The Gift of
WILLIAM H. BUTTS, Ph.D.
A.B. 1878 A.M. 1879
Teacher of Mathematics
1898 to 1922
Assistant Dean, College of Engineering
1908 to 1922
Professor Emeritus
1922
3
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ARTES
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SCIENTIA
LIBRARY
VERITAS
OF THE
UNIVERSITY OF MICHIGAN
ALLURIOUS UNUM
TCEHOR
EQUÆRIS PENINSULAMʼAM
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W

EUCLID'S
ELEMENTS OF GEOMETRY,
CHIEFLY FROM THE TEXT OF DR SIMSON,
WITH EXPLANATORY NOTES;
TOGETHER WITH A SELECTION OF GEOMETRICAL EXERCISES
FROM THE SENATE-HOUSE AND COLLEGE
EXAMINATION PAPERS;
TO WHICH IS PREFIXED AN INTRODUCTION, CONTAINING
A BRIEF OUTLINE OF THE HISTORY OF
GEOMETRY.
DESIGNED FOR THE USE OF THE HIGHER FORMS IN PUBLIC SCHOOLS
AND STUDENTS IN THE UNIVERSITIES.
BY
ROBERT POTTS, M.A.
TRINITY COLLEGE.

T
CAMBRIDGE:
PRINTED AT THE UNIVERSITY PRESS.
LONDON:
JOHN W. PARKER, 445 WEST STRAND.
M.DCCC.XLV.

អូ
រ
L
W.H. Butts
Wheldon
1-14-37
33358
PREFACE.
THIS new edition of Euclid's Elements of Geometry will be
found to differ considerably from those at present in general use
in Academical Education. The text is taken from Dr Simson's
approved edition, with occasional alterations; but so arranged as to
exhibit to the eye of the student the successive steps of the de-
monstrations, and to facilitate his apprehension of the reasoning.
No abbreviations or symbols of any kind are employed in the
text. The ancient Geometry had no symbols, nor any notation
beyond ordinary language and the specific terms of the science.
We may question the propriety of allowing a learner, at the com-
mencement of his Geometrical studies, to exhibit Geometrical
demonstrations in Algebraical symbols. Surely it is not too much
to apprehend that such a practice may occasion serious confusion
of thought. It may be remarked that the practice of exhibiting
the demonstrations of Elementary Geometry in an Algebraical
form, is now generally discouraged in this University. To each
book are appended explanatory notes, in which especial care has
been taken to guard the student against the common mistake of
confounding ideas of number with those of magnitude. The work
contains a selection of problems and theorems from the Senate-
house and College Examination Papers, for the last forty-five years.
These are arranged as Geometrical exercises to the several books
of the Elements, and to a few only in each book the solutions
are given. An Introduction is prefixed, giving a brief outline of
the history and progress of Geometry.
The analysis of language, together with the sciences of number
and magnitude, have been long employed as the chief elements of
intellectual education. At a very early period, the study of
Geometry was regarded as a very important mental discipline, as
may be shewn from the seventh book of the Republic of Plato.
To his testimony may be added that of the celebrated Pascal,
(Euvres, Tom. I. p. 66,) which Mr Hallam has quoted in his
History of the Literature of the Middle Ages. "Geometry,"
Pascal observes, "is almost the only subject as to which we find
truths wherein all men agree; and one cause of this is, that
geometers alone regard the true laws of demonstration." These
iv
PREFACE.
are enumerated by him as eight in number. 1. To define nothing
which cannot be expressed in clearer terms than those in which
it is already expressed. 2. To leave no obscure or equivocal
terms undefined. 3. To employ in the definition no terms not
already known. 4. To omit nothing in the principles from which
we argue, unless we are sure it is granted. 5. To lay down no
axiom which is not perfectly evident. 6. To demonstrate nothing
which is as clear already as we can make it. 7. To prove every
thing in the least doubtful, by means of self-evident axioms, or
of propositions already demonstrated. 8. To substitute mentally
the definition instead of the thing defined. Of these rules he
says, "the first, fourth, and sixth are not absolutely necessary
to avoid error, but the other five are indispensable; and though
they may be found in books of logic, none but the geometers
have paid any regard to them.”
If we consider the nature of Geometrical and Algebraical
reasoning, it will be evident that there is a marked distinction
between them. To comprehend the one, the whole process must
be kept in view from the commencement to the conclusion; while
in Algebraical reasonings, on the contrary, the mind loses the
distinct perception of the particular Geometrical magnitudes com-
pared; the attention is altogether withdrawn from the things
signified, and confined to the symbols, with the performance of
certain mechanical operations, according to rules of which the
rationale may or may not be comprehended by the student. It
must be obvious that greater fixedness of attention is required in
the former of these cases, and that habits of close and patient
observation, of careful and accurate discrimination will be formed
by it, and the purposes of mental discipline more fully answered.
In these remarks it is by no means intended to undervalue the
methods of reasoning by means of symbolical language, which are
no less important than Geometry. It appears, however, highly
desirable that the provinces of Geometrical and Algebraical rea-
soning were more definitely settled than they are at present, at
least in those branches of science which are employed as a means
of mental discipline. The boundaries of Science have been ex-
tended by means of the higher analysis; but it must not be
forgotten that this has been effected by men well skilled in
Geometry and fully able to give a geometrical interpretation of
the results of their operations; and though it may be admitted
that the higher analysis is the more powerful instrument for that
purpose, it may still be questioned whether it be well suited to
PREFACE.
V
form the chief discipline of ordinary intellects without a previous
knowledge of the principles of Geometry, and some skill in their
application. Though the method of Geometrical analysis is very
greatly inferior in power to the Algebraical, yet as supplementary
to the Elements of Euclid, it is of great importance. It may
be added, that a sound knowledge of the ancient geometry is the
best introduction to the pursuits of the higher analysis and its
extensive applications. On this subject the judgment of Sir Isaac
Newton has been recorded by Dr Pemberton, in the preface to his
view of Sir Isaac Newton's Discoveries.
He says:
"Newton
censured the handling of geometrical subjects by algebraical calcu-
lations. He used to commend the laudable attempt of Hugo d'Ome-
rique (in his ‘Analysis Geometrica Nova et Vera,') to restore the
ancient analysis, and very much esteemed the tract of 'Apollonius
De Sectione Rationis,' for giving us a clearer notion of that analysis
than we had before. The taste and mode of geometrical demon-
stration of the ancients he professed to admire, and even censured
himself for not having more closely followed them than he did:
and spoke with regret of his mistake, at the beginning of his mathe-
matical studies, in applying himself to the works of Descartes and
other algebraical writers, before he had considered the Elements of
Euclid with the attention they deserve."
Regarding the study of Geometry as a means of mental
discipline, it is obviously desirable that the student should be
accustomed to the use of accurate and distinct expressions, and
even to formal syllogisms. In most sciences our definitions of
things are in reality only the results of the analysis of our
own imperfect conceptions of the things; and in no science,
except that of number, do the conceptions of the things coincide
so exactly (if we may use the expression) with the things them-
selves, as in Geometry. Hence, în geometrical reasonings, the
comparison made between the ideas of the things, becomes almost
a comparison of the things themselves. The language of pure
Geometry is always precise and definite. The demonstrations are
effected by the comparison of magnitudes which remain unaltered,
and the constant use of terms whose meaning does not on any
occasion vary from the sense in which they were defined. It is
this peculiarity which renders the study so valuable as a mental
discipline for we are not to suppose that the habits of thought
thus acquired, will be necessarily confined to the consideration of
lines, angles, surfaces and solids. The process of deduction pur-
sued in Geometry from certain admitted principles and possible
vi
PREFACE.
constructions to their consequences, and the rigidly exact com-
parison of those consequences with known and established truths,
can scarcely fail of producing such habits of mind as will influence
most beneficially our reasonings on all subjects that may come
before us.
""
In support of the views here maintained, that Geometrical
studies form one of the most suitable and proper introductory
elements of a scientific education, we may add the judgment of
a distinguished living writer, the author of "The History and
Philosophy of the Inductive Sciences," who has shewn, in his
"Thoughts on the Study of Mathematics," that mathematical
studies judiciously pursued, form one of the most effective means
of developing and cultivating the reason: and that "the object
of a liberal education is to develope the whole mental system
of man;—to make his speculative inferences coincide with his
practical convictions;-to enable him to render a reason for
the belief that is in him, and not to leave him in the condition
of Solomon's sluggard, who is wiser in his own conceit than
seven men that can render a reason." To this we may sub-
join that of Mr John Stuart Mill, which he has recorded in
his invaluable System of Logic, (Vol. 1. p. 180) in the following
terms. “The value of Mathematical instruction as a preparation
for those more difficult investigations (physiology, society, govern-
ment, &c.) consists in the applicability not of its doctrines, but of
its method. Mathematics will ever remain the most perfect type
of the Deductive Method in general; and the applications of Mathe-
matics to the simpler branches of physics, furnish the only school
in which philosophers can effectually learn the most difficult and
important portion of their art, the employment of the laws of
simpler phenomena for explaining and predicting those of the more
complex. These grounds are quite sufficient for deeming mathe-
matical training an indispensable basis of real scientific education,
and regarding, with Plato, one who is άyewμétρntos, as wanting
in one of the most essential qualifications for the successful culti-
vation of the higher branches of philosophy."
TRINITY COLLEGE,
October 1, 1845.
R. P.
CONTENTS.
INTRODUCTION
i
On the Algebraical Symbols and Abbreviations used in Geometry... xxxix
First Book of the Elements
1
Notes on the First Book
41
Second Book of the Elements
Notes on the Second Book......
Third Book of the Elements...
Notes on the Third Book
Fourth Book of the Elements
Notes on the Fourth Book
Fifth Book of the Elements
Notes on the Fifth Book
Sixth Book of the Elements
Notes on the Sixth Book
Eleventh Book of the Elements
53
67
75
107
109
124
...
126
159
...
167
203
205
Notes on the Eleventh Book.......
252
Twelfth Book of the Elements......
255
Notes on the Twelfth Book
287
On the Ancient Geometrical Analysis.....
289
Geometrical Exercises on Book I
293
II
305
III..
311
IV
330
....
VI...
XI ...
XII.....
Index to the Geometrical Exercises.....
344
366
371
377
ERRATA.
GREAT care has been taken in the correction of the proofs, and it is believed that
the Student will not meet with any errata of importance in the text to impede his pro-
gress: the following, however, have been discovered on revising the sheets:
PAGE
10
12
LINE
20
14
FOR
CBG
angles
READ
CGB.
angle.
22 In the diagram, Prop. XXIII. D is misplaced.
59
20
80
17
111
125
5
23
BE
equal less
ABC
3-2
3
DE.
less equal.
ACB.
3-2
π.
3
172 In the diagram, Prop. III. A is misplaced.
177 In the diagram, Prop. IX. D and E are interchanged.
192 In the diagram, Prop. XXVI. for D read C.
242
13
323
46
328
1
341
4
AE
arc
centre
of
AQ.
circumference.
circumference.
at.
In the Geometrical Exercises, a few repetitions have occurred, and a few of the
problems, perhaps, with more propriety, might have been arranged under a different
book.
In the Index a few references are omitted, as the Examination Papers, from
which the questions have been taken, were without dates; besides some four or
five others, which were lost.
INTRODUCTION.
B.C.
THE aim of the following brief notices, is to give some account of
the origin and progress of the science of Geometry, together with the
names of the men by whom it has been successively advanced. Respect-
ing the history of science it has been remarked that it serves, at least,
to commemorate the benefactors of mankind; an object which can
scarcely be considered as unworthy or unimportant. It is probable that
this science had its origin, like all others, in the necessities of men.
The word Geometry (yewμerpía), formed from two Greek words, yn
and perpéw, seems to have been originally applied to the measuring of
land. The earliest information on this subject is derived from Hero-
dotus (Book II. c. 109), where he describes the customs of the Egyptians
in the age of Sesostris, who reigned in Egypt from about 1416 to 1357,
The account of Herodotus is to this effect. "I was informed by
the priests at Thebes, that king Sesostris made a distribution of the
territory of Egypt among all his subjects, assigning to each an equal
portion of land in the form of a quadrangle, and, that from these allot-
ments he used to derive his revenue by exacting every year a certain
tax. In cases however where a part of the land was washed away by
the annual inundations of the Nile, the proprietor was permitted to pre-
sent himself before the king, and signify what had happened. The king
then used to send proper officers to examine and ascertain, by admeasure-
ment, how much of the land had been washed away, in order that the
amount of tax to be paid for the future, might be proportional to the
land which remained. From this circumstance I am of opinion, that
Geometry derived its origin; and from hence it was transmitted into
Greece." The natural features and character of the land of Egypt,
where rain is unknown, are such as to give credibility, at least, to the
tradition recorded by Herodotus. In the earliest records of history, the
population of Egypt is represented as numerous, and in the valley of
the Nile, the extent of cultivated land is comparatively small Its
extreme fertility is also placed in close contrast with the barrenness of
the districts beyond the limits of the inundations of the Nile, by which
the boundaries of the land on its margin are annually liable to altera-
tion. There appear therefore some grounds for the belief that the
geometrical allotment of land had its origin on the banks of the
Nile. But independently of the tradition of Herodotus, it seems reason-
able to suppose, that the science of Plane Geometry may have originated
in the necessity of measuring and dividing lands, which must have
arisen as soon as property in land came to be recognized among men.
This recognition is found in the oldest historical records known in any
language. The narrative in the 23rd chapter of the book of Genesis
refers to Palestine, and belongs to a period 1860 years B. C. In Egypt
b
ii
INTRODUCTION.
also, about 160 years later, as we learn from Gen. xlvii. not only was
property in land recognized, but taxes were raised from the possessors
and cultivators of the soil. This necessarily implies that there existed
some method of estimating and dividing land, rude, probably, and inac-
curate at first, but as society advanced and its wants increased, gradually
becoming more exact.
The existence of the pyramids, the ruins of temples, and other archi-
tectural remains, supply evidence of some knowledge of Geometry;
although it is possible that the geometrical properties of figures, neces-
sary for such works, might have been known only in the form of prac-
tical rules, without any scientific arrangement of geometrical truths.
The word Geometry is used in a more extensive sense, as the
science of Space; or that science which discusses and investigates the
properties and relations existing between definite portions of space,
under the fourfold division of lines, angles, surfaces, and volumes, with-
out regard to any properties they may have of a physical nature. Of
the origin and progress of Geometry, in this sense, it is proposed here
to give some short account, as far as can be ascertained.
Whatever geometrical or astronomical science may have been pos-
sessed by the earlier Chaldeans and Egyptians, there are not known to
be any historical records, which supply definite views of its limits or
extent. In the most ancient Jewish writings, there is not the least
allusion from which to infer that scientific Geometry was known and
cultivated by that people. The traditions recorded by Josephus on
this subject (Book 1. c. 3, 9) can scarcely be considered worthy of
being received as historical truth, since the subsequent history of the
Jews does not inform us that they were, at any period, a scientific
people.
Other ancient writers also confirm the tradition of Herodotus, that
from Egypt the knowledge of Geometry passed over into Greece, where
it attained a high degree of cultivation. Proclus, in his Commentary on
Euclid's Elements of Geometry (Book 11. c. 4), attributes to Thales the
merit of having first conveyed the knowledge of Geometry from Egypt
to Greece. Thales was a native of Miletus, at that time, the most flourish-
ing of the Greek colonies of Ionia in Asia Minor. He was born about
640 B.C., and was descended from one of the most distinguished families,
originally of Phoenicia. (Herod. 1. 170. Diog. Laert. 1. 22.) Thales,
from a desire of knowledge, is reported by Diogenes Laertius to have
travelled into Egypt, and to have held a friendly intercourse with the
priests of that country; thus obtaining an acquaintance with the science
of the Egyptians. The same writer also adds that he learned the art of
Geometry among the Egyptians, and suggested a method of ascertaining
the altitude of the pyramids by the length of their shadows. Plutarch
relates this story, and adds, that Amasis, who was then king of Egypt,
was astonished at the sagacity of Thales. If this tradition, recorded
both by Diogenes Laertius and by Plutarch, be deemed worthy of credit,
it would appear that the idea suggested by Thales was, to the Egyptians,
a new application of a geometrical truth. Whatever mathematical
knowledge the Egyptians might possess in the age of Thales, there are
no writings, either cotemporary or of later times, which exhibit its extent.
Thales is also said to have been the discoverer of some geometrical
theorems, and to have left to his successors the principles of many
others. The theorems which stand as the 5th, 15th, and 26th Propo-
INTRODUCTION.
iii
sitions of the first Book of Euclid's Elements of Geometry are attri- Se mech Bay,
buted to him by Proclus, and Prop. 31. Book III, also Props. 2, 3, 4, Aug. 31.1850
5, of Book IV. He is reported by Herodotus (Book 1. c. 74) to have
foretold the year in which an eclipse of the sun would happen. He
also designated the seasons, and found the year to consist of 365 days.
All this implied not only some acquaintance with Geometry, but a con-
siderable knowledge of the motions and periodical revolutions of the
heavenly bodies. Anaximander of Miletus and Anaximenes are men-
tioned as disciples of Thales. The opinions of the latter are discussed
by Aristotle immediately after those of Thales. Ameristus, the brother
of Stesichorus the poet, is named by Hippias of Elis as a celebrated
geometer.
Nearly at the same time with the commencement of speculative phi-
losophy in Ionia, a spirit of enquiry began to shew itself in some of the
Achæan and Dorian colonies in Magna Græcia. The most distinguished
man of these times was Pythagoras. He was born at Samos, about 568
B. C.; and his descent is referred by Diogenes Laertius to the Tyrrhenian
Pelasgi. After having been a disciple of Thales, he is reported to have
visited Egypt, where he became the pupil of Oinuphis at Heliopolis,
(which in the book of Genesis is called On), once a famous city of
Lower Egypt. (Plutarch de Iside et Osiride, s. 10.) After his return
from Egypt, he established a school at Crotona, an Achæan colony in
Magna Græcia, which became very celebrated, and continued for nine-
teen generations. According to the account of Proclus (Book II. c. 4),
Pythagoras was the first who gave to Geometry the form of a deductive
science, by shewing the connexion of the geometrical truths then known,
and their dependence on certain first principles. There are not known
to be extant any particular accounts, or even fragments, of the earliest
attempts to reduce geometrical truths to a system. It is, however,
scarcely possible that any arrangement could have been attempted be-
fore a considerable number of geometrical truths had been discovered,
and their connexion observed. The traditionary account, that Pytha-
goras was the founder of scientific mathematics, is, in some degree,
supported by the statement of Diogenes Laertius, that he was chiefly
occupied with the consideration of the properties of number, weight,
and extension, besides music and astronomy. The passage of Cicero
(De Nat. Deor. III. 36) may be referred to as evidence that later
writers were unable to give any precise account of the mathematical
discoveries of Pythagoras. To Pythagoras, however, is attributed the
discovery of some of the most important elementary properties con-
tained in the first book of Euclid's Elements. The very important
truth contained in Prop. 47, Book 1. is also ascribed to Pythagoras.
Whether his discovery of this truth resulted from geometrical, or from
numerical considerations is not certainly known: Proclus attributes
to him the discovery of that right-angled triangle, the three sides
of which are respectively 3, 4, and 5 units. To Pythagoras also be-
longs the discovery, that there are only three kinds of regular polygons
which can be placed so as to fill up the space round a point; namely,!
six equilateral triangles, four squares, and three regular hexagons.
Proclus attributes to him the doctrine of incommensurables, and the
discovery of the five regular solids, which, if not due to Pythago-
ras, originated in his school. In Astronomy, he is reputed to have
held, that the Sun is the centre of the system, and that the planets
}
b-2
iv
INTRODUCTION.
revolve round it. This has been called, from his name, the Pytha-
gorean System, which was revived by Copernicus, A.D. 1541, and
proved by Newton. As a moral philosopher, many of his precepts
relating to the conduct of life will be found in the verses which bear
the name of the Golden Verses of Pythagoras. It is probable they
were composed by some one of his school, and contain the substance
of his moral teaching. The speculations of the early philosophers did
not end in the investigation of the properties of number and space.
The Pythagoreans attempted to find, and dreamed they had found, in
the forms of geometrical figures and in certain numbers, the principles
of all science and knowledge, whether physical or moral. The figures
of Geometry were regarded as having reference to other truths besides
the mere abstract properties of space. They regarded the unit, as the
point; the duad, as the line; the triad, as the surface; and the tetrac-
tys, as the geometrical volume. They assumed the pentad as the
physical body with its physical qualities. They seem to have been
the first who reckoned the elements to be five in number, on the
supposition of their derivation from the five regular solids. They made
the cube, earth; the pyramid, fire; the octohedron, air; the icosa-
hedron, water; and the dodecahedron, æther. The analogy of the
five senses and the five elements was another favourite notion of the
Pythagoreans.
Pythagoras was followed by Anaxagoras of Clazomene. Aristotle
states that he wrote on Geometry, and Diogenes Laertius reports that
he maintained the sun to be larger than the Peloponnesus. Ænopides
of Chios was somewhat junior to Anaxagoras, both of whom are men-
tioned by Plato in his Rivals. Proclus ascribes to Ænopides the dis-
covery of the truths which form the 12th and 23d Props. of the first Book
of Euclid. Briso and Antipho are mentioned by Aristotle, as dis-
tinguished geometers, but no records are known to be extant of their
writings or their discoveries.
About 450 B.C., Hippocrates of Chios, their cotemporary,,was the most
eminent geometer of his time, and is reported to have written a treatise
on the Elements of Geometry; no fragments of which, however, are
known to be in existence. He discovered the quadrature of the lunes
which bear his name, by describing, in the same direction, semicircles
on the three sides of an isosceles right-angled triangle, and observing
that the sum of the two lunes, between the two quadrants of the larger
semicircle and the two smaller semicircles, was equal to the area of the
triangle. By means of the lunes he attempted the quadrature of the
circle, but without success. The account of his attempts will be found
in the Commentary of Simplicius, on the first book of Aristotle's Physics.
Hippocrates also solved the problem of the duplication of the cube,
which is, to find the length of the edge of a cube which shall be twice
as great as a given cube. This problem, at that time, and for ages
afterwards, excited very great attention among philosophers. He shewed
that the solution depended on finding two geometric mean proportionals
between two given lines.
About 100 years after Pythagoras, Zenodorus wrote a tract, pre-
served in Theon's commentary on the Almagest of Ptolemy, in which
he shews that plane figures having equal perimeters have not equal

areas.
Democritus, a native of Abdera, about the 80th Olympiad was cele-
INTRODUCTION.
#
brated for his knowledge both of Philosophy and the Mathematics.
He is stated to have spent his large patrimony in travelling in distant
countries.
Theodorus of Cyrene was eminent for his knowledge of Geometry,
and is reported to have been one of the instructors of Plato.
We now come to the time of Plato, one of the most distinguished
philosophers that ever lived; whose writings are still read, and regarded
as of inestimable value. Plato visited Egypt, and on his return, founded
his School at Athens, about 390 B.C. Över the entrance he placed
the inscription, Οὐδεὶς ἀγεωμέτρητος εἰσίτω: “Let no one ignorant of
Geometry enter here." This is a plain declaration of Plato's opinion
respecting Geometry. He considered Geometry as the first of the
sciences, and as introductory and preparatory to the pursuit of the
higher subjects of human knowledge. Plato both cultivated and ad-
vanced the science of Geometry, as we learn from the testimony of
Proclus and Pappus. The character of his writings, though not confined
to discourses on Mathematics, affords incontestable evidence how great an
admirer he was of Geometry, and how zealously he cultivated that science.
To Plato is attributed the discovery of the method of the Geometrical
Analysis; but by what means he was led to it, or to the invention of Geo-
metrical loci, is not known. Hippocrates had, before his time, reduced
the problem of the duplication of the cube to that of finding two mean
proportionals between the edge of the given cube, and a line double
that length; Plato attempted a solution of the problem in this form, by
means of the right line and circle only. In this attempt, however, he
failed, but effected a solution by means of two rulers, which could not
be admitted as purely geometrical, since it involved other considerations
besides those of the straight line and circle. It is uncertain whether the
restriction of constructions in Geometry to the right line and circle,
originated in the School of Plato, or at an earlier period. Plato is said
to have discovered the Conic Sections,-curves which result from planes
intersecting the surface of a cone,-and some of their more remarkable
properties. Great attention was given to this subject, both during his
lifetime, and after his death, by his cotemporaries and their successors.
The Conic Sections and their properties were considered a distinct
branch of the science, and called "the Higher Geometry." The tri-
section of an angle, or an arc of a circle, was another famous problem,
which engaged the attention of the School of Plato. This, as well as
the duplication of the cube, was, for many ages, believed possible, by
means of the right line and circle, and has been repeatedly attempted
from the earliest times, without success. It baffled the genius of Archi-
medes and others, who were not aware of the impossibility of its solution
by the method they applied. The problem of the trisection of an angle
or arc is reducible to the following problem: To draw a right line from
a given point, cutting the semi-circumference of a circle in two points,
and the diameter produced, so that the chord intercepted between the
two points of intersection may be equal to the radius. This condition
leads to the algebraical equation of one of the conic sections, whose
properties are not the same as those of the right line and circle; and
hence its impossibility is inferred.
From the Academy of Plato, proceeded many who successfully cul-
tivated, and very considerably extended, the bounds of Geometrical
science. Proclus names thirteen of the disciples and friends of Plato
vi
INTRODUCTION.
Plato
celetă.
who improved and made_additions to the science. Among his cotem-
poraries may be named Leodamas of Thasus, Archytas of Tarentum,
and Theætetus of Athens, by whom some theorems were discovered, and
some improvements made in the methods of demonstration. Diogenes
Laertius reports of Archytas, that he was the first who brought Mecha-
nics into method by the use of mechanical principles, and the first
who applied organic motions to Geometrical figures, and found out
the duplication of the cube in Geometry. (Book VIII.) His solution
of the duplication of the cube is given by Eutocius, in his commen-
tary on the sphere and cylinder of Archimedes. Archytas was also
the writer of a work on the elements of Geometry, which is not ex-
tant. The same writer relates, that Leodamas, by means of the Geo-
metrical Analysis, which he had learned from Plato, discovered the
solution of many problems, and made many discoveries. Theætetus
was celebrated for a treatise on Geometry, which is lost. Plato honoured
this disciple by giving to one of his dialogues the title of Theætetus;
and in another, entitled Phædrus, he ascribes the origin of Geometry to
Thoth, an Egyptian divinity. Aristeas, a disciple and friend of Plato,
composed five books on the Conic Sections, which were highly esteemed.
Another of the scholars of Plato was Neoclides, whose name is cele-
brated by Proclus. Leon was a scholar of Neoclides (B.c. 368), and
was the author of several discoveries in Geometry. To Leon is attri-
buted the invention of a method for discriminating the possibility or
impossibility of a problem. He is also mentioned as the author of a
work on the elements of Geometry. None of his discoveries or writings
have descended to posterity. Amyclas was another friend of Plato;
and the brothers Menæchmus and Dinostratus are both celebrated;
the former for his application of the Conic Sections to solve the problem
of the duplication of the cube, and the latter, for the discovery of a
curve known by the name of the quadratrix. They are also said to have
made some other additions to the science of Geometry, and to have
rendered the whole more perfect. Theudias appears to have excelled
both in mathematics and philosophy, and is said to have composed a
work on Geometry, and to have generalized some theorems. At the
same period, Cyzicenus of Athens, besides other branches of the mathe-
matics, successfully cultivated Geometry. These friends and disciples
of Plato used to resort to the Academy, and employ themselves in pro-
posing, by turns, questions for solution. Eudoxus, a native of Cnidus,
a town of Caria in Asia Minor, was one of the most intimate of the
friends of Plato. He is reported to have written on the Elements, and
to have generalized many results which had originated in the school of
Plato, and to have advanced the science of Geometry by many impor-
tant discoveries. To him is attributed the invention of the doctrine of
proportion as treated in the 5th book of Euclid's Elements. He is said
to have been the first who discovered that the volume of a cone or
pyramid is equal to one third of its circumscribing cylinder or prism;
that is, of a cylinder or prism having the same base and altitude. He
is also reported to have advanced the knowledge of the higher Geo-
metry, by the discovery of several important properties of the Conic
Sections. He died в.c. 368, at 53 years of age. Diogenes Laertius,
in his short memoir of Eudoxus, describes him as an astrologer, a
physician, a legislator, and a geometer. Though none of his writings
have descended to modern times, his name has been celebrated by the

INTRODUCTION.
vii
eminent men, both of Greece and of Rome. Hermotimus wrote on
Loci, and is said to have extended the results of Eudoxus and Theætetus.
Philippus the Mendean, another disciple of Plato, is reported to have
discovered problems, and to have proposed questions, being a great
lover of the mathematical sciences. All these were either disciples of
Plato, or attached themselves to the school he founded at Athens, and
are mentioned by Proclus as having advanced or improved the mathe-
matical sciences. Xenocrates also was a hearer of Plato, and is said to
have been one of the instructors of Aristotle; he was renowned for his
knowledge of the mathematical sciences. Aristotle, though originally a
disciple of Plato, and attached to his school for a period of twenty years,
became the founder of a new sect of philosophers-the Peripatetics,
and opened a school, B.C. 341, at the Lyceum on the banks of the
Ilissus. There he continued twelve years, till the false accusation of
Eurymedon obliged him to flee to Chalcis, where he died at the age of
sixty-three.


•
On this division of the Platonic school, the two sects-both the
Academics and the Peripatetics-continued to hold the same opinion
on the utility of Geometry, as the necessary introductory knowledge
for all who were desirous of proceeding with the study of philosophy.
Thus the science of Geometry continued to be cultivated, and to make
advancement. Among the numerous writings of Aristotle, there is a
treatise on Mechanics, and a collection of Problems in 38 divisions;
the 15th consists of mathematical questions. There is also a tract on
indivisible lines, which has been ascribed by some ancient commentators
to Theophrastus, as we learn from Simplicius. (Rev. J. W. Blakesley's
Life of Aristotle.)
Two of the Peripatetic school are especially celebrated, Theophrastus
and Eudemus, who devoted themselves chiefly to mathematical studies.
Theophrastus was the author of the first history of the mathematical
sciences, from the earliest times to his own. The work consisted of
eleven books, of which there were four on Geometry, six on Astronomy,
and one on Arithmetic. Eudemus also wrote a history of Astronomy
in six books, and a history of Geometry in six books, from which Pro-
clus acknowledges that most of his facts were taken. None of these
writings have been preserved.
Autolycus, of Pitane in Æolis, lived about 300 B. C.
He was
preceptor in mathematics to Arcesilaus, a disciple of Theophrastus the
successor of Aristotle, as we learn from Diogenes Laertius. His treatise
on the moveable sphere is the earliest written on that subject. The
original Greek, with a Latin translation, was published in 1572. He
also was the author of another treatise "On the Rising and Setting of
the Stars," which is still extant, and has been translated and printed.
Aristeas is said to have composed five books on the Conic Sections,
and five books on Solid Loci. He is also said to have been the friend
of Euclid, and his instructor in Geometry.
We come next to the time of Euclid. The birth-place and even the
country of Euclid are unknown, and he has been very frequently con-
founded with another philosopher of the same name, who was a native
of Megara. He studied at Athens, and became a disciple of the Pla-
tonic school. He flourished in the time of Ptolemy Lagus (B.c. 323
to 284), to whom he made the celebrated reply, that "there is no royal
road to Geometry." He is said to have successfully cultivated and
viii
INTRODUCTION.
taught Geometry and the mathematical sciences at Alexandria, shortly
after the school of Philosophy was founded in that city. The school at
Alexandria became most distinguished for the eminent mathematicians
it produced, both in the lifetime of Euclid and afterwards, until the
destruction of the great library at Alexandria, and the subjugation of
Egypt by the Arabians. Euclid has become celebrated chiefly by his
work on the Elements of Geometry, for which his name has become
a synonym. It consists of thirteen books, nine of which are devoted to
the subject of Geometry, and four to the properties of numbers, as
discussed by the Greek Arithmetic, and applied to Geometry. There
are two other books on the five regular solids, usually found appended
to the thirteen books of Euclid. These, however, were subsequently
added to the Elements by Hypsicles of Alexandria. In some editions
a sixteenth book is found, which was added by Flussas. Another
geometrical work of Euclid is the Data, which consists of 100 pro-
positions. This is the oldest specimen of the principles auxiliary to
the Geometrical Analysis. The object of the several propositions of
this book is to shew that, in cases where certain properties or ratios
are given, other properties or ratios are also given, or may be found
geometrically; and thus pointing out what data are essential in order
that Geometrical Problems may be determinate and free from all am-
biguity. Three books on Porisms are attributed to Euclid by Pappus
and Proclus; the former, in the seventh book of his Mathematical Col-
lections, has given some account of them and the general enunciations
of some propositions. This contains all that is known to exist of these
three books. Attempts have been made with some success in later
times for their restoration. Euclid also wrote a Treatise on Fallacies
in geometrical reasoning, and another on Divisions. Pappus makes
mention of another work on Geometry attributed to Euclid under the
title of TÓTшν πpòs éπipáveiav (Coll. Math. Lib. VII. Introd.), which his
Latin translator Commandine has rendered by "Locorum ad Super-
ficiem." Pappus also states that Euclid composed four books on the
Conic Sections, which were afterwards augmented by Apollonius Per-
gæus. Proclus has made no allusion to this work in his account of the
writings of Euclid. Besides these writings on Geometry, Euclid is
reported by Proclus to have been the author of a work on Optics and
Catoptrics, and a work on Harmonics; but it is very questionable
whether the treatise on Harmony which is extant, and attributed to
Euclid, was really composed by him. Pappus mentions a work on
Astronomy entitled "The Phænomena." This treatise contains some
geometrical properties of the sphere.
It has been a question whether Euclid was the author or the com-
piler of the Elements of Geometry, which bear his name. If Euclid
were the discoverer of the propositions contained in the thirteen books
of the Elements, and the author of the demonstrations, he would be a
phenomenon in the history of science. It is by far more probable that
he collected and arranged the books on Geometry in the order in which
they have come down to us, and made a more scientific classification of
the geometrical truths which were known in his time. Euclid may
also have been the discoverer of some new propositions, and may have
amended and rendered more conclusive the demonstrations of others.
From the slow advances of the human mind in making discoveries,
and the general history of the progress of the sciences, it would seem
INTRODUCTION.
ix
:
unreasonable to assign to Euclid a higher place than that of the com-
piler and improver of the Elements of Geometry. This is in complete
accordance with the statement of Proclus, who relates that "Euclid
composed Elements of Geometry, and improved and arranged many
things of Eudoxus, and perfected many things which had been dis-
covered by Theætetus, and gave invincible demonstrations of many
things which had been left loosely or unsatisfactorily demonstrated
before him." The Elements of Geometry, thus arranged and improved
by Euclid, were acknowledged so far superior in completeness and
accuracy to the elementary treatises then existing, that they entirely
superseded them, and in course of time all have disappeared. The
book of Euclid's Elements is therefore the most ancient work on
Geometry known to be extant. The Greek Arithmetical Notation,
employed in the arithmetical portion, has yielded to the more perfect
system of the Indian; but the geometrical portion, from the time it
was first put forth till the present day, a period of upwards of 2000
years, has maintained its high character as an elementary treatise, in
all nations wherever the sciences have been cultivated.
At this part of our subject we cannot forbear making a few remarks
on the comparative claims of the Egyptians and Greeks to the merit of
being the authors of Scientific Geometry. The learned Sir Gardiner
Wilkinson, in his profound work on the Ancient Egyptians, unhesita-
tingly assigns this honour to the Egyptians. In page 342 of Volume I.
he expresses his judgment in the following terms: "Anticlides pretends
that Moeris was the first to lay down the elements of that science, which,
he says, was perfected by Pythagoras; but the latter observation is
merely the result of the vanity of the Greeks, who claimed for their
countrymen the credit of enlightening a people on the very subjects
which they had visited Egypt for the purpose of studying." The vanity
both of the later and the earlier Greeks may be readily admitted,
without allowing that it suggests the true answer to this question. Dio-
genes Laertius, (Book VIII.) in his life of Pythagoras, writes thus:
"Anticlides reports that Maris was the first who invented Geometry,
and Pythagoras brought his imperfect notions to perfection." Moris
was an early king of Egypt (Herod. 11. 13), the son of Amenophis, and
lived before the age of Sesostris. Whatever we learn from Herodotus
respecting Geometry in Egypt, is referred to the age of Sesostris, and
does not carry us beyond such processes as might exist without any
attempts at science. The Geometry which was brought from Egypt to
Greece, appears to have been in its infancy; and all that Pythagoras and
others borrowed from the Egyptians could not have exceeded some
practical rules and their applications. In the learned work referred to,
any positive information as to the existence of Scientific Geometry
among the early Egyptians, has been sought in vain; nor do ancient
writers exhibit any remains of the Scientific Geometry of the Egyptians,
or even make any claims in their favor to the merit of its invention.
The general tenor of all the traditional and probable evidence tends to
shew, that the scientific form of the Elements of Geometry is due to the
acute intellect of the Greeks. And this presumption is reduced almost
to historical certainty by the existing remains of the Greek Geometry.
We may further observe, that in the very brief review we have given of
the earliest commencement of the Greek Geometry and Mathematics,
their simplicity is such as might be expected to characterize the original
X
INTRODUCTION.
attempts of an acute people. The advances were so gradual that any
supposition of the sudden introduction of more perfect science from
foreign sources is completely removed. Moreover, at the time that the
Greek Geometrical science was rapidly advancing towards perfection in
the school of Plato, there is a total absence of any accounts that the
Geometry of the Egyptians or of any other nation had proceeded so far
in its development. Plato truly ascribes to the Egyptians and Phoeni-
cians a certain commercial activity, but distinguishes their native cha-
racter from that of the Greeks, which he represents as remarkable for
its curiosity and desire of knowledge. (De Repub. IV.) Five centuries
afterwards the same characteristic of that people is remarked by St Paul
in his first epistle to the Corinthians.
Archimedes was born at Syracuse, B.C. 287, about the period of the
death of Euclid, and became the most eminent of all the Greek mathe-
maticians. His discoveries in Geometry, Mechanics, and Hydrostatics,
form a distinguished epoch in the history of mathematical science; and
his remaining writings on the pure Mathematics are the most valuable
portion of the ancient Geometry. He was the first who discovered that
the volume of a sphere is two thirds of its circumscribing cylinder, or
of a cylinder having the same diameter and altitude as the sphere, and
that the curved surface of each is equal to four great circles of the
sphere. He also found the relation of the volumes and surfaces of a
hemisphere and cone upon the same base. These and other properties
are investigated in two books, entitled, "On the Sphere and Cylinder,"
which have descended to our times in the original Doric Greek, together
with the Commentary of Eutocius. Another work is still extant on the
measurement of the circle, in which he shews that the area of a circle is
equal to that of a right-angled triangle whose altitude is equal to the
radius and whose base is equal to the circumference. Though he failed
in his attempts to discover the exact proportion of the circumference to
the diameter of a circle, he discovered an useful approximation to that
ratio. He found, by numerical calculation, that the perimeter of a re-
gular polygon of 192 sides, circumscribing a circle, is to the diameter in
a less ratio than 30 to 1; and that the perimeter of the inscribed po-
lygon of 96 sides is to the diameter in a greater ratio than 3 to 1:
whence he concluded that the ratio of the circumference to the diameter
of the circle must lie between these two ratios. The book of Lemmas,
a collection of Problems and Theorems on Plane Geometry, attributed
to Archimedes, is not known to be extant in the original Greek. It
comes to us from the Arabic, of which a translation in Latin was pub-
lished, for the first time, in 1659. On the Higher Geometry, three
tracts of Archimedes are still extant. 1. On the Quadrature of the
Parabola, in which he proves that the area included between the curve
and two ordinates is equal to two thirds of the circumscribing parallel-
ogram. This is the first instance known of the discovery of the quad-
rature of a figure bounded partly by a curved line, if we except that of
the lunes of Hippocrates. 2. His Treatise on Conoids and Spheroids,
which contains many discoveries: among them may be named, the ratio
of the area of an ellipse to a circle having the same diameter as the axis
major of the ellipse; and that the sections of conoids and spheroids are
conic sections. He also first proved that the volume of the cone and
parabolic conoid of the same base and altitude are in the proportion of
2 to 3. This tract also contains the demonstrations of several discoveries

INTRODUCTION.
xi
respecting hyperbolic conoids and spheroids. 3. A Tract on Spirals.
The curve known by the name of the Spiral of Archimedes was ori-
ginally discovered by his friend Conon, whose premature death pre-
vented his completing the investigation of the properties of the curve.
Archimedes completed the investigations, and put them forth as they
appear in this tract. Besides his writings on Geometry, we have a
tract on Arithmetic, entitled Tapμiτns, or Arenarius. The object was to
prove the possibility of expressing the number of grains of sand which
would fill the whole space of the universe considered as a sphere ex-
tending to the stars. In this tract he alludes to a system of numeration
which he had discovered, and which he had described in a work ad-
dressed to Zeuxippus. Two books are also extant on the Equilibrium of
Planes, and on their centres of gravity, in which he has proved the
fundamental property of the Lever, and shewn how to find the centre
of gravity of a triangle, and other figures. Two books, "On Bodies
which are carried in a fluid," in which the general conditions of a body
floating on a fluid are investigated and applied to different forms of
bodies. Archimedes was also the discoverer of the method of finding
the specific gravity of bodies. The story of the crown of king Hiero,
to whom Archimedes was related, is briefly this. Hiero had delivered
to a goldsmith a certain weight of gold to be converted into a votive
crown. The king suspected that the crown he received from the smith
was not of pure gold, though of the proper weight, but that it was
alloyed with silver. He applied to Archimedes to ascertain, without
melting the crown, whether it contained alloy. It was observed by Ar-
chimedes, on going into a bath full of water, that when his body was
immersed in the bath, a quantity of water equal to the bulk of his body
flowed over the edge of the bath. It occurred at once to him, that if a
weight of pure gold equal to the weight of the crown were immersed
in a vessel full of water, and the quantity of water left in the vessel
measured, on the gold being taken out; by doing the same with the
crown in the same vessel, he would be able to ascertain whether the
bulk of the crown were greater than the bulk of an equal weight of
pure gold. For any weight of silver is larger in bulk than an equal
weight of pure gold. According to Vitruvius, as soon as he had dis-
covered the method of solution, he leaped out of the bath, and ran
hastily through the streets to his own house, shouting evρŋka, eйρŋkа !
He was also the inventor of a machine for raising water from lower to
higher levels, which was called the screw of Archimedes. An important
application of the principle of this screw has lately been made in the
propulsion of ships by means of steam power. When Syracuse was
besieged by a land and naval armament under Appius and Marcellus,
the besieged held out a successful resistance for three years chiefly by
the aid of the machines invented by Archimedes. The city was at
length surprised and taken B. C. 212. Archimedes, when seventy-five
years of age, was slain by a soldier, while intent on the solution of a
problem. He is reported to have expressed a desire that a sphere in-
scribed in a cylinder might be engraved on his tomb, to record his
discovery of the relation between the volumes and the surfaces respec-
tively of these two solids. About 200 years after his death, his tomb
was discovered near Syracuse by Cicero, while quæstor in Sicily.
was nearly overgrown with bushes and brambles, which he caused to be
cleared
away. The tomb was identified as the tomb of Archimedes by

•
It
xii
INTRODUCTION.
:
the Sphere and Cylinder engraved upon the stone with the inscription,
the latter part of which was completely effaced. (Cic. Tusc. Quæst.
lib. v.)
Conon was the friend and cotemporary of Archimedes, and is cele-
brated by Virgil in his third Eclogue. In the treatise on the quadra-
ture of the parabola, speaking of his genius, Archimedes exclaims-
"How many theorems in geometry, which to others have appeared
impossible, would Conon have brought to perfection, if he had lived!"
Cotemporary with Archimedes was Eratosthenes, a distinguished
geometrician and astronomer. He is celebrated for his construction in
solving the problem of the duplication of the cube. He was the first
who attempted to measure the circumference of the earth by means of
observations of the Sun at two different places, near the same meridian,
at the time of the solstice. Though he did not completely succeed, on
account of the inaccuracy of his data, he pointed out the method. None
of his works have descended to modern times except a few fragments,
and a list of the names of forty-four constellations, and the principal
stars in each constellation.
Apollonius of Perga, a city of Pamphylia, lived about the same time,
and stands next in fame to Archimedes. He was born at the time when
Ptolemy Euergetes was king of Egypt: he studied the mathematical
sciences at Alexandria in the school which Euclid's disciples had founded,
and passed there the greater part of his life. He was the author of
several works on Geometry, and became so eminent in that science that
he was called by his cotemporaries the Great Geometer. His principal
work is a treatise on the Conic Sections. From the author's dedicatory
epistle to Eudemus, a geometer of Pergamus, it appears that the treatise
consisted of eight books. The first four books still exist in the original
Greek. An Arabic version of seven books, made about the middle of
the thirteenth century, was found in the East about four centuries later
by Golius, a professor of the oriental languages at Leyden; and was
translated into Latin. It has been said that Apollonius appropriated
the discoveries of others in the Conic Sections. It is well known that
although Archimedes discovered many important properties of the curves
which bear that name, it cannot be pretended that he was the origi-
nal discoverer. Long before the time of Archimedes they had been
studied in the school of Plato and at Alexandria, and many properties
of them were well known. It is highly probable that Apollonius, in his
treatise on the Conic Sections, availed himself of the writings of Archi-
medes as well as of others who had gone before him. His treatise on
Conics was most highly esteemed; and to him is justly accorded the
honour of having composed a better treatise on that difficult subject
than any who had written before him. He made important improve-
ments in the problems both of Euclid and of Archimedes. Before
Apollonius, we are informed, by his commentator Eutocius, that writers
on the Conic Sections required three different sorts of cones from which
to cut the three different sections. They used to cut the parabola from a
right-angled cone, the ellipse from an acute-angled cone, and the hyper-
bola from an obtuse-angled cone: because they always supposed the
sections made by the cutting planes to be at right angles to the side of the
cone. But Apollonius cut his sections from any cone by only varying the
inclination and position of the cutting plane. It may be remarked that
Apollonius first gave the names of ellipse and hyperbola to two of the
INTRODUCTION.
xiii
curves: the name of parabola had been already applied to the third by
Archimedes. He also first made the distinction between the diameters
of the sections, and the axes, giving the latter name to the two diameters
which are at right angles to each other in the ellipse and hyperbola, and
restricting the term axis in the parabola, to the line which passes
through the focus and vertex of that curve. The following is a very
brief account of the subject of each book of this treatise.
Book 1. treats of the generation of the Conic Sections and their dis-
tinguishing properties.
Book II. treats of the properties of diameters and axes of these
three curves, and of the asymptotes of the hyperbola.
Book III. consists of Theorems useful in the solution of solid loci.
Book IV. explains his new method of the intersection of the sections
of cones with each other and with the circumferences of circles.
Book v. treats of maxima and minima in the Conic Sections.
Book VI. treats of equal and similar sections of the cone.
Book VII. contains a collection of Theorems useful in the solution of
Problems.
Book VIII. was a collection of Problems with their solutions by
means of the Theorems in Book VII.
Besides the treatise on the Conic Sections, Apollonius was the author
of several treatises relating to the Geometrical Analysis. They bear
the following titles in the translation of Pappus's Collections, and each
treatise consisted of two Books.
1. De Rationis Sectione.
2. De Spatii Sectione.
3. De Sectione Determinatâ.
4.
De Tactionibus.
5. De Inclinationibus...
6. De Planis Locis.
An Arabic Version of the Treatise De Sectionis Ratione was trans-
lated into Latin by Dr Halley and published at Oxford in 1708. The
rest are not known to be extant either in the original Greek or in
Arabic. Attempts however have been made since the revival of learn-
ing in Europe to restore these lost treatises; notices of which will be
found under the names of the respective authors in their proper places.
Proclus, in his commentary on Euclid, informs us that Apollonius
attempted to demonstrate the axioms of Euclid, and cites his method of
proving the first axiom, that things which are equal to the same thing
are equal to one another. Proclus examines his so-called proof, and shews
that properties are assumed not more self-evident than the axiom itself.
Nicomedes lived during the second century before the Christian era,
and is known for the invention of a curve called the conchoid, and for
the application he made of it in finding two mean proportionals between
two given lines. He was also celebrated for the invention of several
useful machines.
About the same time also lived Nitocles, Thrasideus, and Dositheus,
whose discoveries in mathematics, and whose writings, if they left any,
have not descended to our times.
Geminus, a native of Rhodes, and a mathematician of some repute,
lived about 100 years before the Christian era, and is reported to have
been the author of a work entitled 'Enarrationes Geometricæ,' which
is not known to be extant.
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INTRODUCTION.
Hipparchus, a native of Nice, though not a writer on the Elements
of Geometry, is regarded as the first who reduced Astronomy to a
science, and either devised or greatly improved the methods of calcula-
tion in Trigonometry, which form the basis of the science of Astronomy.
His work on the calculation of chords originally consisted of 12 books,
of which a few fragments only are known to be extant. He divided
the circumference of the circle into 360 equal parts, and also the radius
into 60 equal parts, which he likewise called degrees, each degree into
60 parts, and so on. His rules of calculation were derived from the
properties of chords, and were estimated in sexagesimal parts of the
radius, and the lengths of chords were calculated to every half degree
of the semicircumference. It is, however, as an astronomer that his
: name is most celebrated. He was the first who discovered the preces-
sion of the equinoxes, and taught how to foretell eclipses, and form
tables of them. The catalogue of stars which he observed and registered
between the years B. C. 160 and 135, is preserved in Ptolemy's Almagest.
They are arranged according to their longitudes and their apparent
magnitudes. He was also the first who suggested the idea of fixing the
position of places on the earth, as he did in the heavens by means of
their latitude and longitude. He pursued his astronomical studies at
Rhodes, whence he obtained the name of Rhodius, and afterwards in
Bithynia, and at Alexandria. His writings on Astronomy are highly
spoken of by ancient authors, but are not now extant: his commen-
tary, however, on the Phænomena of Aratus, still exists. A full but
rather exaggerated account of the discoveries of Hipparchus will be
found in the work of Delambre, on the ancient Astronomy of the
Greeks.
There is some uncertainty with respect to the exact period when
Hypsicles flourished. He was a native of Alexandria, and, it is said, a
disciple of Isidorus. To him are attributed the 14th and 15th books
of the Elements, which were added to the 13 books of Euclid. In the
introduction, he makes mention of Apollonius, who flourished in the
reign of Ptolemy Euergetes, and probably Hypsicles came after him.
Theodosius of Tripoli flourished about the time of Cicero, and was
the author of a treatise on the sphere in three books, which has come
down to our time, in the original Greek. In this treatise, he investi-
gates the properties of circles, which are made by sections of the surface
of the sphere. It was translated and published by Dr Barrow in 1675:
another edition was published at Oxford, in 1707.
For a period of some hundreds of years after the time of Theo-
dosius, we shall find that few additional discoveries were made in
geometrical and mathematical science. There were, however, some
instances of individuals during that period, not entirely unskilled in
Grecian science. Sosigenes the peripatetic, was a mathematician and
astronomer. He was an Egyptian, and was brought by Julius Cæsar to
Rome, for the purpose of assisting in the reformation of the Roman
Calendar. This philosopher had discovered, by astronomical observa-
tion, that the year consists of 365 days and 6 hours; and to make
allowance for the accumulation of the hours which were above 365
whole days, he invented the intercalation of one day in four years. The
duplication every fourth year of the sixth day before the Calends of
March was the intercalary day; and hence the year in which this took
place consisted of 366 days, and was named bissextile. This was called


INTRODUCTION.
XV
-
the Julian correction of the Calendar, and the reckoning by Julian
years commenced B. c. 45, and continued till the more accurate cor-
rection was made under Pope Gregory XIII. Marcus Vitruvius was
also greatly esteemed by Julius Cæsar, and he was subsequently em-
ployed by Augustus in constructing public buildings and warlike
machines. Vitruvius was the author of a work on Architecture, in ten
books, which was addressed to Augustus. This work is still extant,
and is the only one on the Architecture of the ancients, which has
descended to modern times. It affords evidence in the ninth book, that
the writer was well skilled in Geometry.* Menelaus was born at Alex-
andria, in the time of the Emperor Trajan, and was of Grecian origin.
He composed a Treatise on Trigonometry in six books, and another on
the Sphere in three books, both of which have come down to us
through the medium of translations in Arabic. A Latin translation
of the Spherics was published at Paris in 1664.
Claudius Ptolemæus, one of the most eminent mathematicians and
* The Romans were a nation of warriors, and at no period of their history distin-
guished for their cultivation or advancement of the sciences. Though it must be ad-
mitted that their history exhibits many noble instances of patriotism and the sterner virtues,
the leading principle of the Roman policy was nothing less than universal dominion. In
the life of Agricola, Tacitus has recorded the substance of the address of Calgacus to
the Northern Britons when invaded by the Romans, in which are the following lines:
"Sed nulla jam ultra gens, nihil nisi fluctus et saxa, et infestiores Romani: quorum
superbiam frustra per obsequium et modestiam effugeris: raptores orbis, postquam cuncta
vastantibus defuere terræ, et mare scrutantur: si locuples hostis est, avari: si pauper,
ambitiosi: quos non Oriens, non Occidens, satiaverit: soli omnium, opes atque inopiam
pari affectu concupiscunt: auferre, trucidare, rapere falsis nominibus, imperium; atque,
ubi solitudinem faciunt, pacem appellant." To this may be added the following passage
from the popular work of M. Aimé-Martin; "La règne de Rome fut celui d'un brigand;
elle s'aggrandit par la guerre et la pillage; et aussi elle perit par ses richesses et par la
guerre."
This will scarcely be deemed too highly coloured a description of Roman
policy and practice, when it is compared with the following lines which Virgil puts
into the mouth of Anchises.
"Excudent alii spirantia mollius æra,
Credo equidem; vivos ducent de marmore vultus ;
Orabunt causas melius, cœlique meatus
Describent radio, et surgentia sidera dicent:
Tu regere imperio populos, Romane, memento:
Hæ tibi erunt artes; pacisque imponere morem,
Parcere subjectis, et debellare superbos." (En. vI. 847.)
These lines were written when nearly the whole world was subjected to the Roman
arms, and when the future prospect of undisputed sway appeared to threaten that the
dominion of Rome would be universal and perpetual. It may seem surprising that
even in the golden age of Roman literature and magnificence, there does not appear to
have existed one Roman of original genius, who successfully cultivated and advanced the
mathematical sciences. The slender amount of scientific knowledge attained at that
period, was acquired at Alexandria and at Athens, which seem to have been the chief
places of resort for the philosophers and their hearers. Horace humorously concludes
his description of the course of his education by declaring, that the object for which he
was sent to Athens by his father was merely
"Scilicet ut possem curvo dignoscere rectum."
From this, perhaps, we may infer the opinion of that age, that a little mathematical
knowledge was not deemed wholly useless, or incompatible with the pursuit of literature.
With regard to the knowledge of astronomy which existed among the Romans, it may
be observed, that it was rather cultivated for its supposed utility in relation to astrology
and the prognostication of future events, than for its real value as science.
xvi
INTRODUCTION.
c
astronomers of antiquity, was born at Pelusium, in Egypt, about a.D.
70, and died A.D. 147. Though he is not known to have left any
writings on Geometry, his great works still extant on Astronomy and
Geography, for which he is justly celebrated, supply evidence of their
author's having been deeply skilled in the applications of that science.
His work on Geography consists of seven books, and his great work on
Astronomy of thirteen, entitled "ueɣáλn oúvtağıs.” In the early part
of the 9th century it was translated into Arabic under the title of
Almagest, a word formed from the Arabic article, and a Greek super-
lative. Both the original Greek and the Arabic version are still extant.
Ptolemy adopted that system of the universe which placed the earth
in the centre, and from him it acquired the name of the Ptolemaic
system.
We may pass over the period which intervened between Ptolemy
and Pappus, as no mathematicians of eminence appeared in the school
of Alexandria, which at that period was the chief place where philo-
sophy and the sciences were still cultivated. Pappus lived in the time
of the Emperor Theodosius, who reigned between 379 and 395 a. D.
He was the author of a work entitled "Mathematical Collections,"
which consists, as its title implies, of Problems and Theorems collected
from the works of different mathematicians, with commentaries and
historical notices. This work and the Commentary of Proclus are the
chief repositories of information respecting the ancient Geometry, and
especially the Geometrical Analysis. The work of Pappus originally
consisted of eight books, the whole of which are extant in the original
Greek except the first book, and the first half of the second book.
A translation of the last six books into Latin was made by Commandine,
who also wrote a Commentary on the work. These, after his death,
were published by the Duke of Urbino, at Pisaurum, in 1588. The first
two books of Pappus were not then known to be extant; however, there
were found in a MS. of Pappus, in the Savilian Library at Oxford, the
last twelve Propositions of the second book. They were translated into
Latin by Dr Wallis, and printed at the end of his Aristarchus Samius,
in 1688. This fragment being on Arithmetic, it was conjectured that
the first two books were on the same subject. The next five books are
on Geometry, and the last is chiefly on Mechanics. The following is a
very brief account of the contents of the remaining six books of Pappus.
The third book discusses four general problems. 1. The solution of
the Delian problem, or duplication of the cube by means of the Conic
Sections: besides three other solutions, one by Eratosthenes, another
by Hero, and a third by himself. There is also a fourth by Nicomedes,
by means of the conchoid. 2. A problem respecting the Medietates, a
name given to three lines when they were in arithmetical, geometrical,
or harmonical proportion. 3. To draw two straight lines from two
points in one side of a triangle to a point within it so that they may be
greater than the other two sides of the triangle. 4. To inscribe the
five regular solids in spheres. The fourth book consists of Theorems.
Prop. I. contains an extension of Euclid, 1. 47. Prop. x. is one of the
tangencies of Apollonius, to which the three preceding Props. are
preliminary. Prop. XIII. On the property of the Arbelon. Prop. xIx.
On the Spiral of Conon; also a solution of the Delian problem by means
of the conchoid, and the trisection of an arc of a circle, with the pro-
perties of the quadratrix, and some problems. The fifth book commences
INTRODUCTION.
xvii
with a preface, in which Pappus remarks the instinct of bees whereby
they construct their cells on geometrical principles, employing the figure
whose base is a regular hexagon, which supplies, with the smallest
labour, the greatest possible accommodation. The object of the book
is to prove that, of these plane figures which are equilateral and equi-
angular, and have equal perimeters, the greatest area is contained by
the figure with the greatest number of sides; and that of all plane
figures of equal perimeters, the circle is the greatest. The subject of
isoperimetrical figures is treated in 57 Propositions. It is proved that
of plane figures with equal perimeters, the greatest is that which is
equilateral and equiangular. This principle is extended to solids. The
regular solids are then compared, and it is proved that of those with
equal surfaces, the greatest is that with the greatest number of faces.
These are introductory to the proposition, that of all solids with equal
surfaces, the greatest is the sphere. At the end of the book, it is
shewn that there can be only five regular solids, or that only equi-
lateral triangles, squares, and pentagons, can form the boundaries of
regular solid bodies. The sixth book is employed chiefly in explaining
and connecting some propositions of Theodosius and others in treatises
on the sphere, &c. The object of the book is stated in a short preface
with reference to the three selected propositions: namely, Prop. 6, Book
III. of Theodosius on the Sphere; Prop. 6 of Euclid's Phenomena, and
Prop. 4 of Theodosius on Days and Nights. Eight different treatises
are quoted or alluded to by Pappus in this book, and it may be con-
sidered as peculiarly worthy of notice, that in Props. 31, 32, 33, 34,
which are preliminary to some on the sphere, there are stated, as
examples, some distinctions of magnitudes which may either be in-
creased or diminished without limit, or may be decreased while there
is a limit to the decrease, and conversely. The seventh book is em-
ployed on the ancient geometrical Analysis. The preface contains an
exposition of the method employed by the ancients in the discovery
both of the solution of problems, and the demonstration of theorems.
After that follows a particular description of the object and contents
of some of the most important treatises of the analytical Geometry of
the ancients, the whole of which are said to have consisted of thirty-
three books. The seventh book itself consists of Lemmas or sub-
sidiary propositions assumed or employed in the treatises described
in the preface. The whole of these thirty-three books existed in the
time of Pappus, but the greater part of them have since perished, or
at least, are not known to be in existence, either in the original Greek
or in any translation. The Data of Euclid, in Greek, two books, de
Rationis Sectione in an Arabic version, and seven books of the Conics
of Apollonius, four in Greek and three in Arabic, are all that are pre-
served. The descriptions however of eleven of these books are so parti-
cular and entire, that some eminent mathematicians have attempted the
restoration of them. Of these attempts some short notices will be found
under the names of their respective authors. The treatises of Apollonius,
entitled, De Rationis Sectione, De Spatii Sectione, De Sectione Determi-
natâ, De Tactionibus, and De Inclinationibus, contained the discussion of
general problems of frequent occurrence in geometrical investigations,
completely solved, and all the possible cases distinguished; also of each
case a separate analysis and synthesis were given with determinations in
C
xviii
INTRODUCTION.
all cases which required them. The use of these general problems was,
the more immediate solution of any proposed geometrical problems,
which could be easily reduced to a particular case of some one of them
The other treatises in the list were useful for the same purpose.
The
seventh book contains 238 propositions, some of which exhibit complete
examples of the ancient analysis and synthesis. The eighth book
gives some account of the science of Mechanics, and exhibits the pro-
gress it had made in the age of Pappus: it contains many references
to the mechanical inventions of Archimedes. A considerable part of
the book is employed in describing what are called the five mechanical
powers, and the most obvious combinations of them for raising or draw-
ing large weights. Pappus acknowledges that the substance of the
book is chiefly borrowed from the works of Hero the Elder, who lived
about fifty years after Archimedes. His long preface contains some
statements of the mechanical notions and of the arts of that period, as
well as some observations on the utility of Mechanics, and on the con-
nexion of Mechanics with Geometry. The branches of the science are
distinguished, and some notices are given of treatises which are lost.
Serenus lived about the same time, and is chiefly known for his
treatise, in three books, on the Cone and Cylinder.
Theon, of the same period, was a native of Smyrna: he was a
mathematician of the Platonic school at Alexandria, of which he subse-
quently became president. He wrote a commentary on Euclid's Ele-
ments; the earliest of which we have any notice, and another on the
first eleven books of the Almagest of Ptolemy. The commentary was
translated into Latin by Commandine, and published with his Latin
translation of the Elements of Euclid from the Arabic.
Hypatia, the daughter of Theon, became so well skilled in the
mathematical Sciences as to be chosen to succeed her father in the
school at Alexandria. Her commentaries on the Conic Sections of
Apollonius, and on the Arithmetic of Diophantus, are not known to be
in existence. Though of blameless life, she was assassinated, A.D. 415,
and there are some grounds for the opinion, that Cyril the patriarch of
Alexandria, was not quite exempt from blame in that horrid deed.
Hero the younger was the instructor of Proclus in the mathematical
sciences at Alexandria. He was the author of a work on Mensuration,
entitled Geodæsia, and another on Mechanics, both of which were
published in Latin, at Venice, in 1572. In the Geodesia, there is
given the method of finding the area of a plane triangle in terms of the
sides of the triangle.
Diocles, his cotemporary, discovered the generation of the curve
called the cissoid, which still bears his name, and was applied by him in
finding two mean proportionals between two given lines. Another
solution of this problem was given by Sporus, who lived about the
same time.
We now come down to the latter times of the Greek school of
science and philosophy at Alexandria, which city seems to have been
the chief place of refuge for the Grecian sciences. Proclus was born,
A.D. 412, at Byzantium, and died at Athens at the age of 75 years.
His parents, Patricius and Marcella, were both of Lycian origin, and
are spoken of by Marinus as excelling in virtue. Proclus studied first
at Alexandria under the most eminent Platonic philosophers. He fre-
INTRODUCTION.
xix
quented the discourses of Olympiodorus, for the purpose of learning
the doctrines of Aristotle; and in mathematical science he gave himself
up to Hero, whose constant companion he became. He next studied
at Athens, where he was the pupil of the celebrated Syrianus, and at
length became the chief of the Platonic school established in that city.
Of his numerous writings on the Mathematical sciences, his commen-
tary on the first book of Euclid's Elements of Geometry is still extant in
the original Greek. It was translated into English by Thomas Taylor, in
1788, and inscribed "To the Sacred Majesty of Truth." The commen-
tary of Proclus, though tinged with much of the mysticism of the later
Platonic school, contains some interesting facts relating to the history
of Geometry, and many judicious remarks on the definitions, postulates,
axioms, and propositions of the first book of the Elements. From a
remark at the end, it appears to have been the intention of Proclus, if
he had lived, to write commentaries on the other books of Euclid, in a
similar style. There are extant two other mathematical works ascribed
to him: one is a small treatise on the sphere, which was published in
1620 by Bainbridge, the professor of Astronomy at Oxford: the other is
a compendium of Ptolemy's Almagest, entitled Hypotyposis. The
original Greek was published in 1540, and a Latin translation by Valla
in the following year. Proclus also wrote on many other subjects
commentaries on several dialogues of Plato, of which some are still
extant; lectures on Aristotle, and a commentary on the writings of
Homer and Hesiod. Four Hymns, one to the Sun, one to the Muses,
and two to Venus, are attributed to him. He also wrote on Providence
and Fate, and concerning the existence of Evil; besides numerous other
pieces, of which we may mention his eighteen arguments against Chris-
tianity. These arguments, except the first, are all preserved in the
answer of Philoponus. The Greek was published at Venice in 1535,
and a Latin version at Lyons in 1557.
Marinus of Naples was a disciple of Proclus, and his successor in
the school at Athens. There is a commentary still extant, on Euclid's
Data, of which he was the author: his other writings on mathematical
subjects, though not numerous, have not survived. His most celebrated
work, however, is a life of his master Proclus: it was published in
1700, with a Latin version by Fabricius. Of the mathematicians who
lived about the middle of the sixth century, (the latest period of the
decline of Grecian science,) Eutocius may be regarded as the most
distinguished. He was a native of Ascalon in Palestine, and a disciple
of Isidorus, one of the architects who designed and built the church
at Constantinople, which is now called the Mosque of St Sophia. The
only works of Eutocius which have descended to modern times, are
two commentaries; one on the Sphere and Cylinder of Archimedes,
and the other on the first four books of the Conics of Apollonius Per-
gæus. The latter was published with the Oxford edition of that author
by Dr Halley, in 1710; and the former with the works of Archimedes
at Oxford, in 1792. The commentary on the Sphere and Cylinder con-
tains ten various methods of solving the celebrated Delian problem,
which are of little importance in the present state of mathematical
science. Besides elucidations of difficult passages of the two works,
these commentaries contain many useful observations on the historical
progress of the mathematical sciences.

J
C2
XX
INTRODUCTION.
We must not pass over in silence one writer who lived at the end
of the fifth century, and the early part of the sixth, and who was
almost the latest author of any eminence that wrote in the Latin
language. Boëthius was the most distinguished of the Romans for
his scientific writings; which, however, consisted chiefly of translations
and commentaries. He was a senator and consul in the reigns of Odo-
acer and Theodoric, and was put to death by order of Theodoric,
A. D. 526. He was educated at Athens; and his writings were numerous
on almost every branch of literature and science. He was the author
of a treatise on Arithmetic, and another on Geometry: of the latter
there is an ancient MS. copy preserved in the Library of Trinity
College, Cambridge. He also translated Euclid's Elements, and some
of the writings of Archimedes and Ptolemy. Boëthius, however, is
chiefly celebrated for his work entitled "De Consolatione Philosophiæ,”
which was much read in the middle ages, and has been translated into
almost all the European languages. An Anglo-Saxon version was made
by King Alfred; and ancient MS. copies exist in several public libraries:
it was printed at Oxford, in 1698.
The rise of the Mahommedan power in the seventh century, and
the rapid and desolating conquests which followed, hastened the ex-
inction of the Grecian sciences. In A. D. 640, the Mahommedans
invaded and conquered Egypt. The great Library of Alexandria,
which is said to have contained at that time many thousand volumes,
the writings of geometers, astronomers, and philosophers, was com-
mitted to the flames. As a justification of the act, the Khalif Omar
declared, that, "if they agreed with the Koran, they were useless, and
if they did not, they ought to be destroyed." The learned men who
were congregated at Alexandria for the cultivation of science and philo-
sophy, either fell by the swords of the conquerors, or escaped by flight,
and these carried with them some remains of the sciences. In some-
what more than a century after this event, the Arabians became the
most zealous patrons and cultivators of the science and philosophy of
the Greeks and Hindus. The rapid progress of the Mahommedan power,
both in the East and in the West, led to the foundation of a powerful
empire. The second Abbaside Khalif Almansur ascended the throne
A.D. 753, and shortly after, transferred the seat of his government from
Damascus to the newly-founded city of Bagdad. Haroun Alrashid,
the grandson of Almansur, before his accession to the Khalifat had
overrun the Greek provinces of Asia Minor, and penetrated as far as
the Hellespont. The reigns of Alrashid and his successor Almamun
displayed at Bagdad the highest degree of luxury and splendour, which
are depicted in many scenes of the famous tales of the Arabian Nights'
Entertainment. The Arabians became acquainted with the astronomical
and arithmetical science of the Hindus before they had any knowledge
of the writings of the Greek astronomers and mathematicians. It is
related in the preface to the Astronomical Tables of Ebn Aladami, that
in the reign of Almansur, in the 156th year of the Hegira, A.D. 773,
an Indian astronomer visited the court of the Khalif of Bagdad,
bringing with him astronomical tables, which, he affirmed, had been
computed by an Indian prince whose name was Phighar. The Khalif,
embracing the opportunity thus presented to him, commanded the book
to be translated into Arabic, and to be published for a guide to the
INTRODUCTION.
xxi
Arabians in matters pertaining to the stars: this task was committed
to Alfazari. An abridgment of these tables was made in the succeed-
ing age by Mohammed Ben Musa, under the patronage of Almamun
before his accession to the Khalifat. (Colebrooke). The same Moham-
med Ben Musa is recognized by the Arabians as the first who made
known the Indian Arithmetic and Algebra: his Treatise on Algebra,
the earliest written in Arabic, is still extant: a MS. copy is pre-
served in the Bodleian Library at Oxford, which was translated into
English by the late Dr Rosen, and published at the expense of the
Oriental Translation Fund. In this treatise the rules of the science
are given in prose, and their accuracy is established by geometrical
illustrations. In the Sanscrit treatises on Arithmetic and Algebra, the
rules are given in verse. Almansur ascended the throne of the Khalifs
A.D. 813, and it was the glory of his reign, that he invited learned men
from various countries for the introduction of science and literature
into his dominions. He founded a college at Bagdad, and appointed
Mesuc of Damascus, a famous Christian physician, its president. It was
here, under the auspices and encouragement of Almamun that Arabic
translations of Indian and Greek science commenced, which were con-
tinued in succeeding reigns. The few manuscripts of the mathematical
and philosophical writings of the Greeks which had escaped the general
ruin, were diligently sought for and translated into Arabic. Trans-
lations were made of the writings of Euclid, Archimedes, Apollonius,
and others: besides which, Arabic commentaries were written to eluci-
date and explain these writings. To the Arabians of Bagdad is due
the merit of preserving many writings, which, at present, are not
known to be extant in the original Greek. Almamun died at about
forty-eight years of age, after a reign of more than twenty years, A.D.
833. He is reported to have uttered the following ejaculation just
before his death, "O Thou who never diest, have mercy on me, a
dying man!"
The celebrated Alkindi 'is mentioned among the mathematicians
and astronomers of the age of Almamun: his medical writings, which
are still extant, prove that he sustained a very honourable rank among
Arabian physicians.
Alfragan was a celebrated astronomer, who flourished at the latter
part of the eighth century; he was a native of Fargan, in Samarcand.
In his Elements of Astronomy, a work which consists of thirty chapters,
he adopts Ptolemy's hypothesis, and frequently quotes from his writings.
Professor Golius, of Leyden, translated this treatise into Latin, with
notes on the first nine chapters, which were published with the original
Arabic, in 1669, after his death.
Albategni was an Arabian astronomer of the ninth century: Dr
Halley describes him as a man of great acuteness. His work, entitled
"The Science of the Stars," was founded on his own observations, com-
bined with those of Ptolemy. On his observations were founded the
famous Alphonsine Tables. He died A.D. 888; his work was first
printed in 1537.
Albumazar was a physician and astronomer of the ninth century.
His Introductio ad Astronomiam was printed in 1489, and his work
"De magnis conjunctionibus, annorum revolutionibus ac eorum per-
fectionibus," in 1526, at Venice.
Rhazes was a most distinguished physician, and received the appel-
xxii
INTRODUCTION.
lation of the experimenter. He is also said to have been profoundly
skilled in astronomy and other sciences: he had the reputation of being
a skilful alchemist, and of having been the first to use chemical prepa-
rations in medicine. Many of his numerous works have come down to
us, and some of them have been translated and printed.
Honain was an Arabian physician and a Christian, who lived in the
ninth century. He is reported to have travelled into Greece and
Persia, and afterwards to have settled at Bagdad, where he translated
the Elements of Euclid, and the writings of Hippocrates, into Arabic.
Thabet Ben Korrah lived during the latter part of the ninth and
the early part of the tenth century. Ebn Khallikan relates, in his Bio-
graphical Dictionary, which has been translated into English, that
Thabet Ben Korrah left Harran, and established himself at Kafratutha,
where he remained, till Abu Jafar Mohammed Ben Musa arrived there,
on his return from the Greek dominions, to Bagdad. He became ac-
quainted with Thabet, and on seeing his skill and sagacity, invited him
to Bagdad, made him lodge at his own house there, introduced him to
the Khalif Almotaded, and procured him an appointment amongst the
astronomers. Thabet became secretary to the Khalif, and was dis-
tinguished for his skill in the mathematics and astronomy. To him is
attributed a translation of the Conics of Apollonius, and the Almagest
of Ptolemy.
In the tenth century several treatises on Geometry are ascribed to
Mohammed Bagdedin, among which is one "On the Division of Sur-
faces;" it was translated into Latin, by Commandine, and published in
1570; it was also translated into English, by John Dee. This treatise
on the Division of Surfaces is by some attributed to Euclid.
Avicenna, who lived in the tenth century, has been accounted the
prince of Arabian physicians and philosophers. He was a most volu-
minous writer; his greatest work was an Encyclopædia, in twenty
volumes, with the title of "Utility of Utilities." He is reported to
have written on almost every subject of physics, metaphysics, and the
mathematical sciences. Among his mathematical works was an abridg-
ment of Euclid's Elements of Geometry. Many of his writings are
extant, and some of them have been printed at Venice.
About the middle of the eleventh century, Diophantus was trans-
lated into Arabic, by Abulwafa Buzjani. In the twelfth century,
Nassir Eddin, a Mahommedan, acquired the highest reputation in all
branches of literature and science. He translated the Elements of
Euclid into Arabic, which, with his commentary, was printed at Rome,
in 1594, under the patronage of the Medici. He is also reported to
have written commentaries on the Spherics of Theodosius and Mene-
laus.
The Kholasat al Hisab, is a compendium of Arithmetic and Geo-
metry, composed by Baha Eddin, who died A.D. 1575. The Arabic
text, and a Persian commentary written by Roshan Ali, were printed
at Calcutta, in 1812. In this treatise, Baha Eddin remarks, that
"Learned Hindus have invented the well-known nine figures for
them;" meaning the Arabians.
The Arabian mathematicians are not noted for any very important
discoveries or improvements in Geometry. There is, however, one
improvement in Trigonometry of considerable importance, to which
they have an undoubted claim. The Greeks in their astronomical cal-
INTRODUCTION.
xxill
culations employed the chords of arcs; but the Arabians introduced
the sines, or half the chord of the whole arc, which led to greater sim-
plicity in calculation.
During the middle ages but few names have come down to us of
men who were skilled in the mathematical sciences. Beda, commonly
called the Venerable Bede, was born A. D. 672, near Wearmouth, in the
bishopric of Durham; he was the first of the Anglo-Saxon historians.
His Ecclesiastical History was completed in a. D. 731: his writings
are numerous, and the first collection of them was printed at Paris, in
1544.
Athelard or Adelard, was a monk of Bath, and the first who made
known in England, Euclid's Elements of Geometry. This he did by
making a translation from the Arabic into Latin, long before any Greek
copies of Euclid were known. He is said to have travelled into France,
Germany, Italy, Spain, Egypt and Arabia, to increase his knowledge.
He also wrote numerous works; some MS. copies of which are said by
Vossius to be in some of the College Libraries, at Oxford. The works
of Ptolemy, the astronomer of Alexandria, became known to the learned
of these times through the same language as the works of Euclid.
A Treatise on the Sphere, by John de Sacro Bosco, or John of
Holywood, was first put forth in the early part of the thirteenth cen-
tury. It is asserted by Montucla (1. p. 506), to be only an abridgment
of Ptolemy. MS. copies of it exist, and it has been printed several
times; there is also a commentary on this treatise, written by Clavius.
Roger Bacon, an English monk of the Franciscan order, was born
near Ilchester, in Somersetshire, in 1214. He studied first at Oxford,
and afterwards at Paris, and became the most celebrated philosopher of
his age. Of his writings, which are numerous, some only have been
printed; his tract on Chronology has not been printed. By his great
skill in astronomy, he discovered the error which gave occasion for the
reformation of the Calendar, and his plan was afterwards followed by
Pope Gregory XIII, with this variation, that Bacon would have had the
correction to begin from the birth of our Saviour; whereas Gregory's
amendment reached no higher than the Nicene Council.
His great.
knowledge of the sciences in an ignorant age, no doubt gave rise to the
story of his having dealings with the devil.
John de Basingstoke, who died A. D. 1252, is reported to have intro-
duced into England the knowledge of the Greek numerals. His merits
and learning recommended him to the favour of Robert Grossetete,
then bishop of Lincoln, who is reported by Roger Bacon to have
excelled in Geometry and the other mathematical sciences, and to have
spent many years in the study of them.
Grossetete wrote on many subjects, and some of his pieces are still
extant. He is reported to have studied first at Cambridge, afterwards
at Oxford, and lastly at Paris; he was made Bishop of Lincoln in
1235.
About the year 1261 another translation of Euclid's Elements of
Geometry from the Arabic was made into Latin, by Campanus of
Novara, who also wrote a Commentary on Euclid. It was printed at
Venice, in 1482, without a title-page, by Erhardus Ratdolt, and was
the first printed edition of the Elements in Latin. The translation of
Campanus has been supposed by some to be only a revision of the
translation which had been made by Adelard. It is possible he might
xxiv
INTRODUCTION.
have revised and improved Adelard's translation, and annexed his Com-
mentary; but this could only be determined by a collation of early
MS. copies of the two versions. Campanus was the author of a treatise
on the Quadrature of the Circle, which has been printed, in the Ap-
pendix to the Margarita Filosofica.
In the thirteenth century, Vitello, a native of Poland, displayed an
intimate knowledge of the geometrical writings of the Arabians and
Greeks. He is chiefly celebrated for the treatise on Optics, which he
composed from the writings of the Greeks and Arabians on that
science. It was printed in 1572: he also translated the Spherics of
Theodosius.
As geometers in the fourteenth century, Chaucer and Wallingfort
may be mentioned. Chaucer, the father of English poetry, was born
in London, A.D. 1328, and died at the age of seventy-two. He is said
to have been learned in all the sciences. He was the author of a
treatise on the Astrolabe, in which he describes the methods of astro-
nomical observation known in his day: it has been printed.
Richard Suisset or Swineshead wrote a work called "the Calcu-
lator." It is highly commended by Cardan, and has been printed.
Thomas Bradwardine, Archbishop of Canterbury, in the time of
Edward III. was the author of a work entitled "Geometria Speculativa
cum Arithmetica Speculativa." He also wrote "de Proportionibus,"
and "de Quadratura Circuli." Bradwardine was the most distinguished
geometrician of his time. He died A.D. 1348: his work on Geometry
was printed in 1495.
George of Purbach, commonly called Purbach, was born in 1423,
at a town of that name on the confines of Bavaria and Austria, and
became the most eminent astronomer and mathematician of his time.
He amended the Latin translation of Ptolemy's Almagest, which had
been made, not from the Greek original, but from the Arabic trans-
lation. He also corrected, by means of the Greek text, the translation
of Archimedes, made by Gerrard of Cremona, and wrote commentaries
on those books of Archimedes which Eutocius had omitted. He trans-
lated the Conics of Apollonius, and made a Latin version of the Spherics
of Theodosius and Menelaus; and of the book of Serenus on Cylinders.
He composed an introduction to Arithmetic, and a treatise on Dialling
and Gnomonics. He also made very great improvements in Trigo-
nometry by introducing the table of Sines, and a new division of the
radius into 600,000 instead of 60 equal parts. He thus completely
changed the appearance of that science, so important in Astronomy.
John Muller, or, as he was called, Regiomontanus, from the Latin-
ized name of his native town, Konigsberg, died at Rome, in 1476, at
the early age of forty years: he had been invited thither by Pope
Sixtus IV. to assist in rectifying the Calendar. Muller was the disciple
and friend of Purbach; after whose premature death, he revised and
completed, at Rome, the Latin version of Ptolemy, which Purbach had
left unfinished; he also further improved Purbach's division of the
radius. He rejected the sexagesimal subdivision of the radius, and
made it to consist of 1,000,000 equal parts. With this new division,
he computed the sines of arcs, to every minute of the quadrant, to
seven places of figures; he also annexed a table of secants. His
celebrated treatise on Triangles, in five books, was not published in
his lifetime, and did not appear till 1533, when it was edited and
INTRODUCTION.
XXV
"
published by Schener. The solutions of the more difficult cases of
plane and spherical triangles are to be found in his work; and, with
the exception of what Spherical Trigonometry owes to Napier, that
science may be said to have made but small advances for more than two
centuries after the age of Muller. In his work on Triangles he gave a
table for finding the angle of a right-angled triangle, from the base and
perpendicular, without knowing the hypothenuse. This table, which
he styled "Canon Fœcundus," was calculated for every degree of the
quadrant, and was, in reality, a table of tangents, which he was the first
to introduce into the science of Trigonometry. He added also many
new theorems to that science; and after him few improvements were
made in it till the time of Euler. His fifth book contains numerous
problems concerning rectilinear triangles, of which some are solved by
Algebra. The tables of Regiomontanus were printed in 1490.
The revival of ancient literature in Europe, about the middle of the
fifteenth century, contributed to bring the mathematical writings of
the Greeks into notice; and the discovery of the art of printing, about
the same time, was the commencement of a new era in literature and
science. The writings of the ancients were now no longer confined in
MS. to the religious houses, and to the few who had the means of
purchasing copies of the manuscripts. The fall of the Eastern empire,
and the capture of Constantinople by the Turks, in 1458, drove
many Greeks to seek their safety and subsistence in Italy and other
parts of Europe. These became living instructors in the Greek lan-
guage, and very much facilitated its revival in Italy. As the literature
of the Greeks became known, their mathematical writings also attracted
notice they were printed and translated, and published with commen-
taries. Restorations also of lost treatises were attempted by mathe-
maticians of that and following centuries.
Mr Hallam in a note, p. 157, Vol. 1. of his History of the Middle
Ages, remarks, "It may be considered a proof of the attention paid to
Geometry in England, that two books of Euclid were read at Oxford,
about the middle of the fifteenth century." With respect to the mathe-
matical science of the middle ages, though we do not find any original
writers who made additions to it by their discoveries; there must, in
justice, be conceded to the men of those times, the possession of no
mean or scanty knowledge of Geometry. The splendid ecclesiastical
buildings of the middle ages, which are still standing both in Great
Britain and in the south and west of Europe, evince in their structure
a practical knowledge, at least, of some of the most difficult problems
of Geometry and the science of Equilibrium.
Lucas de Burgo's work, entitled "Summa de Arithmetica, Geo-
metriâ, &c." was published in 1494. Fifteen years later he published
the Latin translation of Euclid, which had been made from the Arabic
by Campanus.
Copernik, or (as he is commonly called) Copernicus, was born at
Thorn in Prussia, in 1473, and his name is immortalized by his dis-
covery of the true Solar system. The motion of the sphere, of the fixed
stars with the sun and moon round the earth, in twenty-four hours,
appeared to him too complex; and he felt persuaded that such could not
be the system of nature. The Pythagoreans and some later philo-
sophers had held the rotation of the earth round its own axis, and its
revolution round the sun. Copernicus collected the writings of pre-
xxvi
INTRODUCTION,
ceding astronomers, and examined all the hypotheses they had devised,
for the explanation of the phænomena of the heavens: the result of his
labours was his great work, "De Revolutionibus Orbium Cœlestium,"
Libri VI. It was completed in 1530, but was not published till a
very few days before his death in 1543. In the title-page is quoted
the admonition of Plato, ἀγεωμέτρητος οὐδεὶς εἰσίτω. Copernicus was
the author of a tract on Plane and Spherical Trigonometry, which
contained a Table of Sines. It was first printed at Nuremberg, and
afterwards at the end of the first book of his great work, De Revolu-
tionibus, &c. The publication of this work was superintended by
George Joachim Rhæticus, the disciple and latterly the assistant of
Copernicus, in his astronomical labours. He was born in 1514, and
died in 1576 at Feldkirk in the Tyrol. With the view of making
astronomical calculations more accurate, he commenced a table of
sines, tangents, and secants for every ten seconds of the quadrant,
to fifteen places of figures; which he did not live to complete. This
work was completed and published by his disciple Valentine Otho in
1596. The table of sines for every ten seconds, and for every second
in the first and last degrees of the quadrant, which he had completed,
was published in 1613 by Pitiscus, who extended the value of some of
the latter sines to twenty-two places of figures.
Nicholas Tartaglia was a celebrated mathematician, born at Brescia
in 1479. He was the original discoverer of the solution of Cubic Equa-
tions, which he first effected in 1530, and which Cardan, who has
generally had the merit of the discovery, surreptitiously obtained from
him. The treatise of Tartaglia on the Theory and Practice of Gun-
nery, is the earliest which treated of the motion of projectiles. He also
published an edition of Euclid's Elements, in Italian, with a Commen-
tary. The last part of his great work, "Trattato de Numeri et Misure,"
was published in 1558.
John Werner of Nuremberg, one of the most distinguished astrono-
mers and geometers of his day, was born in 1468 and died in 1528: he
was the first who attempted to restore the geometrical analysis of the
ancient Greeks. In his "Opera Mathematica," which he published in
1522, will be found what he effected, both in the Conic Sections and in
some solid problems. He also wrote a work on Triangles.
Zamberti made the first translation of Euclid's Elements, from the
original Greek into Latin, which he published at Venice in 1505. The
original Greek was first published at Basle in 1533, edited by Simon
Grynæus. This edition of Euclid was the foundation of Commandine's
translation in 1572, as it has been, in great measure, that of later
editions.
During the sixteenth century Euclid was held in so high estimation
among mathematicians, that no attempts appear to have been made to
advance the science of Geometry, beyond the point at which he left it.
Commentaries and translations seem to have been almost all they at-
tempted.
Erasmus Reinhold wrote commentaries on Euclid's Elements and
the great work of Copernicus. He was born at Salfeldt in Thuringia
in 1511, and died in 1553, and was considered one of the most eminent
mathematicians of his time. He wrote on Plane and Spherical Tri-
angles, improved Muller's Tables of Sines and Tangents, and was the
author of numerous other works on the sciences.
INTRODUCTION.
xxvii
•
F. Maurolyco, of Messina, was born in 1494, and died in 1575; he
made some improvements in Trigonometry, and edited the Spherics of
Theodosius and Autolycus; he also published his "Emendatio et Res-
titutio Conicorum Apollonii Pergæi,” in 1575.
Frederick Commandine was born at Urbino in 1509, and died in
1575: he is justly accounted one of the first geometers of his age.
He composed several original works on the sciences; but is chiefly
known for his translations of several of the Greek geometers, of whose
works some had not previously been translated into Latin. He trans-
lated the geometrical writings of Archimedes, and wrote a com-
mentary upon them; also four books of the Conics of Apollonius
Pergæus, with the Lemmas of Pappus and the Commentaries of Euto-
cius, together with Serenus on Sections of the Cone and Cylinder. He
was the first who translated the last six books of the Mathematical Col-
lections of Pappus into Latin, which were published after his death, in
1588, by the munificence of the duke of Urbino. The edition most
commonly met with is that of Manolessius, printed in 1660. This
translation of Pappus first directed the attention of mathematicians to
the subjects of the lost works of the ancient geometers, and gave rise
to various attempts for the restoration of several, which are described
in the preface to the seventh book. Commandine's translation of
Euclid with a commentary, was put forth in 1572. An Italian version
was published under his direction in 1575. An English translation of
the Latin version was published in 1715, by Dr John Kiel, at that time
Savilian Professor of Astronomy at Oxford. Commandine also trans-
lated into Latin, an Arabic version of Euclid's tract on the Division of
Surfaces.
To this period belongs John Dee, a man distinguished for his
mathematical and astrological knowledge. He was educated at St
John's College, Cambridge, where he chiefly devoted his attention to
mathematical studies. He was made one of the fellows of Trinity
College, at its foundation by Henry VIII., and afterwards, as we learn
from Lilly's Memoirs, read lectures on Euclid at Rheims. He wrote a
learned preface, of fifty folio pages, to Billingsley's Euclid: it bears the
date of February 9, 1570, and was written at Mortlake. He translated
into English the tract on Division of Surfaces, from the Latin version
of Commandine.
1
To Henry Billingsley, a citizen of London, is due the merit of
making the first English translation of Euclid's Elements of Geometry.
It comprises the whole of the 13 books, with the 14th and 15th which
were added by Hypsicles, and a 16th by Flussas. It was chiefly made
from the Latin of Campanus, and contains a commentary, besides the
preface, by John Dee, above-mentioned. It was first published in
1570 in a large folio volume; and a second edition was edited by Leeke
and Serle in 1661.
Francis Vieta was born at Fontenoy in Lower Poitou in 1540, and
was a man of original genius, as is manifest from his discoveries and
improvements in different branches of the mathematical sciences.
He introduced the use of letters into Algebra, and invented many
theorems. He effected great improvements in Geometry, made con-
siderable additions to the science of Trigonometry, and reduced it to a
system. He wrote a treatise on Angular Sections, and restored the
tract of Apollonius on Tangencies, which he published with the title of
xxviii
INTRODUCTION.
Apollonius Gallus. His collected works were published at Leyden by
Schooten.
Galileo Galilei, the cotemporary of Milton and friend of Kepler,
was born at Pisa in Tuscany in 1564, and at a very early age, gave
evidence of great genius for geometrical and philosophical pursuits.
This became more evident while he was under the direction of Guido
Ubaldi at the University of Pisa. From the time of Archimedes, a
period of nearly 2000 years, little or nothing had been done in Mechani-
cal Geometry, till Galileo first extended the bounds of that science by
the application of Geometry to Motion. He first taught the true theory
of uniformly accelerated and retarded motions, and of their composition,
and proved that the spaces described by heavy bodies, falling freely
from the beginning of their motion, are as the squares of the times.
Contrary to the general belief, he maintained that all bodies, whether
light or heavy, fall to the earth, through the same space in equal times;
and attempted to verify the truth of his proposition by experiment. The
two bodies, however, which he let fall from the top of the hanging tower
of Pisa, did not reach the ground exactly at the same instant. The reason
Galileo assigned was that the resistance of the air retarded the lighter
body more than the heavier. He invented the cycloid, and the simple
pendulum which he used in his astronomical experiments. The applica-
tion of the pendulum to clocks was made by his son; and subsequently
brought to perfection by Huygens. Galileo first proved that a body pro-
jected in any direction not perpendicular to the horizon describes a para-
bola; and it may be remarked that the Geometry of Galileo was wholly
applied to explain and advance the science of Motion. The invention, in
1609, of the refracting telescope which still bears his name, disclosed new
views of the solar system. By the aid of this he discovered that the
moon is an opaque body with mountains and vallies, and that she
receives her light by reflection from the sun. He also put forth the
conjecture that the moon might be an inhabited world like the earth.
He observed the different phases of the planet Venus, proving her motion
round the sun. He discovered four of Jupiter's satellites, and caught
an imperfect view of the ring of Saturn, which at the time of his obser-
vation appeared like two small stars, one on each side of the planet's
disk. He was the first who discovered spots on the sun's disk, and
from their varying position he inferred the motion of the sun on its axis.
By his telescope he also discovered that the whiteness of the Milky
Way is caused by innumerable stars apparently more close together
than in the other parts of the heavens. His celebrated work, the "Dia-
logues on the Ptolemaic and Copernican Systems," was published at
Florence, in 1632; and though dedicated to Ferdinand II., brought
Galileo under the hatred of the Jesuits, and the power of the Inqui-
sition. In June 1632, that court condemned him of heresy, for teaching
that the sun is the centre of the solar system, and that the earth re-
volves on its own axis, and moves round the sun. He was obliged, on
his knees, to abjure his belief of all he had advanced in his Dialogues,
and to swear, that for the future, he would never assert or write any
thing in favour of such heretical opinions. He was sentenced to im-
prisonment during the pleasure of the court; and for a certain time to
recite daily the seven penitential psalms. Galileo is reported to have
whispered in the ears of a friend, as he rose from his knees, “E pur
se muove."
INTRODUCTION.
xxix
The influence of his powerful friends no doubt moderated the
sentence of the Inquisition, whose proceedings in what they called
heresy, were always of the most cruel, frequently of the most horrid
description. Galileo was nevertheless kept strictly confined in the
prison of the Inquisition for two years. Even when upwards of 70
years of age, new rigours were exercised against him, on account of
some fresh suspicions of pope Urban VIII., which were inflamed by
the philosopher's inveterate foes, the Jesuits. His health greatly suf-
fered, and he was afterwards released from confinement; but became
blind some years before his death. During this period he finished his
dialogues on Motion. His death took place in 1642, at the age of 78. Most
of the works of Galileo were collected and printed in 1656; they were
translated into English by T. Salusbury, and published in 1661, in his
Mathematical Collections. A more complete collection of his works
was published at Milan, in 1811. At Florence, in 1674, was published
a work of Galileo, under the title of "Quinto Libro de gli Elementi
d'Euclidi, &c.," by Vincenzio Viviani, one of his distinguished pupils.
Christopher Clavius was born at Bamberg in Germany, in 1537,
and died at Rome, in 1612: his writings on the mathematical sciences
were collected and printed at Mayence, in five folio volume, in 1612:
he superintended the reformation of the Calendar under the direction
of pope Gregory XIII. He was skilled in the ancient Geometry, and
edited several of the Greek mathematical writings, and on some of
them he wrote commentaries, among which may be mentioned the
Elements of Euclid. He was the author of a work on Practical Geo-
metry, and of a commentary on Sacro Bosco's treatise on the Sphere,
which works were first published in 1570. He was a zealous cultivator
of the sciences; but no discoveries or improvements are attributed to
him.
Willebrod Snell was a man of original genius: he was born at
Leyden, in 1591, and died in 1626. To him is attributed by Huygens
the discovery of the law of the refraction of light, before it was made
known by Descartes. He was a skilful geometer, and published in
1608, with the title of Apollonius Batavus, his attempted restoration of
three tracts of Apollonius "De Sectione Determinata, "De Sectione
Rationis," and "De Sectione Spatii." The tract "De Sectione Deter-
minata" was translated into English by the Rev. J. Lawson, and pub-
lished in 1772. This tract is imperfectly restored, as there is omitted
the distinction of the situation of the points; and there is no complete
exposition of the determinations. He also wrote a tract on the Circle ;
in which are given various approximations, both geometrical and arith-
metical. Ludolph Van Ceulen is also noted for having calculated the
ratio of the diameter to the circumference of the circle, to thirty-five
places of decimals.
Sir Henry Savile, an accomplished scholar, founded two professorships
at Oxford; one of Geometry, and the other of Astronomy, in 1619.
As the first Savilian professor of Geometry, he delivered thirteen lec-
tures on the first book of Euclid's Elements of Geometry, in 1620,
which he published in Latin during the following year. In 1585 he
was appointed Warden of Merton College, and Provost of Eton in 1596;
the former appointment he held for thirty-six years. In 1613 he pub-
lished the works of Chrysostom in Greek, and was the author and
editor of several other works.
XXX
INTRODUCTION.
1
Leonard Digges was a mathematician of some note, and the author
of a treatise on Geometry, in three books, which he called Pantometria.
It was published by his son in 1591, with a supplementary book on the
five regular solids.
The ancient Geometry of the Greeks was considered perfect, as
indeed within the bounds of its legitimate province it is; and no
attempts were made to improve or extend the methods handed down
from the ancients till the time of Kepler. He was born at Wiel, in
1571, and died in 1630: he introduced the new principle of infinity
into Geometry. He conceived a circle to be made up of an infinite
number of infinitely small triangles with their vertices in the centre,
and their bases coinciding with the circumference of the circle. A
cone, in the same manner, was supposed to consist of an infinite number
of indefinitely small pyramids. This idea of Kepler lies at the foun-
dation of the higher analysis. He published his views in a work
entitled "Nova Stereometria," in 1615, which have been discussed
under the names of "Infinitesimals ;" "Fluxions and Fluents;" "Dif-
ferential and Integral Calculus."
Pierre de Fermat was born in 1595, and died in 1665; though a
person of extraordinary vanity, he was a mathematician of original
genius. He attempted the restoration of the two books of Apollonius
on Plane Loci: he has given the synthesis, but omitted the analysis of
the propositions; he has also omitted the distinction of cases of each
proposition, and has not ascertained the determinations. He wrote a
treatise on Spherical Tangencies, in which he has demonstrated, in the
case of spheres, properties analogous to those which Vieta had before
demonstrated in his restoration of Apollonius on Tangencies. He was
the author of a treatise on Geometric Loci, both plane and solid. Fer-
mat had acquired some general notion of the Porisms of Euclid. He
was the author of a method of maxima and minima, and of the qua-
drature of parabolas of all orders, besides several discoveries in the pro-
perties of numbers, one of which still bears the name of Fermat's
theorem. His collected works were, after his death, published at
Toulouse, in 1679.
Descartes was born in 1596, and died at Stockholm in 1650. He was
the cotemporary of Galileo, Fermat, Roberval, and many other celebrated
mathematicians. He has been cited as the inventor of the New Geo-
metry, or, as it is called, Analytical Geometry; the foundation of
which, however, had certainly been laid before his time. Algebra had
been applied to Geometry by Vieta, and, to some extent, by other
mathematicians. But though he did not originate, he certainly ex-
tended the limits and powers of Analytical Geometry, by the discovery
of a new principle. The use of co-ordinates appears for the first time,
though under a different name, in the second book of his Geometry,
which was published in 1637. He first taught the method of ex-
pressing curves by equations. The simple conception of expressing
curve lines by means of equations between the two variable co-ordi-
nates of a point in the curve; and curve surfaces by means of equations
which contain the three co-ordinates of any variable point in the sur-
face, has led to a new science, and entitles the discoverer to be classed
amongst men of profound genius. This discovery and its applications
by means of the higher analysis, have given a power to Geometry before
unknown. In the Epistles of Descartes, which were printed in 1683, hẹ
INTRODUCTION,
xxxi
remarks (Part II. Ep. 72), "In searching out the solution of geo-
metrical questions, I always make use of lines parallel and perpen-
dicular as much as possible: and I consider no other theorems than
the two following; the sides of similar triangles are proportionals;
and in right-angled triangles the square of the hypothenuse is equal to
the squares of the two sides. And I am not afraid to suppose several
unknown quantities, that I may reduce the proposed equation to such
terms as that it may depend on no other theorems than these two."
Buonaventura Cavalieri, better known by the Latinized appellation,
Cavalerius, was born at Milan in 1598. He was a pupil of the cele-
brated Galileo, and became Professor of Mathematics at Bologna; he
died in 1647. He was the author of several works on the mathematical
sciences, the most important of which is a treatise on Indivisibles in
seven books, which he put forth in 1635: it is important as be-
ing one of the first attempts to extend the powers of the ancient
Geometry. He conceived a line to be made up of an infinite number
of points; a surface to be formed of an infinite number of such lines;
and a solid to be composed of an infinite number of such surfaces.
These elements of geometrical magnitudes he named Indivisibles:
and the principle he assumed in the application of these assumptions
was, that the ratio of the infinite sums of lines, or of planes as com-
pared with the unit of surface or volume, was the same as that of
the surface or volume of which they were the measures. He shewed
that his new principle was, in effect, the same as the method of Ex-
haustions, but a more convenient mode of reasoning, being less tedious
and more direct. In the first six books he explains and applies his
theory of Indivisibles, and in the seventh book he proves the same
results by methods independent of Indivisibles; with the view of shew-
ing the agreement of the results, and the consequent truth of his new
principle. Guldin controverted and wrote against the doctrine of
Indivisibles, and was answered by Cavalerius in the third of his “Ex-
ercitationes Geometrica sex," which were published in 1647. This
work consists of exercises in the Method of Indivisibles, with answers
to Guldin's objections. The method of Cavalerius is not free from
error, as he applies the process of simplification at too early a stage
of the investigation, by which means the strict logic of the reasoning
is violated, while the correctness of the result is not affected.
Roberval was born in 1602, and adopted the theory of Cavalerius,
but improved his modes of expression. His Traité des Indivisibles
was published in 1693, after his death, and contains his application
of the new principles. His method of drawing tangents was an ap-
proximation to that applied afterwards by the principles of Fluxions
and the Differential Calculus.
Albert Girard was a Fleming who displayed great genius in the
Mathematics. He was the first who announced the restoration of the
three books of the Porisms of Euclid, in a work on Trigonometry,
which was printed at the Hague in 1629. To what extent he suc-
ceeded is not known, as the results of his labours have never been
published. To him are due some general theorems for the measuring
of solid angles.
Marinus Ghetaldus was a distinguished geometer at the beginning
of the seventeenth century: he was the author of several works on the
ancient Geometry: he attempted, from what could be gathered from
xxxii
INTRODUCTION.
Pappus, a restoration of the lost book of Apollonius on Inclinations,
and published it in 1607, with the title of Apollonius Redivivus. In
the same year also he put forth a supplement to the Apollonius Gallus
of Vieta, and a collection of problems. Among his other writings
may be mentioned his Archimedes Promotus, which was first published
in 1603.
Blaise Pascal was born at Clermont in 1623, and at an early age
gave proofs of extraordinary ability. He was of an enquiring mind,
desirous of knowing the reasons of every thing. It is reported of him
that whenever he could not obtain from others sound reasons, he used
to seek them out for himself; never giving his assent except on con-
viction. Pascal challenged the mathematicians of his day to prove
some properties of the cycloid; but as the answers he received from
Wallis and other eminent men were unsatisfactory, he himself gave
the complete proofs of all the properties mentioned in his challenge.
He invented the arithmetical machine which bears his name, and was
the author of some pieces on other mathematical subjects. His intense
application to study injured his health. He became a Jansenist, and
retired to the Abbey of Port Royal, near Paris, where he composed
his Provincial Letters, and wrote his thoughts on Religion and other
subjects. These were published after his death, which happened when
he was only thirty-nine years of age. His works were collected and
published at Paris in 1779.
Schooten was Professor of Mathematics at Leyden. In 1649 he
published an edition of Descartes' Geometry, and in 1657 an original
work entitled Exercitationes Mathematica. He attempted the restora-
tion of the Loci Plani of Apollonius. He has given the synthesis
only, omitting the analysis, except in a few instances. He has also
omitted the determinations and the distinction of cases; and in the
preface he acknowledges that his attempted restoration was designed
to be an illustration of the Geometry of Descartes, by furnishing ap-
propriate examples to his method.
Christian Huygens was born at the Hague in 1629, where he
died in 1696. He was one of the most ingenious mathematicians of
his age. He was the author of two treatises, one entitled "Theore-
mata de Quadratura Hyperbolæ, Ellipsis et Circuli, ex dato portionum
gravitatis centro;" and the other, "De Circuli magnitudine inventa;
which, at the time of their publication, were highly esteemed. He
was also the author of some other pieces on Geometry, which were
published at Paris in 1693, and he discovered the theory of Evolutes.
His studies were not confined merely to the speculative portion of the
mathematical sciences, but were extended to questions of practical
utility. Among them may be named his improvements in telescopes,
and his method of rendering the oscillations of pendulums isochronous.
He was the discoverer of Saturn's ring, and a third satellite of that
planet. When nearly sixty years of age he read and admitted the
theory of centripetal forces, and the gravitation of the planets to the
sun, which had been proved in Newton's Principia. His writings are
numerous.
Dr Isaac_Barrow was a distinguished scholar and geometer, and
the tutor of Isaac Newton when an undergraduate at Trinity College.
Dr Barrow was appointed Greek Professor at Cambridge in 1660, and
Gresham Professor of Geometry in 1662: the latter office he resigned
INTRODUCTION.
xxxiii
on his appointment to the new Professorship, founded at Cambridge b
Mr Lucas, in 1663. This also he resigned, in favour of Mr Newton, in
1669. He became Master of Trinity College in 1672, and died in 1677
at the age of 47 years. In 1655 he edited, in Latin, the thirteen books
of Euclid's Elements of Geometry, which were translated into English
and published in 1660. He also put forth Euclid's Data in 1657. His
Lectiones Geometrica were published in 1670, and translated into
English, by Stone, in 1735. They contain his method of drawing tan-
gents to curves, which is similar to that by the method of fluxions or
the differential calculus: the difference being only in the notation. His
Lectiones Optica were published in 1669. He also edited Archimedes,
Apollonius and Theodosius. The Lectures which he delivered, as
Lucasian Professor, were published after his death in 1683, with the
title of Lectiones Mathematica, and were translated into English by
Kirkby, in 1734: they are confined to Euclid's Elements of Geometry.
He also applied the method of indivisibles to the propositions of
Archimedes on the Sphere and Cylinder. His treatise was printed in
1678.
Sir Isaac Newton was one of the greatest philosophers that ever
lived. The inscription on the pediment of the statue of Newton in
the chapel of Trinity College, "Qui genus humanum ingenio supera-
vit," records the unquestioned judgment of posterity. He was the
original inventor of the method of fluxions and fluents; in 1665, his
attention was first directed to the subject, and his method was completed
before 1669. The merit of the invention was claimed for Leibnitz,
who had put forth, in 1684, his view of the principles of infinitesimals
or differentials. The terms employed and the notations adopted by
these two great men were different; they agreed in the substance and
object of the theory, but there was a considerable difference in their
conception of the principles. An angry controversy arose respecting
these claims, and an appeal was at length made to the Royal Society,
by which a committee of enquiry was appointed. The result of their
enquiries and deliberations was a decision in favour of Newton. The
papers relating to this enquiry were printed in 1712, with the title of
Commercium Epistolicum de Analysi promota.
His discoveries in
optics, and his new theory of light and colours for the explanation of
optical phenomena, he did not publish till 1704: nearly thirty years
after his chief discovery in that science.
His greatest discovery, however, was that of universal gravitation.
His thoughts were first led to the subject in 1666, just after he had left
Cambridge for the country, on account of the Plague. As he was
sitting alone in a garden, some apples fell from a tree, and his thoughts
were led to the subject of gravity. He considered that as this power
is not found to be sensibly diminished at the remotest distance from
the surface of the earth to which we can rise, it seemed reasonable to
conclude that it must extend much further than is commonly believed.
He enquired: "Why not as high as the moon? and if so, her motion
must be influenced by the force of gravity: perhaps she is retained in
her orbit by it. Though the power of gravity is not sensibly lessened,
in the little change of distance at which we can place ourselves from
the surface of the earth, yet it is possible that at the height of the
moon this power may differ much in degree from what it is here.'
To make an estimate of the amount of this diminution, he considered,
d
xxxiv
INTRODUCTION.
that if the moon were retained in her orbit by the force of gravity, no
doubt the primary planets are carried round the sun by the like power:
and by comparing the periods of the several planets with their mean
distances from the sun, he found, that if any power like gravity held
them in their courses, its intensity must decrease inversely as the square
of their distances from the sun. He arrived at this conclusion, from con-
sidering them to move in circles concentric with the sun; from which
form the orbits of the greater part of them do not greatly differ. By
supposing therefore the force of gravity, when extended to the moon,
to decrease in the same manner, he computed whether that force would
be sufficient to keep the moon in her orbit. In this computation, taking
sixty miles to a degree, the result at which he arrived did not shew
the power of gravity to decrease as the inverse square of the moon's
distance from the earth: whence he concluded that some other cause
must, at least, combine with the action of gravity on the moon. Being
unable to satisfy himself respecting this, he laid aside, for that time, all
thought upon the subject. In the winter of 1676, he discovered the
two grand propositions, that by a centripetal force varying inversely as
the square of the distance, a planet must revolve in an ellipse about a
centre of force, placed in one of the foci; and by a radius vector drawn
to that focus, describe areas proportional to the times. These are
proved in the second and third sections of the first book of the Prin-
cipia. After the year 1679 his thoughts were again turned to the moon ;
and, by using in his computation the more accurate length of a degree,
he arrived at the conclusion, that that planet appeared, agreeably to his
former conjecture, to be retained in her orbit by the force of gravity,
varying as the inverse square of the distance. On this principle he
proved that the primary planets really moved in such orbits as Kepler
had supposed. He afterwards drew up about twelve propositions
relating to the motion of the primary planets round the sun, and sent
them, at the end of the year 1683, to the Royal Society. Soon after this
communication, Dr Halley became known to Newton, and having
learned from him that the proof of the propositions respecting the
primary planets was completed, earnestly solicited him to finish the
work. Accordingly the first edition was printed under the care of Dr
Halley, with the title of "Philosophiæ Naturalis Principia Mathema-
tica," and published in 1687.
The reader of the Principia cannot fail of perceiving that Sir Isaac
Newton was a most profound geometer. Newton and his cotemporary
Maclaurin were the first who applied the consideration of the degrees
of equations to the discovery of the general properties and characteristics
of curved lines, and to them and Cotes are due the discovery of their
most important general properties. Newton has given the results of
his investigations on this subject, in his "Enumeratio Linearum tertii
ordinis," which, with his tract on the quadrature of curves by the
method of fluxions and fluents, were first printed at the end of the first
edition of his Optics, in 1704. His principal object was the enumeration
of lines contained in an equation of the third degree between two
variables. He discovered seventy-two different species, and four more
were added by Stirling. In his "Arithmetica Universalis," he has
applied the method of Descartes to the solution of geometrical problems
and the construction of the roots of equations. The whole works of
Newton were edited by Dr Horsley in five quarto volumes, in 1779.

