangle 
under AB and half its perimeter. But the perimeter of the 
polygon is greater than the ciixiumference of BC, and therefore 
the area of the polygon is greater than the rectangle AB. D ; 
that is, by hypothesis, than the area of the circle EF, which is 
absurd." What is absurd ? That the circle EF is greater than 
the polygon GHK, etc., or Euclid's argument? The absurdity 
is in considering the area of a circle equal to a larger circle. 
An argument based upon absurdity must necessarily lead to 
absurdity, which in fact has been the case. When Euclid sup- 
posed, i. e., asked to be granted for the cake of argument, AB. 
D= circle EF, it does not prove the area of polygon greater 
than the area of circle EF, because he at the start supposed 
AB. D = circle EF, and consistently with his hypothesis and 
his argument, it must be so to the end ; therefore, consistently 
with his argument and his hypothesis, AB. D is greater than 
the area of polygon GHK, etc. The first part of Euclid's prop- 
osition is nothing more than a demonstration to prove the area 
of a circle is greater than the area of a polygon drawn within 
the circle. And the second part of Euclid's proposition is 
nothing more than a demonstration to prove a circle less than 
the circumscribing polygon. This proposition of Euclid is 
very sophistical, and consequently its fallacy has been imde- 
tected, owing no doubt to the repute of Euclid, and to the sup- 
position that Euclid argued from axioms, and consistently with 
the principles of geometry, which he did ; but in this instance 
he deceived himself, and consequently all those who believe 
him the oracle of geometry. When he attempted to prove 
AB. D=area of circle BC, it was contradictory to his argument 
to suppose AB.'D = area of circle EI*'; because when he based 
his argument upon the premies that AB. D=:area of circle EF, 
consistency demanded that he should stand by his premise, and 
not forsake it as soon as it led to an absurdity, and judge a cir- 
cle less than a polygon within a cii'cle. Tlie absurdity is in 
his own argument, to base it upon a supposition which he knew 
was inconsistent with his proposition, and the inconsistence to 
drop his premiss when he perceived it led to an absurdity ; as 



BOOK v.] EUCLID AND LEGENDRE. 143 

AB. D is less than the circle EF, it is a very fallacious argu- 
ment, when based on the supposition that they are equal, and 
it leads to an absurdity ; and very inconsistent with geometri- 
cal reasoning for Euclid to drop at the conclusion of his argu- 
ment the very premiss upon which he based his argument. 
Every method of demonstration, as well as that method termed 
Heductio adalsurdttm, require that the premiss which is adopt- 
ed at the start be retained to the conclusion. And when Euclid 
adopted AB. D=circle EF at the commencement of his demon- 
stration, consistence of reason and science demanded that he 
fihould have kept it to the conclusion, and then there would 
have been no absurdity, but a demonstration to prove that the 
polygon GHK, etc., is less than circle EF. But Euclid had 
in his mind AB. D = circle BC ; forgetting that he had adopt- 
ed AB. D= circle EF, and had stiU to prove AB. D= circle 
BC. 



END OF BOOK FIFTH. 



BOOK SIXTH. 
ON THE PLANE AND S0LID3. 

DEFINITIONS. 

1. A STRAIGHT line is said to be perpendicular to a plane 
when it makes ri'jrht ansjles with all straiiiht lines meeting' it iu 
that plane. 

2. The inclination of two planes which meet one another is 
the angle contained by two straight lines drawn from any point 
of their common section at right angles to it, one upon each 
plane. The angle which one plane makes >.ith another is 
Boraetimes called a dihedral angle. 

3. If that angle be a right angle, the planes arc perpendicu- 
lar to one another. 

4. Parallel planes are such as do not meet one another, 
though produced ever so far in every direction. 

5. A solid angle is that which is made by more than two 
plane angles meeting in one point, and not lying in the same 
plane. 

If the number of plane angles be three, the solid angle is tri- 
hedral ; if four, tetrahedral ; if more than io\x\\ polyhedral. 

6. Kpolyfiedron is a solid figure contained by plane figures. 

If it be contained by four plane figures, it is called a tetrahe- 
dron ; if by six, a hexahedron ; if by eight, an octahedron ; if 
by twelve, a dodecahedron ; if by twenty, an icosahedron^ etc. 

1. A regular body^ or regular polyhedron^ is a solid con- 
tained by plane figures, which are all equal and similar. 

8. Of solid figures contained by planes^ those are similar 
which have all their solid angles equal, each to each, and which 
are contained by the same number of similar plane figures, simi- 
larly situated. 

9. A pyra^nid is a solid figure contained by one plane figure 
called its base, and by three or more triangles meeting in a 
point without the plane, called the vertex of the pyramid. 



BOOK VI.] EUCLID AND LEGENDEE. 145 

10. A ]jrism is a solid fijjiire, the ends or hases of \vliich are 
parallel, and are equal and similar plane figures, and its otlier 
boundaries are parallelograms. One of tliese parallelograms 
also is sometimes regarded as the base of the j)rism. 

11. Pyramids and prisms are said to be triangular wlien 
their bases are triangles ; quadrangular, when their bases are 
quadrilaterals ; pentagonal, when ihey are pentagons, etc. 

12. The altitude of a pyramid is the perpendicular drawn 
from its vertex to its base; and the altitude of a prism is either 
a perpendicular drawn from any point in one of its ends or 
bases, to the other; or a perpendicular to one of its bounding 
parallelograms from a point in the line opposite. The first of 
these altitudes is sometimes called the length of the prism. 

13. A prism, of which the ends or bases are perpendicular to 
the other sides, is called a right pris7n y any other is an ohluiue 
pristn. 

14. A parallelopiped is a prism of which the bases are par- 
allelograms. 

15. A parallelopiped of which the bases and the other sides 
are rectangles, is said to be rectangular. 

16. A cube is a rectangular parallelopiped, which has all its 
six sides squares. 

17. A sphere is a solid figure described by the revolution of 
a semicircle about its diameter, which remains unmoved. 

18. The axis of a sphere is the fixed straight line about 
which the semicircle revolves. 

19. The center of a sphere is the same as that of the generat- 
ing semicircle. 

20. A diameter of a sphere is any straight line which passes 
through the center, and is terminated both ways by its surface. 

21. A cone is a solid figure described by the revolution of a 
riirht-aniiled triangle about one of the legs, which remains fixed. 

If the fixed leg be equal to the other leg, the cone is called 
a right-angled cone ; if it be less than the other leg, an obtuse- 
angled, and if greater, an acute-angled cone. 

22. The axis of a cone is the fixed straight line about which 
the triangle revolves. 

23. The base of a cone is the circle described by the leg 
■which revolves. 

10 



146 



THE ELEMENTS OF 



[book VI. 



24. A cylinder is a solid figure described by the revolution 
of a rectangle about one of its sides, which remains fixed. 

25. The axis of a cylinder is the fixed straight line about 
which tlie rectano-le revolves. 

26. The bases or ends of a cylinder are the circles described 
by the two revolving opposite sides of the rectangle. 

27. Similar cones and cylinders are those which have their 
axes and the diameters of their bases proportionals. 

PROPOSITIONS. 

Prop. I. — Theor. — One part of a straight line can not be in 
a plane and another part above it. 

J^et EFGH be a plane, then the straight line AB will be 

wholly in the plane. By def 1, 

Book VI., and def 7, Book L, AB, 

being a straight line in the plane 

EFGH, is wholly in that plane, 

and can not have one part in the 

plane and another part above it. 

Cor. 1. Hence two straight lines 

which cut one another are in the same plane ; so also are 

three straight lines which meet one another, not in the same 

point. 

Cor. 2. Hence, if two planes cut one another, their common 
section is a straight line. 




Prop. H. — Theor, — If a straight li?7£ be perpendicular to 
each of two straight lines at their point of intersection^ it is 
also perpendicvlar to the plane in which they are. 

Let the straight line EF be perpendicular to each of the 
straight lines AB, CD at their intersection E ; EF is also per- 
pendicular to the plane passing through AB, CD. 

T.ake the straight lines EB, EC equal to one another, and 
join BC ; in BC and EP" take any points G and F, and join 
EG, FB, FG, FC. Then, in the triangles BEF, CEF, BE is 
equal to CE; EF common; and the angles BEF, CEF are 
equal, being (hyp.) right angles; therefore (I. 3) BF is equal 
to CF. The triangle BFC is therefore isosceles ; and (IL 5, 



BOOK VI.] 



EUCLID AND LKGENDRE. 



IIT 




cor, 5) the square of BF is equivalent to the square of FG and the 
rectangle BG.GC. Foi" the same reason, because (const.) the 
triangle BEG is isosceles, the square of BE 
is equivalent to the square GE and the rect- 
angle BG.GC. To each of these add the 
square of EF ; then tlie squaies of BE, EF 
are equivalent to the squares of GE, EF, 
and the rectangle BG.GC, But (T. 24, cor, 
1) the squares of BE, EFare equivalent to 
the square of V>F, because BEF is a right 
angle; and it has been shown that the square of BF is equiva- 
lent to the .square of FG and the rectangle BG.GC; therefore 
the square of FG and the rectangle BG.GC are equivalent to 
the squares of GE, EF, and the rectangle BG.GC. Take the 
rectangle BG.GC from each, and there remains the square of 
FG, equivalent to the squares of GE, EF ; wherefore (I. 24, cor.) 
FEG is a right angle. In the same manner it would be proved 
that EF is perpendicular to any other straight line drawn 
through E in ihe plane passing through AB, CD. But (VI. 
def 1) a straight line is perpendicular to a plane when it makes 
right angles with all straight lines meeting it in that plane; 
therefore EF is perpendicular to the plane of AB, CD. Where- 
fore, if a straight line, etc. 

Cor. Hence (VI. def. l) if three straight lines meet all in 
one point, and a straight line be perpendicular to each of them 
at that point, the three straight lines are in the same plane, 

Pkop, III. — Theor, — Tf tioo straight lines he perpendicular 
to the same plane, they are parallel to one another. 

Let the straight lines AB, CD be at right angles to the same 
plane BDE; AB is parallel to CD. 

Let them meet the plane in the points B, 
D ; join BD, and draw DE perpendicular to 
BD in the plane BDE ; make DE equal to 
AB, and join BE, AE, AD. Then, because 
AB is perpendicular to the plane, each of the 
angles ABD, ABE is (VL def 1) a right an- 
gle. For the same reason, CDB, CDE are 
right angles. And because AB is equal to 




148 



THE ELKMENTS OF 



[lIOOK VI. 



DE, BD common, and the angle ABD equal to BDE, AD is 
equal (I. 3) to DE. 

A-^ain : in the triangles ABE, ADE, AB is equal to DE, BE 
to AD, and AE common ; therefore (I. 4) the angle ABE is 
equal to EDA ; but ABE is a right angle ; therefore EDA is 
also a right angle, and ED perpendicular to DA ; it is also 
perpendicular to each of the two BD, DC; therefore (VI. 2, 
cor.) these three straight lines DA, DB, DC are all in the same 
plane. But (VI. 1, cor. 1) AB is in the plane in which are BD, 
DA ; therefore AB, BD, DC are in one plane. Now (hyp.) 
each of the angles ABD, BDC is a right angle ; theiefore (L 
16, cor. 1) AB is parallel to CD. Wherefore, etc. 

Cor. 1. Hence (I. def. 11) if two straight lin^s be parallel, 
the straight line drawn from any point in the one to any point 
in the other is in the same plane with the parallels. 

Cor. 2. Hence, also, if one of two parallel straight lines be 
perpendicular to a plane, the other is also perpendicular to it. 

Also, two straight lines which are each of them T)arall(l to 
the same straight line, and are not both in the same plane with 
it, are parallel to one another. 

Scho. The same has been proved (I. 1 7) respecting straight 
lines in the same plane; therefore, universally, straight hues 
•which are parallel to the same straight line, are parallel to one 
another. 



Prop. TV. — Tfieor. — If two^ straight lines meeting one an- 
other be parallel to two others that meet one another^ and are 
not in the same plane with the first two ; tlie first two and the 
other two contain equal angles. 

Let the straight lines AB, BC, which meet one another, be 
parallel to DE, EF, which also meet one 
another, but are not in tlie same ])lane 
with AB, BC ; the angle ABC is equal to 
DEF. 

Take BA, BC, ED, EF, all equal to one 
another, and join AD, CF, BE, AC, DF. 
Because BA is equal and parallel to ED, 
thereft)re AD is (I. 15, cor. l) both equal 
and parallel to BE. For the same reason, CF is equal and 



^ 


N 


E 


\ 



D 



f 



BOOK VI.] 



EUCLID AND LEGKNDKK. 



149 



parallel to BE. Therefore AD and CF Ix'ing each of them 
parallel to BE, are (VI. 3, cor, 2) parallel to one another. 
They are also (I. ax, 1) equal; and AC, DF join them toward 
the same parts; and therefore (I. 15, cor. 1) AC is equal and 
parallel to DF. And because AB, BC are equal to DE, EF, 
and AC to DF, the angle ABC is equal (I. 4) to DEF. There- 
fore, if two stiMight line?, etc. 

Schu. Or supplemental ones, as will be plain after the de- 
monstration here given, if AB be produced through 15. This 
generalizes the third corollary to the sixteenth proposition of 
the first book. 



Prop. V. — Prob. — To clrmo a straight line perpendicular to 
a plane, from a given point above it. 

Let A be the given point above the plane BII ; it is required 
to draw from A a perpendicular to BH. 

In the plane draw any straight line BC, and (I. 8) from A 
draw AD perpendicular to BC. Then, if AD be also perpen- 
dicular to the plane BH, the thing required is done. But if it 
be not, from D (I. 7) draw DE, 
in the i)lane BII, at right angles 
to BC ; from A draw AF perpen- 
dicular to DE; and through F 
draw (I. 18) Gil parallel to^ BC. 
Then, because BC is at right an- 
gles to ED and DA, BC is at 
right angles (VI. 2) to the plane 
passing through ED, DA ; and 

GH being parallel to BC, is also (VT. 3, cor. 2) at right angles 
to the plane through ED, DA ; and it is therefore perjjendicular 
(VI. def 1) to every straight line meeting it in that ])lane; GM 
is consequently perpendicular to AF. Therefore AF is per- 
pendicular to each of the straight lines Gil, DE ; and conse- 
quently (VI. 2) to the plane BHj wherefore AF is the perpen- 
dicular required. 




Prop. VI. — ^Prob. — To draw a straight line perpendicular to 
a given plane from a point given in the plane. 



150 



THE ELEMENTS OF 



[book VI. 



D 



B 



Let A be the point given in the plane ; it is required to draw 
a perpendicular from A to the plane. 

From any point B, above the plane, draw (VI. 5) BC per- 
pendicular to it ; if this pass through A, it is 
the perpendicular required. If not, from A 
draw (I. 18) AD parallel to BC. Then, be- 
cause AD, CB are parallel, and one of them, 
BC, is at right angles to tlie given plane, the 
other, AD, is also (VI. 3, cor. 2) at right angles 
to it. 
Scho. From the same point in a given plane there can not 
be two straight lines drawn perpendicular to the plane upon 
the same side of it ; and there can be but one perpendicular to 
a plane from a point above it. 

Cor. Hence planes to which the same straight line is perpen- 
dicular, are i^arallel to one another. 



E 



Prop. VII. — Theor. — Two planes are parallel^ if tico 
straiglit lines which meet one another on one of them be parallel 
to two which meet on the other. 

Let the straight lines AB, BC meet on the plane AC, and 

DE, EF on the plane DF ; if AB, BC be parallel to DE, EF, 
the plane AC is parallel to DF. 

From B draw (VI. 3, cor. 2) BG perpendicular to the plane 

DF, and let it meet that plane in G ; and through G draw (L 

18) Gil parallel to ED, and GK to 
EF. Then, because BG is perpendicu- 
lar to the plane DF, each of the angles 
BGII, BGK is (VI. def 1) a right an- 
gle ; and because (VI. 3, cor. 2) BA is 
parallel to Gil, each of thom being par- 
allel to DE, the angles GBA, BGII are 
together equal (1. 1 6, cor. 1 ) to two right 

angles. But BGII is a risjht aniile ; 
therefore, also, GBA is a right angle, and GB perpendicular to 
BA. For the same reason, (iB is pei-pendicular to BC. Since, 
therefore, GB is perpendicular to BA, BC, it is perpendicular 
(VL 2) to the plane AC; and (const.) it is perpendicular to the 
plane DF. But (VI. 6, cor.) planes to which the same straight 




BOOK VI.] 



EUCLID AND LEGENDRE. 



151 



line is perpendicular are parallel to one another; therefore the 
planes AC, DF are parallel. Wherefore, two planes, etc. 

Cor. 1. Hence, if two parallel planes be cut by another plane, 
their common sections with it are parallels. 

Cor. 2. If a straight line be perpendicular to a plane, every 
plane which passes through it is perpendicular to that plane. 

C-yr. 3. Hence, if two planes cutting one another be each 
perpendicular to a third plane, their common sectiiBU is perpen- 
dicular to the same plane. 



Paop. VIH. — Theor, — If two straight lines be cut by parallel 
planes, they are cut in the same ratio. 

Let the straight lines AB, CD be cut by the parallel planes 
GH, KL, MN, in the points A, E, B ; C, F, D; as AE : EB : : 
CF : FD. 

Join AC, BD, AD, and let AD meet KL 
in X ; join also EX, XF. Because the 
two parallel planes KL, MN are cut by 
the plane EBDX, the common sections 
EX, BD are (VL 7, cor.) parallel. For 
the same reason, because GH, KL are cut 
by the plane AXFC, the common sections 
AC, XF are parallel. Then (V. 2) because 
EX is parallel to BD, a side of the triangle 
ABD, AE : EB : : AX : XD ; and be- jf 
cause XF is parallel to AC, a side of the triangle ADC, AX : 
XD : : CF : FD ; and it was proved that AX : XD : : AE : 
EB ; therefore (IV. 7) AE : EB : : CF : FD. Wherefore, if 
two straight lines, etc. 




Prop. IX — Theor. — If a solid angle be contained by three 
plane angles, any two of them, are greater than the tliird. 

Let the solid angle at A be contained by the three plane 
angles BAC, CAD, DAB ; any two of these are greater than 
the third. 

If the angles be all equal, it is evident that any two of them 
are greater than the third. But if they be not, let BAC be 
that angle which is not less than either of the other two, and 



153 



THE ELEMENTS OF 



[book VI. 




is rrreator than one of thorn, DAB; aiul make in tlie plane of 
BA, AC the angle BAE equal (I. 13) to DAB; make AE 
equal to AD; through E draw BEC cutting AB, AC in the 
points B, C, and join DB, DC. Then, in tlie triangles BAD, 
BAE, because DA is equal to AE, AB common, and the angle 
DAB is equal to EAB, DB is equal (I. 3) to BE. Again : 
because (I. 2], cor.) BD, I C are greater than CB, and one of 
them, BD, has been pioved equal to BE, a part of CB, there- 
fore the other, DC, is greater (I. ax. 5) than the remaining part, 
EC. Then, because DA is equal (const.) to 
AE, and AC common, but the base DC 
greater than the base EC, therefore (T. 21) 
the angle DAC is greater than EAC, and 
(const.) the angles DAB, BAE are equal ; 
Avherefore (I. ax. 4) the angles DAB, DAC 
are together greater than BAE, EAC, that is, 
than BAC. But BAC is not less than either 
of the angles DAB, DAC ; therefore BAC, 
with either of them, is greater than the 
other. Wherefore, if a solid angle, etc. 

Cor. 1. If every two of three plane angles be greater than 
the third, and if the straight lines which contain them be all 
equal, a ti-iangle may be made, having its sides equal, each to 
each, to the straight lines that join the extremities of those 
equal straight lines. 

Cor. 2. If two solid angles be each contained by three plane 
angles, equal to one another, each to each ; the planes in which 
the equal angles are, have the same inclination. 

Cor. 3. Two solid angles, contained each by three plane an- 
gles which are equal to one another, each to each, and alike 
situated, are equal to one another. 

Cor. 4. Solid figures contained by the same number of equal 
and similar planes, alike situated, and having none of their 
solid angles contained by more than three plane angles, are 
equal and similar to one another. 

Cor. 5. If a solid be contained by six planes, two and two 
of which are parallel, the opposite planes are similar and equal 
parallelograms. 



BOOK VI.] EUCLID AND LEGENDKE. 153 

Prop. X. — Theor. — Ecery solid angle is contained hy plane 
angles^ which are together less than four right angles. 

Let the solid angle <it A be contained hy any nnmher of 
plane angles, BAC,^ CAD, DAE, EAF, FAB ; these together 
are less than four right angles. 

Let the planes in which the angles are be cut by a plane, 
and let the connnon sections of it with those planes be BC, CD, 
DE, EF, FB. Then, because the solid angle at W is contained 
by three plane angles, CBA, ABF, FBC, of which (VL 9) any 
two are greater than the third, CBA, ABF are greater than 
FBC. For the same reason, the two plane angles at each of 
the ])oints C, D, E, F, viz., the angles which are at the bases 
of the triangles having the common vertex A, are greater than 
the third angle at the same point, which is 
one of the ansfles of the figure BCDEF. 
Therefore all the angles at the bases of the 
triangles are together greater than all the 
angles of that figure; and because (L 20) 
all the anoles of the triangles are together 
equal to twice as many right angles as there 
are triangles — that is, as there are sides in 
the figure BCDEF; and that (L 20, cor. 1) all the angles of the 
figure, together with four right angles, are likewise equal to 
twice as many riuht angles as there are sides in thefi<»-ure; 
therefore all the angles of the triangle are equal to all the 
angles of the figure, together with four right angles. But all 
the angles at the bases of the triangles are greater than all the 
angles of the figure, as has been proved ; wherefore the re- 
maining angles of the triangles, viz., those at the vertex, wiiich 
contain the solid angle at A, are less than four ri^-ht angles. 
Therefore, every solid angle, etc. 

Scho. This proposition does not necessarily hold, if any of 
the angles of the rectilineal figure BCDEF be re-entrant; or, 
■which is the same, if any of the planes foi-ming the solid angle 
at A, being produced, pass through that angle. 

Prop. XI. — Prob. — To make a solid angle having the angles 
containing it equal to three given jL>la?ie angles^ any two of 




154 



THE ELEMENTS OF 



[book VI. 



which are greater than the third^ and all three together less than 
four right angles. 

Let B, E, H be given plane angles, any two of which are 
greater than the third, and all of them together less than four 
right angles; it is required to make a solid angle contained by 
plane angles equal to B, E, H, each to each. 

Fjom the lines containing the angles, cut oif BA, BC, ED, 
EF, IIG, HK, all equal to one another, and join AC, DF, GK ; 
then (VI. 9, cor. 1) a triangle may be made of three straight 
lines equal to AC, DF, GK. Let this (L 12) be the triangle 
LMN, AC being equal to LM, DF to MN", and GK to LN. 

About LiMN describe (IIL 25, cor. 2) a circle, and draw the 
radii, LO, JMO, NO ; draw also OP (VI. 6, cor.) perpendicu- 
lar to the plane LMN. Then, any of the radii LO, MO, NO is 
less than AB. Find (I. 24, cor. 3) the side of a square equiva- 
lent to the difference of the squares of AB and LO ; make OP 
equal to that side, and join PL, PM, PN ; the plane angles 
LPM, MPN, and NPL form the solid angle required. 

For, since OP is (const.) perpendicndar to the ])lane LIVIN, 
the angles LOP, MOP, NOP are (VI. def 1) right angles; 
and therefore, since in the triangles OLP, OMP, ONP the 
sides OL, OM, ON are equal, OP common, and the contained 





angles equal, the bases LP, MP, NP are (I. 3) all equal. Also 
(const ) the square of AB is equivalent to the squares of LO, 
OP; and (I. 24, cor. 1) the vsquare of LP is also equivalent to 
the squares of LO, OP, because LOP is a right angle. There- 
fore (I. ax. 1) the square of AB is equal to the square of PL, 
and (I. 23, cor. 3) AB to PL ; and hence all the straight lines 
LP, IMP, NP, liA, BC, ED, etc., are equal. Then, in^lhe tri- 
angles LPM, ABC, the sides AB, BC are equal to LP, PM, 



BOOK VI.] 



EUCLID AND LEGENDRE. 



155 



eacli to each ; and (const.) AC is equal to LM ; therefore (I. 4) 
the angles AI>C, LPM are equal ; and it would be shown in a 
similar manner, that the angle E is equal to ]\IPN, and H to 
NPL. The solid angle at P, therefore, being contained by- 
three plane angles, which are equal to the three given angles, 
B, E, H, each to each, is such as was required. 

Piiop. XTI. — Theor. — A plane cutting a solid., and parallel 
to two of its opposite planes., divides the whole into two solids^ 
the base of one of which is to the base of the other as the one 
solid is to the other. 

Let the solid BC be cut by the plane GF which is parallel to 
the opposite planes BY and IIC, and divides the whole into 
two s( lids, BFand GC; as the base of the first is to the base of 
the second, so is BF to GC. 



N M H G B 



./LTP 



/■ 



V 



LJ-L 



V 



t 



Q 



K 



Produce YC both ways, and take on one side any number of 
straight lines, YK and KL, each equal to YF, and complete 
parallelograms similar and equal to BY. Then, because BY, 
YK, and KL are all equal, the parallelograms on them are also 
equal (I. 15, cor. 5); and for same reason the parallelograms 
on the other side of YC, on the straight lines CQ, QS, each 
equal to FC, are also equal ; therefore three ])lanes of the solid 
XK are equal and similar to three planes of KB, as also to 
three planes of YG. But (VL 9, cor. 5) the three planes oppo- 
site to these three are equal and similar to them hi the several 
solids ; and none of their solid angles are contained bv more 
than three plane angles; therefore (VI. 9, cor. 4) the solids 
XK, KB, and YC are equal. For the same reason, FH, CM 
and QN are also equal. Therefore whatever multiple the base 
LF is of YF, the same multiple is the solid LG of YG. For 
same reason, whatever multiple the base FS is of FC, the same 



156 THE ELEMENTS OF [BOOK TI. 

multiple is the solid FT of FH. And if the base LF be equal 
to SF, the solid LG is equal (VI. 9, cor. 4) to FT ; if greater, 
greater; and if less, less. Therefore (IV. def 5) as the base 
YP^ is to the base FC, so is the solid BF to the solid GC ; 
wherefore, a plane, etc. 

Prop. XTII. — Prob. — At a given point in a given straight 
line, to make a solid angle equal to a given solid angle conr 
tained by three plane angles. 

Let A be a given point in a given straight line AB, and D a 
given solid angle contained by the three plane angles EDO, 
EDF, F'DC; it is required to make at A in the straight line 
AB a solid an le equal to the solid angle D. 

In DF take any point F, from which draw (VI. 5) FG per- 
pendicular to the plane EDO, meeting that plane in G ; join 
DG, and (I. 13) make the angle BAL equal to EDO, and in 
the plane BAL make the angle BAK equal to EDG ; then 
make AK equal to DG, and (VI. 6) draw KH perpendicular to 
the plane BAL, and equal to GF, and join AH. Then the 
solid angle at A, which is contained by the plane angles BAL, 
BAH, HAL, is equal to the given solid angle at D. 

Take AB DE equal to one another; and join IIB, KB, FE, 
GE; and (VI. def 1) because FG is perpendicular to the plane 
EDO, FGD, F'GE are right angles. For the same reason, 
HKA, HKB are right angles; and because KA, AB are equal 
to GD, DE, each to each, and contain equal angles, BK is 
equal (I. 3) to EG ; also KH is equal to GF, and HKB, FGE 
are right angles ; therefore HB is equal to FE. Again : be- 
cause AK, KH are equal to DG, GF, and contain right angles. 





AH 19, equal to DF; also AB is equal to DE, and IIB to FE; 
therefore (L 4) the angles BAH, EDF are equal. Again : since 



BOOK VI.] EUCLID AND LEGENDRE. 157 

(const.) the angle BAL is equal to EDC, and BAK to EDG, 
tlie remaining angles KAL, GDC are (I. ax. 3) eqnal to one 
another; and, by taking AL and DC eqnal, and joining LH, 
LK, CF, CG, it would be proved, as in the foregoing part, that 
the angle HAL is equal (I. 4) to FDC. Therefore, because tlie 
three plane angles BAL, BAII, HAL, which contain the solid 
angle at A, are equal to the three EDC, EDF, FDC, which 
contain the solid angle at D, each to each, and are situated in 
the same order, the solid angle at A is equal (VL 9, cor. 3) to 
the solid angle at D. Therefore, what was required has beeu 
done. 

Prop. XIV. — Theor. — If a parallclopiped he cut by a plane 
passing through the diagonals of two of the opposite planes^ 
it is bisected by that plane. 

Let AB be a parallclopiped, and DE, CF the diagonals of 
the opposite parallelograms AH, GB, viz., those which join the 
equal angles in each. Then (VI. 3, cor. 2) CD, P^E are paral- 
lels, because each of them is parallel to GA ; wherefore (VL 3, 
cor, 1) the diagonals CF, DE are in the plane in which the par- 
allels are, and (VL 8) are themselves parallels. Again: be- 
cause (L 15, cor. 1) the triangle CGF is equal to CBF, and 
DAE to DHE; and that (VL 12) the parallelogram CA is 
equal and similar to the opposite one BE; and 
GE to CH ; therefore the prism contained by 
the two triangles CGF, DAE, and the three 
parallelograms CA, GE, EC, is equal (VL 9, 
cor. 4) to the prism contained by the two tri- 
angles CBF, DHE, and the three parallelo- 
grams BE, CH, EC ; because they are contained by the same 
number of equal and similar planes, alike situated, and none of 
their solid angles are contained by more than three plane an- 
gles. Therefore, if a parallclopiped, etc. 

