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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