INTRODUCTION.
XXXV
Edmund Halley was one of the most eminent geometers and astro-
nomers of his age. In 1703 he succeeded Dr Wallis as Savilian Pro-
fessor of Geometry at Oxford. Soon after his appointment, he com-
menced a translation from the Arabic, of the treatise of Apollonius De
Sectione Rationis, and attempted the restoration of his two books De
Sectione Spatii, from the account given by Pappus in the seventh book
of his Mathematical Collections: the whole of these he published in
1706. The tract de Sectione Rationis, recovered from the Arabic, is a
complete specimen of the ancient method of analysis. He also pub-
lished at Oxford in 1710, an amended translation of seven books of the
Conics of Apollonius, and attempted a restoration of the eighth, which
was lost. This magnificent folio edition contains the first four books
in the original Greek, together with a Latin translation; the next
three in Latin from the Arabic version; and the last book in Latin as
restored by Dr Halley himself, together with the Lemmas of Pappus
and the Commentaries of Eutocius. Halley united a profound know-
ledge of the ancient geometry with the new geometry of Descartes, and
applied the latter to the construction of equations of the third and
fourth degrees by means of a parabola and a circle. His memoir on
this subject was published in the Philosophical Transactions for 1687.
At 63 years of age, he succeeded Flamsteed as Astronomer Royal at
Greenwich; and for eighteen years discharged the duties of that office
without an assistant. He died at the age of 86, in the year 1742.
name.
James Gregory was an eminent mathematician, the cotemporary
and correspondent of Newton, Huygens, Wallis, and others of that
time. He was the inventor of a reflecting telescope which bears his
He discovered a method of drawing tangents to curves, geo-
metrically. His "Geometria Pars Universalis" was first published
at Padua, in 1668, and his "Exercitationes Geometrica" in the same
year.
Dr David Gregory, the nephew of James, was chosen Savilian
Professor of Astronomy, in 1691, in preference to Dr Halley. In 1702
he published his chief work, entitled " Astronomia Physicæ et Geome-
trica Elementa," founded on the Newtonian hypothesis; and in the
following year, the works of Euclid, in Greek and Latin.
cc
Abraham Sharp, a skilful mathematician and expert mechanic, be-
came the amanuensis of Flamsteed, in 1688, and assisted him in his
"Historia Cælestis." He had the chief hand in constructing the mural
arc at the Greenwich Observatory. He published in 1717, Geometry
Improved, by A. S. Philomath; an elaborate treatise, and containing
solutions of many difficult problems: he died in 1742, at the age of 91.
Alexis Claude Clairaut was born in 1713, and died in 1765. To
Clairaut is due the merit of having exhibited methodically the doctrine
of three co-ordinates in space, applied to curve surfaces and the lines
which originate in their intersections. His celebrated treatise on this
subject was published in 1731, entitled “Traité des Courbes à double
Courbure." Before Clairaut, however, M. Pacent, in a memoir read
before the French Academy of Sciences, in 1700, had illustrated the
extension of the principle of Descartes by a curve surface, expressed by
an equation between three variables. Also John Bernouilli had ex-
pressed surfaces by an equation, involving three co-ordinates in his
solution of the problem, "To find the shortest line which can be drawn
on a surface between two given points."
xxxvi
INTRODUCTION.
The expression "curve of double curvature" arises from the con-
sideration, that such a curve partakes of the curvature of two plane
curves, which are in fact its projections on two co-ordinate planes. It
was first suggested by M. Pitot, who employed it in a memoir on the
spiral thread on a right cylinder, and which he read before the French
Academy of Sciences, in 1724.
Euler, in his "Introductio in Analysin Infinitorum," published in
1748, explains the general principles of the analytical theory of curves,
and in the extension of his investigations to geometry of three dimen-
sions, discusses the equation between three variables, which includes
surfaces of the second degree. The treatise of Cramer, first published
in 1750, with the title "Introduction à l'Analyse des lignes courbes
Algébraïques," is one of the most complete on this branch of geometry.
Edmund Stone published, in 1731, an edition of Euclid's Elements,
with an account of his life and writings, and a defence of the Elements
against modern objectors. In 1735 he translated and published Dr
Barrow's Geometrical Lectures. He wrote an account of two new
species of lines of the third order, not noticed by Newton, which was
published in the forty-first volume of the Philosophical Transactions of
the Royal Society, of which he was a fellow he was also the author
of a Mathematical Dictionary and some other works*.
Dr Robert Simson was a native of Ayrshire, and born in 1687. He
was appointed Professor of Mathematics at Glasgow in 1711, where he
continued to discharge his duties as an instructor for nearly fifty years.
During this period his attention was principally directed to the writings
of the ancient Greek geometers. His restoration of the Loci Plani and
the Determinate Section of Apollonius, and his treatise on the Porisms of
Euclid, entitle him to the high reputation he still holds as a geometer.
*The following account of Stone from Dr Hutton, may be cited as an example of
true genius overcoming all the disadvantages of birth, fortune, and education. Edmund
Stone was the son of a gardener of the Duke of Argyle. At eight years of age he was
taught to read; and at eighteen, without further assistance, he had made such advances
in mathematical knowledge as to be able to read the Principia of Newton. As the Duke
was one day walking in his garden, he saw a copy of Newton's Principia lying on the
grass, and called some one near him to take it back to the library. Young Stone, the
gardener, modestly observed, that the book belonged to him. To you! replied the Duke;
do you understand Geometry, Latin, Newton? I know a little of them, replied the
young man, with an air of simplicity. The Duke was surprised, and having himself
a taste for the sciences, he entered into conversation with the young mathematician. He
asked him several questions, and was astonished at the force, the accuracy, and the can-
dour of his answers. But how, said the Duke, came you by the knowledge of all these
things? Stone replied, a servant taught me, ten years since, to read. Does any one
need to know more than the twenty-four letters of the alphabet, in order to learn any
thing else that one wishes? The Duke's curiosity was redoubled: he sat down on a
bank, and requested a detail of all his proceedings. I first learned to read, said Stone;
the masons were then at work upon your house; I went near them one day, and saw
that the architect used a rule and compasses, and that he made calculations. I enquired
what might be the meaning and use of these things, and was informed that there is a
science called Arithmetic. I purchased a book of Arithmetic, and learned it. I was
told there was another science called Geometry; I bought the books, and learned
Geometry. By reading, I found that there were good books in these two sciences in
Latin: I bought a dictionary, and learned Latin. I understood, also, that there were
good books of the same kind in French; I bought a dictionary, and learned French
and this, my Lord, is what I have done. It seems to me that we may learn every
thing when we know the twenty-four letters of the alphabet. The Duke, highly pleased
with the account, brought this wonderful genius out of obscurity, and provided him
with an employment which left him leisure to apply himself to the Sciences.
;
INTRODUCTION.
xxxvii
Dr Simson's first endeavours were directed to improve the defective
restorations of the books on the Geometrical Analysis by preceding
geometers. His restoration of Apollonius is entirely according to the
ancient method; and is more complete than any preceding attempt of
the kind. In his preface to the Sectio Determinata, which he restored,
he points out the defects in Snell's restorations, and notices the solutions
of some of the problems by Alexander Anderson, in his supplement to
Apollonius Redivivus, published at Paris in 1612. He also remarks on
some of the problems in the Treatise on Geometrical Analysis, by Hugo
d'Omerique, and in his work adopts some propositions from these
performances. But Dr Simson is more generally known at the pre-
sent day for his translation of the first six and the eleventh and twelfth
books of Euclid's Elements of Geometry. The first edition was pub-
lished both in Latin and English in 1756. The English translation has
almost superseded every other, and may be regarded as the standard
text of Euclid in English, having maintained its character in this
country for nearly a century. The Data of Euclid was added to the
second edition of the Elements in 1762. Dr Simson's first publication,
except his paper on Porisms in the Philosophical Transactions, was his
Geometrical Treatise on the Conic Sections, which was published in
1735.
The description of the Porisms of Euclid by Pappus is so muti-
lated, that every attempt, before Dr Simson's, to restore them had
failed. Dr Halley, though successful in the restoration of some por-
tions of the ancient Geometry, gave up the Porisms in 1706 as a hope-
less task, as is obvious from his remark in the preface to the seventh
book of Pappus, "Hactenus Porismatum descriptio, nec mihi intellecta,
nec lectori profutura." Dr Simson had been occupied on the subject
of the Porisms in 1715, and perhaps earlier, for he observes that in
that year he had demonstrated the first case of Fermat's fourth Porism,
before he had acquired the knowledge of the true nature of that class of
Propositions. The first object of his researches seems to have been,
to discover the Porisms, from the general description given of them by
Pappus: and when he had failed in this, he tried to discover some of the
individual Porisms, from which he expected to ascertain the distinctive
character of these propositions; but in this attempt he had no better
success. For a considerable time, he informs us, his imagination was
completely occupied by the subject: his mind was harassed by the
constant, but unsuccessful exertion: he lost his sleep, and his health
suffered; all his endeavours were ineffectual, and he finally deter-
mined to banish the subject from his thoughts. For some time he
maintained this resolution, and applied himself to other pursuits; but
afterwards, as he was walking on the banks of the river Clyde, he inad-
vertently fell into a reverie respecting the Porisms. Some new ideas
struck his mind, and having drawn the diagram with chalk on an
adjoining tree, at that moment, for the first time, he acquired a just
notion of one of Euclid's Porisms. This account is given in his pre-
face to the Porisms, p. 319, of his "Opera Reliqua," a volume of his
writings on Geometry, published after his death, by the munificence
of the late Earl Stanhope.
Matthew Stewart was a pupil of Dr Simson, and afterwards became
Professor of Mathematics in the University of Edinburgh. Dr Stewart
was a successful cultivator of the ancient geometry. His "General
xxxviii
INTRODUCTION.
Theorems of considerable use in the higher parts of the Mathematics,"
was published in 1746, and placed him among the first geometers of
his time. In 1761 he published another volume, entitled "Tracts
Physical and Mathematical," and two years after his celebrated work
on geometry, "Propositiones Geometricæ more veterum demonstratæ
ad Geometriam antiquam illustrandam et promovendam idoneæ.”
this work he has given both the analysis and synthesis of a series of
geometrical theorems, many of which were not known before.
In
Dr Waring extended the discoveries of Newton in the theory of
curves much beyond his predecessors. His "Miscellanea Analytica de
æquationibus Algebraicis et curvarum proprietatibus," was published
in 1762, and his "Proprietates Algebraicarum Curvarum," in 1772.
Bishop Horsley was born in 1732, and died in 1806. Though edu-
cated at Cambridge he removed to Oxford, and there in 1769 published
his edition of the Inclinations of Apollonius. Besides his edition of
Newton's Works, he put forth in 1801 his Practical Mathematics, in
three volumes, containing Euclid's Elements, the Data, &c.
William Wales, by his talents and application rose from obscurity
to an eminent position among men of science. He was the person
selected to observe at Hudson's Bay the transit of Venus over the Sun
in 1769, and afterwards accompanied, as astronomer, the celebrated
Captain Cook on his first voyage in the years 1772 to 1774; and again
in his other voyage in the years 1776 to 1779. A short time before he
set sail in 1772, his friend Mr Lawson put forth his restoration of the
two books of Apollonius, De Sectione Determinata, together with an
English translation of Snell's restoration of the same two books. An-
other restoration of these two books was made by Giannini, and pub-
lished in his Opuscula Mathematica in 1773.
Dr Robertson, the late Savilian Professor of Astronomy, to his
treatise entitled "Sectionum Conicarum, Libri v11. &c.,” (1792, Oxon.)
has annexed a learned history of the Conic Sections.
Mascheroni published in Italian, in 1797, a treatise on geometry,
entitled "Geometria del Compassa," in which the solutions of geome-
trical problems are effected by means of the circle only, instead of the
straight line and circle. It was translated into French, and published
in 1798, and again reprinted in 1828. In the twelfth year of the
Republic, a similar treatise by M. Servois, was put forth, in which the
solutions of geometrical problems were effected by means of straight
lines only.
During the period of the French Revolution, at the end of the
eighteenth century, the celebrated Monge discovered and put forth a
new kind of geometry, under the appellation of "Géométrie Descrip-
tive." The discoveries of Monge mark a new era in the history of
geometry, as did the discovery of Descartes. The new geometry of
Monge has two objects in view; first, to represent geometrical solids
on a plane surface; and secondly, to deduce from this method of
representation, the mathematical properties of the figures. It is chiefly
conversant with the determination of the curves in which two or more
surfaces intersect each other, when they are supposed to penetrate one
another. To Monge is also due the theory of projections; and he was
the first who proved that the square of any surface is equal to the sum
of the squares of its projections on the three co-ordinate planes. It
ought to be remarked, that attention had been before directed to this
INTRODUCTION.
xxxix
subject by several who made contributions to it. Among these may be
named Courcier, a jesuit, who published, at Paris, in 1613, a work in
which he investigated the nature, and shewed the description of the
curves which result from the penetration of cylindrical, spherical and
conical surfaces; but to Monge is due the merit of its greatest ex-
tension.
The French school of Monge has produced many eminent geometers.
The "Géométrie de Position," and "L'Essai sur la Théorie des Trans-
versals," by Carnot, are a continuation of the method of Monge. Also
may be mentioned "Les Développemens et les Applications de Géomé-
trie," by Dupin, and the "Traité des Propriétés projectives," of M.
Poncelet, with others, who have amply and skilfully written on the
geometry of correlative figures, and its application to the physical
sciences. To these may be added the names of Leslie, Playfair, Le
Gendre, La Croix, L'Huillier, Vincent, and many others, too numerous
here to mention, who by their writings have furthered the progress
and advancement of scientific geometry.
We must not, before closing this subject, omit naming F. Peyrard,
who published, at Paris, in 1818, an edition of Euclid's Elements from
an ancient Greek manuscript which had not been collated or printed
before. This edition, besides the Greek text, contains a version in
Latin, and a French translation. F. Peyrard is also known as the
editor and translator of the writings of Archimedes and Apollonius.
The short and cursory notices here given of the modern Geometry,
are intended rather to awaken the curiosity of the student than to
afford an ample and satisfactory account. In the "Mémoires Couron-
nés de l'Académie de Bruxelles" for 1837, the student will find in the
"Aperçu Historique" of M. Chasles, a full and particular account of
the history of the methods and developements of the modern Geometry.
ON THE ABBREVIATIONS AND ALGEBRAICAL SYMBOLS
EMPLOYED IN GEOMETRY.
It has been remarked that the ancient geometry of the Greeks ad-
mitted no symbols besides the diagrams and ordinary language. In
later times, after symbols of operation had been devised by writers on
algebra, they were very soon adopted and employed, on account of their
brevity and convenience, in writings purely geometrical. Dr Barrow
was one of the first who introduced algebraical symbols into the lan-
guage of elementary geometry, and distinctly states, in the preface
to his Euclid, that his object is "to content the desires of those who
are delighted more with symbolical than verbal demonstrations." As
algebraical symbols are employed in almost all works on the mathe-
matics, whether geometrical or not, it seems proper in this place to
give some brief account of the marks which may be regarded as the
alphabet of symbolical language.
The mark was first used by Robert Recorde, in his treatise on
Algebra entitled, "The Whetstone of Witte, &c.", for the sign of
equality; "because,” as he remarks, "no two things can be more equal
than a pair of parallels, or gemowe lines of one length." It was em-
ployed by him, in his algebra, simply in the sense of æquatur, or, is
equal to, affirming the equality of two numerical or algebraical ex-

xl
INTRODUCTION.
4
pressions. Geometrical equality is not exactly the same as numerical
equality, and when this symbol is used in geometrical reasonings, it
must be understood as having reference to pure geometrical equality.
The signs of relative magnitude, > meaning, is greater than, and <,
is less than, were first introduced into algebra by Thomas Harriot, in his
"Artis Analyticæ Praxis," which was published after his death in 1631.
The signs and were first employed by Michael Stifel, in his
"Arithmetica Integra," which was published in 1544. The sign +
was employed by him for the word plus, and the sign-, for the word
minus. These signs were used by Stifel strictly as the arithmetical or
algebraical signs of addition and subtraction.
The sign of multiplication x was first introduced by Oughtrede in
his "Clavis Mathematica," which was published in 1631. In algebra-
ical multiplication he either connects the letters which form the factors
of a product by the sign x, or writes them as words without any sign
or mark between them, as had been done before by Harriot, who first
introduced the small letters to designate known and unknown quantities.
However concise and convenient the notation AB × BC may be in
practice for "the rectangle contained by the lines AB and BC;" the
student is cautioned against the use of it, in the early part, at least, of
his geometrical studies, as the use of it is likely to occasion a misap-
prehension of Euclid's meaning. Dr Barrow sometimes expresses "the
rectangle contained by AB and BC" by "the rectangle ABC.”
Michael Stifel was the first who introduced integral exponents
to denote the powers of algebraical symbols of quantity, for which
he employed capital letters. Vieta afterwards used the vowels to
denote known, and the consonants, unknown quantities, but used
words to designate the powers. Simon Stevin, in his treatise on
Algebra, which was published in 1605, improved the notation of
Stifel, by placing the figures that indicated the powers within small
circles. Peter Ramus adopted the initial letters 1, q, c, bq of latus,
quadratus, cubus, biquadratus, as the notation of the first four powers.
Harriot exhibited the different powers of algebraical symbols by repeat-
ing the symbol, two, three, four, &c. times, according to the order of the
power. Descartes restored the numerical exponents of powers, placing
them at the right of the numbers, or symbols of quantity, as at the
present time. Dr Barrow employed the notation ABq, for "the square
of the line AB," in his edition of Euclid. The notations AB², AB³, for
" the square and cube of the line whose extremities are A and B,” are
found in almost all works on the Mathematics, though not wholly
consistent with the algebraical notations a² and a³.
The symbol, being originally the initial letter of the word
radix, was first used by Stifel to denote the square root of the
number, or the symbol, before which it was placed.
The Hindus, in their treatises on Algebra, indicated the ratio of
two numbers, or of two algebraical symbols, by placing one above
the other, without any line of separation. The line was first intro-
duced by the Arabians, from whom it passed to the Italians, and
from them to the rest of Europe. This notation has been employed
for the expression of geometrical ratios by almost all writers on the
Mathematics, on account of its great convenience. Oughtrede first
used points to indicate proportion; thus, a .b:: c. d, means, that a
bears the same proportion to b, as c does to d.
EUCLID'S
ELEMENTS OF GEOMETRY.
BOOK I.
DEFINITIONS.
I.
A POINT is that which has no parts, or which has no magnitude.
II.
A line is length without breadth.
III.
The extremities of lines are points.
IV.
A right line is that which lies evenly between its extreme points.
V.
A superficies is that which has only length and breadth.
VI.
The extremities of superficies are lines.
VII.
A plane superficies is that in which any two points being taken, the
straight line between them lies wholly in that superficies.
VIII.
A plane angle is the inclination of two lines to each other in a plane
which meet together, but are not in the same straight line.
IX.
A plane rectilineal angle is the inclination of two straight lines to one
another, which meet together, but are not in the same straight line.
A
D
E
B
r

1
2
EUCLID'S ELEMENTS.
N. B. When several angles are at one point B, either of them is expressed by
three letters, of which the letter that is at the vertex of the angle, that is, at the point
in which the right lines that contain the angle meet one another, is put between the other
two letters, and one of these two is somewhere upon one of these right lines, and the
other upon the other line. Thus the angle which is contained by the right lines AB,
CB, is named the angle ABC, or CBA; that which is contained by AB, DB, is
named the angle ABD, or DBA; and that which is contained by DB, CB, is called
the angle DBC, or CBD. But, if there be only one angle at a point, it may be
expressed by the letter at that point; as the angle at E.
X.
When a straight line standing on another straight line makes the
adjacent angles equal to each other, each of these angles is called a
right angle; and the straight line which stands on the other is called
a perpendicular to it.

XI.
An obtuse angle is that which is greater than a right angle.

XII.
An acute angle is that which is less than a right angle.
XIII.
A term or boundary is the extremity of any thing.
XIV.
A figure is that which is inclosed by one or more boundaries.
XV.
A circle is a plane figure contained by one line, which is called the
circumference, and is such that all straight lines drawn from a certain
point within the figure to the circumference, are equal to one another.

DEFINITIONS.
3
XVI,
And this point is called the centre of the circle.
XVII.
A diameter of a circle is a straight line drawn through the centre,
and terminated both ways by the circumference.
D
XVIII.
A semicircle is the figure contained by a diameter and the part of
the circumference cut off by the diameter.
XIX.
The centre of a semicircle is the same with that of the circle.
XX.
Rectilineal figures are those which are contained by straight lines.
XXI.
Trilateral figures, or triangles, by three straight lines.
XXII.
Quadrilateral, by four straight lines.
XXIII.
Multilateral figures, or polygons, by more than four straight lines.
XXIV.
Of three-sided figures, an equilateral triangle is that which has
three equal sides.
XXV.
An isosceles triangle is that which has two sides equal.
1-2
4
EUCLID'S ELEMENTS.
XXVI.
A scalene triangle is that which has three unequal sides.
XXVII.
A right-angled triangle is that which has a right angle.
XXVIII.
An obtuse-angled triangle is that which has an obtuse angle.

XXIX.
An acute-angled triangle is that which has three acute angles.
XXX.
Of quadrilateral or four-sided figures, a square has all its sides equal
and all its angles right angles.
XXXI.
An oblong is that which has all its angles right angles, but has not
all its sides equal.
XXXII.
A rhombus has all its sides equal, but its angles are not right angles.
DEFINITIONS.
5
XXXIII.
A rhomboid has its opposite sides equal to each other, but all its
sides are not equal, nor its angles right angles.
XXXIV.
All other four-sided figures besides these, are called Trapeziums.
XXXV.
Parallel straight lines are such as are in the same plane, and whic
being produced ever so far both ways, do not meet.
A.
A parallelogram is a four-sided figure, of which the opposite sides
are parallel: and the diameter or the diagonal is the straight line join-
ing two of its opposite angles.
POSTULATES.
I.
LET it be granted that a straight line may be drawn from any one
point to any other point.
II.
That a terminated straight line may be produced to any length in
a straight line.
III.
And that a circle may be described from any centre at any distance
from that centre.
AXIOMS.
I.
THINGS which are equal to the same thing are equal to one another.
II.
If equals be added to equals, the wholes are equal.
6
EUCLID'S ELEMENTS.
III.
If equals be taken from equals, the remainders are equal.
IV.
If equals be added to unequals, the wholes are unequal.
V.
If equals be taken from unequals, the remainders are unequal.
VI.
Things which are double of the same, are equal to one another.
VII.
Things which are halves of the same, are equal to one another.
VIII.
Magnitudes which coincide with one another, that is, which exactly
fill the same space, are equal to one another.
IX.
The whole is greater than its part.
X.
Two straight lines cannot inclose a space.
XI.
All right angles are equal to one another.
XII.
If a straight line meets two straight lines, so as to make the two
interior angles on the same side of it taken together less than two right
angles, these straight lines being continually produced, shall at length
meet upon that side on which are the angles which are less than two
right angles.
BOOK I. PROP. I, II.
7
PROPOSITION I. PROBLEM.
To describe an equilateral triangle upon a given finite straight line.
Let AB be the given straight line.
It is required to describe an equilateral triangle upon AB.
C

DA
BE
From the centre A, at the distance AB, describe the circle BCD; (post. 3.)
from the centre B, at the distance BA, describe the circle AČE;
and from the point C, in which the circles cut one another,
draw the straight lines CA, CB to the points A, B. (post. 1.)
Then ABC shall be an equilateral triangle.
Because the point A is the centre of the circle BCD,
therefore AC is equal to AB; (def. 15.)
and because the point B is the centre of the angle ACE,
therefore BC is equal to BA;
but it has been proved that CA is equal to AB;
therefore CA, CB are each of them equal to AB;
but things which are equal to the same thing are equal to
one another; (ax. 1.)
therefore CA is equal to CB;
wherefore CA, AB, BC are equal to one another:
and the triangle ABC is therefore equilateral,
and it is described upon the given straight line AB.
Which was required to be done.
PROPOSITION II. PROBLEM.
From a given point, to draw a straight line equal to a given straight line.
Let A be the given point, and BC the given straight line.
It is required to draw from the point A a straight line equal to BC.

K
H
D
B
E
F
From the point A to B draw the straight line AB;
upon AB describe the equilateral triangle DAB,
and produce the straight lines DA, DB to E and F;
(post. 1.)
(1. 1.)
(post. 2.)
from the centre B, at the distance BC describe the circle CGH, (post. 3.)
and from the centre D, at the distance DG, describe the circle GKL.
Then the straight line AL shall be equal to BC.
Because the point B is the centre of the circle CGH,
8
EUCLID'S ELEMENTS.
1
therefore BC is equal to BG; (def. 15.)
and because D is the centre of the circle GKL,
therefore DL is equal to DG;
and DA, DB parts of them are equal; (1. 1.)
therefore the remainder AL is equal to the remainder BG; (ax. 3.)
but it has been shewn that BC is equal to BG,
wherefore AL and BC are each of them equal to BG;
and things which are equal to the same thing are equal to one another;
therefore the straight line AL is equal to BC. (ax. 1.)
Wherefore from the given point A a straight line AL has been drawn
equal to the given straight line BC. Which was to be done.
PROPOSITION III. PROBLEM.
From the greater of two given straight lines to cut off a part equal to the less.
Let AB and C be the two given straight lines, of which AB
is the greater.
It is required to cut off from AB a part equal to C, the less.

D
D
E B
F
From the point A draw the straight line AD equal to C; (1. 2.)
and from the centre A, at the distance AD, describe the
circle DEF. (post. 3.)
Then AE shall be equal to C.
Because A is the centre of the circle DEF,
therefore AE is equal to AD; (def. 15.)
but the straight line C is equal to AD; (constr.)
whence AE and C are each of them equal to AD;
wherefore the straight line AE is equal to C. (ax. 1.)
And therefore from AB the greater of two straight lines, a part AE
has been cut off equal to C the less. Which was to be done.
PROPOSITION IV. THEOREM.
If two triangles have two sides of the one equal to two sides of the
other, each to each; and have likewise the angles contained by those sides
equal to each other; they shall likewise have their bases, or third sides,
equal; and the two triangles shall be equal; and their other angles shall
be equal, each to cach, viz. those to which the equal sides are opposite.
Let ABC, DEF be two triangles, which have the two sides AB, AC
equal to the two sides DE, DF, each to each, viz. AB to DE, and AC
to DF, and the included angle BAC equal to the included angle EDF.
Then shall the base BC be equal to the base EF; and the triangle
ABC to the triangle DEF; and the other angles to which the equal
sides are opposite shall be equal, each to each, viz. the angle ABC to the
angle DEF, and the angle ACB to the angle DFE.
BOOK I. PROP. IV, V.
9
A A
B
F
For, if the triangle ABC be applied to the triangle DEF,
so that the point A may be on D, and the straight line AB on DE;
because AB is equal to DE,
therefore the point B shall coincide with the point E;
and AB coinciding with DE, because the angle BAC is equal
to the angle EDF,
therefore the straight line AC shall fall on DF;
also because AC is equal to DF,
therefore the point C shall coincide with F
but the point B coincides with the point E;
wherefore the base BC shall coincide with the base EF;
because the point B coinciding with E, and C with F,
if the base BC do not coincide with the base EF, the two straight
lines BC and EF would inclose a space, which is impossible. (ax. 10.)
Therefore the base BC does coincide with EF, and is equal to it.
Wherefore the whole triangle ABC coincides with the whole tri-
angle DEF, and is equal to it;
and the other angles of the one coincide with the remaining angles
of the other, and are equal to them,
viz. the angle ABC to the angle DEF,
and the angle ACB to DFE.
Therefore, if two triangles have two sides of the one equal to two sides, &c.
Which was to be demonstrated.
PROPOSITION V. THEOREM.
The angles at the base of an isosceles triangle are equal to each other;
and if the equal sides be produced, the angles on the other side of the base
shall be equal.
Let ABC be an isosceles triangle of which the side AB is equal to AC,
and let the equal sides AB, AC be produced to D and E.
Then the angle ABC shall be equal to the angle ACB,
and the angle DBC to the angle ECB.
A

B
C
F
G
D
E
In BD take any point F;
from AE the greater, cut off AG equal to AF the less, (1. 3.) and
join FC, GB.
Because AF is equal to AG, (constr.) and AB to AC; (hyp.)
the two sides FA, ÂC, are equal to the two GA, AB, each to each;
+
10
EUCLID'S ELEMENTS.
and they contain the angle FAG common to the two triangles AFC, AGB;
therefore the base FC is equal to the base GB, (1. 4.)
and the triangle AFC is equal to the triangle AGB,
also the remaining angles of the one are equal to the remaining angles
of the other, each to each, to which the equal sides are opposite ;
viz. the angle ACF to the angle ABG,
and the angle AFC to the angle AGB.
And because the whole AF is equal to the whole AG,
of which the parts AB, AC, are equal;
remainder BF is equal to the remainder
and FC was proved to be equal to GB;
hence because the two sides BF, FC are equal to the two CG,
GB, each to each;
therefore the
CG; (ax. 3.)
and the angle BFC has been proved to be equal to the angle CGB,
also the base BC is common to the two triangles BFC, CGB;
wherefore these triangles are equal, (1. 4.)
and their remaining angles, each to each, to which the equal sides
are opposite;
therefore the angle FBC is equal to the angle GCB,
and the angle BFC to the angle CBG.
And, since it has been demonstrated,
that the whole angle ABG is equal to the whole ACF,
the parts of which, the angles CBG, BCF are also equal;
therefore the remaining angle ABC is equal to the remaining angle ACB,
which are the angles at the base of the triangle ABC;
and it has been proved,
that the angle FBC is equal to the angle GCB,
which are the angles upon the other side of the base.
Therefore the angles at the base, &c. Q.E.D.
COR. Hence every equilateral triangle is also equiangular.
PROPOSITION VI. THEOREM.
If two angles of a triangle be equal to each other, the sides also which
subtend, or are opposite to, the equal angles, shall be equal to one another.
Let ABC be a triangle having the angle ABC equal to the angle ACB.
Then the side AB shall be equal to the side AC.
D
A
B
C
For, if AB be not equal to AC,
one of them is greater than the other.
Let AB be greater than AC;
and at the point B, from BA cut off BD equal to CA the less,
(1. 3.) and join DC.
Then, in the triangles DBC, ABC,
because DB is equal to AC, and BC is common to both,
the two sides DB, BC are equal to the two sides AC, CB, each to each;
BOOK I. PROP. VI, VII.
11
•
and the angle DBC is equal to the angle ACB; (hyp.)
therefore the base DC is equal to the base AB, (1. 4.)
and the triangle DBC is equal to the triangle ACB,
the less equal to the greater, which is absurd.
Therefore AB is not unequal to AC, that is, AB is equal to AC.
Wherefore, if two triangles, &c.
Q. E.D.
COR. Hence every equiangular triangle is also equilateral.
PROPOSITION VII. .THEOREM.
Upon the same base, and on the same side of it, there cannot be two
triangles that have their sides which are terminated in one extremity of the
base, equal to one another, and likewise those which are terminated in the
other extremity.
If it be possible, on the same base AB, and upon the same side of
it, let there be two triangles ACB, ADB, which have their sides CA,
DA, terminated in the extremity A of the base, equal to one another,
and likewise their sides, CB, DB, that are terminated in B.
C D
A
Join CD.
B
First. When the vertex of each of the triangles is without the
other triangle.
Because AC is equal to AD in the triangle ACD,
therefore the angle ACD is equal to the angle ADC; (1. 5.)
but the angle ACD is greater than the angle BCD; (ax. 9.)
therefore also the angle ADC is greater than BCD;
much more therefore is the angle BDC greater than BCD.
Again, because the side BC is equal to BD in the triangle BCD, (hyp.)
therefore the angle BDC is equal to the angle BCD; (1.5.)
but the angle BDC was proved greater than the angle BCD,
hence the angle BDC is both equal to, and greater than the angle BCD;
which is impossible.
Secondly. Let the vertex D of the triangle ADB fall within the
triangle AČB.
E

F
A
D
Produce AC and AD to E and F.
Then because AC is equal to AD in the triangle ACD,
therefore the angles ECD, FDC upon the other side of the base
CD are equal to one another; ~(1. 5.)
but the angle ECD is greater than the angle BCD; (ax. 10.)
therefore also the angle FDC is greater than the angle BCD;
12
EUCLID'S ELEMENTS.
much more then is the angle BDC greater than the angle BCD.
Again, because BC is equal to BD in the triangle BCD,
therefore the angle BDC is equal to the angle BCD; (1. 5.)
but the angle BDC has been proved greater than BCD,
wherefore the angle BDC is both equal to, and greater than
the angle BCD;
which is impossible.
Thirdly. The case in which the vertex of one triangle is upon a
side of the other needs no demonstration.
Therefore upon the same base and on the same side of it, &c. Q.E.D.
PROPOSITION VIII.
THEOREM.
If two triangles have two sides of the one equal to two sides of the
other, each to each, and have likewise their bases equal; the angle which
is contained by the two sides of the one shall be equal to the angles con-
tained by the two sides equal to them, of the other.
Let ABC, DEF be two triangles, having the two sides AB, AC,
equal to the two sides DE, DF, each to each, viz. AB to DE, and AC
to DF; and also the base BC equal to the base EF.
Then the angle BAC shall be equal to the angle EDF.
A
D G
ΔΑ
B
E
For, if the triangle ABC be applied to DEF, so that the point B be
on E, and the straight line BC on EF;
then because BC is equal to EF, (hyp.)
therefore the point C shall coincide with the point F;
wherefore BC coinciding with EF,
BA and AC shall coincide with ED, DF;
for, if the base BC coincide with the base EF, but the sides BA,
AC do not coincide with the sides ED, DF, but have a different situa-
tion as EG, FG:
Then, upon the same base, and upon the same side of it, there can
be two triangles which have their sides which are terminated in one
extremity of the base equal to one another, and likewise those sides
which are terminated in the other extremity;
but this is impossible. (1. 7.)
Therefore, if the base BC coincide with the base EF,
the sides BA, AC cannot but coincide with the sides ED, DF ;
wherefore likewise the angle BAC coincides with the angle EDF,
and is equal to it, (ax. 8.)
Therefore if two triangles have two sides, &c. Q. E, D.
BOOK I. PROP. IX, X.
13
PROPOSITION IX.
PROBLEM..
To bisect a given rectilineal angle, that is, to divide it into two equal
angles.
Let BAC be the given rectilineal angle.
It is required to bisect it.
A
D
E
F
B
Take any point D in AB ;
from AC cut off AE equal to AD; (1. 3.) and join DE,
describe an equilateral triangle DEF on the side of DE remote from
A, (1. 1.) and join AF.
Then the straight line AF shall bisect the angle BAC.
Because AD is equal to AE, (constr.)
and AF is common to the two triangles DAF, EAF;
the two sides DA, AF, are equal to the two sides EA, AF,
each to each
and the base DF is equal to the base EF; (constr.)
therefore the angle DAF is equal to the angle EAF. (1.8.)
Wherefore the angle BAC is bisected by the straight line AF. Q.E.F.
. PROPOSITION X. PROBLEM.
To bisect a given finite straight line, that is, to divide it into two
equal parts.
Let AB be the given straight line.
It is required to divide AB into two equal parts.
C
A
D B
Upon AB describe the equilateral triangle ABC;_ (1. 1.)
and bisect the angle ACB by the straight line CD meeting AB
in the point D. (1. 9.)
Then AB shall be cut into two equal parts in the point D.
Because AC is equal to CB, (constr.)
and CD is common to the two triangles ACD, BCD;
the two sides AC, CD are equal to BC, CD, each to each;
and the angle ACD is equal to BCD; (constr.)
therefore the base AD is equal to the base DB. (1. 4.)
Wherefore the straight line AB is divided into two equal parts in
the point D.
Q.E.F.
•
14
EUCLID'S ELEMENTS.
PROPOSITION XI. PROBLEM.
To draw a straight line at right angles to a given straight line, from
a given point in the same.
Let AB be the given straight line, and C a given point in it.
It is required to draw a straight line from the point C at right
angles to AB.
F

A D C
E B
In AC take any point D, and make CE equal to CD; (1. 3.)
upon DE describe the equilateral triangle DEF, (1. 1.), and join CF.
Then CF drawn from the point C shall be at right angles to AB.
Because DC is equal to CE, and FC is common to the two tri-
angles DCF, ECF;
the two sides DC, CF are equal to the two sides EC, CF, each to each;
and the base DF is equal to the base EF; (constr.)
therefore the angle DCF is equal to the angle ECF: (1.8.)
and these two angles are adjacent angles.
But when the two adjacent angles which one straight line makes
with another straight line are equal to one another, each of them is
called a right angle: (def. 10.)
Therefore each of the angles DCF, ECF is a right angle.
Wherefore from the given point C, in the given straight line AB,
FC has been drawn at right angles to AB.

Q.E.F.
COR. By help of this problem, it may be demonstrated that two
straight lines cannot have a common segment.
If it be possible, let the segment AB be common to the two straight
lines ABC, ABD.
E

A
B
D
From the point B, draw BE at right angles to AB; (1. 11.)
then because ABC is a straight line,
therefore the angle ABE is equal to the angle EBC;
(def. 10.)
Similarly, because ABD is a straight line,
therefore the angle ABE is equal to the angle EBD;
but the angle ABE is equal to the angle EBC,
wherefore the angle EBD is equal to the angle EBC, (ax. 1.)
the less equal to the greater angle, which is impossible.
Therefore two straight lines cannot have a common segment.
PROPOSITION XII. PROBLEM.
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.
BOOK I. PROP. XII, XIII.
15
It is required to draw a straight line perpendicular to AB from
the point C.
C
E
H
AF
G B
D
Take any point D upon the other side of AB,
and from the centre C, at the distance CD, describe the circle EGF
meeting AB in F and G; (post. 3.)
bisect FG in H (1. 10.), and join CH.
Then the straight line CH drawn from the given point C shall be
perpendicular to the given straight line AB.
Join CF, and CG.
And because FH is equal to HG (constr.), and HC is common to
the triangles FHC, GHC;
the two sides FH, HC, are equal to the two GH, HC, each to each;
and the base CF is equal to the base CG; (def. 15.)
therefore the angle FHC is equal to the angle GHC; (1.8.)
and these are adjacent angles.
But when a straight line standing on another straight line, makes
the adjacent angles equal to one another, each of them is a right angle,
and the straight line which stands upon the other is called a perpen-
dicular to it. (def. 10.)
Therefore from the given point C a perpendicular CH has been
drawn to the given straight line AB.
Q. E. F.
PROPOSITION XIII. THEOREM.
The angles which one straight line makes with another upon one side of
it, are either two right angles, or are together equal to two right angles.
Let the straight line AB make with CD, upon one side of it, the
angles CBA, ABD.
Then these shall be either two right angles, or shall be together,
equal to two right angles.
A
D B
E

A
D
B
C
For if the angle CBA be equal to the angle ABD,
each of them is a right angle. (def. 10.)
But if the angle CBA be not equal to the angle ABD,
from the point B draw BE at right angles to CD. (1.11.)
Then the angles CBE, EBD are two right angles. (def. 10.)
And because the angle CBE is equal to the angles CBA, ABE,
add the angle EBD to each of these equals ;
therefore the angles CBE, EBD are equal to the three angles
CBA, ABE, EBD. (ax. 2.)
16
EUCLID'S ELEMENTS.
Again, because the angle DBA is equal to the two angles DBE, EBA,
add to each of these equals the angle ABC;
therefore the angles DBA, ABC are equal to the three angles
DBE, EBA, ABC.
But the angles CBE, EBD have been proved equal to the same
three angles;
and things which are equal to the same thing are equal to one another;
therefore the angles CBE, EBD are equal to the angles DBA, ABC;
but the angles CBE, EBD are two right angles;
therefore the angles DBA, ABC are together equal to two right
angles. (ax. 1.)
Wherefore when a straight line, &c.
PROPOSITION XIV. THEOREM.
Q.E.D.
If at a point in a straight line, two other straight lines, upon the
opposite sides of it, make the adjacent angles together equal to two right
angles, these two straight lines shall be in one and the same straight line.
At the point B in the straight line AB, let the two straight lines
BC, BD upon the opposite sides of AB, make the adjacent angles
ABC, ABD together equal to two right angles.
Then BD shall be in the same straight line with CB.
A

C B
E
D
For, if BD be not in the same straight line with CB,
let BE be in the same straight line with it.
Then because AB meets the straight line CBE;
therefore the adjacent angles CBA, ABE are equal to two right
angles; (1. 13.)
but the angles CBA, ABD are equal to two right angles; (hyp.)
therefore the angles CBA, ABE are equal to the angles CBA, ABD: (ax. 2.)
take away from these equals the common angle CBA,
therefore the remaining angle ABE is equal to the remaining angle
ABD; (ax. 3.)
the less equal to the greater angle, which is impossible:
therefore BE is not in the same straight line with CB.
And in the same manner it may be demonstrated, that no other can
be in the same straight line with it but BD, which therefore is in the
same straight line with CB.
Wherefore, if at a point, &c.
Q. E.D.
PROPOSITION XV. THEOREM.
If two straight lines cut one another, the vertical, or opposite angles
shall be equal.
Let the two straight lines AB, CD cut one another in the point E.
BOOK I.
17
•
PROP. XV, XVI.
Then the angle AEC shall be equal to the angle DEB, and the
angle CEB to the angle AED.
C
-B
A
E
D
Because the straight line AE makes with CD at the point E, the
adjacent angles CEA, AED;
these angles are together equal to two right angles. (1. 13.)
Again, because the straight line DE makes with AB at the point E
the adjacent angles AED, DEB;
these angles also are equal to two right angles;
but the angles CEA, AED have been shewn to be equal to two
right angles;
wherefore the angles CEA, AED are equal to the angles AED, DEB;
take away from each the common angle AED,
and the remaining angle CEA is equal to the remaining angle DEB. (ax.3.)
In the same manner it may be demonstrated, that the angle CEB is
equal to the angle AED.
Therefore, if two straight lines cut one another, &c. Q. E. D.
COR. 1. From this it is manifest, that, if two straight lines cut each
other, the angles which they make at the point where they cut, are
together equal to four right angles.
COR. 2. And consequently that all the angles made by any number
of lines meeting in one point, are together equal to four right angles.
PROPOSITION XVI. THEOREM.
If one side of a triangle be produced, the exterior angle is greater than
either of the interior opposite angles.
Let ABC be a triangle, and let its side BC be produced to D.
Then the exterior angle ACD shall be greater than either of the
interior opposite angles CBA or BAC.
A
F

E
B
C
G
D
Bisect AC in E, (1. 10.) and join BE;
produce BE to F, making EF equal to BE, (1. 3.) and join FC.
Because AE is equal to EC, and BÈ to EF;
the two sides AE, EB are equal to the two CE, EF, each to each,
in the triangles ABE, CFE;
(1. 15.)
and the angle AEB is equal to the angle CEF,
because they are opposite vertical angles;
therefore the base AB is equal to the base CF, (1.4.)
and the triangle AEВ to the triangle CEF,
2
18
EUCLID'S ELEMENTS.
and the remaining angles of one triangle to the remaining angles of
the other, each to each, to which the equal sides are opposite;
wherefore the angle BAE is equal to the angle ECF;
but the angle ECD or ACD is greater than the angle ECF;
therefore the angle ACD is greater than the angle BAE.
In the same manner, if the side BC be bisected, and AC be pro-
duced to G; it may be demonstrated that the angle BCG, that is, the
angle ACD, (1. 15.) is greater than the angle ABC.
Therefore, if one side of a triangle, &c.
Q.E.D.
PROPOSITION XVII.
THEOREM.
Any two angles of a triangle are together less than two right angles.
Let ABC be any triangle.
Then any two of its angles together shall be less than two right angles.
A
B
C
D
Produce any side BC to D.
Then because ACD is the exterior angle of the triangle ABC;
therefore the angle ACD is greater than the interior and opposite
angle ABC; (1. 16.)
to each of these unequals add the angle ACB;
Therefore the angles ACD, ACB are greater than the angles ABC,
ACB;
but the angles ACD, ACB are equal to two right angles; (1. 13.)
therefore the angles ABC, BCA are less than two right angles.
In like manner it may be demonstrated,
that the angles BAC, ACB are less than two right angles,
as also the angles CAB, ABC.
Therefore any two angles of a triangle, &c.
PROPOSITION XVIII. THEOREM.
Q. E.D.
The greater side of every triangle is opposite to the greater angle.
Let ABC be a triangle, of which the side AC is greater than the
side AB.
Then the angle ABC shall be greater than the angle BCA.
B
A
D
Since the side AC is greater than the side AB,
make AD equal to AB, (1. 3.) and join BD.
Then because AD is equal to AB in the triangle ABD,
therefore the angle ADB is equal to the angle ABD, (1. 5.)
BOOK I PROP. XVIII—XX.
19
but because the side CD of the triangle DBC is produced to A,
therefore the exterior angle ADB is greater than the interior and
opposite angle DCB; (1. 16.)
but the angle ADB has been proved equal to the angle ABD,
therefore the angle ABD is greater than the angle DCB;
wherefore much more is the angle ABC greater than the angle ACB,
Therefore the greater side, &c. Q.E.D.
PROPOSITION XIX. THEOREM.
The greater angle of every triangle is subtended by the greater side,
or, has the greater side opposite to it.
Let ABC be a triangle of which the angle ABC is greater than the
angle BCA.
Then the side AC shall be greater than the side AB.
A
B
For, if AC be not greater than AB,
AC must either be equal to, or less than AB;
if AC were equal to AB,
then the angle ABC would be equal to the angle ACB;
but it is not equal; (hyp.)
therefore the side AC is not equal to AB.
Again, if AC were less than AB,
(1. 5.)
then the angle ABC would be less than the angle ACB; (1. 18.)
but it is not less,
therefore the side AC is not less than AB;
and AC has been shewn to be not equal to AB;
therefore AC is greater than AB.
Wherefore the greater angle, &c.
Q. E.D.
PROPOSITION XX. THEOREM.
Any two sides of a triangle are together greater than the third side.
Let ABC be a triangle.
Then any two sides of it together shall be greater than the third side,
viz. the sides BA, AC greater than the side BC;
AB, BC greater than AC;
and BC, CA greater than AB,
B
A
D
C
Produce the side BA to the point D,
make AD equal to AC, (1. 3.) and join DC.
2-2
20
EUCLID'S ELEMENTS.
Then because AD is equal to AC,
therefore the angle ADC is equal to the angle ACD; (1. 5.)
but the angle BCD is greater than the angle ACD;
therefore also the angle BCD is greater than the angle ADC.
And because in the triangle DBC,
the angle BCD is greater than the angle BDC, (ax. 9.)
and that the greater angle is subtended by the greater side; (1. 19.)
therefore the side DB is greater than the side BC;
but DB is equal to BA and AC,
therefore the sides BA and AC are greater than BC.
In the same manner it may be demonstrated, that the sides AB, BC
are greater than CA;
also that BC, CA are greater than AB.
Therefore any two sides, &c.
Q. E.D.
PROPOSITION XXI. THEOREM.
If from the ends of a side of a triangle, there be drawn two straight
lines to a point within the triangle, these shall be less than the other
two sides of the triangle, but shall contain a greater angle.
Let ABC be a triangle, and from the points B, C, the ends of the
side BC, let the two straight lines BD, CD be drawn to a point D
within the triangle.
Then BD and DC shall be less than BA and AC the other
two sides of the triangle,
but shall contain an angle BDC greater than the angle BAC.
A

E
D
B
Produce BD to meet the side AC in E.
Because two sides of a triangle are greater than the third side, (1.20.)
therefore the two sides BA, AE of the triangle ABE are greater than BE;
to each of these unequals add EC;
therefore the sides BA, AC are greater than BE, EC. (ax. 4.)
Again, because the two sides CE, ED of the triangle CED
are greater than DC; (1. 20.)
add DB to each of these unequals ;
therefore the sides CE, EB are greater than CD, DB. (ax. 4.)
But it has been shewn that BA, AC are greater than BE, EC;
much more then are BA, AC greater than BD, DC.
Again, because the exterior angle of a triangle is greater than
the interior and opposite angle; (1. 16.)
therefore the exterior angle BDC of the triangle CDE is greater
than the interior and opposite angle CED;
for the same reason, the exterior angle CED of the triangle ABE is
greater than the interior and opposite angle BAC;
and it has been demonstrated,
that the angle BDC is greater than the angle CEB;
much more therefore is the angle BDC greater than the angle BAC.
Therefore, if from the ends of the side, &c. Q. E. D.
BOOK I. PROP. XXII, XXIII.
21
PROPOSITION XXII. PROBLEM.
To make a triangle of which the sides shall be equal to three
given straight lines, but any two whatever of these must be greater than
the third.
Let A, B, C be the three given straight lines,
of which any two whatever are greater than the third, (1. 20.)
namely, A and B greater than C;
A and C greater than B;
and B and C greater than A.
It is required to make a triangle of which the sides shall be equal
to A, B, C, each to each.

D
E
G
H
L
A
B
C—
Take a straight line DE terminated at the point D, but unlimited
towards E,
make DF equal to A, FG equal to B, and GH equal to C; (1. 3.)
from the centre F, at the distance FD, describe the circle DKL; (post. 3.)
and from the centre G, at the distance GH, describe the
circle HLK; and join KF, KG.
Then the triangle KFG shall have its sides equal to the three
straight lines A, B, C.
Because the point F is the centre of the circle DKL,
therefore FD is equal to FK; (def. 15.)
but FD is equal to the straight line A ;
therefore FK is equal to A.
Again, because G is the centre of the circle HLK;
therefore GH is equal to GK, (def. 15.)
but GH is equal to C';
therefore also GK is equal to C;
and FG is equal to B;
(ax. 1.)
therefore the three straight lines KF, FG, GK, are respectively
equal to the three, A, B, C:
and therefore the triangle KFG has its three sides KF, FG, GK,
equal to the three given straight lines A, B, C. Q. E. F.
PROPOSITION XXIII. PROBLEM.
At a given point in a given straight line, to make a rectilineal angle.
equal to a given rectilineal angle.
Let AB be the given straight line, and A the given point in it, and
DCE the given rectilineal angle.
It is required to make an angle at the given point A in the given
straight line AB, that shall be equal to the given rectilineal angle DCE.
22
EUCLID'S ELEMENTS.
C
A
E
F
G
B
D
In CD, CE, take any points D, E, and join DE;
make the triangle AFG, the sides of which shall be equal to the
three straight lines CD, DE, EC, so that AF be equal to CD, AG to
CE, and FG to DE. (1. 22.)
Then the angle FAG shall be equal to the angle DCE.
Because FA, AG are equal to DC, CE, each to each,
and the base FG is equal to the base DE;
therefore the angle FAG is equal to the angle DCE. (1.8.)
Wherefore at the given point A in the given straight line AB, the
angle FAG is made equal to the given rectilineal angle DCE. Q.E.F.
PROPOSITION XXIV. THEOREM.
If two triangles have two sides of the one equal to two sides of the
other, each to each, but the angle contained by the two sides of one of them
greater than the angle contained by the two sides equal to them, of the
other; the base of that which has the greater angle, shall be greater than
the base of the other.
#
Let ABC, DEF be two triangles, which have the two sides AB,
AC, equal to the two DE, DF, each to each, namely AB equal to DE,
and AC to DF; but the angle BAC greater than the angle EDF.
Then the base BC shall be greater than the base EF.
A
D
B
C
E
G
F
Of the two sides DE, DF, let DE be not greater than DF,
at the point D, in the straight line DE,
make the angle EDG equal to the angle PAC; (1. 23.)
make DG equal to DF or AC, (1. 3.) and join EG, GF.
Then, because DE is equal to AB, and DG to AC,
the two sides DE, DG are equal to the two AB, AC, each to each,
and the angle EDG is equal to the angle BAC;
therefore the base EG is equal to the base BC. (1.4.)
And because DG is equal to DF in the triangle DFG,
therefore the angle DFG is equal to the angle DGF; (1.5.)
but the angle DGF is greater than the angle EGF; (ax. 9.)
therefore the angle DFG is also greater than the angle EGF;
much more therefore is the angle EFG greater than the angle EGF.
And because in the triangle EFG, the angle EFG is greater than
the angle EGF,
and that the greater angle is subtended by the greater side; (1. 19.)
therefore the side EG is greater than the side EF;
but EG was proved equal to BC;
therefore BC is greater than EF.
Wherefore if two triangles, &c.
Q.E.D.
BOOK I. PROP. XXV, XXVI.
23
1
PROPOSITION XXV. THEOREM.
If two triangles have two sides of the one equal to two sides of the
other, each to each, but the base of one greater than the base of the other ;
the angle contained by the sides of the one which has the greater base,
shall be greater than the angle contained by the sides, equal to them, of the
olher.
Let ABC, DEF be two triangles which have the two sides AB, AC
equal to the two sides DE, DF, each to each, namely, AB equal to DE,
and AC to DF; but the base BC greater than the base EF.
Then the angle BAC shall be greater than the angle EDF.
A
D
B
C
E
F
For, if the angle BAC be not greater than the angle EDF, it must
either be equal to it, or less than it.
If the angle BAC were equal to the angle EDF,
then the base BC would be equal to the base EF; (1. 4.)
but it is not equal,
therefore the angle BAC is not equal to the angle EDF.
Again, if the angle BAC were less than the angle EDF,
then the base BC would be less than the base EF; (1. 24.)
but it is not less,
therefore the angle BAC is not less than the angle EDF;
and it has been shewn, that the angle BAC is not equal to the angle EDF;
therefore the angle BAC is greater than the angle EDF.
Wherefore, if two triangles, &c. Q.E.D.
PROPOSITION XXVI. THEOREM.
If two triangles have two angles of the one equal to two angles of the
other, each to each, and one side equal to one side, viz. either the sides
adjacent to the equal angles in each, or the sides opposite to them; then
shall the other sides be equal, each to each, and also the third angle of the
one equal to the third angle of the other.
Let ABC, DEF be two triangles which have the angles ABC, BCA
equal to the angles DEF, EFD, each to each, namely ABC to DEF,
and BCA to EFD; also one side equal to one side.
First, let those sides be equal which are adjacent to the angles that
are equal in the two triangles, namely BC to EF.
Then the other sides shall be equal, each to each, namely AB to DE,
and AC to DF, and the third angle BAC to the third angle EDF.
A
D
B
C
E
F
24
EUCLID'S ELEMENTS.
For, if AB be not equal to DE, one of them must be greater than
the other.
Let AB be greater than DE,
make BG equal to ED, (1. 3.) and join GC.
Then in the two triangles GBC, DEF,
because GB is equal to DE, and BC to EF, (hyp.)
the two sides GB, BC are equal to the two DE, EF, each to each;
and the angle GBC is equal to the angle DEF;
therefore the base GC is equal to the base DF, (1. 4.)
and the triangle GBC to the triangle DEF,
and the other angles to the other angles, each to each, to which
the equal sides are opposite;
therefore the angle GCB is equal to the angle DFE;
but the angle DFE is, by the hypothesis, equal to the angle ACB;
wherefore also the angle GCB is equal to the angle ACB; (ax. 1.)
the less angle equal to the greater, which is impossible;
therefore AB is not unequal to DE,
that is, AB is equal to DE.
Hence, in the triangles ABC, DEF;
because AB is equal to DE, and BC to EF, (hyp.)
and the angle ABC is equal to the angle DEF; (hyp.)
therefore the base AC is equal to the base DF, (1. 4.)
and the third angle BAC to the third angle EDF.
Secondly, let the sides which are opposite to the equal angles in
each triangle be equal to one another, namely, AB equal to DE.
Then in this case likewise the other sides shall be equal, AC to DF,
and BC to EF, and also the third angle BAC to the third angle EDF.
A
D
B
HC
E
F
For if BC be not equal to EF, one of them must be greater than
the other.
Let BC be greater than EF; make BH equal to EF, (1.3.)
and join AH.
Then in the two triangles ABH, DEF,
because AB is equal DE, and BH to EF,
and the angle ABC to the angle DEF; (hyp.)
therefore the base AH is equal to the base DF, (1.4.)
and the triangle ABH to the triangle DEF,
and the other angles to the other angles, each to each, to which the
equal sides are opposite;
therefore the angle BHA is equal to the angle EFD;
but the angle EFD is equal to the angle BCA; (hyp.)
therefore the angle BHA is equal to the angle BCA, (ax. 1.)
that is, the exterior angle BHA of the triangle AHC, is
equal to its interior and opposite angle BCA; which is impossible; (1.16.)
wherefore BC is not unequal to EF,
that is, BC is equal to EF.
Hence, in the triangles ABC, DEF;
because AB is equal to DE, and BC to EF, (hyp.)
BOOK I. PROP. XXVI—XXVIII.
25
and the included angle ABC is equal to the included angle DEF; (hyp.)
therefore the base AC is equal to the base DF, (1.4.)
and the third angle BAC to the third angle EDF.
Wherefore if two triangles, &c.
PROPOSITION XXVII.
Q.E.D.
THEOREM.
If a straight line, falling on two other straight lines, make the alter-
nate angles equal to each other; these two straight lines shall be parallel.
Let the straight line EF, which falls upon the two straight lines AB,
CD, make the alternate angles AEF, EFD equal to one another.
Then AB shall be parallel to CD.

A E
B
G
C F
D
For, if AB be not parallel to CD,
AB and CD being produced will meet either towards A and C,
or towards B and D.
Let AB, CD be produced and meet towards B and D, in the point G.
Then GEF is a triangle,
and its exterior angle AEF is greater than the interior and
opposite angle EFG; (1.16.)
but the angle AEF is equal to the angle EFG; (hyp.)
therefore the angle AEF is greater than and equal to the angle EFG;
which is impossible.
Therefore AB, CD being produced do not meet towards B, D.
In like manner, it may be demonstrated, that they do not meet
when produced towards A, C.
But those straight lines in the same plane which meet neither way,
though produced ever so far, are parallel to one another; (def. 35.)
therefore AB is parallel to CD.
Wherefore, if a straight line, &c. Q.E.D.
PROPOSITION XXVIII. THEOREM.
If a straight line falling upon two other straight lines, makes the
exterior angle equal to the interior and opposite upon the same side of the
line; or makes the interior angles upon the same side together equal to two
right angles; the two straight lines shall be parallel to one another.
Let the straight line EF, which falls upon the two straight lines AB,
CD, make the exterior angle EGB equal to the interior and opposite
angle GHD upon the same side; or make the two interior angles BGH,
GHD on the same side together equal to two right angles.
Then AB shall be parallel to CD.
E

A
G
B
C
H
D
F
26
EUCLID'S ELEMENTS.
Because the angle EGB is equal to the angle GHD, (hyp.)
and the angle EGB equal to the angle AGH, (1.15.)
therefore the angle AGH is equal to the angle GHD; (ax. 1.)
and they are alternate angles,
therefore AB is parallel to CD.
(1.27.)
Again, because the angles BGH, GHD are together equal to two
right angles, (hyp.)
and that the angles AGH, BGH are also together equal to two
right angles; (1.13.)
therefore the angles AGH, BGH are equal to the angles BGH,
GHD; (ax. 1.)
take away the common angle BGH;
therefore the remaining angle AGH is equal to the remaining
angle GHD; (ax. 3.)
and they are alternate angles;
therefore AB is parallel to CD.
Wherefore, if a straight line, &c. Q.E.D.
(1. 27.)
PROPOSITION XXIX. THEOREM.
If a straight line fall upon two parallel straight lines, it makes the
alternate angles equal to one another; and the exterior angle equal to the
interior and opposite upon the same side; and likewise the two interior
angles upon the same side together equal to two right angles.
Let the straight line EF fall upon the parallel straight lines AB, CD.
Then the alternate angles AGH, GHD shall be equal to one another;
the exterior angle EGB shall be equal to the interior and oppositè
angle GHD upon the same side of the line EF;
and the two interior angles BGH, GHD upon the same side shall
be together equal to two right angles.
E

A
G
C
H
B
D
F
For, if the angle AGH be not equal to the alternate angle GHD,
let AGH be greater than GHD,
then because the angle AGH is greater than the angle GHD,
add to each of these unequals the angle BGH;
therefore the angles AGH, BGH are greater than the angles BGH,
GHD; (ax. 4.)
but the angles AGH, BGH are equal to two right angles; (1. 13.)
therefore the angles BGH, GHD are less than two right angles;
but those straight lines which, with another straight line falling
upon them, make the interior angles on the same side less than two
right angles, will meet together if continually produced; (ax. 12.)
therefore the straight lines AB, CD, if produced far enough, will
meet;
but they never meet, since they are parallel by the hypothesis;
therefore the angle AGH is not unequal to the angle GHD,
that is, the angle AGH is equal to the angle GHD:
but the angle AGH is equal to the angle EGB; (1.15.)
BOOK I. PROP. XXIX-XXXI.
27
therefore likewise the angle EGB is equal to the angle GHD: (ax. 1.)
add to each of them the angle BGH;
therefore the angles EGB, BGH are equal to the angles BGH,
GHD; (ax. 2.)
but EGB, BGH are equal to two right angles; (1. 13.)
therefore also BGH, GHD are equal to two right angles. (ax. 1.)
Wherefore, if a straight line, &c. Q. E.D.
PROPOSITION XXX. THEOREM.
Straight lines which are parallel to the same straight line are parallel
to each other.
Let the straight lines AB, CD be each of them parallel to EF.
Then shall AB be also parallel to CD.
G
A-
B
H
E
F
K
C
D
Let the straight line GHK cut AB, EF, CD.
Then because GHK cuts the parallel straight lines AB, EF,
therefore the angle AGH is equal to the alternate angle GHF. (1. 29.)
Again, because GHK cuts the parallel straight lines EF, CD,
therefore the exterior angle GHF is equal to the interior angle
HKD; (1. 29.)
and it was shewn that the angle AGH is equal to the angle GHF;
therefore the angle AGH is equal to the angle GKD;
and these are alternate angles;
therefore AB is parallel to CD. (1.27.)
Wherefore straight lines which are, &c.
PROPOSITION XXXI.
PROBLEM.
Q. E. D.
To draw a straight line through a given point parallel to a given
straight line.
Let A be the given point, and BC the given straight line.
It is required to draw through the point A a straight line parallel
to the straight line BC.
E
A F
B D
In the line BC take any point D, and join AD,
at the point A in the straight line AD, make the angle DAE equal
to the angle ADC; (1. 23.)
and produce the straight line EA to F.
Then EF shall be parallel to BC.
Because the straight line AD meets the two straight lines EF, BC,
and makes the alternate angles EAD, ADC, equal to one another,
therefore EF is parallel to BC. (1.27.)
Wherefore through the given point A, has been drawn a straight
line EAF parallel to the given straight line BC. Q.E. F.
"

28
EUCLID'S ELEMENTS.
PROPOSITION XXXII.
THEOREM.
If a side of any triangle be produced, the exterior angle is equal to the
two interior and opposite angles; and the three interior angles of every
triangle are together equal to two right angles.
Let ABC be a triangle, and let one of its sides BC be produced to D.
Then the exterior angle ACD shall be equal to the two interior and
opposite angles CAB, ABC:
and the three interior angles ABC, BCA, CAB shall be equal to
two right angles.
A
E

B
C
Through the point C draw CE parallel to the side BA. (1. 31.)
Then because CE is parallel to BA, and AC meets them,
therefore the angle ACE is equal to the alternate angle BAC. (1. 29.)
Again, because CE is parallel to AB, and BD falls upon them,
therefore the exterior angle ECD is equal to the interior and op-
posite angle ABC; (1.29.)
but the angle ACE was shewn to be equal to the angle BAC;
therefore the whole exterior angle ACD is equal to the two interior
and opposite angles CAB, ABC: (ax. 2.)
to each of these equals add the angle ACB,
therefore the angles ACD and ACB are equal to the three angles
CAB, ABC, and ACB; (ax. 2.)
but the angles ACD, ACB are equal to two right angles, (1. 13.)
therefore also the angles CAB, ABC, ACB are equal to two right
angles. (ax. 1.)
Wherefore, if a side of any triangle be produced, &c. Q. E.D.
COR. 1. All the interior angles of any rectilineal figure together
with four right angles, are equal to twice as many right angles as the
figure has sides.
D
E
A
B
For any rectilineal figure ABCDE can be divided into as many tri-
angles as the figure has sides, by drawing straight lines from a point F
within the figure to each of its angles.
Then, because the three interior angles of a triangle are equal to
two right angles, and there are as many triangles as the figure has sides,
therefore all the angles of these triangles are equal to twice as many
right angles as the figure has sides;
but the same angles of these triangles are equal to the interior angles
of the figure together with the angles at the point F:
and the angles at the point F, which is the common vertex of all the
triangles, are equal to four right angles, (1. 15. Cor. 2.)

BOOK I. PROP. XXXII, XXXIII,
29
therefore the same angles of these triangles are equal to the angles
of the figure together with four right angles;
but it has been proved that the angles of the triangles are equal to
twice as many right angles as the figure has sides;
therefore all the angles of the figure together with four right angles
are equal to twice as many right angles as the figure has sides.
COR. 2. All the exterior angles of any rectilineal figure, made by
producing the sides successively in the same direction, are together
equal to four right angles.

D
B
Since every interior angle ABC, together with its adjacent exterior
angle ABD, are equal to two right angles, (1. 13.)
therefore all the interior angles, together with all the exterior angles
of the figure, are equal to twice as many right angles as the figure has
sides;
but it has been proved by the foregoing corollary, that all the inte-
rior angles together with four right angles are equal to twice as many
right angles as the figure has sides;
therefore all the interior angles together with all the exterior angles
are equal to all the interior angles and four right angles, (ax. 1.)
take from these equals all the interior angles,
therefore all the exterior angles of the figure are equal to four right
angles. (ax. 2.)
PROPOSITION XXXIII.
THEOREM.
The straight lines which join the extremities of two equal and pa-
rallel straight lines towards the same parts, are also themselves equal and
parallel.
Let AB, CD be equal and parallel straight lines, and joined towards
the same parts by the straight lines AC, BD.
Then AC, BD shall be equal and parallel.
A
B
C
Join BC.
Ꭰ
Then because AB is parallel to CD, and BC meets them,
therefore the angle ABC is equal to the alternate angle BCD; (1. 29.)
and because AB is equal to CD, and BC common to the two tri-
angles ABC, DCB; the two sides AB, BC, are equal to the two DC,
CB, each to each, and the angle ABC was proved to be equal to the
angle BCD:
therefore the base AC is equal to the base BD, (1. 4.)
and the triangle ABC to the triangle BCD,
i
30
EUCLID'S ELEMENTS.
and the other angles to the other angles, each to each, to which the
equal sides are opposite;
therefore the angle ACB is equal to the angle CBD.
And because the straight line BC meets the two straight lines AC,
BD, and makes the alternate angles ACB, CBD equal to one another;
therefore AC is parallel to BD; (1.27.)
and AC was shewn to be equal to BD.
Therefore, straight lines which, &c. Q. E. D.
PROPOSITION XXXIV. THEOREM.
The opposite sides and angles of parallelograms are equal to one
another, and the diameter bisects them, that is, divides them into two equal
parts.
Let ACDB be a parallelogram, of which BC is a diameter.
Then the opposite sides and angles of the figure shall be equal to
one another; and the diameter BC shall bisect it.
A
B
C
Because AB is parallel to CD, and BC meets them,
therefore the angle ABC is equal to the alternate angle BCD. (1.29.)
And because AC is parallel to BD, and BC meets them,
therefore the angle ACB is equal to the alternate angle CBD. (1. 29.)
Hence in the two triangles ABC, CBD,
because the two angles ABC, BCA in the one, are equal to the two
angles BCD, CBD in the other, each to each;
and one side BC, which is adjacent to their equal angles, common
to the two triangles;
therefore their other sides are equal, each to each, and the third
angle of the one to the third angle of the other, (1. 26.)
namely, the side AB to the side CD, and AC to BD, and the
angle BAC to the angle BDC.
And because the angle ABC is equal to the angle BCD,
and the angle CBD to the angle ACB,
therefore the whole angle ABD is equal to the whole angle ACD; (ax. 2.)
and the angle BAC has been shewn to be equal to BDC;
therefore the opposite sides and angles of a parallelogram are equal
to one another.
Also the diameter BC bisects it.
For since AB is equal to CD, and BC common, the two sides AB,
BC are equal to the two DC, CB, each to each ;
and the angle ABC has been proved to be equal to the angle BCD;
therefore the triangle ABC is equal to the triangle BCD; (1.4.)
and the diameter BC divides the parallelogram ACDB into two equal
parts.
Q. E.D.
BOOK I. PROP. XXXV, XXXVI.
31
PROPOSITION XXXV.
THEOREM.
Parallelograms upon the same base, and between the same parallels,
are equal to one another.
Let the parallelograms ABCD, EBCF be upon the same base BC,
and between the same parallels AF, BC.
Then the parallelogram ABCD shall be equal to the parallelogram
EBCF.
A
D F
A
DE
F
B
C
B
C
AED F
B
C
If the sides AD, DF of the parallelograms ABCD, DBCF, opposite
to the base BC, be terminated in the same point D;
then it is plain that each of the parallelograms is double of the
triangle BDC; (1.34.)
and therefore the parallelogram ABCD is equal to the parallelo-
gram DBCF.
(ax. 6.)
But if the sides AD, EF, opposite to the base BC, be not terminated
in the same point;
Then, because ABCD is parallelogram,
therefore AD is equal to BC; (1.34.)
and for a similar reason, EF is equal to BC;
wherefore AD is equal to EF; (ax. 1.)
and DE is common;
therefore the whole, or the remainder, AE is equal to the whole, or
remainder DF; (ax. 2 or 3.)
and AB is equal to DC; (1. 34.)
hence in the triangles EAB, FDC,
because FD is equal to EA, and DC to AB,
and the exterior angle FDC is equal to the interior and opposite
angle EAB; (1.29.)
therefore the base FC is equal to the base EB, (1. 4.)
and the triangle FDC equal to the triangle EAB.
From the trapezium ABCF take the triangle FDC, and from the
same trapezium take the triangle EAB, and the remainders are
equal, (ax. 3.)
therefore the parallelogram ABCD is equal to the parallelogram EBCF.
Therefore parallelograms upon the same, &c. Q.E. D.
PROPOSITION XXXVI. THEOREM.
Parallelograms upon equal bases, and between the same parallels,
are equal to one another.
Let ABCD, EFGH be parallelograms upon equal bases BC, FG,
and between the same parallels AH, BG.
Then the parallelogram ABCD shall be equal to the parallelogram
EFGH.
A
DE
H

B
C
F
G
32
EUCLID'S ELEMENTS.
Join BE, CH.
Then because BC is equal to FG, (hyp.) and FG to EH, (1.34.)
therefore BC is equal to EH; (ax. 1.)
and these lines are parallels, and joined towards the same parts by
the straight lines BE, CH;
but straight lines which join the extremities of equal and parallel
straight lines towards the same parts, are themselves equal and
parallel; (1.33.)
therefore BE, CH are both equal and parallel;
wherefore EBCH is a parallelogram. (def. A.)
Then since the parallelograms ABCD, EBCH, are upon the same
base BC, and between the same parallels BC, AH;
therefore the parallelogram ABCD is equal to the parallelogram
EBCH. (1.35.)
For a similar reason, the parallelogram EFGH is equal to the
parallelogram EBCH;
therefore the parallelogram ABCD is equal to the parallelogram
EFGH. (ax. 1.)
Therefore parallelograms upon equal, &c. Q.E.D.
PROPOSITION XXXVII. THEOREM.
Triangles upon the same base, and between the same parallels, are
equal to one another.
Let the triangles ABC, DBC be upon the same base BC, and be-
tween the same parallels AD, BC.
Then the triangle ABC shall be equal to the triangle DBC.
E
A D
F

B
C
F:
Produce AD both ways to the points E,
through B draw BE parallel to CA, (1. 31.)
and through C draw CF parallel to BD.
Then each of the figures EBCA, DBCF is a parallelogram;
and EBCA is equal to DBCF, (1.35.) because they are upon the
same base BC, and between the same parallels BC, EF.
And because the diameter AB bisects the parallelogram EBCA,
therefore the triangle ABC is half of the parallelogram EBCA; (1.34.)
also because the diameter BC bisects the parallelogram DBCF,
therefore the triangle DBC is half of the parallelogram DBCF,
but the halves of equal things are equal; (ax. 7.)
therefore the triangle ABC is equal to the triangle DBC.
Wherefore triangles, &c.

Q. E.D.
PROPOSITION XXXVIII. THEOREM.
Triangles upon equal bases, and between the same parallels, are equal
to one another.
BOOK I. PROP, XXXVIII, XXXIX.
33
Let the triangles ABC, DEF be upon equal bases BC, EF, and
between the same parallels BF, AD.
Then the triangle ABC shall be equal to the triangle DEF.
G
A
D
H
ZN

B
CE
F
Produce AD both ways to the points G, H;
through B draw BG parallel to CA, (1. 31.)
and through F draw FH parallel to ED.
Then each of the figures GBCA, DEFH is a parallelogram;
and they are equal to one another, (1. 36.)
because they are upon equal bases BC, EF,
and between the same parallels BF, GH.
And because the diameter AB bisects the parallelogram GBCA,
therefore the triangle ABC is the half of the parallelogram GBCA; (1.34.)
also, because the diameter DF bisects the parallelogram DEFH,
therefore the triangle DEF is the half of the parallelogram DEFH;
but the halves of equal things are equal; (ax. 7.)
therefore the triangle ABC is equal to the triangle DEF.
Wherefore, triangles upon equal bases, &c.
Q.E.D.
PROPOSITION XXXIX. THEOREM.
Equal triangles upon the same base and upon the same side of it, are
between the same parallels.
Let the equal triangles ABC, DBC be upon the same base BC, and
upon the same side of it.
Then the triangles ABC, DBC shall be between the same parallels.
A
E
D
B
Join AD; AD shall be parallel to BC.
For, if it is not, through the point A draw AE parallel to BC, (1. 31.)
meeting BD or BD produced in E, and join EC.
Then the triangle ABC is equal to the triangle EBC, (1.37.)
because they are upon the same base BC,
and between the same parallels BC, AE;
but the triangle ABC is equal to the triangle DBC; (hyp.)
therefore the triangle DBC is equal to the triangle EBC,
the greater equal to the less triangle, which is impossible:
therefore AE is not parallel to BC.
In the same manner it can be demonstrated, that no other line but
AD is parallel to BC;
AD is therefore parallel to BC.
Wherefore, equal triangles upon, &c.
Q.E.D.
3
34
EUCLID'S ELEMENTS.
PROPOSITION XL. THEOREM.
Equal triangles, upon equal bases in the same straight line, and
towards the same parts, are between the same parallels.
Let the equal triangles ABC, DEF be upon equal bases BC, EF,
in the same straight line BF, and towards the same parts.
Then they shall be between the same parallels.
A D

G
B
CE
F
Join AD; AD shall be parallel to BF.
For, if it is not, through A draw AG parallel to BF, (1. 31.)
meeting ED, or ED produced in G, and join GF.
Then the triangle ABC is equal to the triangle GEF,
because they are upon equal bases BC, EF,
and between the same parallels BF, AG;
(1.38.)
but the triangle ABC is equal to the triangle DEF; (hyp.)
therefore the triangle DEF is equal to the triangle GEF, (ax. 1.)
the greater equal to the less triangle, which is impossible:
therefore AG is not parallel to BF.
And in the same manner it can be demonstrated, that there is no
other parallel to it but AD;
AD is therefore parallel to BF.
Wherefore, equal triangles upon, &c.
Q. E.D.
PROPOSITION XLI. THEOREM.
If a parallelogram and a triangle be upon the same base, and between
the same parallels; the parallelogram shall be double of the triangle.
Let the parallelogram ABCD, and the triangle EBC be upon the
same base BC, and between the same parallels BČ, AE.
Then the parallelogram ABCD shall be double of the triangle EBC.
A
DE
B
Join AC.
Then the triangle ABC is equal to the triangle EBC, (1.37.)
because they are upon the same base BC, and between the same
parallels BC, AE.
But the parallelogram ABCD is double of the triangle ABC,
because the diameter AC bisects it; (1. 34.)
wherefore ABCD is also double of the triangle EBC.
Therefore, if a parallelogram and a triangle, &c.

Q.E. D.
BOOK I. PROP. XLII, XLIII.
35
PROPOSITION XLII. PROBLEM.
To describe a parallelogram that shall be equal to a given triangle,
and have one of its angles equal to a given rectilineal angle.
Let ABC be the given triangle, and D the given rectilineal angle.
It is required to describe a parallelogram that shall be equal to the
given triangle ABC, and have one of its angles equal to D.
A F G
BE C
Bisect BC in E, (1. 10.) and join AE;
at the point E in the straight line EC,
make the angle CEF equal to the angle D; (1. 23.)
through Å draw AFG parallel to BC, (1. 31.)
and through C draw CG parallel to ÈF.
Then the figure CEFG is a parallelogram. (def. A.)
And because the triangles ABE, AEC are on the equal bases BE,
EC, and between the same parallels BC, AG;
they are therefore equal to one another; (1. 38.)
and therefore the triangle ABC is double of the triangle AEC;
but the parallelogram FECG is double of the triangle AEC, (1. 41.)
because they are upon the same base EC, and between the same
parallels EC, AG;
therefore the parallelogram FECG is equal to the triangle ABC, (ax. 6.)
and it has one of its angles CEF equal to the given angle D.
Wherefore a parallelogram FECG has been described equal to the
given triangle ABC, and having one of its angles CEF equal to the
given angle D. Q.E.F.


PROPOSITION XLIII. THEOREM.
The complements of the parallelograms which are about the diameter
of any parallelogram, are equal to one another.
Let ABCD be a parallelogram, of which the diameter is AC: and
EH, GF the parallelograms about AC, that is, through which AC passes:
also BK, KD the other parallelograms which make up the whole
figure ABCD, which are therefore called the complements.
Then the complement BK shall be equal to the complement KD.
A H
D

E
K
F
BG
C
Because ABCD is a parallelogram, and AC its diameter,
therefore the triangle ABC is equal to the triangle ADC. (1. 34.)
3—2
36
EUCLID'S ELEMENTS.
Again, because EKHA is a parallelogram, and AK its diameter,
therefore the triangle AEK is equal to the triangle AHK; (1. 34.)
and for the same reason, the triangle KGC is equal to the triangle KFC.
Wherefore the two triangles AEK, KG are equal to the two
triangles AHK, KFC, (ax. 2.
but the whole triangle ABC is equal to the whole triangle ADC;
therefore the remaining complement BK is equal to the remaining
complement KD. (ax. 3.)
Wherefore the complements, &c. Q.E.D.
PROPOSITION XLIV. PROBLEM.
To a given straight line to apply a parallelogram, which shall be
equal to a given triangle, and have one of its angles equal to a given
rectilineal angle.
Let AB be the given straight line, and C the given triangle, and D
the given rectilineal angle.
It is required to apply to the straight line AB a parallelogram equal
to the triangle C, and having an angle equal to D.
FE
K

D
C
G
M
B
HA
L
Make the parallelogram BEFG equal to the triangle C, (1. 42.)
and having the angle EBG equal to the angle D,
so that BE be in the same straight line with AB;
produce FG to H, through A draw AH parallel to BG or
EF, (1. 31.) and join ÍВ.
Then because the straight line HF falls upon the parallels AH, EF,
therefore the angles AHF, HFE are together equal to two
right angles; (1. 29.)
wherefore the angles BHF, HFE are less than two right angles:
but straight lines which with another straight line make the two
interior angles upon the same side less than two right angles, do meet
if produced far enough: (ax. 12.)
therefore HB, FE shall meet, if produced;
let them be produced and meet in K,
through K draw KL parallel to EA or FH,
and produce HA, GB to meet KL in the points L, M.
Then HLKF is a parallelogram, of which the diameter is HK;
and AG, ME, are the parallelograms about HK;
also LB, BF are the complements;
therefore the complement LB is equal to the complement BF; (1. 43.)
but the complement BF is equal to the triangle C; (constr.)
wherefore LB is equal to the triangle C.
And because the angle GBE is equal to the angle ABM, (1. 15.)
and likewise to the angle D; (constr.)
BOOK I. PROP. XLIV, XLV.
37.
therefore the angle ABM is equal to the angle D. (ax. 1.) -
Therefore to the straight line AB, the parallelogram LB is applied,
equal to the triangle C, and having the angle ABM equal to the
angle D. Q.E.F.
PROPOSITION XLV. PROBLEM.
To describe a parallelogram equal to a given rectilineal figure, and
having an angle equal to a given rectilineal angle.
Let ABCD be the given rectilineal figure, and E the given rec-
tilineal angle.
It is required to describe a parallelogram that shall be equal to the
figure ABCD, and having an angle equal to E.
A
D
F GL
E
B
C
KHM
Join DB.
Describe the parallelogram FH equal to the triangle ADB, and
having the angle FKH equal to the angle E; (1. 42.)
to the straight line GH, apply the parallelogram GM equal to the tri-
angle DBC, having the angle GHM equal to the angle E. (1. 44.)
Then the figure FKML shall be the parallelogram required.
Because the angle E is equal to each of the angles FKH, GHM,
therefore the angle FKH is equal to the angle GHM;
add to each of these equals the angle KHG;
therefore the angles FKH, KHG are equal to the angles KHG, GHM;
but FKH, KHG are equal to two right angles; (1. 29.)
therefore also KHG, GĤM are equal to two right angles;
and because at the point H, in the straight line GH, the two straight
lines KH, HM, upon the opposite sides of it, make the adjacent angles
equal to two right angles,
therefore HK is in the same straight line with HM. (1.14.)
And because the line HG meets the parallel KM, FG,
therefore the angle MHG is equal to the alternate angle HGF; (1.29.)
add to each of these equals the angle HGL;
therefore the angles MHG, HGL are equal to the angles HGF, HGL;
but the angles MHG, HGL are equal to two right angles; (1.29.)
therefore also the angles HGF, HGL are equal to two right angles,
and therefore FG is in the same straight line with GL. (1. 14.)
And because KF is parallel to HG, and HG to ML,
therefore KF is parallel to ML; (1.30.)
and KM has been proved parallel to FL,
wherefore the figure FKML is a parallelogram;
and since the triangle ABD is equal to the parallelogram HF,
and the triangle BDC to the parallelogram GM;

.
38
EUCLID'S ELEMENTS.
therefore the whole rectilineal figure ABCD is equal to the whole
parallelogram KFLM.
Therefore the parallelogram KFLM has been described equal to
the given rectilineal figure ABCD, having the angle FKM equal to the
given angle E. Q. E. F.
COR. From this it is manifest how, to a given straight line, to apply
a parallelogram, which shall have an angle equal to a given rectilineal
angle, and shall be equal to a given rectilineal figure; viz. by applying
to the given straight line a parallelogram equal to the first triangle
ABD, (1.44.) and having an angle equal to the given angle.

PROPOSITION XLVI. PROBLEM.
To describe a square upon a given straight line.
Let AB be the given straight line.
It is required to describe a square upon AB.
D
E
A
B
From the point A draw AC at right angles to AB; (1.11.)
make AD equal to AB, (1. 3.)
through the point D draw DE parallel to AB, (1.31).
and through B, draw BE parallel to AD;
therefore ABED is a parallelogram ;
whence AB is equal to DE, and AD to BE; (1.34.)
but BA is equal to AD,
therefore the four lines BA, AD, DE, EB are equal to one another,
and the parallelogram ADEB is equilateral.
It has likewise all its angles right angles
since AD meets the parallels AB, DE,
therefore the angles BAD, ADE are equal to two right angles; (1. 29.)
but BAD is a right angle; (constr.)
therefore also ADE is a right angle.
But the opposite angles of parallelograms are equal; (1.34.)
therefore each of the opposite angles ABE, BED is a right angle;
wherefore the figure ÄDEB is rectangular, and it has been proved
to be equilateral;
therefore the figure ADED is a square, (def. 30.) and it is described
upon the given straight line AB. Q.E.F.
COR. Hence, every parallelogram that has one right angle, has all
its angles right angles.


PROPOSITION XLVII. THEOREM.
In any right-angled triangle, the square which is described upon the
side subtending the right angle, is equal to the squares described upon the
sides which contain the right angle.
BOOK I. PROP. XLVII.
39
Let ABC be a right-angled triangle, having the right angle BAC.
Then the square described upon the side BC, shall be equal to the
squares described upon BA, AC.
F
G

H
K
B
C
D LE
On BC describe the square BDEC, (1. 46.)
and on BA, AC the squares GB, HC;
through A draw AL parallel to BD or CE; (1, 31.)
and join AD, FC.
Then because the angle BAC is a right angle, (hyp.)
and that the angle BAG is a right angle, (def. 30.)
the two straight lines AC, AG upon the opposite sides of AB, make
with it at the point A the adjacent angles equal to two right angles;
therefore CA is in the same straight line with AG. (I.14.)
For the same reason, BA and AH are in the same straight line.
And because the angle DBC is equal to the angle FBA,
each of them being a right angle,
add to each of these equals the angle ABC,
therefore the whole angle DBA is equal to the whole angle FBC. (ax. 2.)
And because the two sides AB, BD, are equal to the two sides FB,
BC, each to each, and the included angle ABD is equal to the included
angle FBC,
therefore the base AD is equal to the base FC, (1. 4.)
and the triangle ABĎ to the triangle FBC.
Now the parallelogram BL is double of the triangle ABD, (1.41.)
because they are upon the same base BD, and between the same
parallels BD, AL;
also the square GB is double of the triangle FBC,
because these also are upon the same base FB, and between the
same parallels FB, GC.
But the doubles of equals are equal to one another; (ax. 6.)
therefore the parallelogram BL is equal to the square GB.
Similarly, by joining AE, BK, it can be proved,
that the parallelogram CL is equal to the square HC.
Therefore the whole square BDEC is equal to the two squares
GB, HC; (ax. 2.)
and the square BEDC is described upon the straight line BC, and
the squares GB, HC, upon AB, AC:
therefore the square upon the side BC is equal to the squares upon
the sides AB, AC.
Therefore, in any right-angled triangle, &c. Q. E. D.