Seho. The insisting litres of a parallclopiped are the sides of 
the parallelograms between the base and the opposite plane. 

Cor. In a parallclopiped, if the sides of two of the opposite 
planes be each bisected, the common section of the planes pass- 
ing through the points of division, and any diagonal of the par- 
allelepiped bisect each other. 




158 



THE ELEMENTS OF 



[book VI. 




Pkop. XV. — Theor. — ParaUelopipeds upon the same base^ 
and of the same altitude, the insisting lines of which are ter- 
minated in the same straight lines of the />^t</ie opposite to the 
base, are equal to one another. 

Let the parallelopipeds All, AK (2fl fig.) be upon tlie same 

base AB, and of tlie same altitude; and let their insisting lines 

AF, AG, LM, LN be terminated in the same straight line F'N, 

and CD, CE, BH, BK in the same DK ; AH is eqlial to AK. 

First, let the parallelograms DG, HN, whicli are opposite to 

the base AB, have a comtnon side 
HG. Then because AH is cut by the 
plane AGHC passing through the di- 
agonals AG, CH of the opposite planes 
ALGF, CBHD, All is bisected (VI. 
14) by the plane AGHC. For the 
A L same reason, AK is bisected by tlie 

plane LGHB through the diagonals 
LG, BH. Therefore the solids AH, AK are equal, each of 
them being double of the prism contained between the trian- 
gles ALG^ CBH. 

But let the parallelograms DM, EX, opposite to the base, 
have no common side. Then, because CH, CK are parallelo- 
grams, CB is equal (I. 15, cor. 1) to each of the ojjposite sides 
DH, EK; wherefore DH is equal to EK. P^'rom 1)K take sep- 
arately EK, DH; then DE is equal to 
HK ; wherefore, also (I. 15, cor. 5), 
the ti'iangles CDE, BHK are equal ; 
and (I. 15, cor. 5) the parallelogram 
DG is equal to HN. For the same 
reason, the triangle AFG is equal to 
LMN; and (VI. 12) the parallelo- 
gram CF is equal to BM, and CG to BN ; for they are oppo- 
site. Therefore (VI. 9, cor. 4) the prism which is contained by 
the two triangles AFG, CDE, and the three jiarallelograms 
AD, DG, GC, is equal to the prism contained by the two tri- 
angles LMN, BHK, and the three parallelograms BM, MK, KL. 
If, therefore, the prism LMNBHK be taken from the solid of 
which the base is the parallelogram AB, and in which FDKN 
is the one opposite to it ; and if from the same solid there be 




BOOK VI.] 



EUCLID AND LEGENDRE. 



150 



taken the prism AFGCDE, the remaining solids AH, AK are 
equal. Therefore, parallelopipeds, etc. 

Cor. 1. Also parallelopipeds upon the same base and of the 
same altitude, the insisting lines of which are not terminated in 
the same straight lines in the plane opposite to the base, are 
equal to one another. 

Cor. 2. Hence parallelopipeds which are upon equal bases, 
and of the same altitude, are equal to one another (I. 15, 
cor. 5). 

Co:: 3. Parallelopipeds which have the same altitude, are to 
one another as their bases (VI. 12). 

Cor. 4. If thei-e be two triangular prisms of the same alti- 
tude, the base of one of which is a parallelogram, and that of 
the other a triangle ; if the parallelogram be double of the tri- 
angle, the prisms are equal (I. 15, cor. 4). 



o 



M 



-o4 



V. 



y\ 



jy 



Vti 



N 



Prop. XVI. — Theor. — Similar solids are one to another in 
the tnplicate ratio of their homologous sides. 

Let AB, CD be similar parallelopipeds, and the side AE 
homologous to CF ; AB has to CD the triplicate ratio of that 
which AE has to CF. 

Produce AE, GE, HE, and in these produced take EK equal 
to CF, EL equal to FN, and 
EM equal to FR ; and com- 
plete the parallelogram KL 
and the solid KO. Because 
IvE, EL are equal to CF, FN, 
and the angle KEL equal to 
the angle CFN, since it is 
equal to the angle AEG, 
which is equal to CFN, be- 
cause the solids AB, CD are 

similar; therefore the parallelogram KL is similar and equal to 
CN. For the same reason, the parallelogram MK is similar 
aii<l equal to CR, and also OE to FD. Therefore three paral- 
lelograms of the solid KO are equal and similar to three paral- 
lelograms of the solid CD; and (VI. 12) the three opposite 
ones in each solid are equal and similar to these. Therefore 
(VI. 9, cor. 4) the solid KO is equal and similar to CD. Com- 



100 



THE FXKMENT8 OF 



[book VI. 




N 



"sC 



Jj 



R 



plete the paralIeloG:rani GK, and the solids EX, LP upon the 
bases GK, KL. so that EH may be an insisting line in each of 
them ; and thus they are of the same altitude with the solid 

>^B. Then, because the sol- 
ids AB, CD are similar (VI. 
def. 8, and alternately), as 
AE is to CF, so is EG to FN, 
and so is EH to FR ; and FG 
is equal to EK, and FN to 
EL, and FR to EM ; there- 
fore, as AE to EK, so is p]G 
to EL ; and so is HE to EM. 
But (V. 1) as AE to EK, so 
is the parallelogram AG to GK ; and as GE to EL, so is GK 
to KL ; and as HE to EM, so is BE to KM ; therefore (IV. 7) 
as the parallelogram AG to GK, so is GK to KL, and PE to 
KM. But (VI. 13) as AG to GK, so is the solid AB to EX ; 
and as GK to KL, so is the solid EX to PL ; and as PE to 
KM, so is the solid PL to KG; and therefore (IV, 7) as the 
solid AB to EX, so is EX to PL, and PL to KG. But if four 
magnitudes be continual proportionals, the first is said to have 
to the fourth the trijtlicate ratio of that which it has to the sec- 
ond; therefore the solid AB has to KG the triplicate ratio of 
that which AB has to EX. But as AB is to EX, so is the par- 
allelogram AG to GK, and the straight line AE to EK. 
Wherefore the solid AB has to the solid KG the triplicate ratio 
of that which AE has to EK; and the solid KG is equal to CD, 
and the straight line EK to CF. Tliei-efore the solid AB has 
to CD the ti-i]tlicate ratio of that which the side AE has to the 
homologous side CF. 

Since (VI- 15) CNF is half of the base of CD, CRF half of CR, 
and NRF half of NR, which planes form the triangular pyramid 
CNF, K ; and since the triangular pyramid AGE, II is formed ia 
a similai" manner, and the parallelo])iped AB has to the paral- 
lcl()pi|)C(l CI) the triplicate ratio of that which the side AE 
has to tlie side CF ; hence (IV, ax. 1) the triangular ])yramid 
CNF, R has to the triangular j)yrainid AGE, II the triplicate 
ratio of tliat which AE has to CF. And (L 20, cor. 1) all 
polygons can be divided into triangles; therefore (V. 14) solids 



BOOK VI.] EUCLID AND LEGENDRE. 131 

on similar bases liave to one another tlie triplicate ratio of 
lioni<)'o<r()iis >i(les. Wherefore, siniihir solids aie, etc. 

Cor. 1. Hence, similar solids of the same or equal bases are 
to one another as their altitudes — and, conversch', those of the 
same ov equal altitudes are to one another as their bases. 

Cor. 2. Hence, also, the bases and altitudes of equivalent 
polids are reciprocally proportional; and conversely, solids 
lia villi? their bases and altitudes reciprocally proportional, are 
equivalent. 

Cor. 3. Hence, cones and cylinders upon equal bases are a3 
their altitudes, and their bases and altitudes are reciprocally 
proportional when the cones and cylinders are equivalent. 
And similar cones and cylinders have to one another the tripli- 
cate ratio of that which the diameters of their bases have ; and 
gpheres have the triplicate ratio of their diameters (V. 14 and 

VI. 1(3). 

Cor. 4. From this it is manifest, that if four straight lines be 
continual proportionals, as the first is to the fourth, so is the 
parallelopiped described from the first to the similar solid simi- 
larly described from the second ; because the first strais^ht line 
has to the fourth the triplicate ratio of that which it has to the 
second. 

Cor. 5. Parallelopipeds contained by parallelograms equi- 
angular to one another, each to each, that is, of which the solid 
angles are equal, each to each, have to one another the ratio 
which is the same with the ratio compounded of the ratios of 
their sides (V. 14, cor. 7). 

Cor. 6. The bases and altitudes of equivalent parallelopipeds 
are reciprocally proportional; and (2) if the bases and altitudes 
be reciprocally proportional, the pai'alleloj)ipeds are equivalent. 

Prop. XVH. — Tiieor. — Every pyramid is one third the 
prism of the same b-fse and altitude, and every co?ie is one 
third of the cylinder with the same base and altitude., or every 
pyramidal solid is one third the solid of the same base and 
altitude. 

Let there be a prism of which the bases are the triangles 
ABC, DEF ; the prism may be divided into three equal tri- 
angular pyrajnids. 
11 



162 



THL ELEMENTS OF 



[book VL- 




Join BD, EC, CD; and because ABED is a parallelogram, 
and BD its diagonal, the triangles ABD, EBD are (I. 15, cor. 
1) equal; therefore (VI. IC, cor. 3) the pyramid of which the 
base is ABD, and vertex C, is equal to tho 
pyramid of which the base is EBD and vertex 
C. But EBC may be taken as the base of 
this pyramid, and D as its veitex. It is there- 
fore equal (VI. 16, cor. 3) to the pyramid of 
which ECF is the base, and D the vertex; for 
they l)ave the same altitude, and (I. 15, cor. 1) 
equal bases ECF, ECB; and it has been al- 
ready proved to be equal to the pyramid 
ABDC. Therefore the prism A13CDEF is 
divided into three equal pyramids having tri- 
angular bases, viz., into the pyramids ABDC, EBDC, ECFD. 
Therefore, every ti'iangular piism, etc. 

Now every polygon can be divided into triangles (I. 20, cor. 
1) ; hence every j)risni with a polygonal base can be divided 
into i)risms havinij trianirular bases : and as each of these tii- 
angu.lar prisms can be divided into three equivalent triangular 
pyramids, therefjre every pyramid is one third the prism of 
Bame base and altitude. As it has been shown (V. 14) that 
nil surfaces are to one another in the duplicate ratio of their 
liomologous sides, and (VI. IC) all solids are to one another in 
the trij)licate ratio of their homologous sides, it follows that all 
Bolids of similar bases and altitudes have the same propoition 
to one another (VI. 16, cor. 3); hence, cones having similar 
bases and altitudes to cylinders, have the same 
proportion to those cylinders which pyramids 
have to prisms of similar bases and altitudes; 
therefore the cones are one thiid the cylinders, 
as it is evident that the section of the cone and 
cylinder is similar to the section of the pyramid 
and prism of whatever regular and similar base. 
For ABC can be the section of a cone and of a 
pyramid of any regular base, and ABED can be 
the section of a prism of any similar base, and 
pcction of a cylinder of similar base with cone ; therefore 
the same proportion which regulates the respective magnitudes 




BOOK VI,] 



EUCLID AND LEGENDRE. 



163 



of tlie pyramid and pi-ism, also regulates the respective mag- 
nitudes of cone and cylinder — as all surfaces are (V. 14) to 
one another in the duplicate ratio of their homologous sides, 
and (VI, 16) all solids are to one another in the triplicate ratio 
of their homologous sides. 

Cor. 1, p]very sphere is two thirds of its circumscribing cyl- 
inder. Let ABCD be a cylinder circumscribing the sphere 
EFGH ; then will the sphere EFGH be two thirds of the cylin- 
der ABCD, For, let the plane AC be a section of the sphere 
and cylinder through the center J, Join AJ, B.T. Also, let 
FJII be paiallel to AD or BC, and OKL be parallel to AB or 
DC, the base of the cylinder, the latter line, KL, meeting BJ 
in M, and the circular section of the sphere in N. 

Then, if the whole plane HFBC be conceived to revolve 
about the line IIF as an axis, the squai'e 
FG will describe a cylinder, AG ; the 
quadrant JFG will describe a hemis- 
phere, EFG; and the triangle JFB will 

describe a cone, JAB. Also, in the rota- El ^^ |o 

tion, the three lines or parts KL, KN, 
KM, as radii, will describe correspond- 
ing circular sections of those solids, viz., D u c 
KI^, a section of the cylinder; KN, a 

«;ection of the sphere; and KM, a section of the cone. Now, 
FB being equal to FJ or JG, and KL parallel to FB, then by 
eimilar triangles JK is equal to KM. And since in the right- 
angled triangle JKN, JN-OJK--|-KN- (L 24, cor. 1), and be- 
cause KL is equal to the radius JG or JN, and KM is equal to 
JK, therefore KL-<;>KM"-|-KN-; and because circles areas 
the squares of their diameters or the squares of their radii, 
therefore (V. 14, cor. 2) circle of KL is equivalent to circles of 
KM and KN, or section of cylinder is equivalent to both cor- 
responding sections of sphere and cone. And as this will 
always be, it follows that the cylinder EB, which is all the 
fonner sections, is equivalent to the hemisphere EFG and cone 
JAB, which are all the latter sections. But JAB is one third 
of the cylinder EB (VI. 1 7) ; therefore (I. ax. 3) the hemisphere 
EFG is two thirds of the cylinder EB. 

Cor. 2. If the parallelogram BEGC be revolved around the 



K^ 


mX 


' \ 


/^ 


J 

V 


y 



16i 



THE ELEMENTS OF 



[book VI. 




fixed axis BC, it will generate a cylincler (VI. def. 24) ; the 
semicircle BNC will generate a sphere (VI. def. 17) ; and the 
triangle BGC will generate a cone (VI. def. 21). The cone 
Avill be one third the cylinder (VI. 17), and the sphere will bo 
two thirds the same cylinder (VI. 17, cor 1). 

The triangle BOP having one half altitnde and one half base 
of the triangle BGC, will generate a cone 
one eighvh of the cone generated by the 
triangle BGC (VI. 16, cor. 3) ; hence, one 
twelfth of the cylinder generated by the 
square BENP;and the cone generated by 
the triangle BNP is one half cone gener- 
ated by the triangle BGC (VI. 16, cor. 1) ; 
hence, four times cone generated by the 
triangle BOP. And the hemisphere gen- 
erated by the quadrant BNP is two thirds 
cylinder generated by the square BENP 
VI. 17, cor. 1), or eight times cone gen- 
erated by the triangle BOP. 

Let the triangle BSN be described on BN, equal to the tri- 
angle BON (I. 23,and 15, cor. 4). Then the trapezium BSNP 
will generate a solid equivalent to the sum of a cylinder one 
half cylinder generated by the square BENP, and a cone one 
sixth of the same cylinder, or eight times the cone generated 
by the triangle BOP, making a solid equivalent to the hemis- 
phere generated by the quadrant BNP on the same radius PN" 
and same altitude BP. But the triangle BNP is common to 
both the trapezium B>NP and the quadiant BNP, and gener- 
ates in each case the solid equivalent to four times cone gener- 
ated by the tiiangle BOP ; therefore the segment BN and the 
triangle BSN generate an equivalence of solid, or four times 
cone generated by the triangle BOP; consequently the seg- 
ment BN and the triangle BSN are equivalent (I. ax. 1). 

Again : the triangle BNP generates a cone one third the 
cylinder generated by the square BENP (VI. 17), and the 
quadrant BNP generates a heniisphere two thiids of the same 
cylinder (VI. 17, cor. 1). The triangle BNP is one half the 
square BlilNP. 

Now, the trapezium BSNP, equivalent to three fourths of the 



BOOK VI.] EUCLID AND LEGENDRE. 165 

gquiire, on same raflius aiul altitude as the square, generates 
a solid two thirds of the solid generated by the sqnuiv, and the 
quadrant BNP witli same radius and altitude as the square 
BENP generates an equivalent solid with the trapezium 
BSNP. That an equivalence of surfaces ujion the same radius 
•will generate an equivalence of solids can be illustrated by- 
taking a trapezium greater than the trapezium BSNP, having 
tlie same radius. It can easily be shown that the greater trape- 
zium generates a greater solid than the less trapezium, and in 
a similar manner it can be shown that a less trapezium than the 
trapezium BSNP generates a less solid ; hence, very evidently, 
when a greater surface upon same radius generates a greater 
solid, and a less surface generates a less solid, equivalent 
surfaces must generate equivalent solids on the same radius; 
and, conversely, when we have equivalent solids generated 
npon the same radius, the generatirig surfaces are equivalent ; 
therefore (I. ax. l) the quadrant BNP is three fourths of the 
square BENP, or the semicircle BNC is three fourths of the 
parallelogram BEGC, or any circle is three fourths of the cir- 
cumscribing square, or Three Times Square of Badius. 
Hence, we have a geometrical confirmation of the mechanical 
construction in scholium to twenty-fifth proposition of book 
fifth. 

< TIIER'WISR : 

Tlie triangle BGC generates a cone one third (VI. 17) of 
the cylinder generated by tlie rectangle BEGC, and the semi- 
cii'cle BXC genei'ates a sphere two thirds ( VI. 1 7, cor. 1 ) of the 
same cylinder ; the s] here is the mean between the cone and 
cylinder; therefore the semicircle is evidently the meanhe- 
tween the triangle BGC and the rectangle BEGC, or three 
fourths of the rectangle BEGC ; or any circle is three fourths 
square of its diameter, or three times square of its radius. 

OTHERWISE : 

Circles are to one another as the squares described on their 
diameters (V. 14) ; consequently squares are to one another as 
the circles described on their sides; therefore there is an 
equality of proportion (V. 24) ; hence, we derived the arith* 
metical proportion : 



166 THE ELEMENTS OF [boOK YI. 

Rectangle BEGC, semicircle BNC, triangle BGC. 

The sum of extremes is equivalent to twice the mean ; there- 
fore we have — 

Rectangle BEGC -f- triangle BGCo2 semicircle BNC; or, 
semicircle BNCOi rectangle BEGC + i triangle BGC. 

Also, the difference between first and second terms of an 
arithmetical proportion is the same as the difference between 
the second and third terms, as the diffei'ence between third and 
fourth terms, and so on ; hence we have — 

Rectangle BEGC— semicircle BNCOsemicircle BNC -tri- 
angle BGC ; therefore we get segment BXOtriangle BSN, 
or one fourth square BENP; consequently, eircle =0 three 
fourth square of diameter, or tliree times square of radius. 

Cor. 3. Archimedes discovered the proi)ortion 1, 2, 3 be- 
tween the cone, sjihere, and cylinder of similar dimensions; but 
from the previous corollary we obtain the proportion 1, 2, 3, 
4 for the cone, sphere, cylinder, and cube of similar diuiensions; 
because the cube is eight times cube of radius of th^ sjyhere 
(Vr. 10) ; the cylinder is six times cube of radius of the sphere 
(VI. 17, cor, 2) ; the sphere infour times cube of radius (f the 
sphere (VI. 17, cors. 1 and 2) ; and the cone is twice ctibe of 
radius of the sphere (VI, 17, cors, 1 and 2). Hence the sphere 
is the mean proportional of the cone, and the cube circum- 
scribing the si)here, or one half the circumscribiug cube; 
therefore the surface of the sphere is fur times the area of 
one of its great circles, or two thirds the surface of the circum- 
scribing cylinder. Hence, there is the identical proportion be- 
tween the surfaces of the sphere and cylinder as there is 
between \}\q solidities of the s[ihere and cylinder. 

Scho. 1. Therefore the second corollary gives the solution 
to the long mooted and much vexed question of the Quadra- 
ture of the Circle, showing that the perplexity of it arose from 
the uii geometrical sitjyposition (V. 25, scho.) that "the circle is 
a regular polygon of an infinite inind)er of sides." Hence it is 
evident that all conclusions derived fi'oni a fdlacious su])])0- 
sition will give ])erplexity so long as the snpposilion is main- 
tained, and must necessarily involve conti'adiction*^ to the rigor 
of geometrical reasoning. And wlicii demonstrations are con- 
ducted consistently with established definitions, axioms, and 



BOOK VI.] EUCLID AND LKGKNDRE. 16T 

propositions, all conclusions derived from tlieni are unimpeach- 
able, and arc valuable to a system of scientific truths. 

iicho. 2. Geometry, like all other sciences, is based upon 
cvviiuuj'undamencal i)rinciples, and a close examination of tliia 
science reveals the fact that, throughout its whole extent and 
its vai'ious applications, tlie principle that the siraljht line is 
the shortest line bcttoeen tv:o fjiven points is the fundamental 
])rinciple of the science ; by this principle the dimensions of 
magnitudes are determined, distances of objects made known, 
and other useful and practical results ascertained. iSince this 
jjrinciple is so important, it would be interesting to inquire the 
reason why it has such manifest usefulness. The angle is a 
magnitude contained l)y the intersection of two straifjht lii.es^ 
and the polygon is another magnitude bounded by three or 
moYG straight lines ^ lience we see how intimate the connec- 
tion between the straight line and the angle and polygon ; 
therefore we find that the functions of the angle are straight 
lines, such as sines, co-sines, tangents, etc. ; and the properties 
of the polygon ai"e defined by straigld lines, such as its jjerime- 
ter and apothem ; therefore in all rectilineal magnitudes we 
discover a use for the straight line above all other lines, and 
evidently the principle of the straight line has a i)eculiar force 
to all rectilineal figures; consequently we adopt x}w straight 
line as a means of measure for all rectilineal magnitudes. The 
adoption of this m^eans of measure constitutes the straight line 
a standard by which all measurements of rectilineal magni- 
tudes are compared. Hence very naturally there is a consist- 
ency between the measurements and other properties of recti- 
lineal magnitudes. 

Now, when w(^ examine the circle or portions of the circle, 
as the segments, sectors, arcs, etc., we at once discover a nou' 
coincidence between the curve which bounds them and the 
straight line which bounds rectilineal magnitudes ; hence, very 
evidently, the superficies of curvilmear spaces require a pecu- 
liar coni-ection between tiiem and the bounding curve, as there 
is a peculiar connection between the superficies of rectilineal 
spaces and the bounding straight lines. In other words, since 
"vve derive the measurements and other properties of rectilineal 
magnitudes from the principle of the straight line^ so we nmst 



V 



168 



THE ELEMKNTS OF 



[book VI. 



determine tlie measurements and other properties of curvilinear 
magnitudes from the prini-iple of tlie curve, and thus we per- 
ceive wh^, when we endeavor to obtain the area of circle by 
the method of exhaustions, using the straight line as a mean$ 
of measure, we can get the a2:>proxirnate area only, and ahy 
it is necessai-y to obtain accurate results to use the [trniciple of 
the curm. Geometry, in its pi'esent state, is the science of iho 
straight line, and the introduction of the principle of t!ie curve 
into geometrical consideration would usher in a distinct science^ 
but eminently useful in solving problems of cui'vilincar spaces 
and boundaries which were before unsolved, inasmuch as the 
approximate results only were given for them. 

The method of exhaustions is applicable to rectilineal magni- 
tudes, and its results are consistent with the ])rinciple of the 
Btraight line, because the straight line is adopted as a means 
and standard oi' measurement; but since the straight line and 
curve do not coincide, the principle and propeities of the 
straight line are not applicable to curvilinear spaces or bound 
aries ; hence, what is true in one case, becomes absurd in the 
other. 



Prop. XVIIT. — Tiieou. — 77te sections of a solid I y parallel 
planes are similar figures. 

Let the prism JMN be cut by the two parallel planes AD, 
FK ; their sections with it are similar figures. 

For (VI. 7, cor.) the sections have parallel sides (T. 15, cor. 
2). The figures AD, FK, thei-efore, have their 
sides similar, each to each. Their several an- 
gles are also (VI. 4) equal ; for they ai"e con- 
tained by straight lines wliich are parallel ; 
and therefore the figures are similar. 

Cor. 1. A section of a prism by a plane par- 
allel to the base is equal and bimilar to the 
base. 

Scho. 1. Since (VI. def. 24) a cylinder is de- 
scribed by the revolution of a rectangle about 
one of its sides, it is plain that any straight 
line in the rectangle perpendicular to the fixed 
line will describe a circle parallel to the base ; and hence 




BOOK VI.] EUCLID AND LEGENDEE. 169 

every section of a cylinder by a plane parallel to the base is a 
circle equal to the base. 

Cor. 2. The section of a pyramid by a plane parallel to its 
base is a fiijure similar to the base. 

Scho. 2. Since (VI. def. 21) a cone is described by the revo- 
lution of a right-angled triangle about one of its legs, it is plain 
that any straight line in the triangle perpendicular to the fixed 
leg will describe a circle parallel to the base; and the radius 
of that circle will be to the radius of the base as the altitude of 
the cone cut off to that of the whole cone. 

Cor. 3. A section of a sphere by a plane is a circle. 

Since the radii of the sphere are all equal, each of them being 
equal to the radius of the describing semicircle, it is plain that 
if the section pass through the center, it is a circle of the same 
radius as the sphere. But if the plane do not pass through the 
center, draw (VI. 5) a perpendicular to it from the center, and 
draw any number of radii of the sphere to the intersection of 
its surface with the plane. These radii, which are equal, are 
the hypothenuses of right-angled triangles, which have the per- 
pendicular from the center as a common leg; and therefore (I. 
24, cor. 2) their other legs are all equal ; wherefore the section 
of the sphere by the plane is a circle, the center of which is the 
point in which the perpendicular cuts the plane. 

8cho. 3. All the sections through the center are equal to one 
another, and are greater than the others. The former are 
therefore called great circles, the latter small or less circles. 

tScho. 4. A straight line drawn through the center of a circle 
of the sphere perpendicular to its plane is a diameter of the 
sphere. The extremities of this diameter are called the poles 
of the circle. It is plain (I. 24, cor. 2) that chords drawn in 
the sphere from either pole of a circle to the circumfeience are 
all equal ; and therefore (III. 16, cor. 3) that arcs of great cir- 
cles between the pole and circumference are likewise equal. 

Scho. 5. The pyramid or cone cut off from another pyramid 
or cone by a plane parallel to the base is similar (VI. defs. 8 
and 27) to the whole pyi-amid or cone. 

Prop. XIX. — Theor. — If the altitixde of a parallelopipedy 
and the length and p rpendicular breadth of its base be all 



170 THE ELKMKNTS OF [bOOK VI. 

d'vided into parts equal to one another^ the continued p oduct 
of the numler of parts in the three lines is the nvmhtr ofctiba 
contained in the parcdldopiped, each cube havinrj the side of its 
base equal to one (f the parts. 

P'irst, suppose the ])nralk'loi)ipefl to 1)8 rcctanccular, Tlien 
])laiies panvlU'l to the base i)assiiii; tliroiiu'h the points ot'seetion 
of the nh.itude will evidently divide the so'id into as many- 
equal solids as there ave parts in the altitude; and each of 
tliese partial solids will l)e composed of as many cuius as the 
base contains squares, each equal to a base of one of the cubes. 
13ut (I. 23, cor. 4) the number of these squares is the product 
cf the length and breadth of the base; and hence the entire 
number of cubes will be equal to the product of the three 
dimensions, the length, breadth, and altitvide. 

If the base be not rectanj,ndar, its area (I. 23, cor. 5) will be 
the product of its lenjxth and j.erpcndicular breadth; and it is 
evident that the product of this by the altitude will be the 
number of cubes as before. 

Lastly, if the iusistiuii lines be not perpendicular to tlie base, 
still the oblique parallelopi])ed is equal (VI. 15, and cor. 1) to 
a rectano-ularone of the same altitude; and therefore the num- 
ber of cubes will be found as before, by multiplying the area 
of the base by the altitude. 

Cor. 1. Hence it is evident that the volume, or numerical 
solid C07itent, or, as it is also called, the solidify, of a parallelo- 
piped is the product of its altittule and the area of its base 
both expressed in numbers; and it is ])lain that the same 
holds in i-egard to any prism whatever, and also in regard to 
cylinders. 

Cor. 2. The content of a pyramid or cone is found by multi- 
plying the area of the base by the altitude, and tak'ug a third 
of the product. For (VI. 17) a pyramid is a third part of a 
prism, and (VI. 17) a cone a third jtart of a cylinder, of the 
same base and altitude. 

An easy method of comjmting the content of a truncated 
pyramid or cone, that is, the frustum which remains when a 
part is cut from the top by a plane parallel to the base, may be 
thus investigated by the help of algebra. The solid cut off is 
(VI. 18, scho. 3) similar to the whole; and tlierefore the areas 



BOOK VI.] EUCLID AND LEOENDRE. 171 

of their bases will be proportional to the squares of their cor- 
respondino^ dimensions, and consequently to the squares of 
tlieir altitudes. Hence puttins^ V to denote the volume or con- 
tent of the frustum, 11 and B the altitudes and base of the 
whole solid, and h and h those of the solid cut oif, if we put 
^IP to denote B, since B : 6 : : H* : A", or B : 6 : : qW \ qh^^ 
we shall have (IV. 2, cor. 1) b = qh'^-^ and therefore (VI. 19, cor. 
2) the contents of the whole cone and the part cut off are equal 
respectively to ^^■IP and ^qlf \ wherelbre \ = ^q(\¥—h')^ ov^ 
by resolvinnf the second member into factors, V=:i'/(H-+ HA 
+ /r-) {\l-h)=^{qW+q\lh + qh') (H-A). Now ^H^ is equal 
to B, qli' to 6, qWh to a mean proportional between them, 
and II — h to the height of the fi'ustum. Hence, to findthe con- 
tent of a truncated pyramid or cone, add to<jet/ter the areas of 
its two bases aud a mean proportional between theni^ mxdtiply 
the sum by the height of the frustum, anddivide the productby 3, 
This admits of convenient modifications in particular cases. 
Thus, if the bases be squares of which S and s are sides, and if 
a be, the altitude of the frustum, we shall have 

V=ia(S'+Ss-f-0=^»(3Ss+S-— 2Ss+5'); 
or, Vz^ia{3S5+(S— s)^}=a{S5+i(S— s)'}. 

Hence, to find the content (f the frustum of a square pyramid^ 
to the rectangle under the sides of its bases add a third of the 
square of their difference, and mxdtiply the sum. by the height. 
It would be shown in like manner (V. 25, scho. and VI. IV, 
cor. 2), that if II and r be the radii of the bases of the frustum 
of a cone, and a its altitude, 

N '^Za{Vxr\-\^—rY\. 

Solidity of cylinder, 3 x 11" x a. 

Solidity of cone, R^Xflf. 

Solidity of sphere, is 4R'. 

Solidity of spherical sector, is 211' X or. 

Solidity of spherical segment, when it has two bases, is 

f R'4-r)xa+ia'; 

and when it has but one base, 

^R-'x« + ^a'. 

Cor. 3. The content of a polyhedron may be found by divid- 



172 



THE ELEMENTS OF 



[book VI. 



ing it into pyramids, and adding togetlicr tlieir contents. The 
division into pyramids may be matle either by phuies passini^ 
throng!) tlie vertex of one of the soliil angles, or by planes 
passing throngh a point within tlie body. 

Piiop. XX. — Theor. — The surfaces of two similar polyhe- 
drons may be divLdedinto the same number cf similar triangles 
similarly situated. 

This fullows immediately from the definition (VT. def. 8) of 
similar bodies bounded by planes, if the sides or faces of the 
polyhedron be triangles; and any face in the one, and the cor- 
responding face in the other, which are not triangles, are yet 
similar, and may be divided (I. 20) into the same number of 
similai- tiiangles similarlj' situated. 

Cor. Hence it would be shown, as in tlie fourteenth proposi- 
tion of the fifth bo()k, that the surfaces of the polyhedions are 
proportional to any two of their ^similar triangles ; and there- 
fore they are to one another in the duplicate ratio of the 
homologous sides of those triangles, that is, of the edges or in- 
tersections of the similar planes. Hence also the surfaces are 
proportional (V. 14, coi'. 2) to the squares of the edges. 

Prop. XXI. — Theor. — Triangular pyramids are similar, if 
two faces in one of them be similar to twj faces in the other^ 
each to each, and their inclinations equ<d. 

Let ABC, abc be the bases, and D, d the vertices of two tn- 
angular pyramids, in which ABC, DBC are respectively similar 
to abc, dbc, and the inclination of ABC, DBC equal to that of 
abc, dbc ^ the pyi'amids are similar. 

To demonstrate this, it is sufficient to show that the triangles 
j^ AI>D, ACD are similar to ahd^ 

acd, i\)Y then the solid angles 
(Vi. 9, cor. 3) will be equal, each 
to each, and (VI. def 8) the pyr- 
amids similar. Since the plane 
angles at B and b are equal, the 
inclinations of ABC, DBC, and 
of vhc, dbc, are (VI. 9, cor, 2) equal ; therefore ABl), abd are 
equal. Then (hyj..) DB : BC : ; c/i : ic, and BC ; BA : : dc : 




BOOK VI.] EUCLID AND LEGENDRE. 173 

ba/ whence, ex ceqnn, DB : BA : : clb : ha ; and therefore (V. 
6) the trianirles ABD, al>d are equiangular, and consequently 
Bimilar ; and it would be proved in the same maimer that 
ACD, acd are similar. Therelure (VI. def. 8) the pyramids 
are similar. 

Cor. Hence triangular pyramids are similar, if three faces of 
one of them be respectively similar to three faces of the other. 

In the triangular pyramids ABCD, ahcd (see the preceding 
figure), let the faces ABC, ABD, DBC be similar to ahc^ abd, 
doc, each to each ; the pyramids are similar. 

For (V. def 1) AD : DB ■.-. ad : dh, and DB : DC : : cf5 : 
dc ; whence, ex aequo, AD : DC : : ad '. dc. Also DC : CB 
: : dc : cb, and CB : CA : : cb : ca ; whence, ex ceqi/o, DC : 
CA : : dc : ca ; and therefore (V. 5) the triangles ADC, adc 
are equiangular, and (VI. 9, cor. 3, and def. 8) the pyramids 
are similar. 

Prop. XXII. — Tiieor. — Similar polyhedrons may be divid- 
ed nfo the same riumfer of triangular pyratnids, similar, each 
to each, and similarly situated. 

Let ABCDEFG and ahcdefg be similar polyhedrons, having 
the solid angles equal which are marked with the corresponding 
large and small letters ; they may be divided into the same 
number of similar triangular jjyramids similarly situated. 

The surfaces of the polygons may be divided (VI. 20) into 
the same number of sim lar triangles, similarly situated; then 
planes passing through any two corresponding solid angles, A, 
a, and through the sides of all these triangles, except those 




forming the solid angles. A, a, will divide the polyhedrons into 
triangular pyramids, similar to one another, and similarly sit- 
uated. 



174 



TnE ELEMENTS OF 



[book VT. 



The pyramids thus formocl have each one solid anole at the 
common vertex A or a ; and these solid angles may be of three 
classes: Is^ those which have two of their faces coincidins 
with faces of one of the polyhedi'ons; 2f/, those which have 
only one face coinciding ; and 3o?, those which lie wholly with- 
in the solid angle A or a. Now those of the first kind in one 
of the polyhedrons are similar to the corresponding ones in the 
other, by the corollary to the twenty-first proposition of this 
Look ; and those of the second kind by the twenty-first 
From the polyhedrons take two of these similar ])yramids, and 
the remaining bodies will be similar, as the boundaries common 
to them and the pyramids are (VI. 21, and cor.) similar trian- 
gles ; and their other boun<laries are similar, being faces of the 
proposed polyhedrons. Also the solid angles of the remaining 
bodies are equal, as some of them are angles of the primitive 
polyhedrons, and the rest are either trihedral angles which are 
contained by equal plane angles, or may be divided into such. 
From these remaining bodies other similar triangular pyramids 
may be taken in a similar manner, and the process may be con- 
tinued till only two similar triangular pyramids remain ; and 
thus the polyhedrons are resolved into the same number of 
eimilar triangular pyramids. 

Prop. XXIII. — Pitoc. — To find the diameter of a given 
sphere. 

Let A he any point in the surface of the civen sphere, and 
take any three jjoints B, C, I) at equal distances from A. De- 
ecribe the triangle hcd having he equal to the distance or chord 
BC, cd equal to CD, and hd to BD. Find e the center of the 





circle described about hcd^ and join he; draw a^perpendicular 
to he^ and make ha eqtial to BA ; draw ^perpendicular to 6a, 
and af is equal to the diameter of the sphere. 



BOOK VI."] EUCLID AND LEGENDRE. 175 

Conceive a circle to be described through BCD, and E to be 
its center; that circle will evidently be the section of the 
sphere by a plane throui;h B, C, D ; and it will be equal to 
the circle described about bed. Conceive the diameter AEF 
to be drawn, and liA, BE, BF to be joined. Then, in the 
right-angled triangles ABE, abe^ the sides AB, BE are respect- 
ively equal to ab^ be, and therefore (I. 24, cor. 2) the angles A, 
a are equal. 

Again : in the right-angled triangles ABF, abf, the angles 
A, a are equal, and also the sides AB, ab/ hence (I. 14) the 
Bides AF, a/' are equal; that is, a/ is equal to the diameter of 
the sphere. 

Pkop. XXIV. — TiiEOR. — T/ie angle of a spherical triangle is 
the angle formed by the tangents of the arcs forming the 
fpherical angle, and is measured by the arc of a great circle 
described from the vertex as apcle, and intercepted by the sides, 
produced if necessary. 

Let BAC be a spherical angle formed by the arcs AB and 
AC, then it is the same as the angle EAD fonncd by the tan- 
gents EA ajid DA, and is measured by the arc 
of a great circle intercepted by the arcs AB 
and AC, produced if necessary. The tangents 
AE and AD are both jierpendicular to the 
common diameter AH (HI. 12), and being in 
the same planes as the arcs AB and AC, form 
an angle EAD equal to the spherical angle 
BAC. ^ 

Again : the radii FB and FC of the great circle described 
from the vertex as a pole, being in the same planes as the arcs 
AB and AC, and perpendicular to AH, are parallel with AE 
and AD respectively, hence the angle EAD is equal to the 
angle BFC. But the angle BFC is measured by the arc BC 
(I. def. 19) ; therefore, also (I. ax. 1), the spherical angle BAC 
is measured by the aic BC. 

^ho. The angles of spli( rical triangles may be compared to- 
gether by means of the arcs of great circles described from 
their vertices as poles aid included between the arcs forming 




176 THE ELEMENTS OP [bOOK TI. 

the angles, and it is easy to make a spherical angle equal to a 
given angle. 

Cor. 1. If from the vertices of the three angles of a spherical 
triangle as poles, arcs be described forming a spherical tiiangle, 
then the veitices of the angles of this second triangle will be 
respectively poles of the sides of the first, and each angle will 
be measured reciprocally by a semicircumference less the side 
of the other triangle opposite to the angle. 

Because A, B, and C are poles respectively of the arcs FE, 
ED, and DF, the distances of the poles from the extremities of 
their respective arcs are, in each case, a quadrant ; hence the 




extremities of the ai'cs FE, ED, and DF are respectively re- 
moved the length of a quadrant from the extremities of the 
arcs AB, BC, and AC ; therefore the extremities of the former 
arcs are the poles of the latter arcs, each to each. 

Since A is the pole of the arc GH, the angle BAG is meas- 
ured by that arc (VI. 24), and F being the pole of AH, FH is 
a quadrant, and E being the pole of AG, GE is a quadrant; 
hence Fllf GEO semicircumference; but FH + GEoFE-f 
GIl, or the arc GH, which measures the angle BAG, is equiva- 
lent to a semicircumference less the arc FE. In like manner, 
the angle ABC can be shown to be measured by a semicircum- 
ference less the arc DE, and the angle ACB to be measured by 
a semicircumference less the arc DF. And, reciprocally, the 
angle FDE is measured by the arc LO. But LO + BCoLC 
4-B0=O semiciivumfcrence; hence L0=O semicircumference 
minus BC ; and a similar condition can be shown for the other 
ambles of the triauijle FED. 



BOOK VI,] EUCLID AND LEGENDRE. 177 

Cor. 2. As each angle of a spherical triangle is less than two 
right angles, the three angles are less than six right angles. 
And as the sum of the sides of a spherical triangle is less than 
the cij'cumference of a great circle, and the angles being meas- 
ured (VI. 24, cor. 1) by three semicircumferences less the three 
sides of the polar triangle, taking away the sides, we have 
the remainder greater than one semicircumference — or the 
three angles greater than two right angles. Hence the angles 
of a spherical triangle vary between two right angles and six 
right angles, without reaching either limit ; therefore two an- 
gles given can not determine the third. 

12 



END OF BOOK SIXTH, 



THE ELEMENTS OF PLANE TRIGONOMETRY. 

DEFIXITIOXS. 

1. TRiGOJfOMETRY is the practical application of geometrical 
principles for the investigation of ratios of the sides of triangles 
in connection with the magnitudes of their angles. Yor perspi- 
cuity, the vertex of the angle is placed in the center of a circle, 
and the arc of the circumference intercepted by the sides con- 
taining the angle is used as 2i measure of the angle (L def 19). 
Let a straight line be supposed to move around a fixed point ; 
it will make with a stationary line angles which will vary as the 
line is moved, and when it has passed around until it coincides 
with the stationary line from which it is supposed to have 
started, it will have gone over the magnitude of four right 
angles (I. 9, cor.), the extremity of the movable line will trace 
the circumference of a circle, and the successive arcs intercept- 
ed between the movable and stationary lines will give the 
magnitude of the angles (I. def 19). 

2. For the purposes of calculation, a right angle is divided 
into an arbitrary number of equal parts; each one of these 
parts is subdivided into other equal parts, and each part 
of this subdivision undergoes a second subdivision of equal 
parts, and when particular nicety and precision are desired, 
there is a third subdivision of equal parts. Thus, the first 
division of a right angle is into degrees. Among the English 
mathematicians, a right angle has ninety degrees, which divi- 
sion is derived from Greek works, and has great antiquity, 
being used by the remotest ancient mathematicians and astron- 
omers of whom we have any account. Among some modem 
French mathematicians, a right angle is divided into one hun- 
dred degrees, which centesimal division is continued through- 
out all the various subdivisions. But in the Greek division, 
which is more generally used on account of the great facility 
with which 360 can be subdivided, each degree has sixty equal 



PLANE TKIGONOMETET, 179 

parts called minntes^ each minute has sixty equal parts called 
seconds, and each second is sometimes subdivided in deci- 
mal parts, and thus the most extreme minuteness can be ob- 
tained. 

3. The symbols for abbreviation in the expression of the 
value of angles are as follows: °, \ "\ thus, 60°, 15', "lb" are 
read, sixty degrees, fifteen minutes, and twenty-five seconds. 

4. The reason why the right angle is assumed for division is 
because that angle preserves an inversion between every angle 
less than a right angle and its complement (L def 20), and a 
similarity between every angle greater than a right angle and 
its supplement (I. def 20) ; thus, in the first case, the functions 
of the angle are inverted in respect to its complement, and in 
the latter case the functions are the same in respect to its 
supplement, as will more readily be seen by the seventh defi- 
nition and following corollaries. 

5. The straight line drawn from one extremity of an arc, 
perpendicular to the diameter passing through the other ex- 
tremity, is the sine of the angle measured by it ; and the part 
of that diameter intercepted between the sine and the arc is 
the versed sine of the angle which it measures. 

6. If a straight line touch a circle at one extremity of an arc, 
the part of it intercepted between that extremity and the 
diameter produced, which passes through the other, is the tati- 
gent of the angle which it measures; and the straight line 
drawn from the center to the remote extremity of the tangent 
is the secant of the angle. 

Y. The cosine of an angle is the sine of its complement. In 
like manner, the coversed sine, cotangent, and cosecant of an 
angle are respectively the versed sine, tangent, and secant of 
its complement. 

The sine, versed sine, tangent, and secant may be denoted 
by the abbreviated expressions, sin, versin (or vs.), tan, and 
sec/ and the cosine, coversed sine, cotangent, and cosecant, 
by cos, coversin (or covs), cotan, and cosec. For the sake of 
eimplicity, the radius of the circle employed for comparing dif- 
ferent angles is generally taken in investigations as unity; 
when this is not done, it is denoted by its initial letter R. 

The sides of a triangle are often conveniently denoted by tha 



180 



THE ELEarENTS OF 



Bmall letters corresponding to the capital ones placed at the 
opposite angles. Thus, a denotes the side opposite to the an- 
gle A, etc. To prevent ambiguity, we may read A, B, C ; 
angle A, angle B, angle C ; while a, 6, c may be called side a, 
side b, side c. 

To illustrate the foregoing definitions, let C be the center of 
a circle, and AB, DE two diameters perpendicular to each 
other. Through any point F in the circumference draw the 
diameter FL; draw FG perpendicular to AB, and FI to DE; 
through A draw AH perpendicular to AB, and therefore (III. 
8, cor.) touching the circle in A ; and let it meet LF produced 
in H ; and, lastly, draw DK perpendicular to DE, meeting FL 







H 
















D 


/^ 


~^^ 


~y 


!^ 


T^N. 


( 


/ 




r 


^ 


\/ 






Q 


C 


\ ; 


k 


J 


^ 


N 


\. 




V 



E 



produced in K. Then the arc AF contains the same number 
of degrees, etc., as the angle ACF ; and FG is the sine of this 
angle ; FI, or its equal CG, the cosine ; AG the versed sine, 
and DI the coversed sine ; AH the tangent, and CH the secant j 
DK the cotangent, and CK the cosecant. 

From these definitions we derive the following corollaries : 

Cor. 1. The sine, of an angle ACF is half the chord of double 
the arc measuring it. For if FG be produced to meet the cir- 
cumference in N, FN is bisected (III. 2) in G, and (III. 17) the 
arc FAN in A. 

Cor. 2. The sine of the right angle ACD is the radius CD. 

Cor. 3. If AF bo half of AD, and consequently ACF half a 
right angle, the tangent AH is equal to the radius. For A be- 
ing a right angle, II must be half a right angle, and (I. 1, cor. 
2) AH equal to AC. 

Cor. 4. Put the angle = A, and the radius=:l. Then (I. 24, 
cor. 1) FG-+CG'=CF^; that is, siu=A+cos'A = l. In like 



PLANE TRIGONOMETRY. 



181 



manner, we find from the right-angled triangles CAH, CDK, 
that Cir=CA^+AH^ and CK^ = CD'+DK^; that is, sec^A= 
l-ftan^A, and cosec^'A^l+cot'^A. 




Cor. 5. In the similar triangles CGF, CAH, CG : CF, or 
CA : : CA : CH ; that is, the cosine of an angle is to the ra- 
dius as the radius to its secant. Hence also (V. 9, cor.) CG.CH 
=:CA^; that is, cosAsecA = l. It would be found in like 
manner from the triangles GIF, CDK, that sin A cosecA=l, 
CI being equal to the sine FG. 

Cor. 6. In the same triangles CGF, CAH, the cosine CG is 
to the sine GF as the radius CA to the tangent AH ; whence 
(V. 8) cos A tanA=sin A. The triangles CIF, CDK give in 
like manner sin A cot A = cos A. 

Cor. 7. The radius is a mean proportional between the tan- 
gent of an angle and its cotangent. For the triangles CAH, 
CDK are similar; and therefore HA : AC : : CD, or CA : 
DK. Hence (V. 9, cor.) tan A cot A=:l. 

Cor. 8. The sine of an angle, and the sine of its supplement 
are equal. So likewise are their cosines, tangents, cotangents, 
secants, and cosecants. 

Let ACF be an angle, FG, AH its sine and tangent, and 
CG, DK its cosine and cotangent. Make the angle BCM equal 
to ACF ; draw the perpendicular MO ; and produce MC both 
ways to meet HA, KD produced in P and Q. Then (I. def. 
20, and I. 9) the angles BCM, ACM, or ACF, ACM are sup- 
plements of each otlier ; as are also the arcs BM, AM, or AF, 
AM, since (HI. 16) BM, AF are equal. Now the triangles 
CGF, COM are equiangular, and have the sides CF, CM equal ; 



182 



THE ELEMENTS OF 



Q 




T) 


/ 


! 





\ 


/ 


^ 


\ 


V 


C 


\ 


J 

4 



H 



therefore (I. 14) MO is equal to FG, and CO to CG; and MO, 

FG are the sines of ACM, ACF, and CO, 
CG their cosines. Again : the triangles 
ACP, ACH are equiangular, and have 
AC common; therefore (I. 14) AP is 
equal to AH, and CP to CH ; and AP, 
AH are the tangents, and CP, CH the 
secants of ACM, ACF. In like manner it 
would be proved, by means of tlie trian- 
gles CDQ, CDK, that DQ, the cotangent 

of ACM, is equal to DK, the cotangent of ACF, and that their 

cosecants CQ, CK are equal. 



PROPOSITIOXS. 

» 

Pkop. I. — TnEOR. — T/i a right-ancjled triangle the hypothe- 
7iuse is to either of the legs as the radius to the sine of the an^ 
gle opposite to that leg, or to the cosine of the adjacent angle ; 
(2) either of the legs is to the other as the radius to the tangent 
of the angle opposite to the latter ; and (3) either of the legs is 
to the hypothenuse as the radius to the secant of the contained 
angle. 

Let ABC be a triangle, right-angled at C ; then (1) c : h :: 
R : sin B, or cos A ; (2) a : 6 : : R : tau B ; and (3) a : c : : 
R : sec B. 

From B as center, with any radius, describe an arc cutting 
AB, BC in D, E; and through D, E draw (I. 8 and V) DF, 
^ EG perpendicular to BC. Then (Trig. 
defs. 5 and 6) FD, EG, and BG are re- 
spectively the sine, tangent, and secant of 
the angle B ; and, since C is a right angle, 
A and B (Trig, def 4) are complements 
of each other; and therefore (Trig, def 7) 
sinB=cosA. Ao-ain : since the an^le B 
is common to the triangles ABC, DBF, 
GBE, and the angles at C, F, E right angles, these triangles (I. 
20, cor. 5) are equiangular. 

Hence (V. 3) in the triangles ABC, DBF, 

BA : AC : : BD : DF ; that is, c : ^< : : R : sin B, or cos A. 
Again : (V. 3) in the triangles AP.C, GBE, 




PLANE TRIGONOMETRY. 



183 



BC : CA : : BE : EG ; that is, a : 5 : : R : tan B ; and 
BC : BA : : BE : BG ; that is, a : c : : R : sec B. 
Cor. Hence (V. 10, cor.) RJ=c sinB=c cos A; that is, the 
'product of either leg and the radius is equal to the product of 
the hypothenuse and the sine of the angle opposite to that leg^ 
or of the hypothenuse and the cosine of the adjacent angle. 
When R=:l, this becomes simply b=c siu B=c cos A. Again : 
J=a tan B and c=:a sec B. 




Prop. II. — Theor. — The sides of a plane triangle are pro- 
portional to the sines of the opposite angles. 

Let ABC be any triangle ; then a : b : : sin A : sin B ; a : 
c : : sin A : sin C ; and b : c : : sin B : sin C. 

Draw AD perpendicular to BC; then AD is a leg of each of 

the right-angled triangles ADB, 
ADC; and therefore (Trig, 1, cor.) 
R.AD=AB sin B, and R.AD^AG 
sin C. Hence (I. ax. 1) AB sin B = 
AC sin C, or c sin B = 5 sin C; 
whence (V. 10, cor.) b : c : : sin B 
: sin C ; and, by drawing perpendic- 
ulars from B and C to the opposite sides, it would be proved 
in a similar manner that a : c : : sin A : sin C, and a : b :: sin 
A : fin B. 

Cor. From B as center M'ith BA as radius, describe an arc 
AD ; and from C as center, with an equal radius, describe an arc 
EF. Draw AG, EH perpendicular to BC ; these (Trig. def. 5) 
are respectively the sines of B and C to equal radii. Then 
the triangles AGC, EHC are equiangular, the angle at C being 
common, and the angles at G and H right angles. Hence (V. 
3) CA : AG : : CE, or (const.) AB : EH ; and, alternately, 
CA : AB : : AG : EH ; that is, 5 : c : : sin B : sin C. 

The demonstration is simplified by taking, as here, one of the 
sides, AB or AC, as radius. This, however, is not essential, as 
arcs may be described from B and C as centers, with equal 
radii of any magnitude, and their sines, and a perpendi-cular 
from A to BC being drawn, the proof will be readily obtained. 
Scho. From one of the foregoing analogies we have, by in- 
version, c : J : : sin C : sin B. If C be a right angle, this 



184 THE ELEMENTS OF 

(Trig. def. cor. 2) becomes c : Z> : : R : sin B, as in Prop. L 
The first part, therefore, of that proposition is a particular case 
of this one. 

Prop. ITT. — Theor. — The sum of any Uco sides of a trian- 
gle is to their difference as the tangent of half the snm of the 
angle opposite to those sides is to the tangent of half their dif- 
ference. 

].et ABC be a triangle, a, h any two of its sides, of which a 
is the greater, and A, B the angles opposite to them ; then 
a+b : a-h :: tan i(A + B) : tan i(A— B). 

From C as center, with the greater side a as radius, describe 

the circle DBE, cutting AC produced 
in D and E, and BA produced in F ; 
join BD, BE, CF ; and draw EG par- 
allel to AB, meeting DB produced in 
G. 

Then because DC and CE are each 
equal to a, DA is equal to a+5, and 
AE to a—h. 

Also (I. 20) the exterior angle DCB is equal to A + B ; and 
DEB, which is at the circumference, is (III. 10) half of DCB, 
•which is at the center ; therefore DlEB=i(A + B). 

Again (I. 1, cor. l) : the angle F is equal to B ; and (I. 20) 
in the triangle ACF, the exterior angle A = ACF+F=ACF+ 
B; and consequently, ACF= A— B ; and (III. 10) ABE, or (I. 
16) its equal, BEG = |(A— B). 

Now, since (III. 11) EBD, being in a semicircle, is a right 
angle, as also (I. 9) EBG ; if a circle were described from E as 
center, with EB as radius, DBG (III. 8, cor.) would touch it, 
and (Trig, def 6) DB would be the tangent of DEB, and BG 
of BEG; and therefore DB, BG will evidently be proportional 
to the tangents of those angles to any other radius. 

Or strictly, EB : BD : : 1 : tan DEB (Trig. 1) and (inver.) 

BD : EB :: tan DEB : 1. Also (Trig. 1) EB : BG :: 1 : 

tan BEG. Hence, ex mquo, BD : BG : : tan DEB : tan BEG. 

Lastly, since BA (const.) is parallel to GE, we have (V. 2) 

DA : AE : : DB : BG ; that is, 

a-\-b : a—b : : tan | (A+B) : tan i(A— B). 




PLANE TRIGONOMETKT. 



185 



Prop. IV — Tiieor. — hi a plane triangle^ the cosine of half 
the difference of any two angles is to the cosine of half their 
sum, as the sum. of the opposite sides to the third side ; and (2) 
the sine of half the difference of any two angles is to the sine 
of half their sum, as the difference of the opposite sides to the 
third side. 

Let ABC (see the last proposition) be any plane triangle ; 

then cosi(A — B) : cos^(A + B) : : a+b : c; 
and sin \{A — B) : sin -KA+B) : : a — b : c. 

For it was shown in the preceding proposition, that BED= 
i(A+B), and ABE=i(A-B) ; and since DBE is a right an- 
gle, DBA is the complement of ABE, and D of BED. But 
(Trig. 2) in the triangle ABD, sin ABD : sin D : : AD : AB ; 
that is, (Trig. def. 7) cosi(A— B) : cosi(A+B) :: a+b : c. 
Again (Trig, 2) : in the triangle ABE, sin ABE : sin AEB : : 
AE : AB J that is, sin ^(A — B) : sin i(A+B) : : a — b : c. 

Prop. V. — Theor. — In any plane triangle the sum of the 
segments of the base made by a perpendicular from the vertex, 
is to the sum of the other sides as the difference of those sides 
to the difference of the segments. 

Let ABC be a triangle, -and AD a perpendicular from the 
vertex to the base ; the sum of the segments BD, DC is to the 
sum of the sides AB, AC, as the difference of AB, AC to the 
difference of BD, DC. 





For (IL 5, cor. 4) the rectangle under the sum and difference 
of AB, AC is equivalent to the rectangle under the sum and 
difference of BD, DC; and therefore (V. 10, cor.) the sum of 
BD, DC is to the sum of AB, AC, as the difference of AB, AC 
to the difference of BD, DC. 

Scho. If the perpendicular fall within the triangle, the seg- 



186 THE ELEMENTS OF 

ments make up the base, and their difference is less than the 
base; but if the perpendicular fall Avithout the triangle, as it 
does (second fig.) when one of the angles at the base is obtuse, 
the base is the difference of the segments, and their sum is 
greater than the base. 

Peop. VI. — Theor. — The rectangle under two sides of a tri- 
angle is to the rectangle under the excesses of half the perimeter 
above those sides, as the square of the radius to the square of 
the sine of half the contained angle. 

Let ABC be a triangle, and let s=i\{a-{-b+c) ; then be : (s 
— b) {s — c) : : R^ : sin'^A. 

Produce the less side AC through C, making AD equal to 
A AB ; join BD ; and draw AE, CF per- 

pendicular, and CG parallel to BD ; 
then (I. 24, cor. 2) AE bisects BD and 
the angle A. Now (II. 5, cor. 4) the 
rectangle under the sum and difference 
of BC, CD is equivalent to tlie rectangle 
under the sum and difference of BF, 
FD, that is, under BD and twice EF ; therefore the rectangle 
under half the sum and half the difference of BC, CD is equiva- 
lent to the rectangle BE.EF. But (Tkig. 1) 

AB : BE : : R : sin 1 A, and 

AC : CG or EF : : R : sin i A; whence (IV. 15) 

AC.AB : BE.EF, or ^ (BC + CD). i (BC-CD) : : R^ : sin^ A ; 

or, be : | (a+c — b). | (a+5— c) : : R^ : sin | A, 
because CD=c—b : and let 2s=a-\-b-\-c ; then, s — a=^ {b-\-c 
— a);s — b—i {a-\-c — b) ; and s — c—i {a+b — c) ; then we have 
be : (s—b) (s—c) : : R^ : suri A. 

Cor. Hence, taking R = l dividing the product of the means 
by the first extreme, and exti acting the square root, we find 

sin^A = Y- ^ 5 and it is plain that we should find in a 

similar manner, sin f o=\/ —- •, and sin 4- C = 

y ac 

V ab 





PLANE TRIGONOMETRY. 187 

Prop. YII. — Theor. — The rectangle under two sides of a 
triangle is to the rectangle under half the perimeter and its ex- 
cess above the third side, as the square of the radius to the 
square of the cosine of half the angle contained by the two 
sides. 

Let ABC be a triangle, and let s=^ {a-^-b+c) ; then be : s 
{s—a) ::RM cos^A. 