40
´EUCLID'S ELEMENTS.
1
PROPOSITION XLVIII. THEOREM.
If the square described upon one of the sides of a triangle, be equal
to the squares described upon the other two sides of it; the angle con-
tained by these two sides is a right angle.
Let the square described upon BC, one of the sides of the triangle
ABC, be equal to the squares upon the other two sides AB, AC.
Then the angle BAỠ shall be a right angle.
A
D
B
From the point A draw AD at right angles to AC, (1. 11.)
make AD equal to AB, and join DC.
Then because AD is equal to AB,
therefore the square of AD is equal to the square of AB;
to each of these equals add the square of AC;
therefore the squares of AD, AC are equal to the squares of AB, AC:
but the squares of AD, AC are equal to the square of DC, (1.47.)
because the angle DAC is a right angle;
and the square of BC, by hypothesis, is equal to the squares of BA, AC;
therefore the square of DC is equal to the square of BC;
and therefore the side DC is equal to the side BC.
And because the side AD is equal to the side AB,
and AC is common to the two triangles DAC, BAC;
the two sides DA, AC, are equal to the two BA, AC, each to each;
and the base DC has been proved to be equal to the base BC;
therefore the angle DAC is equal to the angle BAC; (1.8.)
but DAC is a right angle ;
therefore also BAC is a right angle.
Therefore, if the square described upon, &c. Q. E. D.
NOTES TO BOOK I.
ON THE DEFINITIONS.
GEOMETRY is one of the most perfect of the deductive Sciences, and seems to
rest on the simplest inductions from experience and observation.
The first principles of Geometry are therefore in this view consistent hypotheses
founded on facts cognizable by the senses, and it is a subject of primary importance to
draw a distinction between the conception of things and the things themselves. These
hypotheses do not involve any property contrary to the real nature of the things, and,
consequently, cannot be regarded as arbitrary, but, in certain respects, agree with the
conceptions which the things themselves suggest to the mind through the medium of the
senses. The essential definitions of Geometry therefore being deductions from observa-
tion and experience, rest ultimately on the evidence of the senses.
It is by experience we become acquainted with the existence of individual forms
of magnitudes; but by the mental process of abstraction, which begins with a particular
instance, and proceeds to the general idea of all objects of the same kind, we attain to
the general conception of those forms which come under the same general idea.
The essential definitions of Geometry express generalized conceptions of real existences
in their most perfect ideal forms; the laws and appearances of nature, and the operations
of the human intellect being supposed uniform and consistent.
But in cases where the subject falls under the class of simple ideas, the terms of the
definition so called are no more than merely equivalent expressions. The simple idea
described by a proper term or terms, does not in fact admit of definition properly so
called. The definitions in Euclid's Elements may be divided into two classes, those
which merely explain the meaning of the terms employed, and those, which, besides
explaining the meaning of the terms, suppose the existence of the things described in
the definitions.
Definitions in Geometry cannot be of such a form as to explain the nature and
properties of the figures defined; it is sufficient that they give marks whereby the thing
defined may be distinguished from every other of the same kind. It will at once be
obvious, that the definitions of Geometry, one of the pure sciences, being abstractions of
space, are not like the definitions in any one of the physical sciences. The discovery
of any new physical facts may render necessary some alteration or modification in the
definitions of the latter.
Def. I.
Simson has adopted Theon's definition of a point. Euclid's definition is,
onμeîov ẻœtiv où µépos ovồév, "A point is that, of which there is no part," or which
cannot be parted or divided, as it is explained by Proclus. The Greek term onμetov,
literally means, a visible sign or mark on a surface, in other words, a physical point.
The English term point, means the sharp end of any thing, or a mark made by it.
The word point comes from the Latin punctum, through the French word point. Neither
of these terms, in its literal sense, appears to give a very exact notion of what is to
be understood by a point in Geometry.
Euclid's definition of a point merely expresses a negative property, which excludes
the proper and literal meaning of the Greek term, as applied to denote a physical point,
or a mark which is visible to the senses.
a
42
EUCLID'S ELEMENTS.
μονας θέσιν
Pythagoras defined a point to be povas Déoi exovoa, "a monad having position."
By uniting the positive idea of position, with the negative idea of defect of magnitude,
the conception of a point in Geometry may be rendered perhaps more intelligible. A point
may then be defined to be that which has no magnitude, but position only.
Def. 11. Every visible line has both length and breadth, and it is impossible to
draw any line whatever which shall have no breadth. The definition requires the con-
ception of the length only of the line to be considered, abstracted from, and independently
of, all idea of its breadth.
Def. III. This definition renders more intelligible the exact meaning of the definition
of a point: and we may add, that, in the Elements, Euclid supposes that the intersection
of two straight lines is a point, and that two straight lines can intersect each other in
one point only.
Def. IV.
The straight line or right line is a term so clear and intelligible as to
be incapable of becoming more so by formal definition. Euclid's definition is Evoeta
γραμμὴ ἐστιν, ἥτις ἐξ ἴσου τοῖς ἐφ' ἑαυτῆς σημείοις κεῖται, wherein he states it to
lie evenly, or equally, or upon an equality, (¿¿ loov) between its extremities, and which
Proclus explains as being stretched between its extremities, ἡ ἐπ᾿ ἄκρων τεταμένη.
If the line be conceived to be drawn on a plane surface, the words ég loov may
mean, that no part of the line which is called a straight line deviates either from one
side or the other of the direction which is fixed by the extremities of the line; and thus
it may be distinguished from a curved line, which does not lie, in this sense, evenly
between its extreme points. If the line be conceived to be drawn in space, the words
¿¿ loov, must be understood to apply to every direction on every side of the line between
its extremities.
Every straight line situated in a plane is considered to have two sides; and when the
direction of a line is known, the line is said to be given in position; also, when the
length is known or can be found, it is said to be given in magnitude.
From the definition of a straight line, it follows, that two points fix a straight line in
position, which is the foundation of the first and second postulates. Hence straight lines
which are proved to coincide in two or more points, are called "one and the same straight
line," Prop. 14. Book 1., or, which is the same thing, that, "Two straight lines cannot
have a common segment," as Simson shews in his Corollary to Prop. 11, Book 1.
Archimedes defined "a straight line to be the shortest distance between two points;"
but this is a theorem considered by Euclid as requiring proof.
The following definition of straight lines has also been proposed. "Straight lines are
those which, if they coincide in any two points, coincide as far as they are produced."
But this is rather a criterion of straight lines, and analogous to the eleventh axiom,
which states that, "all right angles are equal to one another," and suggests that all straight
lines may be made to coincide wholly, if the lines be equal; or partially if the lines be
of unequal lengths. A definition should properly be restricted to the description of
the thing defined, as it exists, independently of any comparison of its properties or of
tacitly assuming the existence of axioms.
Def. VII. Euclid's definition of a plane surface is, 'Eπíñedos étiqáveiá éotiv, „tis
ἐξ ἴσου ταῖς ἐφ' ἑαυτῆς εὐθείαις κεῖται, “A plane surface is that which lies evenly
or equally with the straight lines in it;" instead of which Simson has given the definition
which was originally proposed by Hero the Elder. A plane superficies may be supposed
to be situated in any position, and to be continued in every direction to any extent.
Def. VIII. Simson remarks that this definition seems to include the angles formed by
two curved lines, or a curve and a straight line, as well as that formed by two straight lines.
Angles made by straight lines only, are treated of in Elementary Geometry.
Def. IX. It is of the highest importance to attain a clear conception of an angle.
NOTES TO BOOK I.
43
The literal meaning of the term angulus suggests the Geometrical conception of an angle,
which may be regarded as formed by the divergence of two straight lines from a point.
In the definition of an angle, the magnitude of the angle is independent of the lengths of
the two lines by which it is included; their mutual divergence from the point at which
they meet, is the criterion of the magnitude of an angle, as it is pointed out in the
succeeding definitions. The point at which the two lines meet is called the vertex of
the angle, and must not be confounded with the magnitude of the angle itself. The
right angle is fixed in magnitude, and, on this account, it is made the subject with
which all other angles in Geometry are compared.
Two straight lines which actually intersect one another, or which when produced
would intersect, are said to be inclined to one another, and the inclination of the two lines
is determined by the angle which they make with one another.
Def. x.
It may be here observed that in the Elements, Euclid always assumes
that when one line is perpendicular to another line, the latter is also perpendicular to the
former; and always calls a right angle, òp¤n ywvía; but a straight line, evleîa ypaµµn.
Def. XIX. This has been restored from Proclus, as it seems to have a meaning in
the constructions of Prop. 14, Book 11; the first case of Prop. 33, Book III, and
Prop. 13, Book v1. The definition of the segment of a circle is not once alluded to
in Book 1, and is not required before the discussion of the properties of the circle in
Book III.
Proclus remarks on this definition: "Hence you may collect that the centre
has three places. For it is either within the figure, as in the circle; or in its perimeter,
as in the semi-circle; or without the figure, as in certain conic lines."
Def. XXIV-XXIX. Triangles are divided into three classes by reference to the relations
of their sides, and into three other classes by reference to their angles. A further classifi-
cation may be made by considering both the relation of the sides and angles in each triangle.
In Simson's definition of the isosceles triangle, the word only must be omitted, as the
equilateral triangle is considered isosceles in Prop. 15, Book Iv. Objection has been
made to the definition of an acute-angled triangle. It is said that it cannot be admitted
as a definition, that all the three angles of a triangle are acute, which is supposed in
Def. 29. It may be replied, that the definitions of the three kinds of angles point out
and seem to supply a foundation for a similar distinction of triangles.
Def. xxx-xxXIV. The definitions of quadrilateral figures are liable to objection. All
of them, except the trapezium, fall under the general idea of a parallelogram; but as
Euclid has defined parallel straight lines after he had defined four-sided figures, no other
arrangement could be adopted than the one he has followed; and for which there ap-
peared to him, without doubt, some probable reasons. Sir Henry Savile, in his Seventh
Lecture, remarks on some of the definitions of Euclid, "Nec dissimulandum aliquot
harum in manibus exiguum esse usum in Geometrià." A few verbal emendations have
been made in some of them.
A square is a four-sided plane figure having all its sides equal, and one angle a right
angle: because it is proved in Prop. 46, Book 1, that if a parallelogram have one angle
a right angle, all its angles are right angles.
An oblong in the same manner may be defined as a plane figure of four sides having
only its opposite sides equal, and one of its angles a right angle.
A rhomboid is a four-sided plane figure having only its opposite sides equal to one
another and its angles not right angles.
Sometimes an irregular four-sided figure which has two sides parallel, is called a
trapezoid.
Def. xxxv. It is possible for two right lines never to meet when produced, and
not be parallel.
Def. A. The term parallelogram literally implies a figure formed by parallel
44
EUCLID'S ELEMENTS.
X
straight lines, and may consist of four, six, eight, or any even number of sides, where
every two of the opposite sides are parallel to one another.
In the Elements, however, the term is restricted to four-sided figures, and includes
the four species of figures named in the Definitions XXX-XXXIII.
The synthetic method is followed by Euclid not only in the demonstrations of the
propositions, but also in laying down the definitions. He commences with the simplest
abstractions, defining a point, a line, an angle, a superficies, and their different varieties.
This mode of proceeding involves the difficulty, almost insurmountable of defining
satisfactorily the elementary abstractions of Geometry. Simson observes that it is
necessary to consider a solid, that is a magnitude which has length, breadth, and thick-
ness, in order to understand aright the definitions of a point, a line, and a superficies.
A solid or volume considered apart from its physical properties, suggests the idea of
the surfaces by which it is bounded: a surface, the idea of the line or lines which form
its boundaries: and a finite line, the points which form its extremities. A solid is
therefore bounded by surfaces; a surface is bounded by lines; and a line is termi-
nated by two points. A point marks position only: a line has one dimension, length
only, and defines distance: a superficies has two dimensions, length and breadth, and
defines extension: and a solid has three dimensions, length, breadth, and thickness,
and defines some definite portion of space.
It may also be remarked that two points are sufficient to determine the position
of a straight line, and three points not in the same straight line, are necessary to fix
the position of a plane.
ON THE POSTULATES.
THE definitions assume the possible existence of straight lines and circles, and the
postulates predicate the possibility of drawing and of producing straight lines, and of
describing circles. The postulates form the principles of construction assumed in
the Elements; and are, in fact, problems, the possibility of which is admitted to
be self-evident, and to require no proof.
It must, however, be carefully remarked, that the third postulate only admits that
when any line is given in position and magnitude, a circle may be described from
either extremity of the line as a centre, and with a radius equal to the length of the
line, as in Prop. 1, Book 1. It does not admit the description of a circle with any
other point as a centre than one of the extremities of the given line.
Prop. 2, Book 1. shews how, from any given point, to draw a straight line equal
to another straight line which is given in magnitude and position.
ON THE AXIOMS.
AXIOMS are usually defined to be self-evident truths, which cannot be rendered
more evident by demonstration; in other words, the axioms of Geometry are theorems;
the truth of which is admitted without proof. It is by experience we first become
acquainted with the different forms of geometrical magnitudes, and the axioms, or the
fundamental ideas of their equality or inequality appear to rest on the same basis.
The conception of the truth of the axioms does not appear to be more removed from
experience than the conception of the definitions.
These axioms, or first principles of demonstration, are such theorems as cannot be
resolved into simpler theorems, and no theorem ought to be admitted as a first principle
of reasoning which is capable of being demonstrated.
An axiom and its converse should both be of such a nature as that neither of
them should require a formal demonstration.
NOTES TO BOOK I.
45
The first and most simple idea, derived from experience, is, that every magnitude
fills a certain space, and that several magnitudes may fill the same space.
All the knowledge we have of magnitude is purely relative, and the most simple
relations are those of equality and inequality. In the comparison of magnitudes, some
are considered as given or known, and the unknown are compared with the known,
and conclusions are synthetically deduced with respect to the equality or inequality
of the magnitudes under consideration. In this manner we form our idea of equality,
which is thus formally stated in the eighth axiom: "Magnitudes which coincide with
one another, that is, which exactly fill the same space, are equal to one another."
Every specific definition is referred to this universal principle. With regard to a
few more general definitions which do not furnish an equality, it will be found that
some hypothesis is always made reducing them to that principle, before any theory
is built upon them. As for example, the definition of a straight line is to be referred her aprited
to the tenth axiom; the definition of a right angle to the eleventh axiom; and the
definition of parallel straight lines to the twelfth axiom.
It is called the principle of superposition, or, the mental process by which one
Geometrical magnitude may be conceived to be placed on another, so as exactly to
coincide with it, in the parts which are made the subject of comparison. Thus, if
one straight line be conceived to be placed upon another, so that their extremities
are coincident, the two straight lines are equal. If the directions of two lines which
include one angle, coincide with the directions of the two lines which contain another
angle, where the points, from which the angles diverge, coincide, then the two angles
are equal: the lengths of the lines not affecting in any way the magnitudes of the
angles. When one plane figure is conceived to be placed upon another, so that the
boundaries of one exactly coincide with the boundaries of the other, then the two plane
figures are equal. It may also be remarked, that the converse of this proposition is
also true, namely, that when two magnitudes are equal, they coincide with one another.
This explanation of Geometrical equality appears to be out of its proper place.
The definitions of the forms of magnitudes naturally come first, and the criterion of
their equality appears as naturally to follow. If the first seven axioms are to be re-
stricted to Geometrical magnitudes, the eighth ought to have preceded them. Perhaps
Euclid intended that the first seven axioms should be applicable to numbers as well
as to Geometrical magnitudes, and this is in accordance with the words of Proclus,
who calls the axioms, common notions, not peculiar to the subject of Geometry.
The eighth axiom is properly the definition of Geometrical equality.
Axiom v. It may be observed that when equal magnitudes are taken from un-
equal magnitudes, the greater remainder exceeds the less remainder by as much as
the greater of the unequal magnitudes exceeds the less.
Axiom IX. The whole is greater than its part, and conversely, the part is less than
the whole. This axiom appears to assert the contrary of the eighth axiom, namely,
that two magnitudes, of which one is greater than the other, cannot be made to coincide
"with one another.
Axiom x. The property of straight lines expressed by the tenth axiom, namely,
"that two straight lines cannot enclose a space", is obviously implied in the definition
of straight lines; for if they enclosed a space, they could not coincide between their
extreme points, when the two lines are equal.
Axiom XI.
This axiom has been asserted to be a demonstrable theorem. If an
angle be admitted to be a species of magnitude, this axiom is only a particular appli-
cation of the eighth axiom to right angles.
Axiom XII. See the notes on Prop. xxix, Book 1.
JAL
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46
EUCLID'S ELEMENTS.
f
ON THE PROPOSITIONS.
THE deductive truths of Geometry are called propositions, which are divided into
two classes, problems and theorems. A proposition, as the term imports, is something
proposed; it is a problem, when some Geometrical construction is required to be effected:
and it is a theorem when some Geometrical property is to be demonstrated. Every
Proposition is naturally divided into two parts; a problem consists of the data, or
things given; and the quæsita, or things required: a theorem, consists of the premisses,
hypothesis, or the properties admitted; and the conclusion, or predicate, or properties
to be demonstrated.
Hence the distinction between a problem and a theorem is this, that a problem
consists of data and quæsita, and requires solution: and a theorem consists of the
hypothesis and the predicate, and requires demonstration.
The connected course of reasoning by which any Geometrical truth is established
is called a demonstration. It is called a direct demonstration when the predicate of
the proposition is inferred directly from the premisses, as the conclusion of a series
of successive deductions. The demonstration is called indirect, when the conclusion
shews that the introduction of any other supposition contrary to the hypothesis stated
in the proposition, necessarily leads to an absurdity.
The course pursued in the demonstrations of the propositions in Euclid's Elements
of Geometry, is always to refer directly to some expressed principle, to leave nothing
to be inferred from vague expressions, and to make every step of the demonstrations
the object of the understanding.
It has been maintained by some philosophers that a genuine definition contains
some property or properties which can form a basis for demonstration, and that the
science of Geometry is deduced from the definitions, and that on them alone the
demonstrations depend. Others have maintained that a definition explains only the
meaning of a term, and does not embrace the nature and properties of the thing
defined.
If the propositions usually called postulates and axioms are either tacitly assumed
or expressly stated in the definitions; in this view, demonstrations may be said to be
legitimately founded on definitions. If, on the other hand, a definition is simply an
explanation of the meaning of a term, whether abstract or concrete, by such marks
as may prevent a misconception of the thing defined; it will be at once obvious that
some constructive and theoretic principles must be assumed besides the definitions to
form the grounds of legitimate demonstration. These principles we conceive to be the
postulates and axioms. The postulates describe constructions which may be admitted
as possible by direct appeal to our experience; and the axioms assert general theoretic
truths so simple and self-evident as to require no proof, but to be admitted as the
assumed first principles of demonstration. Under this view all Geometrical reasonings
proceed upon the admission of the hypotheses assumed in the definitions, and the un-
questioned possibility of the postulates, and the truth of the axioms.
The general theorems of Geometry are demonstrated by means of syllogisms founded
on the axioms and definitions. The form of syllogism employed in Geometrical reason-
ings is of the simplest character. Every syllogism consists of three propositions, of
which, two are called the premisses, and the third, the conclusion. These propositions
contain three terms, the subject and predicate of the conclusion, and the middle term
which connects the predicate and the conclusion together. The subject of the conclu-
sion is called the minor, and the predicate of the conclusion is called the major term,
of the syllogism. The major term appears in one premiss, and the minor term in
-
་
NOTES TO BOOK I.
47
the other, with the middle term which is in both premisses. That premiss which
contains the middle term and the major term, is called the major premiss; and that
which contains the middle term and the minor term, is called the minor premiss of
the syllogism. As an example, we may take the first syllogism in the demonstration
of Prop. 1, Book 1, wherein it will be seen that the middle term is the subject of the
major premiss and the predicate of the minor.
Major premiss. Because the straight line AB is equal to the straight line AC;
Minor premiss. and, because the straight line BC is equal to the straight line AB;
Conclusion. therefore the straight line BC is equal to the straight line AC.
Here, BC is the subject, and AC the predicate of the conclusion.
BC is the subject, and AB the predicate of the minor premiss.
AB is the subject, and AC the predicate of the major premiss.
Also, AC is the major term, BC the minor term, and AB the middle term of
the syllogism.
In this syllogism, it may be remarked that the definition of a straight line is
assumed, and the definition of the Geometrical equality of two straight lines; also
that a general theoretic truth, or axiom, forms the ground of the conclusion. And
further, though it be impossible to make any point, mark or sign, (onueîov) which
has not both length and breadth, and any line which has not both length and breadth;
the demonstrations in Geometry do not on this account become invalid.
For they
are pursued on the hypothesis that the point has no parts but position only: and the
line has length only, but no breadth or thickness; also that the surface has length
and breadth only, but no thickness: and all the conclusions at which we arrive are
independent of every other consideration.
Every proposition, when complete, may be divided into six parts, as Proclus has
pointed out in his commentary.
1. The proposition or general enunciation which states in general terms the con-
ditions of the problem or theorem.
2. The exposition or particular enunciation which exhibits the subject of the pro-
position in particular terms as a fact, and refers it to some diagram described.
3.
The determination contains the predicate in particular terms, as it is pointed out
in the diagram.
4. The construction applies the postulates to prepare the diagram for the de-
monstration.
5. The demonstration is the connexion of syllogisms, which prove the truth or
falsehood of the theorem, the possibility or impossibility of the problem, in that par-
ticular case exhibited in the diagram.
6. The conclusion is merely the repetition of the general enunciation, wherein the
predicate is asserted as a demonstrated truth.
Prop. I. In Books 1 and 11, the circle is employed as a mechanical instrument, in
the same manner as the straight line, and the use made of it rests entirely on the third
postulate. No properties of the circle are discussed or even alluded to in these books
beyond the definition and the third postulate. One circle may fall within or without
another entirely, or the circumferences may intersect each other, as when the centre of
one circle is in the circumference of the other; and it is obvious from the two circles
cutting each other, in two points, one on each side of the given line, that two equi-
lateral triangles may be formed on the given line.
Prop. II. When the given point is neither in the line, nor in the line produced,
this problem admits of eight different lines being drawn from the given point in dif-
ferent directions, every one of which is a solution of the problem. For 1. The given
line has two extremities, to each of which a line may be drawn from the given point.
•
48
EUCLID'S ELEMENTS.
2. The equilateral triangle may be described on either side of this line. 3. And the
side BD of the equilateral triangle ABD may be produced either way.
But when the given point lies either in the line or in the line produced, the distinction
which arises from joining the two ends of the line with the given point no longer exists,
and there are only four cases of the problem.
Prop. III. This problem admits of two solutions, and it is left undetermined from
which end of the greater line the part is to be cut off.
: Prop. IV. This forms the first case of equal triangles, two other cases are proved
in Props. VIII and XXVI. A distinction ought to be made between equal triangles
and equivalent triangles, the former including those whose sides and angles mutually
coincide, the latter those whose areas only are equivalent.
The term base is obviously taken from the idea of a building, and the same may
be said of the term altitude. In Geometry, however, these terms are not restricted to one
particular position of a figure, as in the case of a building, but may be in any position
whatever.
Prop. v. Proclus has given in his commentary a proof for the equality of the
angles at the base without producing the equal sides. The construction follows the
same order, taking in AB a point D and cutting off from AC a part AE equal to AB,
and then joining CD and BE.
A corollary is a theorem which results from the demonstration of a proposition,
and generally is so obvious as to require no formal proof.
Prop. VI is the converse of one part of Prop. v. One proposition is defined to
be the converse of another when the hypothesis of the former becomes the predicate of
the latter; and vice versâ.
: There is besides this another kind of conversion, when a theorem has several
hypotheses and one predicate; by assuming the predicate and one or more than one of
the hypotheses, some one of the hypotheses may be inferred as the predicate of the
converse. In this manner, Prop. vIII is the converse of Prop. IV. It may here be
.observed, that converse theorems are not universally true: as for instance, the following
direct proposition is universally true; "If two triangles have their three sides respectively
equal, the three angles of each shall be respectively equal." But the converse is not
universally true; namely, "If two triangles have the three angles in each respectively
equal, the three sides are respectively equal." Converse theorems require, in some
instances, the consideration of other conditions than those which enter into the proof
of the direct theorem. Converse and contrary propositions are by no means to be
confounded, the contrary proposition denies what is assumed in the direct proposition,
but the subject and predicate in each are the same.
Prop. vI is the first instance of indirect demonstrations, and they are more suited
for the proof of converse propositions. All those propositions which are demonstrated
ex absurdo, are properly analytical demonstrations, according to the Greek notion of
analysis, which first supposed the thing required to be done, or to be true, and
then shewed the consistency or inconsistency of this construction or hypothesis with
! truths admitted or already demonstrated.
Prop. VII. The enunciation in the text was altered into that form by Simson.
Euclid's is, Επὶ τῆς αὐτῆς εὐθείας, δυσὶ ταῖς αὐταῖς εὐθείαις ἄλλαι δύο εὐθεῖαι
· ἴσαι ἑκατέρα ἑκατέρᾳ οὐ συσταθήσονται, πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ ἐπὶ τὰ αὐτὰ
μέρη, τὰ αὐτὰ πέρατα ἔχουσαι ταῖς ἐξ ἀρχῆς εὐθείαις.
Prop. VIII. When the three sides of one triangle are shewn to coincide with the
three sides of any other, the equality of the triangles is at once obvious. This, however,
is not stated at the conclusion of Prop. vIII or of Prop. xxvI. For the equality of the
areas of two coincident triangles, reference is always made by Euclid to Prop. IV.
NOTES TO BOOK I.,
49
Prop. IX. By means of this problem, any angle may be divided into four, eight,
sixteen, &c. equal angles.
Prop. x. Any finite straight line may, by this problem, be divided into four, eight,
sixteen, &c. equal parts.
Prop. XI. When the point is at the extremity of the line. By the second postulate
the line may be produced, and then the construction applies.
Prop. XIII. It is manifest that the lines which bisect the angles ABC and ABD
are at right angles to each other.
Prop. XIV is the converse of Prop. XII. "Upon the opposite sides of it." If
these words were omitted; it is possible for two lines to make with a third, two angles,
which together are equal to two right angles, in such a manner that the two lines shall not
be in the same straight line.
Prop. xv is the developement of the definition of an angle. If the lines at the
angular point be produced, the produced lines have the same inclination to one an-
other as the original lines, but in a different position.
Prop. XVI. From this Prop. it follows that only one perpendicular can be drawn
from a given point to a given line; and this perpendicular may be shewn to be less
than any other line which can be drawn from the given point to the given line.
The exact
Prop. XVII appears to be only a corollary to the preceding proposition, and it
seems to be introduced to explain Axiom XII, of which it is the converse.
truth respecting the angles of a triangle is proved in Prop. XXXII.
Prop. XIX is the converse of Prop. XVIII. It may be remarked, that Prop. xix
bears the same relation to Prop. XVIII, as Prop. vi does to Prop. v.
C
Prop. XX-XXI. "Proclus, in his commentary, relates, that the Epicureans derided
this proposition, as being manifest even to asses, and needing no demonstration; and his
answer is, that though the truth of it be manifest to our senses, yet it is science which
must give the reason why two sides of a triangle are greater than the third. But the
right answer to this objection against this and Prop. xxi, and some other plain propo-
sitions, is, that the number of axioms ought not to be increased without necessity, as it
must be if these propositions be not demonstrated. Mons. Clairault, in the preface
to his Elements of Geometry, published in French at Paris, 1741, says, that Euclid
has been at the pains to prove, that the two sides of a triangle which is included
within another, are together less than the two sides of the triangle which includes it.'
But he has forgot to add this condition, viz. that the triangles must be upon the same
base: because, unless this be added, the sides of the included triangle may be greater
than the sides of the triangle which includes it, in any ratio which is less than that
of two to one, as Pappus Alexandrinus has demonstrated in Prop. 3, Book III of
his Mathematical Collections." Simson. S. "
Prop. XXII. When the sum of two of the lines is equal to, and when it is less
than, the third line; let the diagrams be described, and they will exhibit the impos-
sibility implied by the restriction laid down in the proposition.
Prop. XXIII. CD might be taken equal to CE and the construction effected
by means of an isosceles triangle. It would, however, be less general than Euclid's.
Prop. XXIV. Simson makes the angle EDG at D in the line ED, the side
which is not the greater of the two ED, DF; otherwise, three different cases would
arise, as may be seen by forming the different figures. The point G might fall below
or upon the base EF produced as well as above it. Prop. XXIV and Prop. xxv bear!
to each other the same relation as Prop. IV and Prop. VIII.
Prop. XXVI. This forms the third case of the equality of two triangles. Every
triangle has three sides and three angles, and when any three of one triangle are given
equal to any three of another, the triangles may be proved to be equal to one another,
X
4
50
EUCLID'S ELEMENTS.
Sex Ba
whenever the three magnitudes given in the hypothesis are independent of one another.
Prop. IV contains the first case, when the hypothesis consists of two sides and the
included angle of each triangle. Prop. VIII contains the second, when the hypothesis
consists of the three sides of each triangle. Prop. XXVI contains the third, when
the hypothesis consists of two angles, and one side either adjacent to the equal angles,
or opposite to one of the equal angles in each triangle. There is another case, not
✰proved by Euclid, when the hypothesis consists of two sides and one angle in each
triangle, but these not the angles included by the two given sides in each triangle.
This case however is only true under a certain restriction.
Prop. XXVII. Alternate angles are defined to be the two angles which two straight
lines make with another at its extremities, but upon opposite sides of it.
Prop. XXVIII. One angle is called "the exterior angle," and another "the interior
and opposite angle,” when they are formed on the same side of a straight line which falls
upon or intersects two other straight lines. It is also obvious that on each side of the
line, there will be two exterior and two interior and opposite angles. The exterior
angle EGB has the angle GHD for its corresponding interior and opposite angle: also
the exterior angle FHD has the angle HGB for its interior and opposite angle.
Prop. XXIX is the converse of Prop. XXVII and Prop. XXVIII.
As the definition of parallel straight lines simply describes them by a statement
of the negative property, that they never meet; it is necessary that some positive property
of parallel lines should be assumed as an axiom, on which reasonings on such lines may
themobe-Thiago be founded.
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Euclid has assumed the statement in the twelfth axiom, which has been objected
to, as not being self-evident. A stronger objection appears to be, that the converse
of it forms Prop. 17, Book 1; for both the assumed axiom and its converse, should
be so obvious as not to require formal demonstration.
Simson has attempted to overcome the objection, not by any improved definition
and axiom respecting parallel lines; but, by considering Euclid's twelfth axiom to
be a theorem, and for its proof, assuming two definitions and one axiom, and then
demonstrating five subsidiary Propositions.
Instead of Euclid's twelfth axiom, the following has been proposed as a more simple
property for the foundation of reasonings on parallel lines; namely, "If a straight line
fall on two parallel straight lines, the alternate angles are equal to one another." In
whatever this may exceed Euclid's definition in simplicity, it is liable to a similar
objection, being the converse of Prop. 27, Book 1.
Professor Playfair has adopted in his Elements of Geometry, that "Two straight
lines which intersect one another cannot be both parallel to the same straight line."
This apparently more simple axiom follows as a direct inference from Prop. 30, Book 1.
But one of the least objectionable of all the definitions which have been proposed
on this subject, appears to be that which simply expresses the conception of equidistance.
It may be formally stated thus: "Parallel lines are such as lie in the same plane, and
which neither recede from, nor approach to, each other." This includes the conception
stated by Euclid, that parallel lines never meet. Dr Wallis observes on this subject,
"Parallelismus et æquidistantia vel idem sunt, vel certe se mutuo comitantur.”
As an additional reason for this definition being preferred, it may be remarked that
the meaning of the terms ypaµµai wapáλλŋλoi, suggests the exact idea of such lines.
Axiom X1 and XII, in some manuscripts of Euclid, are found placed respectively
as the fourth and the fifth postulate.
An account of thirty methods which have been proposed at different times for
avoiding the difficulty in the twelfth axiom, will be found in the appendix to Mr
Thompson's "Geometry without Axioms.”
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NOTES TO BOOK I.
51
Prop. XXXII. The three angles of a triangle may be shewn to be equal to two
right angles without producing a side of the triangle, by drawing through any angle
of the triangle a line parallel to the opposite side, as Proclus has remarked in his
Commentary on this proposition. It is manifest from this proposition, that the third
angle of a triangle is not independent of the sum of the other two; but is known if the
sum of any two is known. Cor. 1 may be also proved by drawing lines from any one
of the angles of the figure to the other angles. If any of the sides of the figure bend
inwards and form what are called re-entering angles, the enunciation of these two corol-
laries will require some modification.
From this proposition, it is obvious that each of the angles of an equilateral triangle,
is equal to two thirds of a right angle, as it is shewn in Prop. 15, Book IV. Also,
if one angle of an isosceles triangle be a right angle, then each of the equal angles is half
a right angle, as in Prop. 9, Book 11.
Prop. XXXIV. If the other diameter be drawn, it may be shewn that the diameters
of a parallelogram bisect each other, as well as bisect the area of the parallelogram.
The converse of this Prop. namely, "If the opposite sides or opposite angles of a
quadrilateral figure be equal, the opposite sides shall also be parallel; that is, the figure
shall be a parallelogram," is not proved by Euclid.aries's Naks
Prop. xxxv. The latter part of the demonstration is not expressed very intelligibly.
Simson, who altered the demonstration, seems in fact to consider two trapeziums of the
same form and magnitude, and from one of them, to take the triangle ABE; and
from the other, the triangle DCF; and then the remainders are equal by the third
axiom: that is, the parallelogram ABCD is equal to the parallelogram EBCF. Other
wise, the triangle, whose base is DE, (fig. 2.) is taken twice from the trapezium, which
would appear to be impossible, if the sense in which Euclid applies the third axiom,
is to be retained here.
It may be observed, that the two parallelograms exhibited in fig. 2 partially lie on
one another, and that the triangle whose base is BC is a common part of them, but that
the triangle whose base is DE is entirely without both the parallelograms. After having
proved the triangle ABE equal to the triangle DCF, if we take from these equals the
triangle whose base is DE, and to each of the remainders add the triangle whose base
is BC; perhaps the proof may appear somewhat more satisfactory.
Prop. XXXVIII. In this proposition, it is to be understood that the bases of the two
triangles are in the same straight line.
Prop. XXXIX. If the vertices of all the equal triangles which can be described
upon the same base, or upon the equal bases as in Prop. 40, be joined, the line thus
formed will be a straight line, and is called the locus of the vertices of equal triangles
upon the same base, or upon equal bases.
A locus in plane Geometry is a straight line or a plane curve, every point of which
and none else satisfies a certain condition.
With the exception of the straight line and the circle, the two most simple loci; all
other loci, perhaps including also the Conic Sections, may be more readily and effectually
investigated algebraically by means of their rectangular or polar equations.
Prop. XLI. The converse of this proposition is not proved by Euclid; viz. If a
parallelogram is double of a triangle, and they have the same base, or equal bases upon
the same straight line, and towards the same parts, they shall be between the same
parallels. Also, it may easily be shewn that if two equal triangles are between the
same parallels; they are either upon the same base, or upon equal bases.
Prop. XLIV.
to that line.
Prop. XLVII.
A parallelogram described on a straight line is said to be applied
In a right-angled triangle, the side opposite to the right angle is
:
*
ค
4-2
52
EUCLID'S ELEMENTS.
called the hypothenuse, and the other two sides, the base and perpendicular, according to
their position.
It is not indifferent on which sides of the lines which form the sides of the triangle
the squares are described. If they were described upon the inner, instead of the outer
sides of the lines, the construction would be found to fail.
By this proposition may be found a square equal to the sum of any given squares,
or equal to any multiple of a given square: or equal to the difference of two given
squares.
The truth of this proposition may be exhibited to the eye in some particular
instances. As in the case of that right-angled triangle whose three sides are 3, 4,
and 5 units respectively. If through the points of division of two contiguous sides
of each of the squares upon the sides, lines be drawn parallel to the sides (see the notes on
Book 11. p. 68), it will be obvious, that the squares will be divided into 9, 16 and 25
small squares, each of the same magnitude; and that the number of the small squares
into which the squares on the perpendicular and base are divided is equal to the number
into which the square on the hypothenuse is divided.
Prop. XLVIII is the converse of Prop. XLVII. In this Prop. is assumed the Corol-
lary that "the squares described upon two equal lines are equal," and the converse;
which properly ought to have been appended to Prop. XLVI.
The first book of Euclid's Elements, it has been seen, is conversant with the con-
struction and properties of rectilineal figures. It first lays down the definitions which
limit the subjects of discussion in the first book, next the three postulates, which restrict
the instruments by which the constructions in plane geometry are effected; and thirdly,
the twelve axioms, which express the principles by which a comparison is made between
the ideas of the things defined.
This Book may be divided into three parts. The first part treats of the origin and
properties of triangles, both with respect to their sides and angles; and the comparison
of these mutually, both with regard to equality and inequality. The second part treats
of the generation and properties of parallelograms. The third part exhibits the con-
nexion of the properties of triangles and parallelograms, and the equality of the squares
on the base and perpendicular of a right-angled triangle to the square on the hypothenuse.
When the propositions of the first book have been read, the student is recommended
to use different letters in the diagrams, and where it is possible, diagrams of a form
somewhat different from those exhibited in the text, for the purpose of testing the
accuracy of his knowledge of the demonstrations. And further, when he is become
sufficiently familiar with the method of geometrical reasoning, he may dispense with the
aid of letters altogether, and acquire the power of expressing in general terms the process
of reasoning in the demonstration of any proposition. Also, should he be not satisfied
with the bare knowledge of the principles of the first book which have been exhibited
synthetically, but be also desirous of knowing how these principles may be applied to the
solution of Problems analytically and the demonstration of Theorems; he may refer to the
Geometrical Exercises on the first book, which will be found at the end of the Elements,
together with some brief account of the Ancient Geometrical Analysis.
i.
BOOK II.
DEFINITIONS.
I.
EVERY right-angled parallelogram is called a rectangle, and is said
to be contained by any two of the straight lines which contain one of
the right angles.
II.
In every parallelogram, any of the parallelograms about a diameter,
together with the two complements, is called a gnomon.
A E
D
F
H
K
B G
C
"Thus the parallelogram HG together with the complements AF, FC, is the
gnomon, which is more briefly expressed by the letters AGK, or EHC, which are
at the opposite angles of the parallelograms which make the gnomon."
t
54
EUCLID'S ELEMENTS.
PROPOSITION I. THEOREM.
If there be two straight lines, one of which is divided into any number
of parts; the rectangle contained by the two straight lines, is equal to the
rectangles contained by the undivided line, and the several parts of the
divided line.
Let A and BC be two straight lines;
and let BC be divided into any parts in the points D, E.
Then the rectangle contained by the straight lines A and BC, shall
be equal to the rectangle contained by A and BD, together with that
contained by A and DE, and that contained by A and EC.
B
DEC

G
KL H
F
A
From the point B, draw BF at right angles to BC, (1. 11.)
and make BG equal to A; (1. 3.)
through G draw GH parallel to BC, (1. 31.)
and through D, E, C, draw DK, EL, CH parallel to BG.
Then the rectangle BH is equal to the rectangles BK, DL, EH.
But BH is contained by A and BC,
for it is contained by GB, BC, and GB is equal to A:
and the rectangle BK is contained by A, BD,
for it is contained by GB, BD, of which GB is equal to A:
also DL is contained by A, DE,
because DK, that is, BG, (1. 34.) is equal to A;
and in like manner the rectangle EH is contained by A, EC:
therefore the rectangle contained by A, BC, is equal to the several
rectangles contained by A, BD, and by A, DE, and by A, EC.
Wherefore, if there be two straight lines, &c. Q.E.D.
PROPOSITION II. THEOREM.
If a straight line be divided into any two parts, the rectangles con-
tained by the whole and each of the parts, are together equal to the square
of the whole line.
Let the straight line AB be divided into any two parts in the point C.
Then the rectangle contained by AB, BC, together with that con-
tained by AB, AC, shall be equal to the square of AB.
A
C B
D
FE
Upon AB describe the square ADEB, (1. 46.)
and through C draw CF parallel to AD or BÈ. (1. 31.)
BOOK II. PROP. II, III, IV.
55
Then AE is equal to the rectangles AF, CE.
And AE is the square of AB;
and AF is the rectangle contained by BA, AC;
for it is contained by DA, AC, of which DA is equal to AB:
and CE is contained by AB, BC,
for BE is equal to AB:
therefore the rectangle contained by AB, AC, together with the
rectangle AB, BC is equal to the square of AB.
If therefore a straight line, &c.
Q.E.D.
PROPOSITION III. THEOREM.
If a straight line be divided into any two parts, the rectangle con-
tained by the whole and one of the parts, is equal to the rectangle contained
by the two parts, together with the square of the aforesaid part.
Let the straight line AB be divided into any two parts in the
point C.
Then the rectangle AB, BC, shall be equal to the rectangle AC, CB,
together with the square of BC.
A C
B

F D
E
Upon BC describe the square CDEB, (1. 46.) and produce ED to F,
through A draw AF parallel to CD or BE. (1. 31.)
Then the rectangle AE is equal to the rectangles AD, CE;
but AE is the rectangle contained by AB, BC,
for it is contained by AB, BE, of which BE is equal to BC:
and AD is contained by AC, CB,
for CD is equal to CB:
and DB is the square of BC:
therefore the rectangle AB, BC, is equal to the rectangle AC, CB,
together with the square of BC.
If therefore a straight line be divided, &c.
PROPOSITION IV. THEOREM.
Q.E. D.
If a straight line be divided into any two parts, the square of the whole
line is equal to the squares of the two parts, together with twice the
rectangle contained by the parts.
Let the straight line AB be divided into any two parts in C.
Then the square of AB shall be equal to the squares of AC, and CB,
together with twice the rectangle contained by AČ, CB.
A a CLB
-
H
1.
a
D
F E
K
56
EUCLID'S ELEMENTS.
Upon AB describe the square ADEB, (1. 46.)
join BD, and through C draw CGF parallel to AD or BE, (1.31.)
and through G draw HGK parallel to AB or DE.
Then, because CF is parallel to AD, and BD falls upon them,
therefore the exterior angle BGC is equal to the interior and
opposite angle ADB; (1. 29.)
but the angle ADB is equal to the angle ABD, (1. 5.)
because BA is equal to AD, being sides of a square;
wherefore the angle CGB is equal to the angle CBG;
and therefore the side BC is equal to the side CG; (1.6.)
but CB is equal also to GK, and CG to BK; (1.34.)
wherefore the figure CGKB is equilateral.
It is likewise rectangular,
for, since CG is parallel to BK, and CB meets them,
therefore the angles KBC, GCB are equal to two right angles; (1. 29.)
but the angle KBC is a right angle; (def. 30. constr.)
wherefore GCB is a right angle:
and therefore also the angles CGK, GKB, opposite to these, are
right angles; (1. 34.)
wherefore CGKB is rectangular:
it is also equilateral, as was demonstrated;
wherefore it is a square, and it is upon the side CB.
For the same reason HF is a square, and it is upon the side HG,
which is equal to AC. (1. 34.)
Therefore the figures HF, CK, are the squares of AC, CB.
And because the complement AG is equal to the complement GE, (1.43.)
and that AG is the rectangle contained by AC, CB,
for GC is equal to CB;
therefore GE is also equal to the rectangle AC, CB;
wherefore AG, GE are equal to twice the rectangle AC, CB;
and HF, CK are the squares of AC, CB;
wherefore the four figures HF, CK, AG, GE, are equal to the
squares of AC, CB, and to twice the rectangle AC, CB:
but HF, CK, AG, GE make up the whole figure ADEB, which
is the square of AB;
therefore the square of AB is equal to the squares of AC, CB, and
twice the rectangle AC, CB.
Wherefore, if a straight line be divided, &c. Q.E. D.
COR. From the demonstration, it is manifest, that the parallelograms
about the diameter of a square are likewise squares.
PROPOSITION V. THEOREM.
If a straight line be divided into two equal parts, and also into two
unequal parts; the rectangle contained by the unequal parts, together
with the square of the line between the points of section, is equal to the
square of half the line.
Let the straight line AB be divided into two equal parts in the
point C, and into two unequal parts in the point D.
Then the rectangle AD, DB, together with the square of CD, shall
be equal to the square of CB.
BOOK II. PROP. V, VI.
57
A
C D B

L
K
H M
E
G F
Upon CB describe the square CEFB, (1.46.)
join BE, and through D draw DHG parallel to CE or BF; (1.31.)
and through H draw KLM parallel to CB or EF;
also through A draw AK parallel to CL or BM.
Then, because the complement CH is equal to the complement HF, (1. 43.)
to each of these equals add DM;
therefore the whole CM is equal to the whole DF;
but because AC is equal to CB,
therefore CM is equal to AL, (1. 36.)
therefore also AL is equal to DF:
to each of these equals add CH,
and therefore the whole AH is equal to DF and CH:
but AH is the rectangle contained by AD, DB, for DH is equal to DB;
and DF together with CH is the gnomon CMG;
therefore the gnomon CMG is equal to the rectangle AD, DB:
to each of these equals add LG, which is equal to the square
of CD; (II. 4. Cor.)
therefore the gnomon CMG, together with LG, is equal to the
rectangle AD, DB, together with the square of CD:
but the gnomon CMG and LG make up the whole figure CEFB,
which is the square of CB;
therefore the rectangle AD, ÓB, together with the square of CD,
is equal to the square of CB.
Wherefore, if a straight line, &c. Q.E.D.
COR. From this proposition it is manifest, that the difference of the
squares of two unequal lines AC, CD, is equal to the rectangle contain-
ed by their sum AD and their difference DB.

PROPOSITION VI. THEOREM.
If a straight line be bisected, and produced to any point; the rect-
angle contained by the whole line thus produced, and the part of it
produced, together with the square of half the line bisected, is equal to
the square of the straight line which is made up of the half and the
part produced.
Let the straight line AB be bisected in C, and produced to the
point D.
Then the rectangle AD, DB, together with the square of CB, shall
be equal to the square of CD.
A
C
B D

L
H
K
M
E
G F
58
EUCLID'S ELEMENTS.
Upon CD describe the square CEFD, (1. 46.) and join DE,
through B draw BHG parallel to CE or DF, (1. 31.)
through H draw KLM parallel to AD or EF,
and through A draw AK parallel to CL or DM.
Then because AC is equal to CB,
therefore the rectangle AL is equal to the rectangle CH, (1. 36.)
but CH is equal to HF; (1. 43.)
therefore AL is equal to HF;
to each of these equals add CM;
therefore the whole AM is equal to the gnomon CMG:
but AM is the rectangle contained by AD, DB,
for DM is equal to DB: (11. 4. Cor.)
therefore the gnomon CMG is equal to the rectangle AD, DB:
add to each of these equals LG which is equal to the square of CB;
therefore the rectangle AD, DB, together with the square of CB,
is equal to the gnomon CMG, and the figure LG;
but the gnomon CMG and LG make up the whole figure CEFD,
which is the square of CD;
therefore the rectangle AD, DB, together with the square of CB,
is equal to the square of CD.
Wherefore, if a straight line, &c.
Q. E.D.
PROPOSITION VII.
THEOREM.
If a straight line be divided into any two parts, the squares of the
whole line, and of one of the parts, are equal to twice the rectangle
contained by the whole and that part, together with the square of the
other part.
Let the straight line AB be divided into any two parts in the
point C.
Then the squares of AB, BC shall be equal to twice the rectangle
AB, BC, together with the square of AC.
H
A
C B
e
K
D
FE
Upon AB describe the square ADEB, (1. 46.) and join BD;
through C draw CF parallel to AD or BE cutting BD in G, (1. 31.)
through G draw HGK parallel to AB or DE.
Then because AG is equal to GE, (1. 43.)
add to each of them CK;
therefore the whole AK is equal to the whole CE;
and therefore AK, CE, are double of AK;
but AK, CE are the gnomon AKF and the square CK;
therefore the gnomon AKF and the square CK is double of AK:
but twice the rectangle AB, BC, is double of AK,
for BK is equal to BC; (II. 4. Cor.)
therefore the gnomon AKF and the square CK, is equal to twice
the rectangle AB, BC;
BOOK II., PROP. VII, VIII.
59
to each of these equals add HF, which is equal to the square of
AC,
therefore the gnomon AKF, and the squares CK, HF, are equal to
twice the rectangle AB, BC, and the square of AC;
but the gnomon AKF, together with the squares CK, HF, make
up the whole figure ADEB and CK, which are the squares of AB
and BC;
therefore the squares of AB and BC are equal to twice the
rectangle AB, BC, together with the square of AC.
Wherefore, if a straight line, &c. Q.E.D.
PROPOSITION VIII.
THEOREM.
If a straight line be divided into any two parts, four times the rect-
angle contained by the whole line, and one of the parts, together with the
square of the other part, is equal to the square of the straight line, which
is made up of the whole and that part.
Let the straight line AB be divided into any two parts in the point C.
Then four times the rectangle AB, BC, together with the square of
AC, shall be equal to the square of the straight line made up of AB and
BC together.
A
C BD

M
P
X
Ul A
G KI
N
R
E HL F
Produce AB to D, so that BD be equal to CB, (1. 3.)
upon AD describe the square AEFD, (1. 46.) and join BE, D
through B, C, draw BL, CH parallel to AE or DF, and cutting
DE in the points K, P respectively;
through K, P, draw MGKN, XPRO parallel to AD or EF.
Then because CB is equal to BD, CB to GK, and BD to KN;
therefore GK is equal to KN;
for the same reason, PR is equal to RO;
and because CB is equal to BD, and GK to KN,
therefore the rectangle CK is equal to BN, and GR to RN; (1.36.)
but CK is equal to RN, (1. 43.)
because they are the complements of the parallelogram CO;
therefore also BN is equal to GR;
and therefore the four rectangles BÑ, CK, GR, RN, are equal to
one another, and so are quadruple of one of them CK.
Again, because CB is equal to BD, and BD to BK, that is, to CG;
and because CB is equal to GK, that is, to GP ;
therefore CG is equal to GP.
And because CG is equal to GP, and PR to RO,
therefore the rectangle AG is equal to MP, and PL to RF;
but the rectangle MP is equal to PL, (1.43.)
because they are the complements of the parallelogram ML;
wherefore also AG is equal to RF;
60
EUCLID'S ELEMENTS.
therefore the four rectangles AG, MP, PL, RF, are equal to one
another, and so are quadruple of one of them AG.
And it was demonstrated that the four CK, BN, GR, and RN are
quadruple of CK;
therefore the eight rectangles which contain the gnomon AOH, are
quadruple of AK.
And because AK is the rectangle contained by AB, BC,
for BK is equal to BC;
therefore four times the rectangle AB, BC is quadruple of AK;
but the gnomon AOH was demonstrated to be quadruple of AK;
therefore four times the rectangle AB, BC is equal to the gnomon AOH;
to each of these equals add XH, which is equal to the square of AC;
therefore four times the rectangle AB, BC, together with the
square of AC, is equal to the gnomon AOH and the square XH;
but the gnomon AOH and XH make up the figure AEFD which
is the square of AD;
therefore four times the rectangle AB, BC together with the square
of AC, is equal to the square of AD, that is, of AB and BC added
together is one straight line.
Vin
Wherefore, if a straight line, &c. Q.E.D.
FROPOSITION IX. THEOREM.
If a straight line be divided into two equal, and also into two unequal
parts; the squares of the two unequal parts are together double of the
square of half the line, and of the square of the line between the points of
section.
Let the straight line AB be divided into two equal parts in the
point C, and into two unequal parts in the point D.
Then the squares of AD, DB together, shall be double of the squares
of AC, CD.
G
E
A
C D B
From the point C draw CE at right angles to AB, (1. 11.)
make CE equal to AC or CB, (1. 3.) and join EA, EB;
through D draw DF parallel to CE, meeting EB in F, (1.31.)
through F draw FG parallel to BA, and join AF.
Then, because AC is equal to CE,
therefore the angle EAC is equal to the angle AEC; (1.5.)
and because ACE is a right angle,
therefore the two other angles AEC, EAC of the triangle are
together equal to a right angle; (1.32.)
and since they are equal to one another
therefore each of them is half of a right angle.
For the same reason, each of the angles CEB, EBC is half a right angle;
and therefore the whole AEB is a right angle.
And because the angle GEF is half a right angle,
and EGF a right angle, for it is equal to the interior and opposite
angle ECB, (1. 29.)
BOOK II. PROP. IX, X.
61
therefore the remaining angle EFG is half a right angle ;
wherefore the angle GEF is equal to the angle EFG,
and the side EG equal to the side GF. (1. 6.)
Again, because the angle at B is half a right angle,
and FDB a right angle, for it is equal to the interior and opposite
angle ECB, (1.29.)
therefore the remaining angle BFD is half a right angle;
wherefore the angle at B is equal to the angle BFD,
and the side DF equal to the side DB. (1. 6.)
And because AC is equal to CE,
the square of AC' is equal to the square of CE;
therefore the squares of AC, CE are double of the square of AC;
but the square of AE is equal to the squares of AC, CE, (1. 47.)
because ACE is a right angle;
therefore the square of AE is double of the square of AC.
Again, because EG is equal to GF,
the square of EG is equal to the square of GF;
therefore the squares of EG, GF are double of the square of GF ;
but the square of EF is equal to the squares of EG, GF; (1. 47.)
therefore the square of EF is double of the square of GF
and GF is equal to CD; (1. 34.)
therefore the square of EF is double of the square of CD;
but the square of AE is double of the square of AC;
therefore the squares of AE, EF are double of the squares of AC, CD;
but the square of AF is equal to the squares of AE, EF,
because AEF is a right angle: (1. 47.)
CD:
therefore the square of AF is double of the squares of AC,
but the squares of AD, DF are equal to the square of AF;
because the angle ADF is a right angle; (1. 47.)
therefore the squares of AD, DF are double of the squares of AC, CD ;
and DF is equal to DB;
therefore the squares of AD, DB are double of the squares of AC, CD.
If therefore a straight line be divided, &c.
PROPOSITION X. THEOREM.
Q. E. D.
If a straight line be bisected, and produced to any point, the square of
the whole line thus produced, and the square of the part of it produced,
are together double of the square of half the line 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 the
point D.
Then the squares of AD, DB, shall be double of the squares of
AC, CD.
E
F

A
B
D
G
From the point C draw CE at right angles to AB, (1. 11.)
make CE equal to AC or CB, (1. 3.) and join AE, EB;
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EUCLID'S ELEMENTS.
through E draw EF parallel to AB, (1. 31.)
and through D draw DF parallel to CE.
Then because the straight line EF meets the parallels CE, FD,
therefore the angles CEF, EFD are equal to two right angles; (1. 29.)
and therefore the angles BEF, EFD are less than two right angles.
But straight lines, which with another straight line make the interior
angles upon the same side less than two right angles, will meet if pro-
duced far enough; (ax. 12.)
therefore EB, FD will meet, if produced towards B, D;
let them meet in G, and join AG.
Then, because AC is equal to CE,
therefore the angle CEA is equal to the angle EAC; (1.5.)
and the angle ACE is a right angle;
therefore each of the angles CEA, EAC is half a right angle. (1. 32.)
For the same reason, each of the angles CEB, EBC is half a right
angle;
therefore the whole AEB is a right angle.
And because EBC is half a right angle,
therefore DBG is also half a right angle, (1. 15.)
for they are vertically opposite;
but BDG is a right angle,
because it is equal to the alternate angle DCE; (1. 29.)
therefore the remaining angle DGB is half a right angle;
and is therefore equal to the angle DBG;
wherefore also the side BD is equal to the side DG. (1.6.)
Again, because EGF is half a right angle, and the angle at F is a
right angle, being equal to the opposite angle ECD, (1.34.)
therefore the remaining angle FEG is half a right angle,
and therefore equal to the angle EGF;
wherefore also the side GF is equal to the side FE. (1.6.)
And because EC is equal to CA ;
the square of EC is equal to the square of CA;
therefore the squares of EC, CA are double of the square of CA
but the square of EA is equal to the squares of EC, CA; (1.47.)
therefore the square of EA is double of the square of AC.
Again, because GF is equal to EF,
the square of GF is equal to the square of EF ;
;
therefore the squares of GF, FE are double of the square of EF;
but the square of EG is equal to the squares of GF, EF ; (1. 47.)
therefore the square of EG is double of the square of EF;
and EF is equal to CD; (1. 34.)
wherefore the square of EG is double of the square of CD ;
but it was demonstrated,
that the square of EA is double of the square of AC;
therefore the squares of EA, EG are double of the squares of AC, CD ;
but the square of AG is equal to the squares of EÂ, EG; (1. 47.)
therefore the square of A G is double of the squares of AC, CD:
but the squares of AD, DG are equal to the square of AG;
therefore the squares of AD, DG are double of the squares of AC, CD ;
but DG is equal to DB;
therefore the squares of AD, DB are double of the squares of AC, CD.
Wherefore, if a straight line, &c. Q.E.D.
BOOK II. PROP. XI.
63
PROPOSITION XI. PROBLEM.
To divide a given straight line into two parts, so that the rectangle
contained by the whole and one of the parts, shall be equal to the square
of the other part.
Let AB be the given straight line.
It is required to divide AB into two parts, so that the rectangle
contained by the whole and one of the parts, shall be equal to the
square of the other part.
F
G

A
H
B
E
C
K D
Upon AB describe the square ACDB; (1. 46.)
bisect AC in E, (1. 10.) and join BE,
produce CA to F, and make EF equal to EB, (1. 3.)
upon AF describe the square FGHA. (1.46.)
Then AB`shall be divided in H, so that the rectangle AB, BH is
equal to the square of AH.
Produce GH to meet CD in K.
Then because the straight line AC is bisected in E, and produced to F,
therefore the rectangle CF, FA, together with the square of AE, is
equal to the square of EF; (II. 6.)
but EF is equal to EB;
therefore the rectangle CF, FA together with the square of AE, is
equal to the square of EB;
but the squares of BA, AE are equal to the square of EB, (1.47.)
because the angle EAB is a right angle;
therefore the rectangle CF, FA, together with the square of AE, is
equal to the squares of BA, AE;
take away the square of AE, which is common to both
therefore the rectangle contained by CF, FA is equal to the square of BA.
But the figure FK is the rectangle contained by CF, FA,
for FA is equal to FG ;
and AD is the square of AB;
therefore the figure FK is equal to AD;
take away the common part AK,
therefore the remainder FH is equal to the remainder HD;
but HD is the rectangle contained by AB, BH,
for AB is equal to BD;
and FH is the square of AH;
therefore the rectangle AB, BH, is equal to the square of AH.
Wherefore the straight line AB is divided in H, so that the rect-
angle AB, BH is equal to the square of AH. Q.E. F.
64
EUCLID'S ELEMENTS.
PROPOSITION XII. THEOREM.
In obtuse-angled triangles, if a perpendicular be drawn from either
of the acute angles to the opposite side produced, the square of the side
subtending the obtuse angle, is greater than the squares of the sides
containing the obtuse angle, by twice the rectangle contained by the side
upon which, when produced, the perpendicular falls, and the straight line
intercepted without the triangle between the perpendicular and the obtuse
angle.
Let ABC be an obtuse-angled triangle, having the obtuse angle
ACB, and from the point A let AD be drawn perpendicular to BC
produced.
Then the square of AB shall be greater than the squares of AC, CB,
by twice the rectangle BC, CD.
A

B
C D
Because the straight line BD is divided into two parts in the point C,
therefore the square of BD is equal to the squares of BC, CD, and
twice the rectangle BC, CD; (II. 4.)
to each of these equals add the square of DA;
therefore the squares of BD, DA are equal to the squares of BC,
CD, DA, and twice the rectangle BC, CD;
but the square of BA is equal to the squares of BD, DA, (1.47.)
because the angle at D is a right angle;
and the square of CA is equal to the squares of CD, DA ;
therefore the square of BA is equal to the squares of BC, CA, and
twice the rectangle BC, CD;
that is, the square of BA is greater than the squares of BC, CA, by
twice the rectangle BC, CD.
Therefore in obtuse-angled triangles, &c.
Q.E.D.
PROPOSITION XIII. THEOREM.
In every triangle, the square of the side subtending either of the acute
angles, is less than the squares of the sides containing that angle, by
twice the rectangle contained by either of these sides, and the straight line
intercepted between the acute angle and the perpendicular let fall upon
it from the opposite angle.
Let ABC be any triangle, and the angle at B one of its acute angles,
and upon BC, one of the sides containing it, let fall the perpendicular
AD from the opposite angle. (1. 12.)
Then the square of AC opposite to the angle B, shall be less than
the squares of CB, BA, by twice the rectangle CB, DB.
First, let AD fall within the triangle ABC.
BOOK II. PROP. XIII.
65
A

B
D
C
Then, because the straight line CB is divided into two parts in D,
the squares of CB, BD are equal to twice the rectangle contained
by CB, BD, and the square of DC; (11.7.)
to each of these equals add the square of AD;
therefore the squares of CB, BD, DÃ, are equal to twice the
rectangle CB, BD, and the squares of AD, DC;
but the square of AB is equal to the squares of BD, DA, (1. 47.)
because the angle BDA is a right angle;
and the square of AC is equal to the squares of AD, DC;
therefore the squares of CB, BA are equal to the square of AC,
and twice the rectangle CB, BD;
that is, the square of AC alone is less than the squares of CB, BA,
by twice the rectangle CB, BD.
Secondly, let AD fall without the triangle ABC.
A

B
C
D
Then, because the angle at D is a right angle,
the angle ACB is greater than a right angle; (1. 16.)
and therefore the square of AB is equal to the squares of AC, CB,
and twice the rectangle BC, CD; (11. 12.)
to each of these equals add the square of BC;
therefore the squares of AB, BC are equal to the square of AC,
twice the square of BC, and twice the rectangle BC, CD;
but because BD is divided into two parts in C,
therefore the rectangle DB, BC is equal to the rectangle BC, CD,
and the square of BC; (11. 3.)
and the doubles of these are equal;
that is, twice the rectangle DB, BC is equal to twice the rectangle
BC, CD and twice the square of BC:
wherefore the squares of AB, BC are equal to the square of AC,
and twice the rectangle DB, BC:
therefore the square of AC alone is less than the squares of AB,
BC, by twice the rectangle DB, BC.
Lastly, let the side AC be perpendicular to BC.
A
}
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 equal to the
square of AC, and twice the square of BC. (1. 47.
Therefore in any triangle, &c. Q. E. D.
5
66
EUCLID'S ELEMENTS.
PROPOSITION XIV. PROBLEM.
To describe a square that shall be equal to a given rectilineal figure.
Let A be the given rectilineal figure.
It is required to describe a square that shall be equal to A.

H
A
B
F
G
E
C
D
Describe the rectangular parallelogram BCDE equal to the rectilineal
figure A. (1. 45.)
Then, if the sides of it, BE, ED, are equal to one another, it is a
square, and what was required is now done.
But if they are not equal,
produce one of them BE to F, and make EF equal to ED,
bisect BF in G; (1.10.)
from the centre G, at the distance GB, or GF, describe the semi-
circle BHF,
The
figure A.
and produce DE to meet the circumference in H.
square described
upon EH shall be equal to the given rectilineal
Join GH.
Then because the straight line BF is divided into two equal parts
in the point G, and into two unequal parts in the point E;
therefore the rectangle BE, EF, together with the square of EG,
are equal to the square of GF; (11. 5.)
but GF is equal to GH; (def. 15.)
therefore the rectangle BE, EF, together with the square of EG,
is equal to the square of GH;
but the squares of HE, EG are equal to the square of GH; (1.47.)
therefore the rectangle BE, EF, together with the square of EG,
are equal to the squares of HE, ÉG;
take away the square of EG, which is common to both;
therefore the rectangle BE, EF is equal to the square of HE.
But the rectangle contained by BE, EF is the parallelogram BD,
because EF is equal to ED ;
therefore BD is equal to the square of EH
but BD is equal to the rectilineal figure A; (constr.)
therefore the square of EH is equal to the rectilineal figure A.
Wherefore a square has been made equal to the given rectilineal
figure A, namely, the square described upon EH. Q.E. F.
NOTES TO BOOK II.
IN Book 1, Geometrical magnitudes of the same kind, lines, angles and surfaces,
more particularly triangles and parallelograms, are compared, either as being absolutely
equal or unequal to, one another.
In Book II, the properties of right-angled parallelograms, but without reference to
their magnitudes are demonstrated, besides an important extension is made of Prop.
47, Book 1, to acute-angled and obtuse-angled triangles. Euclid has given no defini-
tion of a rectangular parallelogram, or rectangle: probably, because the Greek expression
παραλληλόγραμμον ὀρθογώνιον, or ὀρθογώνιον simply, is a definition of the figure.
In English, the term rectangle has no signification, and therefore ought to be defined
before its properties are demonstrated. A rectangle is defined to be a parallelogram
having one angle a right angle; and a square is a rectangle having all its sides equal.
In Prop. 35, Book 1, it may be seen that there may be an indefinite number of
parallelograms on the same base and between the same parallels whose areas are always
equal to one another; but that one of them has all its angles right angles, and the
length of its boundary less than the boundary of any other parallelogram upon the
same base and between the same parallels. The area of this rectangular parallelogram
is therefore determined by the two lines which contain one of its right angles. Hence
it is stated in Def. 1, that every right-angled parallelogram is said to be contained by
any two of the straight lines which contain one of the right angles. No distinction
is made in Book II. between equality and identity, as the rectangle may be said to
be contained by two lines which are equal respectively to the two which contain one
right angle of the figure. It may be remarked that the rectangle itself is bounded by
four straight lines.
It is of primary importance to discriminate the Geometrical conception of a rect-
angle from the Arithmetical or Algebraical representation of it. The subject of Geometry
is magnitude not number, and therefore it would be a departure from strict reasoning
on space, to substitute in Geometrical demonstrations, the Arithmetical or Algebraical
representation of a rectangle for the rectangle itself. It is, however, absolutely neces-
sary that the connection of number and magnitude be clearly understood, as far as
regards the representation of lines and areas.
All lines are measured by lines, and all surfaces by surfaces. Some one line of
definite length is arbitrarily assumed as the linear unit, and the length of every other line
is represented by the number of linear units contained in it. The square is the figure
assumed for the measure of surfaces. The square unit or the unit of area is assumed
to be that square, the side of which is one linear unit in length, and the magnitude
of every surface is represented by the number of square units contained in it. But
here it may be remarked, that Euclid has not informed us that the two lines by which
every rectangle is contained, are capable of being represented by any multiples of the
same linear unit. If, however, in the present case, the sides of rectangles are supposed
to be divisible into an exact number of linear units, a numerical representation for
the area of the rectangle may be deduced.
On two lines at right angles to each other, take AB equal to 4, and AD equal
to 3 linear units.
Complete the rectangle ABCD, and through the points of division of AB, AD,
draw EL, FM, GN parallel to AD; and HP, KQ parallel to AB respectively.
*
5-2
68
EUCLID'S ELEMENTS.
A E F G B
H
P
'R
K
DL MNC
Then the whole rectangle AC is divided into squares, all equal to each other.
And AC is equal to the sum of the rectangles AL, EM, FN, GC; (11. 1.)
also these rectangles are equal to one another, (1. 36.)
therefore the whole AC is equal to four times one of them AL.
Again, the rectangle AL is equal to the rectangles EH, HR, RD, and these
rectangles, by construction, are squares described upon the equal lines AH, HK, KD,
and are equal to one another.
therefore the rectangle AL is equal to 3 times the square AH,
but the whole rectangle AC is equal to 4 times the rectangle AL,
therefore the rectangle AC is 4 × 3 times the square AH, or 12 square units:
that is, the product of the two numbers which express the number of linear units
in the two sides, will give the number of square units in the rectangle, and therefore
will be an arithmetical representation of its area.
And generally, if AB, AD, instead of 4 and 3, consisted of a and b linear units
respectively, it may be shewn in a similar manner, that the area of the rectangle AC
would contain ab square units; and therefore the product ab is a proper representation
for the area of the rectangle AC.
Hence, it follows, that the term rectangle in Geometry corresponds to the term
product in Arithmetic and Algebra, and that a similar comparison may be made between
the products of the two numbers which represent the sides of rectangles, as between the
areas of the rectangles themselves. This forms the basis of what are called Arith-
metical or Algebraical proofs of Geometrical properties.
If the two sides of the rectangle be equal, or if b be equal to a, the figure is a
square, and the area is represented by aa or a².
Also, since a triangle is equal to the half of a parallelogram of the same base
and altitude.
Therefore the area of a triangle will be represented by half the rectangle which has
the same base and altitude as the triangle: in other words, if the length of the base
be a units, and the altitude be ʊ units.
ab
Then the area of the triangle is algebraically represented by ab, or 2
The leading idea which runs through the demonstrations of the first eight pro-
positions, is the obvious axiom, that, "the whole area of every figure in each case,
is equal to all the parts of it taken together.”
Prop. I. For the sake of brevity of expression, "the rectangle contained by the
straight lines AB, BC," is called "the rectangle AB, BC;" and sometimes "the
rectangle ABC.”
The method of reasoning on the properties of rectangles by means of the products
which indicate the number of square units contained in their areas is foreign to Euclid's
ideas of rectangles, as discussed in his second book, which have no reference to any
particular unit of length or measure of surface.
Prop. 1. Algebraically. (fig. Prop. 1.)
Let the line BC contain a linear units, and A, b linear units of the same length.
Also suppose the parts BD, DE, EC to contain m, n, p linear units respectively.
Then a = m + n +P₂,
1
NOTES TO BOOK II.
69
multiply these equals by b,
therefore ab= bm + b n + bp.
That is, the product of two numbers, one of which is divided into any number of
parts, is equal to the sum of the products of the undivided number, and the several parts
of the other;
or, if the Geometrical interpretation of the products be restored.
The number of square units expressed by the product ab, is equal to the number
of square units expressed by the sum of the products bm, bn, bp.
Prop. 11. Algebraically. (fig. Prop. 11.)
Let AB contain a linear units, and AC, CB, m and n linear units respectively.
multiplying these equals by a,
Then m + n = Ag
а,
therefore am+ an = a².
That is, if a number be divided into any two parts, the sum of the products of the
whole and each of the parts is equal to the square of the whole number.
Prop. III. Algebraically. (fig. Prop. 111.)
Let AB contain a linear units, and let BC contain m, and AC, n linear units.
Then a = m + n,
and multiplying these equals by m,
therefore
ma = m² + mn.
That is, if a number be divided into any two parts, the product of the whole number
and one of the parts, is equal to the square of that part, and the product of the two
parts.
Prop. iv. might have been deduced from the two preceding propositions; but Euclid
has rather preferred the method of exhibiting, in all the demonstrations of the second
book, the equality of the spaces compared.
In the corollary to Prop. XLVI. Book 1, it is stated that a parallelogram which has
one right angle, has all its angles right angles. By applying this corollary, the demon-
stration of Prop. iv. may be considerably shortened.
If the two parts be equal, then the square of the whole line is equal to four times
the square of half the line.
Also, if a line be divided into any three parts, the square of the whole line is equal
to the squares of the three parts, and twice the rectangle contained by every two parts.
Prop. iv. Algebraically. (fig. Prop. IV.)
Let the line AB contain a linear units, and the parts of it AC and BC, m and n
linear units respectively.
Then a = M + N,
squaring these equals, ••• a² = (m + n)2,
or
a² = m² + 2mn + n2.
That is, if a number be divided into any two parts, the square of the number is
equal to the squares of the two parts together with twice the product of the two parts.
Prop. v. It must be kept in mind, that the sum of two straight lines in Geometry,
means the straight line formed by joining the two lines together, so that both may be
in the same straight line.
ぶ
​The following simple properties respecting the equal and unequal division of a line ✓
are worthy of being remembered.
2BD÷2DC.
I. Since AB = 2 BC = 2 (BD + DC) = 2BD + 2DC. (fig. Prop. v.)
and AB = AD + DB;
.. 2 CD + 2DB = AD + DB,
Сп
0.
17
70
EUCLID'S ELEMENTS.
and by subtracting 2DB from these equals,
.. 2 CD = AD – DB,
and _CD=}(AD – DB).
That is, if a line AB is divided into two equal parts in C, and into two unequal
parts in D, the part CD of the line between the points of section is equal to half
the difference of the unequal parts AD and DB.
II. Here AD = AC + CD, the sum of the unequal parts, (fig. Prop. v.)
and DB = AC- CD their difference.
Hence by adding these equals together,
.. AD+ DB = 2AC,
or the sum and difference of two lines AC, CD, are together equal to twice the greater line.
And the halves of these equals are equal,
} . AD + }. DB = AC,
or, half the sum of two unequal lines AC, CD added to half their difference is equal
to the greater line AC.
III. Again, since AD = AC + CD, and DB = AC - CD,
by subtracting these equals,
.. AD – DB = 2 CD,
or, the difference between the sum and difference of two unequal lines is equal to
twice the less line.
And the halves of these equals are equal,
•. }.AD-}. DB = CD,
or, half the difference of two lines subtracted from half their sum is equal to the less
of the two lines.
IV.
Since
AC - CD = DB the difference,
.. AC = CD + DB,
and adding CD the less to each of these equals,
.. AC + CD = 2CD + DB,
or, the sum of two unequal lines is equal to twice the greater line together with the
difference between the lines.
Prop. v. Algebraically.
Let AB contain 2a linear units,
its half BC will contain a linear units.
And let CD the line between the points of section contain m linear units.
Then AD the greater of the two unequal parts, contains a + m linear units;
and DB the less contains a m units.
Also m is half the difference of a + m and a
••• (a + m) (a — m) = a² — m²,
to each of these equals add m²;
•. (a+m) (a − m) + m² = a².
-
m;
That is, if a number be divided into two equal parts, and also into two unequal
parts, the product of the unequal parts and the square of half their difference, is equal
to the square of half the number.
Bearing in mind that AC, CD are respectively half the sum and half the difference
of the two lines AD, DB; the corollary to this proposition may be expressed in the
following form: "The rectangle contained by two straight lines is equal to the difference
of the squares of half their sum and half their difference."
NOTES TO BOOK II.
71
The rectangle contained by AD and DB, and the square of BC are each bounded
by the same extent of line, but the spaces enclosed differ by the square of CD.
Prop. VI. Algebraically.
Let AB contain 2a linear units, then its half BC contains a units; and let BD
contain m units,
Then AD contains 2a + m units,
and .. (2a+m) m =
2a+m)m – 2am + m²
to each of these equals add a³,
;
•. (2a + m) m + a² = a² + 2am + m².
But a²+2am + n² = (a + m)³,
•. (2a+m) m + a² = (a + m)².
That is, If a number be divided into two equal numbers, and another number be
added to the whole and to one of the parts; the product of the whole number thus
increased and the other number together with the square of half the given number, is equal
to the square of the number which is made up of half the given number increased.
The Algebraical results of Prop. v and Prop. vi are identical, as it is obvious that
the difference of a+m and am in Prop. v is equal to the difference of 2a + m
and m in Prop. VI, and one algebraical result expresses the truth of both propositions.
This arises from the two ways in which the difference between two unequal lines may
be represented geometrically, when they are in the same direction.
In the diagram (fig. to Prop. v.), the difference DB of the two unequal lines AC
and CD is exhibited by producing the less line CD, and making CB equal to AC the
greater.
Then the part produced DB is the difference between AC and CD,
for AC is equal to CB, and taking CD from each,
the difference of AC and CD is equal to the difference of CB and CD.
In the diagram (fig. to Prop. v1), the difference DB of the two unequal lines CD and
CA is exhibited by cutting off from CD the greater, a part CB equal to CA the less.
Prop. VII. Algebraically.
Let AB contain a linear units, and let the parts AC and CB contain m and n linear
units respectively.
Then a = m + n ;
.. squaring these equals,
.. u² = m² + 2mn + n²,
add n² to each of these equals,
••• a² + n² = m² + 2mn + 2n².
But 2mn + 2 n² = 2 (m + n) n =
••. a² + n² = 2an + m².
2an,
That is, If a number be divided into any two parts, the square of the whole number
and of one of the parts is equal to twice the product of the whole number and that part,
together with the square of the other part.
Prop. VIII. Algebraically.
Let the whole line AB contain a linear units of which the parts AC, CB contain
m, n units respectively.
Then m + n = a,
and subtracting or taking n from each,
m = a — n,
squaring these equals,
... m² = a² - 2an + n²,
and adding 4an to each of these equals,
.. 4an + m² = a² + 2an + n².
72
EUCLID'S ELEMENTS.
But a²+2an + n² = (a + n)²,
... 4an + m² = (a + n)².
That is, If a number be divided into any two parts, four times the product of the
whole number and one of the parts, together with the square of the other part, is equal
to the square of the number made up of the whole and that part.
Prop. IX. Algebraically.
Let AB contain 2a linear units, its half AC or BC will contain a units; and let
CD the line between the points of section contain m units.
Also AD the greater of the two unequal parts contains a + m units,
and DB the less contains a m units.
••
Then (a+m)² = a² + 2am + m²,
and (am)²= a² + 2am + m²,
Hence by adding these equals,
(a + m)² + (a — m)² = 2a² + 2m².
That is, If a number be divided into two equal parts, and also into two unequal
parts, the sum of the squares of the two unequal parts is equal to twice the square
of half the number and twice the square of half the difference of the unequal parts.
Prop. x. Algebraically.
Let the line AB contain 2a linear units, of which its half AC or CB will contain
a units;
and let BD contain m units.
Then the whole line and the part produced will contain 2a+m units,
and half the line and the part produced will contain a+m units,
(2a + m)² = 4a² + 4am + m²,
add m² to each of these equals,
•. (2a + m)² + m² = 4a² + 4am + 2m².
Again (a + m)² = a² + 2am + m²,
add a² to each of these equals,
•. (a + m)² + a² = 2a² + 2am + m²,
2a²+2am
and doubling these equals,
•*. 2 (a + m)² + 2 a² 4a² + 4am + m².
=
But (2a+m)² + m² = 4a² + 4am + m².
Hence .. (2a + m)²+ m² = 2 a² + 2 (a + m)².
(2a+m)²+ 2a²+
That is, If a number be divided into two equal parts, and the whole number and
one of the parts be increased by the addition of another number, the squares of the
whole number thus increased, and of the number by which it is increased, are equal
to double the squares of half the number, and of half the number increased.
The algebraical results of Prop. IX, and Prop. x, are identical, the enunciations of
the two Props. arising, as in Prop. v and Prop. VI, from the two ways of exhibiting the
difference between two lines.
To solve Prop. XI algebraically, or to find the point H in AB such that the rect-
angle contained by the whole line AB and the part HB shall be equal to the square
of the other part AH.
Let AB contain a linear units, and AH one of the unknown parts contain x units,
then the other part HB contains a x units.
-
And •. a (a − x) = x², by the problem,
a², a quadratic equation.
or x² + ax =
Hence
む
​±α √5
2
NOTES TO BOOK II.
73
The former of these values of a determines the point H.
So that λ =
AB
AH, one part,
2
and
a — x = a - AH :
3- √√5
2
AB = HB, the other part.
It may be observed, that the parts AH and HB cannot be numerically expressed
by any rational number. Approximation to their true values in terms of AB, may
be made to any required degree of accuracy, by extending the extraction of the square
root of 5 to any number of decimals.
To ascertain the meaning of the other result x = -
√5 + 1
2
α.
In the equation a (ax) = x²,
for x write-x, then a (a + x) = x²,
which when translated into words gives the following problem.
To find the length a given line must be produced so that the rectangle contained
by the given line and the line made up of the given line and the part produced, may
be equal to the square of the part produced.
Or, the problem may also be expressed as follows:
To find two lines having a given difference, such that the rectangle contained by
the difference and one of them may be equal to the square of the other.
Prop. XII. Algebraically.
Assuming the truth of Prop. 47, Book 1, Algebraically.
Let BC, CA, AB contain a, b, c linear units respectively,
and let CD, DA, contain m, n units,
then BD contains a + m units.
And therefore, c² = (« + m)² + n², from the right-angled triangle ABD,
also b² = m² + n² from ACD;
Prop. XIII.
as Case I.
•'. c² — b² = (a + m)² — m²
=
a² + 2am + m² m²
= a² + 2am,
... c2 b² + a² + 2am,
*
that is, c² is greater than b² + a² by 2am.
Case II may be proved more simply as follows, in the same manner
Since BD is divided into two parts in the point D,
1
therefore the squares of CB, BD are equal to twice the rectangle contained by CB,
BD and the square of CD; (11. 7.)
add the square of AD to each of these equals ;
therefore the squares of CB, BD, DA are equal to twice the rectangle CB, BD, and
the squares of CD and DA,
but the squares of BD, DA are equal to the square of AB, (1. 47.)
and the squares of CD, DA are equal to the squares of AC,
therefore the squares of CB, BA are equal to the square of AC, and twice the
rectangle CB, BD.
Prop. XIII. Algebraically.
That is, &c.
Let BC, CA, AB contain respectively a, b, c linear units, and let BD and AD also
contain m and n units.
Case I.
Then DC contains am units.
74
EUCLID'S ELEMENTS.
Therefore c² = n² + m² from the right-angled triangle ABD,
and b²= n²+(a – m)² from ADC;
•. c²— b² = m² — (a — m)²
m²- a²+2am
=− a³ + 2am,
••• a² + c² = b² + 2am,
or b²+2am = a² + c²,
-
m²
that is, 6² is less than a²+c² by 2am.
Case II. DC
DC = m
- a units,
•. c² = m² + n² from the right-angled triangle ABD,
and b² = (m − a)² + n² from ACD,
•'. c² — b² = m² — (m − a)²
= m² m² + 2am — aº
= 2am - a²,
••• a² + c² = b² + 2am,
Case III.
or
b² + 2am = a² + c²,
that is, b² is less than a² + c² by 2am.
Here m is equal to a.
And b² + a² = c², from the right-angled triangle ABC.
Add to each of these equals a³,
... b² + 2a² = c² + a²,
that is, b² is less than c² + a² by 2a², or 2aa.
BOOK III.
:
DEFINITIONS.
I.
Equal circles are those of which the diameters are equal, or from
the centres of which the straight lines to the circumferences are equal.
This is not a definition, but a theorem, the truth of which is evident; for, if the
circles be applied to one another, so that their centres coincide, the circles must likewise
coincide, since the straight lines from the centres are equal.
II.
++
A straight line is said to touch a circle, when it meets the circle,
and being produced does not cut it.

III.
Circles are said to touch one another, which meet, but do not cut one
another.
IV.
Straight lines are said to be equally distant from the centre of a
circle, when the perpendiculars drawn to them from the centre are
equal.
V.
And the straight line on which the greater perpendicular falls, is
said to be farther from the centre.
VI.
A segment of a circle is the figure contained by a straight line, and
the circumference which it cuts off.
67
EUCLID'S ELEMENTS.
VII.
The angle of a segment is that which is contained by the straight
line and the circumference.
VIII.
An angle in a segment is the angle contained by two straight lines
drawn from any point in the circumference of the segment, to the extre-
mities of the straight line which is the base of the segment.
IX.
An angle is said to insist or stand upon the circumference intercepted
between the straight lines that contain the angle.
X.
A sector of a circle is the figure contained by two straight lines
drawn from the centre, and the circumference between them.
XI.
Similar segments of circles are those in which the angles are equal,
or which contain equal angles.

BOOK III. PROP. I, II.
77
PROPOSITION I. PROBLEM.
To find the centre of a given circle.
Let ABC be the given circle.
It is required to find its centre.
C
F
A
D B
E
Draw within it any straight line AB, and bisect AB in D; (1. 10.)
from the point D draw DC at right angles to AB, (1. 11.)
produce CD to E, and bisect CE in F.
Then the point F shall be the centre of the circle ABC.
For, if it be not, let, if possible, G be the centre, and join GA, GD, GB.
Then, because DA is equal to DB, (constr.) and DG common to the
two triangles ADG, BDG, the two sides AD, DG, are equal to the two
BD, DG, each to each;
and the base GA is equal to the base GB, (1. def. 15.)
because they are drawn from the centre G:
therefore the angle ADG is equal to the angle GDB: (1. 8.)
but when a straight line standing upon another straight line makes
the adjacent angles equal to one another, each of the angles is a right
angle; (1. def. 10.)
therefore the angle GDB is a right angle:
but FDB is likewise a right angle; (constr.)
wherefore the angle FDB is equal to the angle GDB, (ax. 1.)
the greater equal to the less, which is impossible;
therefore G is not the centre of the circle ABC.
In the same manner it can be shewn that no other point out of the
line CE is the centre;
and since CE is bisected in F,
any other point in CE divides CE into unequal parts, and cannot
be the centre.
Therefore no point but F is the centre of the circle ABC.
Which was to be found.
Cor. From this it is manifest, that if in a circle a straight line
bisects another at right angles, the centre of the circle is in the line
which bisects the other.
PROPOSITION II. THEOREM.
If any two points be taken in the circumference of a circle, the straight
line which joins them shall fall within the circle.
Let ABC be a circle, and A, B any two points in the circumference.
The straight line drawn from A to B shall fall within the circle.
78
EUCLID'S ELEMENTS.
▸
C
D
F
A E B
For if AB do not fall within the circle,
let it fall, if possible, without, as AEB;
find D the centre of the circle ABC, (111. 1.) and join DA, DB;
in the circumference AB take any point F, join DF, and produce it
to meet AB in E.
Then, because DA is equal to DB, (1. def. 15.)
therefore the angle DAB is equal to the angle DBA; (1. 5.)
and because AE, a side of the triangle DAE, is produced to B,
the exterior angle DEB is greater than the interior and opposite
angle DAE; (1. 16.)
but DAE was proved equal to the angle DBE;
therefore the angle DEB is greater than the angle DBE;
but to the greater angle the greater side is opposite, (1. 19.)
therefore DB is greater than DE:
but DB is equal to DF; (1. def. 15.)
wherefore DF is greater than DE, the less than the greater, which
is impossible;
therefore the straight line drawn from A to B does not fall without
the circle.
In the same manner, it may be demonstrated that it does not fall
upon the circumference;
therefore it falls within it.
Wherefore, if any two points, &c. Q.E.D.
PROPOSITION III. THEOREM.
If a straight line drawn through the centre of a circle bisect a straight
line in which it does not pass through the centre, it shall cut it at right
angles: and conversely, if it cuts it at right angles, it shall bisect it.
Let ABC be a circle; and let CD, a straight line drawn through
the centre, bisect any straight line AB, which does not pass through the
centre, in the point F.
Then CD shall cut AB at right angles.
A
K
F
C
B
D
Take E the centre of the circle, (111. 1.) and join EA, EB.
Then, because AF is equal to FB, (hyp.) and FE common to the
two triangles AFE, BFE,
there are two sides in the one equal to two sides in the other, each
to each;
and the base EA is equal to the base EB; (1. def. 15.)
BOOK III. PROP. III, IV.
79
therefore the angle AFE is equal to the angle BFE: (1..8.)
but when a straight line standing upon another straight line makes
the adjacent angles equal to one another, each of them is a right angle;
(1. def. 10.)
therefore each of the angles AFE, BFE, is a right angle:
wherefore the straight line CD, drawn through the centre, bisecting
another AB that does not pass through the centre, cuts the same at
right angles.
But let CD cut AB at right angles.
Then CD shall also bisect AB, that is, AF shall be equal to FB.
The same construction being made.
Because EA, EB, from the centre are equal to one another, (1. def. 15.)
the angle EAF is equal to the angle EBF; (1. 5.)
and the right angle AFE is equal to the right angle BFE: (1. def. 10.)
therefore, in the two triangles, EAF, EBF,
there are two angles in the one equal to two angles in the other,
each to each ;
and the side EF, which is opposite to one of the equal angles in
each, is common to both;
therefore the other sides are equal;
(1.26.)
therefore AF is equal to FB.
Wherefore, if a straight line, &c. Q.E.D.
PROPOSITION IV. THEOREM.
If in a circle two straight lines cut one another, which do not both
pass through the centre, they do not bisect each other.
Let ABCD be a circle, and AC, BD two straight lines in it which
cut one another in the point E, and do not both pass through the centre.
Then AC, BD, shall not bisect one another.
A
E
B
F
D
For, if it is possible, let AE be equal to EC, and BE to ED.
If one of the lines pass through the centre,
it is plain that it cannot be bisected by the other which does not
pass through the centre:
but if neither of them pass through the centre,
take F the centre of the circle, (111. 1.) and join EF.
Then because FE, a straight line drawn through the centre, bisects
another AC which does not pass through the centre, (hyp.)
therefore FE cuts AC at right angles: (III. 3.)
wherefore FEA is a right angle.
Again, because the straight line FE bisects the straight line BD,
which does not pass through the centre, (hyp.)
therefore FE cuts BD at right angles: (III. 3.)
wherefore FEB is a right angle:
but FEA was shewn to be a right angle;
80
EUCLID'S ELEMENTS.
therefore the angle FEA is equal to the angle FEB, (ax. 1.)
the less equal to the greater, which is impossible:
therefore AC, BD do not bisect one another.
Wherefore, if in a circle, &c. Q.E.D.
PROPOSITION V. THEOREM.
If two circles cut one another, they shall not have the same centre.
Let the two circles ABC, CDG, cut one another in the points B, C.
They shall not have the same centre.
A
C
D
G
[
CK
F
B
For, if it be possible, let E be their centre:
join EC, and draw any straight line EFG meeting them in F and G.
And because E is the centre of the circle ABC,
therefore EC is equal to EF: (1. def. 15.)
again, because E is the centre of the circle CDG,
therefore EC is equal to EG: (1. def. 15.)
but EC was shewn to be equal to EF;
therefore EF is equal to EG, (ax. 1.)
the equal less to the greater, which is impossible.
Therefore E is not the centre of the circles ABC, CDG.
Wherefore, if two circles, &c. Q. E.D.
PROPOSITION VI. THEOREM.
If one circle touch another internally, they shall not have the same centre.
Let the circle CDE touch the circle ABC internally in the point C.
They shall not have the same centre.
C
A
A
B
E
For, if they have, let it be F:
join FC, and draw any straight line FEB, meeting them in E and B.
And because F is the centre of the circle ABC,
FC is equal to FB; (1. def. 15.)
also, because F is the centre of the circle CDE,
FC is equal to FE: (1. def. 15.)
but FC was shewn to be equal to FB;
BOOK III. PROP. VI, VII,
81
therefore FE is equal to FB, (ax. 1.)
the less equal to the greater, which is impossible:
therefore F is not the centre of the circles ABC, CDĒ.
Therefore, if two circles, &c. Q.E.D.
PROPOSITION VII. THEOREM.
If any point be taken in the diameter of a circle which is not the
centre, of all the straight lines which can be drawn from it to the cir-
cumference, the greatest is that in which the centre is, and the other part
of that diameter is the least; and, of any others, that which is nearer to
the line which passes through the centre is always greater than one more
remote: and from the same point there can be drawn only two equal
straight lines to the circumference, one upon each side of the diameter.
Let ABCD be a circle, and AD its diameter, in which let any point
F be taken which is not the centre:
let the centre be E.
Then, of all the straight lines FB, FC, FG, &c. that can be drawn
from F to the circumference,
FA, that in which the centre is, shall be the greatest, and FD, the
other part of the diameter AD, shall be the least:
and of the others, FB, the nearer to FA, shall be greater than FC,
the more remote, and FC greater than FG.

C
BA
E
K
G D
H
Join BE, CE, GE.
And because two sides of a triangle are greater than the third, (1. 20.)
therefore BE, EF are greater than BF:
but AE is equal to BE; (1. def. 15.)
therefore AE, EF, that is, AF is greater than BF.
Again, because BE is equal to CE,
and FE common to the triangles BEF, CEF,
1
the two sides BE, EF are equal to the two CE, EF, each to each
but the angle BEF is greater than the angle CEF; (ax. 9.)
therefore the base BF is greater than the base CF. (1. 24.)
For the same reason CF is greater than GF.
Again, because GF, FE are greater than EG, (1. 20.)
and EG is equal to ED;
therefore GF, FE are greater than ED:
take away the common part FE,
;
and the remainder GF is greater than the remainder FD. (ax. 5.)
Therefore, FA is the greatest, and FD the least of all the straight
lines from F to the circumference;
and BF is greater than CF, and CF than GF.
Also, there can be drawn only two equal straight lines from the point
F to the circumference, one upon each side of the diameter.
6
82
EUCLID'S ELEMENTS.
At the point E, in the straight line EF, make the angle FEH equal
to the angle FEG, (1. 23.) and join FH.
Then, because GE is equal to EH, (1. def. 15.)
and EF common to the two triangles GEF, HEF;
the two sides GE, EF are equal to the two HE, EF, each to each;
and the angle GEF is equal to the angle HEF; (constr.)
therefore, the base FG is equal to the base FH: (1. 4.)
but, besides FH, no other straight line can be drawn from F to the
circumference equal to FG:
for, if there can, let it be FK:
and, because FK is equal to FG, and FG to FH,
therefore FK is equal to FH; (ax. 1.)
that is, a line nearer to that which passes through the centre, is
equal to one which is more remote;
which has been proved to be impossible.
Therefore, if any point be taken, &c.
Q. E.D.
PROPOSITION VIII. THEOREM.
If any point be taken without a circle, and straight lines be drawn
from it to the circumference, whereof one passes through the centre; of
those which fall upon the concave circumference, the greatest is that which
passes through the centre; and of the rest, that which is nearer to the one
passing through the centre is always greater than one more remote: but
of those which fall upon the convex circumference, the least is that between
the point without the circle and the diameter; and of the rest, that which
is nearer to the least is always less than one more remote: and only two
equal straight lines can be drawn from the same point to the circumference,
one upon each side of the line which passes through the centre.
Let ABC be a circle, and D any point without it, from which let
the straight lines DA, DE, DF, DC be drawn to the circumference,
whereof DA passes through the centre.
Of those which fall upon the concave part of the circumference
AEFC,
the greatest shall be DA, which passes through the centre;
and any line nearer to it shall be greater than one more remote,
viz. DE, shall be greater than DF, and DF greater than DC:
but of those which fall upon the convex circumference HLKG,
the least shall be DG between the point D and the diameter AG;
and any line nearer to it shall be less than one more remote,
viz. DK less than DL, and DL less than DH.
D

XLGB
GB\N
H
F
E
A
M
BOOK III. PROP. VIII, IX.
Take M the centre of the circle ABC, (111. 1.)
and join ME, MF, MC, MK, ML, MH.
And because AM is equal to ME,
add MD to each of these equals,
therefore AD is equal to EM, MD:
(ax. 2.)
but EM, MD are greater than ED; (1. 20.)
therefore also AD is greater than ED.
83
Again, because ME is equal to MF, and MD common to the tri-
angles EMD, FMD;
EM, MD, are equal to FM, MD, each to each:
but the angle EMD is greater than the angle FMD; (ax. 9.)
therefore the base ED is greater than the base FD. (1. 24.)
In like manner it may be shewn that FD is greater than CD.
Therefore, DA is the greatest;
and DE greater than DF, and DF greater than DC.
And, because MK, KD are greater than MD, (1. 20.)
and MK is equal to MG, (1. def. 15.)
the remainder KD is greater than the remainder GD, (ax. 5.)
that is, GD is less than KD:
and because MLD is a triangle, and from the points M, D, the
extremities of its side MD, the straight lines MK, DK are drawn to the
point K within the triangle,
therefore MK, KD are less than ML, LD: (1. 21.)
but MK is equal to ML; (1. def. 15.)
therefore, the remainder DK is less than the remainder DL. (ax. 5.)
In like manner it may be shewn, that DL is less than DÈ.
Therefore, DG is the least, and DK less than DL, and DL less than DH.
Also, there can be drawn only two equal straight lines from the
point D to the circumference, one upon each side of the line through the

centre.
At the point M, in the straight line MD,
make the angle DMB equal to the angle DMK, (1. 23.) and join DB.
And because MK is equal to MB, and MD common to the triangles
KMD, BMD,
the two sides KM, MD are equal to the two BM, MD, each to each;
and the angle KMD is equal to the angle BMD; (constr.)
therefore the base DK is equal to the base DB: (1. 4.)
but, besides DB, there can be no straight line drawn from D to
the circumference equal to DK:
for, if there can, let it be DN:
and because DK is equal to DN, and also to DB,
therefore DB is equal to DN;
that is, a line nearer to the least is equal to one more remote, which
has been proved to be impossible.
If, therefore, any point, &c. Q.E.D.
PROPOSITION IX. THEOREM.
If a point be taken within a circle, from which there fall more than
two equal straight lines to the circumference, that point is the centre of
the circle.
Let the point D be taken within the circle ABC,
from which to the circumference there fall more than two equal
straight lines, viz. DA, DB, DC.
Then the point D shall be the centre of the circle.
6—2
84
EUCLID'S ELEMENTS.

DE
F
G
C
A B
For, if not, let E be the centre:
join DE, and produce it to the circumference in F, G ;
then FG is a diameter of the circle ABC: (1. def. 17.)
and because in FG, the diameter of the circle ABC, there is taken
the point D, which is not the centre,
therefore DG is the greatest line from it to the circumference,
and DC is greater than DB, and DB greater than DA: (111. 7.)
but they are likewise equal, (hyp.) which is impossible:
therefore E is not the centre of the circle ABC.
In like manner it may be demonstrated,
that no other point but D is the centre;
D therefore is the centre.
Wherefore, if a point be taken, &c. Q.E.D.
PROPOSITION X. THEOREM.
One circumference of a circle cannot cut another in more than two points.
If it be possible, let the circumference FAB cut the circumference
DEF in more than two points, viz. in B, G, F.
B
A

H*
AK
D
E
G
F
C
Take the centre K of the circle ABC, (111. 3.) and join KB, KG, KF.
Then because K is the centre of the circle ABC,
therefore KB, KG, KF are all equal to each other: (1. def. 15.)
and because within the circle DEF there is taken the point K, from
which to the circumference DEF fall more than two equal straight lines
KB, KG, KF,
therefore the point K is the centre of the circle DEF: (111. 9.)
but K is also the centre of the circle ABC; (constr.)
therefore the same point is the centre of two circles that cut one
another, which is impossible. (III. 5.)
Therefore, one circumference of a circle cannot cut another in more
than two points. Q.E.D.
PROPOSITION XI. THEOREM.
If one circle touch another internally in any point, the straight line
which joins their centres being produced shall pass through that point of
contact.
Let the circle ADE touch the circle ABC internally in the point A ;
BOOK III. PROP. XI, XII.
.85
and let F be the centre of the circle ABC, and G the centre of
the circle ADE;
then the straight line which joins the centres F, G, being produced,
shall pass through the point A.
A
H
OG T
B
C
For, if FG produced do not pass through the point A,
let it fall otherwise, if possible, as FGDH, and join AF, AG.
Then, because two sides of a triangle are together greater than the
third side, (1. 20.)
therefore FG, GA are greater than FA:
but FA is equal to FH; (1. def. 15.)
therefore FG, GA are greater than FH:
take away from these unequals the common part FG;
therefore the remainder AG is greater than the remainder GH; (ax. 5.)
but AG is equal to GD; (1. def. 15.)
therefore ĜD is greater than GH,
the less than the greater, which is impossible.
Therefore the straight line which joins the points F, G, being pro-
duced, cannot fall otherwise than upon the point A,
that is, it must pass through it.
Therefore, if one circle, &c.
Q. E.D.
PROPOSITION XII. THEOREM.
If two circles touch each other externally in any point, the straight
line which joins their centres shall pass through that point of contact.
Let the two circles ABC, ADE, touch each other externally in the
point A;
and let F be the centre of the circle ABC, and G the centre of ADE.
Then the straight line which joins the points F, G, shall pass
through the point of contact A.

E
B
A
F
:
For, if not, let it pass otherwise, if possible, as FCDG, and join FA, AG.
And because F is the centre of the circle ABC,
FA is equal to FC:
also, because G is the centre of the circle ADE,
GA is equal to GD:
therefore FA, AG are equal to FC, DG; (ax. 2.)
wherefore the whole FG is greater than FÀ, AG:
but FG is less than FA, AG; (1. 20.) which is impossible:
86
EUCLID'S ELEMENTS.
therefore the straight line which joins the points F, G, cannot pass
otherwise than through the point of contact A,
that is, FG must pass through the point A.
Therefore, if two circles, &c. Q.E.D.
PROPOSITION XIII. THEOREM.
One circle cannot touch another in more points than one, whether it
touches it on the inside or outside.
For, if it be possible, let the circle EBF touch the circle ABC in
more points than one,
and first on the inside, in the points B, D.

B
A
E
H
B
F/C
D
H

A
D
G
Join BD, and draw GH bisecting BD at right angles. (1. 11.)
Because the points B, D are in the circumferences of each of the circles,
therefore the straight line BD falls within each of them; (111. 2.)
therefore their centres are in the straight line GH which bisects
BD at right angles; (III. Cor. 1.)
therefore GH passes through the point of contact: (III. 11.)
but it does not pass through it,
because the points B, D are without the straight line GH;
which is absurd:
therefore one circle cannot touch another on the inside in more
points than one.
Nor can two circles touch one another on the outside in more than
one point.
For, if it be possible,
let the circle ACK touch the circle ABC in the points A, C;
join AC.

A
B
K
C
Because the two points A, C are in the circumference of the
circle ACK,
therefore the straight line AC which joins them, falls within the
circle ACK: (III. 2.)
but the circle ACK is without the circle ABC; (hyp.)
therefore the straight line AC is without this last circle:
but, because the points A, C are in the circumference of the circle ABC,
the straight line AC must be within the same circle, (111. 2.)
which is absurd;
BOOK III. PROP. XIII, XIV.
87
therefore one circle cannot touch another on the outside in more
than one point :
and it has been shewn, that they cannot touch on the inside in
more points than one.
Therefore, one circle, &c. Q.E.D.
PROPOSITION XIV. THEOREM.
Equal straight lines in a circle are equally distant from the centre;
and conversely, those which are equally distant from the centre, are equal
to one another.
Let the straight lines AB, CD, in the circle ABDC, be equal to
one another.
Then AB and CD shall be equally distant from the centre.
C
A
A
D
1
B
Take E the centre of the circle ABDC, (111. 1.)
from E draw EF, EG perpendiculars to AB, CD, (1. 12.) and
join EA, EC.
Then, because the straight line EF, passing through the centre,
cuts the straight line AB, which does not pass through the centre,
at right angles,
it also bisects it: (III. 3.)
therefore AF is equal to FB, and AB double of AF.
For the same reason CD is double of CG:
but AB is equal to CD: (hyp.)
therefore AF is equal to CG. (ax. 7.)
And because AE is equal to EC, (1. def. 15.)
the square of AE is equal to the square of EC:
but the squares of AF, FE are equal to the square of AE, (1. 47.)
because the angle AFE is a right angle;
and, for the like reason, the squares of EG, GC are equal to
the square of EC;
therefore the squares of AF, FE are equal to the squares of
CG, GE: (ax. 1.)
but the square of AF is equal to the square of CG,
because AF is equal to CG;
therefore the remaining square of EF is equal to the remaining
square of EG, (ax. 3.)
and the straight line EF is therefore equal to EG:
but straight lines in a circle are said to be equally distant from
the centre, when the perpendiculars drawn to them from the centre
are equal: (III. def. 4.)
therefore AB, CD are equally distant from the centre.
Next, let the straight lines AB, CD be equally distant from the
centre, (III. def. 4.)
that is, let FE be equal to EG;
88
EUCLID'S ELEMENTS.
then AB shall be equal to CD.
For, the same construction being made, it may, as before, be
demonstrated,
that AB is double of AF, and CD double of CG,
and that the squares of EF, FA are equal to the squares of EG, GC:
but the square of FE is equal to the square of EG,
because FE is equal to EG; (hyp.)
therefore the remaining square of AF is equal to the remaining
square of CG: (ax. 3.)
and the straight line AF is therefore equal to CG :
but AB was shewn to be double of AF, and ĈD double of CG;
wherefore AB is equal to CD. (ax. 6.)
Therefore equal straight lines, &c.
Q.E.D.
PROPOSITION XV. THEOREM.
The diameter is the greatest straight line in a circle; and, of all
others, that which is nearer to the centre is always greater than one
more remote: and the greater is nearer to the centre than the less.
Let ABCD be a circle, of which the diameter is AD, and the centre E;
and let BC be nearer to the centre than FG.
Then AD shall be greater than any straight line BC, which is not
a diameter, and BC shall be greater than FG.
A B
F
H
E
C
G
D
From the centre E draw EH, EK perpendiculars to BC, FG, (1. 12.)
and join EB, EC, EF.
And because AE is equal to EB, and ED to EC, (1. def. 15.)
therefore AD is equal to EB, EC: (ax. 2.)
but EB, EC are greater than BC; (1. 20.)
wherefore also AD is greater than BC.
And, because BC is nearer to the centre than FG, (hyp.)
therefore EH is less than EK: (111. def. 5.)
but, as was demonstrated in the preceding proposition,
BC is double of BH, and FG double of FK,
and the squares of EH, HB are equal to the squares of EK, KF:
but the quare of EH is less than the square of EK,
because EH is less than EK;
therefore the square of BH is greater than the square of FK,
and the straight line BH greater than FK,
and therefore BC is greater than FG.
Next, let BC be greater than FG ;
then BC shall be nearer to the centre than FG, that is, the same
construction being made, EH shall be less than EK. (111. def. 5.)
Because BC is greater than FG,
BH likewise is greater than KF:
and the squares of BH, HE are equal to the squares of FK, KE;
BOOK III. PROP. XV, XVI.
of which the square of BH is greater than the square of FK,
because BH is greater than FK:
therefore the square of EH is less than the square of EK,
and the straight line EH less than ĒK:
and therefore BC is nearer to the centre than FG. (111. def. 5.)
Wherefore the diameter, &c. Q. E.D.
89
PROPOSITION XVI. THEOREM.
The straight line drawn at right angles to the diameter of a circle,
from_the_extremity of it, falls without the circle; and no straight line
can be drawn from the extremity between that straight line and the cir
cumference, so as not to cut the circle; or, which is the same thing, no
straight line can make so great an acute angle with the diameter at its
extremity, or so small an angle with the straight line which is at right
angles to it, as not to cut the circle.
Let ABC be a circle, the centre of which is D, and the diameter AB.
Then the straight line drawn at right angles to AB from its extre-
mity A, shall fall without the circle.

C
L
D
A
!
For, if it does not, let it fall, if possible, within the circle, as AC;
and draw DC to the point C, where it meets the circumference.
And because DA is equal to DC, (1. def. 15.)
the angle DAC is equal to the angle ACD: (1. 5.)
but DAC is a right angle; (hyp.)
therefore ACD is a right angle;
and therefore the angles DAC, ACD are equal to two right angles;
which is impossible: (1. 17.)
therefore the straight line drawn from A at right angles to BA,
does not fall within the circle.
In the same manner it may be demonstrated,
that it does not fall upon the circumference
therefore it must fall without the circle, as AE.
Also, between the straight line AE and the circumference, no straight
line can be drawn from the point A which does not cut the circle.
For, if possible, let AF be between them.
FE
H
R
A
D
From the point D draw DG perpendicular to AF, (1. 12.)
and let it meet the circumference in H.
And because AGD is a right angle, and DAG less than a right
angle, (1. 17.)
90
EUCLID'S ELEMENTS.
therefore DA is greater than DG: (1. 19.)
but DA is equal to DH; (1. def. 15.)
therefore DH is greater than DG,
the less than the greater, which is impossible:
therefore no straight line can be drawn from the point A, between
AE and the circumference, which does not cut the circle:
or, which amounts to the same thing, however great an acute angle
a straight line makes with the diameter at the point A, or however
small an angle it makes with AE, the circumference must pass between
that straight line and the perpendicular AE.
"And this is all that is to be understood, when, in the Greek text,
and translations from it, the angle of the semicircle is said to be greater
than any acute rectilineal angle, and the remaining angle less than any
rectilineal angle." Q.E.D.
COR. From this it is manifest, that the straight line which is drawn
at right angles to the diameter of a circle from the extremity of it,
touches the circle; (111. def. 2.) and that it touches it only in one point,
because, if it did meet the circle in two, it would fall within it. (III. 2.)
"Also, it is evident, that there can be but one straight line which
touches the circle in the same point.”
PROPOSITION XVII. PROBLEM.
To draw a straight line from a given point, either without or in the
circumference, which shall touch a given circle.
First, let A be a given point without the given circle BCD;
it is required to draw a straight line from A which shall touch the circle.

G C E4
A
D
F
B
Find the centre E of the circle, (III. 1.) and join AE;
and from the centre E, at the distance EA, describe the circle AFG;
from the point D draw DF at right angles to EA, (1. 11.) and join
EBF, AB.
Then AB shall touch the circle BCD in the point B.
Because E is the centre of the circles BCD, AFG,
therefore EA is equal to EF, (1. def. 15.) 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 at E common to the two triangles AEB, FED;
therefore the base DF is equal to the base AB, (1. 4.)
and the triangle FED to the triangle AEB,
and the other angles to the other angles:
therefore the angle EBA is equal to the angle EDF:
but EDF is a right angle, (constr.)
wherefore EBA is a right angle: (ax. 1.)
and EB is drawn from the centre:
BOOK III.
91
PROP. XVII, XVIII, XIX.
but a straight line drawn from the extremity of a diameter, at right
angles to it, touches the circle: (III. 16. Cor.)
therefore AB touches the circle;
and it is drawn from the given point A.
But if the given point be in the circumference of the circle,
as the point D,
draw DE to the centre E, and DF at right angles to DE:
DF touches the circle. (111. 16. Cor.) Q.E.F.
PROPOSITION XVIII. THEOREM.
If a straight line touches a circle, the straight line drawn from the
centre to the point of contact, shall be perpendicular to the line touching
the circle.
Let the straight line DE touch the circle ABC in the point C;
take the centre F, and draw the straight line FC. (11. 1.)
Then FC shall be perpendicular to DE.
A
F
B
D C GE
For, if it be not, from the point F draw FBG perpendicular to DE.
And because FGC is a right angle,
therefore GCF is an acute angle; (1. 17.)
and to the greater angle the greater side is opposite: (1. 19.)
therefore FC is greater than FG:
but FC is equal to FB; (1. def. 15.)
therefore FB is greater than FG,
the less than the greater, which is impossible:
therefore FG is not perpendicular to DE.
In the same manner it may be shewn,
that no other is perpendicular to DE besides FC,
that is, FC is perpendicular to DE.
Therefore, if a straight line, &c.
Q.E.D.
1
PROPOSITION XIX. THEOREM.
If a straight line touches a circle, and from the point of contact a
straight line be drawn at right angles to the touching line, the centre of
the circle shall be in that line.
Let the straight line DE touch the circle ABC in C,
and from Clet CA be drawn at right angles to DE.
Then the centre of the circle shall be in CA.
A

F
B
D C
E
92
EUCLID'S ELEMENTS.
For, if not, let F be the centre, if possible, and join CF.
Because DE touches the circle ABC, and FC is drawn from the
centre to the point of contact,
therefore FC is perpendicular to DE; (111. 18.)
therefore FCE is a right angle:
but ACE is also a right angle; (hyp.)
therefore the angle FCE is equal to the angle AĆE, · (ax. 1.)
the less to the greater, which is impossible:
therefore F is not the centre of the circle ABC.
In the same manner it may be shewn,
that no other point which is not in CA, is the centre;
that is, the centre of the circle is in CA.
Therefore, if a straight line, &c. Q.E.D.
PROPOSITION XX.
THEOREM.
The angle at the centre of a circle is double of the angle at the cir-
cumference upon the same base, that is, upon the same part of the cir-
cumference.
Let ABC be a circle, and BEC an angle at the centre, and BAC an
angle at the circumference, which have the same circumference BC for
their base.
Then the angle BEC shall be double of the angle BAC.
A
B
F
E
Join AE, and produce it to F.
First, let the centre of the circle be within the angle BAC.
Because EA is equal to EB,
therefore the angle EAB is equal to the angle EBA; (1.5.)
therefore the angles EAB, EBA are double of the angle EAB:
but the angle BEF is equal to the angles EAB, EBA; (1. 32.)
therefore also the angle BEF is double of the angle EAB:
for the same reason, the angle FEC is double of the angle EAC:
therefore the whole angle BEC is double of the whole angle BAC.
Again, let the centre of the circle be without the angle BAC.