Produce tlie less side CA, through A, making AD equal to 
AB ; join BD ; draw AE, CF perpen- d^ 
dicular, and AG parallel to BD. Then 
BD (I. 24, cor. 2) is bisected in E ; and 
the angle BAG being (I. 20) equal to the 
two equal angles D and ABD, each of 
them is equal to half the angle BAG, 

that is, half the angle A in the triangle ABC; and (I. 16) 
GAG is equal to D. Now, it would be shown, as in the pre- 
ceding proposition, that the rectangle under half the sum and 
half the diiference of DC, CB is equivalent to the rectangle 
BE.EF. But (Trig. 1) AB : BE : : R : cos ABE, or cos ^ A ; 
and AG : AG, or EF : : R : cos GAG, or cos ^ A ; whence (IV. 
15), 
AG.AB : BE.EF, or i (DC+GB). i (DG-GB) : : R' : cos^A ; 

ovbc: \ {a+b-\-c). -J {b-\-c — a) : : R^ : cos^^ A, 
because DG=6+c. If 2s be the same as last proposition : 
be : 5 (5 — a) : : R*^ : cos'' ^ A. 

Cor. 1. Hence we find, as in the corollary to the preceding 
proposition, that 

cos-|A=^ — ^ -\ and it would be pi'oved in a similar man- 
ner, that 

costB = i/— ^ -i andcosfG=|/ — —~-. 

Cor. 2. From the sixth corollary to the definitions of trig- 
onometry, it is plain, when the radius is unity, if the sine of an 
angle be divided by its cosine, the quotient is its tangent. 
Hence, by dividing the expression for the sine of -j A in the 
corollary to the preceding proposition, by the value of its co- 

, , ,, , . , . /is — b) (s — e) 
Bine in the last corollary, we obtam tan ^ A=i/ j^ t — ; 



188 



THE ELEMENTS OF 



and we should obviously find in a similar manner, that 



tan|B: 



{s—a) {s—c) 



. As—a) (^ 
'V s {s- 



) and tan 



iC=/ 



(s-a) (s-b) 



-b) '"'-^""2 y s(s—c) ' 

Cor. 3. By dividing the values of tan ^^B, tan 4^C, in the pre- 
ceding corollary, each by that of tan |^A, we obtain 
tan iB 5 — a _ tan AC s — a 

TT= V and : ,— r= . 

tan +A s — o 




tan ^A 5 — c 



Prop. VIII. — Prob. — Given the radius of a circle^ and the 
cosine of an angle^ less than a right angle ; to compute the 
cosine of half the angle. 

Let CD, the cosine of ACB, less than a right angle, be 
given ; it is required to compute the cosine of its half 

Draw the chord AB, and perpendicular to it draw CFE ; 
draw also FG parallel to BB; then (III. 2, 
and Trig. defs. 5 and 7) AF or FB is the sine, 
and CF the cosine of ACE tlie half of ACB. 
Also (V. 2), DG is equal to GA, since BF is 
equal to FA, and DA=2DG; to each of these 
add 2CD; then CA+CDr3 2CG, and conse- 
quently CGi=i(CA + CD) ; or, if the radius be 
taken as unity, and the angle ACB be denoted by A, CG=r| 
(1+cos A). Again, in the similar triangles ACF, CFG, AC : 
CF : : CF : CG; whence (V. 10, cor. 2) CF-r=AC.CG; that is, 
cos 2 ^ A=| (1 +COS A). Hence, to compute cos | A, add 1 to 
cos A, take half the sum, and extract the square root. 

Scho. This proposition and the next afford means by which 
trigonometrical tables can be computed. 

Prop. IX.— Theor. — If A and B be any two angles, R : 
cosB:: sinA : i sin (A-B)+| sin (A-f B). 

Make AKC equal to A, and BKC, CKD 
each equal to B; draw BE, CF, DH, the 
sines of AKB, AKC, AKD ; join BD, and 
through the center K draw KNC ; then 
KN" is evidently the cosine of BKC. Draw 
also BML, NMG parallel to AK, DH. 
Now, in the similar triangles DLB, NMB, 




K n G F E A 

Bince DB is double of NB, DL (V. 3) is double of NM ; to 



PLANE TRIGONOMETRY. 



189 



DL add LII, BE, and to 2NM add what is equivalent, 2MG ; 
tlien DH+BE = 2NG ; wherefore NG is equal to half the sum 
of BE and DH, that is, to i sin (A— B) +^ sin (A-f B). Again, 
in the similar triangles CFK, NGK, we have (V. 3, and altern- 
ately) CK : NK : : CF : NG; that is, R : cos B : : sin A : 
isin (A— B)+isin(A + B). 

Cor. 1. Hence, if R = l, we have, by doubling the second 
and fourth terms, and by taking the products of the extremes 
and means, sin (A— B)4- sin(A + B)=;2 cos B sin A; whence, 
sin (A+B)=i2 cosB sin A— sin (A— B). 

Co7\ 2. If B=:A, the last expression becomes sirapl}"- sin 2 A 
= 2 sin A cos A. 



TRIGONOMETRICAL FORMULA. 

The lines hitherto considered may be computed for every 
conceivable angle, and they will each undergo a change of 
value when the angle passes through the gradations of magni- 
tude, hence they are the functions of the angle, a term imply- 
ing the connection between two varying quantities, that the 
value of the one changes with the value of the other, and they 
receive their values from the ratios or proportions arising from 
them and the angle. We have considered the numerical values 
only of these functions, and the angles from which they were 
deduced were all less than 180 degrees, which relate to plane 
angles and triangles. And we propose now to explain the 
processes for computing the unknown parts of rectilinear trian- 
gles, also the nature and properties of the angular functions, 
together with the methods of deducing all the formulae which 
express relations between them. 

When two diameters are drawn per- 
pendicularly to each other, they divide 
the circle into four equal parts called 
quadrants, which are first, second, 
third, and fourth quadrants, going from 
right to left, and the functions have 
certain algebraic values depending 
upon the particular quadrant in which 
the angle is. For instance, all the 
lines estimated from AC upward are positive, and from CA 




190 THE ELEMENTS OF 

downward are negative ; from DB to the right are positive, 
and from DB to the left are negative. In the formulae the 
algebraic signs + and — are used, the former denoting posi- 
tive, and the latter negative. 

In the diagram it will be seen that the functions are all posi- 
tive in the first quadrant AEB ; that the sine, cosecant, and 
versed sine are positive, and the others negative in the second 
quadrant ; that the tangent, cotangent, and versed sine are 
positive, and the others negative in the third quadrant ; that 
the cosine, secant, and versed sine are positive, and the others 
negative in the fourth quadrant. Hence we can arrange them 
in the following table : 

Third Q. Fourth Q. 
— + 





FmsT Qttai). 


Second Q. 


Thir] 


Sine, 


+ 


+ 




Cosine, 


+ 






Tangent, 


+ 


— 


+ 


Cotangent, 


+ 





+ 


Secant, 


+ 


— 


— 


Cosecant, 


+ 


+ 


— 


Versed sine. 


+ 


+ 


+ 



+ 
+ 

It is convenient to give different signs to the angles also. 
If we suppose the angles to be estimated from left to right, 
they are negative, and the sign of the angle will afiect the 
sign of its sine, but those of its cosine remain the same. 

From corollaries fourth, fifth, sixth, and seventh we can de- 
duce the following equations, when A denotes the angle and 
the radius is unity : 

Sin "A + cos 'A =1 (1) 

Sec=A=l + tan^A (2) 

Cosec^'A^l+cot'A (3) 

r_ . sin A 

TanA=: -r (4) 

cos A ^ ' 

^ . cos A ig,\ 

QoiA=-r—. (5) 

sm A 



TanAxcotA = l (6) 

1 
cos A 



Sec A=-^ (7) 




PLANE TEIOONOMETRY. 191 

Cosec A = -: T- (8) 

Sin A 

Ver, sin A=l — cos A (9) 

By the first proposition we have, in a right-angled triangle, 
radius : cos of either acute angle : : hyp. : side adjacent. 

Hence, in the following diagram : 

CB:0=CA cos C ; and DB:0-DA cosD, 
or CD=c>CA cos C— DA cos D. 

Dividing both members of the equation 
by CD, we have — 

1=0=— =- cosC— ^=^=-cos D; hence (Trig, 2) 
OD kjiJ 

CA sin D - DA sin C , sin D ^ sin C 

pr^O- — T and 7=^^<2>=-^ — r; hence, lo-; — r-cosC — : — -r 
CD^^sin A CD sin A ' sin A sm A 

cos D, or sin Aosin D cos C— sin C cos D ; but the angle A 

is the difference of the angles ADB and ACB (I. 20) ; hence 

sin (D— C)=0=sin D cos C — sin C cos D. 

Thus, from the first and second propositions, by easy pro- 
cesses, we derive the formula for the sine of the difference of 
two angles, which is expressed in the following manner : The 
sine of the difference of any two angles is equivalent to the sine 
of the first into the cosine of the second, minus the cosine of the 
first into the sine of the second.* 

The formula for the sine of the sum of two angles can be 
derived from the preceding by substituting the negative for 
the positive value of the second angle, and bearing in mind 
that in estimating an angle from left to right, the algebraic 
sign of its sine is changed, and we get — The sine of the sum 
of any two angles is equivalent to the sine of the first into the 
cosine of the second, plus the cosine of the first into the sine of 
the second. 

The formula for the cosine of the sum of two angles can be 
derived from the preceding, by substituting the trigonometrical 
values of the functions when the sum of the angles is greater 
than a right angle, and remembering that the sine of an angle 
becomes the cosine of its complement, and we get — The cosine 

* The pupil would be much instructed by converting this and the fol- 
lowing expressions into their equivalent algebraic formula;. 



192 THE ELEMENTS OF 

of the sum of two angles is equivalent to the cosine of the first 
into the cosine of the second, minus the sine of the first into the 
sifie of the second. 

By similar substitutions in the second formula, we get the 
formula for the cosine of the difference of two angles, expressed 
as follows: The cosine of the difference of two angles is equiva- 
le?it to the cosine of the first into the cosine of the second, ^>^ws 
the sine of the first into the sine of the second. 

The other corresponding formuliB are obtained by substitut- 
ing the respective equations for the trigonometrical fmictions 
derived from the fourth, fifth, sixth, and seventh corollaries for 
the values of the various functions ; for instance, to derive the 
formula for the tangent of the sum of two angles, we substi- 

tute, tanA = r in the second and third formulse, getting 

' cos A 

,, „, sin (A+B) sin A cos B 4- cos A sin B „ , ^^ 
t"" (^+^)=co4 At-B) = co.A cosB-sinAs hrB' ""^ '"^ 
ducing, we have — The tangent of the sum of two angles is 
equicalent to the tangent of the first, plus the tangent of the sec- 
ond, divided hy the square of the radius, minus the tangent of 
the first into the tangent of the second. And in a similar way 
we get the other formulae for the various functions. 

When the angle and radius are known, we can get the form- 
ula for the sine of double the angle by making the two angles 
equal in the second formula, and we have — The sine of twice 
an angle is equivalent to twice the sine of the angle into the 
cosine of the angle. 

In similar manner we derive the formula for the cosine of 
double the angle, and substituting the equation sin''A=l — 
cos' A in the third formula, we get — The cosine of twice an 
angle is equivalent to twice the square of the cosine of the an- 
gle, minus the square of the radius. 

Thus by substitution of the several equations in the respec- 
tive formulae, we can derive the other functions of double the 
ano-le when the ansfle and radius are known. 

From the formula for the cosine of double an angle, we can 
get by substitutions and reductions the formula for the sine of 
half an angle, and expressed as follows — The sine of half an 
angle is equivalent to the square root of half the difference of 



PLANE TRIGONOMETRY. 193 

the radius and cosine of the angle. In a similar way we ob- 
tain — The cosine of half an angle is equivalent to the sqf/are 
root of half the sum, of the radius and cosine of the angle ^ 
and — The tangent of half an angle is equivalent to the sine of 
the angle divided hy the sum of the radius and cosine of the 
angle ; and so on for the other functions. 

By adding and subtracting the various formulae already men- 
tioned, we obtain a great number of consequences which are 
useful ; it will suffice to consider a few of them. From the 
first four we obtain — The sine of the sum of two angles added 
to the sine of the difference of the same angles is equivalent to 
twice the sine of the first into the cosine of the second. The 
siiie of the sum of two angles diminished by the sine of the 
difference of the same angles is equivalent to txcice the sine of 
the second into the cosine of the first. TJie cosine of the sum, 
of two angles increased by the cosine of the difference of the 
same angles is equivalent to twice the cos ne of the first into 
the cosine of the second. The cosine of the difference of two 
angles dimhiished by the cosine of the sum of the same angles 
is equioalent to twice the sine of the first into the sine of the 
second. These are very useful, because they change the pro- 
ducts of sines, cosines, and other functions from superficial into 
linear sines, cosines, etc. 

By the substitution of algebraic symbols into the preceding 
equations, we obtain certain analogies which give algebraic ex- 
pressions to so many theorems, as follows : 

Sum of sines : Dif. of sines : : tan of half Sum : tan of half Dif. 

Sum of sines : Sum of cos :: tan of half Sum : radius. 

Sum of sines : Dif of cos :: cot of half Dif : radius. 

Dif of sines : Sum of cos : : tan of half Dif : radius. 

Dif of sines : Dif of cos :: cot of half Sum : radius. 

Sum of cos : Dif of cos :: cot of half Sum : tan of half Dif. 

Sum of sines : sine of Sum : : cos of half Dif : cos of half Sum. 

Dif of sines : sine of Sum :: Sine of half Dif : sine of half Sum. 

INYESTIGATIONS OF THE METHODS OP COMPUTING TABLES OF 
SINES, TANGENTS, AND SECANTS. 

From what has been shown in relation to the previous form- 
12 



194r THE ELEMENTS OF 

nlae, it will be noticed that they all proceed from the first — and 
we derive the ^rst from the first and second propositions, 
namely, from the analogies that in a right-angled triangle : 
hypothennse : radius : : one of the legs : sine of opposite an- 
gle ; hence, when we know the length of the hypothenuse, leg, 
and radius, we can determine the sine of the angle opposite the 
leg. Then (Trig, def V, cor. 4) cos"A=l— sin'A ; that is, the 
square of the radius, minus the square of the sine of the angle, 
IS equivalent to the square of the cosine of the angle; then, 
square root of the difierence between the squares of the radius 
and sine is equivalent to the cosine of the angle; and the other 
functions are derived from the equations resulting from the 
fourth, fifth, sixth, and seventh corollaries of the definitions. 

Now, the chord of 60 degrees (III. 25, cor. 4) is equal to ra- 
dius, and when radius is unity, the cosine of 30° is 0.5 ; hence, 
sine of 30°= ^/(l — cos^30°)=: 4/.75, Bisecting this angle we 
get from the formula of cosine of half an angle, cos 15°= 
■iVl+cos 30°, and seventeen such bisections give cos 1'^= 
.999999299 ; hence, sin V can be obtained. 

Some mathematicians divide 3.14159265358979, etc., into as 
many equal parts as there are seconds or minutes in 180°, and 
express the value of the sine of one second or minute by one of 
these equal parts, contending that the sine, chord, or arc of so 
small an angle differ very imperceptibly from each other; 
when this quantity is used, the cosine of one second becomes 
.999999957. It will thus be seen that the cosines obtained by 
these methods differ very little from each other, and for ordi- 
nary purposes either will give results sufiiciently accui'ate ; but 
when the greatest exactness is desired, the first method should 
be used, because the sine of an angle is a straight line and can 
never coincide with the arc which measures (I. def 19) the an- 
gle, however so small the angle be reduced, and (V. 25, scho.) 
3.1415926, etc., is the approximate relation of the diameter and 
circumference. Hence, the first method is a pure deduction 
from geometrical and trigonometrical principles. When the 
angle is a right angle, the sine and cosine of the angle 
are equal, and the equations of the tangent and cotangent 
will give the values of those functions; and when the angle ex- 
ceed a right angle, the formula for the tangent of the sum of 



PLANE TRIGONOMETKY. 



195 



two angles can be used, making the right angle one of the 
angles and the excess the other angle. 

When the converging series are used, any angle less than 
90'' is expressed by cc, and the formulae for the functions are: 

Sm a;=a; 1 -4- etc. 

1.2.3^1.2.3.4.5 1.2.3.4.5.6.7^ 

x' X* a;* . ^ 

Cos x=:l 1 \- etc. 

1.2^1.2.,3.4 1.2.3.4.5.6^ 

^ , a' 2af 17a;' 62a;' 

Tan x=zx-\ 1 1 — - — -+-^ — --:+ etc. 

. 3 ^3.5^3=.5.7 3'.5.7.9^ 

Cotcc^---- — -^^- — -etc. 

, , a;' 5a;* 61a;* 

Sec a;=lH — --\ 1 \- etc. 

1.2^1.2.3.4^1.2.3.4.5.6 

1 X 1x^ S\x' 

Cosec x= 1 1 1 4- etc. 

a; ^1.2.3^3.4.5.6 3.4.5.6" 

Now, when the base of the Napierian logarithms is used, €= 
2.7182818, and the following formulie will give the sine and co- 
sine of any angle x, from which the other functions can be 
obtained : 

Sm a;= -^ ; and cosa; = 



2 ^^ 2 

Having obtained the sine and cosine of any angle by either 
of the foregoing formulae, we can get the sine of twice the an- 
gle by the consequences fi-om adding and subtracting the first 
four formulae, page 193, from which we derive — 

2 cos a; X sin x — sin = sin 2 ar, 

2 cos X X sin 2 a; — sin a; = sin 3 x, 

2 cos XX sin 3 a; — sin 2 a; = sin 4 a-, 
etc., etc., etc.,^ etc. 

Or by multiplying the first two formulae, page 191, and sub- 
stituting the value of the square of the cosine, we can deter- 
mine new formulae for further computation after having found 
the sines of x and 2 x. 

Sin (a;+2a;) sin (x— 2 x)=sin'a;— sin'2r ; hence, sin (a;+2a-) 



196 THE ELKMENT8 OF 

sin (.c— 2 3-) = (sin x-\-s\n 2 x) (sin a:— sin 2 a;); or, sin (x — 2a'): 
sin X — sin 2x :: sin x+sin 2x : sin {x-{-2 x) ; applying lliia 
proportion, we have, 

Sin X : sin 2 a;— sin x : : sin 2 x + sin x : sin 3 x. 
Sin 2 X : sin 3 J— sin x : : sin 3a: + sin a: : sin 4 x. 
Sin 3 a; : sin 4 x — sin a; : : sin 4 .r+sin a; : sin 5 x. 
etc., etc., etc., etc. 

These last formulje will give the natural functions of the an- 
gles, but to avoid the operations of multiplication and divi- 
sion, and employ the simpler operations of addition and sub- 
traction, tables are constructed giving the logarithmic values 
of the several functions of the ansrles. 

As the sine and cosine of an anorle are the leers of a rieht- 
angled triangle, and the hypothenuse is the radius of the arc 
which measures the angle, for the convenience of logarithms 
the hypothenuse or radius is considered as 10,000,000,000, and 
its logarithm is 10. 

The sines, cosines, tangents, and cotangents are the only- 
functions put in the tables, as the other functions are easily 
found from them. 

TRIGONOMETRICAL PROBLEMS. 

The principles which have been thus established, enable us to 
solve all the elementary cases of plane trigonometry. Now, of 
the three sides and three angles of a triangle, some three, and 
those not the three angles, must be given to determine the tri- 
angle (I, 14, scho.), and the resolution of plane triangles may 
therefore be reduced to the three followinsr cases : 

I. When a side and the opposite angle, and either another 
side or another angle are given ; 

II. When two sides and the contained anijle are aiven ; 

III. When the three sides are c^iven 



o' 



The FIRST CASE is solved on the principle (Trig. 2) that the 
sides are proportional to the sines of the opposite angles. Thus, 
if A, B, a be given, add A and B together, and take the sum 
from 180°; the remainder (I. 20) is C. Then b and c will be 



PLANE TRIGONOMETRY. 197 

found by the following analogies : sin A : sinB : : « : Jy and 
sin A : sin G : : a : c. 

If, again, a, b, A be given, we compute B "by the analogy, 
a : 6 : : sin A : sin B. Then, C is found by subtracting the 
sura of A and B from 180°, and c by the analogy, sin A : sin C 

When in this case two unequal sides, and the angle opposite 
to the less, are given, the angle opposite to the greater (Tkig. 
defs, cor. 8) may be either that which is found in the table of 
Bines or its supplement ; and thus the problem admits of two 
solutions (I. 3, case 4). 

If in this case one of the anHes be a riuht anole, the solution 
is rather easier; as, by the second corollary to the definitions, 
the sine of that angle is equal to the radius. The same con- 
clusion may also be obtained by means of the first propo- 
sition. 

To exemplify the solution of this case,* let az=l3 yards, 5= 
15 yards, and Ar=53° 8', — to resolve the triangle; and the 
operation by means of logarithms will be as follows: 



As a 


13 




1.113943 


: b 


15 




1.176091 


: : sin A 


67° 


8^ 
23^ 


9.903108 


: sinB 


9.965256 


or 


112'=' 


37' 




As sin A 






9.903108 


: sin C 


59» 


29' 


9.935246 


: : a 


14 




1.113943 


: c 


1.146081 


As sin A 






9.903108 


; sin C 


14" 


15' 


9.391206 


:: a 


4 




1.113943 


: c 


0.602041 



* For logarithmic computations the pupil is referred to the Tables 
now in preparation by Prof. Docharty, of the College of the City of New 
York, or the Tables computed by Prof Davies, of tlie United States 



198 THE ELEMENTS OF 

In these operations, to find the fourth term, the second and 
third terms are added together, and the first is taken from the 
sum. This may be done very easily, in a single operation, by 
addins: the fisrures of the second and third terms successively 
to what remains after taking the right-hand figure of the first 
term from 10, and each of the rest from 9, and rejecting 10 
from the final result. Thus, in the first operation, we have 8 
and 1 are 9, and 7 are 16; then 1 and 9 are 10, and 5 are 15, 
etc. It is still easier, however, when the quantity to be sub- 
tracted is a sine, to use the cosecant, and when it is a cosine, 
to use the secant, each diminished by 10, and then to add all 
the terms together. The reason of this is evident from the na- 
ture of logarithms, and from the fifth corollary to the defini- 
tions of Trigonometry. In like manner, when the number to be 
subtracted is a tangent, or cotangent, we may use in the former 
case a cotangent — in the latter, a tangent, subtracting in each 
case 10, either at first or afterward. 

This example evidently belongs to the doubtful case; and 
hence we have two values for each of the quantities B, C, and 
c ; and therefore two analogies are requisite for finding the 
values of c. 

The SECOJTD CASE is solved by means of the third and fourth 
propositions. Thus, if a, h, C be taken, take C from 180°, and 
(I. 20) the remainder is A 4-B. Take the half of this, and then, 
by the third proposition, as a-\-h : a—h : : tan| (A + B) : tan 
\ (A— B). This analogy gives half the difference of A and B ; 
and (II. 12, scho.) by adding this and i (A + B) together, A, 
the greater angle, is obtained, while B is found by taking \ (A 
— B) from-|(A+B). The remaining side will be calculated 
(Tkig. 4) by means of either of the following analogies, and 
"by employing both, an easy verification of the process is ob- 
tained ; 

as cos i (A— B) : cos ^ (A + B) ::a\h:c; and 
sin \ (A— B) : sin | (A + B) wa-h-.c. 

When the given angle C is a right angle, the solution is 
most easily efiected by means of the first proposition of Trig- 
Military Academy, or those of Frof. Loomis, of Yale College, New 
llaven, Connecticut. 



PLANE TRIGONOMETRY. 199 

onometry; the oblique angles being obtained by the analogy, 
a : b :: U : tan B, or cot A ; and the hypotheinise either by 

c. 



the analogy, R : sec B : : a : 


c, or sin A : R : : a : 


As an example, 




As a-\-b 


99.98 1.999913 


: a — b 


14.V8 1.169G74 


: : tan ^ (A+B) 


61° 37'i 10.267498 



:tan^(A— B) 15° 18'i 9.437259 
Hence A = 76° 56', and Bi=46o 19' 

As cos i (A— B) 15° 18'i 9.984311 

: cos i (A+B) 9.676913 

: : a+b 1.999913 



C 49.20 1.692515 

As sin I (A— B) 9.421626 

: sin i (A + B) 9.944411 

:: a-b 1.169674 



: c 49.26 1.692459 

Half the difference of A and B is here taken as 15° 18'^. 
When determined accurately, however, it is found to be 15° 
18' 23", Hence the cause of the slight difference in the loga- 
rithm of c, as obtained by the two different analogies. It is 
plain that after A and B are computed, c might also be found 
by means of the first case, by either of the analogies; sin A : 
Bin C : : a : c, and sin B : sin C : : b : c. The foregoing 
method, however, is much preferable. 

The THIRD CASE may be solved by means of the fifth, sixth, 
or seventh proposition. Thus, assuming a (see the figures for 
the fifth proposition) as base, we have a to b-\-c as b—c,OT- 
c—b to a fourth proportional. If this be less than BC, it is the 
difference of the segments BD, DC, in the first figure ; and if 
half of it and half of the base be added together, the sum will 
be the greater segment, while the less will be found by taking 
half that proportional from half the base. If the fourth pro- 



200 THE ELEMENTS OF 

portional be greater than the base, it is the sura of the seg- 
ments in the second figure, and, as before, the segments are the 
sum and difference of half the proportional and half the base. 
Then, by resolving, by the first case, the two riglit-angled tri- 
angles ADB, ADC, in which there are given the hypothenuses 
AB, AC, and the legs BD, CD, the angles B and ACD will be 
obtained, which, in the first figure, are two of the required an- 
gles ; while in the second, the angle C is the supplement of 
ACD. 

Again : by adding the three sides together, and talcing half 
the sum, the value of 5 is obtained; and, in applying the sixth 
proposition, the sides containing the required angle are to be 
separately taken from sy but, in applying the seventh, only 
the side opposite to the required angle is to be subtracted; 
while if all the three sides be subtracted successively, another 
mode of solution is furnished by the second and third corolla- 
ries to the seventh proposition. This last method is preferable 
to any of the others, when it is necessary to determine all the 
angles ; and if they be all computed by means of it, the cor- 
rectness of the operation is ascertained by trying whether their 
sum is 180°. 

To exemplify the last of these methods, let a=Qld, b—53ly 
and c=429 ; to compute the angles. 

Here, by adding the three sides together, we obtain 1645, 
the half of which, 822.5, is s. Then, by taking from the three 
sides successively, we find s — a= 143.5, s — &= 285.5, and s — 
c=393.5. The rest of the operation, the subtraction in the first 
part of which may be performed in the manner pointed out in 
the example for the first case, is as follows : 



8 

8 — a 


822.5 
143.5 


2.156852 ) 


s h 


285.5 


2.455606 


8 C 


393.5 


2.594945 




2) 


19.978563 


n ^A 44° 


17'i 


9.989281 


A=88° 


35' 





PLANE TRIGONOMETRY. 

tan i A 9.989281 ) ^^^ 

log{s — a) 2.156852) 

12.146133 ) , . 
c subt. 
log{s — b) 2.455606) 



201 



tan IB 26° V'i 9.690527 

B = 520 15' 

12.146133)^^^^ 
log {s — c) 2.594945 ) 



tan i C 19° 35' 9.551188 

C = 39° 10' 

In the first part of this operation, the halving of the loga- 
rithm serves for the extraction of the square root. The remain- 
der of the work consists in adding together tan ^ A and log 
{s — a), and subtractingAog {s — b) and log (s — c) successively 
from the sum. This method of solution is remarkably easy, 
requiring for the entire operation only four logarithms to be 
taken from the tables ; and affording at the same time a 
most satisfactory verification by the addition of the three 
angles, when found. The preparatory part of the process 
also admits of an easy verification, as the sum of the three 
remainders s — «, s — b, s — c is equal to the half sum. 

Prob. I. — Let it be required to find the height of an accessi- 
ble object AB, standing on a horizontal plane. 

On the horizontal plane take a station C, and measure with 
a line, a chain, or any such instrument, the dis- 
tance CB to the base of the object ; and with a 
quadrant, a theodolite, or other angular in tru- 
ment, measure the angle BCA, called the angle 
of elevation. Then, since B is a right angle, the 
height AB will be found (Trig. 1) by the analo- 
gy,^R : tan C : : CB : BA. 

This gives the height of A above CB, the horizontal line 
passing through the eye of the observer ; and tlierefore to find 
the entire height, AB must be increased by the height of hia 





202 THE ELEMENTS OF 

eye above the base or the object. The like addition must be 
made in every problem of this kind, when the angle of eleva- 
tion above the horizontal line is given. 

pROB. n. — To find the height of an object AB, stanaing on 
a horizontal plane, but inaccessible ow account of the uneven- 
ness of the ground near its base, or the intervefition of obsta- 
cles. 

In a straight line passing through the base of the object take 
two stations C, D ; and measure CD, and 
the two angles of elevation BCA, BDA. 
Then (I. 20) CAD is the difference of ACB, 
ADB ; and (TFao. 2) sin CAD : sin D : : 
DC : CA. Again (Trig. 1), R : sin ACB 
: : AC : AB ; whence AB will be found. 
The computation will be rendered rather more easy by mul- 
tiplying together (IV. 15) the terms of the two analogies, and 
dividing the third and fourth terms li^'the result by AC ; as by 
this means we get the analogy li X sin CAD : sin D x sin BCA 
: : DC : AB. Hence, to find the logarithm of AB, to the loga- 
rithm of DC add the logarithmic sines of D and BCA, and from 
the sum take the sine of CAD and the radius. 

Prob. Ill, — To find the distance oftxno objects A a7idB on a 
horizontal plane. 

This may be effected in different ways according to circum- 
stances. 