F
B
E
A
It may be demonstrated, as in the first case,
that the angle FEC is double of the angle FAC,
and that FEB, a part of the first, is double of FAB, a part of the other;
therefore the remaining angle BEC is double of the remaining
angle BAC.
Therefore the angle at the centre, &c. Q.E. D.
BOOK III. PROP. XXI, XXII.
93
PROPOSITION XXI. THEOREM.
The angles in the same segment of a circle are equal to one another.
Let ABCD be a circle,
and BAD, BED angles in the same segment BAED.
Then the angles BAD, BED shall be equal to one another.
First, let the segment BAED be greater than a semicircle.
A
E
F
D
B
C
Take F, the centre of the circle ABCD, (111. 1.) and join BF, FD.
And because the angle BFD is at the centre, and the angle BAD
at the circumference, and that they have the same part of the circum-
ference, viz. BCD for their base;
therefore the angle BFD is double of the angle BAD: (111. 20.)
for the same reason the angle BFD is double of the angle BED:
therefore the angle BAD is equal to the angle BED. (ax. 7.)
Next, let the segment BAED be not greater than a semicircle.
B
A E
F
D
}
C
Draw AF to the centre, and produce it to C, and join CE.
Therefore the segment BADC is greater than a semicircle;
and the angles in it BAC, BEC are equal, by the first case:
for the same reason, because CBED is greater than a semicircle,
the angles CAD, CED, are equal:
therefore the whole angle BAD is equal to the whole angle BED. (ax.2.)
Wherefore the angles in the same segment, &c.
Q. E. D.
PROPOSITION XXII. THEOREM.
The opposite angles of any quadrilateral figure inscribed in a circle,
are together equal to two right angles.
Let ABCD be a quadrilateral figure in the circle ABCD.
Then any two of its opposite angles shall together be equal to two
right angles.
D C
A
B
94
EUCLID'S ELEMENTS.
Join AC, BD.
And because the three angles of every triangle are equal to two
right angles, (1. 32.)
the three angles of the triangle CAB, viz. the angles CAB, ABC,
BCA, are equal to two right angles:
but the angle CAB is equal to the angle CDB, (III. 21.)
because they are in the same segment CDAB;
and the angle ACB is equal to the angle ADB,
because they are in the same segment ADCB:
therefore the two angles CAB, ACB are together equal to the
whole angle ADC: (ax. 2.)
to each of these equals add the angle ABC;
therefore the three angles ABC, CAB, BCA are equal to the
two angles ABC, ADC: (ax. 2.)
but ABC, CAB, BCA, are equal to two right angles;
•
therefore also the angles ABC, ADC are equal to two right
angles. (ax. 1.)
In the same manner, the angles BAD, DCB, may be shewn to be
equal to two right angles.
Therefore, the opposite angles, &c. Q.E.D.
PROPOSITION XXIII. THEOREM.
Upon the same straight line, and upon the same side of it, there
cannot be two similar segments of circles, not coinciding with one another.
If it be possible, upon the same straight line AB, and upon the
same side of it, let there be two similar segments of circles, ACB, ADB,
not coinciding with one another.
D
A
B
Then, because the circle ACB cuts the circle ADB in the two points A, B,
they cannot cut one another in any other point: (III. 10.)
therefore one of the segments must fall within the other :
let ACB fall within ADB:
draw the straight line BCD, and join CA, DA.
And because the segment ACB is similar to the segment ADB, (hyp.)
and that similar segments of circles contain equal angles; (III. def. 11.)
therefore the angle ACB is equal to the angle ADB,
the exterior to the interior, which is impossible. (1. 16.)
Therefore, there cannot be two similar segments of circles upon the
same side of the same line, which do not coincide.
Q. E.D.
PROPOSITION XXIV. THEOREM.
Similar segments of circles upon equal straight lines are equal to one another.
Let AEB, CFD be similar segments of circles upon the equal
straight lines AB, CD.
Then the segment AEB shall be equal to the segment CFD.
BOOK III. PROP. XXIV, XXV.
95
E
F
A
B
C
D
For if the segment AEB be applied to the segment CFD,
so that the point A may be on C, and the straight line AB upon CD,
then the point B shall coincide with the point D,
because AB is equal to CD:
therefore, the straight line AB coinciding with CD,
the segment AEB must coincide with the segment CFD, (111. 23.)
and therefore is equal to it. (ax. 8.)
Wherefore similar segments, &c. Q.E.D.
PROPOSITION XXV. PROBLEM.
A segment of a circle being given, to describe the circle of which it
is the segment.
Let ABC be the given segment of a circle.
It is required to describe the circle of which it is the segment.
Bisect AC in D, (1. 10.) and from the point D draw DB at right
angles to AC, (1. 11.) and join AB.
First, let the angles ABD, BAD be equal to one another:
B
7
A D C
then the straight line BD is equal to DA, (1.6.) and therefore, to DC;
and because the three straight lines DA, DB, DC are all equal,
therefore D is the centre of the circle. (III. 9.)
From the centre D, at the distance of any of the three, DA, DB,
DC, describe a circle;
this shall pass through the other points;
and the circle of which ABC is a segment is described:
and because the centre D is in AC, the segment ABC is a semicircle.
But if the angles ABD, BAD are not equal to one another.
B
C
E
B
E
A D C
At the point A, in the straight line AB, make the angle BAE equal
to the angle ABD, (1. 23.)
and produce BD, if necessary, to meet AE in E, and join EC.
And because the angle ABE is equal to the angle BAE,
therefore the straight line BE is equal to EA: (1.6.)
and because AD is equal to DC, and DE common to the triangles
ADE, CDE,
the two sides AD, DE, are equal to the two CD, DE, each to each ;
and the angle ADÊ is equal to the angle CDE,
96
EUCLID'S ELEMENTS.
for each of them is a right angle; (constr.)
therefore the base AE is equal to the base EC: (1. 4.)
but AE was shewn to be equal to EB;
wherefore also BE is equal to EC; (ax. 1.)
and therefore the three straight lines AE, EB, EĆ are equal to
one another:
wherefore E is the centre of the circle. (111. 9.)
From the centre E, at the distance of any of the three AE, EB, EC,
describe a circle;
this shall pass through the other points;
and the circle, of which ABC is a segment, is described.
And it is evident, that if the angle ABD be greater than the angle
BAD, the centre E falls without the segment ABC, which therefore is
less than a semicircle:
but if the angle ABD be less than BAD, the centre E falls within
the segment ABC, which is therefore greater than a semicircle.
Wherefore a segment of a circle being given, the circle is described
of which it is a segment. Q. E. F.
PROPOSITION XXVI. THEOREM.
In equal circles, equal angles stand upon equal circumferences, whether
they be at the centres or circumferences.
Let ABC, DEF be equal circles,
and let the angles BGC, EHF at their centres,
and BAC, EDF at their circumferences be equal to each other.
The circumference BKC shall be equal to the circumference ELF.
A
D
H
B
K
G
E
C
L
Join BC, EF.
F
And because the circles ABC, DEF are equal,
the straight lines drawn from their centres are equal: (III. def. 1.)
therefore the two sides BG, GC, are equal to the two EH, HF,
each to each:
and the angle at G is equal to the angle at H; (hyp.)
therefore the base BC is equal to the base EF. (1. 4.)
And because the angle at A is equal to the angle at D, (hyp.)
the segment BAC is similar to the segment EDF; (III. def. 11.)
and they are upon equal straight lines BC, ÈF:
but similar segments of circles upon equal straight lines, are equal
to one another, (nr. 24.)
therefore the segment BAĆ is equal to the segment EDF:
but the whole circle ABC is equal to the whole DEF; (hyp.)
therefore the remaining segment BKC is equal to the remaining
segment ELF, (ax. 3.)
and the circumference BKC to the circumference ELF.
Wherefore, in equal circles, &c. Q.E. D.
BOOK III. PROP. XXVII, XXVIII.
97
PROPOSITION XXVII. THEOREM.
In equal circles, the angles which stand upon equal circumferences are
equal to one another, whether they be at the centres or circumferences.
Let ABC, DEF be equal circles,
and let the angles BGC, EHF at their centres,
and BAC, EDF at their circumferences, stand upon the equal
circumferences BC, EF.
The angle BGC shall be equal to the angle EHF, and the angle
BAC to the angle EDF.
A
D

B
C
K
H
F
E\
If the angle BGC be equal to the angle EHF,
it is manifest that the angle BAC is also equal to EDF. (111. 20.
and 1. ax. 7.)
But, if not, one of them must be greater than the other:
let BGC be the greater,
and at the point G, in the straight line BG, make the angle BGK
equal to the angle EHF. (1. 23.)
+
Then because the angle BGK is equal to the angle EHF,
and that equal angles stand upon equal circumferences, when they
are at the centres; (111. 26.)
therefore the circumference BK is equal to the circumference EF:
but EF is equal to BC; (hyp.)
therefore also BK is equal to BC, the less to the greater,
which is impossible: (ax. 1.)
therefore the angle BGC is not unequal to the angle EHF;
that is, it is equal to it:
but the angle at A is half of the angle BGC, (III. 20.)
and the angle at D half of the angle EHF;
therefore the angle at A is equal to the angle at D. (ax. 7.)
Wherefore, in equal circles, &c.
Q.E.D.
PROPOSITION XXVIII. THEOREM.
In equal circles, equal straight lines cut off equal circumferences, the
greater equal to the greater, and the less to the less.
Let ABC, DEF be equal circles,
and BC, EF equal. straight lines in them, which cut off the two
greater circumferences BAC, EDF, and the two less BGC, EHF. ·
Then the greater circumference BAC shall be equal to the greater EDF,
and the less BGC to the less EHF.
B
A
K
D
L
C
E
F
G
H
7
98
EUCLID'S ELEMENTS.
Take K, L, the centres of the circles, (111. 1.) and join BK, KC, EL, LF.
Because the circles ABC, DEF are equal, the straight lines from
their centres are equal: (111. def. 1.)
therefore BK, KČ are equal to EL, LF, each to each:
and the base BC is equal to the base EF; (hyp.)
therefore the angle BKC is equal to the angle ELF: (1. 8.)
but equal angles stand upon equal circumferences, when they are
at the centres; (111. 26.)
therefore the circumference BGC is equal to the circumference EHF:
but the whole circle ABC is equal to the whole EDF; (hyp.)
therefore the remaining part of the circumference,
viz. BAC, is equal to the remaining part EDF. (ax. 3.)
Therefore, in equal circles, &c.
Q.E.D.
PROPOSITION XXIX. THEOREM.
In equal circles, equal circumferences are subtended by equal straight lines.
Let ABC, DEF be equal circles,
and let the circumferences BGC, EHF also be equal;
and join BC, EF.
Then the straight line BC shall be equal to the straight line EF.
A
K
D
B
C
E
F
G
H
Take K, L, (III. 1.) the centres of the circles, and join BK, KC, EL, LF:
and because the circumference BGC is equal to the circumference EHF,
therefore the angle BKC is equal to the angle ELF: (111. 27.)
and because the circles ABC, DEF, are equal,
the straight lines from their centres are equal; (III. def. 1.)
therefore BK, KC, are equal to EL, LF, each to each:
and they contain equal angles;
therefore the base BC is equal to the base EF. (1. 4.)
Therefore, in equal circles, &c.
Q.E.D.
PROPOSITION XXX. PROBLEM.
To bisect a given circumference, that is, to divide it into two equal parts.
Let ADB be the given circumference.
It is required to bisect it.
D
A
B
Join AB, and bisect it in C; (1. 10.)
from the point C draw CD at right angles to AB. (1. 11.)
Then the circumference ADB shall be bisected in the point D.
Join AD, DB.
BOOK III. PROP. XXX, XXXI.
99
And because AC is equal to CB, and CD common to the triangles
ACD, BCD,
the two sides AC, CD are equal to the two BC, CD, each to each ;
and the angle ACD is equal to the angle BCD,
because each of them is a right angle:
therefore the base AD is equal to the base BD. (1. 4.)
But equal straight lines cut off equal circumferences, (III. 28.)
the greater equal to the greater, and the less to the less;
and AD, DB are each of them less than a semicircle ;
because DC passes through the centre: (III. 1. Cor.)
therefore the circumference AD is equal to the circumference DB.
Therefore the given circumference is bisected in D. Q. E. F.
PROPOSITION XXXI. THEOREM.
In a circle, the angle in a semicircle is a right angle; but the angle
in a segment greater than a semicircle is less than a right angle; and the
angle in a segment less than a semicircle is greater than a right angle.
Let ABCD be a circle, of which the diameter is BC, and centre E;
and draw CA, dividing the circle into the segments ABC, ADC,
and join BA, AD, DC.
Then the angle in the semicircle BAC shall be a right angle ;
and the angle in the segment ABC, which is greater than a semi-
circle, shall be less than a right angle;
and the angle in the segment ADC, which is less than a semicircle,
shall be greater than a right angle.
F

D
B
E
C
Join AE, and produce BA to F.
And because BE is equal to EA, (1. def. 15.)
the angle EAB is equal to EBA; (1.5.)
also, because AE is equal to EC,
the angle EAC is equal to ECA;
wherefore the whole angle BAC is equal to the two angles ABC,
ACB: (ax. 2.)
but FAC, the exterior angle of the triangle ABC, is equal to the
two angles ABC, ACB; (1. 32.)
therefore the angle BAC is equal to the angle FAC; (ax. 1.)
and therefore each of them is a right angle: (1. def. `10.)
wherefore the angle BAC in a semicircle is a right angle.
And because the two angles ABC, BAC of the triangle ABC are
together less than two right angles, (1. 17.)
and that BAC has been proved to be a right angle;
therefore ABC must be less than a right angle:
and therefore the angle in a segment ABC greater than a semicircle,
is less than a right angle.
And because ABCD is a quadrilateral figure in a circle,
ta..
7-2
100
EUCLID'S ELEMENTS.
any two of its opposite angles are equal to two right angles: (111. 22.)
therefore the angles ABC, ADC, are equal to two right angles:
and ABC has been proved to be less than a right angle;
wherefore the other ADC is greater than a right angle.
Besides, it is manifest, that the circumference of the greater segment
ABC falls without the right angle CAB; but the circumference of the
less segment ADC falls within the right angle CAF. "And this is all
that is meant, when in the Greek text, and the translations from it, the
angle of the greater segment is said to be greater, and the angle of the
less segment is said to be less, than a right angle."
COR. From this it is manifest, that if one angle of a triangle be
equal to the other two, it is a right angle: because the angle adjacent
to it is equal to the same two; (1.32.) and when the adjacent angles are
equal, they are right angles. (1. def. 10.)
PROPOSITION XXXII. THEOREM.
If a straight line touches a circle, and from the point of contact a
straight line be drawn cutting the circle; the angles which this line makes
with the line touching the circle, shall be equal to the angles which are in
the alternate segments of the circle.
Let the straight line EF touch the circle ABCD in B,
and from the point B let the straight line BD be drawn, cutting
the circle.
Then the angles which BD makes with the touching line EF, shall
be equal to the angles in the alternate segments of the circle;
that is, the angle DBF shall be equal to the angle which is in the
segment DAB,
and the angle DBE shall be equal to the angle in the segment DCB.
A

D
C
E
B
F
A
From the point B draw BA at right angles to EF, (1. 11.)
and take any point C in the circumference DB, and join AD, DC, CB.
And because the straight line EF touches the circle ABCD in the
point B,
and BA is drawn at right angles to the touching line from the point
of contact B,
the centre of the circle is in BA: (111. 19.)
therefore the angle ADB in a semicircle is a right angle: (III. 31.)
and consequently the other two angles BAĎ, ABD, are equal to a
right angle; (1. 32.)
but ABF is likewise a right angle; (constr.)
therefore the angle ABF is equal to the angles BAD, ABD: (ax. 1.)
take from these equals the common angle ABD:
therefore the remaining angle DBF is equal to the angle BAD, (ax. 3.)
which is in the alternate segment of the circle.
And because ABCD is a quadrilateral figure in a circle,
the opposite angles BAD, BCD are equal to two right angles: (III. 22.)
BAU in all we now
ZAD m
LDIZ
C
7
BOOK HI. PROP. XXXII, XXXIII.
101
but the angles DBF, DBE are likewise equal to two right angles; (1. 13.)
therefore the angles DBF, DBE are equal to the angles BAD,
BCD, (ax. 1.)
and DBF has been proved equal to BAD;
therefore the remaining angle DBE is equal to the angle BCD in
the alternate segment of the circle. (ax. 2.)
Wherefore, if a straight line, &c. Q. E.D.
PROPOSITION XXXIII. PROBLEM.
Upon a given straight line to describe a segment of a circle, which shall
contain an angle equal to a given rectilineal angle.
Let AB be the given straight line,
and the angle at C the given rectilineal angle.
It is required to describe upon the given straight line AB, a segment
of a circle, which shall contain an angle equal to the angle C.
First, let the angle at C be a right angle.
H
C
A F
B
Bisect AB in F, (1.10.)
and from the centre F, at the distance FB, describe the semicircle AHB.
Therefore the angle AHB in a semicircle is equal to the right angle
at C. (III. 31.)
But, if the angle C be not a right angle.
H

E
H
C

F
B
B
A
F
G
E
D
D
At the point A, in a straight line AB, make the angle BAD equal
to the angle C, (1. 23.)
and from the point A draw AE at right angles to AD; (1. 11.)
bisect AB in F, (1. 10.)
and from F draw FG at right angles to AB, (1. 11.) and join GB.
And because AF is equal to FB, and FG common to the triangles
AFG, BFG,
the two sides AF, FG are equal to the two BF, FG, each to each:
and the angle AFG is equal to the angle BFG; (1. def. 10.)
therefore the base AG is equal to the base GB; (1.4.)
and therefore the circle described from the centre G, at the dis-
tance GA, shall pass through the point B:
let this be the circle AHB.
The segment AHB shall contain an angle equal to the given rectili-
neal angle C.
Because from the point A, the extremity of the diameter AE, AD is
drawn at right angles to AE,
therefore AD touches the circle: (111. 16. Cor.)
102
EUCLID'S ELEMENTS.
1
and because AB, drawn from the point of contact A, cuts the circle,
the angle DAB is equal to the angle in the alternate segment
AHB: (III. 32.)
but the angle DAB is equal to the angle C; (constr.)
therefore the angle C is equal to the angle in the segment AHB. (ax. 1.)
Wherefore, upon the given straight line AB, the segment AHB of a
circle is described, which contains an angle equal to the given angle
at C. Q.E.F.
PROPOSITION XXXIV. PROBLEM.
From a given circle to cut off a segment, which shall contain an angle
equal to a given rectilineal angle.
Let ABC be the given circle, and D the given rectilineal angle.
It is required to cut off from the circle ABC a segment that shall
contain an angle equal to the given angle D.
DL
E B
F
Draw the straight line EF touching the circle ABC in the point B,
(III. 17.)
and at the point B, in the straight line BF,
make the angle FBC equal to the angle D. (1.23.)
Then the segment BAC shall contain an angle equal to the given
angle D.
Because the straight line EF touches the circle ABC,
and BC is drawn from the point of contact B,
therefore the angle FBC is equal to the angle in the alternate seg-
ment BAC of the circle: (111. 32.)
but the angle FBC is equal to the angle D; (constr.)
therefore the angle in the segment BAC is equal to the angle D. (ax. 1.)
Wherefore from the given circle ABC, the segment BAC is cut off,
containing an angle equal to the given angle D. Q. E.F.
J
PROPOSITION XXXV. THEOREM.
If two straight lines cut one another within a circle, the rectangle con-
tained by the segments of one of them, is equal to the rectangle contained
by the segments of the other.
Let the two straight lines AC, BD, cut one another in the point E,
within the circle ABCD.
Then the rectangle contained by AE, EC shall be equal to the rect-
angle contained by BE, ED.
A
D
E
B
C
BOOK III. PROP. XXXV.
103
centre.
If AC, BD pass each of them through the centre, so that E is the
It is evident that since AE, EC, BE, ED, being all equal, (1. def. 15.)
therefore the rectangle AE, EC, is equal to the rectangle BE, ED.
But let one of them BD pass through the centre, and cut the other
AC, which does not pass through the centre, at right angles, in the
point E.
D

F
C
A
E
B
Then, if BD be bisected in F,
F is the centre of the circle ABCD:
join AF.
And because BD which passes through the centre, cuts the straight
line AC, which does not pass through the centre, at right angles in E,
therefore AE is equal to EC: (III. 3.)
and because the straight line BD is cut into two equal parts in the
point F, and into two unequal parts in the point E,
therefore the rectangle BE, ED, together with the square of EF,
is equal to the square of FB; (11. 5.)
that is, to the square of FA:
but the squares of AE, EF, are equal to the square of FA; (1. 47.)
therefore the rectangle BE, ED, together with the square of EF,
is equal to the squares of AE, EF: (ax. 1.)
take away the common square of EF,
and the remaining rectangle BE, ED is equal to the remaining
square of AE; (ax. 3.)
that is, to the rectangle AE, EC.
Next, let BD, which passes through the centre, cut the other AC,
which does not pass through the centre, in E, but not at right angles.
DA
A
E
C
B
Then, as before, if BD be bisected in F,
F is the centre of the circle.
Join AF, and from F draw FG perpendicular to AC; (1. 12.)
therefore AG is equal to GC; (III. 3.)
wherefore the rectangle AE, EC, together with the square of EG,
is equal to the square of AG: (II. 5.)
to each of these equals add the square of GF;
therefore the rectangle AÊ, EC, together with the squares of EG,
GF, is equal to the squares of AG, GF; (ax. 2.)
but the squares of EG, GF, are equal to the square of EF; (1. 47.)
and the squares of AG, GF are equal to the square of AF:
therefore the rectangle AE, EC, together with the square of EF,
is equal to the square of AF;

that is, to the square of FB:
but the square of FB is equal to the rectangle BE, ED, together
with the square of EF;~ (11. 5.)
104
EUCLID'S ELEMENTS.
r
therefore the rectangle AE, EC, together with the square of
EF, is equal to the rectangle BE, ED, together with the square of
EF; (ax. 1.)
take away the common square of EF,
and the remaining rectangle AE, EC, is therefore equal to the
remaining rectangle BE, ED. (ax. 3.)
Lastly, let neither of the straight lines AC, BD pass through the centre.
H
A
E
G
B
Take the centre F, (111. 1.)
and through E the intersection of the straight lines AC, DB,
draw the diameter GEFÄ.
And because the rectangle AE, EC, is equal as has been shown to
the rectangle GE, EH;
and for the same reason, the rectangle BE, ED is equal to the
same rectangle GE, EH;
therefore the rectangle AE, EC is equal to the rectangle BE, ED. (ax. 1.)
Wherefore if two straight lines, &c.
Q. E.D.
PROPOSITION XXXVI. THEOREM.
If from any point without a circle two straight lines be drawn, one of
which cuts the circle, and the other touches it; the rectangle contained by
the whole line which cuts the circle, and the part of it without the circle,
shall be equal to the square of the line which touches it.
Let D be any point without the circle ABC,
and let DCA, DB be two straight lines drawn from it,
of which DCA cuts the circle, and DB touches the same.
Then the rectangle AD, DC shall be equal to the square of DB.
Either DCA passes through the centre, or it does not:
first, let it pass through the centre E.
D
A
Join EB,
therefore the angle EBD is a right angle. (III. 18.)
And because the straight line AC is bisected in E, and produced to
the point D,
therefore the rectangle AD, DC, together with the square of EC,
is equal to the square of ED: (11. 6.)
but CE is equal to EB;
therefore the rectangle AD, DC, together with the square of EB,
is equal to the square of ED :
but the square of ED is equal to the squares of EB, BD, (1. 47.)
ग


BOOK III. PROP. XXXVI.
105
1
because EBD is a right angle:
therefore the rectangle AD, DC, together with the square of EB,
is equal to the squares of EB; BD: (ax. 1.)
take away the common square of EB;
therefore the remaining rectangle AD, DC is equal to the square
of the tangent, DB. (ax. 3.)
But if DCA does not pass through the centre of the circle ABC.
Take E the centre of the circle, (III. 1.)
D

B
pa
4
E
A
draw EF perpendicular to AC, (1. 12.) and join EB, EC, ED.
And because the straight line EF, which passes through the centre,
cuts the straight line AC, which does not pass through the centre, at
right angles.
Therefore EF also bisects AC; (III. 3.)
therefore AF is equal to FC:
and because the straight line AC is bisected in F, and produced to D,
the rectangle AD, DC, together with the square of FC is equal to
the square of FD: (11.6.)
to each of these equals add the square of FE;
therefore the rectangle AD, DC, together with the squares of CF,
FE is equal to the squares of DF, FE: (ax. 2.)
but the square of ED is equal to the squares of DF, FE, (1.47.)
because EFD is a right angle;
and for the same reason,
the square of EC is equal to the squares of CF, FE;
therefore the rectangle AD, DC, together with the square of EC,
is equal to the square of ED: (ax. 1.)
but CE is equal to EB;
therefore the rectangle AD, DC, together with the square of EB,
is equal to the square of ED:
but the squares of EB, BD are equal to the square of ED, (1.47.)
because EBD is a right angle:
therefore the rectangle AD, DC, together with the square of EB,
is equal to the squares of EB, BD;
take away the common square of EB ;
therefore the remaining rectangle AD, DC is equal to the square
of DB. (ax. 3.)
Wherefore, if from any point, &c. Q. E.D.
COR. If from any point without a circle, there be drawn two
A

D
C
B
:
106
EUCLID'S ELEMENTS.
straight lines cutting it, as AB, AC, the rectangles contained by the
whole lines and the parts of them without the circle, are equal to one
another, viz. the rectangle BA, AE, to the rectangle CA, AF: for each
of them is equal to the square of the straight line AD, which touches
the circle.
PROPOSITION XXXVII. THEOREM.
If from a point without a circle there be drawn two straight lines, one
of which cuts the circle, and the other meets it; if the rectangle contained
by the whole line which cuts the circle, and the part of it without the circle,
be equal to the square of the line which meets it, the line which meets
shall touch the circle.
Let any point D be taken without the circle ABC,
and from it let two straight lines DCA and DB be drawn, of which
DCA cuts the circle, and DB meets it.
If the rectangle AD,DC be equal to the square of DB; then DB
shall touch the circle.
D

B
E
F
A
Draw the straight line DE, touching the circle ABC, in the point
B; (III. 17.)
find F, the centre of the circle, (III. 1.)
and join FE, FB, FD.
Then FED is a right angle: (III. 18.)
and because DE touches the circle ABC, and DCA cuts it,
therefore the rectangle AD, DC is equal to the square of DE: (111. 36.)
but the rectangle AD, DC is, by hypothesis, equal to the square of DB;
therefore the square of DE is equal to the square of DB; (ax. 1.)
and the straight line DE equal to the straight line DB:
and FE is equal to FB; (1. def. 15.)
wherefore DE, EF are equal to DB, BF, each to each;
and the base FD is common to the two triangles DEF, DBF;
therefore the angle DEF is equal to the angle DBF: (1. 8.)
but DEF was shewn to be a right angle ;
therefore also DBF is a right angle: (ax. 1.)
and BF, if produced, is a diameter;
and the straight line which is drawn at right angles to a diameter,
from the extremity of it, touches the circle; (111. 16. Cor.)
therefore DB touches the circle ABC.
Wherefore, if from a point, &c.
Q.E.D.
NOTES TO BOOK III.
:
IN the third Book of the Elements are demonstrated the properties of the circle,
assuming all the properties of figures demonstrated in the first and second books.
A new conception is introduced in the third book, namely, that of similarity, and '
applied to the proof of properties connected with similar segments of circles.
It may be worthy of remark, that the word circle will be found sometimes taken
to mean the surface included within the circumference, and sometimes the circumference
itself. A circle is said to be given in position when the position of its centre is known,
and in magnitude when its radius is known.
Def. I states the criterion of equal circles. Simson calls it a theorem; and Euclid
seems to have considered it as one of those theorems, or definitions involving an axiom,
which might be admitted as a basis for reasoning on the equality of circles.
Def. VI, X.
An arc of a circle is any portion of the circumference; and a chord
is the straight line joining the extremities of an arc. Every chord except a diameter
divides a circle into two unequal segments, one greater than, and the other less than
a semicircle. And in the same manner, two radii drawn from the centre to the cir-
cumference, divide the circle into two unequal sectors, which become equal when the
two radii are in the same straight line. A quadrant is a sector whose radii are per-
pendicular to one another, and contains a fourth part of the circle.
Prop. I. If the point G be in the diameter CE, but not coinciding with the point
F, the demonstration given in the text does not hold good. At the same time, it is
obvious that G cannot be the centre of the circle, because GC is not equal to GE.
Prop. II. In this proposition, the circumference of a circle is proved to be essen-
tially different from a straight line, by shewing that every straight line joining any
two points in the arc falls entirely within the circle, and can neither coincide with
any part of the circumference, nor meet it except in the two assumed points. From
which follows the corollary, that, "a straight line cannot cut the circumference of a
circle in more points than two."
Commandine's direct demonstration of Prop. II depends on the following axiom,
"If a point be taken nearer to the centre of a circle than the circumference, that point
falls within the circle."
Take any point E in AB, and join DA, DE, DB.
Then because DA is equal to DB in the triangle DAB;
therefore the angle DAB is equal to the angle DEB; (v. 1.) D BA….
but since the side AE of the triangle DAE is produced to D, E
therefore the exterior angle DEB is greater than the interior and opposite angle DAE;
but the angle DAE is equal to the angle DBE,
therefore the angle DEB is greater than the angle DBE.
And in every triangle the greater side is subtended by the greater angle;
therefore the side DB is greater than the side DE;
but DB from the centre meets the circumference of the circle,
therefore DE does not meet it.
Wherefore the point E falls within the circle.
And E is any point in the straight line AB,
therefore the straight line AB falls within the circle.
108
EUCLID'S ELEMENTS.
!
1
Prop. VII and Prop. VIII exhibit the same property; in the former, the point is
taken in the diameter, and in the latter, in the diameter produced.
Prop. xI and Prop. XII. In the enunciations it is not asserted that the contact of
two circles is confined to a single point. The meaning appears to be, that supposing
two circles touch each other in any point, the straight line which joins their centres
being produced shall pass through that point in which the circles touch each other.
In Prop. XIII, it is proved that a circle cannot touch another in more points than
one, by assuming two points of contact, and proving that this is impossible.
Prop. xv and xvi. The converse of these propositions are not proved by Euclid.
Prop. XVI may be demonstrated directly by assuming the following axiom; "If
a point be taken further from the centre of a circle than the circumference, that point
falls without the circle."
Prop. XVII. When the given point is without the circumference of the given
circle, it is obvious that two equal tangents may be drawn from the given point to
touch the circle, as may be seen from the diagram to Prop. VIII.
Circles are called concentric circles when they have the same centre.
Prop. XVIII appears to be nothing more than a corollary to Prop. xvI. Because
a tangent to any point of the circumference of a circle is a straight line at right angles
at the extremity of the diameter which meets the circumference in that point.
In Prop. xvi, AE is proved to be perpendicular to AB, and in Prop. XVIII,
AB is proved to be perpendicular to AE; which is the same thing.
Prop. xx. This proposition is proved by Euclid only in the case in which the
angle at the circumference is less than a right angle, and the demonstration is free
from objection. If, however, the angle at the circumference be a right angle, the angle
at the centre disappears by the two straight lines from the centre to the extremities
of the arc becoming one straight line. And, if the angle at the circumference be an
obtuse angle, the angle formed by the two lines from the centre, does not stand on
the same arc, but upon the arc which the assumed arc wants of the whole circumference.
If Euclid's definition of an angle be strictly observed, Prop. xx is geometrically
true only when the angle at the centre is less than a right angle. If, however, the
defect of an angle from four right angles may be regarded as an angle, the proposition
is universally true, as may be proved by drawing a line from the angle in the cir-
cumference through the centre, and thus forming two angles at the centre in Euclid's
strict sense of the term.
In the first case, it is assumed that, if there be four magnitudes, such that the
first is double of the second, and the third double of the fourth, then the first and
third together shall be double of the second and fourth together: also in the second
case, that if one magnitude be double of another, and a part taken from the first be
double of a part taken from the second, the remainder of the first shall be double the
remainder of the second, which is, in fact, a particular case of Prop. 5, Book v.
Prop. xxI. Hence, the locus of the vertices of all triangles upon the same base,
and which have the same vertical angle, is a circular arc.
Prop. xxxi suggests a method of drawing a line at right angles to another when
the given point is at the extremity of the given line.
Prop. xxxv. It is possible to prove the most general case of this proposition, and
from it to deduce the other cases. The converse of Prop. xxxv is not proved by Euclid.
The properties of the circle demonstrated in the third book may be divided under
three heads. 1. Those which relate to the centre. 2. To similar segments. 3. To
the equal rectangles contained by the segments of the lines which intersect each other
within and without the circle.
BOOK IV.
DEFINITIONS.
I.
A RECTILINEAL figure is said to be inscribed in another rectilineal
figure, when all the angles of the inscribed figure are upon the sides of
the figure in which it is inscribed, each upon each.
:
II.
In like manner, a figure is said to be described about another figure,
when all the sides of the circumscribed figure pass through the angular
points of the figure about which it is described, each through each.
III.
A rectilineal figure is said to be inscribed in a circle, when all the
angles of the inscribed figure are upon the circumference of the circle.
IV.
A rectilineal figure is said to be described about a circle, when each
side of the circumscribed figure touches the circumference of the circle.
V.
In like manner, a circle is said to be inscribed in a rectilineal figure,
when the circumference of the circle touches each side of the figure.
VI.
A circle is said to be described about a rectilineal figure, when the
circumference of the circle passes through all the angular points of the
figure about which it is described.
VII.
A straight line is said to be placed in a circle, when the extremities
of it are in the circumference of the circle.
110
EUCLID'S ELEMENTS.
PROPOSITION I. PROBLEM.
In a given circle to place a straight line, equal to a given straight
line which is not greater than the diameter of the circle.
Let ABC be the given circle, and D the given straight line, not
greater than the diameter of the circle.
It is required to place in the circle ABC a straight line equal to D.

E
B
D
Draw BC the diameter of the circle ABC.
Then, if BC is equal to D, the thing required is done;
for in the circle ABĈ a straight line BC is placed equal to D.
But, if it is not, PC is greater than D; (hyp.)
make CE equal to D, (1. 3.)
and from the centre C, at the distance CE, describe the circle AEF,
and join CA.
Then CA shall be equal to D.
Because C is the centre of the circle AEF,
therefore CA is equal to CE: (1. def. 15.)
but CE is equal to D; (constr.)
therefore D is equal to CA. (ax. 1.)
Wherefore in the circle ABC, a straight line CA is placed equal to
the given straight line D, which is not greater than the diameter of the
circle. Q.E.F.
PROPOSITION II. PROBLEM.
In a given circle to inscribe a triangle equiangular to a given triangle.
Let ABC be the given circle, and DEF the given triangle.
It is required to inscribe in the circle ABC a triangle equiangular
to the triangle DEF.

G
H
D
E
F B
Draw the straight line GAH touching the circle in the point A, (III. 17.)
and at the point A, in the straight line AH,
make the angle HAC equal to the angle DEF; (1. 23.)
and at the point A, in the straight line AG,
make the angle GAB equal to the angle DFE;
and join BC: then ABC shall be the triangle required.
Because HAG touches the circle ABC,
and AC is drawn from the point of contact,

BOOK IV. PROP. II, III.
111
therefore the angle HAC is equal to the angle ABC in the alternate
segment of the circle: (111. 32.)
but HAC is equal to the angle DEF; (constr.)
therefore also the angle ABC is equal to DEF: (ax. 1.)
for the same reason, the angle ABC is equal to the angle DFE: CB.
therefore the remaining angle BAC is equal to the remaining angle
EDF: (1. 32 and ax. 1.)
wherefore the triangle ABC is equiangular to the triangle DEF,
and it is inscribed in the circle ABC. Q.E.F.
PROPOSITION III. PROBLEM.
About a given circle to describe a triangle equiangular to a given triangle.
Let ABC be the given circle, and DEF the given triangle.
It is required to describe a triangle about the circle ABC equi-
angular to the triangle DEF.
L

D
A
K
GE FH
M B
N
Produce EF both ways to the points G, H;
find the centre K of the circle ABC, (111. 1.)
and from it draw any straight line KB;
at the point K in the straight line KB,
make the angle BKA equal to the angle DEG, (1. 23.)
and the angle BKC equal to the angle DFH;
and through the points A, B, C, draw the straight lines LAM,
MBN, NCL, touching the circle ABC. (111. 17.)
Then LMN shall be the triangle required.
Because LM, MN, NL touch the circle ABC in the points A, B, C,
to which from the centre are drawn KA, KB, KC,
therefore the angles at the points A, B, C are right angles: (III. 18.)
and because the four angles of the quadrilateral figure AMBK are
equal to four right angles,
for it can be divided into two triangles;
and that two of them KAM, KBM are right angles,
therefore the other two AKB, AMB are equal to two right angles:
(ax. 3.)
but the angles DEG, DEF are likewise equal to two right angles;
(1. 13.)
therefore the angles AKB, AMB are equal to the angles DEG,
DEF; (ax. 1.)
of which AKB is equal to DEG; (constr.)
wherefore the remaining angle AMB is equal to the remaining
angle DEF. (ax. 3.)
In like manner, the angle LNM may be demonstrated to be equal
to DFE;
112
EUCLID'S ELEMENTS. .
and therefore the remaining angle MLN is equal to the remaining
angle EDF: (1. 32 and ax. 3.)
therefore the triangle LMN is equiangular to the triangle DEF:
and it is described about the circle ABC.
Q.E.F.
PROPOSITION IV. PROBLEM.
To inscribe a circle in a given triangle.
Let the given triangle be ABC.
It is required to inscribe a circle in ABC.
A

G
E
D
B F
C
Bisect the angles ABC, BCA by the straight lines BD, CD meeting
one another in the point D, (1.9.)
from which draw DE, DF, DG perpendiculars to AB, BC, CA. (1.12.)
And because the angle EBD is equal to the angle FBD,
for the angle ABC is bisected by BD,
and that the right angle BED is equal to the right angle BFD; (ax. 11.)
therefore the two triangles EBD, FBD have two angles of the one
equal to two angles of the other, each to each
and the side BD, which is opposite to one of the equal angles in
each, is common to both ;
therefore their other sides are equal; (1.26.)
wherefore DE is equal to DF:
for the same reason, DG is equal to DF:
therefore DE is equal to DG: (ax. 1.)
therefore the three straight lines DE, DF, DG are equal to one another;
and the circle described from the center D, at the distance of any of
them, will pass through the extremities of the other two, and touch the
straight lines AB, BC, CA,
because the angles at the points E, F, G are right angles,
(III.
and the straight line which is drawn from the extremity of a dia-
meter at right angles to it, touches the circle: (111. 16.)
therefore the straight lines AB, BC, CA do each of them touch the circle,
and therefore the circle EFG is inscribed in the triangle ABC.
Q.E.F.
PROPOSITION V. PROBLEM.
To describe a circle about a given triangle.
Let the given triangle be ABC.
It is required to describe a circle about ABC.
BOOK IV. PROP. V, VI.
113
A
12
包
​B
B
C
A
E
A
B
C
C
F
Bisect AB, AC in the points D, E, (1. 10.)
and from these points draw DF, EF at right angles to AB, AC; (1. 11.)
DF, EF produced meet one another
for, if they do not meet, they are parallel,
wherefore AB, AC, which are at right angles to them, are parallel;
which is absurd :
let them meet in F, and join FA;
also, if the point F be not in BC, join BF, CF.
Then, because AD is equal to DB, and DF common, and at right
angles to AB,
therefore the base AF is equal to the base FB. (1. 4.
In like manner, it may be shewn that CF is equal to FA;
and therefore BF is equal to FC; (ax. 1.)
and FA, FB, FC are equal to one another:
wherefore the circle described from the centre F, at the distance
of one of them, will pass through the extremities of the other two, and
be described about the triangle ABC. Q.E. F.
COR. And it is manifest, that when the centre of the circle falls
within the triangle, each of its angles is less than a right angle, (111. 31.)
each of them being in a segment greater than a semicircle; but, when
the centre is in one of the sides of the triangle, the angle opposite to
this side, being in a semicircle, (III. 31.) is a right angle; and, if the
centre falls without the triangle, the angle opposite to the side beyond
which it is, being in a segment less than a semicircle, (111. 31.) is greater
than a right angle: therefore, conversely, if the given triangle be acute-
angled, the centre of the circle falls within it; if it be a right-angled
triangle, the centre is in the side opposite to the right angle; and if it
be an obtuse-angled triangle, the centre falls without the triangle, be-
yond the side opposite to the obtuse angle.
PROPOSITION VI. PROBLEM.
To inscribe a square in a given circle.
Let ABCD be the given circle.
It is required to inscribe a square in ABCD.
A
B
D
C
Draw the diameters, AC, BD, at right angles to one another,
(III. 1 and I. 11.)
and join AB, BC, CD, DA.
The figure ABCD shall be the square required.
8
114
EUCLID'S ELEMENTS.
Because BE is equal to ED, for E is the centre, and that EA is
common, and at right angles to BD;
the base BA is equal to the base AD: (1. 4.)
and, for the same reason, BC, CD are each of them equal to BA, or AD;
therefore the quadrilateral figure ABCD is equilateral.
It is also rectangular;
for the straight line BD being the diameter of the circle ABCD,
BAD is a semicircle;
wherefore the angle BAD is a right angle: (III. 31.)
for the same reason, each of the angles ABC, BCD, CDA is a right
angle:
therefore the quadrilateral figure ABCD is rectangular:
and it has been shewn to be equilateral;
therefore it is a square; (1. def. 30.)
and it is inscribed in the circle ABCD.
Q.E. F.
PROPOSITION VII. PROBLEM.
To describe a square about a given circle.
Let ABCD be the given circle.
It is required to describe a square about it.
G A F

13
B
D
H C K
Draw two diameters AC, BD of the circle ABCD, at right angles
to one another,
and through the points A, B, C, D, draw FG, Gh, hk, kF
touching the circle. (III. 17.)
The figure GHKF shall be the square required.
Because FG touches the circle ABCD, and EA is drawn from the
centre E to the point of contact A,
therefore the angles at A are right angles: (III. 18.)
for the same reason, the angles at the points B, C, D are right angles:
and because the angle AEB is a right angle, as likewise is EBG,
therefore GH is parallel to AC: (1. 28.)
for the same reason AC is parallel to FK:
and in like manner GF, HK may each of them be demonstrated
to be parallel to BED:
therefore the figures GK, GC, AK, FB, BK are parallelograms;
and therefore GF is equal to HK, and GH to FK: (1. 34.)
and because AC is equal to BD, and that AC is equal to each of
the two GH, FK;
and BD to each of the two GF, HK;
GH, FK are each of them equal to GF, or HK:
therefore the quadrilateral figure FGHK is equilateral.
It is also rectangular;
for GBEA being a parallelogram, and AEB a right angle,
therefore AGB is likewise a right angle: (1. 34.)
BOOK IV. PROP. VII, VIII, IX.
115
and in the same manner it may be shewn that the angles at H, K,
F are right angles:
therefore the quadrilateral figure FGHK is rectangular:
and it was demonstrated to be equilateral;
therefore it is a square; (1. def. 30.)
and it is described about the circle ABCD. Q.E.F.
PROPOSITION VIII. PROBLEM. Į
To inscribe a circle in a given square.
Let ABCD be the given square.
It is required to inscribe a circle in ABCD.
F
A E
D
K
B H
Bisect each of the sides AB, AD in the points F, E, (1. 10.)
and through E draw EH parallel to AB or DC, (1. 31.)
and through F draw FK parallel to AD or BC:
therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD
is a right-angled parallelogram;
and their opposite sides are equal: (1. 34.)
and because AD is equal to AB, (1. def. 30.) and that AE is the
half of AD, and AF the half of AB,
therefore AE is equal to AF; (ax. 7.)
wherefore the sides opposite to these are equal, viz. FG to GE:
in the same manner it may be demonstrated that GH, GK are each
of them equal to FG or GE:
therefore the four straight lines GE, GF, GH, GK are equal to
one another;
and the circle described from the centre G at the distance of one of
them, will pass through the extremities of the other three, and touch
the straight lines AB, BC, CD, DA ;
because the angles at the points E, F, H, K, are right angles, (1. 29.)
and that the straight line which is drawn from the extremity of a
diameter, at right angles to it, touches the circle: (III. 16. Cor.)
therefore each of the straight lines AB, BC, CD, DA touches the circle,
which therefore is inscribed in the square ABCD. Q.E.F.
PROPOSITION IX. PROBLEM.
To describe a circle about a given square.
Let ABCD be the given square.
It is required to describe a circle about ABCD.
A
E
D
B
1
8--2
116
EUCLID'S ELEMENTS.
Join AC, BD, cutting one another in E:
and because DA is equal to AB, and AC common to the triangles
DAC, BAC, (1. def. 30.)
the two sides DA, AC are equal to the two BA, AC, each to each ;
and the base DĈ is equal to the base BC;
wherefore the angle DAC is equal to the angle BAC, (1.8.)
and the angle DAB is bisected by the straight line AC:
in the same manner it may be demonstrated that the angles ABC,
BCD, CDA are severally bisected by the straight lines BD, AC:
therefore, because the angle DAB is equal to the angle ABC, (1. def. 30.)
and that the angle EAB is the half of DAB, and EBA the half of ABC;
therefore the angle EAB is equal to the angle EBA; (ax. 7.)
wherefore the side EA is equal to the side EB: (1. 6.)
in the same manner it may be demonstrated, that the straight lines
EC, ED are each of them equal to EA or EB:
therefore the four straight lines EA, EB, EC, ED are equal to
one another;
and the circle described from the centre E, at the distance of one of
them, will pass through the extremities of the other three, and be
described about the square ABCD. Q.E.F.
PROPOSITION X. PROBLEM.
To describe an isosceles triangle, having each of the angles at the base
double of the third angle.
Take any straight line AB, and divide it in the point C, (II. 11.)
so that the rectangle AB, BC may be equal to the square of CA;
and from the centre A, at the distance AB, describe the circle BDE,
in which place the straight line BD equal to AC, which is not
greater than the diameter of the circle BDE; (IV. 1.)
and join DA.
Then the triangle ABD shall be such as is required,
that is, each of the angles ABD, ADB shall be double of the angle BAD.
E

B D
Join DC, and about the triangle ADC describe the circle ACD. (Iv. 5.)
And because the rectangle AB, BC is equal to the square of AC,
and that AC is equal to BD, (constr.)
the rectangle AB, BC is equal to the square of BD: (ax. 1.)
and because from the point B, without the circle ACD, two straight
lines BCA, BD are drawn to the circumference, one of which cuts, and
the other meets the circle, and that the rectangle AB, BC, contained by
the whole of the cutting line, and the part of it without the circle, is
equal to the square of BD which meets it;
therefore the straight line BD touches the circle ACD: (111. 87.)
BOOK IV. PROP. X, XI.
117
and because BD touches the circle, and DC is drawn from the
point of contact D,
the angle BDC is equal to the angle DAC in the alternate segment
of the circle: (III. 32.)
to each of these add the angle CDA;
therefore the whole angle BDA is equal to the two angles CDA,
DAC: (ax. 2.)
but the exterior angle BCD is equal to the angles CDA, DAC; (1. 32.)
therefore also BDA is equal to BCD: (ax. 1.)
but BDA is equal to the angle CBD, (1. 5.)
because the side AD is equal to the side AB;
therefore CBD, or DBA, is equal to BCD; (ax. 1.)
and consequently the three angles BDA, DBA, BCD are equal to
one another:
and because the angle DBC is equal to the angle BCD,
the side BD is equal to the side DC: (1. 6.)
but BD was made equal to CA;
therefore also CA is equal to CD,_(ax. 1.)
and the angle CDA equal to the angle DAC; (1. 5.)
therefore the angles CDA, DAC together, are double of the angle DAC:
but BCD is equal to the angles CDA, DAC; (1. 32.)
therefore also BCD is double of DAC:
and BCD was proved to be equal to each of the angles BDA, DBA ;
therefore each of the angles BDA, DBA is double of the angle DAB.
Wherefore an isosceles triangle ABD is described, having each of
the angles at the base double of the third angle. Q.E.F.
PROPOSITION XI. PROBLEM.
To inscribe an equilateral and equiangular pentagon in a given circle.
Let ABCDE be the given circle.
It is required to inscribe an equilateral and equiangular pentagon
in the circle ABCDE.
F
B
A
E
G H
C
Describe an isosceles triangle FGH, having each of the angles at G,
H double of the angle at F; (IV. 10.)
and in the circle ABCDE inscribe the triangle ACD equiangular
to the triangle FGH, (iv. 2.)
so that the angle CAD may be equal to the angle at F,
and each of the angles ACD, CDA equal to the angle at G or H;
wherefore each of the angles ACD, CDA is double of the angle CAD.
Bisect the angles ACD, CDA by the straight lines CE, DB; (1.9.)
and join AB, BC, DE, EA.
Then ABCDE shall be the pentagon required.
Because each of the angles ACD, CDA is double of CAD,
and that they are bisected by the straight lines CE, DB;
118
EUCLID'S ELEMENTS.
•
therefore the five angles DAC, ACE, ECD, CDB, BDA are equal
to one another:
but equal angles stand upon equal circumferences; (111. 26.)
therefore the five circumferences AB, BC, CD, DE, EA are equal
to one another:
and equal circumferences are subtended by equal straight lines; (111. 29.)
therefore the five straight lines AB, BC, CD, DE, EA are equal to
one another.
Wherefore the pentagon ABCDE is equilateral.
It is also equiangular:
for, because the circumference AB is equal to the circumference DE,
if to each be added BCD,
the whole ABCD is equal to the whole EDCB: (ax. 2.)
but the angle AED stands on the circumference ABCD;
and the angle BAE on the circumference EDCB;
therefore the angle BAE is equal to the angle AED: (111. 27.)
for the same reason, each of the angles ABC, BCD, CDE is equal
to the angle BAE, or AED:
therefore the pentagon ABCDE is equiangular;
and it has been shewn that it is equilateral:
wherefore, in the given circle, an equilateral and equiangular
pentagon has been described. Q.E.F.
PROPOSITION XII. PROBLEM.
To describe an equilateral and equiangular pentagon about a given circle.
Let ABCDE be the given circle.
It is required to describe an equilateral and equiangular pentagon
about the circle ABCDE.
G

A
E
H
M
F
B
D
L
Let the angles of a pentagon, inscribed in the circle, by the last
proposition, be in the points A, B, C, D, E,
so that the circumferences AB, BC, CD, DE, EA are equal; (Iv. 11.)
and through the points A, B, C, D, E draw GH, HK, KL, LM,
MG touching the circle; (III. 17.)
the figure GHKLM shall be the pentagon required.
Take the centre F, and join FB, FK, FC, FL, FD.
And because the straight line KL touches the circle ABCDE in the
point C, to which FC is drawn from the centre F,
FC is perpendicular to KL, (III. 18.)
therefore each of the angles at C is a right angle:
for the same reason, the angles at the points B, D are right angles:
and because FCK is a right angle,
the square of FK is equal to the squares of FC, CK: (1. 47.)
BOOK IV. PROP. XII.
119
for the same reason, the square of FK is equal to the squares of
FB, BK:
therefore the squares of FC, CK are equal to the squares of FB,
BK; (ax. 1.)
of which the square of FC is equal to the square of FB;
therefore the remaining square of CK is equal to the remaining
square of BK, (ax. 3.)
and the straight line CK equal to BK:
and because FB is equal to FC, and FK common to the triangles
BFK, CFK,
the two BF, FK are equal to the two CF, FK, each to each;
and the base BK was proved equal to the base KC;
therefore the angle BFK is equal to the angle KFC, (1. 8.)
and the angle BKF to FKC: (1.4.)
wherefore the angle BFC is double of the angle KFC,
and BKC double of FKC:
for the same reason, the angle CFD is double of the angle CFL,
and CLD double of CLF:
and because the circumference BC is equal to the circumference CD,
the angle BFC is equal to the angle CFD; (111. 27.)
and BFC is double of the angle KFC,
and CFD double of CFL;
therefore the angle KFC is equal to the angle CFL: (ax. 7.)
and the right angle FCK is equal to the right angle FCL;
therefore, in the two triangles FKC, FLC, there are two angles of
the one equal to two angles of the other, each to each;
and the side FC, which is adjacent to the equal angles in each, is
common to both;
therefore the other sides are equal to the other sides, and the third
angle to the third angle: (1. 26.)
therefore the straight line KC is equal to CL, and the angle FKC
to the angle FLC:
and because KC is equal to CL,
KL is double of KC.
In the same manner it may be shewn that HK is double of BK:
and because BK is equal to KC, as was demonstrated,
and that KL is double of KC, and HK double of BK,
therefore HK is equal to KL: (ax. 6.)
in like manner it may be shewn that GH, GM, ML are each of
them equal to HK, or KL:
therefore the pentagon GHKLM is equilateral.
It is also equiangular:
for, since the angle FKC is equal to the angle FLC,
and that the angle HKL is double of the angle FKC,
and KLM double of FLC, as was before demonstrated;
therefore the angle HKL is equal to KLM: (ax. 6.)
and in like manner it may be shewn,
that each of the angles KHG, HGM, GML is equal to the angle
HKL or KLM:
therefore the five angles GHK, HKL, KLM, LMG, MGH, being
equal to one another,
the pentagon GHKLM is equiangular:
and it is equilateral, as was demonstrated
and it is described about the circle ABCDE. Q.E.F.
120
EUCLID'S ELEMENTS.
PROPOSITION XIII. PROBLEM.
To inscribe a circle in a given equilateral and equiangular pentagon.
Let ABCDE be the given equilateral and equiangular pentagon.
It is required to inscribe a circle in the pentagon ABCDE.
B
G
A

M
E
H
C K
Bisect the angles BCD, CDE by the straight lines CF, DF, (1. 9.)
and from the point F, in which they meet, draw the straight lines
FB, FA, FE:
therefore since BC is equal to CD, (hyp.) and CF common to the
triangles BCF, DCF,
the two sides BC, CF are equal to the two DC, CF, each to each;
and the angle BCF is equal to the angle DCF; (constr.)
therefore the base BF is equal to the base FD, (1. 4.)
and the other angles to the other angles, to which the equal sides
are opposite;
therefore the angle CBF is equal to the angle CDF:
and because the angle CDE is double of CDF,
and that CDE is equal to CBA, and CDF to CBF;
CBA is also double of the angle CBF;
therefore the angle ABF is equal to the angle CBF;
wherefore the angle ABC is bisected by the straight line BF:
in the same manner it may be demonstrated,
that the angles BAE, AED, are bisected by the straight lines AF, FE.
From the point F, draw FG, FH, FK, FL, FM perpendiculars to
the straight lines AB, BC, CD, DE, EA: (1. 12.)
and because the angle HCF is equal to KCF, and the right angle
FHC equal to the right angle FKC;
therefore in the triangles FHC, FKC, there are two angles of the
one equal to two angles of the other, each to each;
and the side FC, which is opposite to one of the equal angles in
each, is common to both;
therefore the other sides are equal, each to each; (1. 26.)
wherefore the perpendicular FH is equal to the perpendicular FK:
in the same manner it may be demonstrated, that FL, FM, FG are
each of them equal to FH, or FK:
therefore the five straight lines FG, FH, FK, FL, FM are equal
to one another:
wherefore the circle described from the centre F, at the distance of
one of these five, will pass through the extremities of the other four, and
touch the straight lines AB, BC, CD, DE, EA,
because the angles at the points G, H, K, L, M are right angles,
and that a straight line drawn from the extremity of the diameter
of a circle at right angles to it, touches the circle: (III. 16.)
BOOK IV. PROP. XIII, XIV, xv.
121
therefore each of the straight lines AB, BC, CD, DE, EA touches
the circle:
wherefore it is inscribed in the pentagon ABCDE. Q.E.F.
PROPOSITION XIV. PROBLEM.
To describe a circle about a given equilateral and equiangular pentagon.
Let ABCDE be the given equilateral and equiangular pentagon.
It is required to describe a circle about ABCD.
B
F
A
E
C
D
Bisect the angles BCD, CDE by the straight lines CF, FD, (1. 9.)
and from the point F, in which they meet, draw the straight lines
FB, FA, FE, to the points B, A, E.
It may be demonstrated, in the same manner as in the preceding
proposition,
that the angles CBA, BAE, AED are bisected by the straight lines
FB, FA, FE.
And because the angle BCD is equal to the angle CDE,
and that FCD is the half of the angle BCD,
and CDF the half of CDE;
therefore the angle FCD is equal to FDC; (ax. 7.)
wherefore the side CF is equal to the side FD: (1. 6.)
in like manner it may be demonstrated,
that FB, FA, FE, are each of them equal to FC or FD:
therefore the five straight lines FA, FB, FC, FD, FE are equal to
one another;
and the circle described from the centre F, at the distance of one of
them, will pass through the extremities of the other four, and be described
about the equilateral and equiangular pentagon ABCDE. Q.E. F.
PROPOSITION XV. PROBLEM.
To inscribe an equilateral and equiangular hexagon in a given circle.
Let ABCDEF be the given circle.
It is required to inscribe an equilateral and equiangular hexagon in it.
F
A

B
G
C
E
H
Find the centre G of the circle ABCDEF, and draw the diameter
AGD; (III. 1.)
122
EUCLID'S ELEMENTS.
and from D, as a centre, at the distance DG, describe the circle
EGCH,
join EG, CG, and produce them to the points B, F;
and join AB, BC, CD, DE, EF, FA :
the hexagon ABCDEF shall be equilateral and equiangular.
Because G is the centre of the circle ABCDEF,
GE is equal to GD:
and because D is the centre of the circle EGCH,
DE is equal to DG:
wherefore GE is equal to ED, (ax. 1.)
and the triangle EGD is equilateral;
and therefore its three angles EGD, GDE, DEG, are equal to one
another: (1. 5. Cor.)
but the three angles of a triangle are equal to two right angles; (1. 32.)
therefore the angle EGD is the third part of two right angles:
in the same manner it may be demonstrated,
that the angle DGC is also the third part of two right angles:
and because the straight line GC makes with EB the adjacent
angles EGC, CGB equal to two right angles; (1. 13.)
the remaining angle CGB is the third part of two right angles:
therefore the angles EGD, DGC, CGB are equal to one another:
and to these are equal the vertical opposite angles BGA, AGF,
FGE: (1.15.)
therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE, are
equal to one another :
but equal angles stand upon equal circumferences; (111.26.)
therefore the six circumferences AB, BC, CD, DE, EF, FA are
equal to one another:
and equal circumferences are subtended by equal straight lines: (111. 29.)
therefore the six straight lines are equal to one another,
and the hexagon ABCDEF is equilateral.
It is also equiangular:
for, since the circumference AF is equal to ED,
to each of these equals add the circumference ABCD ;
therefore the whole circumference FABCD is equal to the whole
EDCBA:
and the angle FED stands upon the circumference FABCD,
and the angle AFE upon EDCBA;
therefore the angle AFE is equal to FED: (III. 27.)
in the same manner it may be demonstrated
that the other angles of the hexagon ABCDEF are each of them
equal to the angle AFE or FED:
therefore the hexagon is equiangular ;
and it is equilateral, as was shewn ;
and it is inscribed in the given circle ABCDEF.
Q. E. F.
COR.-From this it is manifest, that the side of the hexagon is equal
to the straight line from the centre, that is, to the semi-diameter of the
circle.
And if through the points A, B, C, D, E, F there be drawn straight
lines touching the circle, an equilateral and equiangular hexagon will be
described about it, which may be demonstrated from what has been said
of the pentagon: and likewise a circle may be inscribed in a given
equilateral and equiangular hexagon, and circumscribed about it, by a
method like to that used for the pentagon.
BOOK IV. PROP. XVI.
123
PROPOSITION XVI. PROBLEM.
To inscribe an equilateral and equiangular quindecagon in a given circle.
Let ABCD be the given circle.
It is required to inscribe an equilateral and equiangular quindecagon
in the circle ABCD.
A
B
F
E
C
D
Let AC' be the side of an equilateral triangle inscribed in the circle, (IV. 2.)
and AB the side of an equilateral and equiangular pentagon inscribed
in the same: (IV. 11.)
therefore, of such equal parts as the whole circumference ABCDF
contains fifteen,
the circumference ABC, being the third part of the whole, contains five ;
and the circumference AB, which is the fifth part of the whole,
contains three;
therefore BC, their difference, contains two of the same parts:
bisect BC in E; (III. 30.)
therefore BE, EC are, each of them, the fifteenth part of the whole
circumference ABCD:
therefore if the straight lines BE, EC be drawn, and straight lines
equal to them be placed round in the whole circle, (Iv. 1.) an equilateral
and equiangular quindecagon will be inscribed in it. Q. E. F.
And in the same manner as was done in the pentagon, if through the
points of division made by inscribing the quindecagon, straight lines be
drawn touching the circle, an equilateral and equiangular quindecagon
will be described about it: and likewise, as in the pentagon, a circle may
be inscribed in a given equilateral and equiangular quindecagon, and
circumscribed about it.
:
NOTES TO BOOK IV.
1
1
4
THE fourth book of the Elements contains some particular cases of four general
problems on the inscription, and the circumscription of triangles and regular figures
in and about circles. Euclid has not given any instances of the inscription or circum-
scription of rectilinear figures in and about other rectilinear figures.
Any rectilinear figure, of five sides and angles, is called a pentagon; of seven sides
and angles, a heptagon; of eight sides and angles, an octagon; of nine sides and
angles, a nonagon; of ten sides and angles, a decagon; of twelve sides and angles, a
duodecagon; of fifteen sides and angles, a quindecagon, &c.
These figures are included under the general name of polygons; and are called
equilateral, when their sides are equal; and equiangular, when their angles are equal;
also when both their sides and angles are equal, they are called regular polygons.
Prop. III. An objection has been raised to the construction of this problem. It
is said that in this and other instances of a similar kind, the lines which touch the
circle at A, B, and C, should be proved to meet one another. This may be done by
joining AB, and then since the angles KAM, KBM are equal to two right angles
(III. 18.), therefore the angles BAM, ABM are less than two right angles, and con-
sequently (ax. 12.), AM and BM must meet one another, when produced far enough.
Similarly, it may be shewn that AL and CL, as also CN and BN meet one another.
Prop. v. The corollary to this proposition appears to have been already demon-
strated in Prop. 31, Book III.
Prop. VI, VII. It is obvious that the square described about a circle is equal to
double the square inscribed in the same circle. Also that the circumscribed square is
equal to the square of the diameter, or four times the square of the radius of the circle.
Prop. VII. It is manifest that a square is the only right-angled parallelogram
which can be circumscribed about a circle, but that both a rectangle and a square
may be inscribed in a circle.
Prop. x. By means of this proposition, a right angle may be divided into five
equal parts.
Prop. XVI. The arc subtending a side of the quindecagon, may be found by
placing in the circle from the same point, two lines respectively equal to the sides of
the regular hexagon and pentagon.
The centres of the inscribed and circumscribed circles of any regular polygon are
coincident.
Besides the circumscription and inscription of triangles and regular polygons about
and in circles, some very important problems are solved in the constructions respecting
the division of the circumferences of circles into equal parts.
By inscribing an equilateral triangle, a square, a pentagon, a hexagon, &c. in a
circle, the circumference is divided into three, four, five, six, &c. equal parts. In
Prop. xxvi, Book III, it has been shewn that equal angles at the centres of equal
circles, and therefore at the centre of the same circle, subtend equal arcs; by bisecting
the angles at the centre, the arcs which are subtended by them are also bisected, and
hence, a sixth, eighth, tenth, twelfth, &c. part of the circumference of a circle may
be found.
NOTES TO BOOK IV.
125
By the aid of the first corollary to Prop. 32, Book 1, may be found the magnitude
of an interior angle of any regular polygon whatever.
Let e denote the magnitude of one of the interior angles of a regular polygon
of n sides,
then no is the sum of all the interior angles.
But all the interior angles of any rectilinear figure together with four right angles,
are equal to twice as many right angles as the figure has sides,
that is, if we agree to assume π to designate two right angles,
.. n0 + 2π = NT,
and _no = nπ
2π = (n − 2).π,
(n − 2)
0
=
n
the magnitude of an interior angle of a regular polygon of n sides.
By taking ʼn – 3, 4, 5, 6, &c. may be found the magnitude, in terms of two right
angles, of an interior angle of any regular polygon whatever.
Pythagoras was the first, as Proclus informs us in his commentary, who discovered
that a multiple of the angles of three regular figures only, namely, the trigon, the
square, and the hexagon, can fill up space round a point in a plane.
It has been shewn that the interior angle of any regular polygon of n sides in
terms of two right angles, is expressed by the equation
n 2
Ꮎ
Add
. 7.
N
Let 03 denote the magnitude of the interior angle of a regular figure of 3 sides.
Then 03
•
303
3 2 T
3
= π,
3
and 60₂ = 2π,
3
= one third of two right angles,
that is, six angles each equal to the interior angle of an equilateral triangle are equal
to four right angles, and therefore six equilateral triangles may be placed so as com-
pletely to fill up the space round the point at which they meet in a plane.
In a similar way, it may be shewn that four squares and three hexagons may be
placed so as completely to fill up the space round a point.
Also it will appear from the results deduced, that no other regular figures besides
these three, can be made to fill up the space round a point: for any multiple of the
interior angles of any other regular polygon, will be found to be in excess above, or in
defect from four right angles.
The equilateral triangle or trigon, the square or tetragon, the pentagon, and the
hexagon, were the only regular polygons known to the Greeks, capable of being inscribed
in circles, besides those which may be derived from them.
M. Gauss in his Disquisitiones Arithmeticæ, has extended the number by shewing
that in general, a regular polygon of 2″ + 1 sides is capable of being inscribed in a
circle by means of straight lines and circles, in those cases in which 2" + 1 is a
prime number.
✓
BOOK V.
DEFINITIONS.
I.
'
A LESS magnitude is said to be a part of a greater magnitude when
the less measures the greater; that is, when the less is contained a
certain number of times exactly in the greater.'
II.
A greater magnitude is said to be a multiple of a less, when the
greater is measured by the less, that is, 'when the greater contains the
less a certain number of times exactly.'
III.
"Ratio is a mutual relation of two magnitudes of the same kind to
one another, in respect of quantity."
IV.
Magnitudes are said to have a ratio to one another, when the less
can be multiplied so as to exceed the other.
V.
The first of four magnitudes is said to have the same ratio to the
second, which the third has to the fourth, when any equimultiples
whatsoever of the first and third being taken, and any equimultiples
whatsoever of the second and fourth; if the multiple of the first be
less than that of the second, the multiple of the third is also less than
that of the fourth: or, if the multiple of the first be equal to that of
the second, the multiple of the third is also equal to that of the fourth:
or, if the multiple of the first be greater than that of the second, the
multiple of the third is also greater than that of the fourth.
VI.
Magnitudes which have the same ratio are called proportionals.
'N.B. When four magnitudes are proportionals, it is usually expressed by
saying, the first is to the second, as the third to the fourth.'
VII.
When of the equimultiples 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 the multiple of the
fourth; then the first is said to have to the second a greater ratio than
the third magnitude has to the fourth: and, on the contrary, the third is
said to have to the fourth a less ratio than the first has to the second.
VIII.
Analogy, or proportion, is the similitude of ratios."
IX.
Proportion consists in three terms at least.
DEFINITIONS..
127
1
X.
When three magnitudes are proportionals, the first is said to have to
the third the duplicate ratio of that which it has to the second.
XI.
When four magnitudes are continual proportionals, the first is said
to have to the fourth the triplicate ratio of that which it has to the
second, and so on, quadruplicate, &c. increasing the denomination still
by unity, in any number of proportionals.
Definition A, to wit, of compound ratio.
When there are any number of magnitudes of the same kind, the
first is said to have to the last of them the ratio compounded of the ratio
which the first has to the second, and of the ratio which the second has
to the third, and of the ratio which the third has to the fourth, and so
on unto the last magnitude.
For example, if A, B, C, D be four magnitudes of the same kind, the first A is said
to have to the last D the ratio compounded of the ratio of A to B, and of the ratio of
B to C, and of the ratio of C to D; or, the ratio of A to D is said to be compounded of
the ratios of A to B, B to C, and C to D.
And if A has to B the same ratio which E has to F; and B to C the same ratio that
G has to H; and C to D the same that K has to L; then, by this definition, A is said
to have to D the ratio compounded of ratios which are the same with the ratios of
E to F, G to H, and K to L. And the same thing is to be understood when it is
more briefly expressed by saying, A has to D the ratio compounded of the ratios of
E to F, G to H, and K to L.
In like manner, the same things being supposed, if M has to N the same ratio which
A has to D; then, for shortness sake, M is said to have to N the ratio compounded of
the ratios of E to F, G to H, and K to L.
XII.
In proportionals, the antecedent terms are called homologous to one
another, as also the consequents to one another.
"Geometers make use of the following technical words, to signify
certain ways of changing either the order or magnitude of proportionals,
so that they continue still to be proportionals.'
XIII.
Permutando, or alternando, by permutation or alternately. This
word is used when there are four proportionals, 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 Prop. XVI. of this fifth book.
XIV.
Invertendo, 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. B. Book. v.
XV.
Componendo, by composition; when there are four proportionals,
and it is inferred that the first together with the second, is to the second,
as the third together with the fourth, is to the fourth. Prop. 18, Book v.
XVI.
Dividendo, by division; when there are four proportionals, and it is
inferred, that the excess of the first above the second, is to the second,
}
128
DEFINITIONS.
as the excess of the third above the fourth, is to the fourth. Prop. 17,
Book v.
XVII.
Convertendo, by conversion; when there are four proportionals, and
it is inferred, that the first is to its excess above the second, as the third
to its excess above the fourth. Prop. E. Book v.
XVIII.
Ex æquali (sc. distantiâ), or ex æquo, from equality of distance:
when there is any number of magnitudes more than two, and as many
others, such that they are proportionals when taken two and two of each
rank, and it is inferred, that the first is to the last of the first rank of
magnitudes, as the first is to the last of the others: Of this there are
the two following kinds, which arise from the different order in which
the magnitudes are taken, two and two.'

XIX.
Ex æquali, from equality. This term is used simply by itself, when
the first magnitude is to the second of the first rank, as the first to the
second of the other rank; and as the second is to the third of the first
rank, so is the second to the third of the other; and so on in order: and
the inference is as mentioned in the preceding definition; whence this
is called ordinate proportion. It is demonstrated in Prop. 22, Book v.
XX.
Ex æquali in proportione perturbatâ seu inordinatâ, from equality
in perturbate or disorderly proportion*. This term is used when the
first magnitude is to the second of the first rank, as the last but one is
to the last of the second rank; and as the second is to the third of the
first rank, so is the last but two to the last but one of the second rank;
and as the third is to the fourth of the first rank, so is the third from
the last to the last but two of the second rank; and so on in a cross
order and the inference is as in the 18th definition. It is demonstrated
in Prop. 23, Book v.
AXIOMS.
I.
EQUIMULTIPLES of the same, or of equal magnitudes, are equal to
one another.
II.
Those magnitudes, of which the same or equal magnitudes are
equimultiples, are equal to one another.
III.
A multiple of a greater magnitude is greater than the same multiple
of a less.
IV.
That magnitude, of which a multiple is greater than the same
multiple of another, is greater than that other magnitude.
* Prop. 4. Lib. 11. Archimedis de sphærâ et cylindro.
BOOK V. PROP. I, II.
129
PROPOSITION I. THEOREM.
If any number of magnitudes be equimultiples of as many, each of
each; what multiple soever any one of them is of its part, the same mul-
tiple shall all the first magnitudes be of all the other.
Let any number of magnitudes AB, CD be equimultiples of as many
others E, F, each of each.
Then whatsoever multiple AB is of E,
the same multiple shall AB and CD together be of E and F together.
C
A
G-
E
H- F
Bl
DI
Because AB is the same multiple of E that CD is of F,
as many magnitudes as there are in AB equal to E, so many are
there in CD equal to F.
Divide AB into magnitudes equal to E, viz. AG, GB;
and CD into CH, HD, equal each of them to F:
therefore the number of the magnitudes CH, HD shall be equal to
the number of the others AG, GB:
and because AG is equal to E, and CH to F,
therefore AG and CH together are equal to E and F together: (1. ax. 2.)
for the same reason, because GB is equal to E, and HD to F;
GB and HD together are equal to E and F together;
wherefore as many magnitudes as there are in AB equal to E,
so many are there in AB, CD together, equal to E and F together:
therefore, whatsoever multiple AB is of E,
the same multiple is AB and CD together, of E and F together.
Therefore, if any magnitudes, how many soever, be equimultiples
of as many, each of each; whatsoever multiple any one of them is of
its part, the same multiple shall all the first magnitudes be of all the
others: For the same demonstration holds in any number of magni-
tudes, which was here applied to two.' Q. E.D.
PROPOSITION II. THEOREM.
If the first magnitude be the same multiple of the second that the third
is of the fourth, and the fifth the same multiple of the second that the sixth
is of the fourth; then shall the first together with the fifth be the same
multiple of the second, that the third together with the sixth is of the fourth.
Let AB the first be the same multiple of C the second, that DE the
third is of F the fourth:
and BG the fifth the same multiple of C the second, that EH the
sixth is of F the fourth.
Then shall AG, the first together with the fifth, be the same multiple
of C the second, that DH, the third together with the sixth, is of F the
fourth.
9
130
EUCLID'S ELEMENTS.
A
D
E-
4.3.
G
H F
Because AB is the same multiple of C that DE is of F;
there are as many magnitudes in AB equal to C, as there are in DE
equal to F:
in like manner, as many as there are in BG equal to C, so many
are there in EH equal to F:
therefore as many as there are in the whole AG equal to C,
so many are there in the whole DH equal to F:
therefore AG is the same multiple of C that DH is of F;
that is, AG, the first and fifth together, is the same multiple of the
second C,
that DH, the third and sixth together, is of the fourth F.
If therefore, the first be the same multiple, &c. Q. E.D.
COR. From this it is plain, that if any number of magnitudes AB,
BG, GH be multiples of another C;
and as many DE, EK, KL be the same multiples of F, each of each:
then the whole of the first, viz. AH, is the same multiple of C,
that the whole of the last, viz. DL, is of F.
A
&
D
E.
B-
K-
G-
HCL F
PROPOSITION III.
THEOREM.
If the first be the same multiple of the second, which the third is of the
fourth; and if of the first and third there be taken equimultiples; these
shall be equimultiples, the one of the second, and the other of the fourth.
Let A the first be the same multiple of B the second, that C the
third is of D the fourth;
and of A, C let equimultiples EF, GH be taken.
Then EF shall be the same multiple of B, that GH is of D.
F
K.
L-
H
È ABG C D
Because EF is the same multiple of A, that GH is of C,
there are as many magnitudes in EF equal to A, as there are in
GH equal to Č:
BOOK V. PROP. III, IV.
131
let EF be divided into the magnitudes EK, KF, each equal to A;
and GH into GL, LH, each equal to C:
therefore the number of the magnitudes ÉK, KF shall be equal to
the number of the others GL, LH:
and because A is the same multiple of B, that C is of D,
and that EK is equal to A, and GL equal to C;
therefore EK is the same multiple of B, that GL is of D:
for the same reason, KF is the same multiple of B, that LH is of D:
and so, if there be more parts in EF, GH, equal to A,
therefore, because the first EK is the same multiple of the second
B, which the third GL is of the fourth D,
C:
and that the fifth KF is the same multiple of the second B, which
the sixth LH is of the fourth D;
EF the first, together with the fifth, is the same multiple of the
second B, (v. 2.)
which GH the third, together with the sixth, is of the fourth D.
If, therefore, the first, &c. Q.E.D.
PROPOSITION IV. THEOREM.
If the first of four magnitudes has the same ratio to the second which
the third has to the fourth; then any equimultiples whatever of the first
and third shall have the same ratio to any equimultiples of the second and
fourth, viz. the equimultiple of the first shall have the same ratio to that
of the second, which the equimultiple of the third has to that of the fourth.'
Let A the first have to B the second, the same ratio which the third
C has to the fourth D;
and of A and C let there be taken any equimultiples whatever E, F;
and of B and D any equimultiples whatever G, H.
Then E shall have the same ratio to G, which F has to H.
KEA B G M
LF CDH N
Take of E and F any equimultiples whatever K, L,
and of G, H any equimultiples whatever M, N:
then because E is the same multiple of A, that F is of C;
and of E and F have been taken equimultiples K, L;
therefore K is the same multiple of A, that L is of C: (v. 3.)
for the same reason, M is the same multiple of B, that N is of D.
9—2
132.
EUCLID'S ELEMENTS.
And because, as A is to B, so is C to D, (hyp.)
and of A and C have been taken certain equimultiples K, L,
and of B and D have been taken certain equimultiples M, N;
therefore if K be greater than M, L is greater than N;
and if equal, equal; if less, less: (v. def. 5.)
but K, L are any equimultiples whatever of E, F, (constr.)
and M, N any whatever of G, H;
therefore as E is to G, so is F to H. (v. def. 5.)
Therefore, if the first, &c. Q.E.D.
COR. Likewise, if the first has the same ratio to the second, which
the third has to the fourth, then also any equimultiples whatever of the
first and third shall have the same ratio to the second and fourth; and
in like manner, the first and the third shall have the same ratio to any
equimultiples whatever of the second and fourth.
Let A the first have to B the second the same ratio which the third
C has to the fourth D,
and of A and C let E and F be any equimultiples whatever.
Then E shall be to B as F to D.
Take of E, F any equimultiples whatever K, L,
and of B, D any equimultiples whatever G, H:
then it may be demonstrated, as before, that K is the same multiple
of A, that L is of C:
and because A is to B, as C is to D, (hyp.)
and of A and C certain equimultiples have been taken, viz. K and L;
and of B and D certain equimultiples G, H;
therefore, if K be greater than G, L is greater than H;
and if equal, equal; if less, less: (v. def. 5.)
but K, L are any equimultiples whatever of E, F, (constr.)
and G, H any whatever of B, D;
therefore, as E is to B, so is F to D. (v. def. 5.)
And in the same way the other case is demonstrated.
PROPOSITION V. THEOREM.
If one magnitude be the same multiple of another, which a magnitude
taken from the first is of a magnitude taken from the other; the remainder
shall be the same multiple of the remainder, that the whole is of the whole.
Let the magnitude AB be the same multiple of CD, that AE taken
from the first, is of CF taken from the other.
The remainder EB shall be the same multiple of the remainder FD,
that the whole AB is of the whole CD.
G
A-
C
E-
F.
B D
Take AG the same multiple of FD, that AE is of CF:
therefore AE is the same multiple of CF, that EG is of CD: (v. 1.)
:
BOOK V. PROP. V, VI.
133
but AE, by the hypothesis, is the same multiple of CF, that AB
is of CD;
therefore EG is the same multiple of CD that AB is of CD;
wherefore EG is equal to AB: (v. ax. 1.)
take from each of them the common magnitude AE;
and the remainder AG is equal to the remainder EB.
Wherefore, since AE is the same multiple of CF, that AG is of
FD, (constr.)
and that AG has been proved equal to EB;
therefore AE is the same multiple of CF, that EB is of FD:
but AE is the same multiple of CF that AB is of CD: (hyp.)
therefore EB is the same multiple of FD, that AB is of CD.
Therefore, if one magnitude, &c. Q.E.D.
PROPOSITION VI.
THEOREM.
If two magnitudes be equimultiples of two others, and if equimultiples
of these be taken from the first two; the remainders are either equal to
these others, or equimultiples of them.
Let the two magnitudes AB, CD be equimultiples of the two E, F,
and let AG, CH taken from the first two be equimultiples of the
same E, F.
Then the remainders GB, HD shall be either equal to E, F, or
equimultiples of them.
K
A
3
C-
G-H-
BDEF
First, let GB be equal to E:
HD shall be equal to F.
Make CK equal to F:
and because AG is the same multiple of E, that CH is of F: (hyp.)
and that GB is equal to E, and CK to F;
therefore AB is the same multiple of E, that KH is of F:
but AB, by the hypothesis, is the same multiple of E, that CD is of F;
therefore KH is the same multiple of F, that CD is of F:
wherefore KH is equal to CD: (v. ax. 1.)
take away the common magnitude CH,
then the remainder KC is equal to the remainder HD:
but KC is equal to F; (constr.)
therefore HD is equal to F.
Next let GB be a multiple of E.
Then HD shall be the same multiple of F.
A
K
C-
G-H-
B DE F
1
J
T
134
EUCLID'S ELEMENTS.
Make CK the same multiple of F, that GB is of E:
(hyp.)
and because AG is the same multiple of E, that CH is of F;
and GB the same multiple of E, that CK is of F;
therefore AB is the same multiple of E, that KH is of F: (v. 2.)
but AB is the same multiple of E, that CD is of F; (hyp.)
therefore KH is the same multiple of F, that CD is of F;
wherefore KH is equal to CD: (v. ax. 1.)
take away CH from both;
therefore the remainder KC is equal to the remainder HD:
and because GB is the same multiple of E, that KC is of F, (constr.)
and that KC is equal to HD;
therefore HD is the same multiple of F, that GB is of E.
If, therefore, two magnitudes, &c. Q.E.D.
PROPOSITION A. THEOREM.
If the first of four magnitudes has the same ratio to the second, which
the third has to the fourth; then, if the first be greater than the second,
the third is also greater than the fourth; and if equal, equal; if less, less.
Take any equimultiples of each of them, as the doubles of each :
then, by def. 5th of this book, if the double of the first be greater
than the double of the second, the double of the third is greater than
the double of the fourth:
but, if the first be greater than the second,
the double of the first is greater than the double of the second;
wherefore also the double of the third is greater than the double
of the fourth;
therefore the third is greater than the fourth:
in like manner, if the first be equal to the second, or less than it,
the third can be proved to be equal to the fourth, or less than it.
Therefore, if the first, &c. Q. E. D.
PROPOSITION B. THEOREM.
If four magnitudes are proportionals, they are proportionals also when
taken inversely.
Let A be to B, as C is to D.
Then also inversely, B shall be to A, as D to C.
GABE
HCDF
BOOK V. PROP. B. C.
135
Take of B and D any equimultiples whatever E and F;
and of A and C any equimultiples whatever G and H.
First, let E be greater than G, then G is less than E:
and because A is to B, as C is to D, (hyp.)
and of A and C, the first and third, G and H are equimultiples;
and of B and D, the second and fourth, E and F are equimultiples;
and that G is less than E, therefore H is less than F; (v. def. 5.)
that is, F is greater than H;
if, therefore, E be greater than G,
F is greater than H;
in like manner, if E be equal to G,
F may be shewn to be equal to H;
and if less, less
but E, F, are any equimultiples whatever of B and D, (constr.)
and G, H any whatever of A and C;
therefore, as B is to A, so is D to C. (v. def. 5.)
Therefore, if four magnitudes, &c. Q. E.D.
PROPOSITION C. THEOREM.
If the first be the same multiple of the second, or the same part of it,
that the third is of the fourth; the first is to the second, as the third is to
the fourth.
Let the first A be the same multiple of the second B,
that the third C is of the fourth D.
Then A shall be to B as C is to D.
A B C D
EG FH
Take of A and C any equimultiples whatever E and F;
and of B and D any equimultiples whatever G and H.
Then, because A is the same multiple of B that C is of D; (hyp.)
and that E is the same multiple of A, that F is of C; (constr.)
therefore E is the same multiple of B, that F is of D;` (v.3.)
that is, E and F are equimultiples of B and D:
but G and H are equimultiples of B and D; (constr.)
therefore, if E be a greater multiple of B than G is of B,
F is a greater multiple of D than H is of D;
that is, if E be greater than G,
F is greater than H:
in like manner, if E be equal to G, or less than it,
F may be shewn to be equal to H, or less than it :
136
EUCLID'S ELEMENTS.
but E, F are equimultiples, any whatever, of A, C; (constr.)
and G, H any equimultiples whatever of B, D;
therefore A is to B, as C is to D. (v. def. 5.)
Next, let the first A be the same part of the second B, that the third
Cis of the fourth D.
Then A shall be to B, as C is to D.
A B C D
For since A is the same part of B that C is of D,
therefore B is the same multiple of A, that D is of C:
wherefore, by the preceding case, B is to A, as D is to C;
and therefore inversely A is to B, as C is to D. (V. B.)
Therefore, if the first be the same multiple, &c.
Q. E. D.
PROPOSITION D. THEOREM.
If the first be to the second as the third to the fourth, and if the
first be a multiple, or a part of the second; the third is the same multiple,
or the same part of the fourth.
Let A be to B as C is to D:
and first let A be a multiple of B.
Then C shall be the same multiple of D.
A B C D
E F
Take E equal to A,
and whatever multiple A or E is of B, make F the same multiple of D:
then, because A is to B, as C is to D;
(hyp.)
and of B the second, and D the fourth, equimultiples have been
taken, E and F ;
therefore A is to E, as C to F: (v. 4. Cor.)
but A is equal to E, (constr.)
therefore C is equal to F: (v. A.)
and F is the same multiple of D, that A is of B; (constr.)
therefore C is the same multiple of D, that A is of B.
Next, let 4 the first be a part of B the second.
Then C the third shall be the same part of D the fourth.
Because A is to B, as C is to D; (hyp.)
then, inversely, B is to A, as D to C: (v. B.)
but A is a part of B, therefore B is a multiple of A: (hyp.)
BOOK V. PROP. D, VII, VIII.
137
:
A B C D
therefore, by the preceding case, D is the same multiple of C;
that is, C is the same part of D, that A is of B.
Therefore, if the first, &c. Q. E.D.
PROPOSITION VII.
THEOREM.
Equal magnitudes have the same ratio to the same magnitude: and
the same has the same ratio to equal magnitudes.
Let A and B be equal magnitudes, and C any other.
Then A and B shall each of them have the same ratio to C:
and C shall have the same ratio to each of the magnitudes A and B.
DA
E B
C F
Take of A and B any equimultiples whatever D and E,
and of C any multiple whatever F.
Then, because D is the same multiple of A, that E is of B, (constr.)
and that A is equal to B:
(hyp.)
therefore D is equal to E: (v. ax. 1.)
therefore, if D be greater than F, E is greater than F;
and if equal, equal; if less, less;
but D, E are any equimultiples of A, B,
and F is any multiple of C;
(constr.)
therefore, as A is to C, so is B to C. (v. def. 5.)
Likewise C shall have the same ratio to A, that it has to B.
For having made the same construction,
D may in like manner be shewn to be equal to E:
therefore, if F be greater than D,
it is likewise greater than E;
and if equal, equal; if less, less:
but F is any multiple whatever of C,
and D, E are any equimultiples whatever of A, B;
therefore, C is to A as C is to B. (v. def. 5.)
Therefore, equal magnitudes, &c. Q. E. D.
PROPOSITION VIII. THEOREM.
Of two unequal magnitudes, the greater has a greater ratio to any
other magnitude than the less has: and the same magnitude has a greater
ratio to the less of two other magnitudes, than it has to the greater.
138
EUCLID'S ELEMENTS,
Let AB, BC be two unequal magnitudes, of which AB is the greater,
and let D be any other magnitude.
Then AB shall have a greater ratio to D than BC has to D:
and D shall have a greater ratio to BC than it has to AB.
Fig. 1.
E
Fig. 2.
E
Fig. 3.
E
F.
F.
A
C-
A
F. A
G B
LK HD
G B
C-
LK HD
G B
LK D
If the magnitude which is not the greater of the two AC, CB, be not
less than D,
take EF, FG, the doubles of AC, CB, (as in fig. 1.)
But if that which is not the greater of the two AC, CB, be less than D,
(as in fig. 2 and 3.) this magnitude can be multiplied, so as to
become greater than D, whether it be AC or CB.
Let it be multiplied until it become greater than D,
and let the other be multiplied as often;
and let EF be the multiple thus taken of AC,
and FG the same multiple of CB:
therefore EF and FG are each of them greater than D:
and in every one of the cases,
take H the double of D, K its triple, and so on,
till the multiple of D be that which first becomes greater than FG:
let L be that multiple of D which is first greater than FG,
and K the multiple of D which is next less than L.
Then because L is the multiple of D, which is the first that becomes
greater than FG,
the next preceding multiple K is not greater than FG:
that is, FG is not less than K:
and since EF is the same multiple of AC, that FG is of CB; (constr.)
therefore FG is the same multiple of CB, that EG is of AB; (v. 1.)
that is, EG and FG are equimultiples of AB and CB :
and since it was shewn, that FG is not less than K,
and, by the construction, EF is greater than D;
therefore the whole EG is greater than K and D together:
but K together with D is equal to L; (constr.)
therefore EG is greater than L:
but FG is not greater than L: (constr.)
and EG, FG were proved to be equimultiples of AB, BC;
and L is a multiple of D; (constr.)
therefore AB has to D a greater ratio than BC has to D. (v. def. 7.)
Also D shall have to BC a greater ratio than it has to AB.
For having made the same construction,
it
may
be shewn, in like manner, that L is greater than FG,
but that it is not greater than EG:
and L is a multiple of D; (constr.)
Чем
BOOK V. PROP. VIII, IX.
139
and FG, EG were proved to be equimultiples of CB, AB:
therefore D has to CB a greater ratio than it has to AB. (v. def. 7.)
Wherefore, of two unequal magnitudes, &c.
Q. E.D.
PROPOSITION IX. THEOREM.
Magnitudes which have the same ratio to the same magnitude are equal
to one another: and those to which the same magnitude has the same ratio
are equal to one another.
Let A, B have each of them the same ratio to C.
Then A shall be equal to B.
A
D
c
F
E
For, if they are not equal, one of them must be greater than the other:
let A be the greater:
then, by what was shewn in the preceding proposition,
there are some equimultiples of A and B, and some multiple of C, such,
that the multiple of A is greater than the multiple of C,
but the multiple of B is not greater than that of C.
Let these multiples be taken;
and let D, E be the equimultiples of A, B,
and F the multiple of C,
such that D may be greater than F, but E not greater than A
Then, because A is to C as B is to C, (hyp.)
(hyp.)
and of A, B, are taken equimultiples D, E,
and of C is taken a multiple F;
and that D is greater than F
therefore E is also greater than F: (v. def. 5.)
but E is not greater than F; (constr.) which is impossible:
therefore A and B are not unequal; that is, they are equal.
Next, let C have the same ratio to each of the magnitudes A and B.
Then A shall be equal to B.
For, if they are not equal, one of them must be greater than the other :
let A be the greater :
therefore, as was shewn in Prop. VIII.
there is some multiple F of C,
and some equimultiples E and D of B and A such,
that F is greater than E, but not greater than D:
and because C is to B, as C is to A, (hyp.)
and that F the multiple of the first, is greater than E the multiple
of the second;
therefore F the multiple of the third, is greater than D the multiple
of the fourth: (v. def. 5.)
but F is not greater than D (hyp.); which is impossible:
therefore A is equal to B.
Wherefore, magnitudes which, &c.
Q.E.D.
140
EUCLID'S ELEMENTS.
PROPOSITION X. THEOREM:
That magnitude which has a greater ratio than another has unto the
same magnitude, is the greater of the two: and that magnitude to which
the same has a greater ratio than it has unto another magnitude, is the
lesser of the two.
Let A have to C a greater ratio than B has to C.
Then A shall be greater than B.
A
B
c
D
F
E
For, because A has a greater ratio to C, than B has to C,
there are some equimultiples of A and B, and some multiple of C
such, (v. def. 7.)
that the multiple of A is greater than the multiple of C,
but the multiple of B is not greater than it :
let them be taken;
and let D, E be the equimultiples of A, B, and F the multiple of C;
such, that D is greater than F, but E is not greater than F:
therefore D is greater than E:
and, because D and E are equimultiples of A and B,
and that D is greater than E;
therefore A is greater than B. (v. ax. 4.)
Next, let C have a greater ratio to B than it has to A.
Then B shall be less than A.
For there is some multiple F of C, (v. def. 7.)
and some equimultiples E and D of B and A such,
that F is greater than E, but not greater than D:
therefore E is less than D:
and because E and D are equimultiples of B and A,
and that E is less than D,
therefore B is less than A. (v. ax. 4.)
Therefore, that magnitude, &c.
Q.E.D.
PROPOSITION XI.
THEOREM.
Ratios that are the same to the same ratio, are the same to one another.
Let A be to B as C is to D ;
and as C to D, so let E be to F.
G
H-
K-
A-
C-
E-
B-
D
F
L
M-
N-
Then A shall be to B, as E to F.
Take of A, C, E, any equimultiples whatever G, H, K;
BOOK V. PROP. XI, XII.
141
and of B, D, F, any equimultiples whatever L, M, N.
Therefore, since A is to B as C to D,
and G, H are taken equimultiples of A, C,
and L, M, of B, D;
if G be greater than L, H is greater than M;
and if equal, equal; and if less, less. (v. def. 5.)
Again, because C is to D, as E is to F,
and H, K are taken equimultiples of C, E;
and M, N, of D, F ;
if H be greater than M, K is greater than N;
and if equal, equal; and if less, less:
but if G be greater than L,
it has been shewn that H is greater than M;
and if equal, equal; and if less, less:
therefore, if G be greater than L,
K is greater than N; and if equal, equal; and if less, less:
and G, K are any equimultiples whatever of A, É;
and L, N any whatever of B, F:
therefore, as A is to B, so is E to F. (v. def. 5.)
Wherefore, ratios that, &c. Q.E.D.
PROPOSITION XII. THEOREM.
If any number of magnitudes be proportionals, as one of the antece-
dents is to its consequent, so shall all the antecedents taken together be to
all the consequents.
Let any number of magnitudes A, B, C, D, E, F, be proportionals;
that is, as A is to B, so C to D, and E to F.
Then as A is to B, so shall A, C, E together, be to B, D, F together.
G
H-
K-
A-
C
E-
B
D
F
L
M
N
Take of A, C, E any equimultiples whatever G, H, K;
and of B, D, F any equimultiples whatever, L, M, N.
Then, because A is to B, as C is to D, and as E to F;
and that G, H, K are equimultiples of A, C, E,
and L, M, N, equimultiples of B, D, F;
therefore, if G be greater than L,
H is greater than M, and K greater than N;
and if equal, equal; and if less, less: (v. def. 5.)
wherefore if G be greater than L,
then G, H, K together, are greater than L, M, N together;
and if equal, equal; and if less, less:
but G, and G, H, K together, are any equimultiples of A, and A, C,
E together;
because if there be any number of magnitudes equimultiples of as
many, each of each, whatever multiple one of them is of its part, the
same multiple is the whole of the whole: (v. 1.)
:
142
EUCLID'S ELEMENTS.
for the same reason L, and L, M, N are any equimultiples of B,
and B, D, F:
therefore as A is to B, so are A, C, E together to B, D, F together.
(v. def. 5.)
Wherefore, if any number, &c. Q.E.D.
PROPOSITION XIII. THEOREM.
If the first has to the second the same ratio which the third has to the
fourth, but the third to the fourth a greater ratio than the fifth has to the
sixth; the first shall also have to the second a greater ratio than the fifth
has to the sixth.
Let A the first have the same ratio to B the second, which the
third has to D the fourth, but C the third a greater ratio to D the
fourth, than E the fifth has to F the sixth.
Then also the first A shall have to the second B, a greater ratio than
the fifth E has to the sixth F.
M
G-
H
A-
C-
E
B
D
F
N
K
L
Because C has a greater ratio to D, than E to F,
there are some equimultiples of C and E, and some of D and F
such, that the multiple of C is greater than the multiple of D, but the
multiple of E is not greater than the multiple of F: (v. def. 7.)
let these be taken,
and let G, H be equimultiples of C, E,
and K, L equimultiples of D, F, such that D may be greater than
K, but H not greater than L:
G
and whatever multiple G is of C, take M the same multiple of A;
and whatever multiple K is of D, take N the same multiple of B:
then, because A is to B, as C to D, (hyp.)
and of A and C, M and G are equimultiples;
and of B and D, N and K are equimultiples;
therefore, if M be greater than N, G is greater than K;
and if equal, equal; and if less, less: (v. def. 5.)
but G is greater than K; (constr.)
therefore M is greater than N:
but H is not greater than L: (constr.)
and M, H are equimultiples of A, E;
and N, L equimultiples of B, F;
therefore A has a greater ratio to B, than E has to F. (v. def. 7.)
Wherefore, if the first, &c. Q. E.D.
COR. And if the first have a greater ratio to the second, than the
third has to the fourth, but the third the same ratio to the fourth, which
the fifth has to the sixth; it may be demonstrated, in like manner, that
the first has a greater ratio to the second, than the fifth has to the sixth.
BOOK V.
143
PROP. XIV, XV.
✔
PROPOSITION XIV.
THEOREM.
If the first has the same ratio to the second which the third has to
the fourth; then, if the first be greater than the third, the second shall be
greater than the fourth; and if equal, equal; and if less, less.
Let the first A have the same ratio to the second B which the third
C has to the fourth D.
If A be greater than C, B shall be greater than D.
1
3
lil lil
A B C D
A B C D
A B C D
Because A is greater than C, and B is any other magnitude,
A has to B a greater ratio than C has to B: (v. 8.)
but, as A is to B, so is C to D; (hyp.)
therefore also C has to D a greater ratio than Chas to B: (v. 13.)
but of two magnitudes, that to which the same has the greater ratio
is the less: (v. 10.)
therefore D is less than B;
that is, B is greater than D.
Secondly, if A be equal to C,
then В shall be equal to D.
For A is to B, as C, that is, A to D:
therefore B is equal to D. (v. 9.)
Thirdly, if A be less than C,
then B shall be less than D.
For C is greater than A;
and because C is to D, as A is to B,
therefore D is greater than B, by the first case;
that is, B is less than D.
Therefore, if the first, &c.
Q. E. D.
PROPOSITION XV. THEOREM.
Magnitudes have the same ratio to one another which their equi-
multiples have.
Let AB be the same multiple of C, that DE is of F.
Then C shall be to F, as AB to DE.
A
D
G.
K
H
L
BCEF
Because AB is the same multiple of C, that DE is of F;
there are as many magnitudes in AB equal to C, as there are in
DE equal to F:
let AB be divided into magnitudes, each equal to C, viz. AG, GH, HB;
144
EUCLID'S ELEMENTS.
and DE into magnitudes, each equal to F, viz. DK, KL, LE:
then the number of the first AG, GH, HB, is equal to the number
of the last DK, KL, LE:
and because AG, GH, HB are all equal,
and that DK, KL, LE, are also equal to one another;
therefore AG is to DK as GH to KL, and as HB to LE: (v. 7.)
but as one of the antecedents is to its consequent, so are all the ante-
cedents together to all the consequents together, (v. 12.)
wherefore, as AG is to DK, so is AB to DE:
but AG is equal to C, and DK to F ;
therefore, as C is to F, so is AB to DE.
Therefore, magnitudes, &c. Q.E.D.
PROPOSITION XVI. THEOREM.
If four magnitudes of the same kind be proportionals, they shall also
be proportionals when taken alternately.
Let A, B, C, D be four magnitudes of the same kind, which are
proportionals, viz. as A to B, so C to D.
They shall also be proportionals when taken alternately;
that is, A shall be to C, as B to D.
E
G
A
C-
B
D
F-
H
Take of A and B any equimultiples whatever E and F;
and of C and D take any equimultiples whatever G and H:
and because E is the same multiple of A, that F is of B,
and that magnitudes have the same ratio to one another which their
equimultiples have; (v. 15.)
therefore A is to B, as E is to F:
but as A is to B so is C to D; (hyp.)
wherefore as C is to D, so is E to F: (v. 11.)
again, because G, H are equimultiples of C, D,
therefore as C is to D, so is G to H: (v. 15.)
but it was proved that as C is to D, so is E to F;
therefore, as E is to F, so is G to H. (v. 11.)
But when four magnitudes are proportionals, if the first be greater
than the third, the second is greater than the fourth;
and if equal, equal; if less, less; (v. 14.)
therefore, if E be greater than G, F likewise is greater than H;
and if equal, equal; if less, less:
and E, F are any equimultiples whatever of A, B; (constr.)
and G, H any whatever of C, D:
therefore A is to C as B to D. (v. def. 5.)
If, then, four magnitudes, &c. Q.E. D.
BOOK V. PROP. XVII.
145
PROPOSITION XVII. THEOREM.
If magnitudes, taken jointly, be proportionals, they shall also be pro-
portionals when taken separately: that is, if two magnitudes together have
to one of them, the same ratio which two others have to one of these, the
remaining one of the first two shall have to the other the same ratio which
the remaining one of the last two has to the other of these.
Let AB, BE, CD, DF be the magnitudes, taken jointly which are
proportionals;
that is, as AB to BE, so let CD be to DF.
Then they shall also be proportionals taken separately,
viz. as AE to EB, so shall CF be to FD.
X
P
K
NL
H
B
D M
E
F
GA
Take of AE, EB, CF, FD any equimultiples whatever GH, HK,
LM, MN;
and again, of EB, FD take any equimultiples whatever KX, NP.
Then because GH is the same multiple of AE, that HK is of EB,
therefore GH is the same multiple of AE, that GK is of AB: (v. 1.)
but GH is the same multiple of AE, that LM is of CF;
therefore GK is the same multiple of AB, that LM is of CF.
Again, because LM is the same multiple of CF, that MN is of FD;
therefore LM is the same multiple of CF, that LN is of CD: (v. 1.)
but LM was shewn to be the same multiple of CF, that GK is of AB;
therefore GK is the same multiple of AB, that LN is of CD;
that is, GK, LN are equimultiples of AB, CD.
Next, because HK is the same multiple of EB, that MN is of FD;
and that KX is also the same multiple of EB, that NP is of FD;
therefore HX is the same multiple of EB, that MP is of FD. (v. 2.)
And because AB is to BE as CD is to DF, (hyp.)
and that of AB and CD, GK and LN are equimultiples,
and of EB and FD, HX and MP are equimultiples;
therefore if GK be greater than HX, then LN is greater than MP;
and if equal, equal; and if less, less: (v. def. 5.)
but if GH be greater than KX,
then, by adding the common part HK to both,
GK is greater than HX; (1. ax. 4.)
wherefore also LN is greater than MP;
and by taking away MN from both,
LM is greater than NP; (1. ax. 5.)
therefore, if GH be greater than KX,
LM is greater than NP.
In like manner it may be demonstrated,
that if GH be equal to KX,
10
146
· EUCLID'S ELEMENTS.
LM is equal to NP; and if less, less:
but GH, LM are any equimultiples whatever of AE, CF, (constr.)
and KX, NP are any whatever of EB, FD:
therefore, as AE is to EB, so is CF to FD. (v. def. 5.)
If then, magnitudes, &c. Q. E.D.
:
•
PROPOSITION XVIII. THEOREM.
If magnitudes, taken separately, be proportionals, they shall also be
proportionals when taken jointly: that is, if the first be to the second, as
the third to the fourth, the first and second together shall be to the second,
as the third and fourth together to the fourth.
Let AE, EB, CF, FD be proportionals;
that is, as AE to EB, so let CF be to FD.
Then they shall also be proportionals when taken jointly;
that is, as AB to BE, so shall CD be to DF.
H
0
M
P-
K
N
to
B
DI
E
F.
G
A
L
Take of AB, BE, CD, DF any equimultiples whatever GH, HK,
LM, MN;
and again, of BE, DF, take any equimultiples whatever KO, NP:
and because KO, NP are equimultiples of BE, DF,
and that KH, NM are likewise equimultiples of BE, DF;
therefore if KO, the multiple of BE, be greater than KH, which
is a multiple of the same BE,
then NP, the multiple of DF, is also greater than NM, the multiple
of the same DF;
and if KO be equal to KH,
NP is equal to NM; and if less, less.
First, let KO be not greater than KH;
therefore NP is not greater than NM:
and because GH, HK, are equimultiples of AB, BE, and that AB
is greater than BE,
therefore GH is greater than HK; (v. ax. 3.)
but KO is not greater than KH;
therefore GH is greater than KO.
In like manner it may be shewn, that LM is greater than NP.
Therefore, if KO be not greater than KH,
then GH, the multiple of AB, is always greater than KO, the multiple
of BE;
and likewise LM, the multiple of CD, is greater than NP, the
multiple of DF.
1
BOOK V. PROP. XVIII.
147
Next, let KO be greater than KH;
therefore, as has been shewn, NP is greater than NM.
H
P₁
M
K
B
N
E
AF
D
G
A
And because the whole GH is the same multiple of the whole AB,
that HK is of BE,
therefore the remainder GK is the same multiple of the remainder
AE that GH is of AB. (v. 5.)
which is the same that LM is of CD.
In like manner, because LM is the same multiple of CD, that MN
is of DE,
therefore the remainder LN is the same multiple of the remainder
CF, that the whole LM is of the whole CD: (v. 5.)
but it was shewn that LM is the same multiple of CD, that GK is of AE;
therefore GK is the same multiple of AE, that LN is of CF ;
that is, GK, LN are equimultiples of AE, CF.
And because KO, NP are equimultiples of BE, DF,
therefore if from KO, NP there be taken KH, NM, which are
likewise equimultiples of BE, DF,
the remainders HO, MP are either equal to BE, DF, or equimultiples
of them. (v. 6.)
First, let HO, MP be equal to BE, DF:
then because AE is to EB, as CF to FD, (hyp.)
and that GK, LN are equimultiples of AE, CF;
therefore GK is to EB, as LN to FD: (v. 4. Cor.)
but HO is equal to EB, and MP to FD;
wherefore GK is to HO, as LN to MP:
therefore if GK be greater than HO, LN is greater than MP; (5. A.)
and if equal, equal; and if less, less.
But let HO, MP be equimultiples of EB, FD.
Then because AE is to EB, as CF to FD, (hyp.)
H
P
M
K
B
N-
D
E
F
G A C L
and that of AE, CF are taken equimultiples GK, LN;
and of EB, FD, the equimultiples HO, MP;
if GK be greater than HỒ, LN is greater than MP;
and if equal, equal; and if less, less; (v. def. 5.)
which was likewise shewn in the preceding case.
But if GH be greater than KO,
10—2
148
EUCLID'S ELEMENTS,
taking KH from both, GK is greater than HO: (1. ax. 5.)
wherefore also LN is greater than MP;
and consequently adding NM to both,
LM is greater than NP: (1. ax. 4.)
therefore, if GH be greater than KO,
LM is greater than NP.
In like manner it may be shewn, that if GH be equal to KO,
LM is equal to NP; and if less, less.
And in the case in which KO is not greater than KH,
it has been shewn that GH is always greater than KO,
and likewise LM greater than NP:
but GH, LM are any equimultiples whatever of AB, CD, (constr.)
and KO, NP are any whatever of BE, DF;
therefore, as AB is to BE, so is CD to DF.
If then magnitudes, &c.
Q.E.D.
(v. def. 5.)
PROPOSITION XIX. THEOREM.
If a whole magnitude be to a whole, as a magnitude taken from the
first is to a magnitude taken from the other; the remainder shall be to the
remainder as the whole to the whole.
Let the whole AB be to the whole CD, as AE a magnitude taken
from AB is to CF a magnitude taken from CD.
Then the remainder EB shall be to the remainder FD, as the whole
AB to the whole CD.
A
E-
F
B D
Because AB is to CD, as AE to CF:
therefore alternately, BA is to AE, as DC to CF: (v. 16.)
and because if magnitudes taken jointly be proportionals, they are
also proportionals, when taken separately; (v. 17.)
therefore, as BE is to EA, so is DF to FC;
and alternately, as BE is to DF, so is EA to FC:
but, as AE to CF, so, by the hypothesis, is AB to CD;
therefore also BE the remainder is to the remainder DF, as the
whole AB to the whole CD. (v. 11.)
Wherefore, if the whole, &c.
Q.E.D.
COR.-If the whole be to the whole, as a magnitude taken from the
first is to a magnitude taken from the other; the remainder shall like-
wise be to the remainder, as the magnitude taken from the first to that
taken from the other. The demonstration is contained in the preceding.
PROPOSITION E. THEOREM.
If four magnitudes be proportionals, they are also proportionals by
conversion; that is, the first is to its excess above the second, as the third
to its excess above the fourth.
BOOK V. PROP. E, XX.
149
Let AB be to BE, as CD to DF.
Then BA shall be to AE, as DC to CF.
E
A
-
C
F
BD
Because AB is to BE, as CD to DF,
therefore by division, AE is to EB, as CF to FD; (v. 17.)
and by inversion, BE is to EA, as DF to FC; (v. B.)
wherefore, by composition, BA is to AE, as DC is to CF. (v. 18.)
If therefore four, &c.
Q.E.D.
PROPOSITION XX. THEOREM.
If there be three magnitudes, and other three, which, taken two and
two, have the same ratio; then if the first be greater than the third, the
fourth shall be greater than the sixth; and if equal, equal; and if less,
less.
Let A, B, C be three magnitudes and D, E, F other three, which
taken two and two have the same ratio,
viz. as A is to B, so is D to E;
and as B to C, so is E to F.
If A be greater than C, D shall be greater than F;
and if equal, equal; and if less, less.
A B C
DE F
Q
Because A is greater than C, and B is any other magnitude,
and that the greater has to the same magnitude a greater ratio than
the less has to it; (v. 8.)
therefore A has to B a greater ratio than C has to B:
but as D is to E, so is A to B; (hyp.)
therefore D has to E a greater ratio than C to B: (v. 13.)
and because B is to C, as E to F,
by inversion, C is to B, as F is to E: (v. B.)
and D was shewn to have to E a greater ratio than C to B:
therefore D has to E a greater ratio than F to E: (v. 13. cor.)
but the magnitude which has a greater ratio than another to the
same magnitude, is the greater of the two; (v. 10.)
therefore D is greater than F.
Secondly, let A be equal to C.
Then D shall be equal to F.
150
EUCLID'S ELEMENTS.
A B C
DEF
Because A and C are equal to
A is to B, as C is to B:
but A is to B, as D to E;
and C is to B, as F to E;
one another,
(v. 7.)
(hyp.)
(hyp.)
wherefore D is to E, as F to E; (v. 11. and v. b.)
and therefore D is equal to F. (v. 9.)
Next, let A be less than C.
Then D shall be less than F.
A B C
DEF
For Cis greater than A;
and as was shewn in the first case, C is to B, as F to E,
and in like manner, B is to A, as E to D;
therefore F is greater than D, by the first case;
that is, D is less than F.
Therefore, if there be three, &c.
Q.E.D.
PROPOSITION XXI. THEOREM.
If there be three magnitudes, and other three, which have the same
ratio taken two and two, but in a cross order; then if the first magnitude
be greater than the third, the fourth shall be greater than the sixth; and
if equal, equal; and if less, less.
Let A, B, C be three magnitudes, and D, E, F other three, which
have the same ratio, taken two and two, but in a cross order,
viz. as A is to B, so is E to F,
and as B is to C, so is D to E.
If A be greater than C, D shall be greater than F;
and if equal, equal; and if less, less.
A B C
DE F
BOOK V. PROP. XXI, XXII.
151
Because A is greater than C, and B is any other magnitude,
A has to B a greater ratio than C has to B: (v. 8.)
but as E to F, so is A to B; (hyp.)
therefore E has to F a greater ratio than C to B: (v. 13.)
and because B is to C, as D to E; (hyp.)
(hyp.)
by inversion, C is to B, as E to D:
and E was shewn to have to F a greater ratio than C has to B;
therefore E has to F a greater ratio than E has to D: (v. 13. Cor.)
but the magnitude to which the same has a greater ratio than it has
to another, is the less of the two: (v.10.)
therefore F is less than D;
that is, D is greater than F.
Secondly, Let A be equal to C;
D shall be equal to F.
A B C
DEF
Because A and C are equal,
A is to B, as C is to B: (v. 7.)
but A is to B, as E to F; (hyp.)
and C is to B, as E to D;
wherefore E is to F, as E to D; (v.11.)
and therefore D is equal to F. (v. 9.)
Next, let A be less than C:
D shall be less than F.
A B C
DEF
For C is greater than A;
and, as was shewn, C is to B, as E to D,
and in like manner B is to A, as F to E
therefore F is greater than D, by case first;
that is, D is less than F.
Therefore, if there be three, &c. Q.E.D.
PROPOSITION XXII. THEOREM.
If there be any number of magnitudes, and as many others, which
taken two and two in order, have the same ratio; the first shall have to the
last of the first magnitudes, the same ratio which the first has to the last of
the others. N. B. This is usually cited by the words "ex æquali," or
"ex æquo."
First, let there be three magnitudes A, B, C, and as many others
D, E, F, which taken two and two in order, have the same ratio,
152
EUCLID'S ELEMENTS.
that is, such that A is to B as D to E;
and as B is to C, so is E to F.
Then A shall be to C, as D to F.
АВС
DE F
GK M
HL N
Take of A and D any equimultiples whatever G and H;
and of B and E any equimultiples whatever K and L;
and of C and F any whatever M and N:
then because A is to B, as D to E,
and that G, H are equimultiples of A, D,
and K, L equimultiples of B, E;
therefore as G is to K, so is H to L: (v. 4.)
for the same reason, K is to M as L to N:
and because there are three magnitudes G, K, M, and other three
H, L, N, which two and two, have the same ratio ;
therefore if G be greater than M, H is greater than N;
and if equal, equal; and if less, less; (v. 20.)
but G, H are any equimultiples whatever of A, D,
and M, N are any equimultiples whatever of C, F; (constr.)
therefore, as à is to C, so is D to F. (v. def. 5.)
Next, let there be four magnitudes, A, B, C, D,
and other four E, F, G, H, which two and two have the same ratio,
viz. as A is to B, so is E to F;
and as B to C, so F to G ;
and as C to D, so G to H:
Then A shall be to D, as E to H.
*
A.B.C.D
E.F.G.H
Because A, B, C are three magnitudes, and E, F, G other three,
which taken two and two, have the same ratio;
therefore by the foregoing case, A is to C, as E to G:
but is to D, as G is to H;
wherefore again, by the first case, A is to D, as E to H:
and so on, whatever be the number of magnitudes.
Therefore, if there be any number, &c.
Q.E.D.
PROPOSITION XXIII. THEOREM.
If there be any number of magnitudes, and as many others, which
taken two and two in a cross order, have the same ratio; the first shall
have to the last of the first magnitudes the same ratio which the first
has to the last of the others. N. B. This is usually cited by the words
"ex æquali in proportione perturbatâ ;” or “ex æquo perturbato.”
BOOK V. PROP. XXIII.
153
First, let there be three magnitudes A, B, C, and other three D, E, F,
which taken two and two in a cross order have the same ratio,
that is, such that A is to B, as E to F;
and as B is to C, so is D to E.
Then A shall be to C, as D to F.
A B C
DEF
GH L
K M N
Take of A, B, D any equimultiples whatever G, H, K;
and of C, E, F any equimultiples whatever L, M, N:
and because G, H are equimultiples of A, B,
and that magnitudes have the same ratio which their equimultiples
have; (v. 15.)
therefore as A is to B, so is G to H:
and for the same reason, as E is to F, so is M to N:
but as A is to B, so is E to F;
therefore as G is to H, so is M to N:
and because as B is to C, so is D to E,
and that H, K are equimultiples of B, D, and
(hyp.)
(v. 11.)
(hyp.)
L, M of C, E;
therefore as H is to L, so is K to M: (v. 4.)
and it has been shewn that G is to H, as M to N:
therefore, because there are three magnitudes G, H, L, and other
three K, M, N, which have the same ratio taken two and two in a cross
order;
if G be greater than L, K is greater than N:
and if equal, equal; and if less, less: (v. 21.)
but G, K are any equimultiples whatever of A, D;
and L, N any whatever of C, F ;
(constr.)
therefore as A is to C, so is D to F. (v. def. 5.)
Next, let there be four magnitudes A, B, C, D, and other four E,
F, G, H, which taken two and two in a cross order have the same ratio,
viz. A to B, as G to H;
B to C, as F to G;
and C to D, as E to F.
Then A shall be to D, as E to H.
A.B.C.D
E.F.G.H
Because A, B, C are three magnitudes, and F, G, H other three,
which taken two and two in a cross order have the same ratio;
by the first case, A is to C, as F to H;
but C is to D, as E is to F;
wherefore again, by the first case, A is to D, as E to H;
and so on, whatever be the number of magnitudes.
Therefore, if there be any number, &c.
Q. E.D.
154
EUCLID'S ELEMENTS.
PROPOSITION XXIV. THEOREM.
If the first has to the second the same ratio which the third has to the
fourth; and the fifth to the second the same ratio which the sixth has to
the fourth; the first and fifth together shall have to the second, the same
ratio which the third and sixth together have to the fourth.
Let AB the first have to C the second the same ratio which DE the
third has to F the fourth;
and let BG the fifth have to C the second the same ratio which EH
the sixth has to F the fourth.
Then AG, the first and fifth together, shall have to C the second,
the same ratio which DH, the third and sixth together, has to F the
fourth.
G
H
B
E-
ACDF
Because BG is to C, as EH to F;
by inversion, C is to BG, as F to EH: (V. B.)
and because, as AB is to C, so is DE to F; (hyp.)
and as C to BG, so F to EH ;
ex æquali, AB is to BG, as DE to EH: (v. 22.)
and because these magnitudes are proportionals when taken separately,
they are likewise proportionals when taken jointly; (v. 18.)
therefore as AG is to GB, so is DH to HE:
but as GB to C, so is HE to F: (hyp.)
therefore, ex æquali, as AG is to C, so is DH to F. (v. 22.)
Wherefore, if the first, &c. Q. E. D.
COR. 1.—If the same hypothesis be made as in the proposition, the
excess of the first and fifth shall be to the second, as the excess of the
third and sixth to the fourth. The demonstration of this is the same
with that of the proposition, if division be used instead of composition.
COR. 2.-The proposition holds true of two ranks of magnitudes, what-
ever be their number, of which each of the first rank has to the second
magnitude the same ratio that the corresponding one of the second rank
has to a fourth magnitude: as is manifest.


PROPOSITION XXV. THEOREM.
If four magnitudes of the same kind are proportionals, the greatest
and least of them together are greater than the other two together.
Let the four magnitudes AB, CD, E, F be proportionals,
viz. AB to CD, as E to F;
and let AB be the greatest of them, and consequently F the least.
(v. 14. and a.)
Then AB together with F shall be greater than CD together with E.
BOOK V.
155
PROP. XXV, F.
B
G-D
HI
A CE F
Take AG equal to E, and CH equal to F.
Then because as AB is to CD, so is E to F,
and that AG is equal to E, and CH equal to F,
therefore AB is to CD, as AG to CH: (v. 11, and 7.)
and because AB the whole, is to the whole CD, as AG is to CH,
likewise the remainder GB is to the remainder HD, as the whole
AB is to the whole CD: (v. 19.)
but AB is greater than CD; (hyp.)
therefore GB is greater than HD: (v. a.)
and because AG is equal to E, and CH to F;
AG and F together are equal to CH and E together: (1. ax. 2.)
therefore if to the unequal magnitudes GB, HD, of which GB is
the greater, there be added equal magnitudes, viz. to GB the two AG
and F, and CH and E to HD;
AB and F together are greater than CD and E.
Therefore, if four magnitudes, &c. Q.E. D.
(1. ax. 4.)
PROPOSITION F. THEOREM.
Ratios which are compounded of the same ratios, are the same to one another.
Let A be to B, as D to E; and B to C, as E to F.
Then the ratio which is compounded of the ratios of A to B, and B to C,
which, by the definition of compound ratio, is the ratio of A to C,
shall be the same with the ratio of D to F, which, by the same
definition, is compounded of the ratios of D to E, and E to F.
A.B.C
D.E.F
Because there are three magnitudes A, B, C, and three others D, E, F,
which, taken two and two, in order, have the same ratio;
ex æquali A is to C, as D to F. (v. 22.)
Next, let A be to B, as E to F, and B to C, as D to E:
A.B.C
D.E.F
therefore, ex æquali in proportione perturbatâ, (v. 23.)
A is to C, as D to F;
that is, the ratio of A to C, which is compounded of the ratios of A
to B, and B to C, is the same with the ratio of D to F, which is com-
pounded of the ratios of D to E, and E to F.
And in like manner the proposition may be demonstrated, whatever
be the number of ratios in either case.
156
EUCLID'S ELEMENTS.
PROPOSITION G. THEOREM.
If several ratios be the same to several ratios, each to each; the ratio
which is compounded of ratios which are the same to the first ratios, each
to each, shall be the same to the ratio compounded of ratios which are the
same to the other ratios, each to each.
Let A be to B, as E to F; and C to D, as G to H:
and let A be to B, as K to L; and C to D, as L to M.
Then the ratio of K to M,
by the definition of compound ratio, is compounded of the ratios of
K to L, and L to M, which are the same with the ratios of A to B, and
C to D.
Again, as E to F, so let N be to 0; and as G to H, so let O be to P.
Then the ratio of N to P is compounded of the ratios of N to 0,
and O to P, which are the same with the ratios of E to F, and G to H:
and it is to be shewn that the ratio of K to M, is the same with the
ratio of N to P;
or that K is to M, as N to P.
A.B.C.D. K.L.M.
E.F.G.H. N.O.P.
Because K is to L, as (A to B, that is, as E to F, that is, as) N to 0:
and as L to M, so is (C to D, and so is G to H, and so is) O to P:
ex æquali K is to M, as N to P.
Therefore, if several ratios, &c.
(v. 22.)
Q.E.D.
PROPOSITION H. THEOREM.
If a ratio which is compounded of several ratios be the same to a
ratio which is compounded of several other ratios; and if one of the first
ratios, or the ratio which is compounded of several of them, be the same
to one of the last ratios, or to the ratio which is compounded of several
of them; then the remaining ratio of the first, or, if there be more than
one, the ratio compounded of the remaining ratios, shall be the same to
the remaining ratio of the last, or, if there be more than one, to the
ratio compounded of these remaining ratios.
Let the first ratios be those of A to B, B to C, C to D, D to E,
and E to F;
and let the other ratios be those of G to H, H to K, K to L,
and L to M:
also, let the ratio of A to F, which is compounded of the first ratios,
be the same with the ratio of G to M, which is compounded of the
other ratios;
and besides, let the ratio of A to D, which is compounded of the
ratios of A to B, B to C, C to D, be the same with the ratio of G to K,
which is compounded of the ratios of G to H, and H to K.
Then the ratio compounded of the remaining first ratios, to wit, of
the ratios of D to E, and E to F, which compounded ratio is the ratio
of D to F, shall be the same with the ratio of K to M, which is com-
pounded of the remaining ratios of K to L, and L to M of the other
ratios.
BOOK V.
157
PROP. H, K.
A.B.C.D.E.F.
G.H.K.L.M.
Because, by the hypothesis, A is to D, as G to K,
by inversion, Ď is to A, as K to G ; (v. B.)
and as A is to F, so is G to M; (hyp.)
therefore, ex æquali, D is to F, as K to M. (v. 22.)
,
If, therefore, a ratio which is, &c. Q.E.D.
PROPOSITION K. THEOREM.
If there be any number of ratios, and any number of other ratios
such, that the ratio which is compounded of ratios which are the same
to the first ratios, each to each, is the same to the ratio which is com-
pounded of ratios which are the same, each to each, to the last ratios;
and if one of the first ratios, or the ratio which is compounded of ratios
which are the same to several of the first ratios, each to each, be the
same to one of the last ratios, or to the ratio which is compounded of
ratios which are the same, each to each, to several of the last ratios;
then the remaining ratio of the first, or, if there be more than one, the
ratio which is compounded of ratios which are the same cach to each
to the remaining ratios of the first, shall be the same to the remaining
ratio of the last, or, if there be more than one, to the ratio which is
compounded of ratios which are the same each to each to these remaining
ratios.
Let the ratios of A to B, C to D, E to F, be the first ratios:
and the ratios of G to H, K to L, M to N, O to P, Q to R, be the
other ratios:
and let A be to B, as S to T; and C to D, as T to V; and E to F,
as V to X:
therefore, by the definition of compound ratio, the ratio of S to X is
compounded of the ratios of S to T, T to V, and V to X, which are the
same to the ratios of A to B, C to D, E to F: each to each.
Also, as G to H, so let Y be to Z; and K to L, as Z to a; M to N,
as a to b; 0 to P, as b to c; and Q to R, as c to d:
therefore, by the same definition, the ratio of Y to d is compounded
of the ratios of Y to Z, Z to a, a to b, b to c, and c to d, which are the
same, each to each, to the ratios of G to H, K to L, M to N, 0 to P,
and Q to R:
therefore, by the hypothesis, S is to X, as Y to d.
Also, let the ratio of A to B, that is, the ratio of S to T, which is one
of the first ratios, be the same to the ratio of e to g, which is compounded
of the ratios of e to f, and ƒ to g, which, by the hypothesis, are the same
to the ratios of G to H, and K to L, two of the other ratios;
and let the ratio of h to I be that which is compounded of the ratios
of h to k, and k to l, which are the same to the remaining first ratios,
viz. of C to D, and E to F;
also, let the ratio of m to p, be that which is compounded of the
ratios of m to n, n to o, and o to p, which are the same, each to each, to
the remaining other ratios, viz. of M to N, O to P, and Q to R.
Then the ratio of h to l shall be the same to the ratio of m to p; or
h shall be to l, as m to p.
158
EUCLID'S ELEMENTS.

h, k, l.
A, B; C, D; E, F.
S, T, V. X.
G, H; K, L; M, N; 0, P; Q, R. Y, Z, a, b, c, d.
e, f, g.
m, n, o, p.
Because e is to f, as (G to H, that is, as) Y to Z;
and ƒ is to g, as (K to L, that is, as) Z to a;
therefore, ex æquali, e is to g, as Y to a: (v. 22.)
and by the hypothesis, A is to B, that is, S to T, as e to g;
wherefore S is to T, as Y to a; (v. 11.)
and by inversion T is to S, as a to Y: (v. B.)
but S is to X, as Y to D; (hyp.)
therefore, ex æquali, T is to X, as a to d:
also, because h is to k as (C to D, that is, as) T to V; (hyp.)
and k is to l as (E to F, that is, as) V to X;
therefore, ex æquali, h is to l, as T to x:
in like manner, it may be demonstrated, that m is to p, as a to d;
and it has been shewn, that T is to X, as a to d;
therefore h is to l, as m to p. (v. 11.) Q.E.D.
The propositions G and K are usually, for the sake of brevity,
expressed in the same terms with propositions F and H: and therefore
it was proper to shew the true meaning of them when they are so
expressed; especially since they are very frequently made use of by
geometers.
NOTES TO BOOK V.
In the first four books of the Elements are considered, only the absolute equality and
inequality of Geometrical magnitudes. The fifth book contains an exposition of the
principles whereby a more definite comparison may be instituted of the relation of magni-
tudes, besides their simple equality or inequality.
Def. I, II.
In the first four books the word part is used in the same sense as we find
it in the ninth axiom, "The whole is greater than its part:" where the word part means
any portion whatever of any whole magnitude: but in the fifth book, the word part is
restricted to mean that portion of magnitude which is contained an exact number of times
in the whole. For instance, if any straight line be taken two, three, four, or any number
of times another straight line, by Euc. 1. 3; the less line is called a part, or rather a
submultiple of the greater line; and the greater, a multiple of the less line. The
multiple is composed of a repetition of the same magnitude, and these definitions
suppose that the multiple may be divided into its parts, any one of which is a measure of
the multiple. And it is also obvious that when there are two magnitudes, one of which
is a multiple of the other, the two magnitudes must be of the same kind, that is, they
must be two lines, two angles, two surfaces, or two solids.
Euclid does not seem to consider one triangle a part or multiple of another, except
when both are between the same parallels; and a triangle is doubled, trebled, &c., by
doubling, trebling, &c. the base. The same may be said of parallelograms. Also the
arcs of two circles are not considered as part and multiple, unless the circles have equal
radii. Angles, arcs, and sectors of equal circles may be doubled, trebled, or any multi-
ples found by Prop. XXVI-xxxix, Book 11I.
Two magnitudes are said to be commensurable when a third magnitude of the same
kind can be found which will measure both of them; and this third magnitude is called
their common measure: and when it is the greatest magnitude which will measure both of
them, it is called the greatest common measure of the two magnitudes : also when two
magnitudes of the same kind have no common measure, they are said to be incommen-
surable. The same terms are also applied to numbers.
Unity has no magnitude, properly so called, but may represent that portion of every
kind of magnitude which is assumed as the measure of all magnitudes of the same kind.
The composition of unities cannot produce Geometrical magnitude; three units are more
in number than one unit, but still as much different from magnitude as unity itself.
Numbers may be considered as quantities, for we consider every thing that can be
exactly measured, as a quantity.
Unity is a common measure of all rational numbers, and all numerical reasonings
proceed upon the hypothesis that the unit is the same throughout the whole of any par-
ticular process. Euclid has not fixed the magnitude of any unit of length, nor made
reference to any unit of measure of angles, surfaces, or volumes. Hence arises an essen-'
tial difference between number and magnitude; unity, being invariable, measures all rational
numbers; but though any quantity be assumed as the unit of magnitude, it is impossible
to assert that this assumed unit will measure all other magnitudes of the same kind.
All whole numbers therefore are commensurable; for unity is their common measure :
also all rational fractions proper or improper, are commensurable; for any such frac-
160
EUCLID'S ELEMENTS.
tions may be reduced to other equivalent fractions having one common denominator,
and that fraction whose denominator is the common denominator, and whose numerator
is unity, will measure any one of the fractions. Two magnitudes having a common
measure can be represented by two numbers which express the number of times the
common measure is contained in both the magnitudes.
But two incommensurable magnitudes cannot be exactly represented by any two
whole numbers or fractions whatever; as, for instance, the side of a square is incom-
mensurable to the diagonal of the square. For, it may be shewn numerically, that if
the side of the square contain one unity of length, the diagonal contains more than
one, but less than 2 units of length. If the side be divided into 10 units, the diagonal
contains more than 14, but less than 15 such units. Also if the side contain 100 units,
the diagonal contains more than 141, but less than 142 of such units. It is also obvious,
that as the side is successively divided into a greater number of equal parts, the error in
the magnitude of the diagonal will bę diminished continually, but never can be entirely
exhausted; and therefore into whatever number of equal parts the side of a square be
divided, the diagonal will never contain an exact number of such parts. Thus the
diagonal and side of a square having no common measure, cannot be exactly represented
by any two numbers.
The term equimultiple in Geometry is to be understood of magnitudes of the same
kind, or of different kinds, taken an equal number of times, and implies only a division
of the magnitudes into the same number of equal parts. Thus, if two given lines are
trebled, the trebles of the lines are equimultiples of the two lines: and if a given line
and a given angle or triangle be trebled, the trebles of the line and angle or triangle are
equimultiples of the line and angle or triangle: as (fig. vi. 1.) the straight line HC
and the triangle AHC are equimultiples of the line BC and the triangle ABC: and
(fig. vi. 33.) the arc EN and the angle EHN are equimultiples of the arc EF and the
angle EHF.
Def. III. Λόγος ἐστὶ δύο μεγεθῶν ὁμογενῶν ἡ κατὰ πηλικότητα πρὸς ἄλληλα
ποια σχέσις.
Told σxéois. By this definition of ratio is to be understood the conception of the
mutual relation of two magnitudes of the same kind, as two straight lines, two angles,
two surfaces, or two solids. To prevent any misconception, Def. iv. lays down the
criterion, whereby it may be known what kinds of magnitudes can have a ratio to one
another; namely, Λόγον ἔχειν πρὸς ἄλληλα μεγέθη λέγεται, ἃ δύναται πολλαπλα-
σιαζόμενα ἀλλήλων ὑπερέχειν. Magnitudes are said to have a ratio to one another,
which, when they are multiplied, can exceed one another;" in other words, the magni-
tudes which are capable of mutual comparison must be of the same kind. The former
of the two terms is called the antecedent; and the latter, the consequent of the ratio.
If the antecedent and consequent are equal, the ratio is called a ratio of equality; but
if the antecedent be greater or less than the consequent, the ratio is called a ratio of
greater or less inequality. Care must be taken not to confound the expressions "ratio
of quality" and "equality of ratio :" the former is applied to the terms of a ratio when
they, the antecedent and consequent, are equal to one another, but the latter, to the ratios,
when they are equal.
Arithmetical ratio has been defined to be the relation which one number bears to
another with respect to quotity; the comparison being made by considering what mul-
tiple, part or parts, one number is of the other.
An arithmetical ratio, therefore, is represented by the quotient which arises from
dividing the antecedent by the consequent of the ratio; or by the fraction which has
the antecedent for its numerator and the consequent for its denominator. Hence it
will at once be obvious that the properties of arithmetical ratios will be made to
depend on the properties of fractions.
NOTES TO BOOK V.
161
It must ever be borne in mind that the subject of Geometry is not number, but the
magnitude of lines, angles, surfaces, and solids; and its object is to demonstrate their
properties by a comparison of their absolute and relative magnitudes.
Also, in Geometry, multiplication is only a repeated addition of the same magnitude;
and division is only a repeated subtraction, or the taking of a less magnitude successively
from a greater, until there be either no remainder, or a remainder less than the magnitude
which is successively subtracted.
The Geometrical ratio of any two given magnitudes of the same kind will obviously
be represented by the magnitudes themselves; thus, the ratio of two lines is repre-
sented by the lengths of the lines themselves; and, in the same manner, the ratio of
two angles, two surfaces, or two solids, will be properly represented by the magnitudes
themselves.
In measuring any magnitude, it is obvious that a magnitude of the same kind must
be used; but the ratio of two magnitudes may be measured by every thing which has
the property of quantity. Two straight lines will measure the ratio of two triangles, or
two parallelograms (fig. vi. 1.): and two triangles, or two parallelograms will measure
the ratio of two straight lines. It would manifestly be absurd to speak of the line as
measuring the triangle, or the triangle measuring the line. (See Notes on Book II.
p. 68.)
The ratio of any two quantities depends on their relative and not their absolute mag-
nitudes; and it is possible for the absolute magnitude of two quantities to be changed,
and their relative magnitude to continue the same as before; and thus, the same ratio
may subsist between two given magnitudes, and any other two of the same kind.
In this method of measuring Geometrical ratios, the measures of the ratios are the
same in number as the magnitudes themselves. It has however two advantages; first, it
enables us to pass from one kind of magnitude to another, and thus, independently of
any numerical measure, to institute a comparison between such magnitudes as cannot be
directly compared with one another: and secondly, the ratio of two magnitudes of the
same kind may be measured by two straight lines, which form a simpler measure of
ratios than any other kind of magnitude.
But the simplest method of all would be, to express the measure of the ratio of two
magnitudes by one; but this cannot be done, unless the two magnitudes are commensu-
rable. If two lines AB, CD, one of which AB contains 12 units of any length, and the
other CD contains 4 units; then the ratio of the line AB to the line CD, is the same
as the ratio of the number 12 to 4. Thus, two numbers may represent the ratio of two
lines when the lines are commensurable. In the same manner, two numbers may repre-
sent the ratio of two angles, two surfaces, or two solids.
ད་
Thus, the ratio of any two magnitudes of the same kind may be expressed by two
numbers, when the magnitudes are commensurable. By this means, the consideration of
the ratio of two magnitudes is changed to the consideration of the ratio of two numbers,
and when one number is divided by the other, the quotient will be a single number, or a
fraction, which will be a measure of the ratio of the two numbers, and therefore of the two
quantities. If 12 be divided by 4, the quotient is 3, which measures the ratio of the two
numbers 12 and 4. Again, if besides the ratio of the lines AB and CD which contain
12 and 4 units respectively, we consider two other lines EF and GH which contain
9 and 3 units respectively; it is obvious that the ratio of the line EF to GH is the
same as the ratio of the number 9 to the number 3. And the measure of the ratio of
9 to 3 is 3. That is, the numbers 9 and 3 have the same ratio as the numbers 12 and 4.
But this is a numerical measure of ratio, and can only be applied strictly when the
antecedent and consequent are to one another as one number to another.
And generally, if the two lines AB, CD contain a and b units respectively, and q be
11
162
EUCLID'S ELEMENTS.
1
the quotient which indicates the number of times the number b is contained in a, then q
is the measure of the ratio of the two numbers a and b: and if EF and GH contain
c and d units, and the number d be contained g times in c.
Չ
The number a has to b the
same ratio as the number c has to d.
This is the numerical definition of proportion, which is thus expressed in Euclid's
Elements, Book VII, definition 20.
"Four numbers are proportionals when the first is
the same multiple of the second, or the same part or parts of it, as the third is of the
fourth." This definition of the proportion of four numbers, leads at once to an equation:
a
C
therefore
b
for, since a contains b, q times; =q:
9:
a
b
C
d
and since c contains d, q times; q:
the fundamental equation upon which all reasonings on the pro-
portion of numbers depend.
Sometimes a proportion is defined to be the equality of two ratios.
But we are anticipating the subject of the sixth and eighth definitions.
Def. VIII declares the meaning of the term analogy or proportion. The ratio of
two lines, two angles, two surfaces or two solids, means nothing more than their relative
magnitude in contradistinction to their absolute magnitudes; and a similitude or like-
ness of ratios implies, at least, the two ratios of the four magnitudes which constitute
the analogy or proportion.
Def. IX states that a proportion consists in three terms at least; the meaning of
which is, that the second magnitude is repeated, being made the consequent of the
first, and the antecedent of the second ratio. It is also obvious that when a proportion
consists of three magnitudes, all three are of the same kind. Def. vI appears only to
be a further explanation of what is implied in Def. VIII.
Def. v. Proportion having been defined to be the similitude of ratios, the fifth
definition lays down a criterion by which two ratios may be known to be similar, or
four magnitudes proportionals, without involving any enquiry respecting the four quan
tities, whether the antecedents of the ratios contain or are contained in their consequents
exactly; or whether there are any magnitudes which measure the terms of the two
ratios. The criterion only requires, that the relation of the equimultiples expressed
should hold good, not merely for any particular multiples, as the doubles or trebles,
but for any multiples whatever, whether large or small.
This criterion of proportion may be applied to all Geometrical magnitudes which can
be multiplied, that is, to all which can be doubled, trebled, quadrupled, &c. But it
must be borne in mind, that this criterion does not exhibit a definite measure for either
of the two ratios which constitute the proportion, but only, an undetermined measure
for the sameness or similarity of the two ratios. The nature of the proportion of
Geometrical magnitudes neither requires nor admits of a definite measure of either of
the two ratios, for this would be to suppose that all magnitudes are commensurable.
Though we know not the definite measure of either of the ratios, further than that they
are both similar, and one may be taken as the measure of the other, yet particular
conclusions may be arrived at by this method: for by the test of proportionality here
laid down, it can be proved that one magnitude is greater than, equal to, or less than
another that a third proportional can be found to two, and a fourth proportional to
three straight lines, also that a mean proportional can be found between two straight
lines and further, that which is here stated of straight lines may be extended to other
Geometrical magnitudes.
:
:
With respect to this test or criterion of the proportionality of four magnitudes, it
NOTES TO BOOK V.
163
It
has been objected, that it is utterly impossible to make trial of all the possible equi-
multiples of the first and third magnitudes, and also of the second and fourth.
may be replied, that the point in question is not determined by making such trials,
but by shewing from the nature of the magnitudes, that whatever be the multipliers,
if the multiple of the first exceeds the multiple of the second magnitude, the multiple
of the third will exceed the multiple of the fourth magnitude, and if equal, will be
equal; and if less, will be less, in any case which may be taken.
The Arithmetical definition of proportion in Book VII, Def. 20, even if it were
equally general with the Geometrical definition in Book v, Def. 5, is by means univer-
sally applicable to the subject of Geometrical magnitudes. The Geometrical criterion is
founded on multiplication which is always possible. When the magnitudes are com-
mensurable, the multiples of the first and second may be equal or unequal; but when
the magnitudes are incommensurable, any multiples whatever of the first and second
must be unequal; but the Arithmetical criterion of proportion is founded on division,
which is not always possible. Euclid has not shewn how to take any part of a line or
other magnitude, or that the two terms of a ratio have a common measure, and there-
fore the numerical definition could not be strictly applied, even in the limited way
in which it may be applied.
Number and Magnitude do not correspond in all their relations; and hence the
distinction between Geometrical ratio and Arithmetical ratio; the former is a com-
parison кαтà πŋλıkótηta, according to quantity, but the latter, according to quotity.
The former gives an undetermined, though definite measure, in magnitudes; but the
latter attempts to give the exact value in numbers.
The fifth book exhibits no method whereby two magnitudes may be determined to
be commensurable, and the Geometrical conclusions deduced from the multiples of
magnitudes are too general to furnish a numerical measure of ratios, being all independent
of the commensurability or incommensurability of the magnitudes themselves.
It is the numerical ratio of two magnitudes which will more certainly discover
whether they are commensurable or incommensurable, and hence, recourse must be had
to the forms and properties of numbers. All numbers and fractions are either rational
or irrational. It has been seen that rational numbers and fractions can express the
ratios of Geometrical magnitudes, when they are commensurable. Similar relations of
incommensurable magnitudes may be expressed by irrational numbers, if the Algebraical
expressions for such numbers may be assumed and employed in the same manner as
rational numbers. The irrational expressions being considered the exact and definite,
though undetermined, values of the ratios, to which a series of rational numbers may
successively approximate.
Though two incommensurable magnitudes have not an assignable numerical ratio
to one another, yet they have a certain definite ratio to one another, and two other
magnitudes may have the same ratio as the first two: and it will be found, that, when
reference is made to the numerical value of the ratios of four incommensurable mag-
nitudes, that the same irrational number appears in the two ratios.
The sides and diagonals of squares can be shewn to be proportionals, and though
the ratio of the side to the diagonal is represented Geometrically by the two lines
which form the side and the diagonal, there is no rational number or fraction which
will measure exactly their ratio.
If the side of a square contain a units, the ratio of the diagonal to the side is
numerically as √2 to 1; and if the side of another square contain b units, the ratio
of the diagonal to the side will be found to be in the ratio of √2 to 1. Again, the
two parts of any number of lines which may be divided in extreme and mean ratio
will be found to be respectively in the ratio of the irrational number √5-1 to
A
1
11-2
164
EUCLID'S ELEMENTS.
:
3-√5. Also, the ratios of the diagonals of cubes to the diagonals of one of the faces
will be found to be in the irrational or incommensurate ratio of √3 to √√/2.
Thus it will be found that the ratios of all incommensurable magnitudes which are
proportionals do involve the same irrational numbers, and these may be used as the
numerical measures of ratios in the same manner as rational numbers and fractions.
Def. VII is analogous to Def. v, and lays down the criterion whereby the ratio of
two magnitudes of the same kind may be known to be greater or less than the ratio of
two other magnitudes of the same kind.
Def. XI includes Def. x, as three magnitudes may be continual proportionals, as
well as four or more than four. In continual proportionals, all the terms except the
first and last, are made successively the consequent of one ratio, and the antecedent of
the next; whereas in other proportionals, this is not the case.
A series of numbers or Algebraical quantities in continued proportion, is called a
Geometrical progression, from the analogy they bear to a series of Geometrical magni-
tudes in continued proportion.
Def. A.
The term compound ratio was devised for the purpose of avoiding cir-
cumlocution, and no difficulty can arise in the use of it, if its exact meaning be
strictly attended to.
DC to CE; and the ratio
With respect to the Geometrical measures of compound ratios, three straight lines
may measure the ratio of four, as in Prop. 23, Book VI. For K to L measures the
ratio of BC to CG, and L to M measures the ratio of
of K to M is that which is said to be compounded of the ratios of K to L, and L
to M, which is the same as the ratio which is compounded of the ratios of the sides
of the parallelograms.
Both duplicate and triplicate ratio are species of compound ratio.
Duplicate ratio is a ratio compounded of two equal ratios; and in the case of
three magnitudes which are continual proportionals, means the ratio of the first to a
third proportional to the first and second.
It will be hereafter seen, that the ratio of two triangles, when they have the same
altitude, are to one another as their bases; (Prop. 1, Book vi.) and, when their bases
are equal, they are to one another as their altitudes; but when neither their bases nor
altitudes are equal, they are to each other in a ratio compounded of their bases and
altitudes. And if the ratio of their altitudes be the same as the ratio of their bases,
or the triangles be similar, they are to each other in the duplicate ratio of their alti-
tudes or bases. (Prop. 19, Book VI.) In the same manner, the ratio of two squares
will be to each other in the ratio compounded of the ratios of their sides, and therefore
in the duplicate ratio of their sides. And generally, that all similar figures are in
the duplicate ratio of their homologous sides. (Prop. 20, Book v1.)
Triplicate ratio, in the same manner, is a ratio compounded of three equal ratios;
and in the case of four magnitudes which are continual proportionals, the triplicate ratio
of the first to the second means the ratio of the first to a fourth proportional to the first,
second, and third magnitudes. Instances of the composition of three ratios, and of
triplicate ratio will be found in the eleventh and twelfth books.
The product of the fractions which represent or measure the ratios of numbers,
corresponds to the composition of Geometrical ratios of magnitudes.
It has been shewn that the ratio of two numbers is represented by a fraction, whereof
the numerator is the antecedent, and the denominator the consequent of the ratio; and
if the antecedents of two ratios be multiplied together, as also the consequents, the new
ratio thus formed is said to be compounded of these two ratios; and in the same manner,
if there be more than two. It is also obvious, that the ratio compounded of two equal
ratios is equal to the ratio of the squares of one of the antecedents to its consequent ;
also when there are three equal ratios, the ratio compounded of the three ratios is equal
NOTES TO BOOK V.
165
to the ratio of the cubes of any one of the antecedents to its consequent. And further, it
may be observed, that when several numbers are continual proportionals, the ratio of
the first to the last is equal to the ratio of the product of all the antecedents to the
product of all the consequents.
It may be here remarked, that, though the constructions of the propositions in}
Book v are exhibited by straight lines, the enunciations are expressed of magnitude
in general, and are equally true of angles, triangles, parallelograms, arcs, sectors, &c.
Prop. c. This is frequently made use of by geometers, and is necessary to the
5th and 6th Propositions of the 10th Book. Clavius, in his notes subjoined to
the 8th def. of Book 5, demonstrates it only in numbers, by help of some of the
propositions of the 7th Book; in order to demonstrate the property contained in the
5th definition of the 5th Book, when applied to numbers, from the property of pro-
portionals contained in the 20th def. of the 7th Book: and most of the commentators
judge it difficult to prove that four magnitudes which are proportionals according to
the 20th def. of the 7th Book, are also proportionals according to the 5th def. of the
5th Book. But this is easily made out as follows:
First, if A, B, C, D, be four magnitudes, such that A is the same multiple, or
the same part of B, which C is of D:
Then A, B, C, D, are proportionals:
this is demonstrated in proposition (c).
Secondly, if AB contain the same parts of CD that EF does of GH ;
in this case likewise AB is to CD, as EF to GH.
F
B₁
H
D
K
L
A
C E
G
Let CK be a part of CD, and GL the same part of GH :
and let AB be the same multiple of CK, that EF is of GL :
therefore, by Prop. c, of Book v, AB is to CK, as EF to GL:
and CD, GH, are equimultiples of CK, GL, the second and fourth;
wherefore, by Cor. Prop. 4, Book v, AB is to CD, as EF to GH.
And if four magnitudes be proportionals according to the 5th def. of Book v,
they are also proportionals according to the 20th def. of Book VII.
First, if A be to B, as C to D;
then if A be any multiple or part of B, C is the same multiple or part of D,
by Prop. D, Book v.
Next, if AB be to CD, as EF to GH :
then if AB contains any part of CD, EF contains the same part of GH :
F
B₁
HI
D
K
L
A CE G
M
for let CK be a part of CD, and GL the same part of GH, and let AB be
a multiple of CK:
EF is the same multiple of GL:
take M the same multiple of GL that AB is of CK;
therefore, by Prop. c, Book v, AB is to CK, as M to GL:
166
EUCLID'S ELEMENTS.
!
and CD, GH, are equimultiples of CK, GL;
wherefore, by Cor. Prop. 4, Book v, AB is to CD, as M to GH.
And, by the hypothesis, AB is to CD, as EF to GH;
therefore M is equal to EF by Prop. 9, Book v.
and consequently, EF is the same multiple of GL that AB is of CK.
This is the method by which Simson shews that the Geometrical definition of pro-
portion is a consequence of the Arithmetical definition, and conversely.
a C
and taking ma, mc
ō = d'
It may however be shewn by employing the equation
any equimultiples of a and c the first and third, and nb, nd any equimultiples of b
and d the second and fourth.
Prop. XVIII, being the converse of Prop. XVII, has been demonstrated indirectly.
Let AE, EB, CF, FD be proportionals, that is, as AE to EB, so let CF be to FD.
Then these shall be proportionals also when taken jointly;
that is, as AB to BE, so shall CD be to DF.
B
D
E
F
A
C
For if the ratio of AB to BE be not the same as the ratio of CD to DF;
the ratio of AB to BE is either greater than, or less than the ratio of CD to DF.
First, let AB have to BE a greater ratio than CD has to DF;
and let DQ be taken so that AB has to BE the same ratio as CD to DQ.
And since magnitudes when taken jointly are proportionals,
are also proportionals when taken separately; (v. 17.)
therefore AE has to EB the same ratio as CQ to QD;
but, by the hypothesis, AE has to EB the same ratio as CF to FD;
therefore the ratio of CQ to QD is the same as the ratio of CF to FD. (v. 11.)
And when four magnitudes are proportionals, if the first be greater than the second,
the third is greater than the fourth; and if equal, equal; and if less, less; (v. 14.)
but CQ is less than CF,
therefore QD is less than FD; which is absurd.
Wherefore the ratio of AB to BE is not greater than the ratio of CD to DF;
that is, AB has the same ratio to BE as CD has to DF.
Secondly. By a similar mode of reasoning, it may likewise be shewn, that AB has
the same ratio to BE as CD has to DF, if AB be assumed to have to BE a less ratio
than CD has to DF.
For further information on the very important subject of Ratio and Proportion,
reference may be made to Dr. Barrow's Mathematical Lectures; Professor De Morgan's
Connexion of Number and Magnitude; and the fourth chapter of Professor Peacock's
Algebra.
BOOK VI.
DEFINITIONS.
I.
SIMILAR rectilineal figures are those which have their several angles
equal, each to each, and the sides about the equal angles proportionals.
II.
Reciprocal figures, viz. triangles and parallelograms, are such as
have their sides about two of their angles proportionals in such a
manner, that a side of the first figure is to a side of the other, as the
remaining side of this other is to the remaining side of the first."
III.
A straight line is said to be cut in extreme and mean ratio, when
the whole is to the greater segment, as the greater segment is to the
less.
IV.
The altitude of any figure is the straight line drawn from its vertex
perpendicular to the base.
168
EUCLID'S ELEMENTS.
PROPOSITION I. THEOREM.
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 or BD produced.
As the base BC is to the base CD, so shall the triangle ABC be to
the triangle ACD,
and the parallelogram EC to the parallelogram CF.
E A F

HGB C D K L
Produce BD both ways to the points H, L,
and take any number of straight lines BG, GH, each equal to the
base BC; (1. 3.)
and DK, KL, any number of them, each equal to the base CD;
and join AG, AH, AK, AL.
Then, because CB, BG, GH, are all equal,
the triangles AHG, AGB, ABC, are all equal: (1. 38.)
therefore, whatever multiple the base HC is of the base BC,
the same multiple is the triangle AHC of the triangle ABC:
for the same reason, whatever multiple the base LC is of the base CD,
the same multiple is the triangle ALC of the triangle ADC:
and if the base HC be equal to the base CL,
the triangle AHC is also equal to the triangle ALC: (1. 38.)
and if the base HC be greater than the base CL,
likewise the triangle AHC is greater than the triangle ALC;
and if less, less:
therefore, since there are four magnitudes,
viz. the two bases BC, CD, and the two triangles ABC, ACD;
and of the base BC, and the triangle ABC the first and third, any
equimultiples 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 equimultiples whatever,
viz. the base CL and the triangle ALC;
and since it has been shewn, that, if the base HC be greater than
the base CL,
the triangle AHC is greater than the triangle ALC;
and if equal, equal; and if less, less:
therefore, as the base BC is to the base CD, so is the triangle ABC
to the triangle ACD. (v. def. 5.)
And because the parallelogram CE is double of the triangle ABC, (1.41.)
and the parallelogram CF double of the triangle ACD,
and that magnitudes have the same ratio which their equimultiples
have; (v. 15.)
as the triangle ABC is to the triangle ACD, so is the parallelogram
EC to the parallelogram CF;
BOOK VI. PROP. I, II.
169
and because it has been shewn, that, as the base BC is to the base
CD, so is the triangle ABC to the triangle ACD;
and as the triangle ABC is to the triangle ACD, so is the paral-
lelogram EC to the parallelogram CF;
therefore, as the base BC is to the base CD, so is the parallelogram
EC to the parallelogram CF. (v. 11.)
Wherefore, triangles, &c. Q. E.D.
COR. From this it is plain, that triangles and parallelograms that
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 having drawn perpendiculars from the vertices of
the triangles to the bases, the straight line which joins the vertices is
parallel to that in which their bases are, (1. 33.) because the perpen-
diculars are both equal and parallel to one another. (1. 28.) Then, if
the same construction be made as in the proposition, the demonstration
will be the same.
PROPOSITION II. THEOREM.
If a straight line be drawn parallel to one of the sides of a triangle,
it shall cut the other sides, or these produced, proportionally: and conversely,
if the sides, or the sides produced, be cut proportionally, the straight line
which joins the points of section shall be parallel to the remaining side of
the triangle.
Let DE be drawn parallel to BC, one of the sides of the triangle ABC,
Then BD shall be to DA, as CE to EA.
A
A
E
D

D
E
B
C
B
D
E
B
C
Join BE, CD.
Then the triangle BDE is equal to the triangle CDE, (1.37.)
because they are on the same base DE, and between the same
parallels DE, BC:
but ADE is another triangle;
and equal magnitudes have the same ratio to the same magnitude; (v. 7.)
therefore, as the triangle BDE is to the triangle ADE, so is the
triangle CDE to the triangle ADE:
but as the triangle BDE to the triangle ADE, so is BD to DA, (vI. 1.)
because, having the same altitude, viz. the perpendicular drawn
from the point E to AB, they are to one another as their bases;
and for the same reason, as the triangle CDE to the triangle ADE,
so is CE to EA:
therefore, as BD to DA, so is CE to EA. (v. 11.)
Next, let the sides AB, AC of the triangle ABC, or these sides
produced, be cut proportionally in the points D, E, that is, so that BD
may be to DA as CE to EA, and join DE.
Then DE shall be parallel to BC.
170
EUCLID'S ELEMENTS.
The same construction being made,
because as BD to DA, so is CE to EA;
and as BD to DA, so is the triangle BDE to the triangle ADE; (v1.1.)
and as CE to EA, so is the triangle CDE to the triangle ADE;
·
therefore the triangle BDE is to the triangle ADE, as the triangle
CDE to the triangle ADE; (v. 11)
that is, the triangles BDE, CDE have the same ratio to the
triangle ADE:
therefore the triangle BDE is equal to the triangle CDE: (v. 9.)
and they are on the same base DE:
but equal triangles on the same base and on the same side of it, are
between the same parallels; (1. 39.)
therefore DE is parallel to BC.
Wherefore, if a straight line, &C. Q.E.D.
PROPOSITION III. THEOREM.
If the angle of a triangle be divided into two equal angles, by a straight
line which also cuts the base; the segments of the base shall have the same
ratio which the other sides of the triangle have to one another and con-
versely, if the segments of the base have the same ratio which the other
sides of the triangle have to one another; the straight line drawn from the
vertex to the point of section, divides the vertical angle into two equal angles.
Let ABC be a triangle, and let the angle BAC be divided into two
equal angles by the straight line AD.
Then BD shall be to DC, as BA to AC.
E

A
R
D
Through the point C draw CE parallel to DA, (1. 31.)
and let BA produced meet CE in E.
Because the straight line AC meets the parallels AD, EC,
the angle ACE is equal to the alternate angle CAD: (1. 29.)
but CAD, by the hypothesis, is equal to the angle BAD;
wherefore BAD is equal to the angle ACE. (ax. 1.)
Again, because the straight line BAE meets the parallels AD, EC,
the outward angle BÃD is equal to the inward and opposite angle
AEC: (1.29.)
but the angle ACE has been proved equal to the angle BAD;
therefore also ACE is equal to the angle AEC, (ax. 1.)
and consequently the side AE is equal to the side AC: (1.6.)
and because AD is drawn parallel to EC, one of the sides of the tri-
angle BCE,
therefore BD is to DC, as BA to AE: (v1.2.)
but AE is equal to AC;
therefore, as BD to DC, so is BA to AC. (v.7.)
Next, let BD be to DC, as BA to AC, and join AD.
Then the angle BAC shall be divided into two equal angles by the
straight line AD.
BOOK VI. PROP. III, A.
171
The same construction being made ;
because, as BD to DC, so is BA to AC;
and as BD to DC, so is BA to AE, because AD is parallel to EC; (v1.2.)
therefore BA is to AC, as BA to AE: (v. 11.)
consequently AC is equal to AE,
(v.9.)
and therefore the angle AEC is equal to the angle ACE: (1.5.)
but the angle AEC is equal to the outward and opposite angle BAD;
and the angle ACE is equal to the alternate angle CAD: (1. 29.)
wherefore also the angle BAD is equal to the angle CAD; `(ax. 1.)
that is, the angle BAC is cut into two equal angles by the straight
line AD.
Therefore, if the angle, &c. Q. E. D.
PROPOSITION A. THEOREM.
If the outward angle of a triangle made by producing one of its sides,
be divided into two equal angles, by a straight line, which also cuts the
base produced; the segments between the dividing line and the extremities
of the base, have the same ratio which the other sides of the triangle have
to one another: and conversely, if the segments of the base produced have
the same ratio which the other sides of the triangle have; the straight line
drawn from the vertex to the point of section divides the outward angle of
the triangle into two equal angles.
gles
Let ABC be a triangle, and let one of its sides BA be produced to E;
and let the outward angle CAE be divided into two equal angles
by the straight line AD which meets the base produced in D.
Then BD shall be to DC, as BA to AC.
E

B
C
D
Through C draw CF parallel to AD. (1.31.)
And because the straight line AC meets the parallels AD, FC,
the angle ACF is equal to the alternate angle CAD:
(1.29.)
but CAD is equal to the angle DÃE; (hyp.)
therefore also DAE is equal to the angle ACF. (ax. 1.)
Again, because the straight line FAE meets the parallels AD, FC,
the outward angle DÃE is equal to the inward and opposite angle
CFA: (1. 29.)
but the angle ACF has been proved equal to the angle DAE;
therefore also the angle ACF is equal to the angle CFA; (ax. 1.)
and consequently the side AF is equal to the side AC. (1.6.)
and because AD is parallel to FC, a side of the triangle BCF,
therefore BD is to DC, as BA to AF: (v1.2.)
but AF is equal to AC;
therefore, as BD is to DC, so is BA to AC. (v. 7.)
Next, let BD be to DC, as BA to AC, and join AD.
The angle CAD shall be equal to the angle DAE.
The same construction being made,
because BD is to DC, as BÃ to AC;
172
EUCLID'S ELEMENTS.
and that BD is also to DC, as BA to AF; (v1. 2.)
therefore BA is to AC, as BA to AF: (v. 11.)
wherefore AC is equal to AF, (v. 9.)
and the angle AFC equal to the angle ACF: (1.5.)
but the angle AFC is equal to the outward angle EAD, (1.29.)
and the angle ACF to the alternate angle CAD;
therefore also EAD is equal to the angle CAD. (ax. 1.)
Wherefore, if the outward, &c. Q. E.D.
PROPOSITION IV. THEOREM.
The sides about the equal angles of equiangular triangles are propor-
tionals; and those which are opposite to the equal angles are homolo-
gous sides, that is, are the antecedents or consequents of the ratios.
Let ABC, DCE be equiangular triangles, having the angle ABC
equal to the angle DCE, and the angle ACB to the angle DEC; and
consequently the angle BAC equal to the angle CDE. (1.32.)
The sides about the equal angles of the triangles ABC, DCE shall
be proportionals;
and those shall be the homologous sides which are opposite to the
equal angles.
*
F
D
A
B
CE
Let the triangle DCE be placed, so that its side CE may be contigu-
ous to BC, and in the same straight line with it. (1.22.)
Then, because the angle BCA is equal to the angle CED,
add to each the angle ABC;
(hyp.)
therefore the two angles ABC, BCA are equal to the two angles
ABC, CED: (ax. 2.)
but the angles ABC, BCA are together less than two right angles; (1.17.)
therefore the angles ABC, CED are also less than two right angles :
wherefore BA, ED if produced will meet: (1. ax. 12.)
let them be produced and meet in the point F:
then because the angle ABC is equal to the angle DCE, (hyp.)
BF is parallel to CD; (1. 28.)
and because the angle ACB is equal to the angle DEC,
AC is parallel to FE: (1.28.)
therefore FACD is a parallelogram ;
and consequently AF is equal to CD, and AC to FD: (1. 34.)
and because AC is parallel to FE, one of the sides of the triangle FBE,
BA is to AF, as BC to CE: (v1.2.)
but AF is equal to CD;
therefore, as BA to CD, so is BC to CE:__ (v. 7.)
and alternately, as AB to BC, so is DC to CE; (v. 16.)
again, because CD is parallel to BF,
as BC to CE, so is FD to DE: (vi. 2.)
but FD is equal to AC;
therefore, as BC to CE, so is AC to DE; (v. 7.)
BOOK VI. PROP. IV, V.
173
and alternately, as BC to CA, so CE to ED: (v. 16.)
therefore, because it has been proved that AB is to BC, as DC to CE,
and as BC to CA, so CE to ED,
ex æquali, BA is to AC, as CD to DE. (v. 22.)
Therefore the sides, &c.
Q.E.D.
PROPOSITION V. THEOREM.
If the sides of two triangles, about each of their angles, be proportionals,
the triangles shall be equiangular; and the equal angles shall be those
which are opposite to the homologous sides.
Let the triangles ABC, DEF have their sides proportionals,
so that AB is to BC, as DE to EF;
and BC to CA, as EF to FD;
and consequently, ex æquali, BA to AC, as ED to DF.
Then the triangle ABC shall be equiangular to the triangle DEF,
and the angles which are opposite to the homologous sides shall be equal,
viz. the angle ABC equal to the angle DEF, and BCA to EFD, and also
BAC to EDF.
A
D
E
F
B
C
G
At the points E, F, in the straight line EF, make the angle FEG
equal to the angle ABC, and the angle EFG equal to BCA: (1. 23.)
wherefore the remaining angle EGF, is equal to the remaining
angle BAC, (1. 32.)
and the triangle GEF is therefore equiangular to the triangle ABC:
- consequently they have their sides opposite to the equal angles
proportionals: (VI. 4.)
wherefore, as AB to BC, so is GE to EF
but as AB to BC, so is DE to EF; (hyp.)
therefore as DE to EF, so GE to EF; (v. 11.)
that is, DE and GE have the same ratio to EF,
and consequently are equal; (v. 9.)
for the same reason, DF is equal to FG:
and because, in the triangles DEF, GEF, DE is equal to EG, and
EF is common,
the two sides DE, EF are equal to the two GE, EF, each to each;
and the base DF is equal to the base GF;
therefore the angle DEF is equal to the angle GEF, (1.8.)
and the other angles to the other angles which are subtended by the
equal sides; (1. 4.)
therefore the angle DFE is equal to the angle GFE, and EDF to EGF:
and because the angle DEF is equal to the angle GEF,
and GEF equal to the angle ABC; (constr.)
therefore the angle ABC is equal to the angle DEF: (ax. 1.)
for the same reason, the angle ACB is equal to the angle DFE,
and the angle at A equal to the angle at D:
therefore the triangle ABC is equiangular to the triangle DEF.
Wherefore, if the sides, &c. Q. E.D.
174
EUCLID'S ELEMENTS.
PROPOSITION VI. THEOREM.
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 triangles shall
be equiangular, and shall have those angles equal which are opposite to the
homologous sides.
Let the triangles ABC, DEF have the angle BAC in the one equal
to the angle EDF in the other, and the sides about those angles pro-
portionals;
that is, BA to AC, as ED to DF.
Then the triangles ABC, DEF shall be equiangular, and shall have
the angle ABC equal to the angle DEF, and ACB to DFE.
A
D
G
E
F
B
At the points D, F, in the straight line DF, make the angle FDG
equal to either of the angles BAC, EDF; (1. 23.)
and the angle DFG equal to the angle ACB:
wherefore the remaining angle at B is equal to the remaining angle
at G: (1. 32.)
and consequently the triangle DGF is equiangular to the triangle ABC;
therefore as BA to AC, so is GD to DF: (VI. 4.).
but, by the hypothesis, as BA to AC, so is ED to DF;
therefore as ED to DF, so is GD to DF; (v. 11.)
wherefore ED is equal to DG; (v. 9.)
and DF is common to the two triangles EDF, GDF :
therefore the two sides ED, DF are equal to the two sides GD, DF,
each to each;
and the angle EDF is equal to the angle GDF; (constr.)
wherefore the base EF is equal to the base FG, (1. 4.)
and the triangle EDF to the triangle GDF,
and the remaining angles to the remaining angles, each to each,
which are subtended by the equal sides:
therefore the angle DFG is equal to the angle DFE,
and the angle at G to the angle at E;
but the angle DFG is equal to the angle ACB; (constr.)
therefore the angle ACB is equal to the angle DFE; (ax. 1.)
and the angle BAC is equal to the angle EDF: (hyp.)
wherefore also the remaining angle at B is equal to the remaining
angle at E; (1. 32.)
therefore the triangle ABC is equiangular to the triangle DEF.
Wherefore, if two triangles, &c. Q. E.D.
PROPOSITION VII.
THEOREM.
If two triangles have one angle of the one equal to one angle of the
other, and the sides about two other angles proportionals; then, if each of
the remaining angles be either less, or not less, than a right angle, or if
BOOK VI. PROP. VII.
175
one of them be a right angle; the triangles shall be equiangular, and shall
have those angles equal about which the sides are proportionals.
Let the two triangles ABC, DEF have one angle in the one equal to
one angle in the other,
viz. the angle BAC to the angle EDF, and the sides about two
other angles ABC, DEF proportionals,
so that AB is to BC, as DE to EF ;
and in the first case, let each of the remaining angles at C, F be
less than a right angle.
The triangle ABC shall be equiangular to the triangle DEF,
viz. the angle ABC shall be equal to the angle DEF,
and the remaining angle at C equal to the remaining angle at F.
A
G
D
E
F
B
C
For if the angles ABC, DEF be not equal,
one of them must be greater than the other:
let ABC be the greater,
and at the point B, in the straight line AB,
make the angle ABG equal to the angle DEF; (1. 23.)
and because the angle at A is equal to the angle at D, (hyp.)
and the angle ABG to the angle DEF;
the remaining angle AGB is equal to the remaining angle DFE:
(1. 32.)
therefore the triangle ABG is equiangular to the triangle DEF:
wherefore as AB is to BG, so is DE to EF: (vI. 4.)
but as DE to EF, so, by hypothesis, is AB to BC;
therefore as AB to BC, so is AB to BG: (v. 11.)
and because AB has the same ratio to each of the lines BC, BG,
BC is equal to BG; (v.9.)
and therefore the angle BGC is equal to the angle BCG: (1.5.)
but the angle BCG is, by hypothesis, less than a right angle;
therefore also the angle BGC is less than a right angle;
and therefore the adjacent angle AGB must be greater than a right
angle; (1. 13.)
but it was proved that the angle AGB is equal to the angle at F;
therefore the angle at F is greater than a right angle:
but, by the hypothesis, it is less than a right angle; which is absurd.
Therefore the angles ABC, DEF are not unequal,
that is, they are equal:
and the angle at A is equal to the angle at D:
(hyp.)
wherefore the remaining angle at C is equal to the remaining angle
at F: (1. 32.)
therefore the triangle ABC is equiangular to the triangle DEF.
Next, let each of the angles at C, F be not less than a right angle.
Then the triangle ABC shall also in this case be equiangular to the
triangle DEF.
A
A
B
G
E F
D
176
EUCLID'S ELEMENTS.
it
may be
The same construction being made,
be proved in like manner that BC is equal to BG,
and therefore the angle at C equal to the angle BGC:
but the angle at C is not less than a right angle; (hyp.)
therefore the angle BGC is not less than a right angle:
wherefore two angles of the triangle BGC are together not less than
two right angles:
which is impossible; (1. 17.)
and therefore the triangle ABC may be proved to be equiangular to
the triangle DEF, as in the first case.
Lastly, let one of the angles at C, F, viz. the angle at C, be a right angle:
in this case likewise the triangle ABC shall be equiangular to the
triangle DEF.
A
A
G
B
B C
G
D
E
F
For, if they be not equiangular,
at the point B in the straight line ÂB make the angle ABG equal
to the angle DEF;
then it may be proved, as in the first case, that BG is equal to BC:
and therefore the angle BCG equal to the angle BGC: (1. 5.)
but the angle BCG is a right angle, (hyp.)
therefore the angle BGC is also a right angle; (ax. 1.)
whence two of the angles of the triangle BGC are together not less
than two right angles;
which is impossible: (1.17.)
therefore the triangle ABC is equiangular to the triangle DEF.
Wherefore, if two triangles, &c. Q.E.D.
PROPOSITION VIII. THEOREM.
In a right-angled triangle, if a perpendicular be drawn from the right
angle to the base; the triangles on each side of it are similar to the whole
triangle, and to one another.
Let ABC be a right-angled triangle, having the right angle BAC;
and from the point A let AD be drawn perpendicular to the base BC.
Then the triangles ABD, ADC shall be similar to the whole triangle
ABC, and to one another.
A
B
D C
Because the angle BAC is equal to the angle ADB, each of them
being a right angle, (ax. 11.)
and that the angle at B is common to the two triangles ABC, ABD;
the remaining angle ACB is equal to the remaining angle BAD:
(1. 32.)
BOOK VI.. PROP. VIII, IX, X.
177
therefore the triangle ABC is equiangular to the triangle ABD,
and the sides about their equal angles are proportionals; (v1. 4.)
wherefore the triangles are similar: (vi. def. 1.)
in the like manner it may be demonstrated, that the triangle ADC
is equiangular and similar to the triangle ABC.
And the triangles ABD, ACD, being both equiangular and similar
to ABC, are equiangular and similar to each other.
Therefore, in a right-angled, &c. Q. E.D.
COR. From this it is manifest, that the perpendicular drawn from
the right angle of a right-angled triangle to the base, is a mean propor-
tional between the segments of the base; and also that each of the sides
is a mean proportional between the base, and the segment of it adjacent
to that side: because in the triangles BDA, ADC; BD is to DA,
as DA to DC; (v1. 4.)
and in the triangles ABC, DBA; BC is to BA, as BA to BD; (v1.4.)
and in the triangles ABC, ACD; BC is to CA, as CA to CD. (v1.4.)
PROPOSITION IX, PROBLEM.
From a given straight line to cut off any part required.
Let AB be the given straight line.
It is required to cut off any part from it.
D
A
E
B
From the point A draw a straight line AC, making any angle with AB;
and in AC take any point D,
and take AC the same multiple of AD, that AB is of the part which
is to be cut off from it;
join BC, and draw DE parallel to CB.
Then AE shall be the part required to be cut off.
Because ED is parallel to BC, one of the sides of the triangle ABC,
as CD is to DA, so is BE to EA; (vI. 2.)
and by composition, CA is to AD, as BA to AE:
but CA is a multiple of AD; (constr.)
(v. 18.)
therefore BA is the same multiple of AE: (v. D.)
whatever part therefore AD is of AC, AE is the same part of AB:
wherefore, from the straight line AB the part required is cut off. Q.E.F.
PROPOSITION X. PROBLEM.
To divide a given straight line similarly to a given divided straight
line, that is, into parts that shall have the same ratios to one another which
the parts of the divided given straight line have.
12
178
EUCLID'S ELEMENTS.
Let AB be the straight line given to be divided, and AC the divided line.
It is required to divide AB similarly to AC.
A
F
G
H
E
B K
Let AC be divided in the points D, E;
and let AB, AC be placed so as to contain any angle, and join BC,
and through the points D, E draw DF, EG parallels to BC. (1.31.)
Then AB shall be divided in the points F, G similarly to AC.
Through D draw DHK parallel to AB:
therefore each of the figures, FH, HB is a parallelogram;
wherefore DH is equal to FG, and HK to GB: (1.34.)
and because HE is parallel to KC, one of the sides of the triangle DKC,
as CE to ED, so is KH to HD: (v1.2.)
but KH is equal to BG, and HD to GF;
therefore, as CE to ED, so is BG to GF: (v. 7.)
again, because FD is parallel to GE, one of the sides of the triangle AGE,
as ED to DA, so is GF to FA: (vi. 2.)
therefore, as has been proved, CE is to ED, so is BG to GF,
and as ED to DA, so GF to FA:
therefore the given straight line AB is divided similarly to AC. Q.E.F.


PROPOSITION XI. PROBLEM.
To find a third proportional to two given straight lines.
Let AB, AC be the two given straight lines.
It is required to find a third proportional to AB, AC.
A
B
C
D
E
Let AB, AC be placed so as to contain any angle:
produce AB, AC to the points D, E;
and make BD equal to AC;
join BC, and through D, draw DE parallel to BC. (1. 31.)
Then CE shall be a third proportional to AB and AC.
Because BC is parallel to DE, a side of the triangle ADE,
AB is to BD, as AC to CE: (VI. 2.)
but BD is equal to AC;
therefore as AB is to AC, so is AC to CE. (v. 7.)
Wherefore, to the two given straight lines AB, AC, a third propor-
tional CE is found. Q.E.F.
BOOK VI. PROP. XII, XIII.
179
PROPOSITION XII. PROBLEM.
To find a fourth proportional to three given straight lines.
Let A, B, C be the three given straight lines.
It is required to find a fourth proportional to A, B, C.
Take two straight lines DE, DF, containing any angle EDF:
and upon these make DG equal to A, GE equal to B, and DH
equal to C; (1.3.)
G
E
D
A
B
C
H
F
join GH, and through E draw EF parallel to it. (1. 31.)
Then HF shall be the fourth proportional to A, B, C.
Because GH is parallel to EF, one of the sides of the triangle DEF,
DG is to GE, as DH to HF; (v1.2.)
but DG is equal to A, GE to B, and DH to C;
therefore, as A is to B, so is C to HF. (v. 7.)
Wherefore to the three given straight lines A, B, C, a fourth
portional HF is found. Q.E.F.
pro-
PROPOSITION XIII. PROBLEM.
To find a mean proportional between two given straight lines.
Let AB, BC be the two given straight lines.
It is required to find a mean proportional between them.
D
•
A
B
Place AB, BC in a straight line, and upon AC describe the semi-
circle ADC,
and from the point B draw BD at right angles to AC. (1.11.)
Then BD shall be a mean proportional between AB and BC.
Join AD, DC.
And because the angle ADC in a semicircle is a right angle, (III. 31.)
and because in the right-angled triangle ADC, BD is drawn from
the right angle perpendicular to the base,
DB is a mean proportional between AB, BC the segments of the
base: (VI. 8. Cor.)
therefore between the two given straight lines AB, BC, a mean
proportional DB is found. Q.E.F.
12-2
180
EUCLID'S ELEMENTS.
PROPOSITION XIV. THEOREM.
Equal parallelograms, which have one angle of the one equal to one
angle of the other, have their sides about the equal angles reciprocally pro-
portional: and conversely, parallelograms that have one angle of the one
equal to one angle of the other, and their sides about the equal angles reci-
procally proportional, are equal to one another.
Let AB, BC be equal parallelograms, which have the angles at B
equal.
The sides of the parallelograms AB, BC about the equal angles,
shall be reciprocally proportional
that is, DB shall be to BE, as GB to BF.
A
F

E
D
B
G C
Let the sides DB, BE be placed in the same straight line;
wherefore also FB, BG are in one straight line: (1. 14.)
complete the parallelogram FE.
And because the parallelogram AB is equal to BC, and that FE is
another parallelogram,
AB is to FE, as BC to FE: (v. 7.)
but as AB to FE, so is the base DB to BE, (vi. 1.)
and as BC to FE, so is the base GB to BF;
therefore, as DB to BE, so is GB to BF. (v. 11.)
Wherefore, the sides of the parallelograms AB, BC about their
equal angles are reciprocally proportional.
Next, let the sides about the equal angles be reciprocally propor-
tional,
viz. as DB to BE, so GB to BF:
the parallelogram AB shall be equal to the parallelogram BC.
Because, as DB to BÊ, so is GB to BF ;
and as DB to BE, so is the parallelogram AB to the parallelogram
FE; (vI. 1.)
and as GB to BF, so is the parallelogram BC to the parallelogram FE;
therefore as AB to FE, so BC to FE: (v. 11.)
therefore the parallelogram AB is equal to the parallelogram BC. (v.9.)
Therefore equal parallelograms, &c. Q. E. D.
PROPOSITION XV. THEOREM.
Equal triangles which have one angle of the one equal to one angle of
the other, have their sides about the equal angles reciprocally proportional:
and conversely, triangles which have one angle in the one equal to one
angle in the other, and their sides about the equal angles reciprocally pro-
portional, are equal to one another.
Let ABC, ADE be equal triangles, which have the angle BAC equal
to the angle DAE.
BOOK VI. PROP. XV, XVI.
181
Then the sides about the equal angles of the triangles shall be
reciprocally proportional;
that is, CA shall be to AD, as EA to AB.
B
D
A
C
E
Let the triangles be placed so that their sides CA, AD be in one
straight line;
wherefore also EA and AB are in one straight line; (1. 14.)
and join BD.
Because the triangle ABC is equal to the triangle ADE,
and that ABD is another triangle ;
therefore as the triangle CAB, is to the triangle BAD, so is the
triangle AED to the triangle DAB; (v. 7.)
but as the triangle CAB to the triangle BAD, so is the base CA to
the base AD, (vI. 1.)
and as the triangle EAD to the triangle DAB, so is the base EA to
the base AB; (vI. 1.)
therefore as CA to AD, so is EA to AB: (v. 11.)
wherefore the sides of the triangles ABC, ADE, about the equal
angles are reciprocally proportional.
Next, let the sides of the triangles ABC, ADE about the equal
angles be reciprocally proportional,
viz. CA to AD, as EA to AB.
Then the triangle ABC shall be equal to the triangle ADE.
Join BD as before.
Then because, as CA to AD, so is EA to AB; (hyp.)
and as CA to AD, so is the triangle ABC to the triangle BAD; (vi. 1.)
and as EA to AB, so is the triangle EAD to the triangle BAD; (vi. 1.)
therefore as the triangle BẮC to the triangle BAD, so is the tri-
angle EAD to the triangle BAD; (v. 11.)
that is, the triangles BAC, EAD have the same ratio to the triangle
BAD:
wherefore the triangle ABC is equal to the triangle ADE. (v. 9.)
Therefore, equal triangles, &c.
Q. E.D.
PROPOSITION XVI. THEOREM.
If four straight lines be proportionals, the rectangle contained by the
extremes is equal to the rectangle contained by the means: and conversely,
if the rectangle contained by the extremes be equal to the rectangle con-
tained by the means, the four straight lines are proportionals.
Let the four straight lines AB, CD, E, F be proportionals,
viz. as AB to CD, so E to F.
The rectangle contained by AB, F, shall be equal to the rectangle
contained by CD, E.
182
EUCLID'S ELEMENTS.
E.
G
A
B
H
C D
From the points A, C draw AG, CH at right angles to AB, CD; (1.11.)
and make AG equal to F, and CH equal to E: (1. 3.)
and complete the parallelograms BG, DH. (1. 31.)
Because, as AB to CD, so is E to F;
and that E is equal to CH, and F to AG;
AB is to CD as CH to AG: (v.7.)
therefore the sides of the parallelograms BG, DH about the equal
angles are reciprocally proportional;
but parallelograms which have their sides about equal angles
reciprocally proportional, are equal to one another; (v1. 14.)
therefore the parallelogram BG is equal to the parallelogram DH:
but the parallelogram BG is contained by the straight lines AB, F;
because AG is equal to F;
and the parallelogram DH is contained by CD and E;
because CH is equal to E;
therefore the rectangle contained by the straight lines AB, F, is
equal to that which is contained by CD and E.
And if the rectangle contained by the straight lines AB, F be equal
to that which is contained by CD, E;
these four lines shall be proportional,
viz, AB shall be to CD, as E to F.
The same construction being made,
because the rectangle contained by the straight lines AB, F, is
equal to that which is contained by CD, E,
and that the rectangle BG is contained by AB, F;
because AG is equal to F;
and the rectangle DH by CD, E; because CH is equal to E;
therefore the parallelogram BG is equal to the parallelogram DH; (ax. 1.)
and they are equiangular :
but the sides about the equal angles of equal parallelograms are
reciprocally proportional: (vI. 14.)
wherefore, as AB to CD, so is CH to AG:
but CH is equal to E, and AG to F;
therefore as AB is to CD, so is E to F. (v. 7.)
Wherefore if four, &c. Q.E.D.

PROPOSITION XVII. THEOREM.
If three straight lines be proportionals, the rectangle contained by
the extremes is equal to the square of the mean; and conversely, if the
rectangle contained by the extremes be equal to the square of the mean,
the three straight lines are proportionals.
Let the three straight lines A, B, C be proportionals,
viz. as A to B, so B to C.
The rectangle contained by A, C shall be equal to the square of B.
BOOK VI. PROP. XVII, XVIII.
183
A
B
D—
C
A
C
B
D
Take D equal to B.
And because as A to B, so B to C, and that B is equal to D;
A is to B, as D to C: (v.7.)
but if four straight lines be proportionals, the rectangle contained
by the extremes is equal to that which is contained by the means ;
(v1. 16.)
therefore the rectangle contained by A, C is equal to that contained
by B, D:
but the rectangle contained by B, D, is the square of B,
because B is equal to D;
therefore the rectangle contained by A, C, is equal to the square of B.
And if the rectangle contained by A, C, be equal to the square of B.
Then A shall be to B as B to C.
The same construction being made,
because the rectangle contained by A, C is equal to the square of B,
and the square of B is equal to the rectangle contained by B, D,
because B is equal to D;
therefore the rectangle contained by A, C, is equal to that contained
by B, D:
but if the rectangle contained by the extremes be equal to that
contained by the means, the four straight lines are proportionals :
(vi. 16.)
therefore A is to B, as D to C:
but B is equal to D;
wherefore, as A to B, so B to C.
Therefore, if three straight lines, &c. Q.E.D.
PROPOSITION XVIII. PROBLEM.
Upon a given straight line to describe a rectilineal figure similar, and
similarly situated, to a given rectilineal figure.
Let AB be the given straight line, and CDEF the given rectilineal
figure of four sides.
It is required upon the given straight line AB to describe a rectilineal
figure similar, and similarly situated, to CDEF.
H
G
FE
L
K
A
B
C D
Join DF, and at the points A, B in the straight line AB make the
angle BAG equal to the angle at C, (1.23.)
and the angle ABG equal to the angle CDF;
therefore the remaining angle AGB is equal to the remaining angle
CFD: (1. 32. and ax. 3.)
184
“EUCLID'S ELEMENTS.
therefore the triangle FCD is equiangular to the triangle GAB:
again, at the points G, B, in the straight line GB, make the angle
BGH equal to the angle DFE, (1.23.)
and the angle GBH equal to FDE;
therefore the remaining angle GHB is equal to the remaining
angle FED,
and the triangle FDE equiangular to the triangle GBH:
then, because the angle AGB is equal to the angle CFD, and BGH
to DFE,
the whole angle AGH is equal to the whole CFE; (ax. 2.)
for the same reason, the angle ABH is equal to the angle CDE:
also the angle at A is equal to the angle at C, (constr.)
and the angle GHB to FED:
therefore the rectilineal figure ABHG is equiangular to CDEF:
likewise these figures have their sides about the equal angles pro-
portionals;
because the triangles GAB, FCD being equiangular,
BA is to AG, as DC to CF; (vI. 4.)
and because AG is to GB, as CF to FD;
and as GB is to GH, so is FD to FE,
by reason of the equiangular triangles BGH, DFE,
therefore, ex æquali, AG is to GH, as CF to FE. (v. 22.)
In the same manner it may be proved that AB is to BH, as CD to DE:
and GH is to HB, as FE to ED. (v1. 4.)
Wherefore, because the rectilineal figures ABHG, CDEF are equi-
angular,
and have their sides about the equal angles proportionals,
they are similar to one another. (vI. def. 1.)
Next, let it be required to describe upon a given straight line AB,
a rectilineal figure similar, and similarly situated, to the rectilineal figure
CDKEF of five sides.
Join DE, and upon the given straight line AB describe the rectili-
neal figure ABHG similar, and similarly situated, to the quadrilateral
figure CDEF, by the former case:
and at the points B, H, in the straight line BH, make the angle
HBL equal to the angle EDK,
and the angle BHL equal to the angle DEK;
therefore the remaining angle at L is equal to the remaining angle
at K. (1. 32. and ax. 3.)
And because the figures ABHG, CDEF are similar,
the angle GHB is equal to the angle FED: (VI. def. 1.)
and BHL is equal to DEK;
wherefore the whole angle GHL is equal to the whole angle FEK:
for the same reason the angle ABL is equal to the angle CDK:
therefore the five-sided figures AGHLB, CFEKD are equiangular :
and because the figures AGHB, CFED are similar,
GH is to HB, as FE to ED; (vI. def. 1.)
but as HB to HL, so is ED to EK; (v1.4.)
therefore, ex æquali, GH is to HL, as FE to ÈK: (v. 22.)
for the same reason, AB is to BL, as CD to DK:
and BL is to LH, as DK to KE, (vi. 4.)
because the triangles BLH, DKE are equiangular:
therefore because the five-sided figures AGHLB, CFEKD are
equiangular,
BOOK VI. PROP. XVIII, XIX, XX.
185
and have their sides about the equal angles proportionals,
they are similar to one another.
In the same manner a rectilineal figure of six sides may be described
upon a given straight line similar to one given, and so on. Q. E. F.
PROPOSITION XIX. THEOREM.
Similar triangles are to one another in the duplicate ratio of their
homologous sides.
Let ABC, DEF be similar triangles, having the angle B equal to
the angle E,
and let AB be to BC, as DE to EF,
so that the side BC may be homologous to EF. (v. def. 12.)
Then the triangle ABC shall have to the triangle DEF the duplicate
ratio of that which BC has to EF.
A
D
F
B G C
Take BG a third proportional to BC, EF, (v1. 11.)
so that BC may be to EF, as EF to BG, and join GA.
Then, because, as AB to BC, so DE to EF;
alternately, AB is to DE, as BC to EF; (v. 16.)
but as BC to EF, so is EF to BG; (constr.)
therefore, as AB to DE, so is EF to BG: (v. 11.)
therefore the sides of the triangles ABG, DEF, which are about
the equal angles, are reciprocally proportional:
but triangles, which have the sides about two equal angles recipro-
cally proportional, are equal to one another; (vI. 15.)
therefore the triangle ABG is equal to the triangle DÉF:
and because as BC is to EF, so EF to BG;
and that if three straight lines be proportional, the first is said to
have to the third the duplicate ratio of that which it has to the second;
(v. def. 10.)
therefore BC has to BG the duplicate ratio of that which BC has to EF:
but as BC to BG, so is the triangle ABC to the triangle ABG; (v1.1.)
therefore the triangle ABC has to the triangle ABG the duplicate
ratio of that which BC has to EF:
but the triangle ABG is equal to the triangle DEF;
therefore also the triangle ABC has to the triangle DEF the dupli-
cate ratio of that which BC has to EF.
COR.
Therefore similar triangles, &c.
Q.E.D.
From this it is manifest, that if three straight lines be pro-
portionals, as the first is to the third, so is any triangle upon the first to
a similar and similarly described triangle upon the second.
PROPOSITION XX. THEOREM.
Similar polygons may be divided into the same number of similar tri-
angles, having the same ratio to one another that the polygons have; and
the polygons have to one another the duplicate ratio of that which their
homologous sides have.
186
EUCLID'S ELEMENTS.
Let ABCDE, FGHKL be similar polygons, and let AB be the
side homologous to FG:
the polygons ABCDE, FGHKL may be divided into the same
number of similar triangles, whereof each shall have to each the same
ratio which the polygons have ;
and the polygon ABCDE shall have to the polygon FGHKL the
duplicate ratio of that which the side AB has to the side FG.
A
E
F
B
M
L
G
K H
D C
Join BE, EC, GL, LH.
And because the polygon ABCDE is similar to the polygon FGHKL,
the angle BAE is equal to the angle GFL, (vi. def. 1.)
and BA is to AE, as GF to FL: (v1. def. 1.)
therefore, because the triangles ABE, FGL have an angle in one
equal to an angle in the other, and their sides about these equal angles
p oportionals,
the triangle ABE is equiangular to the triangle FGL; (v1.6.)
and therefore similar to it; (VI. 4.)
wherefore the angle ABE is equal to the angle FGL:
and, because the polygons are similar,
the whole angle ABC is equal to the whole angle FGH; (vi. def. 1.)
therefore the remaining angle EBC is equal to the remaining angle
LGH: (1. 32. and ax. 3.)
and because the triangles ABE, FGL are similar,
EB is to BA, as LG to GF; (VI. 4.)
and also, because the polygons are similar,
AB is to BC, as FG to GH; (v1. def. 1.)
therefore, ex æquali, EB is to BC, as LG to GH; (v.22.)
that is, the sides about the equal angles EBC, LGH are proportionals;
therefore, the triangle EBC is equiangular to the triangle LGH,
(VI. 6.) and similar to it; (v1. 4.)
for the same reason, the triangle ECD likewise is similar to the
triangle LHK:
therefore the similar polygons ABCDE, FGHKL are divided into
the same number of similar triangles.
Also these triangles shall have, each to each, the same ratio which
the polygons have to one another,
the antecedents being ABE, EBC, ECD, and the consequents FGL,
LGH, LHK:
and the polygon ABCDE shall have to the polygon FGHKL
the duplicate ratio of that which the side AB has to the homologous
side FG.
Because the triangle ABE is similar to the triangle FGL,
ABE has to FGL, the duplicate ratio of that which the side BE
has to the side GL: (v1. 19.)
for the same reason, the triangle BEC has to GLH the duplicate
ratio of that which BE has to GL:
therefore, as the triangle ABE is to the triangle FGL, so is the
triangle BEC to the triangle GLH. (v. 11.)
BOOK VI. PROP. XX, XXI.
187
Again, because the triangle EBC is similar to the triangle LGH,
EBC has to LGH, the duplicate ratio of that which the side EC
has to the side LH:
for the same reason, the triangle ECD has to the triangle LHK,
the duplicate ratio of that which EC has to LH:
therefore, as the triangle EBC to the triangle LGH, so is the
triangle ECD to the triangle LHK: (v. 11.)
but it has been proved,
that the triangle EBC is likewise to the triangle LGH, as the
triangle ABE to the triangle FGL:
therefore, as the triangle ABE to the triangle FGL, so is the tri-
angle EBC to the triangle LGH, and the triangle ECD to the triangle
LHK:
and therefore, as one of the antecedents is to one of the consequents,
so are all the antecedents to all the consequents: (v. 12.)
that is, as the triangle ABE to the triangle FGL, so is the polygon
ABCDE to the polygon FGHKL:
but the triangle ABE has to the triangle FGL, the duplicate ratio of
that which the side AB has to the homologous side FG; (v1. 19.)
therefore also the polygon ABCDE has to the polygon FGHKL the
duplicate ratio of that which AB has to the homologous side FG.
Wherefore similar polygons, &c.
Q.E.D.
COR. 1. In like manner it may be proved, that similar four-sided
figures, or of any number of sides, are one to another in the duplicate
ratio of their homologous sides: and it has already been proved in
triangles: (VI. 19.) therefore, universally, similar rectilineal figures are
to one another in the duplicate ratio of their homologous sides.
COR. 2. And if to AB, FG, two of the homologous sides, a third
proportional M be taken, (vi. 11.)
AB has to M the duplicate ratio of that which AB has to FG: (v. def. 10.)
but the four-sided figure or polygon upon AB, has to the four-
sided figure or polygon upon FG likewise the duplicate ratio of that
which AB has to FG; (1. Cor.)
therefore, as AB is to M, so is the figure upon AB to the figure
upon FG: (v. 11.)
which was also proved in triangles: (vI. 19. Cor.)
therefore, universally, it is manifest, that if three straight lines be
proportionals, as the first is to the third, so is any rectilineal figure upon
the first, to a similar and similarly described rectilineal figure upon the
second.
PROPOSITION XXI. THEOREM.
Rectilineal figures which are similar to the same rectilineal figure,
are also similar to one another.
Let each of the rectilineal figures A, B be similar to the rectilineal
figure C.
The figure A shall be similar to the figure B.
A
188
EUCLID'S ELEMENTS.
Because A is similar to C,
they are equiangular, and also have their sides about the equal
angles proportional: (vI. def. 1.)
again, because B is similar to C,
they are equiangular, and have their sides about the equal angles
proportionals: (VI. def. 1.)
therefore the figures A, B are each of them equiangular to C, and
have the sides about the equal angles of each of them and of C pro-
portionals.
Wherefore the rectilineal figures A and C are equiangular, (1. ax. 1.)
and have their sides about the equal angles proportionals: (v. 11.)
therefore A is similar to B. (vi. def. 1.)
Therefore rectilineal figures, &C. Q.E.D.
PROPOSITION XXII. THEOREM.
If four straight lines be proportionals, the similar rectilineal figures
similarly described upon them shall also be proportionals: and conversely,
if the similar rectilineal figures similarly described upon four straight
lines be proportionals, those straight lines shall be proportionals.
Let the four straight lines AB, CD, EF, GH be proportionals,
viz. AB to CD, as EF to GH;
and upon AB, CD let the similar rectilineal figures KAB, LCD
be similarly described;
and upon EF, GH the similar rectilineal figures MF, NH, in
like manner:
the rectilineal figure KAB shall be to LCD, as MF to NH.
K
L
X
A
B C
D
M
N
E
F
G H
S
PR
To AB, CD take a third proportional X; (vI. 11.)
and to EF, GH a third proportional 0:
and because AB is to CD as EF to GH,
therefore CD is to X, as GH to 0; (v. 11.)
wherefore, ex æquali, as AB to X, so EF to 0: (v. 22.)
but as AB to X, so is the rectilineal figure KAB to the rectilineal
figure LCD,
and as EF to O, so is the rectilineal figure MF to the rectilineal
figure NH: (v1. 20. Cor. 2.)
therefore, as KAB to LCD, so is MF to NH. (v. 11.)
And if the rectilineal figure KAB be to LCD, as MF to ÑH ;
the straight line AB shall be to CD, as EF to GH.
Make as AB to CD, so EF to PR, (vI. 12.)
and upon PR describe the rectilineal figure SR similar and simi-
larly situated to either of the figures MF, NH: (vI. 18.)
then, because as AB to CD, so is EF to PR,
and that upon AB, CD are described the similar and similarly
situated rectilineals KAB, LCD,
and upon EF, PR, in like manner, the similar rectilineals MF, SR;
therefore KAB is to LCD, as MF to SR:
but by the hypothesis KAB is to LCD, as MF to NH;
BOOK VI.
189
PROP. XXII, XXIII.
and therefore the rectilineal MF having the same ratio to each of
the two NH, SR,
these are equal to one another; (v. 9.)
they are also similar, and similarly situated;
therefore GH is equal to PR:
and because as AB to CD, so is EF to PR,
and that PR is equal to GH;
AB is to CD, as EF to GH. (v. 7.)
If, therefore, four straight lines, &c.
Q.E.D.
PROPOSITION XXIII. THEOREM.
Equiangular parallelograms have to one another the ratio which is
compounded of the ratios of their sides.
Let AC, CF be equiangular parallelograms, having the angle BCD
equal to the angle ECG.
Then the ratio of the parallelogram AC to the parallelogram CF,
shall be the same with the ratio which is compounded of the ratios of
their sides.
A
DH

G
B
KLM
E F
Let BC, CG be placed in a straight line;
therefore DC and CE are also in a straight line; (1. 14.)
and complete the parallelogram DG;
and taking any straight line K,
make as BC to CG, so K to L; (vi. 12.)
and as DC to CE, so make L to M: (vi. 12.)
therefore, the ratios of K to L, and L to M, are the same with the
ratios of the sides,
viz. of BC to CG, and DC to CE:
but the ratio of K to M is that which is said to be compounded of
the ratios of K to L, and L to M; (v. def. a.)
therefore K has to M the ratio compounded of the ratios of the sides :
and because as BC to CG, so is the parallelogram AC to the paral-
lelogram CH; (vI. 1.)
but as BC to CG, so is K to L;
therefore K is to L, as the parallelogram AC to the parallelogram
CH: (v. 11.)
again, because as DC to CE, so is the parallelogram CH to the
parallelogram CF;
but as DC to CE, so is L to M;
wherefore L is to M, as the parallelogram CH to the parallelogram
CF; (v.11.)
therefore since it has been proved,
that as K to L, so is the parallelogram AC to the parallelogram CH;
190
EUCLID'S ELEMENTS.
and as L to M, so is the parallelogram CH to the parallelogram CF;
ex æquali, K is to M, as the parallelogram AC to the parallelogram
CF: (v. 22.)
but K has to M` the ratio which is compounded of the ratios of the sides;
therefore also the parallelogram AC has to the parallelogram CF
the ratio which is compounded of the ratios of the sides.
Wherefore equiangular parallelograms, &c. Q. E.D.
PROPOSITION XXIV. THEOREM.
Parallelograms about the diameter of any parallelogram, are similar
to the whole, and to one another.
Let ABCD be a parallelogram, of which the diameter is AC;
and EG, HK parallelograms about the diameter.
The parallelograms EG, HK shall be similar both to the whole
parallelogram ABCD, and to one another.

A E
B
ה
G
H
D K
C
Because DC, GF are parallels,
the angle ADC is equal to the angle AGF: (1.29.)
for the same reason, because BC, EF are parallels,
the angle ABC is equal to the angle AEF:
and each of the angles BCD, EFG is equal to the opposite angle
DAB,
(1.34.)
and therefore they are equal to one another :
wherefore the parallelograms ABCD, AEFG, are equiangular:
and because the angle ABC is equal to the angle AEF,
and the angle BAC common to the two triangles BAC, EAF,
they are equiangular to one another;
therefore as AB to BC, so is AE to EF: (VI. 4.)
and because the opposite sides of parallelograms are equal to one
another, (1. 34.)
AB is to AD as AE to AG; (v. 7.)
and DC to CB, as GF to FE;
and also CD to DA, as FG to GA:
therefore the sides of the parallelograms ABCD, AEFG about the
equal angles are proportionals;
and they are therefore similar to one another; (vi. def. 1.)
for the same reason, the parallelogram ABCD is similar to the
parallelogram FHCK:
wherefore each of the parallelograms GE, KH is similar to DB:
but rectilineal figures which are similar to the same rectilineal
figure, are also similar to one another: (VI. 21.)
therefore the parallelogram GE is similar to KH.
Wherefore parallelograms, &c. (Q.E.D.)

{
BOOK VI. PROP. XXV, XXVI.
191
PROPOSITION XXV. PROBLEM.
To describe a rectilineal figure which shall be similar to one, and equal
to another given rectilineal figure.
Let ABC be the given rectilineal figure, to which the figure to be
described is required to be similar, and D that to which it must be equal.
It is required to describe a rectilineal figure similar to ABC, and
equal to D.
A

K
C
B
F
D
L
E M
G
H
;
Upon the straight line BC describe the parallelogram BE equal to
the figure ABC; (1. 45. Cor.)
also upon CE describe the parallelogram CM equal to D, (1. 45. Cor.)
and having the angle FCE equal to the angle CBL:
therefore BC and CF are in a straight line, as also LE and EM:
(1. 29. and 1. 14.)
between BC and CF find a mean proportional GH, (vi. 13.)
and upon GH describe the rectilineal figure KGH similar and
similarly situated to the figure ABC. (vi. 18.)
Because BC is to GH as GH to CF,
and that if three straight lines be proportionals, as the first is to the
third, so is the figure upon the first to the similar and similarly described
figure upon the second; (v1. 20. Cor. 2.)
therefore, as BC to CF, so is the rectilineal figure ABC to KGH:
but as BC to CF, so is the parallelogram BE to the parallelogram
EF; (vI. 1.)
therefore as the rectilineal figure ABC is to KGH, so is the
parallelogram BE to the parallelogram EF: (v. 11.)
and the rectilineal figure ABC is equal to the parallelogram BE; (constr.)
therefore the rectilineal figure KGH is equal to the parallelogram
EF: (v. 14.)
but EF is equal to the figure D; (constr.)
wherefore also KGH is equal to Ď: and it is similar to ABC.
Therefore the rectilineal figure KGH has been described similar to
the figure ABC, and equal to D. Q.E.F.

PROPOSITION XXVI. THEOREM.
If two similar parallelograms have a common angle, and be similarly
situated; they are about the same diameter.
Let the parallelograms ABCD, AEFG be similar and similarly
situated, and have the angle DAB common.
ABCD and AEFG shall be about the same diameter.
192
EUCLID'S ELEMENTS.
:
7
A G
D
K
E
F
B
D
For, if not, let, if possible, the parallelogram BD have its diameter
AHC in a different straight line from AF, the diameter of the paral-
lelogram EG,
and let GF meet AHC in H;
and through H draw HK parallel to AD or BC:
therefore the parallelograms ABCD, AKHG being about the same
diameter, they are similar to one another; (vI. 24.)
wherefore as DA to AB, so is GA to AK: (vi. def. 1.)
but because ABCD and AEFG are similar parallelograms,
as DA is to AB, so is GA to AE;
therefore as GA to AE, so GA to AK; (v. 11.)
(hyp.)
that is, GA has the same ratio to each of the straight lines AE, AK;
and consequently AK is equal to AE, (v. 9.)
the less equal to the greater, which is impossible:
therefore ABCD‍and AKHĞ are not about the same diameter :
wherefore ABCD and AEFG must be about the same diameter.
Therefore, if two similar, &c. Q. E. D.
'To understand the three following propositions more easily, it is
to be observed:
1. 'That a parallelogram is said to be applied to a straight line,
when it is described upon it as one of its sides. Ex. gr. the paral-
lelogram AC is said to be applied to the straight line AB.
2. But a parallelogram AE is said to be applied to a straight
line AB, deficient by a parallelogram, when AD the base of AE is less
than AB, and therefore AE is less than the parallelogram AC described
upon AB in the same angle, and between the same parallels, by the
parallelogram DC; and DC is therefore called the defect of AE.
E C
G
A
D B
F
3. And a parallelogram AG is said to be applied to a straight line
AB, exceeding by a parallelogram, when AF the base of AG is greater
than AB, and therefore AG exceeds AC the parallelogram described
upon AB in the same angle, and between the same parallels, by the
parallelogram BG.
PROPOSITION XXVII. THEOREM.
Of all parallelograms 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.
Let AB be a straight line divided into two equal parts in C;
and let the parallelogram AD be applied to the half AC, which is
therefore deficient from the parallelogram upon the whole line AB by
the parallelogram CE upon the other half CB:
BOOK VI. PROP. XXVII, XXVIII.
193
of all the parallelograms applied to any other parts of AB, and de-
ficient by parallelograms that are similar and similarly situated to CE,
AD shall be the greatest.
Let AF be any parallelogram applied to AK, any other part of AB
than the half, so as to be deficient from the parallelogram upon the
whole line AB by the parallelogram KH similar and similarly situated
to CE:
G
DLE
La
F
H
А СК В
AD shall be greater than AF.
First, let AK the base of AF, be greater than AC the half of AB:
and because CE is similar to the parallelogram HK, (hyp.)
they are about the same diameter: (VI. 26.)
draw their diameter DB, and complete the scheme:
then, because the parallelogram CF is equal to FE, (1. 43.)
add KH to both;
therefore the whole CH is equal to the whole KE:
but CH is equal to CG, (1. 36.)
because the base AC is equal to the base CB;
therefore CG is equal to KE: (ax. 1.)
to each of these equals add CF ;
then the whole AF is equal to the gnomon CHL: (ax. 2.)
therefore CE, or the parallelogram AD is greater than the paral-
lelogram AF.
Next, let AK the base of AF be less than AC:
G FM H
IND

E
A KC B
then, the same construction being made, because BC is equal to CA,
therefore HM is equal to MG; (1.34.)
therefore, the parallelogram DH is equal to the parallelogram DG;
(1. 36.)
wherefore DH is greater than LG:
but DH is equal to DK; (1. 43.)
therefore DK is greater than LG:
to each of these add AL;
then the whole AD is greater than the whole AF.
Therefore, of all parallelograms applied, &c. Q. E.D.
PROPOSITION XXVIII. PROBLEM.
To a given straight line to apply a parallelogram equal to a given
rectilineal figure, and deficient by a parallelogram similar to a given
parallelogram: but the given rectilineal figure to which the parallelogram
to be applied is to be equal, must not be greater than the parallelogram
J
13
194
EUCLID'S ELEMENTS.
applied to half of the given line, having its defect similar to the defect of
that which is to be applied; that is, to the given parallelogram.
Let AB be the given straight line, and C the given rectilineal figure,
to which the parallelogram to be applied is required to be equal, which
figure must not be greater (vi. 27.) than the parallelogram applied to the
half of the line, having its defect from that upon the whole line similar
to the defect of that which is to be applied;
and let D be the parallelogram to which this defect is required to
be similar.
It is required to apply a parallelogram to the straight line AB,
which shall be equal to the figure C, and be deficient from the paral-
lelogram upon the whole line by a parallelogram similar to D.

H G OF
X
FR
L M
C
D
K N
A E SB
Divide AB into two equal parts in the point E, (1. 10.)
and upon EB describe the parallelogram EBFG similar and
similarly situated to D, (vI. 18.)
and complete the parallelogram AG, which must either be equal to
C, or greater than it, by the determination.
If AG be equal to C, then what was required is already done:
for, upon the straight line AB, the parallelogram AG is applied
equal to the figure C, and deficient by the parallelogram EF
similar to D.
But, if AG be not equal to C, it is greater than it :
and EF is equal to AG; (1. 36.)
therefore EF also is greater than C.
Make the parallelogram KLMN equal to the excess of EF above C,
and similar and similarly situated to D: (v1. 25.)
then, since D is similar to ÈF, (constr.)
therefore also KM is similar to EF: (v1. 21.)
let KL be the homologous side to EG, and LM to GF:
and because EF is equal to C and KM together,
EF is greater than KM;
therefore the straight line EG is greater than KL, and GF than LM:
make GX equal to LK, and GO equal to LM, (1. 3.)
and complete the parallelogram XGOP: (1.31.)
therefore XO is equal and similar to KM:
but KM is similar to EF;
wherefore also XO is similar to EF;
and therefore XO and EF are about the same diameter: (VI. 26.)
let GPB be their diameter, and complete the scheme.
Then, because EF is equal to C and KM together,
and XO a part of the one is equal to KM a part of the other,
the remainder, viz. the gnomon ERO, is equal to the remainder
C: (ax. 3.)
and because OR is equal to XS, by adding SR to each, (1. 43.)
the whole OB is equal to the whole XB:
but XB is equal to TE, because the base AE is equal to the base
EB; (1.36.)
BOOK VI. PROP. XXVIII, XXIX.
195
wherefore also TE is equal to OB: (ax. 1.)
add XS to each, then the whole TS is equal to the whole, viz. to
the gnomon ERO:
but it has been proved that the gnomon ERO is equal to C;
and therefore also TS is equal to C.
Wherefore the parallelogram TS, equal to the given rectilineal figure
C, is applied to the given straight line AB, deficient by the parallelogram
SR, similar to the given one Ď, because SR is similar to EF. (VI. 24.)
Q.E. F.
PROPOSITION XXIX. PROBLEM.
To a given straight line to apply a parallelogram equal to a given
rectilineal figure, exceeding by a parallelogram similar to another given.
Let AB be the given straight line, and C the given rectilineal figure
to which the parallelogram to be applied is required to be equal, and D
the parallelogram to which the excess of the one to be applied above
that upon the given line is required to be similar.
It is required to apply a parallelogram to the given straight line
AB which shall be equal to the figure C, exceeding by a parallelogram
similar to D.
F L M
K H
D
A F
BO
NPX
G
Divide AB into two equal parts in the point E, (1. 10.) and upon
EB describe the parallelogram EL similar and similarly situated to D:
(VI. 18.)
and make the parallelogram GH equal to EL and C together, and
similar and similarly situated to D: (VI. 25.)
wherefore GH is similar to EL: (vi. 21.)
let KH be the side homologous to FL, and KG to FE:
and because the parallelogram GH is greater than EL,
therefore the side KH is greater than FL,
and KG than FE:
produce FL and FE, and make FLM equal to KH, and FEN to KG,
and complete the parallelogram MN:
MN is therefore equal and similar to GH:
but GH is similar to EL;
wherefore MN is similar to EL;
and consequently EL and MN are about the same diameter: (v1. 26.)
draw their diameter FX, and complete the scheme.
Therefore, since GH is equal to EL and C together,
and that GH is equal to MN;,
MN is equal to EL and C:
take away the common part EL ;
then the remainder, viz. the gnomon NOL, is equal to C.
And because AE is equal to EB,
the parallelogram AN is equal to the parallelogram NB, (1. 36.) that
is, to BM: (1. 43.)
수
​
13-2
196
EUCLID'S ELEMENTS.
add NO to each ;
therefore the whole, viz. the parallelogram AX, is equal to the
gnomon NOL:
but the gnomon NOL is equal to C;
therefore also AX is equal to C.
Wherefore to the straight line AB there is applied the parallelogram
AX equal to the given rectilineal figure C, exceeding by the parallelo-
gram PO, which is similar to D, because PO is similar to EL. (v1. 24.)
Q. E. F.
PROPOSITION XXX. PROBLEM.
To cut a given straight line in extreme and mean ratio.
Let AB be the given straight line.
It is required to cut it in extreme and mean ratio.
D

E
A
B
C F
Upon AB describe the square BC, (1. 46.)
and to AC apply the parallelogram CD, equal to BC, exceeding by
the figure AD similar to BC: (VI. 29.)
then, since BC is a square,
therefore also AD is a square:
and because BC is equal to CD,
by taking the common part CE from each,
the remainder BF is equal to the remainder AD:
and these figures are equiangular,
therefore their sides about the equal angles are reciprocally propor-
tional: (vI.14.)
therefore, as FE to ED, so AE to EB:
but FE is equal to AC, (1. 34.) that is, to AB; (def. 30.)
and ED is equal to AE ;
therefore as BA to AE, so is AE to EB:
but AB is greater than AE;
wherefore AE is greater than EB: (v. 14.)
therefore the straight line AB is cut in extreme and mean ratio
in E. (v1. def. 3.) Q. E.F.
Otherwise,
Let AB be the given straight line.
It is required to cut it in extreme and mean ratio.
A
¿ B
Divide AB in the point C, so that the rectangle contained by AB,
BC, may be equal to the square of AC. (11. 11.)
Then, because the rectangle AB, BC is equal to the square of AC;
as B.4 to AC, so is AC to CB: (vi. 17.)
therefore AB is cut in extreme and mean ratio in C. (v1. def. 3.) Q.E.F.
BOOK VI. PROP. XXXI, XXXII.
197
PROPOSITION XXXI. THEOREM.
*
In right-angled triangles, the rectilineal figure described upon the side
opposite to the right angle, is equal to the similar and similarly described
figures upon the sides containing the right angle..
Let ABC be a right-angled triangle, having the right angle BAC.
The rectilineal figure described upon BC shall be equal to the
similar and similarly described figures upon BA, AÇ.