1. A base AC may be taken, terminated at one of the ob- 
jects. The angles A and C, and the side AC 
are then measured ; and the required distance 
AB is found by the analogy, sinB : sin C : : 
AC : AB. 

2. This method fails if the objects A and B 
be not visible from one another, as then the 
angle A can not be measured. In this case, a 
station C may be taken as before, from which both A and B 
are visible. Then, having measured the angle C, and the sides 
AC, BC, we compute the distance AB by means of the second 
case of trigonometry. 




PLANE TKIGONOMETKY. 



203 



3. When from inequalities in the ground, or other causes, the 
preceding methods are inapplicable, the solution may be effect- 
ed in the following manner: Measure a base CD, such that A 
and B are both visible from each of its extremities ; measure, 
also, the two angles at C, and the two at D. Then, by the 
first case in trigonometry, -we compute AC in the triangle 
ACD, and BC in the triangle BCD ; from which, and from the 
contained angle ACB, AB is computed by means of the second 
case. The operation may be verified by computing AD, BD, 
by the first case, and thence AB by the second case. 

Prob. IV. — Let AFB be a great circle of the earth, supposed 
to be a sphere; E a point in the diameter B A produced, EF a 
straight line touching the circle in F, and ED a straight line in 
its plane, per2Jendicular to AB; it is required to compute the 
angle DEF, and the straight line EF. 

Draw the radius CF. Then, since (III. 12) CFE is a right 
angle, we have (hyp. and I. 20, cor. 3) 
DEC=CEF+ECF." Take away CEF, 
and there remains DEF=:ECF. Now 
(III. 21) EF- = BE.EA. Hence EF will 
be found by adding AE to AB, multiply- 
ing the sum by EA, and extracting the 
square root. To find CE, add AE to the 

radius AC. Then (Trig. 1) CE : EF : : 

R : sin ECF, or sin DEF ; or CE : CF : : ~ 

Pv : cos ECF, or cos DEF. 

8cho. These examples have been selected from the " Ele- 
ments of Plane Trigonometry," by Prof Thomson, of the Uni- 
versity of Glasgow, since they exhibit in the simplest manner 
the elementary principles of trigonometrical computation. 



4* 




THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

DEFINITIONS. 

1. Spherical Trigonometry explains the processes of cal- 
culating the unknown parts of a spherical triangle when any 
three parts are given; and certain formulae derived from Plane 
Trigonometry are employed to express the relations between 
the six parts of a spherical triangle. 

2. A spherical triangle is that portion of the surface of a 
sphere bounded or contained by the arcs of three great circles 
intersecting each other ; the spherical triangle being formed by 
three planes passing through the sphere, and intersecting each 
other, each angle of the triangle (VI. 24) is contained by 
the tangents of the sides at their point of intersection, and is 
measured by the arcs of great circles described from the ver- 
tices as poles, and limited by the sides of the triangle produced 
if necessary. Also, the angles of a spherical triangle vary 
(VL 24, cor. 2) between two and six right angles. 

3. The spherical angle being contained by the tangents of 
the sides at their point of intersection, the properties of the 
spherical triangle are explained by means of Plane Trigonome- 
try, and its analogies are applied to imaginary rectilineal tri- 
angles, the sides of the spherical triangle being regarded 
functions of rectilineal angles having the sides of the spherical 
triangle as arcs measuring (I. def 19) them. Spherical Trig- 
onometry treats of the angles at the apex of a triangular pyra- 
mid; but Plane Trigonometry treats oi plane angles ^ therefore 
Spherical Trigonometry treats of solid angles. 

4. Let ABC be a spherical triangle, and H the center of the 
sphere ; the angles of the tnangle are equal to the angles in- 
cluded by the planes IIAB, HAC, and IIBC (VI 24),"which 
are the angles formed by the planes at the apex of a triangular 
pyramid, and the arcs AB, BC, and CA measure the angles on 
the planes at the apex of the pyramid AHB, BHC, AHC re- 



SPHERICAL TRIGONOMETRY. 



205 



spcctively. And we can represent the side opposite the angle 
A by a, the side opposite the 
angle B by b, and the side op- 
po>ite the angle C by c. On 
the line HA take any point as 
L, and draw perpendiculars as 
FL, LG to HA. Then, GLF 
will be equal to the angle A, 
LEG equal to B, and FGL 
equal to C ; and the sides CB, 
BA, and AC of the spherical triangle ABC will measure the 
angles CHB, AHB, and AHC respectively ; hence these an- 
gles are denoted by a, c, b. 

5. If FG be joined in the triangles FHG and FLG, we will 
have (Plane Trig., 6 cor.), 

HF'+HG'-FG* 




cos BHC=cos a= 



cos FLG = cos A= 



2HFxHG 
LF=+LG=-FG* 



2LGxLF • 

Reducing and subti-acting second from the first, we will 
have, 

2 [cos a (HFxHG)— cosA (LGxLF)]=.2HLl 

Dividing both members by 2 (HFxHG), we get. 



cos a— cos A 



LGxLF HL HL 



HFxHG" 



HF^HG" 



Since regular and similar polygons have their perimeters 
proportionate to their apothems, and circles have their circum- 
ferences proportionate to their diameters (V. 14, cor. 3), the 
sine of an angle is the ratio of the radius, or the hypothenuse 
of a liglit-angled triangle to the perpendicular from the vertex 
of the right angle to the hypothenuse. 

„ LG , , LF . HL HL 

Hence we get ^- = sin 5, g^ = sm e, g-, = cos c, jj^ = cos b. 

Substituting and ti-ansposing, we derive the formula, 

cos a=cos b cos c+sin b sin c cos A, ) 
cos b = cos. a cos c+sin a sin c cos B, /• (1) 
cos c=cos a cos 6+ sin a sin b cos C. ) 



206 



THE ELEMENTS OF 



Or, The cosine of either side of a spherical triangle is equal 
to the product of the cosiyies of the other two sides increased by 
the product of their sines into the cosine of the an^le included 
hy them. 

The three formulae show the relations between the six parts 
of a spherical triangle such, that if any three of them be given, 
the other three can be determined. 

6. Then (VI. 24, cor. 1), if we denote the angles and sides of 
the spherical triangle polar to ABC, respectively, by A', B'', C^, 
a\ h\ c', we will have, 

«'=180°— A, d'=180°— B, c'=180O— C, 
A'==180°— a, B'r=l80°— 6, C"=180O— c. 
Since any of the formulae (l) is applicable to polar spherical 
triangles, we have, after substituting the respective values 
and changing the signs of the terms, other formulae : 

cos A=sin B sin C cos a — cos B cos C, \ 
cos B=sin A sin C cos h — cos A cos C, \ (2) 
cos C=sin A pin B cos c — cos A cosB. ; 

Or, The cosine of either angle of a spherical triangle is equal 
to the product of the sines of the two other angles into the co- 
sine of their included side, diminished by the product of the 
cosines of those angles. 

V. Transposing the first and second formulae (l) we get, 
cos a — cos h cos c=sin h sin c cos A,^ 
cos b — cos a cos c=:sin a sin c cos B ; 

respectively adding and subtracting, we get, 

cos a+ cos 6— cos c (cos a+ cos 5)=sin c (sin 5 cos A+ sin a 

cos B), 
cos a — cos &+COS c (cos a— cos b) =m\ c (sin b cos A— sin a 

cos B) ; 

which can be put in the forms 

(1 — cose) (cos a -|- cos J) = sin c (sin b cos A -|- sin a cos B), 
(1 +COS c) (cos a— cos J) = sin c (sin b cos A — sin a cos B) ; 

multiplying these equations together, substituting for (1 — cos') 
its value (sin'), and for (cos') its value (1 — sin'), and dividing by 
Bill* c, we have, 



SPHERICAL TRIGONOMETRY. 



207 



COS* a— cos' 5=sin'' 5— sin' h sin" A— sin' a+sin* a sin' B; 
then, since (1— sin'a)— (1— sin= h) =sin' h — sin' «, we have, 

cos' a — cos* 6=sin' h — sin' a, and we get, 
sin" h sin' Am sin' a sin' B ; 

extracting tlie sqiiare root, sin 5 sin A = sin a sin B, or ' 
sin A sin a 



sin B sin b' 
sin A 



derive 



And from first and third formulae (l) we 
From the second and third formu- 



sin C sin c 

sin C sin c 



\i?) 



lae we derive —. — t^_ . ,. 
sin ±> sin 

Or, In every spherical triangle^ the sines of the angles are 
to each other as the sines of their opposite sides. 

8. Taking the third formula (l), and substituting for (cos h) its 
value as expressed in the second, and for (cos' a) its value 
(1— sin' a), and dividing by sin a, we will have, 

cos c sin a=sin c cos a cos B + sin b cos C. 

But (Spher. Trig. 7) we have sin b~ ._ ^, — ; substituting for 



sin C 



cose 



sin b its value, and dividing by sin c, we get, -. sin a=C08 a 



cosB4 



sin B cos C 



sin C 



_, cos 

But -r- =COt, 

sin 



Hence we can derive, by similar processes, 

cot a sin b=cot A sin C + cos b cos C, 
cot a sin c=cot A sin B+cos c cosB, 
cot b sin a=cot B sin C + cos a cosC, 
cot b sin c=cot B sin A+cos c cos A, 
cot c sin a=:cot C sin B+cos a cos B, 
cot c sin b=cot C sin A+cos b cos A. 



} (4) 



The formulae (1) are the fundamental analogies of Spherical 
Trigonometry, from which all the others are derived, which 
others are more adapted for logarithmic computations. 



208 THE ELEMENTS OF 



THE SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES BY 

LOGARITHMS, 

The following equations give the unknown parts of a right- 
angled spherical triangle when C is the right angle and any 
two other parts are known. There are six cases. 

Let C be the right angle, and c be the hypothenuse. 

Case 1. Given a and b to find c, A, and B ; 

, . tan a _, tnn b 
cos e=cos a cos o; tan A= r; tan l3= . 

■^ tan o tan a 

Case 2. Given c and a side b to find a, A, and B ; 

cos c . tan b . „ sin & 

cos a= r; cosA=: — — : sin 13=^ . 

cos b tan c sin c 

Case 3. Given aside a and opposite angle A to find others; 

. , tan a . „ sin a . _ cos A 

sm 0= — r ; sin C=-. — -; sin B= -. 

tan A sm A cos a 

Both acute or both not acute. 

Case 4. Given a side a and adjacent angle B to find others; 

tan 6=sin a tan B ; cot c=:cot a cos B ; cos A = cos a sin B. 

Case 5. Given the hypothenuse c and an angle A to find 
others ; 
sin « = sine sin A ; tan 5i=tan c cos A; cot B = cose tan A. 

Case 6. Given the oblique angles A and B to find others ; 

cos A - cos B . _ 

cos a=—. — ^rr; cos o=— — - ; cos c = cot A cot B. 
sin B sin A 

Napier's circtdar parts are much the simplest method of re- 
solving right-angled spherical triangles; they are the two 
sides about the right angle, the complements of the oblique 
angles, and the complement of the hyjiothenuse. Hence there 
are five circular parts; the right angle not being a circular 
part, is supposed not to separate the two sides adjacent to the 
right angle ; therefore these sides are regarded adjacent to 
each other, so that when any two parts are given, tlieir corre- 
sponding circular ])aits are also known, and these with the re- 
quired part constitute the three parts under consideration ; 
therefore these three paias will lie together, or one of tliem 



SPHERICAL TRIGONOMETRY. 



209 



■will be separated from both tlie others. Hence one part is 
known as the middle part; and when three parts are undei- con- 
sideration, the parts separated by the middle part are caUcd 
t\\G adjacent parts ; and the parts separated liuni tlie middle 
parts are called the opposite parts. IS'ow, assume any part in 




B 



the diagram for the middle part, and using the formulae (1) 
•when the other parts are the ojiposite parts, and we get, 27ie 
sine of the middle part is equal to the product of the cosines of 
the opposite parts. Then, assume again any part for the mid- 
dle, and use the formulae (2) when the other parts are the ad- 
jacent parts, and we get, The sine of the middle pa. t is equal 
to the pyroduct of the tangents of the adjacent parts. Hence 
"We derive the five following equations : 

sin a=tan h tan (90"— B)=cos (90° — A) cos (90° — c;, 
sin 6=tan a tan (90°— A)=:cos (90°— B) cos (90°— c), 
sin (90°— A)=tan b tan (90°— c)=cos (90°— c) cos c/, 
sin (90° — c)=tan (90°— A) tan (90°— B)==cos a cos 6,. 
sin (90°— B)— -tana tan (90°— c)=cos 6 cos (90° — A). 

The a7za7o^/es of Xapier are derived from the formulce (l) by 
eliminating the cosines of any of the sides, reducing and chang- 
ing to linear sines and cosines (Plane Trig. p. 193), and we 
bave, 

cos I {a-\-h) : cos | (a — b) : : cot | C : tan | (A + B), 
sin \ (a+b) : sin -^ (a—b) : : cot -^ C : tan ^ (A— B), 
cos -^ (a-f c) : cos | {a—c) : : cot ^ B : tan ^ (A+C), 
sin ^ {a+c) : sin -^ (a—c) : : cot | B : tan ^ (A — C), 
cos ^ {b-hc) : cos -^ (b—c) : : cot ^ A : tan ^ (B + C), 
sin ^ {b+c) : sin -^ {b—c) : : cot | A : tan | (B — C), 
14 



210 THI. ELEMENTS OF 

The same proportions applied to tlic triangle polar to ABC, 
with accents omitted, we have, 

cos I (A + B) : cos i (A— B) : : tan | c : tan i {a+b), 
sin i (A+B) : sin i (A— B) : : tan -J c : tan i (a-b), 
cos i (A+C) : cos i (A— C) : : cot ^ & : tan ^ (a+c), 
sin i (A+C) : sin | (A— C) : : cot i 6 : tan ^ (a— c), 
cos i (B + C) : cos ^ (B— C) : : cot i a : tan i (6+c), 
sin i (B + C) : sin i (B— C) : : cot i a : tan | (*— c). 

The same ambiguity that there is between plane triangles 
(I 3, Fourth Case) exists also between spherical triangles, 
which may be avoided by remembering that every angle and 
gide of a spherical triangle are each less than two right angles, 
and that the greater angle is opposite to the greater side, and 
the least angle is opposite to the least side; and conversely. 

Quadrantal spherical triangles are such which have one side 
equal to ninety degrees ; hence they can very easily be solved 
by formulae for right-angled spherical triangles. 

SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES BT 

LOGARITHMS. 

Case 1. Given the three sides to find the angles. 

Ain {s—fi) sin (5— 5) sin (»— c) 
Find s=i (a+6+c) ;let M=|/ ■ ^-'^ ^ 5 

then, tan iA = ^-^^^^3:^; tan i B=^^^^^-; tan i C= 

]\I 

ein (.« — c)' 

Case 2. Given two sides a and b and the included angle C, 

to find others. 

Tan^(A+B)=^^^^i^4NotiC; tan^A-B^'-i^^-^ 
■^^"^^^^^^ cosi(a+6) ' ' Biui(u+6) 

cot \ C. 

But A=i (A+B)+i (A-B) ; B=| (A+B)-i (A-B) ; 

sin c=sin a^!^=sin b ^^, or find an angle cot 9=tan o 
sin A sin 15 

cos a sin (J+(j)) 

cos C ; cos 0= -. • 

' sin (J) 



BPOEKICAL TKIGONOMETKY. 



211 



Case 3. Given the sides a and b and an angle opposite to 
one of them, to find others. 

Find cot (p=tan b cos A, and tanj^^cos b tan A; then, sin 

, , . cos a sin (p . ^ . . sin b . , , . . 

{c+(p)=i — ; sm JL> = sm A -; — : sin (c +>-)=: cot a tan 

^ ' cos 6 ' sin a ^ ^' 

fiinx- 

Case 4. Given the angles A and B and the inchiaed side, to 

find others. 

Find tan (p=cos c tan A, and tan p^=cos c tan B ; 

, tan c sin (p , tan c sin y 

then, tan a— . .,, , — -; tan 6 = - . , . , t; 

p_cos A cos (B + (p) _ cosB cos (A+y) 
cos 9 cos -/^ 

Case 5. Given A and B and a side opposite one of them, to 
find others. 

Find tan (p=:tan a cos B ; cot ;^=:C08 a tan B; 



ein irzsin a 



sin B 
sin A ■ 



sin (<3— (p)=cot A tan B sin 9, 



• //-I \ cos A sin Y 

sm (C — y) = — ^. 

^ ^' cosB 

Case 6. Given the three angles A, B, C, to find the sides. 
Take • 

S=|(A+B + C);andN=,4/ — - ""'*''' ^ 

' cos 10 — J 



cos (S— A) cos (S— B) cos (tS— C) ; 

then, tan \ a—1^ cos (S — A) ; 
tan| &=N' cos (S — B); 
tan \ c=N cos (S — C). 



* THE SURFACE OP A SPHERICAL TRIANGLE. 

Let ABC be a spherical triangle, AC=: 
DF, BC=FE, ABC=DEF. 

S=surface of ACB, 5=surface of hemis- 
phere— BHDC—AGEC—DCE=0 R'— 
(lune AHD — S) — (lune BGE— S)— (lune 
CDFE-S) 



=eR'(,-j 



80 



180""" W + ^^- 




212 BPHKRICAL TRIGONOMETKT. 

180*^, or equivalent to the sum of the three angles above 180°; 
hence, its spherical excess is sometimes taken as the measure 
of the ti'iangle. 

Or in terms of its sides, formula given by L'Huiller, 

tan ^ E= V [tan ^ s tan ^ {s — a) tan ^ (s — b) tan ^ {s — c)]. 



EXERCISES IN 

ELEMENTARY GEOMETRY, 

AND m 

PLANE AND SPIIEKICAL TKIGONOMETRY.* 



DEFI^nXIOXS. 



1. Lines^ avgUs^ and spaces are said to be given in magni- 
tude, when they are either exhibited, or when the method of 
finding them is known. 

2. Points, lines, and spaces are said to be given in position, 
■which have always the same situation, and which are actually 
exhibited or can be found. 

3. A circle is said to be given in magnitude when its radius 
is oiven ; and \w position, when its center is given. 

Mao-nitudes, instead of being said to be given in magnitudey 
or given in position, are often said simply to be given, when no 
ambiguity arises from the omission. 

* For these Exercises I am indebted to Thomson's Euclid (Belfast), they 
being judiciously selected by that eminent writer, and their presentation 
here is a valuable acquisition to an American school text-book. I 
would gladly acknowledge my obligations for many propositions in 
this volume, but they being culled for more than two thousand years 
from the best writers on Geometry, and being so much modified by each 
succeeding age, that it is impossible at this day to attribute tliem to their 
rightful authors, and many of them being more or less contained in 
every work on the subject, they have become public property. What I 
have introduced myself will be well recognized by every student of 
Geometry, and my only apolugy is, the desire to advance the cause of 
Truth. 



214 EXEBCISK8. 

4. A ratio is said to be giveti when it is the same as that of 
two cciven maGrnitudes. 

5. A rectilineal figure is said to be given in species, when its 
several anarles and the ratios of its sides are iriven. 

6. When a series of unequal magnitudes, unlimited in num- 
ber, agree in certain relations, the greatest of them is called a 
ma/x'mum ^ the least, a minimum. 

Thus, of chords in a given circle, the diameter is the maxi- 
mum ; and of straiiiht lines drawn to a given straight line, 
from a given point without it, the minimum line is the perpen- 
dicular. 

7. A line which is such that any point whatever in it fulfills 
certain conditions, is called the locus of that point. 

8cho. 1. Several instances of loci have already occurred in 
the preceding books. 

1. Tlius, it was stated in the fifth corollary to the fifteenth 
proposition of the first book, that all triangles on tlie same 
base, aiid between the same pai'allels, are equivalent in area; 
and hence, if only the base and area of a triangle be given, its 
vertex may be at any point in a straight line parallel to the 
base, and at a distance from it which may be determined by 
ap])lying (11. 5, scho.) to half the base a parallelogram equiva- 
lent to the given area ; and therefore the parallel is the locus 
of the vertex. Here the conditions fulfilled are, that straight 
lines drawn from any point in the parallel to the extremities of 
the given line, form with it a triangle having a given area. 

2. It was stated in the first corollary to the eighteenth proposi- 
tion of the third book, that all anHes in the same segment of a 
cinOe are equal ; and hence, if only the base and vertical angle 
of a triangle be given, the vertex may be at any point of the 
arc of a segment described on the base, in the manner pointed 
out in the nineteenth pi'oposition of the third book; that arc, 
therefore, is the locus of the vei-tex. 

3. It will be seen in the sixth proposition of these Exercises that 
straight lines drawn from any point whatever in the circumfer- 
ence of the circle ABGC to the points E, F, have the same 
ratio — that of EA to AF. Hence, therefore, if the base of a 
triangle, and the ratio of the sides be given, the locus of the 
■vertex is the circumference of the circle described in the man- 



EXEUCISES. 215 

ner pointed out in the corollary to this proposition ; unless 
the ratio be that of equality, in which case the locus is evi- 
dently a perpendicular bisecting the straight line joining the 
points. 

4, It follows likewise, from the fifth corollary to the twcnty- 
fomth proposition of the first book, that when the base of a tri- 
angle and the difference of the squares of the sides are given, 
if the point D be found (II. 12, scho.) in the base BC, or its 
continuation, such that the difference of the squares of BD, CD 
is equivalent to the difference of the squares of the sides; and 
if through D a perpendicular be drawn to ]3C, straight lines 
drawn from any point of the perpendicular to B, C will have 
the difference of their squares equivalent to the given difference; 
and hence the perpendicular is the locus of the vertex, when the 
base and the difference of the squares of the sides are given. 

5. It will appear in a similar manner from the corollary to the 
twelfth proposition of the second book, that if BC the base of a 
triangle, and the sum of the squai'es of the other sides AB, AG 
be given, the locus of the vertex is the circumference of a circle 
described from D, the middle point of the base as center, and 
with the radius T>A. To find DA, take the diagonal of the 
square of BD as one leg of a right-angled triangle, and for its 
bypothenuse take the side of a square equivalent to the given 
sum of the squares of AB, AC ; then the diagonal of the square 
described on half the remaininoj le<x of that rigcht-aniiled trian- 
gle will be the radius of DA. The proof of this is easy, de- 
pending on the third and fourth coi'ollaries to the twenty-fourth 
jDroposition of the first book, and on the corollai-y to the twelfth 
proposition of the second book. 

Scho. 2. In discovering loci, as well as in other investigations 
in geometry, the stuflent is assisted by what is termed geomet- 
rical analysis ; of the nature of which it may be proper here 
to give some explanation. 

Take this proposition : If a chord of a given circle have one 
extremity given in positior), and if a segment terminated at 
that extremity he taken on the ch rd, produced if necessary^ 
such that the rectangle under the segment and chord may be 
equivalent to a given space ; the locus of the point of section is 
a straight I i?ie given inp)osition. 



216 EXERCISES. 

Let AB be the diameter of the circle and AC a chord of the 
circle. 

If, in the proposition, instead of being informed that the locns 
is a strai!j;ht line, we were required to find what tlie locus is, 
we might proceed in the following manner: Lft D be any point 
in the required line, so that the rectangle ACAD is equivalent 
to the given space ; and having drawn the diameter AB, find 
E, so that the rectangle AB.AE may be equal to ACAD, and 
therefore E a point in the required line ; and join DE, BC 
Then (V. 10, cor.) AB : AC : : AD : AE. Hence (V. 6) tho 
triangles DAE, BAC, having the angle A common, are equi- 
angular; and thereiore AED is equal to ACB, which is a right 
angle. The point D is therefore in a perpendicular passing 
through E; and in the same manner it would be t^hown, that 
any other point in the required line is in the perpendicular ; 
that is, the perpendicular is its locus. 

The investigation just given is called the analysis of the 
projjosition, while the solutions hitherto given are called the 
synthesis or compositio7i. In analysis we commence by sup- 
posing that to be effected which is to be done, or that to be 
true which is to be proved ; and, by a regular succession of con- 
sequences founded on that supposition, and on one another, 
we arrive at something which is known to be true, or which 
we know the means of effecting. Thus, in the second corollary 
to the seventeenth proposition of the sixth book, the conclusion 
obtained for the area of the circle is shown by the third corol- 
lary to be consistent with the proportion established by Ar- 
chimedes between the cone, sphere, and cylinder, and also 
consistent with the geometrical truth in regard to the sur- 
faces of the sphere and cylinder. Plence, analysis takes into 
consideration this consistence, and confirms the second corol- 
lary from its agreement with established truths of geometr3^ 

Again: in the corollary to the twenty-fourth proposition of 
the fifth book, since circles are in propoition to the squares of 
tlieir radii, the quadrant ACB is equivalent to the semicircle 
ADC, we have (I. ax. 3) the triangle ABC equivalent to th^ 
crescent ADC. Now, when we api)ly the conclusion derived 
by the second corollary to the seventeenth proposition of the 
sixth book to the above, we find a perfect agreement ; taking 



EXERCISES. ^17 

the circle as three times square of radius, we have quadrant 
ACB equivalent to J AB'; hence (I. ax. 1) the semiciicle ADO 
is equivalent to | AW. But (VI. 17, cor, 2) the segment AC 
of the quadrant ACB is equivalent toi AB"; therefore (I. ax. 3) 
we have the triangle ABC and the crescent ADC each equiva- 
lent to ^ AB', thus showing the agreement between the second 
corollary to the seventeenth proposition of the sixth book, and 
the established truth relating to the crescent or lune. 

Also, we have (VI. 17, cor. 2) the hemisphere generated by 
the quadrant BNP equivalent to the solid generated by the 
trapezium BSNP, and we have the solid generated by the 
figure BTNP common ; thei-efore (I. ax. 3) the solid generated 
by the segment BT is equivalent to the solid generated by the 
figure TSN. Now, the solid generated by the segment BT is 
a part of the hemisphere; hence its contents are computed by 
the same radius as the hemisphere ; the solid generated by the 
figure TSN is a part of the solid generated by the trapezium 
BSNP ; hence its contents are also computed by the same 
radius as the hemisphere. Therefore we obtain by analysis 
the ti-uth, that vchen equivalent solids are generated by equiva- 
lent surfaces, the generating surfaces are up n equal radii, a 
truth corresponding to the truth established by the second 
coroUaiy to the seventeenth proposition of the sixth book, tJiat 
equivalent surfaces upon the same radius loill generate equiva- 
lent solids. The synthesis then commences with the conclusion 
of the analysis, and retraces its sevei-al steps, making that pre- 
cede which before followed, till we arrive at the required con- 
clusion. Therefore the demonstrations given in the second 
corollary to the seventeenth proposition, book sixth, obtain the 
conclusion from which the analyses precede. From this it ap- 
pears that analysis is the instrument of investigation ; while 
svnthesis affords the means of communicatino- what is already 
known ; and hence, in the Elements of Euclid, the synthetic 
method is followed throughout. What is now said will receive 
further illustration from the solution of the following easy 
problem. 

Given the perimeter and angles of a triangle, to construct it. 

Analysis. — Suppose ABC to be the required triangle, and 
produce BC both ways, making BD equal to BA, and CE to 



218 



EXERCISES. 




CA ; then DE is given, for it is equal to the sum of the throo 
sides AB, BC, CA ; that is, it is equal to the given perimeter. 
j^ Join AD, AE. Then (I. 1, cor. 

1) the angles D and DAB are 
equal, and therefore each of 
them is half of ABC, because (T. 
20) ABC is equal to both. The 
angle D therefore is given ; and 
in the same manner it may be shown that E is given, being 
half of ACB. Hence the triangle ADE is given, because the 
base DE, and the angles D, E are given ; and ADE being 
given, ABC is also given, the angle DAB being equal to D, 
and EAC equal to E. 

Composition. — Make DE equal to the given perimeter, 
the angle D equal to the half of one of the given angles, and E 
equal to the half of another; draw AB, AC, maldng the angle 
DAB equal to D, and EAC to E; ABC is the triangle re- 
quired. 

For (T. 1, cor. 2) AB is equal to BD, and AC to CE. To 
these add BC, and the three, AB, BC, CA, are equal to DE, 
that is, to the given perimeter. Also (I. 20) the angle ABC is 
equal to D and DAB, and is therefore double of D, since D 
and DAB are equal. . But D is equal to the half of one of the 
given angles; therefore ABC is equal to that angle; and, in 
the same manner, ACB may be proved to be equal to another 
of the given angles. ABC therefore is the triangle required, 
since it has its perimeter equal to the given perimeter, and its 
angles equal to the given angles. 

It is impossible to give rules for effecting analyses that will 
answer in all cases. It may be stated, however, in a general 
way, that when sums or differences are concerned, the corre- 
sponding sums or differences should be exhibited in the as- 
sumed figure ; that in many cases remarkable points should be 
joined ; or that tlirough them lines may be drawn perpendicu- 
lar or parallel to remarkable lines, or making given angles with 
them ; and that circles may be described with certain radii, 
and from certain points as centers; or touching certain lines, 
or passing tlirough certain points. Some instances of analysis 
will be given in subsequent propositions ; and the student will 



EXERCISES. 219 

find it useful to make analyses of many other propositions, such 
as several in the Exercises. 

8. A jyorism is a proposition affirming the possibility of find- 
ing such conditions as will render a certain problem indeter- 
minate, or capable of innumerable solutions. 

Scho. 3. Porisms may be regarded as liaving their origin in 
the solution of j^roblenis, which, in particular cases, on account 
of pccidiar relations in their data, admit of innumerable solu- 
tions ; and the proposition announcing the property or relation 
which renders the problem indeterminate, is called a porism. 
This will be illustrated by the solution of the following easy 
problem. 

Through a given point A, let it be required to draw a straight 
line bisecting a given parallelogram BCDE. 

Suppose AFG to be the required line, and let it cut the sides 
BE, CD in F, G, and the diagonal CE in H. Then from tho 
equivalent figures EBC, FBCG take FBCII, and the remaining 
triangles EHF, ClIG are equal. Now, since (I. 16 and 11) 
these triangles are equiangular, it is evident that they can be 
equal in area only when their sides are equal ; wherefore II is 
the middle point of the diagonal. The construction, therefore, 
is effected by bisecting the diagonal EC in H, and drawiu"' 
AFHG. For the triangles CHG, EIIF are 
equiangular, and since CH, HE are equal, the 
triangles are equal. To each of them add the 
figure FBCH ; then the figure FBCG is equiv- 
alent to the triangle EBC, that is, to half the 
parallelogram BD. 

Now, since the diagonal CE is given in 
magnitude and position, its middle point II 
is given in position, and therefore H is always a point in the 
required line, wherever A is taken. Hence, so long as A and 
II are different points, the straight line AHG (I. post 1) is de- 
termined. This, however, is no longer so, if the given point A 
be the intersection of the diagonals, that is, the point H, as in 
that case only one point of the required line is known, and the 
problem becomes indetermhiate, any straight line whatever, 
through H, equally answering the conditions of the problem; 
and we are thus led by the solution of the problem to the con- 




220 EXERCISES. 

elusion, that in a parallelogram a point may he found, such 
that a)iy straight line wliatever drawn through it, bisects the 
parallelogra7n / and this is a porism. 

The seventy-sixtii pi'oposition of the Exercises, when con- 
sidered in a particular manner, affords another instance of a 
porism ; as it appears that if a circle and a point D or E be 
given, another point E or D may be found, such that any circle 
whatever, desciibed through D and E, will bisect the circum- 
ference of the given circle; and this may be regarded as the 
indeterminate case of the problem, in which it is required, 
through two given points, to describe a circle bisecting tlie cir- 
cumference of another given circle, — a problem which is always 
determinate, except when the points are situated in the man- 
ner supposed in the proposition. 

9. Isopjeritnetrical figures are such as have their perimeters, 
or bounding lines, equal. 

10, The general problem of the tangencies, as understood by 
the ancients, is as follows: Of three points, three straight lines, 
and three circles of given radii, any three being given in posi- 
tion ; it is required to describe a circle passing through the 
points, and touching the straight lines and circles. This gen- 
eral problem comprehends ten subordinate ones, the data of 
which are as follows: (l.) three points; (2.) two points and a 
straight line ; (3.) two points and a circle ; (4.) a point and two 
straight lines ; (5.) a point, a straight line, and a circle ; (G.) a 
point and two circles; (7.) three straight lines; (8.) two 
straight lines and a circle ; (9.) a straight line and two circles; 
and (10.) three circles. The first and seventh of these are the 
second and fifth corollaries of the twenty-fifth proposition of 
the third book. 

If a circle be continually diminished, it may be regarded as 
becoming ultimately a point. By being continually enlarged, 
on the contrary, it may have its curvature so much diminished 
that any portion of its circumference may be made to differ in 
as small a degree as we please from a straight line. Viewing 
the subject in this light, we may regard the first nine of the 
problems now mentioned, as comprehended in the tenth. Thus, 
we shall have the first, by supposing the circles to become in- 
finitely small ; the seventh, by supposing them infinitely great; 



EXEHCISES. 



221 



the fifth, by taking one of them infinitely small, one infinitely 
great, and one as a circle of finite magnitude; and so on with 
regard to the othei'S. These views of tlie subject tend to illus- 
trate it ; but they do not assist in the solution of the problems. 
Hcho. 4. In tiie fifth problem, the straight line may fall with- 
out the circle, may cut it, or may touch it ; the point may be 
without the circle, withfn it, or in its circumference ; or it may 
be in the given straight line, or on either side of it ; and it will 
be an interesting exercise for the student, in this and many 
other problems, to consider the variations arising in the solu- 
tion from such changes in the relations of the data, and to de- 
termine what relations make the solution possible, and what 
render it impossible. It may also be remarked, that in many 
problems there will be slight variations in the proofs of differ- 
ent solutions of the same problem, even when there is no 
chi"ige in the method of solution; such as in the present in- 
stance, when the required circle is touched externally, and 
Avhen internally. Thus, while in one case angles may coincide, 
in another the corresponding ones may be vertically ojiposite; 
and the reference luay sometimes be to the coJiverse of the first 
corollary and sometimes to the converse of the second corollary 
of the eighteenth proposition, book third. It is, in general, un- 
necessary to point out these variations, as, though they merit 
the attention of the student, they occasion no difficulty. 

PROPOSITIONS. 

Prop. I. — Tiieor. — If an angle of a triangle, he bisected by a 
straight line^ which likewise cuts the base,, the rectangle con- 
tained by the sides of the triangle is equivalent to the rectangle 
contained by the segments of the base,, together with the square 
of the straight line bisecting the angle. 

Let ABC be a triangle, and let the angle BAC be bisected 
by AD; the rectangle BA.AC is equal to the rectangle BD.DC, 
together with the square of AD. 

Describe the circle (III. 25, cor.) ACB about the triangle ; 
produce AD to meet the circumference in E, and join EC. 
Then (hyp.) the angles BAD, CAE are equal ; as are also (IIL 
18, cor. 1) the angle B and E, for they are in the same segment; 



222 EXERCISES. 

thej-efore (V. 3, cor.) in the ti-iangles ABD, AEC, as BA : AD 
:: EA : AC; and consequently (V. 10, cor.) the rectangle BA. 
AC is equivalent to EA.AD, that is (II. 3), ED.DA, together 
with the square of AD. But (III. 20) the rectangle ED.DA ia 




equivalent to the rectangle BD.DC ; therefore the rectangle 
BA.AC is equivalent to BD.DC, together with the square of 
AD ; wherefore, if an angle, etc. 

Scho. From this proposition, in connection -with the fourth 
proposition of fifth book, we have the means of computing AD, 
when the sides are given in numbers. For, by the fourth, BA: 
AC :: BD : DC; and, by composition, BA+AC : AC : : 
BC : DC. This analogy gives DC, and BD is then found by 
taking DC from BC. But by this proposition BA. AC = BD.DC 
+ AD-; therefore from BA.AC take BD.DC, and the square 
root of the remainder will be AD, 

In a similar manner, from the fourth proposition of the fifth 
book, and the second proposition of these Exercises, the line 
bisecting the exterior vertical angle may be computed; and 
from the third proposition of these Exercises, in connection 
with the eleventh or twelfth of the second book, the diameter 
of the circumscribed circle may be computed, when the sides 
of the triangle are given in numbers. 

Prop. II. — Theor. — If an exterior angle of a triangle be 
bisected by a straight line, which cuts tlie base produced ; 
the rectangle contained by the sides of the triangle, and the 
square of the bisecting line are together equivalent to the rect- 
angle contained by the segments of the base intercepted between 
its extremities and the bisecting line. 

Let, in the foregoing diagram, the exterior angle ACF of the 
triangle BAC be bisected by HC ; the rectangle BC. AC and 



EXERCISES. 223 

the square of HC are together equivalent to the rectangle 
BH.AII. 

Describe the circle (III. 25, cor.) ABEC about the triangle 
BAG; produce HC (I. post. 2) to E, and join EA. Then, 
since (l>yp.) the angles FCII and HCA are equal, their supple- 
ments FCE and ACE (T. def. 20 and ax. 3) are also equal ; and 
(III. 18, cor. 1) B and E are equal. Therefore (V. 3, cor.) in 
the triangles BCH and EAC, BC : CH : : EC : AC, and con- 
sequently (V. 10, cor.) the rectangle BC. AC is equivalent to 
EC.CII. To each add square of CII; therefore BC.AC+CIP 
are equivalent to EC.CH+Cff; or (II. 3) BC.AC + CIP are 
equivalent to EH.CH ; or (III. 21, cor.) BC.AC + CIP are 
equivalent to BH. AH. Therefore, if an exterior angle, etc. 

Prop. III. — Theor. — If from an angle of a triangle a per- 
pendicular be drawn to the basej the rectangle contained by 
the sides of the triangle is equivalejit to the rectangle contained 
by the perpendicular and iAe diameter of the circle described 
about the triangle. 

Also, in the foregoing diagram, let ABC be a triangle, AL 
the perpendicular from the angle A to BC ; and AE a diameter 
of the circumscribed circle ABEC; the rectangle BA.AC is 
equivalent to the rectangle AL.AE. 

Join EC. Then the right angle BLA is equal (III. 11) to the 
angle EC A in a semicircle, and (HI. 18, cor. 1) the angle B to 
the angle E in the same segment ; therefore (V. 3, cor.) as 
BA : AL : : EA : AC; and consequently (V. 10, cor.) the 
rectangle BA.AC is equivalent to the rectangle EA.AL. If, 
therefore, from an angle of a triangle, etc. 

Prop. IV. — Theor. — The rectam^le contained by the diago- 
nals of a quadrilateral inscribed in a circle, is equivalent to 
both the rectangles contained by its opposite sides. 

Let ABCD be a quadrilateral inscribed in a circle, and join 
AC, BD ; the rectangle AC.BD is equivalent to the two rect- 
angles AB.CD and AD.BC. 

Make the angle ABE equal to DBC, and take each of them 
from the whole angle ABC ; then the remaining angles CBE, 
ABD are equal; and (HI. 18, cor. l) the angles ECB, ADB 
are equal. Therefore (V. 3, cor.) in the triangles ABD, EBC, 



224 



EXERCISES. 



as BC : CE 
BD.CE 



Again 




BD : DA; wlience (V. 10, cor.) BC.DA= 
in the triangles BAE, BDC, because (const.) 
the angles ABE, DBC are equal, as also 
(III. 18, cor. 1) BAE, BDC; therefore 
(V. 3, cor.) as BA : AE : : BD : DC; 
^\ hence (V. 10, cor.) BA.DC==BD.AE. 
' Add these equivalent rectangles to the 
equivalents BC.DA and BD.CE; then 
BA.DC + BC.DA = BD.CE + BD.AE, or 
(II. 1) BA.DC + BC.DA=BD.AC. There- 
fore, the rectangle, etc. 
Cor. 1. If the sides AD, DC, and consequentlj^ (TIT. 16, cor. 
1) the arcs AD, DC, and the angles ABD, CBD be equal, the 
rectangle BD.AC is equivalent to AB.AD together with BC. AD, 
or (II. 1) to the rectangle under AD, and the sum of ABand BC. 
Hence (V. 10, cor.) AB-f-BC : BD : : AC : AD or DC. 

Cor. 2. If AC, AD, CD be all equal, the last analogy be- 
comes AB+BC : BD : : AD : AD ; whence AB + BC=BD. 
Hence in an equilateral triangle inscribed in a circle, a straight 
line drawn from the vertex to a point in the arc cut oiF by the 
base is equal to the sum of the chords drawn from that point 
to the extremities of the base. 



Prop. V. — Theor. — The diagonals of a qtiadrilateral in- 
scribed in a circle, are proportional to the sums of the rectan- 
gles contained by the sides meet'nr/ at their extremities. 

Let ABCD be a quadi-ilateral inscribed in a circle, and AC, 
BD its diagonals; AC : BD : : BA.AD-h 
BC.CD : AB.BC+AD.DC. 

If AC, BD cut one another perpendicu- 
larly in L, then (Ex. 2) AK being the diame- 
ter of the circle, AL. AK = BA.AD, and CL. 
AK=:BC.CD; whence, by addition (I. ax. 
2), AL.AK + CL.AK, or (II. 1) AC.AKz= 
B.\.AD+BC.CD. In a similar manner, 
it would be proved that BD.AK = AB.BC 
Hence ACIAK : BD.AK, or (V. 1) AC : BD : • 
BA.AD + BC.CD : AB.BC + AD.DC. 

But if AC be not perpendicular to BD, draw AEF perpen- 




+ AD.DC. 



EXERCISES. 



225 




dicnlar and CF parallel to BD, and DGH perpendicular and 
BH parallel to AC. Then, because EF is equal to the perpen- 
dicular drawn from C to BI), and GH equal to the one drawn 
from B to AC ; it would be prov- 
ed as before, that AP\AIv = BA. 
AD + BC.CD, and I)n.AK = 
AB.BC + AD.DC. Hence, AF. 
AK : DH.AK, or (V. l) AF : 
DII :: BA.AD-^BC.CD : AB. 
BC + AD.DC. But the triani^les 
AFC, DIIB are equiangular, liav- 
in<x the rio;ht anojes Fand H, and 
the angles ACF, DBH, each equal 
(I. 16) to ALD; therefore (V. 3) 

AF : AC : : DH : DB, and alternately AF : T3IT : : AC : 
DB. Hence (IV. 7) the foregoing analogy becomes AC : BD 
: : BA.AD + BC.CD : AB.BC-hAD.DC. Wherefore, the 
diagonals, etc. 

Scho. From this proposition and the last, when the sides of 
a quadrilateral inscribed in a circle are given, we can find the 
ratio of the diagonals and their rectangle, and thence (V, 15) 
the diagonals themselves. Also, if the sides be given in num- 
bers, we can compute the diagonals. Thus, let the sides taken 
in succession round the figure be 50, 78, 104, and 120. Then, 
the ratio of the diagonals will be that of 50 x 784-104 x 120 to 
60x120-1-78x104; that is, 16880 to 14112, or 65 to 56, by- 
dividing by 252. Again, the rectangle of the diagonals is 50 x 
104-1-78x120, or 14560. But similar rectilineal figures are as 
the squares of the corresponding sides, and consequently the 
sides are as the square roots of the areas ; therefore, taking 65 
and 56 as the sides of a rectangle, we have its area equal to 
3640; and ^ 3640 is to 4/14560, or -^Z 3640 is to |/ (4x3640), 
that is, 1 : 2 : : 65 : 130 : : 56 



112 ; so that 130 and 112 



are the diagonals. 



Prop. VI. — Theor. — If in a straight line drawn through 

the center of a circle^ and on the same side of (he center, (wo 

povits be taken so that the radius is a mean proportional her- 

tween their distances from, the cei^ter / two straight lines drawn 

15 




226 EXEECISES. 

from, those points to any point whatever in the circumference^ 
are proportional to the segments into which the circumference 
divides the straight line intercepted leticeen the same points. 

Let ABC l)e a circle, and CAE a straight line drawn throngh 
its center D ; if ED : DA : : DA : DF, and if BE, BF be 
drawn from any point B of the circumference ; EB : BF : : 
EA : AF. 

Join AB, BD, Then, since DB is equal to DA, we have 

(hyp.) ED : DB :: DB : DF. 
The two triangles EDB, BDF, 
therefore, have their sides about 
the common angle D proportion- 
al ; wherefore (V. 6) the angle 
E is equal to FBD. Now the 
angle liAD is equal (I. 20) to the 
two angles E and EBA, and also 
(I. 1, cor.) to ABD ; wherefore E and EBA are (I. ax. 1) equal 
to ABD. From these take the equal angles E and FBD, and 
(I. ax. 3) the remaining angles EBA, ABF are equal ; and 
therefore (V. 4) in the triangle EBF, EB : BF : : EA : AF. 
If, therefore, in a straight line, etc. 

Cor. Join BC, and produce EB to G. Then, since ABC is 
(III. 11) a right angle, it is equal to the two EBA, CBG. 
From these equals take the equal angles ABF, EBA, and the 
remainders FBC, CBG are equal ; and therefore (V. 4, 2d case) 
EB : BF : : EC : CF. But it has been proved that EB : 
BF : : EA : AF ; therefore (IV. 1) EA : AF : : EC : CF. 
Hence, if the segments EA, AF be given, the point C may be 
determined by the method shown in the third corollary to the 
sixteenth proposition of the first book; and the circle ABC 
may then be described, its diameter AC being determined. 

Scho. The circle may also be determined in the following 
manner: Since (hyp.) ED : DA : : DA : DF, by division, 
EA : DA : : AF : DF ; whence, alternately and by division, 
EA— AF : AF : : AF : DF. Hence DF is a third proper- 
tional to the difference of E A, AF, and to AF, the less ; and 
thus the center D is determined. From the last analogy also 
we obtained (IV. 11) EA— AF : EA : : AF : AD; an anal- 
ogy which serves the same purpose, since it shows that the ra- 



EXERCISES. 227 

dius of the circle is a fourth proportional to the difference of 
the segments EA, AF, and to those segments themselves. 

Prob. VII, — TnEOR. — The perpendiculars drawn from the 
three angles of any triangle to the ojjposite sides, intersect one 
another in the same point. 

If the triangle be right-angled, it is plain that all the perpen- 
diculars pass through the right angle. But if it be not right- 
angled, let ABC be the triangle, and about it describe a circhi; 
then, B and C being acute angles, draw ADE perpendicular to 
BC, cutting BC in D, and the circumference in E; and make 
DF equal to DE ; join BF and produce it, if necessary, to cut 
AC, or AC produced in G ; BG is perpeixlicular to AC. Join 
BE ; and because FD is equal to DE, the angles at D right 
angles, and DB common to the two triangles FDB, EDB, the 
angle FBD is equal (I. 3) to EBD ; but 
(III. 18, cor. 1) CAD, EBD are also equal, 
because they are in the same segment ; 
therefore CAD is equal to FBD or GBC, 
But the angle ACB is common to the tv o 
triangles ACD, BCG; and therefore (I, 
20, cor. 5) the remaining angles ADC, 
BGC are equal ; but (const.) ADC is a 
right angle; therefore also BGC is a right angle, and BG is 
perpendicular to AC, In the same manner it would be shown 
that a straight line CII, drawn through C and F, is perpendicu- 
lar to AB. The three perpendiculars therefore all pass through 
F ; wherefore, the perpendiculars, etc, 

Scho. This limitation prevents the necessity of a different 
ease, which would arise if the perpendicular AD fell without 
the triangle. If the angle A be obtuse, the point F lies with- 
out the circle, and BF, not produced, cutsAC produced. The 
proof, however, is the same, and it is very easy and obvious. 
Another easy and elegant proof, of which the following is an 
outline, is given in Garnier's "Reciproques," etc., Theor. III., 
page 78 : Draw BG and CH perpendicular to AC and AB ; 
join GH, and about the quadrilaterals AHFG and BHGC de- 
8cribe circles, which can be done, as is easily shown ; draw 
also AFD. Then the angles BAD, BCH are equal, each of 




223 EXERCISKS. 

them being equal (TIT, 18, cor. 1) to HGF ; and the angle ABO 
"being coniinoii, ADB is equal (I. 20) to BHC, and is therefore 
a right angle. 

Prop. VTTT. — Theor. — From AB, the greater side of the tri- 
angle ABC, cut off AD equal to AC, and join DC; draw AE 
bisecting the vertical angle BAC, andjoiji DE ; d aw also AF 
perpendicular to BC, and DG parallel to AE. Then (1.) the 
angle DEB is eqtdvalent to the difference of the angles at the 
base, ACB, ABC ; or of BAF, CAF, or of AEB, AEC ; and 
DE is equal to EC ; (2.) the angles BCD, EAF are each equiv- 
alent to half the same difference ; (3.) ADC or ACD is equiva- 
lent to half the sum of the angles at the base, or to the comple- 
ment of half the vertical angle ; (4.) BG is equivalent to the 
difference of the segments BE, EC, made by the line bisecting 
the vertical angle. ^ 

1. In the triangles AED, AEC, AD, AC are equal, AE com- 
mon, and the contained angles equal ; therefore (I. 3) DE is 
equal to EC, the angle ADE to ACE, and AED to AEC. But 

(I. 20) because BD is produced, the 
angle ADE is equivalent to B and 
BED; therefore BED is the differ- 
ence of B and ADE, or of B and 
ACB. Also BED is the difference 
of AEB, AED, or of AEB, AEC. 
Again : ABF, BAF are equivalent 
to ACF, CAF, each pair being (L 
20, cor. 3) equivalent to a right angle. Take away ABF; 
then, because the difference of ACF, ABF is BED, there re- 
mains BAF equivalent to BED, CAF ; that is, BED is the 
difference of BAI\ CAF. 

2. The difference BED is equivalent (T. 20) to the two angles 
ECD, EDC, which (I. 1, cor.) are equal ; therefore ECD is half 
of BED. Again : in the triangles AlID, AIIC, because AD, 
AC are equal, AIT common, and the contained angles equal, 
DII is equal (I. 3) to IIC, and the angles at H are equal, and 
are therefore right angles. Then, in the triangles AEF, CEII, 
the angles AFE, CUE are equal, and AEC common ; therefore 




EXERCISES. 229 

(T. 20, cor. 5) the article EAF is equal to ECU, which has been 
proved to be equivalent to the half of BED, 

3, Since the anole BAG is common to the triangjles ABC, 
ADC, the angles ADC, ACD are (I. 20) equal to ABC, ACB ; 
and therefore, since ADC, ACD are equal, each of them is 
equivalent to half the sum of ABC, ACB ; also either of them, 
ADC, is the complement ofDAIi, half the vertical angle, since 
AHD is a right angle. 

4. Because DH, HC are equal, and HE, DG parallel, GE is 
equal (V. 2) to EC; and therefore BG, the difference of BE, 
GE, is also the difference of BE, EC. 

Sclw. 1. It is easy to prove without proportion, that if AB (in 
the figure for the above proposition) be bisected in D, the 
straight line DE parallel to BC bisects AC, and that the trian- 
gle ADE is a fourth of ABC. For (I. 15, cor. 5) the triangles 
BDC, BEC are equal. But (I. 15, cor. 5) BDC is half of ABC ; 
and therefore BEC is half of ABC, and is equal to BEA. 
Hence the bases AE, EC are equal ; for if they were not, the 
triangles ABE, CBE (I. 15, cor. 6) would be unequal. Again : 
because AD, DB are equal, the triangle ADE is (I. 15, cor. 5) 
half of ABE, and therefore a fourth of ABC. 

Conversely, if DE bisect AB, AC, it is parallel to BC. For 
(T. 15, cor.) the triangles BDC, BEC are each half of ABC; 
and these being therefore equal, DE is parallel (I. 15, cor. 5) to 
BC. 

Hence, it is plain (I. 14) that the straight lines joining DE, 
DF, EF divide the triangle ABC into four equal triangles, 
similar to the whole and to one another; and that each of 
these lines is equal to half the side to which it is parallel. 

Scho. 2. Instead of cutting off AD equal to AC, AC may be 
produced through C, and by cutting off, on AC thus produced, a 
part terminated at A, and equal to AB, and by making a con- 
struction similar to that of the foregoing proposition, it will be 
easy to establish the same properties as those above demon- 
strated, or ones exactly analogous. 

Prop. IX. — Prob. — Giveii the base of a triangle, the differ- 
ence of the S'des, and the difference of the angles at the base; 
to construct it. 




230 EXERCISES. 

Make BC equal to the given base, and CBD equal to half 
the difference of the angles at the base ; from C as center, at a 
distance equal to the difference of the sides, describe an arc 
cutting BD in D ; join CD and produce it ; make the angle 
DBA equal to BDA ; ABC is the required triangle. 

For (I. 1, cor.) AD is equal to AB, and the difference of AC, 
AD, or of AC, AB, is CD ; and (Ex. 8) 
since AD is equal to AB, CBD is equal 
to half the difference of the angles at the 
base. The triangle ABC, therefore, is 
the one required, as it has its base equal 
to the given base, the difference of its 
sides equal to the given difference, and 
the difference of the angles at the base equal to the given dif- 
ference of those angles. 

Method of C\miputation. In the triangle BCD there are 
given BC, CD, and the angle CBD ; whence the angle C can 
be computed ; and the sum of this and twice CBD is the angle 
ABC. Then, in the whole triangle ABC, the angles and the 
Bide BC are giveti ; whence the other sides may be computed ; 
or, one of them being computed, the other will be found by 
means of the given difference CD. 

Prop. X.— Prob. — Given the segments into which the base 
of a triangle is divided by the line bisecting the vertiecd anglQ 
and the difference of the sides ; to construct the triangle. 

Construct the triangle CED, having the sides CE, ED equal 
to the given segments, and CD equal to the given dillerence of 
the sides; produce CE, and make EB equal to ED; bisect the 
angle BED by EA, meeting CD produced in A, and join AB ; 
ABC is the required, triangle. 

A For, in the triangles AEB, AED, BE 

is equal to ED, EA common, and the 
angle BEA equal to DEA ; therefore (I, 
3) BA is equal to DA, and the angle 
EAB to EAD. Hence, ABC is the re- 
quired trianale; for CD, the difference 
of its sides, is equal to the given differ- 
ence, and BE, EC, the segments into which the base is 




FXERCISE8. 



231 



divided by the line bisecting the vertical angle, are equal to 
the given segments. 

Meth d of Computation. The sides of the triangle CDE 
are given, and therefore its angles may be computed; one of 
which and the sni)plement of the other are tlie angles C and 
B. Then, in the triangle ABC, the angles and BC are given, 
to compute the remaining sides. 

OTHERWISE, 

Since (V. 4) CE : EB : : CA : AB, we have, by division, 
CE— EB : EB : : CA— AIJ : AB ; which, therefore, becomea 
known, since the first three terms of the analogy are given; 
and thence AC will be found by adding to AB the given dit- 
ference of the sides. 

Prop. XT. — I V.ob. — Given the base of a triangle^ the vertical 
angle, and the difference of the sides ; to construct the tri- 
angle. 

Let MNO be the given vertical angle; produce ON" to P, 
and bisect the angle MNP by NQ. Then, make BD equal to 
the difference of the sides, and the 
angle ADC equal to QNP; from B 
as center, with the given base as ra- 
dius, describe an arc cutting DC in 
C ; and make the angle DCA equal 
to ADC ; ABC is the required tri- 
angle. For (const.) the angles ACD, 
ADC are equal to MNP, and there- 
fore (I. 20 and 9) the angle A is 
equal to MN). But (I. 1, cor.) AD 
is equal to AC, and therefore BD is 
the difference of the sides AB, AC ; 
and the base BC is equal to the 
given base ; wherefore ABC is the triangle required. 

Method of Computation. In the triangle CBD, the sides 
BC, BD, and the angle BDC, the supplement of ADC or MNQ 
are given ; whence the other angles can be computed. The 
rest of the operation will proceed as in the ninth proposition of 
these Exercises. 




Q 



N 





232 EXERCISES. 

Prop. XII. — Prob. — Given one of the angles at the base of 
a triangle, the difference of the sides, and the difference of the 
segments into which the base is divided by the line bisecting the 
vertical angle; to construct the triangle. 

Construct the triangle DBG, liaving the angle B equal to 
the given angle, BD equal to the difference of the sides, and 

BG equal to the difference of the segments ; 
draw DC perpendicular to DG, and meet- 
^'^ '^ ing BG produced in C; produce BD, and 

make the angle DCA equal to CDA ; 
B G E c ABC is the triangle required. For it has 
B equal to the given angle, and the differ- 
ence of its sides BD equal to the given difference; and if AHE 
be drawn bisecting the angle BAC, it bisects (I. 3) CD, and is 
perpendicular to it ; it is therefore parallel to DG, one side of 
the triangle CDG; and, bisecting CD in II, it also (V, 2) 
bisects GC in E. Hence BG, the difference of BE, GE is also 
the difference of BE, EC, the segments into which the base is 
divided by the line bisecting the vertical angle. 

Method of Computation. In the triangle DBG, BD, BG, 
and the angle B are given ; whence (Trig. 3) we find half the 
difference of the angles BGD, BDG, which is equal to half the 
angle C. Then (by the same proposition) we have in the tri- 
angle ABC, tan I (C-B) : tan ^ (C+B) : : c-b or BD : c+ 
b ; whence the sides c and b become known, and thence BC 
by the first case. 

For (I. 20) BEA=r| A+C, and consequently BEA— | A = C. 
But (I. 16) BGD = BEA, and BDG^BAE^^A; and there- 
fore BGD— BDG =C. 

Prop. XIII. — Prob. — Given the base of a triangle, the verti- 
cal angle, and the sutn of the sides ; to construct it. 

Make BD equal to the sum of the sides, and the angle D 
equal to half the vertical angle ; from B as center, with the 
base as radius, describe an arc meeting DC in C ; and make 
the angle DCA equal to D ; ABC is the required triangle. 

For (I. 1, cor.) AD is equal to AC, and therefore BA, AC are 
equal to BD, the given sum. Also (I. 20) the exterior angle 
BAC is equal to the two D and ACD, or to the double of D, 



EXERCISES. 



233 




because D and ACD are equal ; therefore, since D is half the 

given vertical angle, BAG is equal 

to that ansjle. The triancjle ABC, 

therefore, has its base equal to the 

given base, its vertical angle equal 

to the given one, and the sum of its 

sides equal to the given sum ; it is 

therefore the triangle required. 

Scho. Should the circle neither cut nor touch DC, the prob- 
lem would be impossible with the proposed data. If the circle 
meet DC in two points, there will be two triangles, each of 
which will answer the conditions of the problem. These tii- 
angles, however, will differ only in position, as they will be on 
equal bases, and will have their remaining sides equal, each to 
each. This problem might also be solved by describing (III. 
19) on the given base 15C a segment of a circle containing an 
angle equal to half the vertical angle ; by inscribing a chord 
BD equal to the sura of the sides; by joining DC; and then 
proceeding as before. The construction given above is prefer- 
able. / 

Method of Computation. In the triangle BDC, the angle 
D, and the sides BC, BD are given ; whence the remaining 
angles can be compute 1 ; and then, in the triangle ABC, the 
angles and the side BC are given, to compute the other sides. 