B D
C
Draw the perpendicular AD: (1. 12.)
therefore, because in the right-angled triangle ABC,
AD is drawn from the right angle at A perpendicular to the base BC,
the triangles ABD, ĂDC are similar to the whole triangle ABC,
and to one another: (vI. 8.)
and because the triangle ABC is similar to ADB,
as CB to BA, so is BA to BD: (v1.4.)
and because these three straight lines are proportionals,
as the first is to the third, so is the figure upon the first to the
similar and similarly described figure upon the second: (vI. 20. Cor. 2.)
therefore as CB to BD, so is the figure upon CB to the similar and
similarly described figure upon BA:
and inversely, as DB to BC, so is the figure upon BA to that upon
BC: (v. B.)
for the same reason, as DC to CB, so is the figure upon CA to that
upon CB:
therefore as BD and DC together to BC, so are the figures upon
BA, AC to that upon BC: (v. 24.)
but BD and DC together are equal to BC;
therefore the figure described on BC is equal to the similar and
similarly described figures on BA, AC. (v. a.)
Wherefore, in right-angled triangles, &c.
Q.E. D.
PROPOSITION XXXII. THEOREM.
If two triangles which have two sides of the one proportional to two
sides of the other, be joined at one angle, so as to have their homologous
sides parallel to one another; the remaining sides shall be in a straight line.
Let ABC, DCE be two triangles which have the two sides BA, AC
proportional to the two CD, DE,
viz. BA to AC, as CD to DE;
and let AB be parallel to DC, and AC to DE.
A

D
B
E
198
EUCLID'S ELEMENTS.
}
1
Then BC and CE shall be in a straight line.
Because AB is parallel to DC, and the straight line AC meets them,
the alternate angles BAC, ACD are equal; (1. 29.)
for the same reason, the angle CDE is equal to the angle ACD ;
wherefore also BAC is equal to CDE: (ax. 1.)
and because the triangles ABC, DCE have one angle at A equal to
one at D, and the sides about these angles proportionals,
viz. BA to AC, as CD to DE,
the triangle ABC is equiangular to DCE: (vi. 6.)
therefore the angle ABC is equal to the angle DCE:
and the angle BAC was proved to be equal to ACD;
therefore the whole angle ACE is equal to the two angles ABC,
BAC: (ax. 2.)
add to each of these equals the common angle ACB,
then the angles ACE, ACB are equal to the angles ABC, BAC, ACB:
but ABC, BAC, ACB are equal to two right angles: (1.32.)
therefore also the angles ACE, ACB are equal to two right angles:
and since at the point C, in the straight line AC, the two straight
lines BC, CE, which are on the opposite sides of it, make the adjacent
angles ACE, ACB equal to two right angles;
therefore BC and CE are in a straight line. (1. 14.)
Wherefore, if two triangles, &c.
Q.E.D.
PROPOSITION XXXIII. THEOREM.
In equal circles, angles, whether at the centres or circumferences, have
the same ratio which the circumferences on which they stand have to one
another: so also have the sectors.
Let ABC, DEF be equal circles; and at their centres the angles
BGC, EHF, and the angles BAC, EDF, at their circumferences.
As the circumference BC to the circumference EF, so shall the
angle BGC be to the angle EHF, and the angle BAC to the
angle EDF;
and also the sector BGC to the sector EHF.


A
D
G
H
N
K
M
B
E F
Take any number of circumferences CK, KL, each equal to BC,
and any number whatever FM, MN, each equal to EF:
and join GK, GL, HM, HN.
Because the circumferences BC, CK, KL are all equal,
the angles BGC, CGK, KGL are also all equal: (III. 27.)
therefore what multiple soever the circumference BL is of the cir-
cumference BC, the same multiple is the angle BGL of the angle
BGC:
for the same reason, whatever multiple the circumference EN is of
the circumference EF, the same multiple is the angle EHN of the
angle EHF:
and if the circumference BL be equal to the circumference EN,
BOOK VI. PROP. XXXIII.
199
the angle BGL is also equal to the angle EHN; (111. 27.)
and if the circumference BL be greater than EN,
likewise the angle BGL is greater than EHN; and if less, less:
therefore since there are four magnitudes, the two circumferences
BC, EF, and the two angles BGC, EHF; and that of the circum-
ference BC, and of the angle BGC, have been taken any equimultiples
whatever, viz. the circumference BL, and the angle BGL; and of the
circumference EF, and of the angle EHF, any equimultiples whatever,
viz. the circumference EN, and the angle EHN;
and since it has been proved, that if the circumference BL be greater
than EN;
the angle BGL is greater than EHN;
and if equal, equal; and if less, less;
therefore as the circumference BC to the circumference EF, so is
the angle BGC to the angle EHF: (v. def. 5.)
but as the angle BGC is to the angle EHF, so is the angle BAC
to the angle EDF: (v. 15.)
for each is double of each; (III. 20.)
therefore, as the circumference BC is to EF, so is the angle BGC
to the angle EHF, and the angle BAC to the angle EDF.
Also, as the circumference BC to EF, so shall the sector BGC be to
the sector EHF.

A
D
G
L
H
N
K
BX C
E
M
F
Join BC, CK, and in the circumferences, BC, CK take any points X,
O, and join BX, XC, CO, OK.
Then, because in the triangles GBC, GCK
the two sides BG, GC are equal to the two CG, GK each to each,
and that they contain equal angles;
the base BC is equal to the base CK, (1. 4.)
and the triangle GBC to the triangle GCK:
and because the circumference BC is equal to the circumference CK,
the remaining part of the whole circumference of the circle ABC, is
equal to the remaining part of the whole circumference of the same
circle: (ax. 3.)
therefore the angle BXC is equal to the angle COK; (111. 27.)
and the segment BXC is therefore similar to the segment COK;
(III. def. 11.)
and they are upon equal straight lines, BC, CK:
but similar segment of circles upon equal straight lines, are equal
to one another; (111. 24.)
therefore the segment BXC is equal to the segment COK:
and the triangle BGC was proved to be equal to the triangle CGK;
therefore the whole, the sector BGC, is equal to the whole, the
sector CGK:
for the same reason, the sector KGL is equal to each of the sectors
BGC, CGK:
in the same manner, the sectors EHF, FHM, MHN
proved equal to one another:
may be
200
EUCLID'S ELEMENTS.
therefore, what multiple soever the circumference BL is of the cir-
cumference BC, the same multiple is the sector BGL of the sector
BGC:
and for the same reason, whatever multiple the circumference EN
is of EF, the same multiple is the sector EHN of the sector EHF:
and if the circumference BL be equal to EN, the sector BGL is
equal to the sector EHN;
and if the circumference BL be greater than EN, the sector BGL
is greater than the sector EHÑ;
and if less, less ;
since, then, there are four magnitudes, the two circumferences BC,
EF, and the two sectors BGC, EHF, and that of the circumference BC,
and sector BGC, the circumference BL and sector BGL are any equi-
multiples whatever; and of the circumference EF, and sector EHF,
the circumference EN, and sector EHN are any equimultiples whatever;
and since it has been proved, that if the circumference BL be
greater than EN, the sector BGL is greater than the sector EHN;
and if equal, equal; and if less, less:
therefore, as the circumference BC is to the circumference EF, so
is the sector BGC to the sector EHF. (v. def. 5.)
Wherefore, in equal circles, &c. Q.E.D.
PROPOSITION B. THEOREM.
If an angle of a triangle be bisected by a straight line, which likewise
cuts the base; the rectangle contained by the sides of the triangle is equal
to the rectangle contained by the segments of the base, together with the
square of the straight line which bisects the angle.
Let ABC be a triangle, and let the angle BAC be bisected by the
straight line AD.
The rectangle BA, AC shall be equal to the rectangle BD, DC,
together with the square of AD.
A

B
C
D
E
Describe the circle ACB about the triangle, (IV. 5.)
and produce AD to the circumference in E, and join EC.
Then because the angle BAD is equal to the angle ĈAE, (hyp.) and
the angle ABD to the angle AEC, (11. 21.)
for they are in the same segment;
the triangles ABD, AEC are equiangular to one another: (1. 32.)
therefore as BA to AD, so is EA to AC; (vI. 4.)
and consequently the rectangle BA, AC is equal to the rectangle
EA, AD, (vi. 16.)
that is, to the rectangle ED, DA, together with the square of AD ;
(11. 3.)
BOOK VI. PROP. B, C, D.
201
but the rectangle ED, DA is equal to the rectangle BD, DC; (111. 35.)
therefore the rectangle BA, AC is equal to the rectangle BD, DC,
together with the square of AD.
Wherefore, if an angle, &c. Q.E.D.
PROPOSITION C.
THEOREM.
If from any angle of a triangle a straight line be drawn perpendicular
to the base; the rectangle contained by the sides of the triangle is equal to
the rectangle contained by the perpendicular and the diameter of the circle
described about the triangle.
Let ABC be a triangle, and AD the perpendicular from the angle A
to the base BC.
The rectangle BA, AC shall be equal to the rectangle contained by
AD and the diameter of the circle described about the triangle.
B
A

C
E
Describe the circle ACB about the triangle, (Iv. 5.) and draw its
diameter AE, and join EC.
because the right angle BDA is equal to the angle ECA in a semi-
circle, (III. 31.)
and the angle ABD equal to the angle AEC in the same segment;
(III. 21.)
the triangles ABD, AEC are equiangular:
therefore as BA to AD, so is EA to AC; (vI. 4.)
and consequently the rectangle BA, AC is equal to the rectangle
EA, AD. (vi. 16.)
If therefore from an angle, &c.
Q. E. D.
PROPOSITION D. THEOREM.
The rectangle contained by the diagonals of a quadrilateral figure in-
scribed in a circle, is equal to both the rectangles contained by its opposite
sides.
Let ABCD be any quadrilateral figure inscribed in a circle, and
join AC, BD.
The rectangle contained by AC, BD shall be equal to the two
rectangles contained by AB, CD, and by AD, BC.

B
D
A
202
EUCLID'S ELEMENTS.
Make the angle ABE equal to the angle DBC: (1.23.)
add to each of these equals the common angle EBD,
then the angle ABD is equal to the angle EBC:
and the angle BDĂ is equal to the angle BCE, because they are in
the same segment: (111. 21.)
therefore the triangle ABD is equiangular to the triangle BCE:
wherefore, as BC is to CE, so is BD to DA; (VI. 4.)
and consequently the rectangle BC, AD is equal to the rectangle
BD, CE: (VI. 16.)
again, because the angle ABE is equal to the angle DBC, and the
angle BAE to the angle BDC,¯ (111. 21.)
the triangle ABE is equiangular to the triangle BCD:
therefore as BÃ to AE, so is BD to DC ;
wherefore the rectangle BA, DC is equal to the rectangle BD, AE:
but the rectangle BC, AD has been shewn equal to the rectangle
BD, CE;
therefore the whole rectangle AC, BD is equal to the rectangle AB,
DC, together with the rectangle AD, BC. (11. 1.)
Therefore the rectangle, &c. Q.E.D.
This is a Lemma of Cl. Ptolemæus, in page 9 of his Meyaλŋ Σúvtağıs.
NOTES TO BOOK VI.
In this book, the theory of proportion exhibited in the fifth book, is applied to
the comparison of the sides and areas of plane rectilineal figures, both to those which are
similar, and those which are not similar.
Def. I.
In defining similar triangles, one condition is sufficient, namely, that
similar triangles are those which have their three angles respectively equal; as in Prop. 4,
Book VI, it is proved that the sides about the equal angles of equiangular triangles are
proportionals. But in defining similar figures of more than three sides, both of the
conditions stated in Def. 1, are requisite, as it is obvious, for instance, in the case of a
square and a rectangle, which have their angles respectively equal, but have not their
sides about their equal angles proportionals.
The following definition has been proposed: “Similar rectilineal figures of more
than three sides, are those which may be divided into the same number of similar
triangles." This definition would, if adopted, require the omission of a part of
Prop. 20, Book vi.
Def. III. To this definition may be added the following:
X
A straight line is said to be divided harmonically, when it is divided into three
parts, such that the whole line is to one of the extreme segments, as the other extreme
segment is to the middle part. Three lines are in harmonical proportion, when the first
is to the third, as the difference between the first and second, is to the difference between
the second and third; and the second is called a harmonic mean between the first and
third.
6
The expression. harmonical proportion' is derived from the following fact in the
Science of Acoustics, that three musical strings of the same material, thickness and
tension, when divided in the manner stated in the definition, or numerically as 6, 4,
and 3, produce a certain musical note, its fifth, and its octave.
Def. IV. The term altitude, as applied to the same triangles and parallelograms,
will be different according to the sides which may be assumed as the base.
Prop. I. In the same manner may be proved, that triangles and parallelograms
upon equal bases, are to one another as their altitudes.
Prop. A. When the triangle ABC is isosceles, the line which bisects the exterior
angle at the vertex is parallel to the base. In all other cases;
If the line which bisects the angle BAC cut the base BC in the point G.
Then the straight line BD is harmonically divided in the points G, C.
·
For BG is to GC as BA is to AC;
and BD is to DC as BA is to AC,
(v1. 3.)
(vi. a.)
therefore BD is to DC as BG is to GC,
but BG BD - DG and GC GD - DC.
=
Wherefore BD is to DC as BD - DG is to GD - DC.
Hence BD, DG, DC, are in harmonical proportion.
Prop. Iv is the first case of similar triangles, and corresponds to the third case of |
equal triangles, Prop. 26, Book 1.
Sometimes the sides opposite to the equal angles in two equiangular triangles, are
called the corresponding sides, and these are said to be proportional, which is simply
taking the proportion in Euclid alternately.
The term homologous (óµóλoyos), has reference to the places the sides of the triangles
have in the ratios, and in one sense, homologous sides may be considered as corresponding
$
204
EUCLID'S ELEMENTS.
sides. The homologous sides of any two similar rectilineal figures will be found to be
those which are adjacent to two equal angles in each figure.
Prop. v, the converse of Prop. IV, is the second case of similar triangles, and
corresponds to Prop. 8, Book 1, the second case of equal triangles.
Prop. VI is the third case of similar triangles, and corresponds to Prop. 4, Book 1.
The property of similar triangles, and that contained in Prop. 47, Book 1, are the
most important theorems in Geometry.
Prop. VII is the fourth case of similar triangles, and corresponds to the fourth case
of equal triangles pointed out in the note to Prop. 26, Book 1, p. 49. . -
Prop. XIII may be compared with Prop. xv1, Book II.
T
It may be observed, that half the sum of AB and BC is called the Arithmetic mean
between these lines; also that BD is called the Geometric mean between the same lines.
To find two mean proportionals between two given lines is impossible by the straight
line and circle. Pappus has given several solutions of this problem in Book 111, of his
Mathematical Collections; and Eutocius has given, in his Commentary on the Sphere and
Cylinder of Archimedes, ten different methods of solving this problem.
Prop. XIV, depends on the same principle as Prop. xv, and both may easily be
demonstrated from one diagram. Join DF, FE, EG in the fig. to Prop. xiv, and
the figure to Prop. xv is formed. We may add, that there does not appear any
reason why the properties of the triangle and parallelogram should be here separated
and not in the first proposition of the sixth book.
Prop. xv, holds good when one angle of one triangle is equal to the defect from what
the corresponding angle in the other wants of two right angles.
Prop. XVII is only a particular case of Prop. XVI, and more properly, might
appear as a corollary.
Algebraically, Let AB, CD, E, F, contain a, b, c, d units respectively.
Then, since a, b, c, d are proportionals, ..
α с
Multiply these equals by bd, .. ad = bc,
or, the product of the extremes is equal to the product of the means.
And conversely, If the product of the extremes be equal to the product of the means,
or ad = bc,
then dividing these equals by bd, ..
с
or the ratio of the first to the second number, is equal to the ratio of the third to the
fourth.
Similarly may be shewn, that if ; then ad = b².
a
b
b
And conversely, if ad = b²; then
a b
ī = ā
Prop. XVIII. Similar figures are said to be similarly situated when their homologous
sides are parallel.
Prop. xx. It may easily be shewn, that the perimeters of similar polygons are
proportional to their homologous sides.
Prop. XXXI. This proposition is an extension of Prop. 47, Book 1, to similar
rectilineal figures, and may be deduced from Prop. 22, Book v1, and Prop. 47, Book 1.
Prop. B. The converse of this proposition does not hold good when the triangle is
isosceles.
The seventh, eighth, ninth and tenth books of the Elements treat of numbers, and
employ the Greek numerical notation.
BOOK XI.
DEFINITIONS.
I.
A SOLID is that which hath length, breadth, and thickness.
II.
That which bounds a solid is a superficies.
III.
A straight line is perpendicular, or at right angles, to a plane,
when it makes right angles with every straight line meeting it in that
plane.
IV.
A plane is perpendicular to a plane, when the straight lines drawn
in one of the planes perpendicular to the common section of the two
planes, are perpendicular to the other plane.
V.
The inclination of a straight line to a plane is the acute angle
contained by that straight line, and another drawn from the point in
which the first line meets the plane, to the point in which a perpen-
dicular to the plane drawn from any point of the first line above the
plane, meets the same plane.
VI.
The inclination of a plane to a plane is the acute angle contained
by two straight lines drawn from any the same point of their common
section at right angles to it, one upon one plane, and the other upon
the other plane.
VII.
Two planes are said to have the same, or a like inclination to one
another, which two other planes have, when the said angles of inclination
are equal to one another.
VIII.
Parallel planes are such as do not meet one another though produced.
IX.
A solid angle is that which is made by the meeting, in one point,
of more than two plane angles, which are not in the same plane.
X.
Equal and similar solid figures are such as are contained by similar
planes equal in number and magnitude.
206
EUCLID'S ELEMENTS.
XI.
Similar solid figures are such as have all their solid angles equal,
each to each, and are contained by the same number of similar planes.
XII.
A pyramid is a solid figure contained by planes that are constituted
betwixt one plane and one point above it in which they meet.
XIII.
A prism is a solid figure contained by plane figures, of which two
that are opposite are equal, similar, and parallel to one another; and
the others parallelograms.
XIV.
A sphere is a solid figure described by the revolution of a semicircle
about its diameter, which remains unmoved.
XV.
The axis of a sphere is the fixed straight line about which the
semicircle revolves.
XVI.
The centre of a sphere is the same with that of the semicircle.
XVII.
The diameter of a sphere is any straight line which passes through
the centre, and is terminated both ways by the superficies of the sphere.
XVIII.
A cone is a solid figure described by the revolution of a right-angled
triangle about one of the sides containing the right angle, which side
remains fixed.
If the fixed side be equal to the other side containing the right
angle, the cone is called a right-angled cone; if it be less than the
other side, an obtuse-angled; and if greater, an acute-angled cone.
XIX.
The axis of a cone is the fixed straight line about which the triangle
revolves.
XX.
The base of a cone is the circle described by that side containing
the right angle, which revolves.
XXI.
A cylinder is a solid figure described by the revolution of a right-
angled parallelogram about one of its sides which remains fixed.
XXII.
The axis of a cylinder is the fixed straight line about which the
parallelogram revolves.
DEFINITIONS.
207
XXIII.
The bases of a cylinder are the circles described by the two revolving
opposite sides of the parallelogram.
XXIV.
Similar cones and cylinders are those which have their axes and the
diameters of their bases proportionals.
XXV.
A cube is a solid figure contained by six equal squares.
XXVI.
A tetrahedron is a solid figure contained by four equal and equilateral
triangles.
XXVII.
An octahedron is a solid figure contained by eight equal and
equilateral triangles.
XXVIII.
A dodecahedron is a solid figure contained by twelve equal pentagons
which are equilateral and equiangular.
XXIX.
An icosahedron is a solid figure contained by twenty equal and
equilateral triangles.
Def. A.
A parallelopiped is a solid figure contained by six quadrilateral
figures, whereof every opposite two are parallel.
208
EUCLID'S ELEMENTS.
PROPOSITION I. THEOREM.
One part of a straight line cannot be in a plane, and another part above it.
If it be possible, let AB, part of the straight line ABC, be in the
plane, and the part BC above it:
C
A B D
and since the straight line AB is in the plane, it can be produced
in that plane:
let it be produced to D;
and let any plane pass through the straight line AD, and be turned
about it until it pass through the point C:
and because the points B, C are in this plane,
the straight line BC is in it: (1. def. 7.)
therefore there are two straight lines ABC, ABD in the same plane
that have a common segment AB; (1. 11. Cor.)
which is impossible.
Therefore, one part, &c. Q.E.D.
PROPOSITION II. THEOREM.
Two straight lines which cut one another are in one plane, and three
straight lines which meet one another are in one plane.
Let two straight lines AB, CD cut one another in E;
then AB, CD shall be in one plane:
and three straight lines EC, CB, BE, which meet one another,
shall be in one plane.
A
D
E
C
B
Let any plane pass through the straight line EB,
and let the plane be turned about EB, produced if necessary, until
it pass through the point C.
Then, because the points E, C are in this plane,
the straight line EC is in it: (1. def. 7.)
for the same reason, the straight line BC is in the same:
and by the hypothesis, EB is in it:
therefore the three straight lines EC, CB, BE are in one plane ;
but in the plane in which EC, EB are, in the same are CD, AB: (x1. 1.)
therefore, AB, CD are in one plane.
Wherefore two straight lines, &C. Q.E.D.
BOOK XI. PROP. III, IV.
209
PROPOSITION III.
THEOREM.
If two planes cut one another, their common section is a straight line.
Let two planes AB, BC cut one another, and let the line DB be
their common section.
Then DB shall be a straight line.
B
E
F
A
D
If it be not, from the point D to B, draw, in the plane AB, the
straight line DEB, (post. 1.)
and in the plane BC, the straight line DFB:
then two straight lines DEB, DFB have the same extremities,
and therefore include a space betwixt them;
which is impossible: (1. ax. 10.)
therefore BD, the common section of the planes AB, BC, cannot
but be a straight line.
Wherefore, if two planes, &c. Q.E.D.
PROPOSITION IV. THEOREM.
If a straight line stand at right angles to each of two straight lines
in the point of their intersection, it shall also be at right angles to the
plane which passes through them, that is, to the plane in which they are.
Let the straight line EF stand at right angles to each of the straight
lines AB, CD, in E the point of their intersection.
Then EF shall also be at right angles to the plane passing
through AB, CD.
A
F

G
E
H
B
Take the straight lines AE, EB, CE, ED all equal to one another;
and through E draw, in the plane in which are AB, CD, any
straight line GEH;
and join AD, CB;
then from any point F, in EF, draw FA, FG, FD, FC, FH, FB.
And because the two straight lines AE, ED are equal to the two
BE, EC, each to each,
and that they contain equal angles AED, BEC, (1. 15.)
the base AD is equal to the base BC,(1. 4.)
and the angle DAE to the angle EBC:
14
210
EUCLID'S ELEMENTS.
and the angle AEG is equal to the angle BEH: (1. 15.)
therefore the triangles AÉG, BEH have two angles of the one
equal to two angles of the other, each to each,
and the sides AE, EB, adjacent to the equal angles, equal to
one another:
wherefore they have their other sides equal: (1. 26.)
therefore GE is equal to EH, and AG to BH:
and because AE is equal to EB, and FE common and at right
angles to them,
the base AF is equal to the base FB; (1. 4.)
for the same reason, CF is equal to FD:
and because AD is equal to BC, and AF to FB,
the two sides FA, AD are equal to the two FB, BC, each to each ;
and the base DF was proved equal to the base FC;
therefore the angle FAD is equal to the angle FBC: (1. 8.)
again, it was proved that GA is equal to BH, and also AF to FB;
therefore FA and AG are equal to FB and BH, each to each;
and the angle FAG has been proved equal to the angle FBH;
therefore the base GF is equal to the base FH: (1.4.)
again, because it was proved that GE is equal to EH, and EF is common;
therefore GÊ, EF are equal to HE, EF, each to each ;
and the base GF is equal to the base FH;
therefore the angle GEF is equal to the angle HEF; (1.8.)
and consequently each of these angles is a right angle. (1. def. 10.)
Therefore FE makes right angles with GH, that is, with any straight
line drawn through E in the plane passing through AB, CD.
In like manner, it may be proved, that FE makes right angles with
every straight line which meets it in that plane.
But a straight line is at right angles to a plane when it makes right
angles with every straight line which meets it in that plane: (x1. def. 3.)
therefore EF is at right angles to the plane in which are AB, CD.
Wherefore, if a straight line, &c.
Q. E.D.
PROPOSITION V. THEOREM.
If three straight lines meet all in one point, and a straight line stands
at right angles to each of them in that point; these three straight lines
are in one and the same plane.
Let the straight line AB stand at right angles to each of the straight
lines BC, BD, BE, in B the point where they meet:
Then BC, BD, BE shall be in one and the same plane.
A
C
F
B
D
E
If not, let, if it be possible, BD and BE be in one plane, and BC be
above it;
and let a plane pass through AB, BC, the common section of which,
with the plane in which BD and BE are, is a straight line; (xI. 3.)
BOOK XI. PROP. V, VI.
211
let this be BF:
therefore the three straight lines AB, BC, BF are all in one plane,
viz. that which passes through AB, BC.
And because AB stands at right angles to each of the straight lines
BD, BE,
it is also at right angles to the plane passing through them: (xI. 4.)
and therefore makes right angles with every straight line meeting
it in that plane : (XI. def. 3.)
it;
but BF, which is in that plane, meets it ;
therefore the angle ABF is a right angle:
but the angle ABC, by the hypothesis, is also a right angle;
therefore the angle ABF is equal to the angle ABC,
and they are both in the same plane, which is impossible; (1. ax. 9.)
therefore the straight line BC is not above the plane in which are
BD and BE:
wherefore the three straight lines BC, BD, BE are in one and the
same plane.
Therefore, if three straight lines, &c. Q.E. D.
PROPOSITION VI. THEOREM.
If two straight lines be at right angles to the same plane, they shall be
parallel to one another.
Let the straight lines AB, CD be at right angles to the same plane.
Then AB shall be parallel to CD.
A
B
D
E
Let them meet the plane in the points B, D,
and draw the straight line BD, to which draw DE at right angles,
in the same plane; (1. 11.)
and make DE equal to AB, (1. 3.) and join BE, AE, AD.
Then, because AB is perpendicular to the plane,
it makes right angles with every straight line which meets it, and
is in that plane: (x1. def. 3.)
but BD, BE, which are in that plane, do each of them meet AB;
therefore each of the angles ABD, ABE is a right angle;
for the same reason, each of the angles CDB, CDE is a right angle:
and because AB is equal to DE, and BD common,
the two sides AB, BD are equal to the two ED, DB, each to each;
and they contain right angles :
therefore the base AD is equal to the base BE: (1.4.)
again, because AB is equal to DE, and BE to AD;
AB, BE are equal to ED, DA, each to each;
and, in the triangles ABE, EDÁ, the base AE is common:
therefore the angle ABE is equal to the angle EDA: (1. 8.)
but ABE is a right angle;
14-2
212
EUCLID'S ELEMENTS.
therefore EDA is also a right angle, and ED perpendicular to DA:
but it is also perpendicular to each of the two BD, DC;
wherefore ED is at right angles to each of the three straight lines
BD, DA, DC in the point in which they meet:
therefore these three straight lines are all in the same plane : (XI. 5.)
but AB is in the plane in which are BD, DA, (XI. 2.)
because any three straight lines which meet one another are in one plane:
therefore AB, BD, DC are in one plane:
and each of the angles ABD, BDC is a right angle;
therefore AB is parallel to CD. (1. 28.)
Wherefore, if two straight lines, &c.
Q. E.D.
PROPOSITION VII. THEOREM.
If two straight lines 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.
Let AB, CD be parallel straight lines, and take any point E in the
one, and the point F in the other.
Then the straight line which joins E and F shall be in the same
plane with the parallels.
A E
B
G
H
C
F D
If not, let it be, if possible, above the plane, as EGF;
and in the plane ABCD in which the parallels are,
draw the straight line EHF from E to F.
And since EGF also is a straight line, the two straight lines EHF,
EGF include a space between them, which is impossible. (1. ax. 10.)
Therefore the straight line joining the points E, F is not above the
plane in which the parallels AB, CD are,
and is therefore in that plane.
Wherefore, if two straight lines, &c.
Q. E.D.
PROPOSITION VIII. THEOREM.
If two straight lines be parallel, and one of them is at right angles to
a plane; the other also shall be at right angles to the same plane.
Let AB, CD be two parallel straight lines, and let one of them AB
be at right angles to a plane.
Then the other CD shall be at right angles to the same plane.
IC
B
D
E
BOOK XI. PROP. VIII, IX.
213
Let AB, CD meet the plane in the points B, D, and join BD:
therefore AB, CD, BD are in one plane. (XI. 7.)
•
In the plane to which AB is at right angles, draw DE at right
angles to BD, (1.11.)
and make DE equal to AB, (1. 3.) and join BE, AE, AD.
And because AB is perpendicular to the plane,
it is perpendicular to every straight line which meets it, and is in
that plane; (x1. def. 3.)
therefore each of the angles ABD, ABE is a right angle:
and because the straight line BD meets the parallel straight lines
AB, CD,
the angles ABD, CDB are together equal to two right angles: (1. 29.)
and ABD is a right angle ;
therefore also CDB is a right angle, and CD perpendicular to BD:
and because AB is equal to DE, and BD common,
the two AB, BD are equal to the two ED, DB, each to each;
and the angle ABD is equal to the angle EDB, because each of
them is a right angle ;
therefore the base AD is equal to the base BE: (1. 4.)
again, because AB is equal to DE, and BE to ÅD,
the two AB, BE are equal to the two ED, DA, each to each;
and the base AE is common to the triangles ABE, EDA ;
wherefore the angle ABE is equal to the angle EDA: (1.8.)
but ABE is a right angle;
and therefore EDA is a right angle, and ED perpendicular to DA:
but it is also perpendicular to BD; (constr.)
therefore ED is perpendicular to the plane which passes through
BD, DA; (xI. 4.)
and therefore makes right angles with every straight line meeting
it in that plane: (xI. def. 3.)
but DC is in the plane passing through BD, DA,
because all three are in the plane in which are the parallels AB, CD;
wherefore ED is at right angles to DC;
and therefore CD is at right angles to DE:
but CD is also at right angles to DB;
therefore CD is at right angles to the two straight lines DE, DB
in the point of their intersection D;
and therefore is at right angles to the plane passing through DE,
DB, (x1. 4.)
which is the same plane to which AB is at right angles.
Therefore, if two straight lines, &c. Q. E. D.

PROPOSITION IX. THEOREM.
Two straight lines which are each of them parallel to the same straight
line, and not in the same plane with it, are parallel to one another.
Let AB, CD be each of them parallel to EF, and not in the same
plane with it.
Then AB shall be parallel to CD.
214
EUCLID'S ELEMENTS.
AH B
:
E
F
C K
D
In EF take any point G, from which draw, in the plane passing
through EF, AB, the straight line GH at right angles to EF; (1. 11.)
and in the plane passing through EF, CD draw GK at right angles
to the same EF.
And because EF is perpendicular both to GH and GK,
EF is perpendicular to the plane HGK passing through them: (XI. 4.)
and EF is parallel to AB
therefore AB is at right angles to the plane HGK. (x1. 8.)
For the same reason, CD is likewise at right angles to the plane HGK.
Therefore AB, CD are each of them at right angles to the plane HGK.
But if two straight lines are at right angles to the same plane, they
are parallel to one another: (XI. 6.)
therefore AB is parallel to CD.
Wherefore, two straight lines, &c.
Q. E.D.
PROPOSITION X. THEOREM.
If two straight lines meeting one another be parallel to two others that
meet one another, and are not in the same plane with the first two; the
first two and the other two shall contain equal angles.
Let the two straight lines AB, BC, which meet one another, be
parallel to the two straight lines DE, EF, that meet one another, and
are not in the same plane with AB, BC.
The angle ABC shall be equal to the angle DEF.
B
A
C
D
D
F
Take BA, BC, ED, EF all equal to one another;
and join AD, CF, BÊ, AC, DF.
Then, because BA is equal and parallel to ED,
therefore AD is both equal and parallel to BE. (1. 33.)
For the same reason, CF is equal and parallel to BE.
Therefore AD and CF are each of them equal and parallel to BE.
But straight lines that are parallel to the same straight line, and not
in the same plane with it, are parallel to one another: (x1. 9.)
therefore AD is parallel to CF; and it is equal to it; (1. ax. 1.)
and AC, DF join them towards the same parts;
and therefore AC is equal and parallel to DF. (1. 33.)
And because AB, BC are equal to DE, EF, each to each,
and the base AC to the base DF;
the angle ABC is equal to the angle DEF. (1. 8.)
Therefore, if two straight lines, &c. Q.E.D.
BOOK XI. PROP. XI, XII.
215
PROPOSITION XI. PROBLEM.
To draw a straight line perpendicular to a plane, from a given point
above it.
Let A be the given point above the plane BH.
It is required to draw from the point A a straight line perpen-
dicular to the plane BH.
In the plane draw any straight line BC,
and from the point A draw AD perpendicular to BC. (1. 12.)
If then AD be also perpendicular to the plane BH, the thing
required is already done;
but if it be not, from the point D draw, in the plane BH, the
straight line DE at right angles to BC; (1. 11.)
and from the point A draw AF perpendicular to DE.
Then AF shall be perpendicular to the plane BH.
E
G
F
A

H
B
D
C
Through F draw GH parallel to BC. (1. 31.)
And because BC is at right angles to ED and DA,
BC is at right angles to the plane passing through ED, DA: (xI. 4.)
and GH is parallel to BC:
but, if two straight lines be parallel, one of which is at right angles
to a plane,
the other is at right angles to the same plane; (xI. 8.)
wherefore GH is at right angles to the plane through ED, DA;
and is perpendicular to every straight line meeting it in that plane:
(XI. def. 3.)
but AF, which is in the plane through ED, DA, meets it;
therefore GH is perpendicular to AF;
and consequently AF is perpendicular to GH;
and AF is perpendicular to DE;
therefore AF is perpendicular to each of the straight lines GH, DE.
But if a straight line stand at right angles to each of two straight
lines in the point of their intersection, it is also at right angles to the
plane passing through them: (XI. 4.)
but the plane passing through ED, GH is the plane BH;
therefore AF is perpendicular to the plane BH:
therefore, from the given point A, above the plane BH, the straight
line AF is drawn perpendicular to that plane. Q.E.F.
PROPOSITION XII. PROBLEM.
To erect a straight line at right angles to a given plane, from a point
given in the plane.
Let A be the point given in the plane.
It is required to erect a straight line from the point A at right
angles to the plane.
་
216
EUCLID'S ELEMENTS.
D B
A
From any point B above the plane draw BC perpendicular to it; (xI. 11.)
and from A draw AD parallel to BC. (1. 31.)
Because, therefore, AD, CB are two parallel straight lines,
and one of them BC is at right angles to the given plane,
the other AD is also at right angles to it:
: (XI. 8.)
therefore a straight line has been erected at right angles to a given
plane, from a point given in it. Q. E. F.
PROPOSITION XIII. THEOREM.
From the same point in a given plane, there cannot be two straight lines
at right angles to the plane, upon the same side of it: and there can be
but one perpendicular to a plane from a point above the plane.
For, if it be possible, let the two straight lines AB, AC be at right
angles to a given plane from the same point A in the plane, and upon
the same side of it.
Let a plane pass through BA, AC;
the common section of this with the given plane is a straight line
passing through A: (x1.3.)
B
C
DA E
let DAE be their common section:
therefore the straight lines AB, AC, DAE are in one plane :
and because CA is at right angles to the given plane,
it makes right angles with every straight line meeting it in that
plane: (XI. def. 3.)
but DAE, which is in that plane, meets CA;
therefore CAE is a right angle.
For the same reason, BAE is a right angle.
(ax. 11.)
Wherefore the angle CAE is equal to the angle BAE;
and they are in one plane, which is impossible.
Also, from a point above a plane, there can be but one perpendi-
cular to that plane:
for, if there could be two, they would be parallel to one another,
which is absurd. (x1. 6.)
Therefore, from the same point, &c.
Q. E.D.
PROPOSITION XIV. THEOREM.
Planes to which the same straight line is perpendicular, are parallel to
one another.
BOOK XI. PROP. XIV, XV.
217
Let the straight line AB be perpendicular to each of the planes
CD, EF.
These planes shall be parallel to one another.

G
H
E
D
B
F
If not, they shall meet one another when produced:
let them meet; their common section is a straight line GH, in
which take any point K, and join AK, BK.
Then, because AB is perpendicular to the plane EF,
it is perpendicular to the straight line BK which is in that
plane: (XI. def. 3.)
therefore ABK is a right angle.
For the same reason, BAK is a right angle:
wherefore the two angles ABK, BAK of the triangle ABK are
equal to two right angles, which is impossible: (1. 17.)
therefore the planes CD, EF, though produced, do not meet one
another;
that is, they are parallel. (xI. def. 8.)
Therefore, planes, &c.
Q. E.D.
PROPOSITION XV. THEOREM.
If two straight lines meeting one another be parallel to two other
straight lines which meet one another, but are not in the same plane with
the first two; the plane which passes through these is parallel to the plane
passing through the others.
Let AB, BC, two straight lines meeting one another, be parallel to
DE, EF, two other straight lines that meet one another, but are not in
the same plane with AB, BC.
The planes through AB, BC, and DE, EF shall not meet, though
produced.
E

G
B
F
K
D
H
From the point B draw BG perpendicular to the plane which passes
through DE, EF, (x1.11.)
and let it meet that plane in G;
and through G draw GH parallel to ED, and GK parallel to EF. (1.31.)
And because BG is perpendicular to the plane through DE, EF,
it makes right angles with every straight line meeting it in that
plane: (xI. def. 3.)
but the straight lines GH, GK in that plane meet it;
therefore each of the angles BGH, BGK is a right angle:
218
EUCLID'S ELEMENTS.
(XI. 9.)
and because BA is parallel to GH (for each of them is parallel to
DE, and they are not both in the same plane with it),
the angles GBA, BGH are together equal to two right angles: (1. 29.)
and BGH is a right angle ;
therefore also GBA is a right angle, and GB perpendicular to BA.
For the same reason, GB is perpendicular to BC.
Since therefore the straight line GB stands at right angles to the
two straight lines BA, BC that cut one another in B;
GB is perpendicular to the plane through BA, BC: (x1. 4.)
and it is perpendicular to the plane through DE, EF; (constr.)
therefore BG is perpendicular to each of the planes through AB,
BC, and DE, EF:
but planes to which the same straight line is perpendicular, are
parallel to one another; (XI. 14.)
therefore the plane through AB, BC is parallel to the plane through
DE, EF.
Wherefore, if two straight lines, &c. Q. E.D.

PROPOSITION XVI. THEOREM.
If two parallel planes be cut by another plane, their common sections
with it are parallels.
Let the parallel planes AB, CD be cut by the plane EFHG, and let
their common sections with it be EF, GH.
Then EF shall be parallel to GH.
K

F
H
B
D
A
E
For, if it is not, EF, GH shall meet, if produced, either on the side
of FH, or EG.
First, let them be produced on the side of FH, and meet in the point K.
Therefore, since EFK is in the plane AB,
every point in EFK is in that plane : (xI. 1.)
and K is a point in EFK;
therefore K is in the plane AB:
for the same reason, K is also in the plane CD:
wherefore the planes AB, CD produced, meet one another:
but they do not meet, since they are parallel by the hypothesis;
therefore the straight lines EF, GH, do not meet when produced
on the side of FH.
In the same manner it may be proved, that EF, GH do not meet
when produced on the side of EG.
But straight lines which are in the same plane, and do not meet,
though produced either way, are parallel;
therefore EF is parallel to GH.
Wherefore, if two parallel planes, &c. Q.E.D.
BOOK XI. PROP. XVII, XVIII,
219
PROPOSITION XVII. THEOREM.
If two straight lines be cut by parallel planes, they shall be 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 is to EB, so shall CF be to FĎ.
H

G
C
I
K
N
M
B
D
Join AC, BD, AD, and let AD meet the plane KL in the point X;
and join EX, XF.
Because the two parallel planes KL, MN are cut by the plane EBDX,
the common sections EX, BD are parallel: (x1. 16.)
for the same reason, because the two parallel planes GH, KL are
cut by the plane AXFC,
the common sections AC, XF are parallel :
and because EX is parallel to BD, a side of the triangle ABD;
as AE to EB, so is AX to XD: (vi. 2.)
again, because XF is parallel to AC, a side of the triangle ADC;
as AX to XD, so is CF to FD:
and it was proved that AX is to XD, as AE to EB;
therefore, as AE to EB, so is CF to FD. (v. 11.)
Wherefore, if two straight lines, &c.
Q.E.D.
PROPOSITION XVIII. THEOREM.
If a straight line be at right angles to a plane, every plane which
passes through it shall be at right angles to that plane.
Let the straight line AB be at right angles to the plane CK.
Every plane which passes through AB shall be at right angles to
the plane CK.
D G A H

K
C FB E
Let any plane DE pass through AB,
and let CE be the common section of the planes DE, CK;
take any point F in CE, from which draw FG in the plane DE at
right angles to CE. (1. 11.)
And because AB is perpendicular to the plane CK,
therefore it is also perpendicular to every straight line in that plane
meeting it; (xi. def. 3.)
220
EUCLID'S ELEMENTS.
and consequently it is perpendicular to CE:
wherefore ABF is a right angle:
but GFB is likewise a right angle; (constr.)
therefore AB is parallel to FG: (1. 28.)
and AB is at right angles to the plane CK;
therefore FG is also at right angles to the same plane. (x1. 8.)
But one plane is at right angles to another plane when the straight
lines drawn in one of the planes, at right angles to their common section,
are also at right angles to the other plane; (x1. def. 4.)
and any straight line FG in the plane DE, which is at right angles
to CE, the common section of the planes, has been proved to be per-
pendicular to the other plane CK;
therefore the plane DE is at right angles to the plane CK.
In like manner, it may be proved that all planes which pass through
AB are at right angles to the plane CK.
Therefore, if a straight line, &c. Q.E.D.
PROPOSITION XIX. THEOREM.
If two planes which cut one another be each of them perpendicular
to a third plane; their common section shall be perpendicular to the
same plane.
Let the two planes AB, BC be each of them perpendicular to a third
plane, and let BD be the common section of the first two.
Then BD shall be perpendicular to the third plane.
B

EF
D
A
C
If it be not, from the point D draw, in the plane AB, the straight
line DE at right angles to AD the common section of the plane AB
with the third plane; (1. 11.)
•
and in the plane BC draw DF at right angles to CD the common
section of the plane BC with the third plane.
And because the plane AB is perpendicular to the third plane, and
DE is drawn in the plane AB at right angles to AD, their common
section,
DE is perpendicular to the third plane. (x1. def. 4.)
In the same manner, it may be proved, that DF is perpendicular to
the third plane.
Wherefore, from the point D two straight lines stand at right angles
to the third plane, upon the same side of it, which is impossible: (xI. 13.)
therefore, from the point D there cannot be any straight line at
right angles to the third plane, except BD the common section of the
planes AB, BC:
therefore BD is perpendicular to the third plane.
Wherefore, if two planes, &c. Q.E.D.
BOOK XI. PROP. XX, XXI.
221
PROPOSITION XX. THEOREM.
If a solid angle be contained by three plane angles, any two of them
are greater than the third.
Let the solid angle at A be contained by the three plane angles
BAC, CAD, DAB.
Any two of them shall be greater than the third.
A
D
B
E C
If the angles BAC, CAD, DAB be all equal,
it is evident, that any two of them are greater than the third.
But if they are not, let BAC be that angle which is not less than
either of the other two, and is greater than one of them DAB;
and at the point A in the straight line AB,
make, in the plane which passes through BA, AC, the angle BAE
equal to the angle DAB; (1. 23.)
and make AE equal to AD, and through E draw BEC cutting AB,
AC in the points B, C, and join DB, DC.
And because DA is equal to AE, and AB is common,
the two DA, AB are equal to the two EA, AB, each to each;
and the angle DAB is equal to the angle EAB:
therefore the base DB is equal to the base BE:
and because BD, DC are greater than CB, 1. 20.)
1. 4.)
and one of them BD has been proved equal to BE a part of CB,
therefore the other DC is greater than the remaining part EC: (1. ax. 5.)
and because DĂ is equal to AE, and AC common,
but the base DC greater than the base EC;
therefore the angle DAC is greater than the angle EAC; (1. 25.)
and, by the construction, the angle DAB is equal to the angle BAÉ;
wherefore the angles DAB, DAC are together greater than BAE,
EAC, that is, than the angle BAC: (1. ax. 4.)
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, &c.
Q.E. D.
PROPOSITION XXI. THEOREM:
Every solid angle is contained by plane angles, which together are
less than four right angles.
First, let the solid angle at A be contained by three plane angles
BAC, CAD, DAB.
These three together shall be less than four right angles.
D
B
222
EUCLID'S ELEMENTS.
Take in each of the straight lines AB, AC, AD, any points B, C, D,
and join BC, CD, DB.
Then, because the solid angle at B is contained by the three plane
angles CBA, ABD, DBC,
any two of them are greater than the third; (x1. 20.)
therefore the angles CBA, ABD are greater than the angle DBC:
for the same reason, the angles BCA, ACD are greater than the
angle DCB;
and the angles CDA, ADB greater than BDC:
wherefore the six angles CBA, ABD, BCA, ACD, CDA, ADB
are greater than the three angles DBC, BCD, CDB:
but the three angles DBC, BCD, CDB are equal to two right
angles; (1. 32.)
therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are
greater than two right angles :
and because the three angles of each of the triangles ABC, ACD,
ADB are equal to two right angles,
therefore the nine angles of these three triangles, viz. the angles
CBA, BAC, ACB, ACD, CDA, DAC, ADB, DBA, BAD are equal to
six right angles;
of these the six angles CBA, ACB, ACD, CDA, ADB, DBA are
greater than two right angles;
therefore the remaining three angles BAC, CAD, DAB, which
contain the solid angle at A, are less than four right angles.
Next, let the solid angle at A be contained by any number of plane
angles BAC, CAD, DAË, EAF, FAB.
These shall together be less than four right angles.
A

B
K
D
Let the planes in which the angles are, be cut by a plane,
and let the common sections of it with those planes be BC, CD,
DE, EF, FB.
And because the solid angle at B is contained by three plane angles
CBA, ABF, FBC, of which any two are greater than the third, (x1. 20.)
the angles CBA, ABF, are greater than the angle FBC:
for the same reason, the two plane angles at each of the points C, D,
E, F, viz. those 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 angles of the polygon BCDEF:
therefore all the angles at the bases of the triangles are together
greater than all the angles of the polygon:
and because all the angles of the triangles are together equal to
twice as many right angles as there are triangles; (1. 32.)
that is, as there are sides in the polygon BCDEF;
and that all the angles of the polygon, together with four right
angles, are likewise equal to twice as many right angles as there are
sides in the polygon; (1. 32. Cor. 1.)
BOOK XI. PROP. XXI, XXII.
223
therefore all the angles of the triangles are equal to all the angles
of the polygon together with four right angles: (1. ax. 1.)
but all the angles at the bases of the triangles are greater than all
the angles of the polygon, as has been proved;
wherefore the remaining angles of the triangles, viz. those of the
vertex, which contain the solid angle at A, are less than four right
angles.
Therefore, every solid angle, &c. Q.E.D.
PROPOSITION XXII. THEOREM.
If every two of three plane angles be greater than the third, and if
the straight lines which contain them be all equal; a triangle may be made
of the straight lines that join the extremities of those equal straight lines.
Let ABC, DEF, GHK be the three plane angles, whereof every
two are greater than the third, and let them be contained by the equal
straight lines AB, BC, DE, EF, GH, HK:
if their extremities be joined by the straight lines AC, DF, GK, a
triangle may be made of three straight lines equal to AC, DF, GK;
that is, every two of them shall together be greater than the third.
B
E
H
A
C
D
G K
If the angles at B, E, H are equal,
AC, DF, GK are also equal, (1. 4.) and any two of them greater
than the third:
but if the angles are not all equal,
let the angle ABC be not less than either of the two at E, H;
therefore the straight line AC is not less than either of the other
two DF, GK;~~ (1. 4. or 24.)
and therefore it is plain that AC, together with either of the other
two, must be greater than the third.
Also, DF with GK shall be greater than AC.
For, at the point B in the straight line AB make the angle ABL
equal to the angle GHK, (1. 23.)
and make BL equal to one of the straight lines AB, BC, DE, EF,
GH, HK, and join AL, LC.
.
Then, because AB, BL are equal to GH, HK, each to each,
and the angle ABL to the angle GHK,
the base AL is equal to the base GK: (1. 4.)
and because the angles at E, H are greater than the angle ABC,
(hyp.) of which the angle at H is equal to ABL,
therefore the remaining angle at E is greater than the angle LBC: (ax. 5.)
and because the two sides LB, BC are equal to the two DE, EF,
each to each,
and that the angle DEF is greater than the angle LBC,
the base DF is greater than the base LC: (1. 24.)
and it has been proved that GK is equal to AL;
therefore DF and GK are greater than AL and LC: (1. ax. 4.)
224
EUCLID'S ELEMENTS.
but AL and LC are greater than AC; (1. 20.)
much more then are DF and GK greater than AC.
Wherefore every two of these straight lines AC, DF, GK are greater
than the third;
and, therefore, a triangle may be made, the sides of which shall be
equal to AC, DF, GK. (1. 22.) Q.E.D.
PROPOSITION XXIII. PROBLEM.
To make a solid angle which shall be contained by three given plane
angles, any two of them being greater than the third, (x1. 20.) and all
three together less than four right angles. (x1. 21.)
Let the three given plane angles be ABC, DEF, GHK, 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 three plane angles
equal to ABC, DEF, GHK, each to each.
From the straight lines which contain the angles cut off AB, BC,
DE, EF, GH, HK, all equal to one another; and join AC, DF, GK:
then a triangle may be made of three straight lines equal to AC,
DF, GK. (x1. 22.)

A
B
E
H
R
I
X
M
N
D
F
G
K
Let this be the triangle LMN, so that AC be equal to LM, DF to
MN, and GK to LN; (1. 22.)
and about the triangle LMN describe a circle, (IV. 5.) and find its
centre X, which will be either within the triangle, or in one of its sides,
or without it. (III. 1.)
First, let the centre X be within the triangle, and join LX, MX, NX.
AB shall be greater than LX.
If not, AB must either be equal to, or less than LX.
First, let it be equal:
then, because AB is equal to LX, and that AB is also equal to BC,
and LX to XM,
AB and BC are equal to LX and XM, each to each ;
and the base AC is, by construction, equal to the base LM;
wherefore the angle ABC is equal to the angle LXM. (1.8.)
For the same reason, the angle DEF is equal to the angle MXÑ, and
the angle GHK to the angle NXL:
therefore the three angles ABC, DEF, GHK are equal to the three
angles LXM, MXN, NXL:
but the three angles LXM, MXN, NXL are equal to four right
angles; (1. 15. Cor. 2.)
therefore also the three angles ABC, DEF, GHK, are equal to four
right angles:
BOOK XI. PROP. XXIII.
225
but, by the hypothesis, they are less than four right angles; which
is absurd:
therefore AB is not equal to LX.
But neither can AB be less than LX:
for, if possible, let it be less;
and upon the straight line LM, on the side of it on which is the
centre X, describe the triangle LOM, the sides LO, OM of which are
equal to AB, BC: (1.22.)
and because the base LM is equal to the base AC,
the angle LOM is equal to the angle ABC. (1. 8.)
And AB, that is, LO is by the hypothesis less than LX:
wherefore LO, OM fall within the triangle LXM ;
for if they fell upon its sides, or without it, they would be equal to,
or greater than LX, XM: (1. 21.)
therefore the angle LOM, that is, the angle ABC, is greater than
the angle LXM. (1. 21.)
In the same manner it may be proved that the angle DEF is greater
than the angle MXN, and the angle GHK greater than the angle NXL:
therefore the three angles ABC, DEF, GHK are greater than the
three angles LXM, MXN, NXL; that is, than four right angles:
(1. 15. Cor. 2.)
but the same angles ABC, DEF, GHK are less than four right
angles; which is absurd: (hyp.)
therefore AB is not less than LX:
and it has been proved that it is not equal to LX;
wherefore AB is greater than LX.
Next, let the centre X of the circle fall in one of the sides of the
triangle, viz. in MN, and join XL.
In this case also, AB shall be greater than LX.
I
R

M
N
If not, AB is either equal to LX, or less than it.
First, let it be equal to LX:
therefore AB and BC, that is, DE and EF, are equal to MX and
XL, that is, to MN:
but, by the construction, MN is equal to DF;
therefore DE, EF are equal to DF, which is impossible: (1. 20.)
wherefore AB is not equal to LX:
nor is it less; for then, much more, an absurdity would follow:
therefore AB is greater than LX.
But, let the centre X of the circle fall without the triangle LMN,
and join LX, MX, NX.
In this case likewise AB shall be greater than LX.
If not, it is either equal to or less than LX.
First, let it be equal:
it may be proved, in the same manner as in the first case, that the
angle ABC is equal to the angle MXL, and GHK to LXN:
therefore the whole angle MXN is equal to the two angles ABC, GHK;
15
226
EUCLID'S ELEMENTS.
but ABC and GHK are together greater than the angle DEF; (hyp.)
therefore also the angle MXN is greater than DEF:
and because DE, EF are equal to MX, XN, each to each,
and the base DF to the base MN,
the angle MXN is equal to the angle DEF: (1. 8.)
but it has been proved, that it is greater than DEF, which is absurd.
Therefore AB is not equal to LX:
neither is it less;
for then, as has been proved in the first case,
the angle ABC is greater than the angle MXL, and the angle GHK
greater than the angle LXN.
At the point B, in the straight line CB, make the angle CBP equal
to the angle GHK,
and make BP equal to HK, and join CP, AP.
R

B
E
H
M
N
A
D
F
G
K
And because CB is equal to GH,
CB, BP are equal to GH, HK, each to each; and they contain
equal angles;
wherefore the base CP is equal to the base GK, that is, to LN.
And in the isosceles triangles ABC, MXL,
because the angle ABC is greater than the angle MXL,
therefore the angle MLX at the base is greater than the angle ACB
at the base. (1. 32.)
For the same reason, because the angle GHK or CBP is greater
than the angle LXN,
the angle XLN is greater than the angle BCP:
therefore the whole angle MLN is greater than the whole angle ACP.
And because ML, LN are equal to AC, CP, each to each,
but the angle MLN is greater than the angle ACP,
the base MN is greater than the base AP: (1. 24.)
but MN is equal to DF;
therefore also DF is greater than AP.
Again, because DE, EF are equal to AB, BP, each to each,
but the base DF greater than the base AP,
the angle DEF is greater than the angle ABP: (1.25.)
but ABP is equal to the two angles ABC, CBP, that is, to the two
angles ABC, GHK;
therefore the angle DEF is greater than the two angles ABC, GHK:
but it is also less than these, which is impossible. (hyp.)
Therefore AB is not less than LX:
and it has been proved that it is not equal to it;
therefore AB is greater than LX.
From the point X erect XR at right angles to the plane of the circle
LMN. (XI. 12.)
And because it has been proved in all the cases, that AB is greater
than LX,
BOOK XI. PROP. XXIII, A.
227
find a square equal to the excess of the square of AB above the
square of LX,
and make RX equal to its side, and join RL, RM, RN.
The solid angle at R shall be the angle required.
Because RX is perpendicular to the plane of the circle LMN,
it is perpendicular to each of the straight lines LX, MX, NX. (x1. def. 3.)
And because LX is equal to MX, and XR common, and at right
angles to each of them,
the base RL is equal to the base RM. (1. 4.)
For the same reason, RN is equal to each of the two RL, RM:
therefore the three straight lines RL, RM, RN, are all equal.
And because the square of XR is equal to the excess of the square
of AB above the square of LX;
therefore the square of AB is equal to the squares of LX, XR:
but the square of RL is equal to the same squares, because LXR
is a right angle; (1. 47.)
therefore the square of AB is equal to the square of RL,
and the straight line AB to RL.
But each of the straight lines BC, DE, EF, GH, HK is equal to AB,
and each of the two RM, RN is equal to RL;
therefore AB, BC, DE, EF, GH, HK are each of them equal to
each of the straight lines RL, RM, RN.
And because RL, RM are equal to AB, BC, each to each,
and the base LM to the base AC;
the angle LRM is equal to the angle ABC. (1. 8.)
For the same reason, the angle MRN is equal to the angle DEF,
and NRL to GHK.
Therefore there is made a solid angle at R, which is contained by
three plane angles LRM, MRN, NRL, which are equal to the three
given plane angles ABC, DEF, GHK, each to each. Q.E.F.
PROPOSITION A. THEOREM.
If each of two solid angles be contained by three plane angles, which
are equal to one another, each to each; the planes in which the equal
angles are, have the same inclination to one another.
Let there be two solid angles at the points A, B;
and let the angle at A be contained by the three plane angles
CAD, CAE, EAD;
and the angle at B by the three plane angles FBG, FBH, HBG;
of which the angle CAD is equal to the angle FBG, and CAE to
FBH, and EAD to HBG.
Then the planes in which the equal angles are, shall have the same
inclination to one another.
In the straight line AC take any point K,
and from K draw in the plane CAD the straight line KD at right
angles to AC, (1.11.)
and in the plane CAE the straight line KL at right angles to the
same AC:
therefore the angle DKL is the inclination of the plane CAD to the
plane CAE. (XI. def. 6.)
In BF take BM equal to AK,
*
15-2
228
EUCLID'S ELEMENTS.
and from the point M draw in the planes FBG, FBH, the straight
lines MG, MN at right angles to BF;
therefore the angle GMN is the inclination of the plane FBG to
the plane FBH. (x1. def. 6.)
A
B
K
M
H
D
F
G
C E
H
?
:
(
Join LD, NG.
And because in the triangles KAD, MBG, the angles KAD, MBG
are equal, as also the right angles AKD, BMG, and that the sides AK,
BM, adjacent to the equal angles, are equal to one another; (hyp.)
therefore KD is equal to MG, and AD to BG: (1. 26.)
for the same reason, in the triangles KAL, MBN,
KL is equal to MN, and AL to BN :
therefore in the triangles LAD, NBG, LA, AD are equal to NB,
BG, each to each; and they contain equal angles:
therefore the base LD is equal to the base NG.
Lastly, in the triangles KLD, MNG, the sides DK, KL are equal to
GM, MN, each to each, and the base LD to the base NG;
(1.4.)
therefore the angle DKL is equal to the angle GMN: (1.8.)
but the angle DKL is the inclination of the plane CAD to the
plane CAE,
and the angle GMN is the inclination of the plane FBG to the
plane FBH,
which planes have therefore the same inclination to one another,
(XI. def. 7.)
And in the same manner it may be demonstrated, that the other
planes in which the equal angles are, have the same inclination to one
another.
Therefore if each of two solid angles, &c.
PROPOSITION B. THEOREM,
Q.E.D.
If two solid angles be contained, each by three plane angles which are
equal to one another, each to each, and alike situated; these solid angles
are equal to one another.
Let there be two solid angles at A and B, of which the solid angle
at A is contained by the three plane angles CAD, CAE, EAD;
and that at B, by the three plane angles FBG, FBH, HBG; of which
CAD is equal to FBG; CAE to FBH; and EAD to HBG.
Then the solid angle at A shall be equal to the solid angle at B.
Let the solid angle at A be applied to the solid angle at B:
A
B
C
E
D
F
H
G
´and first, the plane angle CAD being applied to the plane angle FBG,
so that the point A may coincide with the point B, and the straight
line AC with BF;
BOOK XI. PROP. B, C.
229.
then AD coincides with BG,
because the angle CAD is equal to the angle FBG:
and because the inclination of the plane CAE to the plane CAD is
equal to the inclination of the plane FBH to the plane FBG, (XI. A.)
the plane CAE coincides with the plane FBH,
because the planes CAD, FBG coincide with one another:
and because the straight lines AC, BF coincide,
and that the angle CAE is equal to the angle FBH;
therefore AE coincides with BH:
and AD coincides with BG;
wherefore the plane EAD coincides with the plane HBG:
therefore the solid angle A coincides with the solid angle B, and
consequently they are equal to one another. (1. ax. 8.) Q.E. D.
PROPOSITION C. THEOREM.
Solid figures which are contained by the same number of equal and
similar planes alike situated, and having none of their solid angles con-
tained by more than three plane angles, are equal and similar to one·
another.
Let AG, KQ be two solid figures contained by the same number of
similar and equal planes, alike situated,
viz. let the plane AC be similar and equal to the plane KM; the
plane AF to KP; BG to LQ; GD to QN; DE to NO; and, lastly,
FH similar and equal to PR.
The solid figure AG shall be equal and similar to the solid figure KQ.
H
G
R
Q
E
F
P
C
IM
D
A
B
K
L
Because the solid angle at A is contained by the three plane angles
BAD, BAE, EAD, which, by the hypothesis, are equal to the plane
angles LKN, LKO, OKN, which contain the solid angle at K, each to
each;
therefore the solid angle at A is equal to the solid angle at K. (XI. B.)
In the same manner, the other solid angles of the figures are equal
to one another.
Let then the solid figure AG be applied to the solid figure KQ;
first, the plane figure AC being applied to the plane figure KM,
so that the straight line AB may coincide with KL,
the figure AC must coincide with the figure KM, because they are
equal and similar:
therefore the straight lines AD, DC, CB coincide with KN, NM,
ML, each with each;
and the points A, D, C, B with the points K, N, M, L:
and the solid angle at A coincides with the solid angle at K; (XI. B.)
wherefore the plane AF coincides with the plane KP, and the
figure AF with the figure KP,
because they are equal and similar to one another:
therefore the straight lines AE, EF, FB coincide with KO, OP, PL;
230
EUCLID'S ELEMENTS.
and the points E, F with the points O, P.
In the same manner, the figure AH coincides with the figure KR,
and the straight line DH with NR, and the point H with the point R.
And because the solid angle at B is equal to the solid angle at L,
it may be proved, in the same manner,
that the figure BG coincides with the figure LQ, and the straight
line CG with MQ, and the point G with the point Q.
Therefore, since all the planes and sides of the solid figure AG coin-
cide with the planes and sides of the solid figure KQ,
AG is equal and similar to KQ.
And in the same manner, any other solid figures whatever 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, may be proved to be equal and similar to one another.
Q.E. D.
PROPOSITION XXIV. THEOREM.
If a solid be contained by six planes, two and two of which are paral-
lels, the opposite planes are similar and equal parallelograms.
Let the solid CDGH be contained by the parallel planes AC, GF;
BG, CE; FB, AE.
Its opposite planes shall be similar and equal parallelograms.
B
H
A
G
€
F
D
E
Because the two parallel planes BG, CE are cut by the plane AC,
their common sections AB, CD are parallel: (xI. 16.)
again, because the two parallel planes BF, AE are cut by the plane AC,
their common sections AD, BC are parallel: (XI. 16.)
and AB is parallel to CD;
therefore AC is a parallelogram.
In like manner it may be proved, that each of the figures CE, FG,
GB, BF, AE is a parallelogram.
Join AH, DF;
And because AB is parallel to DC, and BH to CF;
the two straight lines AB, BH, which meet one another, are paral-
lel to DC and CF, which meet one another, and are not in the same
plane with the other two:
wherefore they contain equal angles; (x1. 10.)
therefore the angle ABH is equal to the angle DCF :
and because AB, BH are equal to DC, CF, each to each,
and the angle ABH equal to the angle DCF;
therefore the base AH is equal to the base DF, (1. 4.)
and the triangle ABH to the triangle DCF:
but the parallelogram BG is double of the triangle ABH, (1. 34.)
and the parallelogram CE double of the triangle DCF:
therefore the parallelogram BG is equal and similar to the paral-
lelogram CE.

BOOK XI. PROP. XXIV, XXV.
231
In the same manner it may be proved, that the parallelogram AC is
equal and similar to the parallelogram GF, and the parallelogram AE
to BF.
Therefore, if a solid, &c. Q.E.D.
PROPOSITION XXV. THEOREM.
If a solid parallelopiped be cut by a plane parallel to two of its opposite
planes; it divides the whole into two solids, the base of one of which shall
be to the base of the other, as the one solid is to the other.
Let the solid parallelopiped ABCD be cut by the plane EV, which
is parallel to the opposite planes AR, HD, and divides the whole into
the two solids ABFV, EGCD.
As the base AEFY of the first is to the base EHCF of the other,
so shall the solid ABFV be to the solid EGCD.
L
X
Z
B
G
R
I
T
K
EHMF
YFC
Q
S
Produce AH both ways, and take any number of straight lines HM,
MN, each equal to EH,
and any number AK, KL, each equal to EA,
and complete the parallelograms LO, KY, HQ, MS,
and the solids LP, KR, HU, MT.
Then, because the straight lines LK, KA, AE are all equal,
the parallelograms LO, KY, AF are equal; (1. 36.)
and likewise the parallelograms KX, KB, AG:
also the parallelograms LZ, KP, AR are equal, because they are
opposite planes: (XI. 24.)
for the same reason, the parallelograms EC, HQ, MS are equal, (1.36.)
and the parallelograms HG, HI, IN:
as also HD, MU, NT: (XI. 24.)
therefore three planes of the solid LP are equal and similar to three
planes of the solid KR, as also to three planes of the solid AV:
but the three planes opposite to these three are equal and similar to
them in the several solids, and none of their solid angles are contained
by more than three plane angles; (x1. 24.)
therefore the three solids LP, KR, AV are equal to one another: (xI. c.)
for the same reason, the three solids ED, HU, MT are equal to
one another:
therefore, what multiple soever the base LF is of the base AF,
the same multiple is the solid LV of the solid AV;
and whatever multiple the base NF is of the base HF,
the same multiple is the solid NV of the solid ED:
and if the base LF be equal to the base NF,
the solid LV is equal to the solid NV; (x1. c.)
and if the base LF be greater than the base NF,
the solid LV is greater than the solid NV; and if less, less.
Since then there are four magnitudes, viz. the two bases AF, FH,
and the two solids AV, ED;
+

232
EUCLID'S ELEMENTS.
and that of the base AF and solid AV, the base LF and solid LV
are any equimultiples whatever;
and of the base FH and solid ED, the base FN and solid NV are
any equimultiples whatever:
and since it has been proved, that if the base LF is greater than the
base FN, the solid LV is greater than the solid NV:
and if equal, equal; and if less, less;
therefore as the base AF is to the base FH, so is the solid AV
to the solid ED. (v. def. 5.)
Wherefore, if a solid, &c. Q.E. D.
PROPOSITION XXVI.
PROBLEM.
At a given point in a given straight line, to make a solid angle equal
to a given solid angle contained by three plane angles.
Let AB be a given straight line, A a given point in it, and D a
given solid angle contained by the three plane angles EDC, EDF,
FDC.
It is required to make at the point A, in the straight line AB, a solid
angle equal to the solid angle D.
In the straight line DF take any point F, from which draw FG
perpendicular to the plane EDC, meeting that plane in G, (xI. 11.)
and i oin DG:
j
at the point A, in the straight line AB, make the angle BAL equal
to the angle EDC, (1.23.)
and in the plane BAL make the angle BAK equal to the angle EDG;
then make AK equal to DG, and from the point K erect KH
at right angles to the plane BAL, (x1. 12.) and make KH equal to
GF, and join AH.
The solid angle at A which is contained by the three plane angles
BAL, BAH, HAL shall be equal to the solid angle at D contained by
the three plane angles EDC, EDF, FDC.
A
D
B
L
E
C
K H
G
Take the equal straight lines AB, DE, and join HB, KB, FE, GE
And because FG is perpendicular to the plane EDC, it makes right
angles with every straight line meeting it in that plane: (x1. def. 3.)
therefore each of the angles FGD, FGE is a right angle.
For the same reason, HKA, HKB are right angles.
And because KA, AB are equal to GD, DE, each to each,
and that they contain equal angles,
therefore the base BK is equal to the base EG; (1. 4.)
and KH is equal to GF, and HKB, FGE are right angles, (constr.)
therefore HB is equal to FE. (1. 4.)
Again, because AK, KH are equal to DG, GF, each to each, and
contain right angles,
the base AH is equal to the base DF;
and AB is equal to DE,
BOOK XI. PROP. XXVI, XXVII.
233
therefore, HA, AB are equal to FD, DE, each to each ;
and the base HB is equal to the base FE;
therefore the angle BAH is equal to the angle EDF. (1. 8.)
For the same reason, the angle HAL is equal to the angle FDC:
because if AL and DC be made equal, and KL, HL, GC, FC be joined;
since the whole angle BAL is equal to the whole EDC, and the
parts of them BAK, EDG are, by the construction, equal;
therefore the remaining angle KAL is equal to the remaining
angle GDC:
and because KA, AL are equal to GD, DC, each to each, and
contain equal angles,
the base KL is equal to the base GC; (1. 4.)
and KĤ is equal to GF;
so that LK, KH are equal to CG, GF, each to each ;
and they contain right angles; (xI. def. 3.)
therefore the base HL is equal to the base FC: (I. 4.)
again, because HA, AL are equal to FD, DC, each to each, and
the base HL to the base FC,
the angle HAL is equal to the angle FDC. (1. 8.)
Therefore, because the three plane angles BAL, BAH, HAL, which
contain the solid angle at A, are equal to the three plane angles 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 to the solid angle at D. (XI. B.)
Therefore at a given point in a given straight line a solid angle
has been made equal to a given solid angle contained by three plane
angles. Q.E.F.
PROPOSITION XXVII. PROBLEM.
To describe, from a given straight line, a solid parallelopiped similar
and similarly situated to one given.
Let AB be the given straight line, and CD the given solid
parallelopiped.
It is required from AB to describe a solid parallelopiped similar and
similarly situated to CD.
At the point A of the given straight line AB, make a solid angle
equal to the solid angle at C, (xI. 26.)
and let BAK, KAH, HAB be the three plane angles which contain it,
so that BAK be equal to the angle ECG, and KAH to GCF, and
HAB to FCE:
and as EC to CG, so make BA to AK; (v1. 12.)
and as GC to CF, so make KA to AH; (vI. 12.)
wherefore, ex æquali, as EC to CF, so is BA to AH: (v. 22.)
complete the parallelogram BH, and the solid AL.
Then AL shall be similar and similarly situated to CD.
L
H
D
M
F
K
A
B
C
E
234
EUCLID'S ELEMENTS.
Because, as EC to GC, so is BA to AK,
the sides about the equal angles ECG, BAK, are proportionals;
therefore the parallelogram BK is similar to EG.* (v1. def. 1.)
For the same reason, the parallelogram KH is similar to GF, and
HB to FE;
wherefore three parallelograms of the solid AL are similar to three
of the solid CD:
and the three opposite ones in each solid are equal and similar to
these, each to each. (XI. 24.)
Also, because the plane angles which contain the solid angles of the
figures are equal, each to each, and situated in the same order,
the solid angles are equal, each to each. (XI. B.)
Therefore the solid AL is similar to the solid CD. (x1. def. 11.)
Wherefore, from a given straight line AB, a solid parallelopiped AL
has been described similar and similarly situated to the given one
CD. Q.E.F.
PROPOSITION XXVIII. THEOREM.
If a solid parallelopiped be cut by a plane passing through the diagonals
of two of the opposite planes; it shall be cut into two equal parts.
Let AB be a solid parallelopiped, and DE, CF the diagonals of the
opposite parallelograms AH, GB, viz. those which are drawn betwixt
the equal angles in each.
And because CD, FE are each of them parallel to GA, and not in
the same plane with it,
CD, FE are parallel; (x1. 9.)
wherefore the diagonal CF, DE, are in the plane in which the
parallels are, and are themselves parallels: (x1. 16.)
and the plane CDEF shall cut the solid AB into two equal parts.
C
B

G
n
D
H
A
E
Because the triangle CGF is equal to the triangle CBF, (1.34.)
and the triangle DAE to DHE;
and that the parallelogram CA is equal and similar to the opposite
one BE, (xI. 24.)
and the parallelogram GE to CH;
therefore the prism contained by the two triangles CGF, DAE, and
the three parallelograms, CA, GE, EC, is equal to the prism contained
by the two triangles CBF, DHE, and the three parallelograms BE,
CH, EC; (x1. c.)
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 angles.
Therefore the solid ÃB is cut into two equal parts by the plane
CDEF.
Q.E.D.
BOOK XI. PROP. XXIX.
235
PROPOSITION XXIX. THEOREM.
Solid parallelopipeds upon the same base, and of the same altitude, the
insisting straight lines of which are terminated in the same straight lines
in the plane opposite to the base, are equal to one another.
Let the solid parallelopipeds AH, AK be upon the same base AB,
and of the same altitude,
and let their insisting straight lines AF, AG, LM, LN be terminated
in the same straight line FN,
and CD, CE, BH, BK be terminated in the same straight line DK.
The solid AH shall be equal to the solid AK.
First, let the parallelograms DG, HN, which are opposite to the
base AB, have a common side HG.
D
H
K

12
G
N
B
A
L
Then, because the solid AH is cut by the plane AGHC passing
through the diagonals, AG, CH, of the opposite planes ALGF, CBHD,
AH is cut into two equal parts by the plane AGHC; (x1. 28.)
therefore the solid AH is double of the prism which is contained
betwixt the triangles ALG, CBH:
for the same reason, because the solid AK is cut by the plane
LGHB, through the diagonals LG, BH of the opposite planes AĹNG,
CBKH,
the solid AK is double of the same prism which is contained betwixt
the triangles ALG, CBH:
therefore the solid AH is equal to the solid AK. (1. ax. 6.)
Next let the parallelograms DM, EN, opposite to the base, have no
common side.
D
HE K
M
F
G
N
C
B
A
L
DEH K

G
N
બ
F
B
A
L
Then, because CH, CK, are parallelograms,
CB is equal to each of the opposite sides DH, EK; (1.34.)
wherefore DH is equal to EK:
add or take away the common part HE;
then DE is equal to HK: (1. ax. 2 or 3.)
wherefore also the triangle CDE is equal to the triangle BHK, (1. 38.)
and the parallelogram DG is equal to the parallelogram HN: (1. 36.)
for the same reason, the triangle AFG is equal to the triangle LMN:
and the parallelogram CF is equal to the parallelogram BM, and
CG to BN; for they are opposite. (xI. 24.)
Therefore the prism which is contained by the two triangles AFG,
CDE, and the three parallelograms AD, DG, GC is equal to the prism
contained by the two triangles LMN, BHK, and the three parallelograms
BM, MK, KL. (x1. c.)
If therefore the prism LMN, BHK be taken from the solid of
:
236.
EUCLID'S ELEMENTS.
which the base is the parallelogram AB, and in which FDKN is the
one opposite to it; and if from this same solid there be taken the prism
AFG, CDE;
the remaining solid, viz. the parallelopiped AH, is equal to the
remaining parallelopiped AK. (1. ax. 3.)
Therefore solid parallelopipeds, &c. Q.E.D.
PROPOSITION XXX. THEOREM.
Solid parallelopipeds upon the same base, and of the same altitude, the
insisting straight lines of which are not terminated in the same straight
lines in the plane opposite to the base, are equal to one another.
Let the parallelopipeds CM, CN be upon the same base AB, and of
the same altitude, but their insisting straight lines AF, AG, LM, LN,
CD, CE, BH, BK not terminated in the same straight lines.
The solids CM, CN shall be equal to one another.
N K

E
G
M H
P
Q
R
F
B
A
C
Produce FD, MH, and NG, KE, and let them meet one another in
the points O, P, Q, R; and join AO, LP, BQ, CR.
And because the plane LBHM is parallel to the opposite plane ACDF,
and that the plane LBHM is that in which are the parallels LB,
MHPQ, in which also is the figure BLPQ;
and the plane ACDF is that in which are the parallels AC, FDOR,
in which also is the figure CAOR;
therefore the figures BLPQ, CAOR are in parallel planes:
in like manner, because the plane ALNG is parallel to the opposite
plane CBKE,
and that the plane ALNG is that in which are the parallels AL,
OPGN, in which also is the figure ALPO;
and the plane CBKE is that in which are the parallels CB, RQEK,
in which also is the figure CBQR;
therefore the figures ALPO, CBQR are in parallel planes:
and the planes ACBL, ORQP are parallel; (hyp.)
therefore the solid CP is a parallelopiped:
but the solid CM is equal to the solid CP, (x1. 29.)
because they are upon the same base ACBL, and their insisting
straight lines AF, AO; CD, CR; LM, LP; BH, BQ are in the same
straight lines FR, MQ:
and the solid CP is equal to the solid CN, (xI. 29.)
for they are upon the same base ACBL, and their insisting straight
lines AO, AG; LP, LN; CR, CE; BQ, BK are in the same straight
lines ON, RK:
therefore the solid CM is equal to the solid CN.
Wherefore solid parallelopipeds, &c. Q.E.D.
BOOK XI. PROP. XXXI.
237
PROPOSITION XXXI. THEOREM.
Solid parallelopipeds, which are upon equal bases, and of the same
altitude, are equal to one another.
Let the solid parallelopipeds AE, CF be upon equal bases AB, CD,
and be of the same altitude.
The solid AE shall be equal to the solid CF.
First, let the insisting straight lines be at right angles to the bases
AB, CD,
and let the bases be placed in the same plane, and so that the sides
CL, LB may be in a straight line;
therefore the straight line LM, which is at right angles to the plane
in which the bases are, in the point L, is common to the two solids
AE, CF: (xI. 13.)
let the other insisting lines of the solids be AG, HK, BE; DF, OP, CN:
and first, let the angle ALB be equal to the angle CLD:
then AL, LD are in a straight line. (1. 14.)
P F
Ꭱ

N M
E
כן
K
B
A S
HT
Produce OD, HB, and let them meet in Q,
and complete the solid parallelopiped LR, the base of which is the
parallelogram LQ, and of which LM is one of its insisting straight
lines.
Therefore, because the parallelogram AB is equal to CD,
as the base AB is to the base LQ, so is the base CD to the base
LQ. (v. 7.)
And because the solid parallelopiped AR is cut by the plane LMEB,
which is parallel to the opposite planes AK, DR;
as the base AB is to the base LQ, so is the solid AE to the solid
LR: (XI. 25.)
for the same reason, because the solid parallelopiped CR is cut by
the plane LMFD, which is parallel to the opposite planes CP,BR;
as the base CD to the base LQ, so is the solid CF to the solid LR:
but as the base AB to the base LQ, so the base CD to the base LQ,
as before was proved;
therefore, as the solid AE to the solid LR, so is the solid CF to the
solid LR: (v. 11.)
and therefore the solid AE is equal to the solid CF. (v. 9.)
But let the solid parallelopipeds SE, CF be upon equal bases SB,
CD, and be of the same altitude, and let their insisting straight lines be
at right angles to the bases;
and place the bases SB, CD in the same plane, so that CL, LB
may be in a straight line;
and let the angles SLB, CLD be unequal:
the solid SE shall be equal to the solid CF.
Produce DL, T'S until they meet in A, and from B draw BH parallel
to DA;
238
EUCLID'S ELEMENTS.
P
F
R

ENT
E
X
K
Q
B
L
A S
HT
and let HB, OD produced meet in Q, and complete the solids AE, LR;
therefore the solid AE is equal to the solid SE; (xI. 29.)
because they are upon the same base LE, and of the same altitude,
and their insisting straight lines, viz. LA, LS, BH, BT; MG, MV, EK,
EX, are in the same straight lines AT, GX:
and because the parallelogram AB is equal to SB, (1. 35.)
for they are upon the same base LB, and between the same parallels
LB, AT;
and that the base SB is equal to the base CD;
therefore the base AB is equal to the base CD ;
and the angle ALB is equal to the angle CLD;
therefore, by the first case, the solid AE is equal to the solid CF:
but the solid AE is equal to the solid SE, as was demonstrated;
therefore the solid SE is equal to the solid CF.
But, if the insisting straight lines AG, HK, BE, LM; CN, RS, DF,
OP be not at right angles to the bases AB, CD;
in this case likewise the solid AE shall be equal to the solid CF.
M E
G
K
B
L
A
H Q
T
P
F

אן
U
I
C R
Y Z
From the points G, K, E, M; N, S, F, P draw the straight lines
GQ, KT, EV, MX; NY, SZ, FI, PU, perpendicular to the plane in
which are the bases AB, CD; (x1. 11.)
and let them meet it in the points Q, T, V, X; Y, Z, I, U;
and join QT, TV, VX, XQ; YZ, ZI, IU, UY.
Then, because GQ, KT are at right angles to the same plane,
they are parallel to one another: (x1. 6.)
and MG, EK are parallels;
therefore the planes MQ, ET, of which one passes through MG, GQ,
and the other through EK, KT, which are parallel to MG, GQ, and not
in the same plane with them, are parallel to one another; (XI. 15.)
for the same reason, the planes MV, GT are parallel to one another;
therefore the solid QE is a parallelopiped.
In like manner it may be proved, that the solid YF is a parallelopiped.
But, from what has been demonstrated, the solid EQ is equal to the
solid FY,
because they are upon equal bases MK, PS, and of the same altitude,
and have their insisting straight lines at right angles to the bases:
and the solid EQ is equal to the solid AE, (XI. 29. or 30.)
and the solid FY to the solid CF,
because they are upon the same bases and of the same altitude;
therefore the solid AE is equal to the solid CF.
Wherefore, solid parallelopipeds, &c.
Q.E.D.
BOOK XI. PROP, XXXII, XXXIII.
239
PROPOSITION XXXII. THEOREM.
Solid parallelopipeds which have the same altitude, are to one another
as their bases.
Let AB, CD be solid parallelopipeds of the same altitude.
They shall be to one another as their bases;
that is, as the base AE to the base CF, so shall the solid AB be to
the solid CD.
B
D K
P
EN
E
E 35
C
G H
A M
To the straight line FG apply the parallelogram FH equal to AE,
(1. 45. Cor.)
so that the angle FGH may be equal to the angle LCG;
and upon the base FH complete the solid parallelopiped GK, one of
whose insisting lines is FD, whereby the solids CD, GK must be of the
same altitude.
Therefore the solid AB is equal to the solid GK, (x1. 31.)
because they are upon equal bases AE, FH, and are of the same altitude:
and because the solid parallelopiped CK is cut by the plane DG,
which is parallel to its opposite planes,
the base HF is to the base FC, as the solid HD to the solid DC: (x1. 25.)
but the base HF is equal to the base AE, and the solid GK to the
solid AB;
therefore, as the base AE to the base CF, so is the solid AB to the
solid CD.
Q.E.D.
Wherefore, solid parallelopipeds, &c.
COR. From this it is manifest, that prisms upon triangular bases, of
the same altitude, are to one another as their bases.
Let the prisms, the bases of which are the triangles AEM, CFG, and
NBO, PDQ the triangles opposite to them, have the same altitude:
they shall be to one another as their bases.
Complete the parallelograms AE, CF, and the solid parallelopipeds
AB, CD, in the first of which let MO, and in the other let GQ be one
of the insisting lines.
And because the solid parallelopipeds AB, CD have the same altitude,
they are to one another as the base AE is to the base CF:
wherefore the prisms, which are their halves, are to one another,
as the base ÃE to the base CF; (x1. 28.)
that is, as the triangle AEM to the triangle CFG.
PROPOSITION XXXIII. THEOREM.
Similar solid parallelopipeds are one to another in the triplicate ratio
of their homologous sides.
Let AB, CD be similar solid parallelopipeds, and the side AE homo-
logous to the side CF.
240
EUCLID'S ELEMENTS.
The solid AB shall have to the solid CD the triplicate ratio of that
which AE has to CF.
B X

P
H
G
D
K
R
A
E
L
M
C
Produce AE, GE, HE, and in these produced take EK equal to CF,
EL equal to FN, and EM equal to FR ;
and complete the parallelogram KL, and the solid KO.
Because KE, EL are equal to CF, FN, each to each,
and the angle KEL equal to the angle CFN,
because it is equal to the angle AEG, which is equal to CFN,
by reason that the solids AB, CD are similar;
therefore the parallelogram KL is similar and equal to the paral-
lelogram CÑ.
For the same reason the parallelogram MK is similar and equal to
CR, and also OE to FD.
Therefore three parallelograms of the solid KO are equal and similar
to three parallelograms of the solid CD:
and the three opposite ones in each solid are equal and similar to
these: (XI. 24.)
therefore the solid KÓ is equal and similar to the solid CD. (x1. c.)
Complete the parallelogram GK; and upon the bases GK, KL,
complete the solids EX, LP, so that EH be an insisting straight line in
each of them, whereby they must be of the same altitude with the
solid AB.
And because the solids AB, CD are similar,
and, by permutation, as AE is to CF, so is EG to FN, and so is
EH to FR:
but FC is equal to EK, and FN to EL, and FR to EM ;
therefore, as AE to EK, so is EG to EL, and so is HE to EM:
but as AE to EK, so is the parallelogram AG to the parallelogram
GK; (vI. 1.)
and as GE to EL, so is GK to KL; (vI. 1.)
and as HE to EM, so is PE to KM: (v1. 1.)
therefore, as the parallelogram AG to the parallelogram
GK to KL, and PE to KM:
GK, so is
(x1. 25.)
(XI. 25.)
(x1. 25.)
but as AG to GK, so is the solid AB to the solid EX;
and as GK to KL, so is the solid EX to the solid PL;
and as PE to KM, so is the solid PL to the solid KO;
and therefore as the solid AB to the solid EX, so is EX to PL, and
PL to KO:
but if four magnitudes be continual proportionals, the first is said to
have to the fourth, the triplicate ratio of that which it has to the second;
(v. def. 11.)
therefore the solid AB has to the solid KO, the triplicate ratio of
that which AB has to EX:
but as AB is to EX, so is the parallelogram AG to the parallelogram
GK, and the straight line AE to the straight line EK;
BOOK XI. PROP. XXXIII, D.
241
wherefore the solid AB has to the solid KO, the triplicate ratio of
that which AE has to EK:
but the solid KO is equal to the solid CD,
and the straight line EK is equal to the straight line CF;
therefore the solid AB has to the solid CD, the triplicate ratio of
that which the side AE has to the homologous side CF.
Therefore, similar solid parallelopipeds, &c. Q. E.D.
COR.-From this it is manifest, that if four straight lines be continual
proportionals, as the first is to the fourth, so is the solid parallelopiped
described from the first to the similar solid similarly described from the
second; because the first straight line has to the fourth the triplicate
ratio of that which it has to the second.
PROPOSITION D. THEOREM.
Solid parallelopipeds which are contained by parallelograms equian-
gular to one another, each to each, that is, of which the solid angles are
equal, each toʻeach, have to one another the ratio which is the same with
the ratio compounded of the ratios of their sides.
Let AB, CD be solid parallelopipeds, of which AB is contained by
the parallelograms AE, AF, AG, which are equiangular, each to each,
to the parallelograms CH, CK, CL, which contain the solid CD.
The ratio which the solid AB has to the solid CD, shall be the same
with that which is compounded of the ratios of the sides AM to DL,
AN to DK, and AO to DH.

D L
B G

F
K
H
Y
E
N
C
a.
M A
S
b-
T
R
X
C
d-
Produce MA, NA, OA to P, Q, R, so that AP be equal to DL, AQ
to DK, and AR to DH;
and complete the solid parallelopiped AX contained by the parallelo-
grams AS, AT, AV similar and equal to CH, CK, CL, each to each.
Therefore the solid AX is equal to the solid CD. (x1. c.)
Complete likewise the solid AY, the base of which is AS, and AO
one of its insisting straight lines.
Take any straight line a,
and as MA to AP, so make a to b; (vI. 12.)
and as NA to AQ, so make b to c;
and as AO to AR, so c to d.
Then, because the parallelogram AE is equiangular to AS,
AE is to AS, as the straight line a to c, as is demonstrated in Prop.
23, Book VI.:
and the solids AB, AY, being betwixt the parallel planes BOY,
EAS, are of the same altitude;
therefore the solid AB is to the solid AY, as the base AE to the
base AS; that is, as the straight line a is to c. (XI. 32.)
16
242
EUCLID'S ELEMENTS.
And the solid AY is to the solid AX, as the base OQ is to the base
QR; (XI. 25.)
that is, as the straight line OA to AR;
that is, as the straight line c to the straight line d.
And because the solid AB is to the solid AŸ, as a is to c,
and the solid AY to the solid AX, as c is to d
ex æquali, the solid AB is to the solid AX, or CD which is equal to
it, as the straight line a is to d.
But the ratio of a to d is said to be compounded of the ratios of a to
b, b to c, and c to d, (v. def. A.)
which are the same with the ratios of the sides MA to AB, NA to AQ,
and OA to AR, each to each :
and the sides, AP, AE, AR are equal to the sides DL, DK, DH,
each to each:
therefore the solid AB has to the solid CD the ratio which is the
same with that which is compounded of the ratios of the sides AM to
DL, AN to DK, and AO to DH. Q.E.D.
PROPOSITION XXXIV. THEOREM.
The bases and altitudes of equal solid parallelopipeds, are reciprocally
proportional: and conversely, if the bases and altitudes be reciprocally
proportional, the solid parallelopipeds are equal.
Let AB, CD be two solid parallelopipeds:
and first, let the insisting straight lines AG, EF, LB, HK;
CM, NX, OD, PR be at right angles to the bases.
If the solid AB be equal to the solid CD, their bases shall be
reciprocally proportional to their altitudes;
that is, as the base EH is to the base NP, so shall CM be to AG.
If the base EH be equal to the base NP,
K B
G
R
D
M
F
X
H
P
A E
C
N
then because the solid AB is likewise equal to the solid CD,
CM shall be equal to AG:
because if the bases EH, NP be equal, but the altitudes AG, CM
be not equal, neither shall the solid AB be equal to the solid CD:
but the solids are equal, by the hypothesis;
therefore the altitude CM is not unequal to the altitude AG; that
is, they are equal.
Wherefore, as the base EH to the base NP, so is CM to AG.
Next, let the bases EH, NP not be equal, but EH greater than
the other:
R D
K B
M
X
F
T
P
0
H
A
K
C
N
BOOK XI. PROP. XXXIV.
243
then since the solid AB is equal to the solid CD,
CM is therefore greater than AG:
for, if it be not, neither also in this case would the solids AB, CD
be equal, which, by the hypothesis, are equal.
Make then CT equal to AG, and complete the solid parallelopiped
CV, of which the base is NP, and altitude CT.
Because the solid AB is equal to the solid CD,
therefore the solid AB is to the solid CV, as the solid CD to the
solid CV: (v. 7.)
but as the solid AB to the solid CV, so is the base EH to the
base NP; (XI. 32.)
for the solids AB,
CV are of the same altitude:
and as the solid CD to CV so is the base MP to the base PT,
(XI. 25.) and so is the straight line MC to CT; (vI. 1.)
and CT is equal to AG:
therefore, as the base EH to the base NP, so is MC to AG.
Wherefore the bases of the solid parallelopipeds AB, CD are
reciprocally proportional to their altitudes.
Let now the bases of the solid parallelopipeds AB, CD be reci-
procally proportional to their altitudes, viz. as the base EH is to the
base NP, so let CM be to AG.
The solid AB shall be equal to the solid CD.
H
K B
G
F
Ꭱ
D
M
X
P
C
N
A E
If the base EH be equal to the base NP,
then since EH is to NP, as the altitude of the solid CD is to the
altitude of the solid AB,
therefore the altitude of CD is equal to the altitude of AB: (v. A.)
but solid parallelopipeds upon equal bases, and of the same altitude,
are equal to one another; (XI. 31.)
therefore the solid AB is equal to the solid CD.
But let the bases EH, NP be unequal, and let EH be the greater
of the two.
R D
K
B
M
G
F
સ
T
H
I
P
0
A E
CN
Therefore, since, as the base EH to the base NP, so is CM the
altitude of the solid CD to AG the altitude of AB,
CM is greater than AG. (v. A.)
Therefore, as before, take CT equal to AG, and complete the solid CV.
And because the base EH is to the base NP, as CM to AG,
and that AG is equal to CT,
therefore the base EH is to the base NP, as MC to CT.
But as the base EH is to NP, so is the solid AB to the solid CV; (xI. 32.)
for the solids AB, CV are of the same altitude:
and as MC to CT, so is the base MP to the base PT, (vI. 1.) and
the solid CD to the solid CV: (x1.25.)
16-2
244
EUCLID'S ELEMENTS.
therefore as the solid AB to the solid CV, so is the solid CD to
the solid CV;
that is, each of the solids AB, CD has the same ratio to the solid CV;
and therefore the solid AB is equal to the solid CD. (v. 9.)
Second general case. Let the insisting straight lines FE, BL, GA,
KH; XN, DO, MC, RP not be at right angles to the bases of the solids.
In this case, likewise, if the solids AB, CD be equal, their bases
shall be reciprocally proportional to their altitudes, viz. the base EH
shall be to the base NP, as the altitude of the solid CD to the altitude
of the solid AB.
K B
G
F
R D


M
X
P
I
H
A
E
C N
From the points F, B, K, G; X, D, R, M, draw perpendiculars to
the planes in which are the bases EH, NP, meeting those planes in the
points S, Y, V, T; Q, I, U, Z; and complete the solids FV, XU, which
are parallelopipeds, as was proved in the last part of Prop. 31, of this book.
Because the solid AB is equal to the solid CD, and that the solid AB
is equal to the solid BT, for they are upon the same base FK, and of
the same altitude; (xI. 30. or 29.)
and that the solid CD is equal to the solid DZ, being upon the same
base XR, and of the same altitude; (x1. 30. or 29.)
therefore the solid BT is equal to the solid DZ:
but the bases are reciprocally proportional to the altitudes of equal
solid parallelopipeds of which the insisting straight lines are at right
angles to their bases, as before was proved;
therefore as the base FK to the base XR, so is the altitude of the
solid DZ to the altitude of the solid BT:
and the base FK is equal to the base EH,
and the base XR to the base NP;
wherefore, as the base EH to the base NP, so is the altitude of the
solid DZ to the altitude of the solid BT:
but the altitudes of the solids DZ, DC, as also of the solids BT,
BA, are the same ;
therefore as the base EH to the base NP, so is the altitude of the
solid CD to the altitude of the solid AB;
that is, the bases of the solid parallelopipeds AB, CD are recipro-
cally proportional to their altitudes.
Next, let the bases of the solids AB, CD be reciprocally proportional
to their altitudes,
viz. the base EH to the base NP, as the altitude of the solid CD to
the altitude of the solid AB.
The solid AB shall be equal to the solid CD.
K
B
G
F
H
T
A
E
S
R
D


M
X
P
C
N
BOOK XI. PROP. XXXIV, xxxv.
245
The same construction being made;
because, as the base EH to the base NP, so is the altitude of the
solid CD to the altitude of the solid AB;
and that the base EH is equal to the base FK, and NP to XR ;
therefore the base FK is to the base XR, as the altitude of the solid
CD to the altitude of AB:
but the altitudes of the solids AB, BT are the same, as also of
CD and ᎠᏃ ;
therefore as the base FK to the base XR, so is the altitude of the
solid DZ to the altitude of the solid BT:
wherefore the bases of the solids BT, DZ are reciprocally propor-
tional to their altitudes :
and their insisting straight lines are at right angles to the bases;
wherefore, as was before proved, the solid BT is equal to the solid DZ:
but BT is equal to the solid BA, and DZ to the solid DC, (x1.30. or 29.)
because they are upon the same bases, and of the same altitude;
therefore the solid AB is equal to the solid CD.
Therefore, the bases, &c.
Q.E.D.
PROPOSITION XXXV. THEOREM.
If, from the vertices of two equal plane angles, there be drawn two
straight lines elevated above the planes in which the angles are, and con-
taining equal angles with the sides of those angles, each to each; and if
in the lines above the planes there be taken any points, and from them per-
pendiculars be drawn to the planes in which the first-named angles are;
and from the points in which they meet the planes, straight lines be drawn
to the vertices of the angles first named: these straight lines shall contain
equal angles with the straight lines which are above the planes of the
angles.
Let BAC, EDF be two equal plane angles :
and from the points A, D let the straight lines AG, DM be elevated
above the planes of the angles, making equal angles with their sides,
each to each,
viz. the angle GAB equal to the angle MDE, and GAC to MDF;
and in AG, DM let any points G, M be taken,
and from them let perpendiculars GL, MN be drawn to the planes
BAC, EDF, (XI. 11.) meeting these planes in the points L, N; and
join LA, ND.
The angle GAL shall be equal to the angle MDN.
A
B
H
G
D
C
F
K
E
L
N
M
Make AH equal to DM, and through H draw HK parallel to GL:
but GL is perpendicular to the plane BAC;
wherefore HK is perpendicular to the same plane. (XI. 8.)
From the points K, N, to the straight lines AB, AC, DE, DF, draw
perpendiculars KB, KC, NE, NF; and join HB, BC, ME, EF.
246
EUCLID'S ELEMENTS.
Because HK is perpendicular to the plane BAC,
the plane HBK which passes through HK is at right angles to the
plane BAC; (xI. 18.)
and AB is drawn in the plane BAC at right angles to the common
section BK of the two planes;
therefore AB is perpendicular to the plane HBK, (x1. def. 4.)
and makes right angles with every straight line meeting it in that
plane: (xI. def. 3.)
but BÍ meets it in that plane;
therefore ABH is a right angle:
for the same reason DEM is a right angle, and is therefore equal to
the angle ABH:
and the angle HAB is equal to the angle MDE: (hyp.)
therefore in the two triangles HAB, MDE there are two angles in
one, equal to two angles in the other, each to each, and one side
equal to one side, opposite to one of the equal angles in each, viz. HA
equal to DM;
therefore the remaining sides are equal, each to each: (1.26.)
wherefore AB is equal to DE.
In the same manner, if HC and MF be joined,
it may be demonstrated that AC is equal to DF:
therefore, since AB is equal to DE, BA and AC are equal to ED
and DF, each to each ;
and the angle BAC is equal to the angle EDF; (hyp.)
wherefore the base BC is equal to the base EF,
I. 4.)
and the remaining angles to the remaining angles:
therefore the angle ABC is equal to the angle DEF :
and the right angle ABK is equal to the right angle DEN;
whence the remaining angle CBK is equal to the remaining angle FEN:
for the same reason, the angle BCK is equal to the angle EFN:
therefore, in the two triangles BCK, EFN, there are two angles in
one equal to two angles in the other, each to each, and one side equal
to one side adjacent to the equal angles in each, viz. BC equal to EF ;
therefore the other sides are equal to the other sides;
BK then is equal to EN, and AB is equal to DE;
wherefore AB, BK are equal to DE, EN, each to each;
and they contain right angles;
wherefore the base AK is equal to the base DN.
And since AH is equal to DM,
the square of AH is equal to the square of DM:
but the squares of AK, KH are equal to the square of AH, because
AKH is a right angle; (1. 47.)
and the squares of DN, NM are equal to the square of DM, for
DNM is a right angle:
wherefore the squares of AK, KH are equal to the squares of DN, NM:
and of these the square of AK is equal to the square of DN;
therefore the remaining square of KH is equal to the remaining
square of NM;
and the straight line KH to the straight line NM:
and because HA, AK are equal to MD, DN, each to each,
and the base HK to the base MN, as has been proved;
therefore the angle HAK is equal to the angle MÍN. (1.8.)
Therefore, if from the vertices, &c.
Q.E.D.
BOOK XI. PROP. XXXV, XXXVI.
247
COR. From this it is manifest, that if from the vertices of two equal
plane angles, there be elevated two equal straight lines containing equal
angles with the sides of the angles each to each; the perpendiculars
drawn from the extremities of the equal straight lines to the planes of
the first angles are equal to one another.
Another demonstration of the Corollary.
Let the plane angles BAC, EDF be equal to one another, and let
AH, DM be two equal straight lines above the planes of the angles,
containing equal angles with BA, AC; ED, DF, each to each, viz. the
angle HÃB equal to MDE, and HAC equal to the angle MDF;
and from H, M, let HK, MN be perpendiculars to the planes
BAC, EDF:
HK shall be equal to MN.
Because the solid angle at A is contained by the three plane angles
BAC, BAH, HAC, which are, each to each, equal to the three plane
angles EDF, EDM, MDF, containing the solid angle at D;
the solid angles at A and D are equal, and therefore coincide with
one another;
to wit, if the plane angle BAC be applied to the plane angle EDF,
the straight line AH coincides with DM, as was shewn in Prop. B.
of this book:
and because AH is equal to DM,
the point H coincides with the point M:
wherefore HK, which is perpendicular to the plane BAC, coincides
with MN which is perpendicular to the plane EDF, (x1. 13.)
because these planes coincide with one another.
Therefore, HK is equal to MN. Q.E.D.
PROPOSITION XXXVI. THEOREM.
If three straight lines be proportionals, the solid parallelopiped
described from all three, as its sides, is equal to the equilateral paral-
lelopiped described from the mean proportional, one of the solid angles of
which is contained by three plane angles equal, each to each, to the three
plane angles containing one of the solid angles of the other figure.
Let A, B, C be three proportionals, viz. A to B, as B to C.
The solid described from A, B, C shall be equal to the equilateral
solid described from B, equiangular to the other.
0
H
N
M
G
03
K
E
D
B
A
Take a solid angle D contained by three plane angles EDF, FDG,
GDE:
and make each of the straight lines ED, DF, DG equal to B,
and complete the solid parallelopiped DH;
1
248
EUCLID'S ELEMENTS.

make LK equal to A, and at the point K in the straight line LK,
make a solid angle contained by the three plane angles LKM, MKN,
NKL, equal to the angles EDF, FDG, GDE, each to each; (XI. 26.)
and make KN equal to B, and KM equal to C; and complete the
solid parallelopiped KO..
And because, as A is to B, so is B to C,
and that A is equal to LK, and B to each of the straight lines DE,
DF, and C to KM;
therefore LK is to ED, as DF to KM;
that is, the sides about the equal angles are reciprocally proportional;
therefore the parallelogram LM is equal to EF: (vi. 14.)
and because EDF, LKM are two equal plane angles, and the two
equal straight lines DG, KN are drawn from their vertices above their
planes, and contain equal angles with their sides;
therefore the perpendiculars from the points G, N to the planes
EDF, LKM are equal to one another: (xI. 35. Cor.)
therefore the solids KO, DH are of the same altitude:
and they are upon equal bases LM, EF
and therefore they are equal to one another: (xI. 31.)
but the solid KO is described from the three straight lines A, B, C,
and the solid DH from the straight line B.
Therefore if three straight lines, &c.