Prop. XIV. — Prob. — Given the vertical angle of a triangle^ 
and the segments into which the line iisectiiig it divides the 
base ; to construct it. 

In the straight line BC, take BH and CII equal to the seg- 
ments of the base ; on BC describe (III. 
19) the segment BAC containing an angle 
equal to the vertical angle, and complete 
the circle ; bisect the arc BEC in E ; draw 
EHA, and join BA, CA ; ABC is the re- 
quired triangle. For (III. 16, cor. 1) the 
angles BAII, CAH are equal, because the 
arcs BE, EC are equal ; and therefore the 
triangle ABC manifestly answers the conditions of the ques- 
tion. 




234 EXEECISF.8. 

Scho. The construction might also be effected by describing 
on BH and CH segments each containing an angle equal to half 
the vertical angle, and joining their point of intersection A 
■with B and C. Another solution may be obtained by the 
principle (V. 4) that BA • AC : : BII : HC. For if a trian- 
gle be constructed having its vertical angle equal to the given 
one, and the sides containing it equal to the given segments, 
or having the same ratio, that triangle will be similar to the 
required one ; and therefore on CB construct a triangle equi- 
angular to the one so obtained. 

Method of Computation. Join BE, and draw ED perpen- 
dicular to BC. Then BD or DC is half the sum of the sear- 
nients BII, HC, and DH half their difference ; and BD is to 
DM, or twice BD to twice DH, as tan DEB to tan DEH. 
Now it is easy to show that BED is half the sum of the angles 
ABC, ACB, and HED half their difference ; and therefore these 
angles become known; and BC being given, the triangle ABC 
is then resolved by the method for the Hrst case. 

Cor. Hence, we have the method of solving the problem in 
wliich the base, the vertical angle, and the ratio of the sides of 
a triangle are given^ to construct it. For (V. 4) the sides be- 
ing proportional to the segments BH, HC, it is only necessary 
to divide the given base into segments proportional to the 
sides and then to proceed as above. * 

Prop. XV. — Pbob. — Given the base^ the perpendicular^ and 
the vertical angle of a triangle / to construct it. 

Make BC equal to the given base, and (HI. 19) on it de- 
scribe a segment capable of containing 
an angle equal to the vertical angle ; 
draw AK parallel to BC, at a distance 
from it equal to the given perpendicular 
and meeting the arc in A ; join AB, 
AC; A13C is evidently the tiiangle re- 
quired. 

Method of Compiitat'on. Draw the 

perpendicular AD, and parallel to it 

draw LGH, through the center G; join BG, AG, AH. Now, 

iiuce AH evidently bisects the anglu BAC, the angle HAD or 




EXERCISES. 



235 



n is (Ex. 8) equal to half the difference of the ansrles ABC, 
ACB, and therefore (III. 1 0) AGK is the wliole difference of 
those angles. Then, in the riglit-angled triangle BP'G, the 
angles and BF being known, FG can be computed ; from 
which and from AD or KF, KG becomes known. Now, to the 
radius BG or AG, FG is the cosine of BGF, or BAG, and KG 
the cosine of AKG; and therefore FG : KG : : cos liAC : 
cos AGK. Hence the angles ABC, ACB become known, and 
thence the remaining sides. 

iSc/io. Should the parallel AK not meet the circle, the solu- 
tion would be impossible, as no triangle could be constructed 
having its base, perpendicular, and vevlic:il angle of the given 
magnitudes. If the parallel cut the cinrle, there will be two 
triangles, either of which will answer the conditi(m of the ques- 
tion. They will differ, however, only in position, as their sides 
will be equal, each to each. If the parallel touch the circle, 
there will be only one triangle ; and it will be isosceles. 



Pkop. XVI. — Prob. — Given the base of a triangle, the ver- 
tical angle, and the radius of the inscribed circle ; to construct 
the triangle. 

Let GHK be the given angle ; produce GH to L, and bisect 
LHK by HM; on the given base BC desciibe (III. 19) the seg- 
ment BDC, containing an angle equal to GHM ; draw a straight 
line parallel to BC, at a distance equal to the given ra<lius, and 
meeting the arc of the segment in D; join I)B, DC; and make 
the angles DBA, DCA equal to DBC, DCB, each to each ; 
ABC is the required triangle. L ii g 

Produce BD to E, and drawDF perpen- 
dicular to BC. Then, since (const.) the 
angle BDC is equal to GHM, the two jj 
DBC, DCB are equal (I. 20 and 9) to 
LHM, and therefore (const.) ABD, ACD 
are equal to KHM. But (1. 20) BDC is 
equal to BEC, ECD, or to BAC, ABD, 
ACD, because (I. 20) BEC is equal to 
BAC, ABD. Therefore BAC, ABD, ACD 
are equal to GHM; from the former take 
ABD, ACD, and from the latter KHM, which is equal to 




236 



KXERCISE8. 



them, and the remainders BAG, GHK are equal. It is plain, 
also (I. 14), tliat perpendiculars drawn IVoni D to AB and AC 
would be each equal to DF ; and therefore a circle described 
from D as center, with DF as radius, would be inscribed in the 
trian<ijle ABC ; and BC being the given base, and A being 
equal to the given vertical angle, ABC is the required tri- 
angle. 

The method of computation is easily derived from that of the 
preceding proposition. 

Scho. Should the parallel to BC not meet the arc of the seg- 
ment, the solution would be impossible, as there would be no 
triangle which could have its base, its vertical angle, and its 
inscribed circle of the given magnitudes. If the parallel be a 
tangent to the arc, the radius of the inscribed circle would be 
a maximum. Hence, to solve the problem in which the base 
and the vertical angle are given, to construct the triangle, so 
that the inscribed circle may be a maximum, describe the seg- 
ment as belbre, and to find D bisect the arc of the segment. 
The rest of the construction is the same as before ; and the tri- 
angle will evidently be isosceles. 



Prop. XVII. — Prob. — Given the three lines drawn from the 
vertex of a triangle, one of them 'perpend cnlorto the base, one 
bisecting the base, and one bisecting the vertical angle y to con- 
struct the triitngle. 

Take any straight line BC and draw DA perpendicular to it, 
and equal to the given perpendicular; fi-om A as center, with 
radii equal to the lines bisecting the vertical angle and the 

base, describe arcs cutting BC in E and 
F, and draw AEH and AF ; through F 
draw GFII perpendicular to BC, and 
draw AG making the angle IIAG equal 
to II, and cutting HG in G; from G as 
center, with GA as radius, describe a 
circle cutting BC in B and C; join AB, 
AC ; ABC is the triangle required. 
For (HI. 2) since GFII is perpendicu- 
lar to BC, BC is bisected in F ; and (III. 17) the arcs BII, HO 
are equal. Therefore (III. 16, cor. 1) the angles BAII, CAH 




EXERCISES. 237 

are equal. Hence, in the triangle ABC, the perpendicular AD, 
the line AE bisecting the veitical angle, and the line AF 
bisecting the base, are equal to the given lines. Therefore 
ABC is the triangle required. 

Scho. It" the three given lines be equal, the problem is inde- 
terminate ; as any isosceles triangle, having its altitude equal 
to one of the given lines, will answer the conditions. 

3fethod of Computation. Through A draw a parallel to BC, 
meeting HG ))rodnced in K, Then, in the right-angled triangle 
ADE, AE, AD being given, DAE, or H, may be computed ; 
the double of which is AGK ; and AK or FD being given, AG, 
GK can be found, and thence GF. Hence, if GB were drawn, 
it and GF being known, BF, and the angle BGF, or BAC, can 
be computed. The rest is easy; DAE, half the difference of B 
and C, being known. 

Analysis. Let ABC be the required triangle, AD the per- 
pendicular, and AE, AF the lines bisecting the vertical angle 
and the base. About ABC describe (HI, 25, cor. 2) a circle, 
and join its center G, with A and F, and produce GF to meet 
the circumference in H. Then (HI. 2) GFH is perpendicular 
to BC, and (IH. \1) the arcs BH, HC are equal. But (HI. 16) 
the equal angles BAE, CAE at the circumference stand on 
equal arcs; and therefore AE being produced, will also pass 
through H; and the point H, and the angle GHA and its equal 
HAG are given. Hence also the center G and the circle are 
given, and the method of solution is plain. 

Prop. XVIH. — Prob. — Given the base of a triangle^ the ver- 
tical angle, and the straight line bisecting that angle y to con- 
struct the triangle. 

On the given base BC describe (HI. 19) the segment BAC 
capable of containing an angle equal to the given vertical an- 
gle, and complete the circle ; bisect the arc BEC in E, and join 
EC; perpendicular to this, draw CF equal to half the line 
bisecting the vertical angle, and from F as center, with FC as 
radius, describe the circle CGII, cutting the straight line pass- 
in through E and F in G and H ; make ED equal to EG, and 
draw EDA; lastly, join AB, AC, and ABC is the required 
triangle. 



238 EXEKCI8E8. 

For the triangles CEA, CED are equiangular, the angle CEA 
being tomnion, and BCE, CAE being each equal to BAE. 
Therefuie AE : EC : : EC : ED, and (V. 9, cor.) AE.ED= 







EC=. But (ITI. 16, cor. 3, and 21) HE.EG or HE.ED=EC 
and therefore AE.ED^HE.ED; whence AE=:HE, and (I. ax. 
3) AD=GH = '2CF. AD is therefore equal to the given bisect- 
ing line, and it bisects the angle BAC. Hence ABC is the re- 
quired triangle. 

Method of Computation. Draw EL perpendicular to BC, 
and join CH. Then BCE is equal to BAE, half the vertical 
angle A ; and therefore, to the radius EC ; CL is the cosine of 
■§ A, and CF is the tangent of CEF to the same radius ; where- 
fore, to any radius, CL : CF, or BC : AD : : cos ^ A : tan 
CEF or cot EFC ; and hence the angle H, being half of EEC, 
is known. Also ECU is the complement of II, because ECF is 
a right angle, and FCH equal to H. But (Tgig. 2) EC : EH or 
EA : : sin II : sin ECU, or cosH ; or (Trig. defs. cor. 6) EC : 
EA : : It : cot H. Also in the triangle ACE, EC : EA : : 
Bin -^ A : sin ACE ; whence (IV. 7) R : cot H : : sin ^ A : sin 
ACE ; whence ACE may be found ; and if from it, and from 
ABE, its supplement (BE being supposed to be joined) BCE 
be taken, the remainders are the angles at the base. 

Analysis. Let ABC be the required triangle, and let AD, 
the line bisecting the vertical angle, be produced to meet the 
circumference of the circumscribed circle in E; join also EC. 
Then (III. 19) the circumscribed circle is given, since the base 
and vertical angle are given ; and the arc BEC is given, as are 
also its half EC, and the chord EC. Now the triangles AEO, 



EXERCISES. 239 

DEC are equiangular ; for the angle CEA is common, and (IH. 
18, cor. 1) BCE is equal to BAE or EAC. Hence AE : EC : : 
EC : ED, and theiefore AE.ED=ECl Hence (IH. 21) it is 
evident that if EC be made a tangent to a circle, and if 
through the extremity of the tangent a line be drawn cutting 
the circle, so that the part within the circle may be equal to 
AD, DE will be equal to the external part ; whence the con- 
struction is manifest. 

Prop. XIX. — Proh. — Given the straight liius drawn from 
the three angles of a trian;^ le to the points of bisection of the 
opposite sides / to construct the triangle. 

Trisect the three given lines, and describe the triangle ABC 
having its three sides respectively equal to two thirds of the 
three given lines ; complete the parallelogram ABEC, and 
draw the diagonal AE ; produce also 
CB, making BE equal to BC ; and 
join FA, FE; AFE is the required 
triangle. 

Produce AB, EB to G, H. Then 
(H. 14, cor.) AE, BC bisect each other 
in D, and therefore FD is equal to 
one of the given lines, for BD is one 

third of it, and FB two thirds. Again : because FB, BC are 
equal, and HB pai-allel to AC, FA is bisected in H, and HB is 
half of AC or BE. Hence, HE is equal to another of the given 
lines, and it bisects FA. In the same manner it would be 
proved, that AG is equal to the remaining line, and that it 
bisects FE. Hence FAE is the triangle required. 

Method of Computation. BD, which is a third of one of 
the given lines, bitects AE, a side of the triangle ABE, in 
•which the sides AB, BE are respectively two thirds of the two 
remaining lines. Then (I. 24, cor. 6, and II. 12, cor.) 2AD^= 
AB'+BE*— 2BD'; whence AD, and consequently AE may be 
found ; and in the same manner the other sides may be com- 
puted. 

Prop. XX.— Prob, — Given the three perpendiculars of a tri- 
angle y to construct it. 




240 



EXERCI8E8. 



A- 
B- 
C- 
D- 



E 




Let A, B, C be three given straight lines; it is required to 
describe a triangle having its three perpendiculars respectively 
eqnJil to A, B, C. 

Take any straight line D, and describe a triangle EFG. hav- 
ing the sides FG, P'E, EG third i)ro- 
portionals to A and 1), B and D, C 
and D; and draw the perpendiculars 
EH, GL, FK. 

Then the rectangles rG.A,EF.Bare 
equal, each being equal to the square 
ofD; and therefore EF : FG : : A : 
B. But in the similar triangles EHF, 
GLF, EF : FG : : EH : GL; where- 
fore EH : GL : : A : B ; and in the 
same manner it would be proved, that 
EH : FK : : A : C. Hence (IV. 7) 
if EH be equal to A, GL is equal to B, and FK to C ; and 
EFG is the triangle requii-ed. 

But if EH be not equal to A, make EM equal to it, and draw 
NMO parallel to FG, and meeting EF, EG, produced, if neces- 
sary, in N and O ; ENO is the required triangle. Draw OP 
perpendicular to EN. Then EM : EH : : EO : EG, and OP 
GL : : EO : EG ; whence (IV. 7 and alternately) EM : OP : 
EH : GL ; or, by the foregoing part, EM : OP : : A : B 
wherefore (IV. 7) since EM is equal to A, OP is equal to B 
and it would be proved in a similar manner, that the perpen- 
dicular from N to EO is equal to C. 

Method of Computation. By dividing any assumed num- 
ber successively by A, B, C, we find the sides of the triangle 
EFG, and thence its angles, or those of ENO; whence, since 
the perpendicular EM is given, the sides are easily found. 

Or, when the sides of EFG are found, its perpendicular EH 
may be comj)Uted in the manner pointed out in the note to tlie 
twelfth proposition of the second book. Then EH : A : : FG: 
NO : : EF : EN : : EG : EO. 



Prop. XXI. — Prob, — Given the sum of the legs of a right- 
angled triangle, and the sum of the hypothenuse, and the per- 
pendicular to it from the right angle; to construct the triangle. 




EXERCISES. 241 

Let tho sum of the legs of a right-angled triangle be equal 
to the straight line A, and the sum of the hypothenuse and 
perpendicular equal to BC ; it is required to construct the tri- 
angle. 

Find (I. 24, cor. 4) a straight line the square of which is equal 
to the excess of the square of BC 

above that of A, and cut off BD ~ 

equal to that line ; on DC as diam- 
eter describe a semicircle, and draw 
EF parallel to DC ate a distance 
equal to BD ; join either point of 
intersection, E, with D and C ; 
DEC is the required triangle. 

Draw the perpendicular EG, which (const.) is equal to BD. 
Then (II. 4) BC^ or AHEG^ = (DC-fEG)^=DC'^ + EG=+ 
2DC.EG; whence A^=DC^ + 2DC.EG. Also (DE + EC)^= 
DE^ + EC^+2DE.EC=DC^-|-2DC.EG, because (III. 11, and I. 
24, cor. 1) DC'=DE^+EC^ and DC.GE^DE.EC, each being 
equal to twice the area of the triangle DEC. Hence (DE4- 
EC)- = A-; wherefore (I. 23, cor. 3)bE + EC=A; and DEC 
is the triangle required. 

Method of Computatiofi. From the construction, we have 
BD or EG= V(BC'—A'). Then DC = BC-BD; by halving 
which we get the radius of the semicircle ; and if from the 
square of the radius drawn from E, the square of EG be taken, 
and if the square root of the remainder be successively taken 
from the radius, and added to it, the results will be the seg- 
ments DG, GC ; from which, and from EG, the sides (I. 24, cor. 
1) are readily computed. 

Prop. XXII. — Prob. — Given the base of a triangle, the per- 
pendicular, and the difference of the sides ; to construct it. 

Make AB equal to the given base, and parallel to it draw 
CD, at a distance equal to the given perpendicular; dravvBDP 
perpendicular to CD, and make DF equal to DB ; from A as 
center, with a radius AE equal to the given difference, describe 
the circle ELN ; through B, F describe any circle cutting ELN" 
in L, N, and let G be the point in which a straight line drawn 
through L, N cuts FB produced ; draw the tangent GK, and 
16 



242 



EXEECISES. 



draw AKM cutting CD in M; join BM; and it k ^ident 

from the second corollary to tbe 
ninth proposition of the third book, 
that AMB is the required triangle* 
Method of Computation. From 
M as center, with MK as radius, 
describe a circle, and by the corol- 
lary referred to, it will pass through 
B and F, Join AG and produce 
BA to H. Then the rectangle 
FG.GB=AG^-AK^ each being 
equal to the square of GK ; that is, 
FB.BG+BG==AB^-fBG^— AK^ 
Take awayBG^; the:i FB.BG=AB^-AK==(AB+AK) (AB 
— AK)=HB.EB. Hence BG becomes known. Then, in the 
two right-angled triangles GBA, GKA, the angles at A can be 
computed, and their difference is the angle MAB in the re- 
quired triangle. The rest is easy, if the perpendicular from 
M to AB be drawn. 




Prop. XXIII. — Prob. — Given the base, the area, and the 
•ratio of the sides of a triangle; to construct it. 

Let AB be the given base, and (V. 3, and scho.) find the 
points C, D, so that AC, CB, and AD, DB are in the ratio of 

the sides ; on CD as diame- 
ter describe the semicircle 
CED, and (II. 5, scho.) to 
AB apply a parallelogram 
BF double of the given area ; 
let FG, the side of this oppo- 
site to AB, produced if neces- 
sary, cut the semicJircle in E, and join EA, EB j EAB is the re- 
quired triangle. 

For (Ex. 6, cor.) AE is to EB as AC to CB ; that is, in the 
given ratio. Also (1. 15, cor. 4) AEB is half of the parallelo- 
gram BF, which is double of the given area. Therefore AEB 
is on the given base, is of the given area, and has its sides in 
the given ratio. 

Scho. If FG, or FG produced, do not meet the semicircle, 




EXERCISE8. 243 

the problem is impossible ; if it cut it in E and E^, there will 
be two triangles essentially different, each of which will answer 
the conditions of the problem ; if it touch the semicircle, there 
will be only one triangle, and it will be the greatest possible 
with the base of the given magnitude, and the sides in the 
given ratio ; and hence we have the means of solving the 
problem in which it is required to construct a triangle on a 
given base, having its sides in a given ratio, and its area a 
maximu7n. 

Method of Computation. Join the center H with E, and 
draw the perpendicular EK. Then, let m '. n : '. AC : CB, 
and consequently m, : n : : AD : DB ; then, from these, by 
composition and division, we get m+n : 7i : : AB : CB, and 
m—n '. n : : AB : DB; whence DC and its half, the radius 
of the circle, become known. EK also is found by dividing 
double of the area by AB. Then, in the triangle EKH, KH 
can be found, and thence AK and KB ; and, by means of them 
and EK, the sides EA, EB may be computed. If E'' be taken 
as the vertex, the method of computation is almost the same, 
and is equally easy. 

Prob. XXIV. — Prob. — Given the base of a triangle, the ver- 
tical angle, and the rectangle of the sides ; to construct it. 

On the given base describe a segment containing an angle 
€qual to the given angle; to the diameter of the circle of which 
this segment is a part, and to the lines containing the given 
rectangle, find a fourth proportional; this proportional (Ex. 3) 
is the perpendicular of the triangle ; and the rest of the solu- 
tion is effected by means of the fifteenth proposition of these 
Exercises. 

Prop. XXV. — Prob, — To divide a given triangle into two 
parts in a given ratio, by a straight line parallel to one of the 
sides. 

Let ABC be a given triangle ; it is required to divide it into 
two parts in the ratio of the two straight lines, m, n, by a 
straight line parallel to the side BC. 

Divide (V. 3, scho.) AB in G, so that BG : GA : : m : w, 
and between AB, AG find (V. 11) the mean proportional AH' 



244 



EXERCISES. 




draw HK parallel to BC ; ABC is divided by HK in the man- 
ner required. 

For (V. 14, scho.) since the three 
straight lines AB, All, AG are propor- 
tionals, AB is to AG as the triangle ABC 
to AHK ; whence, by division, BG is to 
AG, or (const.) m is to w, as the quadrilat- 
eral BCKH to the triangle AHK. 

In practice, the construction is easily 

and elegantly effected, when the triangle 

is to be divided either into two or more 

"^ parts proportional to given lines, by divid- 

n ing AB into parts proportional to those 

lines, and through the points of section 

drawing perpendiculars to AB, cutting the arc of a semicircle 

described on AB as diameter; then by taking lines on AB, 

terminated at A, and severally equal to the chords drawn from 

A to the points of section of the arc, the points on AB wall be 

obtained through which the parallels to BC are to be drawn. 

The reason is evident from the foregoing proof in connection 

with the corollary to the eighth proposition of the fifth book. 

Cor. Hence a given triangle can be divided into two parts 
in a given ratio, by a straight line parallel to a given straight 
line. 

Also, a triangle can be divided into two parts in a given ra- 
tio, by a straight line drawn through a given point in one of 
the sides; and a given quadrilateral can be divided into tw^o 
parts in a given ratio, by a straight line jsarallel to one of its 
sides. 



Prop. XXVI. — Peob. — From a given point in one of the 
sides of a given triangle, to draw two straight lines trisecting 
the triangle. 

. Let ABC be the given triangle, and 

D the given point ; then, if BD be not 
less than one third of BC, to BD, to a 
third part of BC, and to the perpendic- 
ular from A to BC, find a fourth pro- 
portional, and at a distance equal to it 




EXERCISES. 245 

draw a parallel to BC, cutting BA in E, and join ED ; the tri- 
angle BED is evidently (V. 10, cor., and I. 15, cor. 4) a third 
part of ABC ; and CDF will be constructed in a similar man- 
ner, if CD be not less than a third of BC. 

If DC, one of the segments, be less than a third of BC, the 
triangle BDE is constructed as before, but the rest of the pre- 
ceding solution fails, as the second parallel would fall above 
the triangle. In this case, cut off BA between E and A, a part 
equal to BE, and call it EG; then, if DG be joined, the trian- 
gle ABC is trisected by DE, DG. 

Scho, It is easy to see that this method may be readily ex- 
tended to the division of a triangle into more equal parts than 
three, or into parts proportional to given magnitudes, by 
straight lines drawn from a given point in one of the sides. 

Prop. XXVII. — Theor. — If the sides of a right-angled tri- 
angle be continual proportionals, the hypothenure is divided in 
extreme and mean ratio by the perpendicular to it from the 
right angle / and the greater segment is equal to the less or re- 
mote side of the triangle. 

Let ABC be a triangle right-angled at A, and let AD be per- 
pendicular to BC ; then if CB : BA 
: : BA : AC, BC is divided in ex- ■*■ 

treme and mean ratio in D, and BD 
is equal to AC. 

For (hyp.) CB : BA : : BA : AC, 
and (V. 8, cor.) CB : BA : : BA : 
BD ; therefore BA : AC : : BA : 

BD, and AC is equal to BD. Again (V. 8, cor.), BC : CA : : 
CA : CD, or BC : BD : : BD : DC, and therefore (V. def 3) 
BC is divided in extreme and mean ratio in D. 

Scho. Conversely, if BC : BD : : BD : CD, and if BAC be a 
right angle, and DA perpendicular to BC; CB : BA : : BA : 
AC, and BD is equal to AC. For (hyp.) CB : BD : : BD : 
CD, and (V. 8, cor.) CB : CA : : CA : CD ; wherefore BD^ is 
equal to CA', each (V. 9, cor.) being equal to the rectangle 
BC.CD, and therefore BD is equal to CA. Again (V. 8), CB : 
BA : : BA : BD or AC. 




246 EXERCISES. 

Prop. XXVllL — Prob. — Given the angles and diagonals 
of a parallelogram ; to construct it. 

On one of the diagonals describe a segment of a circle con- 
taining an angle equal to the given angle at either extremity of 
the other ; from the middle point of this diagonal as center, 
with half the other diagonal as radius, describe an arc cutting 
the arc of the segment ; through the extremities of the first 
diagonal draw four straight lines, two to the intersection of the 
arcs, and two parallel to these ; the parallelogram thus formed 
is easily proved to be the one required. 

Prop. XXIX. — Prob. — Given the vertical angle of a trian- 
gle^ and the radii of the circles inscribed in the parts into which 
the triangle is divided by the perpendicular y to construct the 
triangle. 

Take any straight line ABC, and through any point B draw 
the perpendicular BD ; make BA, BC equal to the given radii, 
and let E, F be the angular points, remote from B, of squares 

described on AB, BC ; join EF, and 
on it describe the segment EDF, 
containing an angle equal to half the 
given vertical angle ; let the perpen- 
dicular cut the arc EDF in D, and 
join DE, DF ; draw DG, DH making 
the angles EDG, FDH respectively 
equal to EDB, FDB; DGH is the 
required triangle. 

For (I. 14) perpendiculars drawn 
from E to DB, DG are equal, and 
each of them is equal (const.) to the perpendicular from E to 
GB. Each of them therefore is equal to the given radius AB 
and a circle described from E at the distance of one of these is 
inscribed in the triangle DGB. In the same manner it would 
be shown, that a circle described from F as center, with the 
other given radius, would be inscribed in DBII. Hence, since 
the angle GDH is double of EDF, GDII is equal to the given 
vertical angle, and the triangle GDH answers the conditions 
of the question. 

Scho. The preceding solution is strictly in accordance with 




EXEECI8E8. 24:T 

the enunciation, taken in its limited sense. There will be in- 
teresting variations, however, if we regard the given circles, 
not merely as inscribed., but as those which touch all the sides 
of each of the right-angled triangles, either internally or exter- 
nally. These variations will be obtained by giving the squares 
on the radii every possible variety of position in the four right 
angles formed by the intersection of AC, DB ; and the solution 
will obtain complete generality by taking into consideration 
both the points in which BD cuts the circle of which EF is a 
chord. 

Prop. XXX. — Theo^, — The area of a triangle ABC ts 
equal to half the continued product of two of its sides, AB, 
BC, a?id the sine of their contained angle B, to the radius I. 

Draw the perpendicular AD. Then (Trig. 1, cor.) AD = 
AB X sin B. Multiply by BC, and take 
Liiif the product ; and (I. 23, cor. 6) we 
have the area equal to ^ AB x BC X sin B. 

Cor. Hence (I. 15, cor. 1) the area of a 
parallelogram is equal to the continual 
product of two contiguous sides, and the 
sine of the contained angle. 

jScho. From this proposition, and from the third and sixth 

corollaries to the twenty-fifth proposition of the third book, we 

can derive neat algebraic expressions for the radii of the four 

circles, each touching the three sides of a triangle. Thus, by 

dividing the expression for the area by s, we find, according to 

the third corollary, that the radius of the inscribed circle is 

/(s — a) (s—b) (s—c) T ... , T 'T 

equal to ,{/ —^ —-^ ~ . In like manner, by dividing 

the expression for the area successively by s — a, s—b, s — c, we 
find, according to the sixth corollary, that the radii of the cir- 
cles touching a, b, c, externally, are respectively, 

/s{s-b){s-c)/s{s-a){s_-c)^ and ^ ^ (s-a) (s-b) ^ 
r s—a y s—b r s — o 

By taking the continual product of these four expressions, 
and contracting the result, we get s {s — a) (s — b) (s — c), which. 
is equal to the square of the area ; and hence, by expressing 




248 EXERCISES. 

this inwards, we have the following remarkable theorem : The 
continual product of the radii of the four circles^ each of which 
touches the three sides of a triangle, or their prolongations, is 
equal to the second power of the area, which is proven by the 
next proposition. 

Prop. XXXI. — Theor. — Let a, b, c he the sides of a trian- 
gle, and s half their sum ; the area is equal to the square root 
of the continual product ofs, s— a, s — b, and s — c. 

It was proved in the second corollary to the ninth proposi- 
tion, Plane Trigonometry, that the sine of twice any angle is 
twice the product of the sine and cosine of the angle. Hence, 
by multiplying together the values of sin-^A and cos^A, 
given in the corollaries to the sixth and seventh propositions, 
Plane Trigonometry, and doubling the result, we get sin A= 

2 \/\s (s-a) (s-b) (s-c)] ^^ , , 

-j^ -. Now, by the precedmg proposi- 
tion, the area of a triangle is found by multiplying the sine of 
one of its angles by the sides containing it, and taking half of 
the product; multiplying, therefore, the value now found for 
sin A, by be, and taking half the product, we find the area to 
be V[s {s — a) {s — b) (s — c)]. This proposition is much used in 
surveying coasts and harbors. 

Prop. XXXTI. — Prob, — A semicircle ACB being given, and 
other semicircles being described as in the diagram, / it is re- 
quired to find the sum of the areas of all those inscribed semi- 
circles. 

Cii'cles (V. 14), and consequently semicircles, are as the 

squares of their diameters or of 
their radii. Now the square of 
GD is half the square of DF or 
CF, and therefore the semicircle 
DFE is half of ACB. For the 
same reason HGK is half of DFE; 
and universally, each semicircle is 
half of the one in which it is in- 
scribed. Hence the entire amount will be the sum of the 
infinite series iACB-f-J ACB+i ACB+^ ACB+etc. ; and 




EXERCISES. 