Q. E.D.
PROPOSITION XXXVII. THEOREM.
If four straight lines be proportionals, the similar solid parallelopipeds
similarly described from them shall also be proportionals. And if the
similar parallelopipeds similarly described from four straight lines be pro-
portionals, the straight lines shall be proportionals.
Let the four straight lines AB, CD, EF, GH be proportionals,
viz. as AB to CD, so EF to GH;
and let the similar parallelopipeds AK, CL, EM, GN be similarly
described from them.
AK shall be to CL, as EM to GN.
K
L
A
B
C
D
P
M
N
V
E
G H S
R
Make AB, CD, O, P continual proportionals,
as also EF, GH, Q, R: (vi. 11.)
and because as AB is to CD, so EF to GH;
and that CD is to 0, as GH to Q, (v. 11.)
and O to P, as Q to R;
therefore, ex æquali, AB is to P, as EF to R: (v. 22.)
but as AB to P, so is the solid AK to the solid CL; and as EF to
R, so is the solid EM to the solid GN; (x1. 33. Cor.)
therefore as the solid AK to the solid CL, so is the solid EM to the
solid GN. (v. 11.)
BOOK XI. PROP. XXXVII, XXXVIII.
249
Next, let the solid AK be to the solid CL, as the solid EM to the
solid GN.
The straight line AB shall be to CD, as EF to GH.
Take as AB to CD, so EF to ST,
and from ST describe a solid parallelopiped SV similar and simi-
larly situated to either of the solids EM, GN. (x1.27.)
And because AB is to CD, as EF to ST,
and that from AB, CD the solid parallelopipeds AK, CL are simi-
larly described;
and in like manner the solids EM, SV from the straight lines EF, ST;
therefore AK is to CL, as EM to SV;
but, by the hypothesis, AK is to CL, as EM to GN;
therefore GN is equal to SV: (v. 9.)
but it is likewise similar and similarly situated to SV;
therefore the planes which contain the solids GN, SV are similar and
equal, and their homologous sides GH, ST equal to one another:
and because as AB to CD, so EF to ST,
and that ST is equal to GH,
therefore AB is to CD, as EF to GH.
Therefore, if four straight lines, &c. Q. E. D.
PROPOSITION XXXVIII. THEOREM.
"If a plane be perpendicular to another plane, and a straight line be
drawn from a point in one of the planes perpendicular to the other plane,
this straight line shall fall on the common section of the planes.
"Let the plane CD be perpendicular to the plane AB, and let AD
be their common section:
if any point E be taken in the plane CD, the perpendicular drawn
from E to the plane AB shall fall on AD.
C
E
G
A
D
B
For, if it does not, let it, if possible, fall elsewhere, as EF;
and let it meet the plane AB in the point F;
and from F draw, in the plane AB, a perpendicular FG to DA,
(1. 12.) which is also perpendicular to the plane CD; (xI. def. 4.) and
join ÉG.
Then, because FG is perpendicular to the plane CD, and the straight
line EG, which is in that plane, meets it;
therefore FGE is a right angle: (x1. def. 3.)
but EF is also at right angles to the plane AB;
and therefore EFG is a right angle;
wherefore two of the angles of the triangle EFG are equal together
to two right angles; which is absurd: (1.17.)
therefore the perpendicular from the point E to the plane AB does
not fall elsewhere than upon the straight line AD;
it therefore falls upon it.
If therefore a plane, &c. Q.E. D.”
*
250
EUCLID'S ELEMENTS.
:
PROPOSITION, XXXIX. THEOREM.
In a solid parallelopiped, if the sides of two of the opposite planes be
divided, each into two equal parts, the common section of the planes pass-
ing through the points of division, and the diameter of the solid parallelo-
piped, cut each other into two equal parts.
Let the sides of the opposite planes CF, AH, of the solid parallelo-
piped AF, be divided each into two equal parts in the points K, L, M,
N; X, O, P, R; and join KL, MN, XO, PR:
and because DK, CL are equal and parallel,
KL is parallel to DC: (1. 33.)
for the same reason, MN is parallel to BA:
and BA is parallel to DC;
therefore, because KL, BA are each of them parallel to DC, and
not in the same plane with it, KL is parallel to BA: (x1.9.)
and because KL, MN are each of them parallel to BA, and not in
the same plane with it,
KL is parallel to MN: (x1.9.)
wherefore KL, MN are in one plane.
In like manner, it may be proved, that XO, PR are in one plane.
Let YS be the common section of the planes KN, XR;
and DG the diameter of the solid parallelopiped AF:
YS and DG shall meet, and cut one another into two equal parts.
D
K
F

C
I
E
T
B
H
P
A
N
G
Join DY, YE, BS, SG.
Because DX is parallel to OE, the alternate angles DXY, YOE are
equal to one another: (1. 29.)
+
and because DX is equal to OE, and XY to YO, and that they
contain equal angles,
the base DY is equal to the base YE, and the other angles are
equal; (1.4.)
therefore the angle XYD is equal to the angle OYE,
and DYE is a straight line: (1.14.)
for the same reason BSG is a straight line, and BS equal to SG.
And because CA is equal and parallel to DB, and also equal and
parallel to EG;
therefore DB is equal and parallel to EG: (x1.9.)
and DE, BG join their extremities;
therefore DE is equal and parallel to BG: (1.33.)
and DG, YS are drawn from points in the one, to points in the other;
and are therefore in one plane:
whence it is manifest, that DG, YS must meet one another :
let them meet in T.
BOOK XI. PROP. XXXIX, XL.
251
And because DE is parallel to BG,
the alternate angles EDT, BGT are equal; (XI. 29.)
and the angle DIY is equal to the angle GTS: (1. 15.)
therefore in the triangles DTY, GTS there are two angles in the
one equal to two angles in the other, and one side equal to one side,
opposite to two of the equal angles, viz. DY to GS;
for they are the halves of DE, BG:
therefore the remaining sides are equal, each to each: (1. 26.)
wherefore DT is equal to TG, and YT equal to TS.
Wherefore, if in a solid, &c. Q.E. D.
PROPOSITION XL. THEOREM.
If there be two triangular prisms of the same altitude, the base of one
of which is a parallelogram, and the base of the other a triangle; if the
parallelogram be double of the triangle, the prisms shall be equal to one
another.
Let the prisms ABCDEF, GHKLMN be of the same altitude, the
first whereof is contained by the two triangles ABE, CDF, and the
three parallelograms AD, DE, EC; and the other by the two triangles
GHK, LMN, and the three parallelograms LH, HN, NG; and let one
of them have a parallelogram AF, and the other a triangle GHK, for
its base.
If the parallelogram AF be double of the triangle GHK, the prism
ABCDEF shall be equal to the prism GHKLMN.
B
D
X
M
0
N
C
H
E
F
G K
Complete the solids AX, GO:
and because the parallelogram AF is double of the triangle GHK;
and the parallelogram HK double of the same triangle;
(1. 34.)
therefore the parallelogram AF is equal to HK:
but solid parallelopipeds upon equal bases, and of the same alti-
tude, are equal to one another; (xI. 31.)
therefore the solid AX is equal to the solid GO:
and the prism ABCDEF is half of the solid AX; (x1. 28.)
and the prism GHKLMN half of the solid GO: (x1. 28.)
therefore the prism ABCDEF is equal to the prism GHKLMN.
Wherefore, if there be two, &c. Q. E. D.
•
1
NOTES TO BOOK XI.
THE Eleventh Book of the Elements commences with the definitions of the Geo-
metry of Planes and Solids, and then proceeds to demonstrate the most elementary pro-
perties of straight lines and planes, solid angles and parallelopipeds.
The solids considered in the eleventh and twelfth books are Geometrical solids,
portions of space bounded by surfaces which are supposed capable of penetrating and
intersecting one another.
In the first six books, all the diagrams employed in the demonstrations are supposed
to be in the same plane, which may lie in any position whatever, and be extended in
every direction, and there is no difficulty in representing them roughly on any plane
surface; this, however, is not the case with the diagrams employed in the demonstra-
tions in the eleventh and twelfth books, which cannot be so intelligibly represented
on a plane surface on account of the perspective. A more exact conception may be
attained, by adjusting pieces of paper to represent the different planes, and drawing
lines upon them as the constructions may require, and by fixing pins to represent
the lines which are perpendicular to, or inclined to any planes.
Any plane may be conceived to move round any fixed point in that plane, either
in its own plane, or in any direction whatever; and if there be two fixed points in the
plane, the plane cannot move in its own plane, but may move round the straight line
which passes through the two fixed points in the plane, and may assume every possible
position of the planes which pass through that line, and every different position of the
plane will represent a different plane; thus, an indefinite number of planes may be
conceived to pass through a straight line which will be the common intersection of all
the planes. Hence, it is manifest, that though two points fix the position of a straight
line in a plane, neither do two points nor a straight line fix the position of a plane in
space. If however, three points, not in the same straight line, be conceived to be fixed
in the plane, it will be manifest, that the plane cannot be moved round, either in its
own plane or in any other direction, and therefore is fixed.
Also any conditions which involve the consideration of three fixed points not in
the same straight line, will fix the position of a plane in space; as two straight lines
which meet or intersect one another or two parallel straight lines in the plane.
Def. v.
When a straight line meets a plane, it is inclined at different angles to
the different lines in that plane which may meet it; and it is manifest that the incli-
nation of the line to the plane is not determined by its meeting any line in that plane.
The inclination of the line to the plane can only be determined by its inclination to
some fixed line in the plane. If a point be taken in the line different from that point
where the line meets the plane, and a perpendicular be drawn to meet the plane in
another point; then these two points in the plane will fix the position of the line
which passes through them in that plane, and the angle contained by this line and
the given line, will measure the inclination of the line to the plane; and it will be
found to be the least angle which can be formed with the given line and any other
straight line in the plane.
If two perpendiculars be drawn upon a plane from the extremities of a straight
line which is inclined to that plane, the straight line in the plane intercepted between
the perpendiculars is called the projection of the line on that plane; and it is obvious
that the inclination of a straight line to a plane is equal to the inclination of the
straight line to its projection on the plane. If however, the line be parallel to the
NOTES TO BOOK XI.
253
plane, the projection of the line is of the same length as the line itself; in all other
cases the projection of the line is less than the line, being the base of a right-angled
triangle, the hypothenuse of which is the line itself.
The inclination of two lines to each other, which do not meet, is measured by
the angle contained by two lines drawn through the same point and parallel to the two
given lines.
Def. VI.
Planes are distinguished from one another by their inclinations, and the
inclinations of two planes to one another will be found to be measured by the acute
angle formed by two straight lines drawn in the planes, and perpendicular to the straight
line which is the common intersection of the two planes.
It is also obvious that the inclination of one plane to another will be measured
by the angle contained between two straight lines drawn from the same point, and per-
pendicular, one on each of the two planes.
The intersection of two planes suggests a new conception of the straight line.
Def. IX.
When a solid angle is contained by three plane angles, each plane which
contains one plane angle, is fixed by the position of the other two, and consequently, only
one solid angle can be formed by three plane angles. But when a solid angle is formed
by more than three plane angles, if one of the planes be considered fixed in position, there
are no conditions which fix the position of the rest of the planes which contain the solid
angle, and hence, an indefinite number of solid angles, unequal to one another, may be
formed by the same plane angles, when the number of plane angles is more than three.
Def. x is restored, as it is found in the editions of Euclid, by Dr Barrow and
others. It appears to be universally true, supposing the planes to be similarly situated,
in which are contained the corresponding equal plane angles of each figure.
Def. A. Parallelopipeds are solid figures in some respects analogous to parallelo-
grams, and remarks might be made on parallelopipeds similar to those which were made
on parallelograms in the notes to Book 11, p. 67; and every right-angled parallelopiped
may be said to be contained by any three of the straight lines which contain the three
right angles by which any one of the solid angles of the figure is formed; or more
briefly, by the three adjacent edges of the parallelopiped.
As all lines are measured by lines, and all surfaces by surfaces, so all solids are
measured by solids. The cube is the figure assumed as the measure of solids or volumes,
and the unit of volume is that cube, the edge of which is one unit in length.
If the edges of a rectangular parallelopiped can be divided into units of the same
length, a numerical expression for the number of cubic units in the parallelopiped may
be found, by a process similar to that by which a numerical expression for the area of a
rectangle was found.
Let AB, AC, AD be the adjacent edges of a rectangular parallelopiped AG, and let
AB contain 5 units, AC, 4 units, and AD, 3 units in length.
Then if through the points of division of AB, AC, AD, planes be drawn parallel
to the faces BG, BD, AE respectively, the parallelopiped will be divided into cubic
units, all equal to one another.

D
A
B
C
E
G
254
EUCLID'S ELEMENTS.
And since the rectangle ABEC contains 5 x 4 square untis, (note, p. 68.) and
that for every linear unit in AD there is a layer of 5 × 4 cubic units corresponding
to it;
consequently, there are 5 × 4 × 3 cubic units in the whole parallelopiped AG.
That is, the product of the three numbers which express the number of linear units
in the three edges, will give the number of cubic units in the parallelopiped, and therefore
will be the arithmetical representation of its volume.
And generally, if AB, AC, AD; instead of 5, 4 and 3, consisted of a, b and c
linear units, it may be shewn, in a similar manner, that the volume of the parallelopiped
would contain abc cubic units, and the product abc would be a proper representation of
the volume of the parallelopiped.
If the three sides of the figure were equal to one another, or b and c each equal to a,
the figure would become a cube, and its volume would be represented by aaa, or a³.
Prop. VI. From the diagram, the following important construction may be made.
If from B a perpendicular BF be drawn to the opposite side DE of the triangle
DBE, and AF be joined; then AF shall be perpendicular to DE, and the angle AFB
measures the inclination of the planes AED and BED.
Prop. XIX. It is also obvious, that if three planes intersect one another; and if the
first be perpendicular to the second, and the second be perpendicular to the third; the
first shall be perpendicular to the third; also the intersections of every two shall be
perpendicular to one another.
Prop. XXXIII. Algebraically.
Let the adjacent edges of the solid AB contain a, b, c units,
and those of the solid CD contain a', b', c' units respectively.
Also, let V, V' denote their volumes.
Then V = abc, and V' a'b' c'.
But since the parallelopipeds are similar, ..
a
818
=
•
Hence
V abc
V' a'b'c'
a' b'c'
a b c a α a a³
a''b'' Z
b3
c3
=
•
a'3 b'8
C'3
In a similar manner, it may be shewn that the volumes of all similar solid figures
bounded by planes, are proportional to the cubes of their homologous edges.
BOOK XII.
:
LEMMA I.
If from the greater of two unequal magnitudes, there be taken more
than its half, and from the remainder more than its half; and so on:
there shall at length remain a magnitude less than the least of the pro-
posed magnitudes. (Book x. Prop. 1.)
Let AB and C be two unequal magnitudes, of which AB is the greater.
If from AB there be taken more than its half,
and from the remainder more than its half, and so on;
there shall at length remain a magnitude less than C.
D
A
K
F.-
H.
G
BC E
For may be multiplied so as at length to become greater than AB.
Let it be so multiplied, and let DE its multiple be greater than AB,
and let DE be divided into DF, FG, GE, each equal to C.
From AB take BH greater than its half,
and from the remainder AH take HK greater than its half, and so on,
until there be as many divisions in AB as there are in DE:
and let the divisions in AB be AK, KH, HB;
and the divisions in DE be DF, FG, GE.
And because DE is greater than AB,
and that EG taken from DE is not greater than its half, but BH
taken from AB is greater than its half;
therefore the remainder GD is greater than the remainder HA.
Again, because GD is greater than HA, and that GF is not greater
than the half of GD, but HK is greater than the half of HA;
therefore the remainder FD is greater than the remainder AK:
and FD is equal to C,
therefore C is greater than AK;
that is, AK is less than C. Q. E. D.
And if only the halves be taken away, the same thing may in the
same way be demonstrated.
PROPOSITION I.
THEOREM.
Similar polygons inscribed in circles, are to one another as the squares
of their diameters.
256
EUCLID'S ELEMENTS.
Let ABCDE, FGHKL be two circles, and in them the similar
polygons ABCDE, FGHKL;
and let BM, GN be the diameters of the circles:
as the polygon ABCDE is to the polygon FGHKL, so shall the
square of BM be to the square of GN.
A
B
F
E
G
L
M
N
H
K
D
Join BE, AM, GL, FN. ·
And because the polygon ABCDE is similar to the polygon FGHkl,
the angle BAE is equal to the angle GFL,
and as BA to AË, so is GF to FL:
therefore the two triangles BAE, GFL having one angle in one
equal to one angle in the other, and the sides about the equal angles
proportionals, are equiangular;
and therefore the angle AEB is equal to the angle FLG:
but AEB is equal to AMB, because they stand upon the same
circumference: (III. 21.)
and the angle FLG is, for the same reason, equal to the angle FNG:
therefore also the angle AMB is equal to FNG:
and the right angle BAM is equal to the right angle GFN; (III. 31.)
wherefore the remaining angles in the triangles ABM, FGN are equal,
and they are equiangular to one another:
therefore as BM to GN, so.is BA to GF; (vI. 4.)
and therefore the duplicate ratio of BM to GN, is the same with
the duplicate ratio of BA to GF: (v. def. 10. and v. 22.)
but the ratio of the square of BM to the square of GN, is the
duplicate ratio of that which BM has to GN; (VI. 20.)
and the ratio of the polygon ABCDE to the polygon FGHKL is
the duplicate of that which BA has to GF: (VI. 20.)
therefore as the polygon ABCDE is to the polygon FGHKL, so is
the square of BM to the square of GN.
Wherefore, similar polygons, &c. Q.E.D.
PROPOSITION II. THEOREM.
Circles are to one another as the squares of their diameters.
Let ABCD, EFGH be two circles, and BD, FH their diameters.
As the square of BD to the square of FH, so shall the circle ABCD
be to the circle EFGH.
For, if it be not so, the square of BD must be to the square of FH,
as the circle ABCD is to some space either less than the circle EFGH,
or greater than it.
A
X
R
E

K
N
H
B
F
D
S
LG
M
P
BOOK XII. PROP. II.
257
First, if possible, let it be to a space S less than a circle EFGH;
and in the circle EFGH inscribe the square EFGH. (1v. 6.)
This square is greater than half of the circle EFGH;
because, if through the points E, F, G, H, there be drawn tangents
to the circle,
the square EFGH is half of the square described about the circle: (1. 47.)
and the circle is less than the square described about it;
therefore the square EFGH is greater than half of the circle.
Divide the circumferences EF, FG, GH, HE, each into two equal
parts in the points K, L, M, N, and join Ek, kf, fl, LG, ĠM,
HM, HN, NE;
therefore each of the triangles EKF, FLG, GMH, HNE, is greater
than half of the segment of the circle in which it stands;
because, if straight lines touching the circle be drawn through the
points K, L, M, N, and the parallelograms upon the straight lines EF,
FG, GH, HE be completed,
each of the triangles EKF, FLG, GMH, HNE is the half of the
parallelogram in which it is: (I. 41.)
but every segment is less than the parallelogram in which it is;
wherefore each of the triangles EKF, FLG, GMH, HNE is greater
than half the segment of the circle which contains it.
Again, if the remaining circumferences be divided each into two
equal parts, and their extremities be joined by straight lines, by con-
tinuing to do this, there will at length remain segments of the circle,
which together are less than the excess of the circle EFGH above the
space S;
because, by the preceding Lemma, if from the greater of two unequal
magnitudes there be taken more than its half, and from the remainder
more than its half, and so on, there shall at length remain a magnitude
less than the least of the proposed magnitudes.
Let then the segments EK, KF, FL, LG, GM, MH, HN, NE be
those that remain, and are together less than the excess of the circle
EFGH above S:
therefore the rest of the circle, viz. the polygon EKFLGMHN is
greater than the space S.
Describe likewise in the circle ABCD the polygon AXBOCPDR
similar to the polygon EKFLGMHN:
as therefore the square of BD is to the square of FH, so is the
polygon AXBOCPDR to the polygon EKFLGMHN: (XII. 1.)
but the square of BD is also to the square of FH, as the circle
ABCD is to the space S; (hyp.)
therefore as the circle ABCD is to the space S, so is the polygon
AXBOCPDR to the polygon EKFLGMHN: (v. 11.)
but the circle ABCD is greater than the polygon contained in it;
wherefore the space S is greater than the polygon EKFLGMÍN:
(v. 14.)
but it is likewise less, as has been demonstrated; which is impossible.
Therefore the square of BD is not to the square of FH, as the circle
ABCD is to any space less than the circle EFGH.
In the same manner, it may be demonstrated, that neither is the
square of FH to the square of BD, as the circle EFGH is to any space
less than the circle ABCD.
Nor is the square of BD to the square of FH, as the circle ABCD
is to any space greater than the circle EFGH.
17
258
EUCLID'S ELEMENTS.
For, if possible, let it be so to T, a space greater than the circle EFGH:
A
E

K
N
R
H
B
D
ន
L
M
G
C
T
therefore, inversely, as the square of FH to the square of BD, so is
the space T to the circle ABCD ;
but as the space T is to the circle ABCD, so is the circle EFGH
to some space, which must be less than the circle ABCD, (v. 14.)
because the space T is greater, by hypothesis, than the circle EFGH;
therefore as the square of FH is to the square of BD, so is the circle
EFGH to a space less than the circle ABCD, which has been demon-
strated to be impossible;
therefore the square of BD is not to the square of FH as the circle
ABCD is to any space greater than the circle EFGH:
and it has been demonstrated, that neither is the square of BD to
the square of FH, as the circle ABCD to any space less than the circle
EFGH:
wherefore, as the square of BD to the square of FH, so is the circle
ABCD to the circle EFGH.
Circles, therefore, are, &c. Q.E.D.
PROPOSITION III. THEOREM.
Every pyramid having a triangular base, may be divided into two
equal and similar pyramids having triangular bases, and which are
similar to the whole pyramid; and into two equal prisms which together
are greater than half of the whole pyramid.
Let there be a pyramid of which the base is the triangle ABC, and
its vertex the point D.
The pyramid ABCD may be divided into two equal and similar
pyramids having triangular bases, and similar to the whole; and into
two equal prisms which together shall be greater than half of the whole
pyramid.
D

K
L
B F C
Divide AB, BC, CA, AD, DB, DC, each into two equal parts in
the points E, F, G, H, K, L,
and join EH, EG, GH, HK, KL, LH, EK, KF, FG.
Because AE is equal to EB, and AH to HD,
HE is parallel to DB: (vI. 2.)
for the same reason, HK is parallel to AB:
BOOK XII. PROP. III.
259
therefore HEBK is a parallelogram, and HK equal to EB: (1. 34.)
but EB is equal to AE;
therefore also AE is equal to HK:
and AH is equal to HD;
wherefore EA, AH are equal to KH, HD, each to each;
and the angle EAH is equal to the angle KHD; (1. 29.)
therefore the base ÈH is equal to the base KD,
and the triangle AEH equal and similar to the triangle HKD. (1. 4.)
For the same reason, the triangle AGH is equal and similar to the
triangle HLD.
Again, because the two straight lines EH, HG, which meet one
another, are parallel to KD, DL, that meet one another, and are not in
the same plane with them,
they contain equal angles; (xI. 10.)
therefore the angle EHG is equal to the angle KDL;
and because EH, HG are equal to KD, DL, each to each,
and the angle EHG equal to the angle KDL;
therefore the base EG is equal to the base KL,
and the triangle EHG equal and similar to the triangle KDL. (1. 4.)
For the same reason, the triangle AEG is also equal and similar to
the triangle HKL.
Therefore the pyramid of which the base is the triangle AEG, and of
which the vertex is the point H, is equal and similar to the pyramid,
the base of which is the triangle KHL, and vertex the point D. (xI. c.)
And because HK is parallel to AB, a side of the triangle ADB,
the triangle ADB is equiangular to the triangle HDK, and their
sides are proportionals: (v1.4.)
therefore the triangle ADB is similar to the triangle HDK:
and for the same reason, the triangle DBC is similar to the triangle DKL ;
and the triangle ADC to the triangle HDL;
and also the triangle ABC to the triangle AEG;
but the triangle AEG is similar to the triangle HKL, as before
was proved;
therefore the triangle ABC is similar to the triangle HKL: (vI. 21.)
and therefore the pyramid of which the base is the triangle ABC,
and vertex the point D, is similar to the pyramid of which the base is
the triangle HKL, and vertex the same point D: (XI. B. & XI. def. 11.)
but the pyramid of which the base is the triangle HKL, and vertex
the point D, is similar, as has been proved, to the pyramid the base of
which is the triangle AEG, and vertex the point H;
wherefore the pyramid, the base of which is the triangle ABC, and
vertex the point D, is similar to the pyramid of which the base is the
triangle AEG, and vertex H:
therefore each of the pyramids AEGH, HKLD is similar to the
whole pyramid ABCD.
And because BF is equal to FC,
the parallelogram EBFG is double of the triangle GFC: (1.41.)
but when there are two prisms of the same altitude, of which one
has a parallelogram for its base, and the other a triangle that is half of
the parallelogram,
these prisms are equal to one another; (xI. 40.)
therefore the prism having the parallelogram ÈBFG for its base,
and the straight line KH opposite to it, is equal to the prism having
the triangle GFC for its base, and the triangle HKL opposite to it;
17-2
260
EUCLID'S ELEMENTS.
for they are of the same altitude, because they are between the
parallel planes ABC, HKL: (XI. 15.)
and it is manifest that each of these prisms is greater than either of
the pyramids of which the triangles AEG, HKL are the bases, and the
vertices the points H, D;
because, if EF be joined, the prism having the parallelogram EBFG
for its base, and KH the straight line opposite to it, is greater than the
pyramid of which the base is the triangle EBF, and vertex the point K:
but this pyramid is equal to the pyramid, the base of which is the
triangle AEG, and vertex the point H; (xI. c.)
because they are contained by equal and similar planes:
wherefore the prism having the parallelogram EBFG for its base,
and opposite side KH, is greater than the pyramid of which the base is
the triangle AEG, and vertex the point H:
and the prism of which the base is the parallelogram EBFG, and
opposite side KH, is equal to the prism having the triangle GFC for its
base, and HKL the triangle opposite to it;
and the pyramid of which the base is the triangle AEG, and vertex
H, is equal to the pyramid of which the base is the triangle HKL, and
vertex D:
therefore the two prisms before mentioned are greater than the two
pyramids of which the bases are the triangles AEG, HKL, and vertices
the points H, D.
Therefore the whole pyramid of which the base is the triangle ABC,
and vertex the point D, is divided into two equal pyramids similar to
one another, and to the whole pyramid; and into two equal prisms;
and the two prisms are together greater than half of the whole pyramid.
Q.E.D.
PROPOSITION IV. THEOREM.
If there be two pyramids of the same altitude, upon triangular bases,
and each of them be divided into two equal pyramids similar to the whole
pyramid, and also into two equal prisms; and if each of these pyramids
be divided in the same manner as the first two, and so on: as the base of
one of the first two pyramids is to the base of the other, so shall all the
prisms in one of them be to all the prisms in the other, that are produced
by the same number of divisions.
Let there be two pyramids of the same altitude upon the triangular
bases ABC, DEF, and having their vertices in the points G, H; and
let each of them be divided into two equal pyramids similar to the
whole, and into two equal prisms; and let each of the pyramids thus
made be conceived to be divided in the like manner, and so on.
7
As the base ABC is to the base DEF, so shall all the prisms in the
pyramid 4BCG be to all the prisms in the pyramid DEFH made by
the same number of divisions.
C
M
G
N
H


D
Y
R
B X
E
F
BOOK XII. PROP. IV.
261
Make the same construction as in the foregoing proposition:
and because BX is equal to XC, and AL to LC,
therefore XL is parallel to AB, (vi. 2.)
and the triangle ABC similar to the triangle LXC.
For the same reason, the triangle DEF is similar to RVF.
And because BC is double of CX, and EF double of FV,
therefore BC is to CX, as EF to FV: (v. c.)
and upon BC, CX are described the similar and similarly situated
rectilineal figures ABC, LXC;
and upon EF, FV, in like manner, are described the similar figures
DEF, RFV:
therefore, as the triangle ABC is to the triangle LXC, so is the
triangle DEF to the triangle RVF, (v1. 22.)
and, by permutation, as the triangle ABC to the triangle DEF,
so is the triangle LXC to the triangle RVF.
And because the planes ABC, OMN, as also the planes DEF, STY,
are parallel, (xI. 15.)
the perpendiculars drawn from the points G, H to the bases ABC,
DEF, which, by the hypothesis, are equal to one another, shall be cut
each into two equal parts by the planes OMN, STY, (x1. 17.)
because the straight lines GC, HF are cut into two equal parts in
the points N, Y by the same planes:
therefore the prisms LXCOMN, RVFSTY are of the same altitude;
and therefore, as the base LXC to the base RVF, that is, as the
triangle ABC to the triangle DEF, so is the prism having the triangle
LXC for its base, and OMN the triangle opposite to it, to the prism of
which the base is the triangle RVF, and the opposite triangle STY:
(XI. 32. Cor.)
and because the two prisms in the pyramid ABCG are equal to
one another,
and also the two prisms in the pyramid DEFH equal to one another;
as the prism of which the base is the parallelogram KBXL and
opposite side MO, to the prism having the triangle LXC for its base,
and OMN the triangle opposite to it; so is the prism of which the base
is the parallelogram PEVR, and opposite side TS, to the prism of
which the base is the triangle RVF, and opposite triangle STY: (v.7.)
therefore, componendo, as the prisms KBXLMO, LXCOMN, to-
gether, are to the prism LXCOMN, so are the prisms PEVRTS,
RVFSTY to the prism RVFSTY;
and permutando, as the prisms KBXLMO, LXCOMN are to the
prisms PEVRTS, RVFSTY, so is the prism LXCOMN to the prism
RVFSTY:
but as the prism LXCOMN to the prism RVFSTY, so is, as has
been proved, the base ABC to the base DEF;
therefore, as the base ABC to the base DEF, so are the two prisms
in the pyramid ABCG to the two prisms in the pyramid
DEFH:
and likewise if the pyramids now made, for example, the two
OMNG, STYH, be divided in the same manner;
as the base OMN is to the base STY, so are the two prisms in the
pyramid OMNG to the two prisms in the pyramid STYH;
but the base OMN is to the base STY, as the base ABC to the
base DEF;
262
EUCLID'S ELEMENTS.
therefore, as the base ABC to the base DEF, so are the two prisms
in the pyramid ABCG to the two prisms in the pyramid DEFH;
and so are the two prisms in the pyramid OMNG to the two
prisms in the pyramid STYH;
and so are all four to all four:
and the same thing may be shewn of the prisms made by dividing
the pyramids AKLO and DPRS, and of all made by the same number
of divisions.
Q.E. D.
PROPOSITION V. THEOREM.
Pyramids of the same altitude which have triangular bases, are to one
another as their bases.
Let the pyramids of which the triangles ABC, DEF are the bases,
and of which the vertices are the points G, H, be of the same altitude.
As the base ABC to the base DEF, so shall the pyramid ABCG be
to the pyramid DEFH.
G
H


M
N
T
Y
D
K
R
E V
F
Q
Z
B X C
For, if it be not so, the base ABC must be to the base DEF, as the
pyramid ABCG to a solid either less than the pyramid DEFH, or
greater than it.
First, if possible, let it be to a solid less than it, viz. to the solid Q:
and divide the pyramid DEFH into two equal pyramids, similar
to the whole, and into two equal prisms ;
therefore these two prisms are greater than the half of the whole
pyramid. (XII. 3.)
And again, let the pyramids made by this division be in like manner
divided, and so on until the pyramids which remain undivided in the
pyramid DEFH be, all of them together, less than the excess of the
pyramid DEFH above the solid Q: (XII. Lem. 1.)
let these, for example, be the pyramids DPRS, STYH:
therefore the prisms, which make the rest of the pyramid DEFH,
are greater than the solid Q.
Divide likewise the pyramid ABCG in the same manner, and into
as many parts, as the pyramid DEFH.
Therefore, as the base ABC to the base DEF, so are the prisms in
the pyramid ABCG to the prisms in the pyramid DEFH: (XII. 4.)
but as the base ABC to the base DEF, so, by hypothesis, is the
pyramid ABCG to the solid Q;
and therefore, as the pyramid ABCG to the solid Q, so are the prisms
in the pyramid ABCG to the prisms in the pyramid DEFH;
but the pyramid ABCG is greater than the prisms contained in it;
wherefore also the solid Q is greater than the prisms in the pyramid
DEFH; (v. 14.)
BOOK XII. PROP. V, VI.
263
but it is also less, which is impossible.
Therefore the base ABC is not to the base DEF, as the pyramid
ABCG to any solid which is less than the pyramid DEFH.
In the same manner it may be demonstrated, that the base DEF is
not to the base ABC, as the pyramid DEFH to any solid which is less
than the pyramid ABCG.
Nor can the base ABC be to the base DEF, as the pyramid ABCG
to any solid which is greater than the pyramid DEFH.
For, if it be possible, let it be so to a greater, viz. the solid Z.
And because the base ABC is to the base DEF as the pyramid
ABCG to the solid Z ;
by inversion, as the base DEF to the base ABC, so is the solid Z
to the pyramid ABCG:
but as the solid Z is to the pyramid ABCG, so is the pyramid
DEFH to some solid, which must be less than the pyramid ABCG,
(v. 14.)
because the solid Z is greater than the pyramid DEFH;
and therefore, as the base DEF to the base ABC, so is the pyra-
mid DEFĤ to a solid less than the pyramid ABCG;
the contrary to which has been proved:
therefore the base ABC is not to the base DEF, as the pyramid
ABCG to any solid which is greater than the pyramid DEFH.
And it has been proved, that neither is the base ABC to the base
DEF, as the pyramid ABCG to any solid which is less than the pyra-
mid DEFH.
Therefore, as the base ABC is to the base DEF, so is the pyramid
ABCG to the pyramid DEFH.
Wherefore, pyramids, &c. q.E.D.
PROPOSITION VI. THEOREM.
Pyramids of the same altitude which have polygons for their bases, are
to one another as their bases.
Let the pyramids which have the polygons ABCDE, FGHKL for
their bases, and their vertices in the points M, N, be of the same
altitude.
As the base ABCDE is to the base FGHKL, so shall the pyramid
ABCDEM be to the pyramid FGHKLN.
M
N
A
D
F
K
B
G
H
Divide the base ABCDE into the triangles ABC, ACD, ADE,
and the base FGHKL into the triangles FGH, FHK, FKL;
and upon the bases ABC, ACD, ADE let there be as many pyra-
mids of which the common vertex is the point M,
and upon the remaining bases as many pyramids having their
common vertex in the point N.
264
EUCLID'S ELEMENTS.
Therefore, since the triangle ABC is to the triangle FGH, as the
pyramid ABCM to the pyramid FGHN; (x11. 5.)
and the triangle ACD to the triangle FGH, as the pyramid ACDM
to the pyramid FGHN;
and also the triangle ADE to the triangle FGH, as the pyramid
ADEM to the pyramid FGHN;
as all the first antecedents to their common consequent, so are all the
other antecedents to their common consequent; (v. 24. Cor. 2.)
that is, as the base ABCDE to the base FGH, so is the pyramid
ABCDEM to the pyramid FGHN:
and for the same reason, as the base FGHKL to the base FGH, so
is the pyramid FGHKLN to the pyramid FGHN:
and, by inversion, as the base FGH to the base FGHKL, so is the
pyramid FGHN to the pyramid FGHKLN:
then, because, as the base ABCDE to the base FGH, so is the
pyramid ABCDEM to the pyramid FGHN;
and as the base FGH to the base FGHKL, so is the pyramid
FGHN to the pyramid FGHKLN;
therefore, ex æquali, as the base ABCDE to the base FGHKL, so
the pyramid ABCDEM to the pyramid FGHKLN. (v. 22.)
Therefore, pyramids, &c. Q.E.D.
PROPOSITION VII. THEOREM.
Every prism having a triangular base may be divided into three
pyramids that have triangular bases, and are equal to one another.
Let there be a prism of which the base is the triangle ABC, and
DEF the triangle opposite to it.
The prism ABCDEF may be divided into three equal pyramids
having triangular bases.
F
D
E
A
B
Join BD, EC, CD.
And because ABED is a parallelogram of which BD is the diameter,
the triangle ABD is equal to the triangle EBD; (1.34.)
therefore the pyramid of which the base is the triangle ABD, and
vertex the point C, is equal to the pyramid of which the base is the
triangle EBD, and vertex the point C: (XII. 5.)
but this pyramid is the same with the pyramid the base of which
is the triangle EBC, and vertex the point D;
for they are contained by the same planes:
therefore the pyramid of which the base is the triangle ABD, and
vertex the point C, is equal to the pyramid, the base of which is the
triangle EBC, and vertex the point D.
Again, because FCBE is a parallelogram of which the diameter is CE,
the triangle ECF is equal to the triangle ECB; (1. 34.)
therefore the pyramid of which the base is the triangle ECB, and

BOOK XII. PROP. VII, VIII.
265
vertex the point D, is equal to the pyramid, the base of which is the
triangle ECF, and vertex the point D:
+
but the pyramid of which the base is the triangle ECB, and vertex
the point D, has been proved equal to the pyramid of which the base is
the triangle ABD, and vertex the point C:
therefore the prism ABCDEF is divided into three equal pyra-
mids having triangular bases, viz. into the pyramids ABDC, EBDC,
ECFD.
And because the pyramid of which the base is the triangle ABD,
and vertex the point C, is the same with the pyramid of which the base
is the triangle ABC, and vertex the point D,
for they are contained by the same planes;
and that the pyramid of which the base is the triangle ABD, and
vertex the point C, has been demonstrated to be a third part of the
prism, the base of which is the triangle ABC, and DEF the opposite
triangle;
therefore the pyramid of which the base is the triangle ABC, and
vertex the point D, is the third part of the prism which has the same
base, viz. the triangle ABC, and DEF its opposite triangle. Q.E.D.
COR. 1. From this it is manifest, that every pyramid is the third
part of a prism which has the same base, and is of an equal altitude with
it: for if the base of the prism be any other figure than a triangle, it
may be divided into prisms having triangular bases.
COR. 2. Prisms of equal altitudes are to one another as their bases;
because the pyramids upon the same bases, and of the same altitude, are
to one another as their bases. (XII. 6.)
PROPOSITION VIII. THEOREM.
Similar pyramids, having triangular bases, are one to another in the
triplicate ratio of that of their homologous sides.
Let the pyramids having the triangles ABC, DEF for their bases,
and the points G, H for their vertices, be similar and similarly situated.
The pyramid ABCG shall have to the pyramid DEFH, the triplicate
ratio of that which the side BC has to the homologous side EF.
K
L
N
M
A
B
C
X O

H
R
P
D
E
F
Complete the parallelograms ABCM, GBCN, ABGK, and the solid
parallelopiped BGML, contained by these planes and those opposite to
them:
and, in like manner, complete the solid parallelopiped EHPO con-
tained by the three parallelograms DEFP, HEFR, DEHX, and those
opposite to them.
And because the pyramid ABCG is similar to the pyramid DEFH,
the angle ABC is equal to the angle DEF, (x1. def. 11.)
and the angle GBC to the angle HEF, and ABG to DEŃ:
and AB is to BC, as DE to EF; (v1. def. 1.)
266
EUCLID'S ELEMENTS.
that is, the sides about the equal angles are proportionals:
wherefore the parallelogram BM is similar to EP:
for the same reason, the parallelogram BN is similar to ER, and
BK to EX:
therefore the three parallelograms BM, BN, BK are similar to the
three EP, ER, EX:
but the three BM, BN, BK are equal and similar to the three
which are opposite to them, (xI. 24.)
and the three ÉP, ER, EX equal and similar to the three opposite
to them:
wherefore the solids BGML, EHPO are contained by the same
number of similar planes:
and their solid angles are equal; (XI. B.)
and therefore the solid BGML is similar to the solid EHPO:
(XI. def. 11.)
but similar solid parallelopipeds have the triplicate ratio of that
which their homologous sides have; (xI. 33.)
therefore the solid BGML has to the solid EHPO the triplicate
ratio of that which the side BC has to the homologous side EF:
but as the solid BGML is to the solid EHPO, so is the pyramid
ABCG to the pyramid DEFH; (v. 15.)
because the pyramids are the sixth part of the solids, since the
prism, which is the half of the solid parallelopiped, (x1. 28.) is triple
of the pyramid: (XII. 7.)
wherefore, likewise the pyramid ABCG has to the pyramid DEFH,
the triplicate ratio of that which BC has to the homologous side EF.
Q.E.D.
COR. From this it is evident, that similar pyramids which have
multangular bases, are likewise to one another in the triplicate ratio of
their homologous sides. For they may be divided into similar pyramids
having triangular bases, because the similar polygons, which are their
bases, may be divided into the same number of similar triangles
homologous to the whole polygons: therefore, as one of the triangular
pyramids in the first multangular pyramid is to one of the triangular
pyramids in the other, (v. 12.) so are all the triangular pyramids in the
first to all the triangular pyramids in the other; that is, so is the first
multangular pyramid to the other: but one triangular pyramid is to its
similar triangular pyramid, in the triplicate ratio of their homologous
sides; and therefore the first multangular pyramid has to the other, the
triplicate ratio of that which one of the sides of the first has to the
homologous side of the other.

PROPOSITION IX. THEOREM.
The bases and altitudes of equal pyramids having triangular bases
are reciprocally proportional: and conversely, triangular pyramids, of
which the buses and altitudes are reciprocally proportional, are equal
to one another.

Let the pyramids of which the triangles ABC, DEF are the bases,
and which have their vertices in the points G, H, be equal to one
another.
BOOK XII. PROP. IX.
267:
The bases and altitudes of the pyramids ABCG, DEFH shall be
reciprocally proportional,
viz. the base ABC shall be to the base DEF, as the altitude of the
pyramid DEFH to the altitude of the pyramid ABCG.
L
N
O R
G
H
K
X
C
M
A
B
P
F
D
E
Complete the parallelograms AC, AG, GC, DF, DH, HF;
and the solid parallelopipeds BGML, EHPO contained by these
planes and those which are opposite to them.
And because the pyramid ABCG is equal to the pyramid DEFH,
and that the solid BGML is sextuple of the pyramid ABCG, (XI.
28. and XII. 7.)
and the solid EHPO sextuple of the pyramid DEFH;
therefore the solid BGML is equal to the solid EHPO: (v. ax. 1.)
but the bases and altitudes of equal solid parallelopipeds are reci-
procally proportional; (x1. 34.)
therefore as the base BM to the base EP, so is the altitude of the
solid EHPO to the altitude of the solid BGML:
but as the base BM to the base EP, so is the triangle ABC to the
triangle DEF; (v. 15.)
therefore as the triangle ABC to the triangle DEF, so is the alti-
tude of the solid EHPO to the altitude of the solid BGML:
but the altitude of the solid EHPO is the same with the altitude of
the pyramid DEFH;
and the altitude of the solid BGML is the same with the altitude
of the pyramid ABCG;
therefore, as the base ABC to the base DEF, so is the altitude of
the pyramid DEFH to the altitude of the pyramid ABCG:
wherefore, the bases and altitudes of the pyramids ABCG, DEFH
are reciprocally proportional.
Again, let the bases and altitudes of the pyramids ABCG, DEFH
be reciprocally proportional,
viz. the base ABC to the base DEF, as the altitude of the pyramid
DEFH to the altitude of the pyramid ABCG.
The pyramid ABCG shall be equal to the pyramid DEFH.
The same construction being made,
because as the base ABC to the base DEF; so is the altitude of the
pyramid DEFH to the altitude of the pyramid ABCG;
and as the base ABC to the base DEF, so is the parallelogram BM
to the parallelogram EP:
therefore the parallelogram BM is to EP, as the altitude of the
pyramid DEFH to the altitude of the pyramid ABCG:
but the altitude of the pyramid DEFH is the same with the alti-
tude of the solid parallelopiped EHPO;
and the altitude of the pyramid ABCG is the same with the alti-
tude of the solid parallelopiped BGML:
therefore, as the base BM to the base EP, so is the altitude of the
;
.
268
EUCLID'S ELEMENTS.
solid parallelopiped EHPO to the altitude of the solid parallelopiped
BGML.
But solid parallelopipeds having their bases and altitudes recipro-
cally proportional, are equal to one another; (xI. 34.)
therefore the solid parallelopiped BGML is equal to the solid
parallelopiped EĦPO:
and the pyramid ABCG is the sixth part of the solid BGML,
and the pyramid DEFH is the sixth part of the solid EHPO;
therefore the pyramid ABCG is equal to the pyramid DEFH. (v. ax.
Therefore the bases, &c. Q. E.D.
x. 2.)
PROPOSITION X. THEOREM.
Every cone is the third part of a cylinder which has the same base and
is of an equal altitude with it.
Let a cone have the same base with a cylinder, viz. the circle ABCD,
and the same altitude.
The cone shall be the third part of the cylinder; that is, the
cylinder shall be triple of the cone.
If the cylinder be not triple of the cone,
it must either be greater than the triple, or less than it.
First, if possible, let it be greater than the triple;
B
A
E
H
D
F
G
C
and inscribe the square ABCD in the circle:
this square is greater than the half of the circle ABCD.
Upon the square ABCD erect a prism of the same altitude with the
cylinder;
this prism shall be greater than half of the cylinder :
for let a square be described about the circle, and let a prism be
erected upon the square, of the same altitude with the cylinder:
then the inscribed square is half of that circumscribed;
and upon these square bases are erected solid parallelopipeds, viz.
the prisms of the same altitude;
therefore the prism upon the square ABCD is the half of the prism
upon the square described about the circle;
because they are to one another as their bases: (xI. 32.)
and the cylinder is less than the prism upon the square described
about the circle ABCD:
therefore the prism upon the square ABCD of the same altitude
with the cylinder, is greater than half of the cylinder.
Bisect the circumferences AB, BC, CD, DA, in the points E, F, G, H;
and join AE, EB, BF, FC, CG, GD, DH, HA:
then, each of the triangles AEB, BFC, CGD, DHA is greater than
the half of the segment of the circle in which it stands, as was shewn in
Prop. 11. of this book.
BOOK XII. PROP. X.
269
Erect prisms upon each of these triangles, of the same altitude with
the cylinder;
each of these prisms shall be greater than half of the segment of
the cylinder in which it is;
because if through the points E, F, G, H, parallels be drawn to
AB, BC, CD, DA,
and parallelograms be completed upon the same AB, BC, CD, DA,
and solid parallelopipeds be erected upon the parallelograms;
the prisms upon the triangles AEB, BFC, CGD, DÍA, are the
halves of the solid parallelopipeds; (xII. 7. Cor. 2.)
and the segments of the cylinder which are upon the segments of
the circle cut off by AB, BC, CD, DA, are less than the solid paral-
lelopipeds which contain them;
therefore the prisms upon the triangles AEB, BFC, CGD, DHA,
are greater than half of the segments of the cylinder in which they are:
therefore, if each of the circumferences be divided into two equal
parts, and straight lines be drawn from the points of division to the
extremities of the circumferences, and upon the triangles thus made
prisms be erected of the same altitude with the cylinder, and so on,
there must at length remain some segments of the cylinder which
together are less than the excess of the cylinder above the triple of
the cone: (XII. Lem. 1.)
let them be those upon the segments of the circle AE, EB, BF,
FC, CG, GD, DH, HA;
therefore the rest of the cylinder, that is, the prism of which the
base is the polygon AEBFCGDH, and of which the altitude is the
same with that of the cylinder, is greater than the triple of the cone:
but this prism is triple of the pyramid upon the same base, of which
the vertex is the same with the vertex of the cone; (xII. 7. Cor. 1.)
therefore the pyramid upon the base AEBFCGDH, having the
same vertex with the cone, is greater than the cone, of which the base
is the circle ABCD:
but it is also less, for the pyramid is contained within the cone ;
which is impossible:
therefore the cylinder is not greater than the triple of the cone.
Nor can the cylinder be less than the triple of the cone.
Let it be less, if possible;
therefore, inversely, the cone is greater than the third part of the
cylinder.


EK
B
H
D
G
F
In the circle ABCD inscribe a square :
this square is greater than the half of the circle:
and upon the square ABCD erect a pyramid, having the same
vertex with the cone;
this pyramid is greater than the half of the cone;
because, as was before demonstrated, if a square be described about
the circle, the square ABCD is the half of it:
$
270
EUCLID'S ELEMENTS.
*
and if upon these squares there be erected solid parallelopipeds of
the same altitude with the cone, which are also prisms,
the prism upon the square ABCD is the half of that which is upon
the square described about the circle ;
for they are to one another as their bases; as are also the third
parts of them: (XI. 32.)
therefore the pyramid, the base of which is the square ABCD, is
half of the pyramid upon the square described about the circle:
but this last pyramid is greater than the cone which it contains;
therefore the pyramid upon the square ABCD, having the same
vertex with the cone, is greater than the half of the cone.
Bisect the circumferences AB, BC, CD, DA in the points E, F, G, H,
and join AE, EB, BF, FC, CG, GD, DH, HA:
therefore each of the triangles AEB, BFC, CGD, DHA is greater
than half of the segment of the circle in which it is:
upon each of these triangles erect pyramids having the same vertex
with the cone:
therefore each of these pyramids is greater than the half of the
segment of the cone in which it is, as before was demonstrated of the
prisms and segments of the cylinder:
and thus dividing each of the circumferences into two equal parts,
and joining the points of division and their extremities by straight lines,
and upon the triangles erecting pyramids having their vertices the same
with that of the cone, and so on, there must at length remain some
segments of the cone, which together are less than the excess of the
cone above the third part of the cylinder: (XII. Lem. 1.)
let these be the segments upon AE, EB, BF, FC, CG, GD, DH, HA:
therefore the rest of the cone, that is, the pyramid of which the base
is the polygon AEBFCGDH, and of which the vertex is the same with
that of the cone, is greater than the third part of the cylinder:
but this pyramid is the third part of the prism upon the same base
AEBFCGDH, and of the same altitude with the cylinder;
therefore this prism is greater than the cylinder of which the base
is the circle ABCD:
but it is also less, for it is contained within the cylinder; which is
impossible:
therefore the cylinder is not less than the triple of the cone.
And it has been demonstrated that neither is it greater than the triple:
therefore the cylinder is triple of the cone, or, the cone is the third
part of the cylinder.
Wherefore, every cone, &c.
Q.E.D.
PROPOSITION XI. THEOREM.
Cones and cylinders of the same altitude, are to one another as their bases.
Let the cones and cylinders, of which the bases are the circles
ABCD, EFGH, and the axes KL, MN, and AC, EG the diameters of
their bases, be of the same altitude.
As the circle ABCD to the circle EFGH, so shall the cone AL be
to the cone EN.
BOOK XII.
271
PROP. XI.
L
N


D
H
S
R
E
I
C
M G
K
P
F
1
B
If it be not so, the circle ABCD must be to the circle EFGH, as the
cone AL to some solid either less than the cone EN, or greater than it.
First, let it be to a solid less than EN, viz. to the solid X;
and let Z be the solid which is equal to the excess of the cone EN
above the solid X;
therefore the cone EN is equal to the solids X, Z together.
In the circle EFGH inscribe the square EFGH;
therefore this square is greater than the half of the circle:
upon the square EFGH erect a pyramid of the same altitude with
the cone;
this pyramid shall be greater than half of the cone:
for, if a square be described about the circle, and a pyramid be
erected upon it, having the same vertex with the cone, the pyramid
inscribed in the cone is half of the pyramid circumscribed about it,
because they are to one another as their bases: (XII. 6.)
but the cone is less than the circumscribed pyramid;
therefore the pyramid of which the base is the square EFGH, and
its vertex the same with that of the cone, is greater than half of
the cone.
Divide the circumferences EF, FG, GH, HE, each into two equal
parts in the points O, P, R, S,
and join EO, OF, FP, PG, GR, RH, HS, SE:
therefore each of the triangles EOF, FPG, GRH, HSE is greater
than half of the segment of the circle in which it is:
upon
each of these triangles erect a pyramid having the same vertex
with the cone;
each of these pyramids is greater than the half of the segment of
the cone in which it is:
and thus dividing each of these circumferences into two equal parts,
and from the points of division drawing straight lines to the extremities
of the circumferences, and upon each of the triangles thus made erecting
pyramids having the same vertex with the cone, and so on, there must
at length remain some segments of the cone which are together less than
the solid Z; (Lemma.)
let these be the segments upon EO, OF, FP, PG, GR, RH, HS, SE:
therefore the remainder of the cone, viz. the pyramid of which the
base is the polygon EOFPGRHS, and its vertex the same with that of
the cone, is greater than the solid X.
In the circle ABCD inscribe the polygon ATBYCVDQ similar to
the polygon EOFPGRHS, and upon it erect a pyramid having the
same vertex with the cone AL:
and because as the square of AC is to the square of EG, so is the
polygon ATBYCVDQ to the polygon EOFPGRHS; (x11. 1.)
and as the square of AC to the square of EG, so is the circle ABCD
to the circle EFGH; (XII. 2.)
therefore the circle ABCD is to the circle EFGH, as the polygon
ATBYCVDQ to the polygon EOFPGRHS: (v. 11.)
272
EUCLID'S ELEMENTS.
but as the circle ABCD to the circle EFGH, so is the cone AL to
the solid X;
and as the polygon ATBYCVDQ to the polygon EOFPGRHS, so
is the pyramid of which the base is the first of these polygons, and
vertex L, to the pyramid of which the base is the other polygon, and
its vertex N: (XII. 6.)
therefore, as the cone AL to the solid X, so is the pyramid of which
the base is the polygon ATBYCVDQ, and vertex L, to the pyramid
the base of which is the polygon EOFPGRHS, and vertex N:
but the cone AL is greater than the pyramid contained in it;
therefore the solid X is greater than the pyramid in the cone EN: (v. 14.)
but it is less, as was shewn; which is absurd:
therefore the circle ABCD is not to the circle EFGH, as the cone
AL to any solid which is less than the cone EN.
In the same manner it may be demonstrated, that the circle EFGH
is not to the circle ABCD, as the cone EN to any solid less than the
cone AL.
Nor can the circle ABCD be to the circle EFGH, as the cone AL
to any solid greater than the cone EN.
For, if it be possible, let it be so to the solid I, which is greater than
the cone EN:
therefore, by inversion, as the circle EFGH to the circle ABCD, so
is the solid I to the cone AL:
but as the solid I to the cone AL, so is the cone EN to some solid,
which must be less than the cone AL; (v. 14.)
because the solid I is greater than the cone EN;
therefore, as the circle EFGH is to the circle ABCD, so is the cone
EN to a solid less than the cone AL, which was shewn to be im-
possible:
therefore the circle ABCD is not to the circle EFGH, as the cone
AL is to any solid greater than the cone EN.
And it has been demonstrated, that neither is the circle ABCD to
the circle EFGH, as the cone AL to any solid less than the cone EN:
therefore the circle ABCD is to the circle EFGH, as the cone AL
to the cone EN:
but as the cone is to the cone, so is the cylinder to the cylinder, (v. 15.)
because the cylinders are triple of the cones, each of each, (XII. 10.)
therefore as the circle ABCD to the circle EFGH, so are the
cylinders upon them of the same altitude.
Wherefore, cones and cylinders of the same altitude are to one
another as their bases. Q.E.D.
PROPOSITION XII. THEOREM.
Similar cones and cylinders have to one another the triplicate ratio of
that which the diameters of their bases have.
Let the cones and cylinders of which the bases are the circles
ABCD, EFGH, and the diameters of the bases AC, EG, and KL,
MN the axes of the cones or cylinders, be similar.
The cone of which the base is the circle ABCD and vertex the
point L, shall have to the cone of which the base is the circle EFGH
and vertex N, the triplicate ratio of that which AC has to EG.
BOOK XII. PROP. XII.
273
L
N


V
S
R
A
C
K
E
G
T
M
B
P
F
N
x
For, if the cone ABCDL has not to the cone EFGHN the triplicate
ratio of that which AC has to EG,
the cone ABCDL must have the triplicate of that ratio to some
solid which is less or greater than the cone EFGHN.
First, if possible, let it have it to a less, viz. to the solid X.
Make the same construction as in the preceding proposition,
and it may be demonstrated the very same way as in that proposition,
that the pyramid of which the base is the polygon EOFPGRHS,
and vertex N, is greater than the solid X.
Inscribe also in the circle ABCD the polygon ATBYCVDQ similar
to the polygon EOFPGRHS, upon which erect a pyramid having the
same vertex with the cone;
and let LAQ be one of the triangles containing the pyramid upon
the polygon ATBYCVDQ, the vertex of which is L;
and let ÑES be one of the triangles containing the pyramid upon
the polygon EOFPGRHS of which the vertex is N, and join KQ,
MS.
Then, because the cone ABCDL is similar to the cone EFGHN,
AC is to EG as the axis KL to the axis MN; (x1. def. 24.)
and as AC to EG, so is AK to EM; (v. 15.)
therefore as AK to EM, so is KL to MN;
and alternately, AK to KL, as EM to MN:
and the right angles AKL, EMN are equal:
therefore, the sides about these equal angles being proportionals,
the triangle AKL is similar to the triangle EMN.
Again, because AK is to KQ, as EM to MS,
(vi. 6.)
and that these sides are about equal angles AKQ, EMS, because
these angles are, each of them, the same part of four right angles at the
centres K, M;
therefore the triangle AKQ is similar to the triangle EMS. (v1. 6.)
And because it has been shewn that as AK to KL, so is EM to MN,
and that AK is equal to KQ, and EM to MS:
therefore as QK to KL, so is SM to MN:
and therefore, the sides about the right angles QKL, SMN being
proportionals,
the triangle LKQ is similar to the triangle NMS.
And because of the similarity of the triangles AKL, EMN,
as LA is to AK, so is NE to EM ;
and by the similarity of the triangles AKQ, EMS,
as KA to AQ, so ME to ES:
therefore, ex æquali, LA is to AQ, as NE to ES. (v. 22.)
Again, because of the similarity of the triangles LQK, NSM,
as LQ to QK, so NS to SM;
and from the similarity of the triangles KAQ, MES,
18
274
EUCLID'S ELEMENTS.
as KQ to QA, so MS to SE:
therefore, ex æquali, LQ is to QA, as NS to SE: (v. 22.)
and it was proved that QA is to AL, as SE to EN:
therefore, again, ex æquali, as QL to LA, so is SN to NE:
wherefore the triangles LQA, NSE, having the sides about all
their angles proportionals, are equiangular and similar to one another:
(VI. 5.)
and therefore the pyramid of which the base is the triangle AKQ,
and vertex L, is similar to the pyramid the base of which is the triangle
EMS, and vertex N,
because their solid angles are equal' to one another, and they are
contained by the same number of similar planes: (XI. B.)
but similar pyramids which have triangular bases have to one another
the triplicate ratio of that which their homologous sides have; (XII. 8.)
therefore the pyramid AKQL has to the pyramid EMSN the tri-
plicate ratio of that which AK has to EM.
In the same manner, if straight lines be drawn from the points D,
V, C, Y, B, T to K,
and from the points H, R, G, P, F, O to M,
and pyramids be erected upon the triangles having the same vertices
with the cones,
it may be demonstrated that each pyramid in the first cone has to
each in the other, taking them in the same order, the triplicate ratio of
that which the side AK has to the side EN;
that is, which AC has to EG:

but as one antecedent to its consequent, so are all the antecedents
to all the consequents; (v. 12.)
therefore as the pyramid AKQL to the pyramid EMSN, so is the
whole pyramid the base of which is the polygon DQATBYCV, and
vertex L, to the whole pyramid of which the base is the polygon
HSEOFPGR, and vertex N:
wherefore also the first of these two last-named pyramids has to the
other the triplicate ratio of that which AC has to EG:
but, by the hypothesis, the cone of which the base is the circle
ABCD, and vertex L, has to the solid X, the triplicate ratio of that
which AC has to EG;
therefore, as the cone of which the base is the circle ABCD, and
vertex L, is to the solid X, so is the pyramid the base of which is the
polygon DQATBYCV, and vertex L, to the pyramid the base of which
is the polygon HSEOFPGR, and vertex N:
but the said cone is greater than the pyramid contained in it;
therefore the solid X is greater than the pyramid, the base of which
is the polygon HSEOFPGR, and vertex N: (v. 14.)
but it is also less; which is impossible:
therefore the cone, of which the base is the circle ABCD and vertex
L, has not to any solid which is less than the cone of which the base is
the circle EFGH, and vertex N, the triplicate ratio of that which AC
has to EG.
In the same manner it may be demonstrated, that neither has the
cone EFGHN to any solid which is less than the cone ABCDL, the
triplicate ratio of that which EG has to AC.
Nor can the cone ABCDL have to any solid which is greater than
the cone EFGHN, the triplicate ratio of that which AC has to EG.
BOOK XII. PROP. XII, XIII.
275
For, if it be possible, let it have it to a greater, viz. to the solid Z:
therefore, inversely, the solid Z has to the cone ABCDL, the tripli-
cate ratio of that which EG has to AC:
but as the solid Z is to the cone ABCDL, so is the cone EFGHN
to some solid, which must be less than the cone ABCDL, (v.14.)
because the solid Z is greater than the cone EFGHN;
therefore the cone EFGHN has to a solid which is less than the
·cone ABCDL, the triplicate ratio of that which EG has to AC,
which was demonstrated to be impossible:
therefore the cone ABCDL has not to any solid greater than the
cone EFGHN, the triplicate ratio of that which AC has to EG:
and it was demonstrated, that it could not have that ratio to any
solid less than the cone EFGHN;
therefore the cone ABCDL has to the cone EFGHN, the tripli-
cate ratio of that which AC has to EG:
but as the cone is to the cone, so the cylinder to the cylinder; (v. 15.)
for every cone is the third part of the cylinder upon the same base,
and of the same altitude: (XII. 10.)
therefore also the cylinder has to the cylinder, the triplicate ratio
of that which AC has to EG.
Wherefore, similar cones, &c.
Q. E.D.
PROPOSITION XIII. THEOREM.
If a cylinder be cut by a plane parallel to its opposite planes, or bases;
it divides the cylinder into two cylinders, one of which is to the other as
the axis of the first to the axis of the other.
Let the cylinder AD be cut by the plane GH parallel to the oppo-
site planes AB, CD, meeting the axis EF in the point K,
and let the line GH be the common section of the plane GH and
the surface of the cylinder AD.
Let AEFC be the parallelogram in any position of it, by the revo-
lution of which about the straight line EF the cylinder AD is described;
and let GK be the common section of the plane GH, and the plane
AEFC.
And because the parallel planes AB, GH are cut by the plane
AEKG,
their common sections AE, KG, with it, are parallel: (x1.16.)
wherefore AK is a parallelogram,
and GK equal to EA the straight line from the centre of the circle AB:
for the same reason, each of the straight lines drawn from the point
K to the line GH may be proved to be equal to those which are drawn
from the centre of the circle AB to its circumference, and are therefore
all equal to one another;
therefore the line GH is the circumference of a circle of which the
centre is the point K: (1. def. 15.)
therefore the plane GH divides the cylinder AD into the cylinders
AH, GD;
for they are the same which would be described by the revolution
of the parallelograms AK, GF about the straight lines EK, KF:
and it is to be shewn, that the cylinder AH is to the cylinder HC,
as the axis EK to the axis KF.
+
•
18-2
276
EUCLID'S ELEMENTS.

L
R
N
A
E
B
K
H
F
D
T
X
Y
M Q
Produce the axis EF both ways:
and take any number of straight lines EN, NL, each equal to EK;
and any number FX, XM, each equal to FK;
and let planes parallel to AB, CD, pass through the points L, N, X, M:
therefore the common sections of these planes with the cylinder
produced are circles, the centres of which are the points L, N, X, M, as
was proved of the plane GH;
and these planes cut off the cylinders PR, RB, DT, TQ.
And because the axes LÑ, NE, EK, are all equal,
therefore the cylinders PR, RB, BG, are to one another as their
bases: (XII. 11.)
but their bases are equal, and therefore the cylinders PR, RB, BG,
are equal:
and because the axes LN, NE, EK, are equal to one another,
as also the cylinders PR, RB, BG,
and that there are as many axes as cylinders:
therefore, whatever multiple the axis KL is of the axis KE, the
same multiple is the cylinder PG of the cylinder GB:
for the same reason, whatever multiple the axis MK is of the axis
KF, the same multiple is the cylinder QG of the cylinder GD:
and if the axis KL be equal to the axis KM, the cylinder PG is
equal to the cylinder GQ;
and if the axis KL be greater than the axis KM, the cylinder PG
is greater than the cylinder GQ; and if less, less:
therefore, since there are four magnitudes, viz. the axes EK, kf,
and the cylinders BG, GD;
and that of the axis EK and cylinder BG there have been taken
any equimultiples whatever, viz. the axis KL and cylinder PG,
and of the axis KF and cylinder GD, any equimultiples whatever,
viz. the axis KM and cylinder GQ;
and since it has been demonstrated, that if the axis KL be greater
than the axis KM,
the cylinder PG is greater than the cylinder GQ;
and if equal, equal; and if less, less:
therefore as the axis EK is to the axis KF, so is the cylinder BG
to the cylinder GD.
Wherefore, if a cylinder, &c. Q.E.D.
BOOK XII. PROP. XIV, xv.
277.
PROPOSITION XIV. THEOREM.
Cones and cylinders upon equal bases are to one another as their
altitudes.
Let the cylinders EB, FD, be upon the equal bases AB, CD.
As the cylinder EB to the cylinder FD, so shall the axis GH be
to the axis KL.
Produce the axis KL to the point N, and make LN equal to the
axis GH,
and let CM be a cylinder of which the base is CD, and axis LN.
Then because the cylinders EB, CM, have the same altitude, they
are to one another as their bases: (XII. 11.)
but their bases are equal, therefore also the cylinders EB, CM,
are equal:
F K
EG
C
LD
A
H
B
N M
:
and because the cylinder FM is cut by the plane CD parallel to its
opposite planes,
as the cylinder CM to the cylinder FD, so is the axis LN to the
axis KL: (XII. 13.)
but the cylinder CM is equal to the cylinder EB, and the axis LN
to the axis GH;
therefore as the cylinder EB to the cylinder FD, so is the axis GH
to the axis KL :
and as the cylinder EB to the cylinder FD, so is the cone ABG to
the cone CDK, (v.15.)
because the cylinders are triple of the cones: (XII. 10.)
therefore also the axis GH is to the axis KL, as the cone ABG to
the cone CDK, and as the cylinder EB to the cylinder FD.
Wherefore cones, &c. Q.E.D.
PROPOSITION XV. THEOREM.
The bases and altitudes of equal cones and cylinders are reciprocally
proportional; and conversely, if the bases and altitudes be reciprocally
proportional, the cones and cylinders are equal to one another.
Let the circles ABCD, EFGH, the diameters of which are AC, EG,
be the bases, and KL, MŃ, the axes, as also the altitudes, of equal cones
and cylinders;
and let ALC, ENG be the cones, and AX, EO the cylinders:
the bases and altitudes of the cylinders AX, EO shall be reciprocally
proportional;
that is, as the base ABCD to the base EFGH, so shall the altitude
MN be to the altitude KL.
27.8
EUCLID'S ELEMENTS.
R
Ο
A
I
X
Y
/H
E
G
F
K
B
Either the altitude MN is equal to the altitude KL, or these altitudes
are not equal.
First let them be equal;
and the cylinders AX, EO being also equal,
and cones and cylinders of the same altitude being to one another
as their bases, (XII. 11.)
therefore the base ABCD is equal to the base EFGH: (v.a.)
and as the base ABCD is to the base EFGH, so is the altitude MN
to the altitude KL.
But let the altitudes KL, MN, be unequal, and MN the greater of
the two,
:
and from MN take MP equal to KL,
and through the point P cut the cylinder EO by the plane TYS,
parallel to the opposite planes of the circles EFGH, RO;
therefore the common section of the plane TYS and the cylinder
EO is a circle, and consequently ES is a cylinder, the base of which is
the circle EFGH, and altitude MP:
and because the cylinder AX is equal to the cylinder EO,
as AX is to the cylinder ES, so is the cylinder EO to the same ES: (v. 7.)
but as the cylinder AX to the cylinder ES, so is the base ABCD to
the base EFGH; (x11. 11.)
for the cylinders AX, ES are of the same altitude;
and as the cylinder EO to the cylinder ES, so is the altitude MN
to the altitude MP, (xII. 13.)
because the cylinder EO is cut by the plane TYS parallel to its
opposite planes;
therefore as the base ABCD to the base EFGH, so is the altitude
MN to the altitude MP:
but MP is equal to the altitude KL;
wherefore as the base ABCD to the base EFGH, so is the altitude
MN to the altitude KL:
that is, the bases and altitudes of the equal cylinders AX, EO, are
reciprocally proportional.
But let the bases and altitudes of the cylinders AX, EO be recipro-
cally proportional,
viz. the base ABCD to the base EFGH, as the altitude MN to the
altitude KL:
the cylinder AX shall be equal to the cylinder EO.
First, let the base ABCD be equal to the base EFGH:
then because as the base ABCD is to the base EFGH, so is the
altitude MN to the altitude KL;
MN is equal to KL; (v. A.)
and therefore the cylinder AX is equal to the cylinder EO. (XII. 11.)
But let the bases ABCD, EFGH be unequal,
and let ABCD be the greater:
and because as ABCD is to the base EFGH, so is the altitude MN
to the altitude KL;
BOOK XII PROP. XV, XVI,
279
therefore MN is greater than KL. (v. a.)
Then, the same construction being made as before,
because as the base ABCD to the base EFGH, so is the altitude
MN to the altitude KL;
and because the altitude KL is equal to the altitude MP;
therefore the base ABCD is to the base EFGH, as the cylinder AX
to the cylinder ES; (x11. 12.)
and as the altitude MN to the altitude MP or KL, so is the cylinder
EO to the cylinder ES:
therefore the cylinder AX is to the cylinder ES, as the cylinder EO
is to the same ES:
whence the cylinder AX is equal to the cylinder EO:
and the same reasoning holds in cones.
Q.E.D.
PROPOSITION XVI. PROBLEM.
In the greater of two circles that have the same centre, to inscribe a
polygon of an even number of equal sides, that shall not meet the lesser
circle.
Let ABCD, EFGH be two given circles having the same centre K.
It is required to inscribe in the greater circle ABCD, a polygon of
an even number of equal sides, that shall not meet the lesser circle.
A
B
E
H
L
KGMD
F
CN
Through the centre K draw the straight line BD,
and from the point G, where it meets the circumference of the
lesser circle,
draw GA at right angles to BD, and produce it to C;
therefore AC touches the circle EFGH: (111. 16. Cor.)
then, if the circumference BAD be bisected, and the half of it be
again bisected, and so on, there must at length remain a circumference
less than AD: (XII. Lem. 1.)
let this be LD;
and from the point L draw LM perpendicular to BD, and produce
it to N;
and join LD, DN:
therefore LD is equal to DN:
and because LN is parallel to AC, and that AC touches the
circle EFGH;
therefore LN does not meet the circle EFGH
and much less shall the straight lines LD, DN, meet the circle EFGH:
so that if straight lines equal to LD be applied in the circle ABCD
from the point L around to N,
there shall be inscribed in the circle a polygon of an even number
of equal sides not meeting the lesser circle. Q. E. F.
280
EUCLID'S ELEMENTS.
LEMMA II.
If two trapeziums ABCD, EFGH be inscribed in the circles, the
centres of which are the points K, L; and if the sides AB, DC be
parallel, as also EF, HG; and the other four sides AD, BC, EH, FG,
be all equal to one another; but the side AB greater than EF, and DC
greater than HG: the straight line KA from the centre of the circle in
which the greater sides are, is greater than the straight line LE drawn
from the centre to the circumference of the other circle.
If it be possible, let KA be not greater than LE;
then KA must be either equal to it, or less than it.

A
D
K
C

B
E
H
G
PO
L
First, let KA be equal to LE:
therefore, because in two equal circles AD, BC, in the one, are equal
to EH, FG in the other, the circumferences AD, BC, are equal to the
circumferences EH, FG; (III. 28.)
but because the straight lines AB, DC are respectively greater
than EF, GH;
the circumferences AB, DC are greater than EF, HG;
therefore the whole circumference ABCD is greater than the
whole EFGH:
but it is also equal to it, which is impossible:
therefore the straight line KA is not equal to LE.
But let KA be less than LE,
and make LM equal to KA,
and from the centre L, and distance LM, describe the circle MNOP,
meeting the straight lines LE, LF, LG, LH, in M, N, O, P ;
and join MN, NO, OP, PM, which are respectively parallel to, and
less than EF, FG, GH, HE: (VI. 2.)
then because EH is greater than MP,
AD is greater than MP;
and the circles ABCD, MNOP are equal;
therefore the circumference AD is greater than MP:
for the same reason, the circumference BC is greater than NO:
and because the straight line AB is greater than EF, which is
greater than MN,
much more is AB greater than MN:
therefore the circumference AB is greater than MN;
and for the same reason, the circumference DC is greater than PO:
therefore the whole circumference ABCD is greater than the whole
MNOP:
but it is likewise equal to it, which is impossible:
therefore KA is not less than LE:
nor is it equal to it;
therefore the straight line KA must be greater than LE. Q. E. D.
BOOK XII. PROP. XVII.
281
COR. And if there be an isosceles triangle, the sides of which are
equal to AD, BC, but its base less than AB, the greater of the two
sides AB, DC; the straight line KA may, in the same manner, be
demonstrated to be greater than the straight line drawn from the centre
to the circumference of the circle described about the triangle.
PROPOSITION XVII. PROBLEM.
In the greater of two spheres which have the same centre, to inscribe a
solid polyhedron, the superficies of which shall not meet the lesser sphere.
Let there be two spheres about the same centre A.
It is required to inscribe in the greater a solid polyhedron, the
superficies of which shall not meet the lesser sphere.

D
X
R
ན
P
I
I
Τ
E
M
E
S
G
eb
N
B
Let the spheres be cut by a plane passing through the centre;
the common sections of it with the spheres shall be circles;
because the sphere is described by the revolution of a semicircle
about the diameter remaining unmoveable;
so that in whatever position the semicircle be conceived, the common
section of the plane in which it is with the superficies of the sphere is
the circumference of a circle;
and this is a great circle of the sphere, because the diameter of the
sphere, which is likewise the diameter of the circle, is greater than any
straight line in the circle or sphere. (III. 15.)
Let then the circle made by the section of the plane with the greater
sphere be BCDE, and with the lesser sphere be FGH;
and draw the two diameters BD, CE at right angles to one another;
and in BCDE, the greater of the two circles, inscribe a polygon
of an
even number of equal sides not meeting the lesser circle
FGH; (x11. 16.)
and let its sides in BE, the fourth part of the circle, be BK, KL,
LM, ME;
join KA, and produce it to N;
and from A draw AX at right angles to the plane of the circle BCDE,
meeting the superficies of the sphere in the point X: (xI. 12.)
and let planes pass through AX, and each of the straight lines BD,
KN, which from what has been said, shall produce great circles on the
282
EUCLID'S ELEMENTS.
superficies of the sphere, and let BXD, KXN be the semicircles thus
made upon the diameters, BD, KN:
therefore, because XA is at right angles to the plane of the
circle BCDE,
every plane which passes through XA is at right angles to the
plane of the circle BCDE; (xI. 18.)
wherefore the semicircles BXD, KXN are at right angles to that plane:
and because the semicircles BED, BXD, KXŇ upon the equal
diameters BD, KN, are equal to one another;
their halves BE, BX, KX, are equal to one another,
therefore as many sides of the polygon as are in BE, so many are
there in BX, KX, equal to the sides BK, KL, LM, ME:
let these polygons be described, and their sides be BO, OP, PR,
RX; KS, ST, TY, YX;

and join OS, PT, RY;
and from the points O, S, draw OV, SQ perpendiculars to AB, AK:
and because the plane BOXD is at right angles to the plane BCDE,
and in one of them BOXD, OV is drawn perpendicular to AB the
common section of the planes,
therefore OV is perpendicular to the plane BCDE: (x1. def. 4.)
for the same reason SQ is perpendicular to the same plane,
because the plane KSX is at right angles to the plane BCDE.
Join VQ:
and because in the equal semicircles BXD, KXN,
the circumferences BO, KS are equal,
and OV, SQ are perpendicular to their diameters,
therefore OV is equal to SQ, and BV equal to KQ: (1.26.)
but the whole BA is equal to the whole KA,
therefore the remainder VA is equal to the remainder QA ;
therefore as BV is to VA, so is KQ to QA;
wherefore VQ is parallel to BK: (vI. 2.)
and because OV, SQ are each of them at right angles to the plane
of the circle BCDE,
OV is parallel to SQ: (x1. 6.)
and it has been proved, that it is also equal to it;
therefore QV, SÕ are equal and parallel: (1. 33.)
and because QV is parallel to SO, and also to KB;
OS is parallel to BK; (x1.9.)
and therefore BO, KS, which join them are in the same plane in
which these parallels are,
and the quadrilateral figure KBOS is in one plane:
and if PB, TK be joined, and perpendiculars be drawn from the
points P, T, to the straight lines AB, AK,
it may be demonstrated, that TP is parallel to KB in the very same
way that SO was shewn to be parallel to the same KB;
wherefore TP is parallel to SO, (x1. 9.)
and the quadrilateral figure SOPT is in one plane:
for the same reason the quadrilateral TPRY is in one plane:
and the figure YRX is also in one plane: (xI. 2.)
therefore, if from the points O, S, P, T, R, Y, there be drawn
straight lines to the point A, there will be formed a solid polyhedron
between the circumferences BX, KX, composed of pyramids, the bases
of which are the quadrilaterals KBOS, SOPT, TPRY, and the triangle
YRX, and of which the common vertex is the point A:

BOOK XII. PROP. XVII.
283
}
and if the same construction be made upon each of the sides KL,
LM, ME, as has been done upon BK,
and the like be done also in the other three quadrants, and in the
other hemisphere;
there will be formed a solid polyhedron inscribed in the sphere,
composed of pyramids, the bases of which are the aforesaid quadrilateral
figures, and the triangle YRX, and those formed in the like manner in
the rest of the sphere, the common vertex of them all being the point A.
Also the superficies of this solid polyhedron shall not meet the
lesser sphere in which is the circle FGÌ.
For, from the point A draw AZ perpendicular to the plane of the
quadrilateral KBOS, meeting it in Z, and join BZ, ZK: (xI. 11.)
and because AZ is perpendicular to the plane KBOS,
it makes right angles with every straight line meeting it in that plane;
therefore AZ is perpendicular to BZ and ZK:
and because AB is equal to ÂK, and that the squares of AZ, ZB are
equal to the square of AB, and the squares of AZ, ZK to the square of
AK; (1.47.)
therefore the squares of AZ, ZB are equal to the squares of AZ, ZK:
take from these equals the square of AZ,
and the remaining square of BZ is equal to the remaining square of ZK,
and therefore the straight line BZ is equal to ZK:
in the like manner it may be demonstrated, that the straight lines
drawn from the point Z to the points O, S, are equal to BZ or ZK;
therefore the circle described from the centre Z, and distance ZB,
will pass through the points K, O, S,
and KBOS will be a quadrilateral figure in the circle:
and because KB is greater than QV, and QV equal to SO,
therefore KB is greater than SO:
but KB is equal to each of the straight lines BO, KS;
wherefore each of the circumferences cut off by KB, BO, KS, is
greater than that cut off by OS;
and these three circumferences, together with a fourth equal to one
of them, are greater than the same three together with that cut off
by OS;
that is, than the whole circumference of the circle;
therefore the circumference subtended by KB is greater than the
fourth part of the whole circumference of the circle KBOS,
and consequently the angle BZK at the centre is greater than a
right angle:
and because the angle BZK is obtuse,
the square of BK is greater than the squares of BZ, ZK; (11. 12.)
that is, greater than twice the square of BZ.
Join KV:
and because in the triangle KBV, OBV,
KB, BV are equal to OB, BV, and that they contain equal angles ;
the angle KBV is equal to the angle OVB:
and OVB is a right angle;
therefore also KVB is a right angle:
and because BD is less than twice DV;
(1.4.)
the rectangle contained by BD, BV is less than twice the rectangle
DV, VB;
that is, the square of KB is less than twice the square of KV: (v1.8.)
284
EUCLID'S ELEMENTS.
1
but the square of KB is greater than twice the square of BZ;
therefore the square of KV is greater than the square of BZ:
and because BA is equal to AK,
and that the squares of BZ, ZA are equal together to the square of BA,
and the squares of KV, VA to the square of AK;
therefore the squares of BZ, ZA are equal to the squares of KV, VA;
and of these the square of KV is greater than the square of BZ;
therefore the square of VA is less than the square of ZA,
and the straight line AZ greater than VA:
much more then is AZ greater than AG;
because, in the preceding proposition,
it was shewn that KV falls without the circle FGH:
and AZ is perpendicular to the plane KBOS,
and is therefore the shortest of all the straight lines that can be
drawn from A, the centre of the sphere, to that plane.
Therefore the plane KBOS does not meet the lesser sphere.
And that the other planes between the quadrants BX, KX, fall
without the lesser sphere, is thus demonstrated.
From the point A draw AI perpendicular to the plane of the quad-
rilateral SOPT, and join 10;
and, as was demonstrated of the plane KBOS and the point Z, in
the same way it may be shewn that the point I is the centre of a circle
described about SOPT;

and that OS is greater than PT;
and PT was shewn to be parallel to OS:
therefore, because the two trapeziums KBOS, SOPT inscribed in
circles, have their sides BK, OS parallel, as also OS, PT;
and their other sides BO, KS, OP, ST all equal to one another,
and that BK is greater than OS, and OS greater than PT,
therefore the straight line ZB is greater than IO. (x11. Lem. 2.)
Join AO, which will be equal to AB;
and because AIO, AZB are right angles,
the squares of AI, 10 are equal to the square of AO or of AB ;
that is, to the squares of AZ, ZB ;
and the square of ZB is greater than the square of IO,
therefore the square of AŽ is less than the square of AI;
and the straight line AZ less than the straight line AI:
and it was proved, that AZ is greater than AG;
much more then is AI greater than AG:
therefore the plane SOPT falls wholly without the lesser sphere.
In the same manner it may be demonstrated,
that the plane TPRY falls without the same sphere, as also the
triangle YRX, viz. by x11. Lem. 2. Cor.
And after the same way it may be demonstrated, that all the planes
which contain the solid polyhedron, fall without the lesser sphere.
Therefore in the greater of two spheres, which have the same centre,
a solid polyhedron is inscribed, the superficies of which does not meet
the lesser sphere. Q.E.F.
But the straight line AZ may be demonstrated to be greater than
AG otherwise, and in a shorter manner, without the help of Prop. xvI.,
as follows.
From the point G draw GU at right angles to AG, and join AU.
If then the circumference BE be bisected, and its half again
BOOK XII. PROP. XVII, XVIII.
285
bisected, and so on, there will at length be left a circumference less
than the circumference which is subtended by a straight line equal to
GU, inscribed in the circle BCDE:
let this be the circumference KB:
therefore the straight line KB is less than GU:
and because the angle BZK is obtuse, as was proved in the preceding,
therefore BK is greater than BZ:
but GU is greater than BK;
much more then is GU greater than BZ,
and the square of GU than the square of BZ:
and AU is equal to AB;
therefore the square of AU, that is, the squares of AG, GU are
equal to the square of AB, that is, to the squares of AZ, ZB:
but the square of BZ is less than the square of GU;
therefore the square of AZ is greater than the square of AG,
and the straight line AZ consequently greater than the straight line AG.
COR.-And if in the lesser sphere there be inscribed a solid poly-
hedron, by drawing straight lines betwixt the points in which the
straight lines from the centre of the sphere drawn to all the angles of
the solid polyhedron in the greater sphere meet the superficies of the
lesser; in the same order in which are joined the points in which the
same lines from the centre meet the superficies of the greater sphere:
the solid polyhedron in the sphere BCDE shall have to this other solid
polyhedron the triplicate ratio of that which the diameter of the sphere
BCDE has to the diameter of the other sphere. For if these two solids
be divided into the same number of pyramids, and in the same order,
the pyramids shall be similar to one another, each to each: because
they have the solid angles at their common vertex, the centre of the
sphere, the same in each pyramid, and their other solid angles at the
bases, equal to one another, each to each, (XI. B.) because they are con-
tained by three plane angles, each equal to each; and the pyramids are
contained by the same number of similar planes; and are therefore
similar to one another, each to each: (x1. def. 11.) but similar pyramids
have to one another the triplicate ratio of their homologous sides: (XII.
8. Cor.) therefore the pyramid of which the base is the quadrilateral
KBOS, and vertex A, has to the pyramid in the other sphere of the
same order, the triplicate ratio of their homologous sides, that is, of that
ratio, which AB from the centre of the greater sphere has to the straight
line from the same centre to the superficies of the lesser sphere. And
in like manner, each pyramid in the greater sphere has to each of the
same order in the less, the triplicate ratio of that which AB has to the
semi-diameter of the less sphere. And as one antecedent is to its
consequent, so are all the antecedents to all the consequents. Where-
fore the whole solid polyhedron in the greater sphere has to the whole
solid polyhedron in the other, the triplicate ratio of that which AB the
semi-diameter of the first has to the semi-diameter of the other; that is,
which the diameter BD of the greater has to the diameter of the other
sphere.

PROPOSITION XVIII. THEOREM.
Spheres have to one another the triplicate ratio of that which their
diameters have.
1
286
EUCLID'S ELEMENTS.
Let ABC, DEF be two spheres, of which the diameters are BC, EF,
The sphere ABC shall have to the sphere DEF the triplicate ratio
of that which BC has to EF.
B
A
E
D
G
L

K
F
M
N
For, if it has not, the sphere ABC must have to a sphere either less
or greater than DEF, the triplicate ratio of that which BC has to EF.
First, if possible, let it have that ratio to a less, viz. to the sphere GHK;
and let the sphere DEF have the same centre with GHK:
and in the greater sphere DEF inscribe a solid polyhedron, the super-
ficies of which does not meet the lesser sphere GHK; (x11. 17.)
and in the sphere ABC inscribe another similar to that in the
sphere DEF:
therefore the solid polyhedron in the sphere ABC has to the solid
polyhedron in the sphere DEF, the triplicate ratio of that which BC
has to EF. (x11. 17. Cor.)
But the sphere ABC has to the sphere GHK, the triplicate ratio of
that which BC has to EF;
therefore, as the sphere ABC to the sphere GHK, so is the solid
polyhedron in the sphere ABC to the solid polyhedron in the sphere
DEF:
but the sphere ABC is greater than the solid polyhedron in it;
therefore also the sphere GHK is greater than the solid polyhedron
in the sphere DEF: (v. 14.)
but it is also less, because it is contained within it, which is impossible:
therefore the sphere ABC has not to any sphere less than DEF,
the triplicate ratio of that which BC has to EF.
In the same manner, it may be demonstrated, that the sphere DEF
has not to any sphere less than ABC, the triplicate ratio of that which
EF has to BČ.
Nor can the sphere ABC have to any sphere greater than DEF, the
triplicate ratio of that which BC has to EF:
for, if it can, let it have that ratio to a greater sphere LMN:
therefore, by inversion, the sphere LMN has to the sphere ABC, the
triplicate ratio of that which the diameter EF has to the diameter BC.
But as the sphere LMN to ABC, so is the sphere DEF to some
sphere, which must be less than the sphere ABC, (v. 14.)
because the sphere LMN is greater than the sphere DEF;
therefore the sphere DEF has to a sphere less than ABC the
triplicate ratio of that which EF has to BC;
which was shewn to be impossible:
therefore the sphere ABC has not to any sphere greater than DEF,
the triplicate ratio of that which BC has to EF:
and it was demonstrated, that neither has it that ratio to any sphere
less than DEF.
Therefore the sphere ABC has to the sphere DEF, the triplicate
ratio of that which BC has to EF. Q. E.D.
NOTES TO BOOK XII.
THIS book treats of the properties of prisms and cylinders, pyramids and cones.
A new principle is introduced called "the method of Exhaustions," which may be
applied for the purpose of finding the areas and ratios of circles, and the relations of
the surfaces and of the volumes of cones, spheres and cylinders.
The first comparison of rectilinear areas is made in the first book of the Elements
by the principle of superposition, where two triangles are coincident in all respects;
next, comparison is made between triangles and other rectilinear figures when they are
not coincident.
In the sixth book, similar triangles are compared by shewing that they are in the
duplicate ratio of their homologous sides, and then by dividing similar polygons into
the same number of similar triangles, and shewing that the polygons are also in the
duplicate ratio of any of their homologous sides. In the eleventh book, similar
rectilinear solids are compared by shewing that their volumes are to one another in
the triplicate ratio of their homologous sides.
"The method of Exhaustions" is founded on the principle of exhausting a mag-
nitude by continually taking away a certain part of it, as is explained in the tenth
book of the Elements, where Euclid states, that two quantities are equal, whose dif-
ference is less than any assignable quantity. If A and A' be two magnitudes of the
same kind, and if d and d be any other magnitudes of the same kind, such that
A' + d is always greater than, and A'-d' always less than A, however small d and
d' may be made, then A' is equal to A.
The method of exhaustions may be applied to find the circumference and area of
a circle. A rectilinear figure may be inscribed in the circle and a similar one circum-
scribed about it, and then by continually doubling the number of sides of the inscribed
and circumscribed polygons, by this principle, it may be demonstrated, that the area
of the circle is less than the area of the circumscribed polygon, but greater than the
area of the inscribed polygon; and that as the number of sides of the polygon is in-
creased, and consequently the magnitude of each diminished, the differences between
the circle and the inscribed and circumscribed polygons are continually exhausted, and
at length become less than any assignable difference; and the area of the circle is the
limit to which the inscribed and circumscribed polygons continually approach, as the
number of sides is increased.
Also in comparing two unequal circles, two similar polygons are inscribed in the
circles, and then by doubling the number of sides continually, it is shewn that the limit
to which the ratio of the areas of the rectilineal figures continually approach is the same
as the ratio of the circles.
In a similar way it will be seen, that the principle is applied to the surfaces and
volumes of cones, cylinders and spheres.
Prop. II. (1.) For there is some square equal to the circle ABCD; let P be the
side of it, and to three straight lines BD, FH, and P, there can be a fourth propor-
tional; let this be Q: therefore (v1. 22.) the squares of these four straight lines are
proportionals; that is, to the squares of BD, FH, and the circle ABCD it is possible
there may be a fourth proportional. Let this be S. And in like manner are to be
understood some things in some of the following propositions.
C
288
EUCLID'S ELEMENTS.
•
(2.) For as, in the foregoing note, it was explained how it was possible there could
be a fourth proportional to the squares of BD, FH, and the circle ABCD, which was
named S; so, in like manner, there can be a fourth proportional to this other space,
named T, and the circles ABCD, EFGH. And the like is to be understood in some
of the following propositions.
(3.) Because, as a fourth proportional to the squares of BD, FH, and the circle
ABCD, is possible, and that it can neither be less nor greater than the circle EFGH, it
must be equal to it. SIMSON.
Prop. IV. Because GO is equal to OA, and GM to MB, therefore (v1. 2.) OM is
parallel to AB; in the same manner ON is parallel to AC; therefore (XI. 15.) the plane
MON is parallel to the plane BAC. SIMSON.
Prop. XI. Vertex is put in place of altitude, which is in the Greek, because the
pyramid, in what follows, is supposed to be circumscribed about the cone, and so
must have the same vertex. And the same change is made in some places following.
SIMSON.
The thirteenth book of the Elements relates to equilateral and equiangular figures
inscribed in circles, and to the five regular solids.
Two books which treat of the inscriptions of the five regular solids in one another
and in spheres, are frequently found annexed to the Elements as the fifteenth and six-
teenth books; these however, were composed by Hypsicles of Alexandria.
There is a continuation of the same subject by Flussas, which has been appended to
the Elements, and called the sixteenth book of the Elements.
ON THE
ANCIENT GEOMETRICAL ANALYSIS.
SYNTHESIS, or the method of composition, is a mode of reasoning
which begins with something given, and ends with something required,
either to be done or to be proved. This may be termed a direct process,
as it leads from principles to consequences.
Analysis, or the method of resolution, is the reverse of synthesis,
and thus may be considered an indirect process, a method of reasoning
from consequences to principles.
The synthetic method is pursued by Euclid in his elements of Geo-
metry. He commences with certain assumed principles, and proceeds
to the solution of problems and the demonstration of theorems by unde-
niable and successive inferences from them.
The Geometrical Analysis was a process employed by the ancient
Geometers, both for the discovery of the solution of problems and for
the investigation of the truth of theorems. In the analysis of a problem,
the quæsita, or what is required to be done, is supposed to have been
effected, and the consequences are traced by a series of geometrical con-
structions and reasonings, till at length they terminate in the data of the
problem, or in some previously demonstrated or admitted truth, whence
the direct solution of the problem is deduced.
In the Synthesis of a problem, however, the last consequence of the
analysis is assumed as the first step of the process, and by proceeding
in a contrary order through the several steps of the analysis until the
process terminate in the quæsita, the solution of the problem is effected.
But if, in the analysis, we arrive at a consequence which contra-
dicts any truth demonstrated in the Elements, or which is inconsistent
with the data of the problem, the problem must be impossible: and
further, if in certain relations of the given magnitudes the construc-
tion be possible, while in other relations it is impossible, the disco-
very of these relations will become a necessary part of the solution of the
problem.
In the analysis of a theorem, the question to be determined, is,
whether by the application of the geometrical truths proved in the
Elements, the predicate is consistent with the hypothesis. This
point is ascertained by assuming the predicate to be true, and by
deducing the successive consequences of this assumption combined with
proved geometrical truths, till they terminate in the hypothesis of the
theorem or some demonstrated truth. The theorem will be proved
synthetically by retracing, in order, the steps of the investigation pur-
sued in the analysis, till they terminate in the predicate, which was
assumed in the analysis. This process will constitute the demonstration
of the theorem.
If the assumption of the truth of the predicate in the analysis lead
to some consequence which is inconsistent with any demonstrated truth,
19
290
ANCIENT GEOMETRICAL ANALYSIS.
the false conclusion thus arrived at indicates the falsehood of the predi-
cate; and by reversing the process of the analysis, it may be demon-
strated, that the theorem cannot be true.
It may be here remarked, that the geometrical analysis is more
extensively useful in discovering the solution of problems than for in-
vestigating the demonstration of theorems.
From the nature of the subject, it must be at once obvious, that no
general rules can be prescribed, which will be found applicable in
all cases, and infallibly lead to the solution of every problem. The
conditions of problems must suggest what constructions may be pos-
sible; and the consequences which follow from these constructions and
the assumed solution, will shew the possibility or impossibility of
arriving at some known property consistent with the data of the
problem.
Though the data of a problem may be given in magnitude and
position, certain ambiguities will arise, if they are not properly re-
stricted. Two points may be considered as situated on the same side,
or one on each side of a given line; and there may be two lines drawn
from a given point making equal angles with a line given in position;
and to avoid ambiguity, it must be stated on which side of the line
the angle is to be formed.
A problem is said to be determinate when, with the prescribed con-
ditions, it admits of one definite solution; the same construction which
may be made on the other side of any given line, not being considered
a different solution: and a problem is said to be indeterminate when it
admits of more than one definite solution. This latter circumstance
arises from the data not absolutely fixing but merely restricting the
quæsita, leaving certain points or lines not fixed in one position only.
The number of given conditions may be insufficient for a single deter-
minate solution; or relations may subsist among some of the given
conditions from which one or more of the remaining given condi-
tions may be deduced.
If the base of a right-angled triangle be given, and also the dif-
ference of the squares of the hypothenuse and perpendicular, the tri-
angle is indeterminate. For though apparently here are three things
given, the right angle, the base, and the difference of the squares of the
hypothenuse and perpendicular, it is obvious that these three apparent
conditions are in fact reducible to two: for since in a right-angled
triangle, the sum of the squares on the base and on the perpendicular
are equal to the square on the hypothenuse, it follows that the differ-
ence of the squares of the hypothenuse and perpendicular is equal to
the square of the base of the triangle, and therefore the base is known
from the difference of the squares of the hypothenuse and perpendicular
being known. The conditions therefore are insufficient to determine a
right-angled triangle; an indefinite number of triangles may be found
with the prescribed conditions, whose vertices will lie in the line which
is perpendicular to the base.
If a problem relate to the determination of a single point, and the
data be sufficient to determine the position of that point, the problem
is determinate: but if one or more of the conditions be omitted, the data
which remain may be sufficient for the determination of more than
one point, each of which satisfies the conditions of the problem; in that
case, the problem is indeterminate, and in general such points are found
ANCIENT GEOMETRICAL ANALYSIS.
291
to be situated in some line, and hence such line is called the locus of p. 51.
the point which satisfies the conditions of the problem.
If any two given points A and B (fig. Prop. 5, Book Iv.) be
joined by a straight line AB, and this line be bisected in D, then if
a perpendicular be drawn from the point of bisection, it is manifest
that a circle described with any point in the perpendicular as a cen-
tre, and a radius equal to its distance from one of the given points,
will pass through the other point, and the perpendicular will be the
locus of all the circles which can be described passing through the
two given points.
Again, if a third point C be taken, but not in the same straight line
with the other two, and this point be joined with the first point A;
then the perpendicular drawn from the bisection E of this line will be
the locus of the centres of all circles which pass through the first and
third points A and C. But the perpendicular at the bisection of the
first and second points A and B is the locus of the centres of circles
which pass through these two points. Hence the intersection F of
these two perpendiculars, will be the centre of a circle which passes
through the three points, and is called the intersection of the two loci.
Sometimes this method of solving geometrical problems may be pursued
with advantage, by constructing the locus of every two points sepa-
rately, which are given in the conditions of the problem. In the Geo-
metrical Exercises which follow, only those local problems are given
where the locus is either a straight line or a circle.
There is another class of Propositions called Porisms. The exact
meaning of the term as applied to this class of propositions, appears to
be as uncertain as the derivation of the word itself. In the original
Greek of Euclid's Elements, the corollaries to the propositions are
called porisms (Tоρioμатa); but this scarcely explains the nature of
porisms, as it is manifest that they are different from simple deductions
from the demonstrations of propositions. Some analogy, however, we
may suppose them to have to the porisms or corollaries in the Elements.
Pappus (Coll. Matth. Lib. VII. pref.) informs us that Euclid wrote three
books on Porisms, and gives an obscure account of them. He defines
"a porism to be something between a problem and a theorem, or that
in which something is proposed to be investigated." Dr. Simson, to
whom is due the merit of having restored the porisms of Euclid, gives
the following definition of that class of propositions: "Porisma est pro-
positio in qua proponitur demonstrare rem aliquam, vel plures datas
esse, cui, vel quibus, ut et cuilibet ex rebus innumeris, non quidem
datis, sed quæ ad ea quæ data sunt eandem habent relationem, con-
venire ostendendum est affectionem quandam communem in pro-
positione descriptam." Professor Dugald Stewart defines a porism to
be "A proposition affirming the possibility of finding one or more of
the conditions of an indeterminate theorem." Professor Playfair in a
paper (from which the following account is taken) on Porisms, printed
in the Transactions of the Royal Society of Edinburgh, for the year
1792, defines a porism to be "A proposition affirming the possibility of
finding such conditions as will render a certain problem indeterminate
or capable of innumerable solutions.”
It may without much difficulty be perceived that this definition
represents a porism as almost the same with an indeterminate problem.
There is a large class of indeterminate problems which are, in general, loci,

19-2
292
ANCIENT GEOMETRICAL ANALYSIS.
and satisfy certain defined conditions. Every indeterminate problem
containing a locus may be made to assume the form of a porism, but
not the converse. Porisms are of a more general nature than indeter-
minate problems which involve a locus.
It is highly probable that the ancient geometers arrived at all geo-
metrical truths in their attempts at the solution of problems. The first
mathematical enquiries must have occurred in the form of questions,
where something was given, and something required to be done; and
by reasonings necessary to answer these questions, or to discover the
relations between the things that were given, and those that were to be
found, many truths were suggested which came afterwards to be the
subjects of separate demonstration.
Such
The ancient geometers appear to have undertaken the solution of
problems with a scrupulous and minute attention, which would scarcely
allow any of the collateral truths to escape their observation. They never
considered a problem as solved till they had distinguished all its
varieties, and evolved separately every different case that could occur,
carefully distinguishing whatever change might arise in the construction
from any change that was supposed to take place among the magni-
tudes which were given. This cautious method of proceeding soon led
them to see that there were circumstances in which the solution of a
problem would cease to be possible; and this always happened when
one of the conditions of the data was inconsistent with the rest.
instances would occur in the simplest problems; but in the analysis
of more complex problems, they must have remarked that their con-
structions failed, for a reason directly contrary to that assigned. In-
stances would be found where the lines, which, by their intersection, were
to determine the thing sought, instead of intersecting one another, as
they did in general, or of not meeting at all, would coincide with one
another entirely, and consequently leave the question unresolved.
The confusion thus arising would soon be cleared up, by observing,
that a problem before determined by the intersection of two lines would
This was
now become capable of an indefinite number of solutions.
soon perceived to arise from one of the conditions of the problem in-
volving another, or from two parts of the data becoming one, so that
there was not left a sufficient number of independent conditions to
confine the problem to a single solution, or any determinate number of
solutions. It was not difficult afterwards to perceive, that these cases
of problems formed very curious propositions, of an indeterminate nature
between problems and theorems, and that they admitted of being enun-
ciated separately. It was to such propositions so enunciated that the
ancient geometers gave the name of Porisms.
Besides, it will be found, that some problems are possible within
certain limits, and that certain magnitudes increase while others de-
crease within those limits; and after having reached a certain value the
former begin to decrease, while the latter increase. This circumstance
gives rise to questions of maxima and minima, or the greatest and least
values which certain magnitudes may admit of in indeterminate pro-
blems.
In the following collection of problems and theorems, many will be
found to be of so simple a character, (being almost obvious deductions
from propositions in the Elements) as scarcely to admit of the principle
of the Geometrical Analysis being applied, in their solution.
50, 10.
GEOMETRICAL EXERCISES ON BOOK I.
PROBLEM I.
From two given points on the same side of a straight line given in Lese B.1,2.3.
position, draw two straight lines which shall meet in that line, and make
equal angles with it; also prove, that the sum of these two lines is less than B.,P.33.
the sum of any other two lines drawn to any other point in the line.
alon Senge.
ANALYSIS. Let A, B be the two given points, and CD the given line.
Suppose G the required point in the line, such that AG and BG being
joined, the angle AGF is equal to the angle BGD.
A
B

C-
D
F
Н
E
Draw AF perpendicular to CD and meeting BG produced in E.
Then, because the angle BGD is equal to AGF, (hyp.)
and also to the vertical angle FGE, (1. 15.)
therefore the angle AGF is equal to EGF;
also the right angle AFG is equal to the right angle EFG,
and the side FG is common to the two triangles AFG, EFG,
therefore AG is equal to EG, and AF to FE.
Hence the point E being known, the point G is determined by the
intersection of CD and BE.
Synthesis. From A draw AF perpendicular to CD, and produce it
to E, making FE equal to AF, and join BG cutting CD in G. Join
also AG.
Then AG and BG make equal angles with CD.
For since AF is equal to FE, and FG is common to the two triangles
AGF, CGF, and the included angles AFG, EFG are equal;
therefore the base AG is equal to the base EG, and the angle AGF
to the angle EGF,
but the angle EGF is equal to the vertical angle BGD,
therefore the angle AGF is equal to the angle BGD;
that is, the straight lines AG and BG make equal angles with the
straight line CD.
Also the sum of the lines AG, GB is a minimum.
For take any other point H in CD, and join EH, HB.
Then since any two sides of a triangle are greater than the third side,
therefore EH, HB are greater than EB in the triangle EHB.
But EG is equal to AG;
therefore EH, HB are greater than AG, GB.
That is, AG, GB are less than any other two lines which can be
drawn from A, B, to any other point H in the line CD.
294
GEOMETRICAL EXERCISES
PROBLEM II.
To trisect a given straight line.
Analysis. Let AB be the given straight line, and suppose it divided
into three equal parts in the points D, E.
C

F
A DE B
On DE describe an equilateral triangle DEF,
then DF is equal to AD, and FE to EB.
On AB describe an equilateral triangle ABC,
and join AF, FB.
Then because AD is equal to DF,
therefore the angle DAF is equal to the angle DFA,
and the two angles DAF, DFA are double of one of them DAF.
But the angle FDE is equal to the angles DAF, DFA,
and the angle FDE is equal to DAC, each being an angle of an
equilateral triangle;
therefore the angle DAC is double the angle DAF;
wherefore the angle DAC is bisected by AF.
Also because the angle FAC is equal to the angle FAD, and FAD
to DFA;
therefore the angle CAF is equal to the alternate angle AFD:
and consequently FD is parallel to AC.
Synthesis. Upon AB describe an equilateral triangle ABC,
bisect the angles at A and B by the straight lines AF, BF, meeting in F;
through F draw FD parallel to AC, and FE parallel to BČ.
Then AB is trisected in the points D, E.
For since AC is parallel to FD and FA meets them,
therefore the alternate angles FAC, AFD are equal;
but the angle FAD is equal to the angle FAC,
hence the angle DAF is equal to the angle AFD,
and therefore DA is equal to DF.
But the angle FDE is equal to the angle CAB, and FED to CBA;
(1. 29.)
and therefore the remaining angle DFE is equal to the remaining
angle ACB.
Hence the three sides of the triangle DFE are equal to one another,
and DF has been shewn to be equal to DA,
therefore AD, DE, EB are equal to one another.
Hence the following theorem.