249 



therefore (IV. 19) i ACB-i ACB : ^ACB :: iACB : ACB, 
llie required sum ; and it thus appears that the sum of all the 
inscribed semicircles is equivalent to the given semicircle. 

Pkop. XXXIII— Theor.— J;z any triangle, the center of the 
circumscribed circle, the point in which the three perpendiculars 
intersect one another, and the point of intersection of the 
straight lines drawn from the angles to bisect the opposite sides, 
lie all in the same straight line. 

Let ABC be a triangle, and let the two perpendiculars AD, 
CE intersect in F ; bisect AB, BC in H, G, and draw AG, CH 
intersecting in K ; draw also GI, HI perpendicular to BC, BA, 
and intersecting in I. Then (Ex. 
7) F is the intersection of the three 
perpendiculars, K (III. 1, cor. 2) 
the intersection of the three lines 
drawn from the angles to bisect 
the opposite sides, and (III. 25, 
cor. 2) I is the center of the cir- 
cumscribed circle. Join FK, KI ; 
FKI is a straight line. 

Join HG; it is (V. 2 and 3) 
parallel to AC, and is half of it. Also the triangles ACF, 
GHI are (I. 16, cor. 3) equiangular, and therefore GI is half of 
AF. So likewise (III. 1, scho.) is KG of KA. Hence the two 
triangles AKF, GKI have the alternate angles KAF, KGI 
equal, and the sides about them proportional ; therefoi-e (V. 6) 
the angles AKF, GKI are equal, and (1. 10, cor.) since AKG is 
a straight line, FKI is also a straight line. 

Scho. It is plain (V. 3) that FK is double of KI. We have 
also seen that AF is double of GI. Hence it appears, that the 
distance between any of the angles and the point of intersec- 
tion of the three perpendiculars is double of the perpendicular 
drawn from the center of the circumscribed circle to the side 
opposite to that angle. 

Prop. XXXIV. — Theor. — Straight lines drawn from the 
angles of a triangle to the points in which the opposite sides 
touch the inscribed circle, all pass through the same point. 




250 



EXERCISES. 




Let ABC be a triangle, and D, E the points in which the 
sides AB, AC touch the inscribed circle ; draw BFE, CFD ; 
draw also AFG cutting BC in G ; G is the point in which BC 
touches the inscribed circle. 

If possible, let another point K be the point of contact, and 

draw DH,DI parallel to BC, 
CA. Then in the similar tri- 
angles FDI, FCE, FD : DI : : 
FC : CE, or CK ; and in the 
similar triangles FDH, FGC, 
DH : DF : : GC : FC ; from 
which and from the preced- 
ing analogy we get, ex cequo^ 
DH : DI : : CG : CK. Again, 
BD : DI : : BA : AE or AD 
: : BG : DH. Hence, altern- 
ately, and by inversion, BG : BD : : DH : DI : whence (IV. 7) 
BG : BD or BK : : CG : CK, or alternately, BG : CG : : BK : 
CK; and by composition, BC : CG :: BC : CK ; and there- 
fore CG, CK are equal ; that is, G and K coincide, and AFG 
passes through the point in which BC touches the circle. 

Prop. XXXV. — Theor. — 171 a triangle, the sum of the per- 
pendiculars draion from the center of the circumscribed circle 
to the three sides is equal to the sum of the radii of the in- 
scribed and circumscribed circles. 

Let ABC be a triangle, having its sides bisected in D, E, F, 
by perpendiculars meeting in G, the center of the circum- 
scribed circle ; the sura of GD, 
GE, GF is equal to the sum of 
the radii of the inscribed and 
circumscribed circles. 

Join GA, GB, GC, and DE, 
EF, FD. Then, putting a, b, c 
to denote the sides opposite to 
the angles A, B, C, we have 
(V. 2 and 3) FE^^a, FD=i6, 
and DE=ic; and (HI. 11) 
Bince AEG, AFG are right angles, a circle may be described 




EXERCISER. 



251 



about the quadrilateral AEGF. For a like reason circles may- 
be described about BDGF and CDGE. Hence (Ex. 4) FE. 
AG=AF.GE + AE.FG; or, by doubling, a.AG=c.GEi-b.FG. 
In the same manner, it would be shown, since AG, BG, CG are 
equal, that ^>.AG=c.GD + a.FG, and c.AG = a.GE + ^..GD. 
Hence, by addition, (a-|-5+c)AG = (a+c)GE+(ff + 5)GF + 
(54-c)GD, Now 5.GE is evidently equal to twice the triangle 
AGO, c.GF equal to twice AGE, and a.GD equal to twice 
BGC ; also, denoting the radius of the inscribed circle by r, 
we have (IH. 25, cor. 3), r{a+b-\-c) equal to twice the area of 
the triangle ABC, and consequently r{a-\-b-i-c)=b.GE + c.GF 
+ a.GD. Hence, by addition, {a+b-\-c)AG-\-r{aA-b-t-c) = 
(a+5+c)GE+(a+5+c)GF+(a+6+c)GD; and consequently 
AG4-r=GE+GF+GD. 

Cor. Since, by the scholium to proposition thirty-third of 
these Exercises, the parts of the three perpendiculars of the 
trianffle, between their common intersection and the three an- 
gles, are respectively double of GD, GE, GF, the sum of those 
parts of the perpendiculars is equal to the sum of the diameters 
of the inscribed and circumscribed circles. 



Prop. XXXYI. — Theor, — If on the three sides of any tri- 
angle equilateral triangles be described^ either all externally^ or 
all internally, straight lines joining the centers of the circles 
inscribed in those three triangles form an equilateral triangle. 

On the three sides of any triangle ABC, let'^the equilateral 
triangles ABD, BCF, CAE be de- 
scribed externally, and find G, H, 
K, the centers of the circles de- 
scribed in those triangles ; draw 
GH, HK, KG ; GHK is an equilat- 
eral triangle. 

Join GA, GB, HB, HC, HF, KC, 
AF. Then the angle FBC is two 
thirds of a right angle, and the an- 
gles GAB, GBA, FBH, BFH each 
one third. The triangles FBH, 
ABG are therefore similar, and (V. 
3) FB : BH : : BA : BG ; whence, alternately, FB : BA : : 




252 EXERCISES. 

HB : BG; that is, in the triangles FBA, HBG the sides about 
the angles FBA, HBG are proportional; and these angles aj-e 
equal, each of thorn being equal to the sum of the angle ABC 
and two thirds of a right angle. Hence (V. 6) these triangles 
are equiangular ; and therefore (V. 3) FB or BC : FA : : BH 
: HG; and it would be shown in the same manner, by means 
of the triangles ACF, KCH, that FC or BC : FA : : CH : 
HK ; therefore (IV. 1) BH : HG : : CH : HK. But BH is 
equal to CH, and therefore (IV. 6) HG is equal to HK ; and it 
would be demonstrated in a similar manner, that HG, HK are 
each equal to GK. 

If the equilateral triangles were described on the other sides 
of the lines AB, BC, CA, the angles ABF, GBH would be the 
difference between ABC and two thirds of a right angle ; but 
the rest of the proof is the same. 

Sc/io. If ABC exceed aright angle and a third, the sum of it 
and two thirds of a right angle is greater than two right angles. 
In that case, the angles ABF, GBH, understood in the ordinary- 
sense, are each the difference between that sum and four right 
angles. If ABC be a right angle and a third, the sum becomes 
two right angles, and FB, BA are in the same straight line, as 
are also HB, BG. In this case it is proved as before, that FB : 
BA : : HB : BG ; and then (IV. 11) FB or BC : FA :: HB : 
HG. The rest of the proof would proceed as above. 

It may be remarked that if an equilateral triangle be de- 
Bcribed on a stj'aight line, and if on the two parts into which it 
is divided at any point, other equilateral triangles be described, 
lying in the opposite direction, the lines joining the centers of 
the three equilateral triangles will also form an equilateral tri- 
angle. The connection of this and the proposition will be per- 
ceived by supposing two angles of the triangle continually to 
diminish, till they vanish, as the triangle may thus be conceived 
to become a straight line. 



TO BE PROVEN. 

1. The least straight line that can be drawn to another 
etraight line from a point without it, is the perpendicular to it ; 
of others, that which is nearer to the perpendicular is less than 



EXERCISES. 253 

one more remote ; and only two equal straight hnes can be 
drawn, one on each side of the perpendicular. 

2. Of the triangles formed by drawing straight Hnes from a 
point within a parallelogram to the several angles, each pair 
that have opposite sides of the parallelogram as bases, are half 
of it. 

3. If, in proceeding round an equilateral triangle, a square, 
or any regular polygon, in the same direction, points be taken 
on the sides, or the sides produced, at equal distances from the 
several angles, a similar rectilineal figure will be formed by 
joining each point of section with those on each side of it. 

4. If the three sides of one triangle be perpendicular to the 
three sides of another, each to each, the triangles are equi- 
angular. 

5. A trapezoid, that is, a trapezium having two of its sides 
parallel, is equivalent to a triangle which has its base equal to 
the sum of the parallel sides, and its altitude equal to their per- 
pendicular distance. 

6. Given the segments into which the line bisecting the ver- 
tical angle divides the base, and the difference of the angles at 
the base ; to construct the triangle, and compute the sides. 

7. Within or without a triangle, to draw a straight line par- 
allel to the base, such that it may be equivalent to the parts 
of the other sides, or of their continuations, between it and the 
base. 

8. Given the perpendicular of a triangle, the diffei*ence of the 
segments into which it divides the base,and the difference of 
the angles at the base ; to construct the triangle. 

9. In the figure for the first corollary to the twenty-fourth 
proposition of the first book, prove that CD is perpendicular to 
AH, or a line from C to F is perpendicular to CD. 

10. The angle made by two chords of a circle, or by their 
continuations, is equal to an angle at the circumference stand- 
ing on an arc equivalent to the sum of the arcs intercepted be- 
tween the chords, if the point of intersection be within the 
circle, or to their difference, if it be without ; (2) the angle 
made by a tangent and a line cutting the circle, is equal to aa 
angle at the circumference on an arc equivalent to the differ- 
ence of the arcs intercepted between the point of contact and 



254 EXEKCIPHa, 

the other line; and (3) the angle made by two tangents is 
equal to an angle at the circumference standing on an arc 
equivalent to the difference of those into which the circumfer- 
ence is divided at the points of contact. 

Cor. If a tangent be parallel to a chord, the arcs between 
the point of contact and the extremities of the chord are equal. 

11. Given the sum of the perimeter and diagonal of a square ; 
to construct it. 

12. On a given straight line describe a square, and on the 
side opposite to the given line describe equilateral triangles 
lying in opposite directions ; circles described throiigh the ex- 
tremities of the given line, and through the vertices of these 
triangles, are equal. 

13. To inscribe an equilateral ti'iangle in a given square. 

14. Given the angles and the two opposite sides of a trape- 
zium ; to construct it. 

15. In a given circle to place two chords of given lengths, 
and inclined at a given angle. 

16. In the figure for the twenty-fifth proposition of the third 
book, if AM, BM be joined, the angle AMB is half of ACB. 

17. If, in proceeding in the same direction round any trian- 
gle, as in 3, points be taken at distances from the several 
angles, each equal to a third of the side, the triangle formed by 
joining the points of section is one third of the entire triangle. 

18. To describe a square having two of its angular points on 
the circumference of a given circle, and the other two on two 
given straight lines drawn through the center. Show that 
there may be eight such squares. 

19. Given the vertical angle of a triangle, and the segments 
into which the base is divided at the point of contact of the in- 
scribed circle ; to describe the triangle, and compute the sides. 

20. If any three angles of an equilateral pentagon be equal, 
all its angles are equal. 

21. Given two sides of a triangle, and the difference of the 
sefrments into which the third side is divided by the perpen- 
dicular from the opposite angle ; to construct the triangle. 

22. Given the vertical angle of a triangle, the line bisecting 
it, and the perpendicular; to construct the triangle. 

23. From a given center to describe a circle, from which a 



EXERCISES. 265 

straight line, given in position, will cut off a segment contain- 
ing an angle equal to a given angle. 

24. In a given triangle to inscribe a semicircle having its 
center in one of the sides. 

25. Through three given points to draw three parallels, 
two of which may be equally distant from the one between 
them. 

26. Given an angle of a triangle, and the radii of the circles 
touching the sides of the triangles into which the straight line 
bisecting the given angle divides the triangle ; to construct it. 

27. In a rhombus to inscribe a square. 

28. Given the lengths of the two parallel sides of a trapezoid, 
and the lengths of the other sides ; to construct it. 

29. Given one of the angles at the base, and the segments 
into which the base is divided at the point of contact of the in- 
scribed circle ; to describe the triangle. 

30. To draw a tangent to a given circle, such that the part 
of it intercepted between the continuations of two given diam- 
eters may be equal to a given straight line. 

31. To draw a tangent to a given circle, such that the part 
of it between two given tangents to the cii'cle may be equal to 
a given straight line. 

32. Given the vertical angle of a triangle, the difference of 
the sides, and the difference of the segments into which the 
line bisecting the vertical angle divides the base j to construct 
the triangle. 

33. Given any three of the circles mentioned in the fifth 
■corollary to the twenty-fifth proposition of the third book ; to 
describe the triangle. 

34. A straight line and a point being given in position, it is 
required to draw through the point two straight lines inclined 
at a given angle, and inclosing with the given line a space of 
given magnitude. 

35. From two given straight lines to cut off equal parts, each 
of which will be a mean proportional between the remainders. 

36. A square is to a regular octagon described on one of its 
sides, as 1 to 2 (1-f V2). 

37. In a given triangle to inscribe a parallelogram of a given 
area. 



256 EXEECI8E8. 

38. In a given circle to inscribe a parallelogram of a given 
area. 

39. Through a given point between the lines forming a 
given angle, to draw a straight line cutting off the least possi- 
ble triangle. 

40. To divide a given straight line into two parts, such that 
the square of one of them may be double of the square of the 
other, or may be in any given ratio to it. 

41. To produce a given straight line, so that the square of 
the whole line thus produced may be double of the square of 
the part added, or in any given ratio to it. 

42. Given the area of a right-angled triangle, and the sura of 
the legs ; to construct it. 

43. Given the area and the difference of the legs of a right- 
angled triangle ; to construct it. 

44. Given one leg of a right-angled triangle, and the remote 
segment of the hypothenuse, made by a perpendicular from the 
right angle ; to construct the triangle. 

45. Given the base of a triangle, the vertical angle, and the 
side of the inscribed square standing on the base ; to describe 
the triangle. 

46. Given the base of a triangle, and the radii of the 
inscribed and circumscribed circles ; to construct the tri- 
angle. 

47. To draw a chord in a circle which will be equal to one 
of the segments of the diameter that bisects it. 

48. In a given circle to draw a chord which will be equal to 
the difference of the parts into which it divides the diameter 
that bisects it. 

49. If two sides of a regular octagon, between which two 
others lie, be produced to meet, each of the produced parts is 
equivalent to a side of the octagon, together with the diagonal 
of a pquare, described on the side. 

50. The perimeter of a triangle is to the base as the perpen- 
dicular to the radius of the inscribed circle. 

51. From a given point without a given circle to draw a 
straight line cutting the circle, so that the external and inter- 
nal parts may be in a given ratio. 

52. Each of the complements of the parallelograms, about 



EXERCISES. 257 

the diagonal of a parallelogram, is a mean proportional between 
those parallelograms. 

53. Given the ratio of two straight lines, and the difference 
of their squares ; to find them. 

54. The square of the perimeter of a right-angled triangle is 
equivalent to twice the rectangle under the sura of the hy- 
pothenuse and one leg, and the sum of the hypothenuse and the 
other. 

55. The quadrilateral formed by straight lines bisecting each 
pair of adjacent sides of a quadrilateral is a parallelogram, 
which is half of the quadrilateral ; and straight lines, joining 
the points in which the sides of that parallelogram are cut by 
the diagonals of the primitive figure, form a quadrilateral simi- 
lar to that figure and equivalent to a fourth of it. 

66. From three given points as centers, and not in the same 
straight line, to describe three circles each touching the other 
two. Show that this admits of four solutions. 

57. To add a parallelogram to a rhombus, such that the 
M'hole figure may be a parallelogram similar to the one added. 

58. A straight line being given in position, and a circle in 
magnitude and position, it is required to describe two equal 
circles touching one another, and each touching the straight 
line and the circle. 

59. The squares of the diagonals of a quadrilateral are to- 
gether double of the squares of the straight lines joining the 
points of bisection of the opposite sides. 

60. In a given rhombus to inscribe a rectangle having its 
sides in a given ratio. 

61. If, throu2rh the vertex and the extremities of the base of 
a triangle, two circles be described intersecting one another in 
the base or its continuation, their diameters are proportional to 
the sides of the triangle. 

62. To draw a straight line cutting two given concentric 
circles, so that the parts of it within them may be in a given 
ratio. 

63. From a given point, within a given circle, or without it, 
to draw two straight lines to the circumference, perpendicular 
to one another, and in a given ratio. "When will this be im- 
possible ? 

It 



258 EXERCISES. 

64. If a straight line be divided in extreme and mean ratio, 
the squares of the whole and the less part are together equiva- 
lent to thi-ee times the square of the greater, 

65. If a straight line be cut in extreme and mean ratio, and 
be also bisected, the square of the intermediate part, and three 
times the square of half the line are equivalent to twice the 
square of the greater part. 

66. If the hypothenuse of a right-angled triangle be given, 
the side of the greatest inscribed square, standing on the hy- 
pothenuse, is one third of the hypothenuse. 

67. To divide a given semicircle into two parts by a perpen- 
dicular to the diameter, so that the radii of the circles inscribed 
in them may be in a given ratio. 

68. To draw a straight line parallel to the base of a triangle, 
making a segment of one side equal to the remote segment of 
the other. 

69. In the figure for the tenth proposition of the second book, 
the square of the diameter of the circle, passing through the 
points F, H, D, is six times the square of the straight line join- 
ing P^D. 

VO. Given the base, the area, and the sum of the squares of 
the sides of a triangle ; to construct it. 

71. If, from the extremities of the hypothenuse of a right- 
angled triangle as centers, arcs be described passing through 
the right angle, the hypothenuse is divided into three segments, 
such that the square of the middle one is equivalent to twice 
the rectangle of the others. 

72. On a given hypothenuse to describe a right-angled tri- 
angle, such that the difference between one leg and the adja- 
cent segment of the hypothenuse made by a perpendicular from 
the right angle, may be a maximum, 

73. On a given hypothenuse to construct a right-angled tri- 
angle, such that one segment of the hypothenuse made by the 
perpendicular from the right angle, may be equivalent to the 
Bum of the perpendicular and the other segment. 

74. The square of DH (see figure for III. 24) is equivalent to 
the rectangle CF.BG ; and if the circles touch one another ex- 
ternally, DH is a mean proportional between their diameters. 
Also the square of DH is equivalent to the rectangle CG.BF. 



EXERCISES. 



259 



75. On a given straight line to describe an isosceles triangle, 
having the vertical angle treble of each of the angles at the 
base. 

V6. If in the diameter of a circle and its continuation two 
points be taken on the opposite sides of the center, such that 
the rectangle under their distances from the center may be 
equivalent to the square of the radius, any circle whatever de- 
scribed through these points bisects the circumference of the 
otlier circle. 

'77. To find a point from which, if straight lines be drawn to 
three given points, they will be proportional to three given 
straight lines. 

78. Given the segments into which the base of a tnangle is 
divided by two straight lines trisecting the vertical angle ; to 
construct it. 

79. If one diagonal of a quadrilateral inscribed in a circle be 
bisected by the other, the square of the latter is equivalent to 
half the sum of the squares of the four sides. 

80. To divide a straight line into- two parts, such that the 
squares of the whole and one of the parts may be double of the 
square of the other part. 

81. Given the segments into which the base of a triangle is 
divided by the straight line bisecting the vertical angle, to 
construct the triangle so that its angle adjacent to the greater 
segment may be either of a given magnitude, or a maximum, 
and in each case to compute the remaining sides and angles. 

82. To draw a straight line bisecting a given parallelogram, 
so that if it be produced to meet the sides produced, the ex- 
ternal triangles will have a given ratio to the parallelogram. 

83. Through a given point to draw a straight line, which, 
if continued, would pass through the point of intersection of 
two given inclined sti'aight lines, without producing those lines 
to meet. 

84. If, from any point in the circumference of the circle de- 
scribed about an equilateral triangle, chords be drawn to its 
three angles, the sum of their squares is equivalent to six times 
the square of the radius of the same circle, or to twice the 
square of a side of the triangle. 

85. Given the difference of the angles at the base of a trian- 



260 EXERCISES. 

gle, the difference of the segments into which the hase is 
divided by the perpendicular, and the ratio of the sides ; to 
construct the triangle. 

86. Given the base and vertical angle of a triangle, to con- 
struct it so that the line bisecting the vertical angle may be a 
mean proportional between the segments into which it divides 
the base. 

87. Given two sides of a triangle, and the ratio of the base 
and the line bisecting tlie vertical angle j to construct the tri- 
angle. 

88. If the vertical angle of a triangle be double of one of the 
angles at the base, the rectangle under the sides is equivalent 
to the rectangle under the base, and the line bisecting the ver- 
tical angle. 

89. If a straight line be divided into parts, which, taken in 
succession, are continual proportionals, and if circles be de- 
scribed on the several parts as diameters, a straight line which 
touches two of the circles on the same side of the straight line 
joining their centers, will touch all the others. 

90. Given the segments into which the base is divided by 
the straight line bisecting the vertical angle, and the angles 
which that straight line makes with the base ; to construct the 
triangle. 

91. To divide a given circle into two segments, such that 
the squares inscribed in them may be in a given ratio. 

92. Through a given point in the base of a given isosceles 
triangle, or its continuation, to draw a straight line such that 
the lines intercepted on the equal sides, or their continuations 
between that line and the extremities of the base, may have 
one of the equal sides as a mean proportional between them. 

93. Through a point in tlie circumference of a given circle, to 
draw two chords, such that their rectangle may be equivalent 
to a given space, and the chord joining their other extremities 
equal to a given straight line. 

94. In any triangle the radius of the circvmiscribed circle is 
to the radius of the circle which is the locus of the vertex, 
when the base and the ratio of the sides are given, as the dif- 
ference of the squares of those sides is to four times the area. 

95. The difference of the sides of a triangle is a mean pro- 



KXERCI8K8. 261 

portional between the clifFerence of the segments into which 
the base is divided by the perpendicular, and the difference of 
those into which it is divided by the line bisecting the vertical 
angle. 

96. Let the angles of a pai*allelogram which has unequal 
sides be bisected by straight lines cutting the diagonals, and 
let the points of intersection be joined ; the figure thus formed 
is a parallelogram, which has to the proposed parallelogram the 
duplicate ratio of that which the difference of the unequal sides 
of the latter has to their sum. 

97. A circle and a point being given, it is required to de- 
scribe a triangle similar to a given one, having its vertex at the 
given point, and its base a chord of the given circle. 

98. Given the three points in which the sides of a triangle 
are cut by the perpendiculars from the opposite angles ; to con- 
struct the triangle. 

99. Given the ansjles of a triansrle, and the leno;ths of three 
straight lines drawn from the angular points to meet in an- 
other point ; to construct the triangle. 

100. Given the base of a triangle, and the ratio of its sides; 
to construct it, so that the distance of its vertex from a given 
point may be a maximum or minimum. 

101. To divide {^circle into two segments, such that the sum 
of the squares inscribed in them may be equivalent to a given 
space. 

102. Through a given point, with a given radius, to describe 
a circle bisecting the circumference of a given circle. 

103. With a given radius to describe a cn-cle bisecting the 
circumferences of two given circles. 

104. In a right-angled triangle, the rectangle under the 
radius of the inscribed circle, and the radius of the circle 
touching the hypothenuse and the legs produced, is equivalent 
to the area. So, likewise, is the rectangle under the circles 
touching the legs externally, and the continuations of the other 
sides. 

105. If three straight lines be continual proportionals, the 
6um of the extremes, their difference, and double the mean will 
be the hypothenuse and legs of a right-angled triangle. 

106. From two given centers, to describe circles having their 



263 EXERCISES. 

radii in a given ratio, and the part of their common tangent, 
between the points of contact, equal to a given straiglit line. 

107. A straiglit line and two points equally distant from it, 
on the same side, being given in position, it is required to draw 
through the points two straight lines forming with the given 
line the least isosceles triangle possible, on the side on which 
the points ai"e. 

108. To describe a circle touching a diameter of a given cir- 
cle in a given point, and having its circumference bisected by 
that of the given one. 

109. If an angle of a triangle be 60^, the square of the oppo- 
site side is less than the squares of the other two by their rect- 
angle ; but if an angle be 1 20°, the square of the opposite side 
is greater than the squares of the others by their rectangle. 

110. In the figure on page 251, prove that the three straight 
lines joining AF, BE, CD are all equal. 

111. The chord of 120° is equal to the tangent of 60°. 

112. The sines of the parts into which the vertical angle of a 
triangle is divided by the straight line bisecting the base, are 
reciprocally proportional to the adjacent sides. Show from 
this how a given angle may be divided iuto two parts, having 
their sines in a given ratio. 

113. The diameter of the circle described about any triangle 
is equivalent to the product of any side and the cosecant of the 
oj^posite angle. 

114. In any triangle ABC, the radius of the inscribed circle 

. , sin A B sin A C . sin i A sin ^^ C 

IS equivalent to a. t—: > to o. r-f, — > or to 

^ cos Y A cos ^- 1> 

sin 4- A sin 4 B £ n . -i i * /• 

c. r-^ — ; or, finally, to the cube root oi 

cos -J O 

abc sin ^ A sin ^ B sin -J C tan ^ A tan | B tan ^ C. 

115. Given the sum of the tangents, and the ratio of the se- 
cants, of two angles to a given radius ; to determine the angles 
geometrically and by computation. 

116. Find an angle, such that its tangent is to the tangent 
of its double, in a given ratio ; suppose that of 2 to 5. 

117. If a spherical triangle be right-angled at C, and any two 
of the other parts be made s -.ccessively 108° 42' and 87° 33' 
19", what will be the values o the remaining parts? 



ADDENDUM. 



263 



118. Let any three parts of an oblique-angled spherical trian- 
gle be successively 87° 45^ 96° 57' 48'^, and 106° 53' 13''; what 
will be the values of the other parts? 

119. When the three sides of a spherical triangle be respect- 
ively, 34° 39' 44", 78° 27' 49", and 134° 15' 23", what will be 
the surface of the triangle when the radius of the sphere is 16 ? 
"What will be the base of the triangular pyramid subtended by 
those sides, and what will be the surface and. solidity of the 
spherical pyramid with its apex at the center of the sphere ? 

120. If the foregoing values be the magnitudes of the angles 
of a spherical triangle when the radius of the sphere is 16, what 
will be the base of the triangular pyramid subtended by the 
sides of the spherical triangle ? What will be the surface of 
the spherical triangle, and the surface and solidity of the 
spherical j^yramid with its apex at the center of the sphere ? 

121. If the earth be regarded a sphere with 7973.8798-f miles 
for its diameter, and a spherical pentagon be measured on its 
surface having 341.78 miles, 309.25 miles, 278.64 miles, 173.97 
miles, and 97 miles for its sides ; and the angles contained by 
those sides be respectively 74° 34' 19", 107° 09' 51", 41° 0' 
n'", 85° 17' 09", and 76° 41' 35", what will be the surface of 
the spherical pentagon ? and what will be the solidity of the 
pyramid having the spherical pentagon for its base and its apex 
at the center of the earth ? 



ADDENDUM, 

Illustrating the Third Proof for the Second Corollary 
TO THE Seventeenth Proposition of the tixTii Book, 
Elements of Euclid and Legendre. 

Circles are to one another as the squares described upoft 
their diameters (V. 14), consequently squares are to one an- 
other as the circles described (V. 14, cor. 2) upon their sides; 
that is, there is a ratio of equality between the circle and 
squares which have the same straight line for their respective 
diameter, side, and diagonal • therefore the circle has the same 



264 ADDENDUM. 

arithmetical proportion to the inscribed square having the 
diameter for its diagonal, as the circumscribed square having 
the same diameter for its side has to the circle. If 10 be the 
diameter of the circle, and x ^6 the area of the circle, 100 will 
be the area of the circumscribed square, and 50 will be the area 
of the inscribed square — and we have the arithmetical propor- 
tion, 

100, X, 50; 

and the geometrical proportion, 

100 : Y : : 50 : -5-. 

From the first proportion, we derive 

100— x=x— 50. 
From the second proportion, we have 



100-x = 2 (50- I). 



.-. x-50 = 2 (50-|x) = 100-x; 
or, 2p(;=:150, hence p^ = 75. 
The arithmetical proportion gives, 

2x=150 or pf=75 
Substituting this value for -/^ in the above proportions, \,lpl 
get 

100 : 75 : : 50 : 371; 
and 100—75 = 75—50; 
and 100— 75 = 2(50— 371); 
and 100, 75, 50; 

/. s ^ 150 ^ 
or, 2 (75) = 150, or 75= =75. 

Li 

Hence, the circle is the arithmetical mean between the 
squares, circumscribed and inscribed about it ; or three fourths 
of the circumscribed square; or three times square of the ra- 
dius. (See Exercises, def 7, scho. 2). 

Thus we have the area of the circle expressed by a finite 
quantity instead of the irrational quantity (V. 25, scho. 1), 
giving the approximate area only of the circle. 



THE- END. 



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