If the angles at the base of an equilateral triangle be bisected by
two lines which meet at a point within the triangle; the two lines
drawn from this point parallel to the sides of the triangle, divide the
base into three equal parts.
ON BOOK I.
295
THEOREM I.
If two opposite sides of a parallelogram be bisected, and two lines
be drawn from the points of bisection to the opposite angles, these two
lines trisect the diagonal.
Let ABCD be a parallelogram of which the diagonal is AC.
Let AB be bisected in E and DC in F,
also let DE, FB be joined cutting the diagonal in G, H.
Then AC is trisected in the points G, H.

A
I
G
E
B

K
H
F
C
D
Through E draw EK parallel to AC and meeting FB in K.
Then because EB is the half of AB, and DF the half of DC;
therefore EB is equal to DF;
and these equal and parallel straight lines are joined towards the
same parts by DE and FB;
therefore DE and FB are equal and parallel. (1. 33.)
And because AEB meets the parallels EK, AC,
therefore the exterior angle BEK is equal to the interior angle EAG.
For a similar reason, the angle EBK is equal to the angle AEG.
Hence in the triangles AEG, EBK, there are the two angles GAE,
AEG in the one, equal to two angles KEB, EBK in the other, and one
side adjacent to the equal angles in each triangle, namely AE equal
to EB:
therefore AG is equal to EK, (1. 26.)
but EK is equal to GH, (1. 34.)
therefore AG is equal to GH.
By a similar process, it may be shewn that GH is equal to HC.
Hence AG, GH, HC are equal to one another,
and therefore AC is trisected in the points G, H.
PROBLEM III.
To bisect a triangle by a line drawn from a given point in one of 143.9M
the sides. (Euclidis de Divisionibus.)
Analysis. Let ABC be the given triangle, and D the given point
in the side AB.
D
A

B
E F
C
Suppose DF the line drawn from D which bisects the triangle;
therefore the triangle DBF is half of the triangle ABC.
i
296
GEOMETRICAL EXERCISES
Bisect BC in E and join AE, DE, AF,
then the triangle ABE is half of the triangle ABC:
hence the triangle ABE is equal to the triangle DBF;
take away from these equals the triangle DBE,
therefore the remainder ADE is equal to the remainder DEF.
But ADE, DEF are equal triangles upon the same base DE, and on
the same side of it,
they are therefore between the same parallels,
that is, AF is parallel to DE,
therefore the point F is determined.
Synthesis. Bisect the base BC in E, join DE,
from A, draw AF parallel to DE, and join DF.
Then because DE is parallel to AF,
therefore the triangle ADE is equal to the triangle DEF;
to each of these equals, add the triangle BDE,
therefore the whole triangle ABE is equal to the whole DBF,
but ABE is half of the whole triangle ABC;
therefore ABF is also half of the triangle ABC.
Siam'p. Alg,
ifp. Prob. 1.
PROBLEM IV.
Given one angle, a side opposite to it, and the sum of the other two
sides, construct the triangle.
Analysis. Suppose BAC the triangle required, having BC equal to
the given side, BAC equal to the given angle opposite to BC, also BD
equal to the sum of the other two sides.
B
A
Join DC.
D
C
Then since the two sides BA, AC are equal to BD,
by taking BA from these equals,
the remainder AC is equal to the remainder AD.
Hence the triangle ACD is isosceles, and therefore the angle ADC is
equal to the angle ACD.
But the exterior angle BAC of the triangle ADC is equal to the
two interior and opposite angles ACD and ADC:
Wherefore the angle BAC is double the angle BDC, and BDC is
the half of the angle BAC.
Hence the synthesis.
At the point D in BD, make the angle BDC equal to half the given
angle,
and from B the other extremity of BD, draw BC equal to the given
side, and meeting DC in C,
at C in CD make the angle DCA equal to the angle CDA, so that
CA may meet BD in the point A.
Then the triangle ABC shall have the required conditions.
ON BOOK I. PROBLEMS.
297
A
PROBLEMS.
5. Given the base and one of the sides of an isosceles triangle, to
describe the triangle.
6. Describe an isosceles triangle, each of the sides of which shall
be double of the base.
7. In a given straight line, find a point equally distant from two
given points; one in, and the other above or below, the given straight
line.
8. AB, AC are straight lines cutting one another in A, D is a
given point. Draw through D a straight line cutting off equal parts
from AB and AC.
9. Draw through a given point, between two straight lines not
parallel, a straight line, which shall be bisected in that point.
10. Divide a given right angle into three equal angles.
11. One of the acute angles of a right-angled triangle is three
times as great as the other; trisect the smaller of these.
12. From a given point in a given straight line, it is required to
erect a perpendicular by the help of straight lines only.
13. From a given point without a given straight line, to draw
a line making an angle with the given line equal to a given rectilineal
angle.
14. Through a given point draw a straight line which shall make
equal angles with two straight lines given in position.
15. To determine that point in a straight line from which the
straight lines drawn to two other given points shall be equal, provided
the line joining the two given points is not perpendicular to the given
line.
16. To place a straight line in a triangle (terminated by the two
sides) which shall be equal to one straight line and parallel to an-
other.
17. Determine the shortest path from one given point to another,
subject to the condition, that it shall meet two given lines.
18. In the base of a triangle, find the point from which lines, drawn
parallel to the sides of the triangle and limited by them, are equal.
19. From a given point in either of the equal sides of an isosceles
triangle, to draw a straight line to the other side produced, which shall
make with these sides a triangle equal to the given triangle. Prove
that the line thus drawn will be greater than the base of the isosceles
triangle.
20. From one of the obtuse angles of a rhomboid draw a straight
line to the opposite side, which shall be bisected by the diagonal drawn
through its acute angles.
21. Upon a straight line as a diagonal describe a parallelogram
having an angle equal to a given angle.
22. On a given line to describe a parallelogram, having two of its
opposite angles double of the other two, and all its sides equal. By
means of this problem, trisect a right angle.
23. Describe a parallelogram equal to a given square, and having
an angle equal to half a right angle.
24. To describe a rhombus which shall be equal to any given
quadrilateral figure.

298
GEOMETRICAL EXERCISES
H
pe
7.
25. Describe a parallelogram equal in area and perimeter to a given
triangle.
26. Describe on a given straight line a triangle which shall be equal
to a given rectilineal figure, and have its vertical angle equal to a given
angle.
27. Transform a given rectilinear figure into a triangle whose
vertex shall be in a given angle of the figure and whose base be in one
of the sides.
28. Straight lines are drawn from a fixed point to the several points
of a straight line given in position, and on each base is described an
equilateral triangle. Determine the locus of the vertices.
29. Upon a given base to describe an isosceles triangle, which
shall be equal to a given triangle.
30. Given the base and one side of a triangle, to find the third
side, so that the area may be the greatest possible.
31. Describe a right-angled triangle upon a given hypothenuse,
so that the hypothenuse and one side shall be together double of the
third side.
32. Having given two lines, which are not parallel, and a point
between them; describe a triangle having two of its angles in the
respective lines, and the third at the given point; and such that the
sides shall be equally inclined to the lines which they meet.
33. Bisect a triangle by a straight line drawn parallel to one of
its sides.
34. Bisect a triangle by a straight line drawn through a point
within or without the triangle.
35. It is required to bisect any triangle by a line perpendicular to
the base.
36. It is required to determine a point within a given triangle,
from which lines drawn to the several angles, will divide the triangle
into three equal parts.
37. Bisect a parallelogram, (1) by a line drawn from a point in
one of its sides: (2) by a line drawn from a given point within or with-
out it: (3) by a line perpendicular to one of the sides.
38. To bisect a trapezium (1) by a line drawn from one of its
angular points:.(2) by a line drawn from a given point in one side.
39. Divide a triangle into three equal parts, (1) by lines drawn
from a point in one of the sides:.(2) by lines drawn from the angles to
a point within the triangle: V(3) by lines drawn from a given point
within the triangle. In how many ways can the third case be done?
To trisect a parallelogram by lines drawn from a given point
in one of its sides.

~ 40.
41. To divide a parallelogram into four equal portions by straight
lines drawn from a given point in one of its sides.
42. To divide a square into four equal portions by three straight
lines drawn from any point in one of its sides.
43. From a given isosceles triangle, cut off a trapezium which shall
have the same base as the triangle, and shall have its remaining three
sides equal to each other.
44. Divide an equilateral triangle into nine equal parts.
45. To find a point in the side or side produced of any paral-
lelogram, such that the angle it makes with the line joining the point
and one extremity of the opposite side, may be bisected by the line
joining it with the other extremity.

ON BOOK I. PROBLEMS.
299
46. Find a point in the diagonal of a square produced, from which
if a straight line be drawn parallel to any side of the square, and
meeting another side produced, it will form together with the produced
diagonal and produced side, a triangle equal to the square.
47. A trapezium is such, that the perpendiculars let fall on a
diagonal from the opposite angles are equal. Divide the trapezium
into four equal triangles, by straight lines drawn to the angles from a
point within it.
48. Given one side of a right-angled triangle, and the difference
between the hypothenuse and the sum of the other two sides, to con-
struct the triangle.
49. In a right-angled triangle, given the sums of the base and the
hypothenuse, and of the base and the perpendicular; to determine the
triangle.
50. Given half the perimeter and the vertical angle of an isosceles
triangle, it is required to find the sides.
51. Given the perimeter and the angles of a triangle, to con-
struct it.
52. Given one of the angles at the base of a triangle, the base
itself, and the sum of the two remaining sides, to construct the triangle.
53. Given the base, an angle adjacent to the base, and the differ-
ence of the sides of a triangle, to construct it.
54. Given one angle, a side opposite to it, and the difference of
the other two sides; to construct the triangle.
55. Given the base, the perpendicular and the sum of the sides;
to construct the triangle.
56. Given the base, the altitude, and the difference of the two
remaining sides; construct the triangle.
57. To find a point in the base of a triangle, such that if perpen-
diculars be drawn from it upon the sides, their sum shall be equal to a
given line.
58. Determine the locus of the vertices of all the equal triangles,
which can be described on the same base, and upon the same side of it.
59. To describe a square upon a given straight line as a diameter.
60. Shew how the squares upon the sides of a right-angled tri-
angle may be dissected, so as exactly to cover the square of the
hypothenuse.
61. Find a square equal to 3, 5, or any number of squares.
62. Construct a square whose area shall be 8 or n times that of a
given square.
63. Construct all right-angled triangles whose sides shall be rational,
upon a given line as their base.
64. Describe a square which shall be equal to the difference between
two given squares.
65. In a right-angled triangle, it is required to find analytically the
base and perpendicular, their difference being 1, and the hypothenuse
equal to 5.
66. Any two parallelograms having been described upon two sides
of a given triangle, apply to the third side a parallelogram equal to
their sum.
67. Given a square of one inch, shew how a rhombus may be con-
structed, whose area shall be equal to it, and each of its sides a mile
long.
300
GEOMETRICAL EXERCISES
Not futis
See Mapella
THEOREMS.
* 2. In the fig. 1. 5. If FC and BG meet in H, then prove that
AH bisects the angle BAC.
3. In the fig. 1. 5. If the angle FBG be equal to the angle
ABC, and BG, CF intersect in O; the angle BOF is equal to twice
the angle BAC.
4. If a straight line drawn bisecting the vertical angle of a triangle,
also bisects the base, the triangle is isosceles.
5. In the base BC of an isosceles triangle ABC take a point D,
and in CA take CE equal to CD, let ED produced meet AB produced
in F; then 3. AEF=2 right angles + AFE.
6. The difference between any two sides of a triangle is less than
the third side.
7. ABC is a triangle right-angled at B, and having the angle A
double the angle C; shew that the side C is less than double the
side AB.
B
8. The difference of the angles at the base of any triangle, is
double the angle contained by a line drawn from the vertex perpen-
dicular to the base, and another bisecting the angle at the vertex.
9. If from the right angle of a right-angled triangle two straight
lines be drawn, one perpendicular to the base, and the other bisecting
it, they will contain an angle equal to the difference of the two acute
angles of the triangle.
10. If one angle at the base of a triangle be double of the other,
the less side is equal to the sum or difference of the segments of the
base made by the perpendicular from the vertex, according as the angle
is greater or less than a right angle.
11. If one angle of a triangle be equal to the sum of the other
two, the greatest side is double of the distance of its middle point from
the opposite angle.
12. If two exterior angles of a triangle be bisected, and from the
point of intersection of the bisecting lines a line be drawn to the oppo-
site angle of the triangle, it will bisect that angle.
13. In an obtuse-angled triangle if perpendiculars be drawn from
the points bisecting the sides, prove that they all will pass through the
same point.
14. In an obtuse-angled triangle, if perpendiculars be drawn from
the angles to the opposite sides, produced if necessary, they will pass
through the same point: required a proof.
v15. Let ACB be a triangle; and let AD, CG, BE respectively
bisect the exterior angles HAE, ECB, CBG, and meet BC, AB, AC
produced in the points D, E, G. It is required to demonstrate that
these three points are in the same straight line.
16. If two sides of a triangle be produced, the three straight lines
which bisect the two exterior angles and the third interior angle shall
all meet in the same point.
17. ABC is a triangle in which the angle ABC is twice the
angle ACB; shew that if the point D in BC which divides it into
segments, whose difference is equal to the side opposite to the angle
ACD, be joined with the point A, AD is perpendicular to BC.
18. In Prop. 35, Book 1, shew that the two parallelograms can be
ON BOOK I. THEOREMS.
301
divided into the same number of triangles which are actually equal, each
to each, so that the divided figures may be superposed.
19. If from the vertex of a triangle, two straight lines be drawn
to the base, one bisecting the vertical angle, and the other bisecting the
base, prove that the latter is the greater of the two straight lines.
20. If from the vertical angle of a triangle three straight lines be
drawn, one bisecting the angle, another bisecting the base, and the
third perpendicular to the base, the first is always intermediate in mag-
nitude and position to the other two.
21. In a right-angled isosceles triangle, the lines drawn from any
of the angles to the opposite angle of the square described
opposite side are all equal.
upon the
22. Shew that the perimeter of the triangle, formed by joining the
feet of the perpendiculars dropped from the angles upon the opposite
sides of a triangle, is less than the perimeter of any other triangle,
whose angular points are on the sides of the first.
23. From a given point there can be drawn only two equal straight
lines to a given line, one on each side of the shortest line; and the
shortest line is the perpendicular.
24. A, B are two fixed points: if two straight lines AC, BC be
drawn making a given angle C, prove that the straight line bisecting C
passes through a fixed point, and determine the point geometrically.
25. From every point of a given straight line, the straight lines
drawn to each of two given points on opposite sides of the line are
equal: prove that the line joining the given points will cut the given
line at right angles.
26. From a given point without the angle contained by two straight
lines given in position, draw a straight line in such a direction that the
part of it intercepted between the given point and the nearest straight
line, shall be equal to the part intercepted between the two straight lines.
27. If a straight line be drawn from a given point, and making
a given angle with a given straight line, its length and the points of its
intersection with the given line are given. egal,
28. Shew that if there be two rectilinear figures on the same base,
one of which wholly envelopes the other, the perimeter of the envelop-
ing figure is greater than the perimeter of the other.
29. Shew that we may draw to a point within a triangle_two
straight lines which shall be greater than the sides of the triangle, if
one of these two straight lines be not terminated in the extremity of
the base.
• 30. If parallel lines be defined to be "lines in the same plane
which make the same angle with any straight line which meets them,"
prove the following propositions respecting them.
(a) The alternate angles are equal.
(b) The two interior angles on the same side of the cutting line are
equal to two right angles.
(c) Parallel lines never meet however far they are produced.
31. Can it be properly predicated of any two straight lines that
they never meet if indefinitely produced either way, antecedently to our
knowledge of some other property of such lines which makes the por-
perty first predicated of them à necessary conclusion from it?
•
32. Lines which are perpendicular to parallel lines are also parallel.
. 33. If the line joining two parallel lines be bisected, all the lines
302
GEOMETRICAL EXERCISES
drawn through the point of bisection and terminated by the parallel
lines are also bisected in that point.
34. A quadrilateral figure whose sides are equal will be a paral-
lelogram.
35. If the opposite angles of a quadrilateral figure be equal the
opposite sides will be equal and parallel.
36. The diagonals of a parallelogram bisect each other.
37. The perimeter of a square is less than that of any other paral-
lelogram of equal area.
38. Any straight line which bisects the diagonal of a parallelogram
will also bisect the parallelogram; and no straight line can bisect a
parallelogram unless it cut or meet the opposite sides.
39. The diagonals of a square and of a rhombus bisect each other
at right angles.
40. If from any point in the diagonal of a parallelogram straight
lines be drawn to the angles, the parallelogram will be divided into
two pairs of equal triangles.
41. ABCD is a parallelogram of which the angle C is opposite
to the angle A. If through A any straight line be drawn, then the
distance of C is equal to the sum or difference of the distances of B
and of D from that straight line according as it lies without or within
the parallelogram.
42. If in a parallelogram two lines be drawn parallel to adjacent
sides, and meeting the other sides of the figure, the lines joining their
extremities, if produced, will meet the diameter in the same point.
43. ABCD is a parallelogram; draw the diagonal BC, and from D
draw DE at right angles to BC, then if perpendiculars be drawn from
B and C, they shall intersect in the line DE, produced if necessary.
44. If ABCD be a parallelogram, and E any point in the diagonal
AC, or AC produced; shew that the triangles EBC, EDC are equal.
45. If from a point without a parallelogram, lines be drawn to the
extremities of two adjacent sides, and of the diagonal which they in-
clude. Of the triangles thus formed, that, whose base is the diagonal,
is equal to the sum of the other two.
46. It is impossible to divide a quadrilateral figure (except it be a
parallelogram) into equal triangles by lines drawn from a point within
it to its four corners.
47. If of the four triangles into which the diagonals divide a
trapezium, any two opposite ones are equal, the trapezium has two of
its opposite sides parallel.
48. If two sides of a trapezium be parallel, the triangle contained
by either of the other sides and the two straight lines drawn from its
extremities to the bisection of the opposite sides is equal to half the
trapezium.
49. The sum of the diagonals of a trapezium is less than the sum
of any four lines which can be drawn to the four angles, from any point
within the figure, except their intersection.
50. When the corner of a leaf of a book is turned down a second
time, so that the lines of folding are parallel and equidistant, the space
in the second fold is equal to three times that in the first.
51. If the sides of a quadrilateral figure be bisected and the points
of bisection joined, the included figure is a parallelogram, and equal in
area to half the original figure.



ON BOOK I. THEOREMS.
303
52. Along the sides of a parallelogram ABCD taken in order,
measure AA' - BB' = CC' = DD'; the figure A'B'C'D' will be a paral-
lelogram.
=
53. If the points of bisection of the sides of a triangle be joined,
the triangle so formed shall be one-fourth of the given triangle.
54. Prove that two lines drawn to bisect the opposite sides of a
trapezium will also bisect each other.
55. Upon stretching two chains AC, BD, across a field ABCD, I
find that BD and AC make equal angles with DC, and that AC makes
the same angle with AD, that BD does with BC; hence prove that AB
is parallel to CD.
56. If a line intercepted between the extremity of the base of an
isosceles triangle, and the opposite side (produced if necessary) be equal
to a side of the triangle, the angle formed by this line and the base
produced is equal to three times either of the equal angles of the
triangle.
57. AD, BC are two parallel straight lines, cut obliquely by AB
and perpendicularly by AC; BED is drawn cutting AC in E so that
ED is equal to twice BA; prove that the angle DBC is equal to one-
third of the angle ABC.
58. AB, BC, DE, EF are rods joined at B, F, E, and D, capable
of angular motion in the same plane, and so placed that FBDE is a
parallelogram. If, when the rods are in any given position, points A,
G and C be taken in the same line, shew that these points will always be
in the same line, whatever be the angle the rods make with each other.
59. If upon the sides of a triangle as diagonals, parallelograms be
described, having their sides parallel to two given lines, the other diago-
nals of the parallelograms will intersect in a point.
60. Prove that the perimeter of an isosceles triangle is greater than
that of an equal rectangle of the same altitude.
61. If the areas of any triangle and of a square be equal, the
perimeter of the triangle will be the greater.
62. If from the extremities A and B of the base of any triangle
ABC, and on the same side of it, two straight lines AD, BE be drawn
perpendicular to the base, each being double the altitude of the triangle,
and straight lines DF, EG be drawn from D and E to the middle points
of AC, BC; the sum or difference of the triangles ADF, BEG will be
equal to the triangle ABC according as the angles, at the base of the
latter, are acute or one of them is obtuse.
63. The perimeter of an isosceles triangle is less than that of any
other equal triangle upon the same base.
64. Of all triangles having the same base and the same perimeter,
that is the greatest which has the two undetermined sides equal.
65. Of all triangles having the same vertical angle, and whose
bases pass through a given point, the least is that whose base is bisected
in the given point.
66. If from the base to the opposite sides of an isosceles triangle
three straight lines be drawn, making equal angles with the base, viz.
one from its extremity, the other two from any other point in it, these
two shall be together equal to the first.
67. From the extremities of the base of an isosceles triangle
straight lines are drawn perpendicular to the sides, the angles made
by them with the base are each equal to half the vertical angle.

304
GEOMETRICAL EXERCISES
68. If each of the equal angles of an isosceles triangle be one-
fourth of the third angle, and from one of them a perpendicular be.
drawn to the base meeting the opposite side produced; then will the
part produced, the perpendicular, and the remaining side, form an
equilateral triangle.
69. If two sides of a triangle be given, the triangle will be
greatest when they contain a right angle.
70. The area of any two parallelograms described on the two
sides of a triangle is equal to that of a parallelogram on the base, whose
side is equal and parallel to the line drawn from the vertex of the
triangle to the intersection of the two sides of the former parallelograms
produced to meet.
AE.
71. In the figure to Prop. 47, Book 1,
(a) If BG and CH be joined, those lines will be parallel.
(b) If FC and BK be joined, they will cut off equal portions AD,
(c) If perpendiculars be let fall from F and K on BC produced,
the parts produced will be equal; and the perpendiculars together will
be equal to BC.
(d) Shew that AL, BK, CF, intersect each other in the same point.
(e) Join GH, KE, FD, and prove that each of the triangles so
formed equals the given triangle ABC.
(f) The sum of the squares of GH, KE, and FD will be equal
to eight times the square of the hypothenuse.
(g) If the exterior angular points of the squares be joined, an
irregular hexagon will be formed, whose area is equal to the area of
the square described upon the hypothenuse of a right-angled triangle,
one of whose sides is equal to the hypothenuse of the original triangle,
and the other is equal to the sum of its sides.
·
72. A point is taken within a square, and straight lines drawn
from it to the angular points of the square, and perpendicular to the
sides; the squares on the first are double the sum of the squares on the
last. Shew that these sums are least when the point is in the centre of
the square.
73. If from the vertex of a plane triangle, a perpendicular fall upon
the base or the base produced, the difference of the squares of the sides
is equal to the difference of the squares of the segments of the base.
74. If from the middle point of one of the sides of a right-angled
triangle a perpendicular be drawn to the hypothenuse, the difference of
the squares of the segments into which it is divided, is equal to the
square of the other side.
75. If a straight line be drawn from one of the acute angles of a
right-angled triangle, bisecting the opposite side, the square upon that
line is less than the square upon the hypothenuse by three times the
square upon half the side bisected.
76. If one angle of a triangle be equal to a right angle, and
another equal to two-thirds of a right angle, prove from Euclid, Book 1,
that the equilateral triangle described on the hypothenuse, is equal
to the sum of the equilateral triangles described upon the sides which
contain the right angle.
GEOMETRICAL EXERCISES ON BOOK II.
THEOREM I.
The square of the excess of one straight line above another is less than
the squares of the two straight lines by twice their rectangle.
Let AB, BC be the two straight lines, whose difference is AC.
Then the square of AC is less than the squares of AB and BC by
twice the rectangle contained by AB and BC.
:
A
C B
H
K
D
FE
Constructing as in Prop. 4. Book 11,
Because the complement AG is equal to GE,
add to each CK,
therefore the whole AK is equal to the whole CE;
and AK, CE together are double of AK;
but AK, CE are the gnomon AKF and CK,
and AK is the rectangle contained by AB, BC;
therefore the gnomon AKF and CK are equal to twice the rectangle
AB, BC.
But AE, CK are equal to the squares of AB, BC;
hence taking the former equals from these equals,
therefore the difference of AE, and the gnomon AKF is equal to
the difference between the squares AB, BC, and twice the rectangle
AB, BC;
but the difference AE and the gnomon AKF is the figure HF which
is equal to the square of AC.
Wherefore the square of AC is equal to the difference between the
squares AB, BC, and twice the rectangle AB, BC.
THEOREM II.
If straight lines be drawn from each angle of a triangle bisecting the
opposite side, four times the sum of the squares of these lines is equal to
three times the sum of the squares of the sides of the triangle.
Let ABC be any triangle, and let AD, BE, CF be drawn from
A, B, C, to D, E, F, the bisections of the opposite sides of the triangle:
draw AE perpendicular to BC.
A

E
F
B
DE C
20
306
GEOMETRICAL EXERCISES
Then the square of AB is equal to the squares of BD, DA together
with twice the rectangle BD, DE, (11. 12.)
and the square of AC is equal to the squares of CD, DA diminished
by twice the rectangle CD, DE; (II. 13.)
therefore the squares of AB, AC are equal to twice the square of
BD, and twice the square of AD; for DC is equal to BD:
and twice the squares of AB, AC are equal to the square of BC, and
four times the square of AD: for BC is twice BD.
Similarly, twice the squares of AB, BC are equal to the square of
AC, and four times the square of BE:
and twice the squares of BC, CA are equal to the square of AB,
and four times the square of FC:
hence, by adding these equals,
four times the squares of AB, AC, BC are equal to four times the
squares of AD, BE, CF together with the squares of AB, AC, BC:
and taking the squares of ÄB, AC, BC from these equals,
therefore three times the squares of AB, AC, BC are equal to four
times the squares of AD, BE, CF.
PROBLEM I.
Divide a given straight line into two such parts, that the rectangle
contained by them may be three-fourths of the greatest which the case will
admit.
Analysis. Let AB be the given line, and let AB be bisected in
D: then the rectangle AD, DB, or the square of DB is the greatest
possible rectangle.
E
G

A
C D F
B
Let C' be the point required, such that the rectangle AC, CB is equal
to three-fourths of the square of DB.
On AB describe a semicircle, and draw CE perpendicular to AB,
then the square of CE is equal to the rectangle AC, CB.
Again, bisect DB in F; on FB describe the right-angled triangle
FBG, having the hypothenuse FG equal to DB.
Then BG is the line, the square of which is three-fourths of the
square of DB.
Hence GB is equal to EC: join GE.
Therefore the point E, and also the point C is found.
Synthesis. Bisect AB in D and DB in F: on FB make the right-
angled triangle FBG having the hypothenuse FG equal to DB.
On AB describe a semicircle, and through G draw GE parallel to
AB, and draw EC parallel to GB.
Then C is the point required, such that the rectangle contained by
AC, CB is equal to three-fourths of the square of half the line AB.
ON BOOK II.
307
PROBLEMS.
PROBLEMS.
2. Divide a straight line into two parts, such, that their rectangle
may be equal to a given square; and determine the greatest square
that the rectangle can equal.
3. Divide a straight line into two such parts that the difference
of the squares of the two parts shall be equal to twice the rectangle
contained by them.
4. Divide a straight line into two parts such, that the rectangle
contained by them may be equal to the square of their difference.
5. To divide a straight line so that the rectangle under its seg-
ments may equal a given rectangle.
6. Divide a given straight line so that the rectangle under the
parts may be equal to a given square, and point out the limit which the
side of the given square must not exceed so that the problem may be
possible.
7. Divide a given straight line into two parts, such that the
squares of the whole line and one of the parts shall be equal to twice
the square of the other part.
8. Divide a given line, so that the square of the greater part
may equal twice the rectangle of the whole and the less part.
9. Divide algebraically a given line (a) into two parts, such that
the rectangle contained by the whole and one part may be equal to the
square of the other. Deduce Euclid's construction from one solution,
and explain the other.
10. Divide a given straight line into three parts, such that the
square of the whole line may be equal to the squares of the extreme
parts together with twice the rectangle contained by the whole and the
middle part.
11. To produce a straight line AB to C, so that the rectangle con-
tained by the sum and difference of AB and AC may be equal to a given
square.
12. If a straight line be divided into any two parts, produce it so
that the rectangle contained by the whole line produced and the part
produced, may be equal to the rectangle contained by the given line
and one segment.
13. Construct a rectangle that shall be equal to a given square
and the difference of whose adjacent sides shall be equal to a given line.
14. Construct a rectangle equal to a given square, and having the
sum of its sides equal to a given line.
15. Find a square which shall be equal to the sum of two given
rectilineal figures.
16. Shew how to divide a given rectangle into parts which
together will form a rectangle of any proposed length.
17. A given line AB is divided into two parts in the point C.
Find the position of any point D above the line, so that the sum of
the squares on the lines DA, DC, DB may be equal to the square on
AB, diminished by the rectangle AC, AB.
18. Construct a triangle with three sides a, b and c, such that
ab=c², and a + b = 4c.
20-2
308
GEOMETRICAL EXERCISES
3.
THEOREMS.
The area of a rhombus is equal to half the rectangle con-
tained by the diagonals.
4. Prove that the area of a trapezium whose bases are parallel is
half of the rectangles contained by each of the bases, and the perpen-
dicular distance.
5. The area of any right-angled triangle is equal to the rectangle
of the semiperimeter and excess of the semiperimeter above the hypo-
thenuse. Required proof.
6. Any rectangle is the half of the rectangle contained by the
diameters of the squares on its two sides.
7. The sum of the squares of two lines is never less than twice
their rectangle.
8. If a straight line be divided into two equal and into two
unequal parts, the squares of the two unequal parts are equal to twice
the rectangle contained by the two unequal parts, together with four
times the square of the line between the points of section.
9. If the points C, D be equidistant from the extremities of the
straight line AB, shew that the squares constructed on AD and AC
exceed twice the rectangle AC, AD by the square constructed on CD.
10. In a right-angled triangle, the square on that side which is
the greater of the two containing the right angle is equal to the rect-
angle by the sum and difference of the other sides.
11. Shew that the first of the algebraical propositions,
(u + x) (α − x) + x² = a²,
(a + x)² + (a− x)² = 2a² + 2x²,
is equivalent to the two Propositions v and v1, and the second of them
to the two Propositions Ix and x of the second book of Euclid.
12. Shew how in all the possible cases, a straight line may be
geometrically divided into two such parts, that the sum of their squares
shall be equal to a given square.
13. ABCD is a rectangular parallelogram, of which A, C are
opposite angles, E any point in BC, F any point in CD. Prove that
the area of the triangle AEF together with the rectangle BE, DF is
equal to the parallelogram AC.
14. A, B, C, D are four points in the same straight line, E a
point in that line equally distant from the middle of the segments AB,
CD; F is any other whatever in AD; then AF² + BF² + CF² + DF²
= AE² + BE² + CE² + DE² + 4 EF².
15. The sum of the perpendiculars let fall from any points within
an equilateral triangle, will be equal to the perpendicular let fall from
one of its angles upon the opposite side.
Is this proposition true when the point is in one of the sides of the
triangle?
In what manner must the proposition be enunciated when the point
is without the triangle?
16. If a line AB be divided into two parts AC and CB in the
point C (Prop. 11, Book 11.), so that the rectangle AB x BC= AC²;
and if AC be divided in D, so that CD=BC, prove that AC× AD=BC².
ON BOOK II. THEOREMS.
309
-
17. All plane rectilineal figures admit of quadrature. Point `out
the succession of steps by which Euclid establishes the truth of this
proposition.
18. The hypothenuse (c) of a right-angled triangle ABC is tri-
sected in the points D, E; prove that if CD, CE be joined, the sum
of the squares of the sides of the triangle CDE = .c².
19. If from the right angle C of a right-angled triangle ABC,
straight lines be drawn to the opposite angles of the square described
on the hypothenuse AB; shew that the difference of the squares
described on these lines is equal to the difference of the squares
described on the two sides AC, BC.
20. If the sides of the triangle be as the numbers 2, 4, 5, shew
whether it will be acute or obtuse-angled.
21. If an angle of a triangle be two-thirds of two right angles,
shew that the square of the side subtending that angle is equal to the
squares of the sides containing it, together with the rectangle contained
by those sides.
22. In any triangle the squares of the two sides are together
double of the two squares of half the base and of the straight line
joining its bisection with the opposite angle.
23. The square described on a straight line drawn from one of
the angles at the base of a triangle to the middle point of the opposite
side, is equal to the sum or difference of the square of half the side
bisected and the rectangle contained between the base and that part of
it, or of it produced, which is intercepted between the same angle and a
perpendicular drawn from the vertex.
24. If perpendiculars be drawn from the extremities of the base.
of a triangle on a straight line which bisects the angle opposite to the
base, the area of the triangle is equal to the rectangle contained by either
of the perpendiculars and the segment of the bisecting line between the
angle and the other perpendicular.
25. Upon the sides AB, BC, CA of the triangle ABC, or upon
these produced, let fall the perpendiculars DE, DF, DG, from the
point D within or without the triangle. Then AE2+ BF²+ GC2
BE² + CF² + AG. Required a demonstration.
26. If from the three angles of a triangle, lines be drawn to the
points of bisection of the opposite sides, the squares of the distances
between the angles and the common intersection are together one-third
of the squares of the sides of the triangle.
of a
27. Prove that the square of any straight line drawn from the
vertex of an isosceles triangle to the base, is less than the square
side of the triangle by the rectangle contained by the segments of
the base.
28. If from one of the equal angles of an isosceles triangle a per-
pendicular be drawn to the opposite side, the rectangle contained by
that side and the segment of it intercepted between the perpendicular
and base is equal to half the square described upon the base.
29. If in an isosceles triangle a perpendicular be let fall from one
of the equal angles to the opposite side, the square of the perpendicular
is equal to the square of the line intercepted between the other equal
angle and the perpendicular, together with twice the rectangle con-
tained by the segments of that side.
30. The square on the base of an isosceles triangle whose ver-
310
GEOMETRICAL EXERCISES
AT
tical angle is a right angle, is equal to four times the area of the
triangle.
31. If ABC be an isosceles triangle, and CD be drawn perpendicular
to AB; the sum of the squares of the three sides = AD² + 2.BĎ² + 3.CD³.
32. If ABC be an isosceles triangle, and DE be drawn parallel to
the base BC, and EB joined; prove that BE² = BC × DE+ ĈE².
33. If ABC be an isosceles triangle of which the angles at B and
C are each double of A; then the square of AC is equal to the square
of BC together with the rectangle contained by AC and BC.
34. Shew that in a parallelogram the squares of the diagonals are
equal to the sum of the squares of all the sides.
35. If ABCD be any rectangle, A and C being opposite angles,
and O any point either within or without the rectangle: OA² + OC²
= OB² + OD².
36. In any quadrilateral figure, the sum of the squares of the
diagonals together with four times the square of the line joining their
middle points is equal to the sum of the squares of all the sides.
37. In any trapezium, if the opposite sides be bisected, the sum of
the squares of the two other sides, together with the squares of the
diagonals, is equal to the sum of the squares of the bisected sides
together with four times the square of the line, joining the points of
bisection.
38. The squares of the diagonals of a trapezium are together
double the squares of the two lines joining the bisections of the oppo-
site sides.
39. In any trapezium two of whose sides are parallel, the squares
of the diagonals are together equal to the squares of its two sides which
are not parallel, and twice the rectangle contained by the sides which
are parallel.
40. If squares be described on the sides of any triangle and the
angular points of the squares be joined; the sum of the squares of the
sides of the hexagonal figure thus formed is equal to four times the sum
of the squares of the sides of the triangle.
ON BOOK III.
311
GEOMETRICAL EXERCISES ON BOOK III.
THEOREM I.
If AB, CD be chords of a circle at right angles to each other, prove
that the sum of the arcs AC, BD is equal to the sum of the arcs AD, BC.
(Archimedis, Lemm. Prop. 9.)
Draw the diameter FHG parallel to AB, and cutting CD in H.
D

B
E
F
G
H
C
Then the arcs FDG and FCG are each half the circumference.
Also since CD is bisected in the point H,
the arc FD is equal to the arc FC,
and the arc FD is equal to the arcs FA, AD, of which, AF is equal to BG,
therefore the arcs AD, BG are equal to the arc FC;
add to each CG,
therefore the arcs AD, BC are equal to the arcs FC, CG which make
up the half circumference.
Hence also the arcs AC, DB are equal to half the circumference.
Wherefore the arcs AD, BC are equal to the arcs AC, DB.
PROBLEM I.
The diameter of a circle having been produced to a given point, it
is required to find in the part produced a point, from which if a tan-
gent be drawn to the circle, it shall be equal to the segment of the part
produced, that is, between the given point and the point found.
Analysis. Let AEB be a circle whose centre is C and whose
diameter AB is produced to the given point D.
F

E
D
B G
Suppose that G is the point required, such that the segment GD is
equal to the tangent GE drawn from G to touch the circle in E.
Join DE and produce it to meet the circumference again in F
join also CE and CF.
Then in the triangle GDF, because GD is equal to GE,
therefore the angle GED is equal to the angle GDE;
and because CE is equal to CF,
the angle CEF is equal to the angle CFE;
312
GEOMETRICAL EXERCISES
therefore the angles CEF, GED are equal to the angles CFE, GDE:
but since GE is a tangent at E,
therefore the angle CEG is a right angle, (III. 18.)
hence the angles CEF, GEF are equal to a right angle,
and consequently, the angles CFE, EDG are also equal to a right angle,
wherefore the remaining angle FCD of the triangle CFD is a
right angle.
and therefore CF is perpendicular to AD.
Synthesis. From the centre C, draw CF perpendicular to AD
meeting the circumference of the circle in F:
join DF cutting the circumference in E,
join also CE, and at E draw EG perpendicular to CE and
intersecting BD in G.
Then G will be the point required.
For in the triangle CFD, since FCD is a right angle, the angles
CFD, CDF are together equal to a right angle;
also since CEG is a right angle,
therefore the angles CEF, GED are together equal to a right angle ;
therefore the angles CEF, GED are equal to the angles CFD, CDF ;
but because CE is equal to CF,
the angle CEF is equal to the angle CFD,
wherefore the remaining angle GED is equal to the remaining
angle CDF,
and the side GD is equal to the side GE of the triangle EGD,
therefore the point G is determined according to the required
conditions.
PROBLEM II.
Given the base, the vertical angle, and the perpendicular in a plane
triangle, to construct it.
Upon the given base AB describe a segment of a circle containing
an angle equal to the given angle. (111. 33.)

D
E
Á
B
At the point B draw BC perpendicular to AB, and equal to the
altitude of the triangle. (1. 11, 3.)
Through C draw CDE parallel to AB, and meeting the circum-
ference in D and E. (1. 31.)
Join DA, DB; also EA, EB.
Then EAB or DAB is the triangle required.
It is also manifest, that if CDE touch the circle, there will be only
one triangle which can be constructed on the base AB with the given
altitude.
ON BOOK III.
313
THEOREM II.
If a chord of a circle be produced till the part produced be equal
to the radius, and if from its extremity a line be drawn through the
centre and meeting the convex and concave circumferences, the convex is
one third of the concave circumference. (Archimedis, Lemm. Prop. 8.)
Let AB any chord be produced to C, so that BC is equal to the
radius of the circle:
and let CE be drawn from C through the centre D, and meeting
the convex circumference in F, and the concave in E.
Then the arc BF is one third of the arc AE,

B
A
C
D
F
E
G
Draw EG parallel to AB, and join DB, DG.
Since the angle DEG is equal to the angle DGE; (1.5.)
and the angle GDF is equal to the angles DEG, DGE, (1. 32.)
therefore the angle GDC is double of the angle DEG.
But the angle BDC is equal to the angle BCD, (1. 5.)
and the angle CEG is equal to the alternate angle ACE; (1. 29.)
therefore the angle GDC is double of the angle CDB,
add to these equals the angle CDB,
therefore the whole angle GDB is treble of the angle CDB,
but the angles GDB, CDB at the centre D, are subtended by the
arcs BF, BG, of which BG is equal to AE.
Wherefore the circumference AE is treble of the circumference BF,
and BF is one third of AE.
Hence may be solved the following problem:
AE, BF are two arcs of a circle intercepted between a chord and
a given diameter. Determine the position of the chord, so that one arc
shall be triple of the other.
THEOREM III.
AB, AC and ED are tangents to the circle CFB; at whatever point
between C and B the tangent EFD is drawn, the three sides of the tri-
angle AED are equal to twice AB or twice AC: also the angle subtended
by the tangent EFD at the centre of the circle, is a constant quantity.

B
E
D
C
314
GEOMETRICAL EXERCISES
Take G the centre of the circle, and join GB, GE, GF, GD, GC.
Then EB is equal to EF, and DC to DF; (III. 37.)
therefore ED is equal to EB and DC;
to each of these add AE, AD,
wherefore AD, AE, ED are equal to AB, AC;
and AB is equal to AC,
therefore AD, AE, ED are equal to twice AB, or twice AC;
or the perimeter of the triangle AED is a constant quantity.
Again, the angle EGF is half of the angle BGF,
and the angle DGF is half of the angle CGF,
therefore the angle DGE is half of the angle CGB,
or the angle subtended by the tangent ED at G, is half of the
angle contained between the two radii which meet the circle at the
points where the two tangents AB, AC meet the circle.
THEOREM IV.
If two chords of a circle intersect each other at right angles, the
sum of the square described upon the four segments is equal to the
square described upon the diameter. (Archimedis, Lemm. Prop. 11.)
Let the chords AB, CD intersect at right angles in E.
A

D
C
B F
Draw the diameter AF, and join AC, AD, CF, DB.
Then the angle ACF in a semicircle is a right angle, (III. 31.)
and equal to the angle AED: also the angle ADC is equal to the
angle AFC. (III. 21.)
Hence in the triangles ADE, AFC, there are two angles in the one
respectively equal to two angles in the other;
consequently, the third angle CAF is equal to the third angle DAB,
therefore the arc DB is equal to the arc CF, (III. 26.)
and therefore also the chord DB is equal to the chord CF. (III. 29.)
Because AEC is a right-angled triangle,
the squares of AE, EC are equal to the square of AC; (1. 47.)
similarly, the squares of DE, EB are equal to the square of DB;
therefore the squares of AE, EC, DE, EB, are equal to the
squares of AC, DB;
but DB was proved equal to FC,
and the squares of AC, FC are equal to the square of AF,
wherefore the squares of AE, EC, DE, EB, are equal to the
square of AF, the diameter of the circle.
ON BOOK III. PROBLEMS.
315
PROBLEMS.
3. Given the centre of a circle; find its diameter by means of
the compasses alone.
4. Through a given point within a circle, to draw a chord which
shall be bisected in that point.
5. Through a point in a circle which is not the centre, to draw
the least chord.
6. To draw that diameter of a given circle which shall pass at a
given distance from a given point.
7. Draw through one of the points in which any two circles cut
one another, a straight line which shall be terminated by their circum-
ferences and bisected in their point of section.
8. Determine the distance of a point from the centre of a given
circle, so that if tangents be drawn from it to the circle, the concave
part of the circumference may be double of the convex.
9. Find two points in a given straight line from each of which
if two tangents be drawn to two given points on the same side of the
given circle, they shall make an angle equal to a given angle.
Is any limitation required?
10. Find a point without a given circle, such that the sum of the
two lines drawn from it touching the circle, shall be equal to the line
drawn from it through the centre to meet the circle.
11. From a given point without a circle, a straight line is drawn
cutting a circle. Draw from the same point another line so as to
intercept two arcs which together shall subtend an angle equal to a
given angle.
12. In a chord of a circle produced, it is required to find a point,
from which if a straight line be drawn touching the circle, the line so
drawn shall be equal to a given straight line.
13. Determine the point without a circle, from which, if two
straight lines be drawn touching the circle, they may form an equi-
lateral triangle with the chord which joins the points of contact.
14. A, B, C, are three given points, find the position of a circle
such that all the tangents to it drawn from the points A, B, C shall be
equal to one another. What is that circle which is the superior limit to
those that satisfy the above condition?
15. Two parallel chords in a circle are respectively six and eight
inches in length, and are one inch apart; how many inches is the
diameter in length?
16. The radius of the circle ABDE whose centre is C, is equal to
five inches. The distance of the line AB from the centre is four inches.
The distance of the line DE from the centre is three inches. Required
the lengths of the straight lines AB, DE.
17. Required the locus of the vertices of all triangles upon the
same base, having the sum of the squares of their sides equal to a given
square.
18. Draw a straight line which shall touch a given circle, and
make a given angle with a given straight line.
19. Draw a straight line which shall touch two given circles:
(1) on the same side; (2) on the alternate sides.
20. Through a given point without a given circle, draw a straight
line which shall cut off a quadrantal arc of that circle.
316
GEOMETRICAL EXERCISES
21. Three given straight lines are in the same straight line; find
a point from which lines drawn to their extremities shall contain equal
angles.
22. Draw through a given point in the diameter of a circle à
chord, which shall form with the lines joining its extremities with either
extremity of the diameter, the greatest possible triangle.
23. ADB, ACB, are the arcs of two equal circles cutting one
another in the straight line AB, draw the chord ACD cutting the inner
circumference in Cand the outer in D, such that AD and DB together
may be double of AC and CB together.
24. From a given point without a given circle a line is drawn
cutting the circle. It is required to draw from the same point another
line also cutting the circle, so that the sum of the arcs intercepted be-
tween these two lines shall be equal to a given arc.
25. A given straight line being divided in a given point, to find a
point at which each segment of the given straight line shall subtend an
angle equal to half a right angle.
26. Divide a circle into two parts such that the angle contained
in one segment shall equal twice the angle contained in the other.
27. Any segment of a circle being described on the base of a
triangle; to describe on the other sides segments similar to that on the
base.
28. Through a given point within or without a circle, it is re-
quired to draw a straight line cutting off a segment containing a given
angle.
29. Through two given points to describe a circle bisecting the
circumference of a given circle.
30. A segment of a circle being described upon AB, it is required
to draw a chord AC, such that CK being drawn perpendicular to AB,
AC+ CK shall be a maximum.
31. In the circumference of a given circle, to determine a point
to which two straight lines drawn to two given points shall contain an
angle equal to a given angle, pointing out the limitations within which
the problem is possible.
32. One side of a trapezium capable of being inscribed in a given
circle is given, the sum of the remaining three sides is given; and also
one of the angles opposite to the given side: construct it.
33. To find a point P, so that tangents drawn from it to the out-
sides of two equal circles which touch each other, may contain an angle
equal to a given angle.
34. Given two circles: it is required to find a point from which
tangents may be drawn to each, equal to two given straight lines.
35. Between two given circles to place a straight line terminated
by them, such that it shall equal a given straight line, and be inclined
at a given angle to the straight line joining their centres.
36. Two circles being given in position and magnitude, draw a
straight line cutting them, so that the chords in each circle may be equal
to a given line not greater than the diameter of the smaller circle.
37. Describe two circles with given radii which shall cut each
other, and have the line between the points of section equal to a given
line.
If two circles cut each other; to draw from one of the points
of intersection a straight line meeting the circles, so that the part of
38.
ON BOOK III. PROBLEMS.
317
it intercepted between the circumferences may be equal to a given
line.
39. Three points being in the same plane, find a fourth, where
lines drawn from the former three shall make given angles with one
another.
40. Two given circles touch each other internally. Find the
semichord drawn perpendicularly to the diameter passing through the
point of contact, which shall be bisected by the circumference of the
inner circle.
41. The circumference of one circle is wholly within that of
another. Find the greatest and the least straight lines that can be
drawn touching the former and terminated by the latter.
42.
Draw a straight line through two concentric circles, so that
the chord terminated by the exterior circumference may be double that
terminated by the interior. What is the least value of the radius of the
interior circle for which the problem is possible?
43. To draw a straight line cutting two concentric circles, so that
the part of it which is intercepted by the circumference of the greater
may be triple the part intercepted by the circumference of the less.
44. Find the greatest of all triangles having the same vertical
angle and equal distances between that angle and the bisection of the
opposite sides.
45. If a string of a given length be fixed at each end to two
given points A and B, and be pulled downwards, so as to form a
triangle with the line joining A and B, determine the lowest point
that it may be made to reach.
46. From a given point without a circle, at a distance from the
circumference of the circle not greater than its diameter, draw a straight
line to the concave circumference which shall be bisected by the convex
circumference.
47. Find a point in the circumference of a circle, from whence a
line drawn, making a given angle with a given radius, may be equal to
a given straight line.
48. To find within an acute-angled triangle a point from which,
if straight lines be drawn to the three angles of the triangle, they shall
make equal angles with each other.
49. From two lines, including a given angle, cut off by a line of
given length, a triangle equal to a given rectilineal figure.
50. From the extremities of the diameter of a given semicircle,
draw two chords to meet in the circumference, which shall intercept a
given length on a given oblique chord.
51. The positions of three stations, A, B, and C, have been laid
down on a map, and an observer at D (a station in the same horizontal
plane as A, B, and C,) determines the angles ADB and BDC; give a
geometrical construction for laying down D on the map.
52. In an acute-angled triangle, to find a point from which if
three lines be drawn to the three angles, the sum of these lines shall be
-the least possible.
53. Draw lines from the angles of a triangle to the points of
bisection of the opposite sides and terminated in these points. If from
the extremities of any one of them, lines be drawn parallel to the re-
maining two and produced to meet, a triangle will be formed whose
sides are equal to the three lines first drawn.
54. It it required within an isosceles triangle to find a point such,
318
GEOMETRICAL EXERCISES
that its distance from one of the equal angles may be double its distance
from the vertical angle.
55. To construct an isosceles triangle equal to a scalene triangle
and having an equal vertical angle with it.
56. Given the angle at the base of an isosceles triangle, and the
perpendicular from it on the opposite side, to construct the triangle.
57. Given the base, the vertical angle, and the differences of the
sides, to construct the triangle.
58. Describe a triangle with a given vertical angle, so that the
line which bisects the base shall be equal to a given line, and the angle
which the bisecting line makes with the base shall be equal to a given
angle.
59. Given the perpendicular height, the vertical angle and the
sum of the sides, to construct the triangle.
60. Construct a triangle in which the vertical angle and the
difference of the two angles at the base shall be respectively equal
to two given angles, and whose base shall be equal to a given straight
line.
61. Given the vertical angle, the difference of the two sides con-
taining it, and the difference of the segments of the base made by a
perpendicular from the vertex; construct the triangle.
62. On a given straight line to describe a triangle having its
vertical angle equal to a given angle, and the difference of its sides equal
to a given line.
63. Given the vertical angle, and the lengths of two lines drawn
from the extremities of the base to the points of bisection of the sides,
to construct the triangle.
64. Given the base, and vertical angle, to find the triangle whose
area is a maximum.
65. Find a triangle of which the vertical angle, the sum of the
squares of the two sides containing it and the area are given.
66. The base, vertical angle, and rectangle under the sum of the
other sides and one of them are given. Construct the triangle.
67. Describe a circle the circumference of which shall pass
through a given point and touch a given circle in a given point.
68. Describe a circle which shall pass through a given point and
which shall touch a given straight line in a given point.
69. Describe a circle to touch two right lines given in position
and such that a tangent drawn to it from a given point may be equal to
a given line.
70. Let AB, AC be any two lines given in position; DE a line
of given length; find the position of that circle which is touched by
both the lines AB, AC and whose diameter is equal to DE.
71. Describe a circle to touch two right lines given in position,
so that lines drawn from a given point to the points of contact shall
contain a given angle.
72. Describe a circle through a given point, and touching a given
straight line, so that the chord joining the given point and point of con-
tact may cut off a segment containing a given angle.
73. To describe a circle through two given points to cut a straight
line given in position, so that a diameter of the circle drawn through
the point of intersection shall make a given angle with the line.
74. The straight lines drawn from the same point, and touching
the same circle, are equal. Having proved this, exhibit a construction
ON BOOK III. PROBLEMS.
319
that shall include all the triangles which can be described with a given
perimeter and given vertical angle.
75. A flag-staff of a given height is erected on a tower whose
height is also given: at what point on the horizon will the flag-staff
appear under the greatest possible angle.
76. Find a point from which, if straight lines be drawn to touch
three given circles, none of which lies within the other, the tangents so
drawn shall be equal.
77. The centres of three circles (A, B, C,) are in the same
straight line, B and C touch each other externally and A internally, if
a line be drawn through the point of contact of B and C, making any
angle with the common diameter, then the portion of this line inter-
cepted between C and A, is equal to the portion intercepted between
B‍and A.
78. If P be a point without a circle whose centre is O, and AOB
a diameter perpendicular to PO: draw a line PMEC cutting the circle
in M and C and the diameter in E, so that the rectangle PM, PC, may
be four times the rectangle AE, EB.
79. From a given point without a circle draw a straight line
cutting the circle, so that the rectangle contained by the part of it
without and the part within the circle shall be equal to a given square.
80. Let AP be a tangent to any circle, and AB a diameter. To
determine the point P, so that PCB being drawn, cutting the circum-
ference in C, the rectangle contained by PC, CB, shall be equal to a
given square; and shew in what cases this is impossible.
81. The diameter ACD of a circle, whose centre is C, is produced
to P, determine a point F in the line AP such that the rectangle
PF.PC may be equal to the rectangle PD. PA.
82. Through a given point draw a line terminating in two lines
given in position, so that the rectangle contained by the two parts may
be equal to a given rectangle.
83. A ladder is gradually raised against a wall; find the locus of
its middle point.
84. A, B, C, D are four points in order in a straight line, find a
point E between B and C, such that AE.EB = ED. EC by a geome-
trical construction.
85. Determine the locus of the extremities of any number of
straight lines drawn from a given point, so that the rectangle contained
by each, and a segment cut off from each by a line given in position,
may be equal to a given rectangle.
86. Find a point without a given circle from which if two tan-
gents be drawn to it, they shall contain an angle equal to a given angle,
and shew that the locus of this point is a circle concentric with the
given circle.
·
£
87. Find the locus of the vertex of a triangle described on a given
base; (1) when the sum of the angles at the base is given; (2) when
one of them is always double of the other.
88. Find the locus of the centres of all circles which cut off from
the directions of two sides of a triangle, chords equal to two given
straight lines.
Hence describe a circle that shall cut off from the direction of three
sides of a triangle, chords respectively equal to three given straight
lines.
320
GEOMETRICAL EXERCISES
1
89. In a given straight line to find a point at which two other
straight lines, being drawn to two given points, shall contain a right
angle. Shew that if the distance between the two given points be
greater than the sum of their distances from the given line, there will
be two such points; if equal, there may be only one; if less, the pro-
blem may be impossible.
90. Produce a given straight line so that the rectangle under the
given line, and the whole line produced may equal the square of the
part produced.
91. To produce a given straight line, so that the rectangle con-
tained by the whole line thus produced, and the part of it produced,
shall be equal to a given square.
92. To determine a right-angled triangle whose hypothenuse may
be equal to a given straight line, and the rectangle contained by whose
sides may be equal to the square of their difference.
93. Given the lengths of the three lines drawn from the angles of
a triangle to the points of bisection of the opposite sides, construct the
triangle.
94. Describe a triangle whose sides shall be bisected by three
given straight lines, and one of whose sides shall pass through a given
point.
95. Find the locus of a point, such that if straight lines be drawn
from it to the four corners of a given square, the sum of the squares
shall be invariable.
THEOREMS.
5. The arcs intercepted between any two parallel chords in a
circle are equal. Also the sum of the arcs subtending the vertical angle
made by any two chords that intersect, is the same, as long as the angle
of intersection remains the same.
6. From the extremities A and C of a given circular arc AC, equal
arcs AB, CD are measured in opposite directions: prove that the chords
AC, BD are parallel.
7. Two circles cut each other, and from the points of intersection
straight lines are drawn parallel to one another, the portions intercepted
by the circumferences are equal.
8. A, B, C, A', B', C' are points on the circumference of a circle;
if the lines AB, AC be respectively parallel to A'B', A'C', shew that
BC' is parallel to B'C.
9. C, C' are the centres of two circles of unequal radii, CR, C'R'
any pair of parallel radii, join RR': then shall all such lines produced
meet in one point. Prove this property, and from it deduce à method
of drawing a common tangent to two circles.
10. If from a given point a straight line be drawn touching a
circle given in position, the straight line is given in position and mag-
nitude.
11. DF is a straight line touching a circle, and terminated by
AD, BF, the tangents at the extremities of the diameter AB, shew that
the angle which DF subtends at the centre is a right angle.
ON BOOK III. THEOREMS.
321
12. The circles described on the sides of any triangle as diameters
will intersect in the sides, or sides produced, of the triangle.
13. The circles which are described upon the sides of a right-
angled triangle as diameters, meet the hypothenuse in the same point;
and the line drawn from the point of intersection to the centre of either
of the circles will be a tangent to the other circle.
14. If on the sides of a triangle circular arcs be described con-
taining angles whose sum is equal to two right angles, the triangle
formed by the lines joining their centres has its angles equal to those in
the segments.
15. If AO, BO be the bounding radii of a quadrant, and in OB
any point Q be taken, and QC drawn meeting the circumference in C,
and making BQC=<AQO, shew that 2.QAO+≤ COB = ≤ AOB.
/
16. Any two chords of a circle which cut a diameter in the
same point and at equal angles, are equal to each other.
17. If two equal chords be drawn in a circle, and another chord
be drawn through their middle points, the portions of this last chord
intercepted between the middle points and the circumference are equal.
18. If through the middle point of any chord of a circle two
chords be drawn, the lines joining their extremities shall intersect the
first chord at equal distances from its extremities.
19. If from any point in a circular arc, perpendiculars be let fall
on its bounding radii, the distance of their feet is invariable.
20. Of two circular segments upon the same base, the larger is
that which contains the smaller angle.
21. ABCD is a semicircle; AC, BD are chords drawn from the
extremities of the diameter AD, cutting each other in F; join AB, DC,
and produce them till they meet in E; join EF, and produce it to meet
the diameter in G. Prove that EG is perpendicular to AD.
22. Let N be any point in the diameter of a circle, whose centre
is S, PNQ a chord drawn through N, and join SP; shew geometrically
and analytically that PQ is a minimum, and the angle SPQ a maximum,
when PQ is perpendicular to the diameter.
23. If the diameter of a circle be one of the equal sides of an
isosceles triangle, the base will be bisected by the circumference.
24. If two diameters to a circle be drawn, and through their
extremities any other circles pass, the portions of the diameters lying
between any two of these circles are equal to each other.
25. Two diameters AOA', BOB' of a circle are at right angles to
each other; P is a point in the circumference, the tangent at P meets
BOB' produced in Q, and AP, A'P meet the same line in C, C' respec-
tively. Prove that CQ' is equal to C'Q.
26. ADB, ACB are the arcs of two equal circles cutting one
another in the straight line AB; draw any chord ACD cutting both
arcs, and join CB, DB; CB is equal to DB.
27. If from a point without a circle lines be drawn touching the
circle, the angle included is measured by half the difference of the
concave and convex parts of the circumference.
28. If the chord AB be produced to D, the angle which DB
makes with the chord BC, is equal to half the sum of the angles which
the arcs AB, BC subtend at the centre of the circle.
29. The perpendiculars let fall from the three angles of any
triangle upon the opposite sides, intersect each other in the same point.
21
322
GEOMETRICAL EXERCISES
30. Through a given point within a circle two chords are drawn,
and the extremities of one joined to different extremities of the other;
shew that two triangles are formed which are equiangular.
31. The lines, which bisect the vertical angles of all triangles on
the same base and with the same vertical angle, all intersect in one point.
32. A number of triangles are described upon the same base and
having the same vertical angle; and from a given point, at the same
distance from the two extremities of the base, lines are drawn to the
vertices of the triangles; shew that if this point be so chosen that one
of these lines shall bisect the angle at the vertex to which it is drawn,
all the lines will bisect the angles at the vertices to which they are
drawn.
33. Of all triangles on the same base, having equal perimeters,
the equilateral has the greatest area.
34. The middle point of the hypothenuse of a right-angled triangle
is equally distant from the three angular points.
35. Of all triangles on the same base and between the same
parallels, the isosceles has the greatest vertical angle.
36. Of all straight lines which can be drawn from two given
points to meet in the convex circumference of a given circle, the sum
of those two will be the least, which make equal angles with the tangent
at the point of concourse.
37. If from any point without a circle, lines be drawn cutting the
circle and making equal angles with the longest line, they will cut off
equal segments.
38. From a point without a circle two straight lines are drawn
cutting the convex and concave circumferences, and also respectively
parallel to two radii of the circle. Prove that the difference of the
concave and convex arcs intercepted by the cutting lines, is equal to twice
the arc intercepted by the radii.
39. If round any point within the circumference of a circle, (not
being the centre) equal adjoining angles be described; of the circum-
ferences on which they stand, that which is nearer the diameter pass-
ing through the point is less than the more remote.
40. If two equal chords of a circle cut one another either within
or without a circle, the segments of the one between the point of inter-
section and the circumference, shall be equal to the segments of the
other, each to each.
41. If in a given circle a diameter be drawn, and in it a point
be taken which is not the centre of the circle, and a line be drawn
through it to meet the circumference both ways, only one line equal to
it can pass through that point and meet the circle both ways, and of
all other lines which pass through the same point, those which make
smaller angles with the diameter are greater than those which make
a greater.
42. Two straight lines which at one point make equal angles
with the diameter of a circle or the diameter produced, are equally
distant from the centre.
43. If from a point without a circle two tangents be drawn ;
the straight line which joins the points of contact will be bisected at
right angles by a line drawn from the centre to the point without the
circle.
44. Tangents to a circle at the extremities of any chord, contain
ON BOOK III. THEOREMS.
323
an angle which is twice the angle contained by the same chord and
a diameter drawn from either of the extremities.
45. If tangents be drawn at the extremities of any two diame-
ters of a circle, and produced to intersect one another; the straight
lines joining the opposite points of intersection will both pass through
the centre.
46. If any chord of a circle be produced equally both ways,
and tangents to the circle be drawn, on opposite sides of it, from its
extremities, the line joining the points of contact bisects the given chord.
47. When the vertical angle and the sum of the sides of a plane
triangle are given, prove that the locus of the middle of the base is a
line given in position.
48. If a tangent be drawn at any point of a circle, and from the
point of contact a perpendicular be let fall on a diameter of the circle,
and from the extremities of the diameter perpendiculars be let fall on
the tangent, these perpendiculars will be respectively equal to the seg-
ments of the diameter.
49. If from two fixed points in the circumference of a circle,
straight lines be drawn intercepting a given arc and meeting without
the circle, the locus of their intersections is a circle.
50. Two straight lines stand at right angles to each other, one of
which passes through the centre of a given circle, and from any point
in the other, tangents are drawn to the circle. Prove that the chord
joining the points of contact cuts the first line in the same point, what-
ever be the point in the second from which the tangents are drawn.
51. If from any point without a circle two lines be drawn
touching the circle, and from the extremities of any diameter lines be
drawn to the point of contact cutting each other within the circle, the
line drawn from the point without the circle to the point of intersection
shall be perpendicular to the diameter.
52. If from a point without a circle two lines be drawn touching
the circle, and from one of the points of contact a perpendicular be let
fall on the diameter passing through the other, it will be bisected by a
straight line drawn from the point without the circle to the farthest
extremity of the diameter.
53. If from any point without a circle, lines be drawn touching it,
the angle contained by the tangents is double the angle contained by the
line joining the points of contact and the diameter drawn through any
one of them.
54. Let ACB be a quadrant of a circle, of which the centre is C,
and the terminating radii CA, CB; join AB, and on it describe a semi-
circle: from the point A draw a straight line AED cutting the quadrant
in E, and terminated by the semicircle in D; and join DB; it is required
to prove that DE is always equal to DB, and that the greater only of
the two lines AD, DB can cut the quadrant.
1
55. With any point A in the arc of the circle ABF as a centre,
and any radius AB less than the diameter, and greater than half the
radius of ABF, describe a circle cutting ABF in F; in it place the
chords BD, DE, EC each equal to the chord AB; with radius CF, and
centres A and C, describe two circles intersecting in G; with the same
radius and centre G, describe a circle intersecting the circle BDF in
H; the chord HB is equal to the radius of the circle ABF.
56. If from a point in the circumference of a circle any number
21-2
324
GEOMETRICAL EXERCISES
of chords be drawn, the locus of their points of bisection will be a
circle.
57. In a circle with centre O, any two chords AB, CD are drawn
cutting in E, and OA, OB, OC, OD are joined. Prove that the angles
AOC+BOD=2. AEC, and AOD+BOČ=2.AED.
58. Any two points A, B being taken within the circle CDE, it
is required to find a point D so that the difference of the angles BDE,
ADC may equal a given angle.
59. If from the point where a common tangent to two circles
meets the line joining their centres, any line be drawn cutting the circles,
it will cut off similar segments.
60. ACB, ADB are two segments of circles on the same base AB,
take any point C in segment ACB; join AC, BC, and produce them to
meet the segment ADB in D and E respectively: shew that the arc DE
is constant.
61. If there be two circles in the same plane, not cutting each
other, and two lines be drawn (one of them meeting the line joining the
centres) to touch both circles and terminate at the points of contact;
prove that the difference of the squares of these lines will be equal to
the rectangle of the diameters of the circles.
62. If on two lines containing an angle segments of circles be
described containing angles equal to it, the lines produced will touch
the segments.
63. If two circles touch each other externally, and two parallel
lines be drawn, so touching the circles in points A and B respectively
that neither circle is cut, then a straight line AB will pass through the
point of contact of the circles.
64. A common tangent is drawn to two circles which touch each
other externally; if a circle be described on that part of it which lies
between the points of contact, as diameter, this circle will pass through
the point of contact of the two circles, and will touch the line which
joins their centres.
65. If two circles do not touch each other, and a segment of the
line joining their centres be intercepted between the convex circum-
ferences, any circle whose diameter is not less than that segment may
be so placed as to touch both the circles.
66. If lines be drawn from a given point in the circumference of
a circle, such that the rectangles between the whole lines and the parts
of them within the circle be constant, the locus of their extremities will
be a straight line.
67. C, c, are the centres of two unequal circles; CA, CB, CD,
&c., any radii in the one, and ca, cb, cd, &c., radii in the other respec-
tively parallel to the former, and on the same side of the line Cc.
Join Aa, Bb, &c.; these lines being produced all meet in the same
point.
68. If two equal circles cut each other, and if through one of the
points of intersection a line be drawn terminated by the circles, the
lines joining its extremities with the other point of intersection are
equal.
69. Shew that, if two circles cut each other, and from any point
in the straight line produced which joins their intersections two tan-
gents be drawn, one to each circle, they shall be equal to one another.
70. If any two circles, the centres of which are given, intersect
ON BOOK III. THEOREMS.
325
each other, the greatest line which can be drawn through either point
of intersection and terminated by the circle, is independent of the dia-
meters of the circles.
71. If two circles intersect each other in A, B; any two parallel
lines CD, EF, drawn through A, B, respectively, and cutting the circles
in C, D; E, F; are equal. Required proof.
72. If two circles cut each other, and from their points of inter-
section, lines be drawn to any point in the circumference of one of them,
the arc of the other circle intercepted by these lines, (produced if neces-
sary) is invariable.
73.
If two circles intersect, the common chord produced bisects
the common tangent.
74. Two circles intersect in the points A and B; through A and
B any two straight lines CAD, EBF, are drawn cutting the circles in
the points C, D, E, F; prove that CE is parallel to DF.
75. If two circles intersect, the straight line joining their centres
bisects the intercepted circumferences.
76. If two circles cut each other, the straight line joining their
centres, will bisect their common chord at right angles.
77. If two equal circles cut each other, and from either point of
intersection a circle be described cutting them, the points where this
circle cuts them and the other point of intersection of the equal circles,
are in the same straight line.
78. Two circles intersect in the points A and C, the centre of one
being in the circumference of the other; draw any chord COH common
to both circles, and join AH, then shall AH = HỖ.
79. If two circles touch each other internally, and from any two
points in the circumference of the interior circle, straight lines be drawn
to the point of contact, they will contain a greater angle than any other
two lines drawn from the same points to any other point in the exterior
circumference.
80. Let one circle touch another internally, and let straight lines
touch the inner circle, and be terminated by the outer: shew that the
greatest of these lines is the one parallel to the common tangent at the
point of contact.
81. Two circles touch each other externally, the diameter of one
being double of the diameter of the other; through the point of contact
any line is drawn to meet the circumferences of both; shew that the part
of the line which lies in the larger circle is double of that in the smaller.
82. If two circles touch each other externally and parallel diame-
ters be drawn, the straight line joining the extremities of these diameters
will pass through the point of contact.
83. If two circles touch each other internally, and any circle be
described touching both, prove that the sum of the distances of its centre
from the centres of the two given circles will be invariable.
84. If two circles touch each other externally, and a straight line
which touches both of them intersect another straight line passing
through their centres, at a point whose distance from the nearer circle is
equal to its diameter, the radius of one of the circles will be twice as
great as that of the other.
85. If two circles touch each other, any straight line passing
through the point of contact cuts off similar parts of their circum-
ferences.
}
326
GEOMETRICAL EXERCISES
86. If a straight line be drawn cutting any number of concentric
circles, shew that the segments so cut off are similar.
87. If from any point in the circumference of the exterior of two
concentric circles, two straight lines be drawn touching the interior and
cutting the exterior; the distance between the points of contact will be
half that between the points of intersection.
88. If three equal circles have a common point of intersection,
prove that a straight line joining any two of the points of intersection
will be perpendicular to the straight line joining the other two points of
intersection.
89. If three circles touch each other in one point, and if from a
point external to all, a pair of tangents be drawn to each circle, the
three chords joining the points of contact will all pass through one
point.
90. Two equal circles cut one another, and a third circle touches
each of these two equal circles externally; the straight line which joins
the points of section will, if produced, pass through the centre of the
third circle.
91. If on any three chords drawn through the same point in the
circumference of a circle, as diameters, three circles be described: the
points of intersection of these circles two and two lie in the same straight
line.
92. Three equal circles whose centres are in one straight line are
so placed that the two extreme circles touch, and that point of contact
is the centre of the middle one: shew that by joining the points of in-
tersection of the circles, a rectangle will be formed whose diagonals in-
tersect in the centre of the middle circle.
93. A number of circles touch each other at the same point, and
a straight line is drawn from it cutting them: the straight lines joining
each point of intersection with the centre of the circle will be all
parallel.
94. A fixed circle is cut by a series of circles, all of which pass
through two given points: shew that the lines which join the points of
intersection of the fixed circle with each circle of the series, all converge
to one point.
95. A series of circles touching each other at one point are cut by
a fixed circle; shew that the intersections of the pairs of tangents to the
latter, at the points where it is cut by each of the other circles, lie in a
straight line.
96. If a common tangent be drawn to any number of circles,
which touch each other internally; and from any point in this tangent
as a centre, a circle be described cutting the others, and from this centre
lines be drawn through the intersections of the circles respectively; the
segments of those within each circle will be equal.
97. If in the diameter or diameter produced of a circle, two
points be taken equally distant from the centre; the sum of the squares
of the distances of any point in the circumference from these two points
is constant.
98. If from any point in the diameter of a circle, straight lines be
drawn to the extremities of a parallel chord, the squares of these lines
are together equal to the squares of the segments into which the diame-
ter is divided.
99. If any two chords be drawn in a circle perpendicular to each

ON BOOK III. THEOREMS.
327
other, the sum of their squares is equal to twice the square of the dia-
meter diminished by four times the square of the line joining the centre
with their point of intersection.
100. Let ABCD be any quadrilateral figure, let the line joining
the middle points of the diagonals be bisected in E, and with centre E
any circle be described. Prove that for all points P in this circle,
PA²+ PD² + PC² + PD²
2
is the same, and that it = EA² + EB² + ED²+4.EP².
101. ABCD is a quadrilateral figure having the angles at A and B
right angles; CD is bisected in E, and with centre E and distance
EC a circle is drawn cutting AD and BC, or these lines produced in
H, C. AF is a tangent to the circle, and FEG a diameter. Join AG,
and shew that the squares on the four sides of the quadrilateral figure
are together equal to twice the square on AG.
102. If two circles be concentric, and any chord in the interior
circle be produced either way to meet the circumference of the exterior
circle, the rectangle contained by the whole line thus produced and the
part produced, is a constant quantity.
103. If there be two concentric circles, and any chord of the greater
circle cut the less in any point, this point will divide the chord into
two segments whose rectangle is invariable.
104. Prove that if a straight line could be drawn from any assumed
point in the curve of a semi-circle to meet the diameter produced, so
that the part of the line without the curve should be equal to the radius,
any angle might be trisected.
105. If AOB be a chord to the outer, and a tangent to the inner of
two concentric circles, whose diameters are D and d; and if any two
chords EOF, COD be drawn through 0: and E, D, and C, F be joined,
cutting AB in M and N respectively, then
4. CN.NF+ MN²= D²- d².
106. If a rectangle ABKI be described on the diameter of the
circle AB, cutting the perimeter in C, and from C, CA, CB be drawn,
and AD, BD parallel to BC, AC, the rectangle ADBC is equal to the
rectangle ABKI.
107. Two points are taken in the diameter of a circle at any equal
distances from the centre; through one of these draw any chord, and
join its extremities and the other point. The triangle so formed has
the sum of the squares of its sides invariable.
108. If chords drawn from any fixed point in the circumference of
a circle, be cut by another chord which is parallel to the tangent at that
point, the rectangle contained by each chord, and the part of it inter-
cepted between the given point and the given chord, is constant.
109. ABC is a triangle whose acute vertex is A and base BC,
shew that BC is less than AC²+AB² by twice the square of a line
drawn from A to touch the circle of which BC is the diameter.
110. A chord POQ cuts the diameter of a circle in O, in an angle
equal to half a right angle;
PO²+ OQ² = 2 (rad.)³.
111. Let ACDB be a semicircle whose diameter is AB; and AD,
BC any two chords intersecting in P; prove that
AB²=DA. AP + CB.BP.
328
GEOMETRICAL EXERCISES
*
112. The centre of the circle ACE is divided into six equal parts
in the points A, B, C, D, E, F; G is the centre of ACE; with radius
AC, and centres A, D, describe two circles intersecting in H; with radius
AC, and centres C, E, describe two circles intersecting in K;
AK. AG=KG².
113. If on the radius AC of a circle another semicircle be described
and a perpendicular BDE be drawn to the diameter cutting the circles
in B and D, and AB, AD be joined, then the square on AB is double
of the square on AD.
114. If ABDC be any parallelogram, and if a circle be described
passing through the point A, and cutting the sides AB, AC, and the
diagonal AD, in the points F, G, H respectively, shew that
AB. AF + AC.AG=AD. AH.
115. ACB is an isosceles triangle having a right angle at C; with
centre C and distance CA, describe a circle; if, from a point Q in the
circumference of this circle, QRr be drawn parallel to AB meeting AC,
BC, in R, r, respectively, prove that
QR²+ Qr² = AB².
116. If any two circles touch each other in the point O, and lines be
drawn through O at right angles to each other, the one line cutting the
circles in P, P', the other in Q, Q': and if the line joining the centres
of the circles cut them in A, A',
PP/2+QQAA'³.
117. If from the intersection of any two tangents to a circle, any
straight line be drawn cutting the chord which joins the two points of
contact and again meeting the circumference, it shall be divided by the
circumference and the chord into three segments, such that the rectangle
contained by the whole and the middle point shall be equal to the rect-
angle contained by the extreme parts.
118. The straight line AB joining A and B, the centres of two
circles, whose radii are R and r respectively, is divided in C, so that
AC² — BC²= R² — f³, and a straight line is drawn from C perpendicular
to AB; prove that the tangents drawn to both circles from any point in
this line are equal.
119. Two circles whose centres are A and B, touch one another in
the point C; shew that if D be a point without the circles such that
DA, DB make equal angles with DC, the rectangle contained by the
tangents drawn from D is equal to the square of DC.
120. If to any point Cin a circle, a tangent ACB be drawn, and
from A, C, B perpendiculars be drawn to the diameter DE, and CF be
produced to I; the rectangle contained by AB, CI is equal to that con-
tained by DE, GH.
121. If A be a point within a circle, BC the diameter, and through
A, AD be drawn perpendicular to the diameter, and BAE meeting the
circumference in E, then BA.BE = BC.BD.
122. FEI is a common tangent to two circles, meeting the line
which joins the centres produced in the point I; IDCGH is any
straight line cutting the circles. Prove that
IE × IF = ID × IH = IC × IG.
123. If the sides of a quadrilateral figure inscribed in a circle be
produced to meet, and from each of the points of intersection a straight
ON BOOK III.
329
THEOREMS.
line be drawn touching the circle, the squares of these tangents are
together equal to the square of the straight line joining the points of
intersection.
124. If from a point without a circle two straight lines be drawn
to touch or cut the circle, and from the same point a secant be drawn
through the line joining the points of contact or intersection, and
terminating in the circumference, it will be so divided, that the
rectangle under the whole line and its middle segment shall be equal
to the rectangle under the extreme segments.
125. If two chords of a circle intersect each other without the
circle at right angles, the sum of the squares of the segments is equal
to the square of the diameter.
126. If a circle roll within another of twice its size, any point in
its circumference will trace out a diameter of the first.
127. If from a given point S, a perpendicular Sy be drawn to the
tangent Py at any point P of a circle whose centre is C, and in the
line MP drawn perpendicular CS, or in MP produced, a point Q be
always taken such that MQ = Sy, then the locus of Q is a straight line.
128. If from a fixed point P, any two straight lines be drawn cut-
ting a circle, and BD, AC be drawn intersecting in O, prove that O is
always in the straight line joining the points where the tangents from
P meet the circle.
129. If a rod move betwixt two fixed straight lines CP, CQ; and
the perpendiculars from P, Q upon CP, CQ respectively meet in R,
whilst those from P, Q upon CQ, CP respectively meet in S: the loci
of R and S will be two circles having their common centre in C.
130. In all triangles upon the same base and with equal angles at
the vertex, the perpendiculars drawn from the angles at the base to the
opposite sides in each triangle shall intersect each other in the circum-
ference of a circle; prove this, and describe the circle.
330
GEOMETRICAL EXERCISES
GEOMETRICAL EXERCISES ON BOOK IV.
PROBLEM I.
To describe a circle which shall touch a straight line given in position,
and pass through two given points. (Apollonii de Tactionibus, Prob. 7.).
Analysis. Let AB be the given straight line, and C, D the two
given points.
Suppose the circle required which passes through the points C, D,
to touch the line AB in the point E.
A
E
F B

K
H
D
CG
Join C, D, and produce DC to meet AB in F,
and let the circle be described having the centre L,
join also LE, and draw LH perpendicular to CD.
Then CD is bisected in H, and LE is perpendicular to AB.
Also, since from the point F without the circle, are drawn two
straight lines, one of which FE touches the circle, and the other FDC
cuts it; the rectangle contained by FC, FD, is equal to the square of
FE. (III. 36.)
Synthesis. Join C, D, and produce CD to meet AB in F,
take the point E in FB, such that the square of FE, shall be equal
to the rectangle FD, FC.
Bisect CD in H, and draw HK perpendicular to CD;
then HK passes through the centre. (III. 1. Cor. 1.)
At E draw EG perpendicular to FB,
then EG passes through the centre, (III. 19.)
consequently L, the point of intersection of these two lines is the
centre of the circle.
It is also manifest, that another circle may be described passing
through C, D, and touching the line AB on the other side of the point
F; and this circle will be equal to, greater than, or less than the other
circle, according as the angle CFB is equal to, greater than, or less than
the angle CFA.
ON BOOK IV.
331
PROBLEM II.
Inscribe a circle in a given sector of a circle.
Analysis. Let CAB be the given sector, and let the required circle
whose centre is O, touch the radii in P, Q, and the arc of the sector
in D.
C

P
B
A
E
D
F
Join OP, OQ, these lines are equal to one another.
Join also CO.
Then in the triangles CPO, CQO, the two sides PC, CO are equal
to QC, CO, and the base OP is equal to the base OQ;
therefore the angle PCO is equal to the angle QCO;
and the angle ACB is bisected by CO:
also CO produced will bisect the arc AB in D. (111. 26.)
If a tangent EDF be drawn to touch the arc AB in D ;
and CA, CB be produced to meet it in E, F:
the inscription of the circle in the sector is reduced to the inscrip-
tion of a circle in a triangle. (IV. 4.)
PROBLEM III.
ABCD is a rectangular parallelogram. Required to draw EG, FG
parallel to AD, DC, so that the rectangle EF may be equal to the figure
EMD, and EB equal to FD.
Analysis. Let EG, FG be drawn, as required, bisecting the rect-
angle ABCD.
Draw the diagonal BD cutting EG in H and FG in K.
Then BD also bisects the rectangle ABCD;
and therefore the area of the triangle KGH is equal to that of the
two triangles EHB, FKD.
A
E B

H
F
D
N C
M
Draw GL perpendicular to BD, and join GB,
also produce FG to M, and EG to N.
If the triangle LGH be supposed equal to the triangle EHB, by
adding HGB to each,
the triangles LGB, GEB are equal, and they are upon the same
base GB, and on the same side of it;
therefore they are between the same parallels,
that is, if LE were joined, LE would be parallel to GB;
332
GEOMETRICAL EXERCISES
and if a semicircle were described on GB as a diameter, it would
pass through the points E, L; for the angles at E, L are right angles:
also LE would be a chord parallel to the diameter GB;
therefore the arcs intercepted between the parallels LE, GB are equal,
and consequently the chords EB, LG are also equal;
•
but EB is equal to GM, and GM to GN;
wherefore LG, GM, GN, are equal to one another;
hence G is the centre of the circle inscribed in the triangle BDC.
Synthesis. Draw the diagonal BD.
Find G the centre of the circle inscribed in the triangle BDC;
through G draw EGN parallel to BC, and FKM parallel to AB.
Then EG and FG bisect the rectangle ABCD.
Draw GL perpendicular to the diagonal BD.
In the triangles GLH, EHB, the angles GLH, HEB are equal,
each being a right angle, and the vertical angles LHG, EHB, ‘also
the side LG is equal to the side EB;
therefore the triangle LHG is equal to the triangle EHB.
Similarly, it may be proved, that the triangle GLK is equal to the
triangle KFD;
therefore the whole triangle KGH is equal to the two triangles
EHB, KFD;
and consequently EG, FG bisect the rectangle ABCD.

THEOREM I.
If an equilateral triangle be inscribed in a circle, the square of the
side of the triangle is triple the square of the radius, or of the side of
the regular hexagon inscribed in the same circle. (Euclid's Elements,
XIII. 12.)
Let ABD be an equilateral triangle inscribed in the circle ABD,
of which the centre is C.
A

E
C
D
B
Join BC, and produce BC to meet the circumference in E, also
join AE.
And because ABD is an equilateral triangle inscribed in the circle;
therefore AED is one third of the whole circumference,
and therefore AE is one sixth of the circumference,
and consequently, the straight line AE is the side of a regular
hexagon (IV. 15.), and is equal to EC.
And because BE is double of EC or AE,
therefore the square of BE is quadruple of the square of AE,
but the square of BE is equal to the squares of AB, AE;
therefore the squares of AB, AE are quadruple of the square of AE,
and taking from these equals the square of AE,
therefore the square of AB is triple the square of AE.
ON BOOK IV. PROBLEMS.
333
PROBLEMS.
4. In a given circle, place a straight line equal and parallel to a
given straight line not greater than the diameter of the circle.
5. From two given points in the circumference of a given circle,
inflect straight lines to a point in the opposite circumference, such as to
intercept on a given chord, a segment which may be equal to a given
line.
6. In a given circle to place a straight line equal to a given
straight line, not greater than the diameter of the circle, which shall
pass through a given point, either within or without the circle.
7.
Trisect a given circle by dividing it into three equal sectors.
8. In a given circle place a straight line cutting two radii which
are perpendicular to each other, in such a manner, that the line itself may
be trisected.
9. Draw through the angles of an equilateral triangle three
straight lines which shall form by their intersection another equilateral
triangle; shew that there may be an infinite number of such triangles,
and describe the greatest.
10. Describe, about a given circle, a triangle having each of the
angles at the base double the vertical angle.
11. Inscribe three circles in an isosceles triangle touching each
other, and each of them touching two of the three sides of the triangle.
12. In a given triangle inscribe two isosceles triangles of equal
areas, having one of the equal sides of the one equal to one of the equal
sides of the other, and each of these equal sides equal to half one of the
sides of the given triangle.
13. In a given segment of a circle inscribe an isosceles triangle,
such that its vertex may be in the middle of the chord, and the base
and perpendicular together equal to a given line.
14. Describe a circle which shall pass through three given points.
What condition is necessary in order that it may pass through four
given points?
15. Describe a circle about a given triangle, and the sides of the
triangle being 6, 8, 10, find the radius of the circle.
16. Inscribe in a circle a triangle whose sides or sides produced
shall pass through three given points in the same plane.
17. In a given triangle having inscribed a circle, inscribe another
circle in the space thus intercepted at one of the angles.
18. Inscribe a square in a given square.
number can be inscribed, and find the least.
19.
Inscribe a circle in a rhombus.
Shew that an infinite
20. In a given circle inscribe a rectangle equal to a given recti-
lineal figure not exceeding half of the square upon the diameter.
21. Inscribe the greatest quadrilateral figure in a given circle.
Can a circle always be inscribed in a proposed quadrilateral figure, or
described about one?
22. Let two straight lines be drawn from any point within a circle
to the circumference: describe a circle, which shall touch them both,
and the arc between them.
23. From a given point without a given circle two tangents are
334
GEOMETRICAL EXERCISES
drawn to the circle; describe another circle which shall touch the first
circle and its tangents.
24. Having given the distances of the centres of two equal circles
which cut one another, inscribe a square in the space included between
the two circumferences.
25. If two circles cut each other, inscribe in the space between
them a parallelogram of which one side is given.
26. Find a point which is equally distant from three given points,
the three points not being in the same straight line.
27. Describe a circle about a figure formed by constructing an
equilateral triangle upon the base of an isosceles triangle, the vertical
angle of which is four times the angle at the base. State also in what
case it is possible to circumscribe any trapezium with a circle.
28. To describe a circle about the quadrilateral figure formed by
two tangents drawn from the same points, and the lines drawn from the
circumference to the centre.
29. The centre of the circle which touches the two semicircles
described on the sides of a right-angled triangle is the middle point of
the hypothenuse.
30. If a straight line AB is bisected in the point C, and upon the
lines AB, AC, and CB, and towards the same side of AB, three semi-
circles are described; it is required to describe a circle which shall
touch each of these semicircles, and to compare its diameter with the
line AB.
31. If on one of the bounding radii of a quadrant a semicircle be
described, and on the other, another semicircle be described, so as to
touch the former and the quadrantal arc; find the centre of the circle
inscribed in the figure bounded by the three curves.
32. In a given triangle inscribe a parallelogram which shall be
equal to one half the triangle. Is there any limit to the number of such
parallelograms?
33. In a given triangle to inscribe a triangle, the sides of which
shall be parallel to the sides of a given triangle.
34. Inscribe an equilateral triangle in a square. (1) When the
vertex of the triangle is in an angle of the square. (2) When the
vertex of the triangle is in the point of bisection of a side of the square.
35. By means of the compasses alone, it is required to inscribe in
a square an equilateral triangle having one angle in an angle of the
square.
36. Given a circle and two points A, B, exterior to it, find a point
X in the circle such that if XA, XB, be drawn cutting the circle in P,
Q, PQ shall be parallel to AB.
37. Two points being given without a given circle, determine a
point in the circumference, from which lines being drawn to the two
given points shall contain the greatest possible angle.
38. Given two points A, B, within a circle. Determine the
point P in the circumference such that the angle APB shall be a
maximum.
39. Determine that point in the arc of a quadrant from which
two lines being drawn, one to the centre and the other bisecting the
radius, the included angle shall be the greatest possible.
40. The given straight line AC is inclined at any angle to the
straight line CD; to find a point D in the straight line CD, such, that
ON BOOK IV. PROBLEMS.
335
the angle subtended by the given part AB of AC may be the greatest
possible.
41. Find a point in a triangle from which two straight lines
drawn to the extremities of the base shall contain an angle equal to
twice the vertical angle of the triangle. Within what limitations is this
possible?
42. Given the base of a triangle, and the point from which the
perpendiculars on its three sides are equal; construct the triangle. To
what limitation is the position of this point subject in order that the
triangle may lie on the same side of the base?
43. Describe a circle touching three straight lines.
44. Draw a circle touching two given straight lines, and passing
through a given point.
45. To describe a circle which shall touch a given circle in a
given point, and also a given straight line.
46. To describe a circle touching a given circle in a given point,
and passing through a given point not in the circumference of the given
circle. In what case is this impossible?
47. Describe a circle touching a given straight line, and also two
given circles.
48. Describe a circle which shall touch a given circle, and each
of two given straight lines.
49. Two points are given, one in each of two given circles;
describe a circle passing through both points and touching one of
the circles.
50. Shew how to describe a circle that shall have its centre in
a given straight line, which shall pass through a given point, and also
touch another given straight line.
51. Describe a circle which shall touch two sides and pass through
one angle of a given square.
52. If two circles touch each other externally, describe a circle
which shall touch one of them in a given point, and also touch the other.
In what case does this become impossible?
53. Describe a circle touching a straight line in a given point,
and also touching a given circle. When the line cuts the given circle,
shew that your construction will enable you to obtain six circles touch-
ing the given circle and the given line, but not necessarily in the given
point.
54. Describe three circles touching each other and having their
centres at three given points. In how many different ways may this
be done?
55. Describe three equal circles touching each other, and each
passing through the angle of a given equilateral triangle.
56. Having given the hypothenuse of a right-angled triangle, and
the radius of the inscribed circle, to construct the triangle.
57. Having given the vertical angle of a triangle, and the radii of
the inscribed and circumscribed circles, to construct the triangle.
58. Given the base and vertical angle of a triangle, and also the
radius of the inscribed circle, required to construct it.
59. Given the three angles of a triangle, and the radius of the
inscribed circle, to construct the triangle.
60. A straight line is drawn from the extremity of the diameter
(2R) of a circle, and making an angle with it equal to one third of a
336
GEOMETRICAL EXERCISES
3
right angle; another circle whose radius is 2. R, touches the former
circle externally and the straight line. Find its position.
61. If the base and vertical angle of a plane triangle be given,
prove that the locus of the centres of the inscribed circle is a circle, and
find its position and magnitude.
62. To describe two circles, each having a given semidiameter,
which shall touch the same straight line, and also each other.
63. In a given circle inscribe three equal circles touching each
other and the given circle.
64. Find the centre and diameter of a circle that touches three
given circles, each of which touches the other two.
65. Describe in a given circle three circles which shall touch one
another and the given circle.
66. If there be three concentric circles, whose radii are 1, 2, 3;
determine how many circles may be described round the interior one,
having their centres in the circumference of the circle, whose radius is
2, and touching the interior and exterior circles, and each other.
67. Shew how a right angle may be divided into five equal parts.
68. Upon a given straight line to construct a regular pentagon.
Would an equilateral stellated pentagon answer the conditions of
Euclid's definition?
69. Describe a triangle which shall be equal to a given equilateral
and equiangular pentagon, and of the same altitude.
70. Inscribe a regular pentagon in a given square so that four
angles of the pentagon may touch respectively the four sides of the
square.
71. Inscribe a regular decagon in a given circle.
72. Express the side of a regular decagon inscribed in a circle, in
terms of the radius.
73. Having an equilateral triangle inscribed in a given circle,
shew how a regular hexagon may be described about the same circle.
74. Compare the area of an equilateral and equiangular hexagon
inscribed in a circle, with that of one circumscribing the same circle.
75. Inscribe a regular hexagon in a given equilateral triangle.
76. Inscribe a hexagon equilateral, but not equiangular in a given
square.
77. To inscribe a regular dodecagon in a given circle, and shew
that its area is equal to the square of the side of an equilateral triangle
inscribed in the circle.
78. Determine the distance between the opposite sides of an
equilateral and equiangular hexagon inscribed in a circle; and compare
the area of the hexagon with the area of the rectangle formed by joining
corresponding extremities of the opposite sides.
79. In a given pentagon describe an equilateral and equiangular
hexagon, so that five of its angles may touch the five sides of the
pentagon.
80. On a given straight line describe an equilateral and equian-
gular octagon.
81. To inscribe an equilateral and equiangular octagon in a circle.
82. Find the area of an equilateral and equiangular octagon.
83. Compare the area of the regular hexagon inscribed in a circle
with the area of the circumscribing octagon.
84. Describe a regular octagon equal to a given regular pentagon.
ON BOOK IV.
337
PROBLEMS.
4
•
85. Find the value of the interior angle of a regular decagon, and
of a regular quindecagon.
86. The angles of a ten-sided figure are in arithmetic progression,
and the common difference is 10°. Required the angles.
£
87. The interior angles of a rectilineal figure are in arithmetic
progression; the least angle is 120°, and the common difference 5º.
Required the number of sides.
88. The angles of a polygon of 16 sides are in arithmetic pro-
gression, the smallest angle is 30°, required the common difference.
89. If the angles of a rectilinear figure be in arithmetic progression
and the least angle be 30º, and the common difference 15º, find the
number of sides.
90. The angles of an octagon are in arithmetic progression, and
the common difference is 10°, find the angles.
91. How many sides has each of the regular polygons when one
of the interior angles of each polygon is 104°, 135, 156° respectively?
92. Shew that a regular polygon can be described which has its
sides each equal to a given line, and each angle equal to 140°, and
construct it.
93. Find the centre of the circle which will circumscribe a given
regular polygon of any number of sides.
94. Given a polygon traced upon a plane, describe a triangle that
shall have an equivalent area.
95. To convert a given regular polygon into another which shall
have the same perimeter, but double the number of sides.
96. Produce the sides of a given heptagon both ways, till they
meet, forming seven triangles; required the sum of their vertical angles;
also the sum of the vertical angles when the figure has (n) sides.

1
97. Shew what regular figures will fill up the plane space round
a point. What advantage arises from the honey-comb consisting of
hexagonal cells ?
98. Into what number of equal parts may a right angle be divided
geometrically? What connection has the solution of this problem with
the possibility of inscribing regular figures in circles?
99. Find the least triangle which can be circumscribed about a
given circle.
100. On the sides of an equilateral triangle three squares are de-
scribed. Compare the area of the triangle formed by joining the centres
of these squares with the area of the equilateral triangle.
101. If three equal circles touch each other; to compare the area
of the triangle formed by joining their centres with the area of the tri-
angle formed by joining the points of contact.
102. Four equal circles touch each other and a given circle ex-
ternally compare the radius of one of the four circles with the radius
of the given circle.
103. A circle having a square inscribed in it being given, to find a
circle in which a regular octagon of a perimeter equal to that of the
square, may be inscribed.
-
104. What plane rectilineal figure is that in which a similar recti-
lineal figure can be inscribed, whose area shall be one nth part of the
area of the former? Ex. n=2, n=4.
22
338
GEOMETRICAL EXERCISES
THEOREMS.
2. The centres of the circles inscribed in, and circumscribed
about, an equilateral triangle coincide; and the diameter of one is twice
the diameter of the other.
3. An equilateral triangle is inscribed in a circle, and through
the angular points another is circumscribed: compare the sides and
areas of the two triangles, and shew that the line joining any two points
of contact is parallel to the remaining side.
4. If an equilateral triangle be inscribed in a circle, and the ad-
jacent arcs cut off by two of its sides be bisected, the line joining the
points of bisection shall be trisected by the sides.
5. If perpendiculars Aa, Bb, Cc be drawn from the angular
points of a triangle ABC upon the opposite sides, shew that they will
bisect the angles of the triangle abc, and thence prove that the peri-
meter of abc will be less than that of any other triangle which can be
inscribed in ABC.
6. ABC is an equilateral triangle, and upon BC, AC, AB, equi-
lateral triangles Ba C, AbC, AcB, are described; prove that the tri-
angle acb is equilateral.
7. If on the sides of any triangle, three equilateral triangles be
described, and circles inscribed in each of these triangles, the straight
lines joining the centres of the circles will form an equilateral tri-
angle.
8. If an equilateral triangle be inscribed in a circle, and a
straight line be drawn from the vertical angle to meet the circumference,
it will be equal to the sum or difference of the straight lines drawn from
the extremities of the base to the point where the line meets the circum-
ference, according as the line does or does not cut the base.
9. ABC is an equilateral triangle; AF, DE, are drawn perpen-
dicular to the sides BC, AC, intersecting each other in D: shew that
if FS be drawn to the middle point of AB, it will be a tangent to the
circle described about CEDF.
10. The perimeter of an equilateral triangle inscribed in a circle
is greater than the perimeter of any other isosceles triangle inscribed in
the same circle.
11. If an equilateral triangle be turned about its centre in its own
plane, any two positions of the altitude will always make the same
angle as those of the sides; and there will be three positions of coin-
cidence of the triangle.
12.
If an equilateral triangle be inscribed in a circle, any of its
sides will cut off one-fourth part of the diameter drawn through the
opposite angle.
13. The perpendicular from the vertex on the base of an equi-
lateral triangle, is equal to the side of an equilateral triangle inscribed in
a circle whose diameter is the base. Required proof.
14. If a circle be inscribed in a right-angled triangle, the excess
of the sides containing the right angle above the hypothenuse is equal
to the diameter of the inscribed circle.
15. If two circles be drawn one within, and the other about, a
given right-angled triangle, the sum of their diameters will equal the
sum of the sides containing the right angle.
ON BOOK IV. THEOREMS.
339
16. ABC is a triangle inscribed in a circle, the line joining the
middle points of the arcs AB, AC, will cut off equal portions of the
two contiguous sides measured from the angle A.
17. If in a given triangle a circle be inscribed, and tangents to it
be drawn parallel to the sides; the sum of the perimeters of the three
small triangles cut off by these tangents will equal the perimeter of the
given triangle.
18. Let the three perpendiculars from the angles of a triangle
ABC on the opposite sides meet in P, a circle described so as to pass
through P and any two of the points A, B, C, is equal to the cir-
cumscribing circle of the triangle.
19. If from the centre of the circle circumscribing any triangle,
perpendiculars be let fall upon the sides, the sum of these perpen-
diculars is equal to the sum of the radii of the inscribed and circum-
scribed circles.
20. The line joining the centres of the inscribed and circum-
scribed circles of a triangle, subtends at any one of the angular points
an angle equal to the semi-difference of the other two angles.
21. If ACB be any plane triangle, GCF its circumscribing circle,
and GEF a diameter perpendicular to the base AB, then if CF be
joined, the angle GFO is equal to half the difference of the angles at the
base of the triangle.
22. The straight line which bisects any angle of a triangle in-
scribed in a circle, cuts the circumference in a point which is equi-
distant from the extremities of the side opposite to the bisected angle,
and from the centre of a circle inscribed in the triangle.
23. In a triangle ABC let AD bisecting the angle A meet BC in
D: from O the centre of the inscribed circle draw ŎE perpendicular
to BC; then is the angle BOE equal to the angle DOC.
24. Any number of triangles having the same base and the same
vertical angle, will be circumscribed by one circle.
•
25. From any point B in the radius CA of a given circle whose
centre is C, a straight line is drawn at right angles to CA meeting the
circumference in D; the circle described round the triangle CBD
touches the given circle in D.
26. The diameter of a semicircle is divided into two parts, on each
of which as diameter a semicircle is described, and on the same side as
the given one. If a circle be described touching each of these three
semicircles, the distance of its centre from their common diameter is
equal to twice its radius.
27. The locus of the centres of the circles, which are inscribed in
all right-angled triangles on the same hypothenuse, is the quadrant
described on the hypothenuse.
28. If a circle be described, touching the base of a triangle and
the sides produced, and a second circle be inscribed in the triangle;
prove that the points where the circles touch the base are equidistant
from its extremities, and that the distance between the points where
they touch either one of the sides, is equal to the base.
29. If a circle be described about a triangle ABC, and perpen-
diculars be let fall from the angular points A, B, C, on the opposite
sides, and produced to meet the circle in D, E, F, respectively, the
circumferences EF, FD, DE, are bisected in the points A, B, C.
30. If from the angles of a triangle, lines be drawn to the points
22-2
340
GEOMETRICAL EXERCISES
where the inscribed circle touches the sides; these lines shall intersect
in the same point.
31. A square is inscribed in another, the difference of the areas is
twice the rectangle contained by the segments of the side which are
made at the angular point of the inscribed square.
32. The sides of a given square are bisected, and by joining the
bisections another is inscribed; another in this, and so on, ad infinitum.
Shew that the sum of all the inscribed squares is equal to the area of
the original one.
33. The square inscribed in a circle is equal to half the square
described about the same circle.
34. Shew that no parallelogram can be inscribed in a circle except
a rectangle.
35. The square is greater than any oblong inscribed in the same
circle.
36. If two triangles have one side of the one equal to one side of
the other, and the angles opposite these sides equal, the circumscribing
circles are equal.
37. If a quadrilateral rectilineal figure be described about a circle,
the angles subtended at the centre of the circle by any two opposite
sides of the figure, are together equal to two right angles.
38. If a quadrilateral figure be described about a circle, the sums
of the opposite sides are equal; and each sum equal to half the perimeter
of the figure.
39. Shew that a circle may be inscribed within any quadrilateral
figure if the sums of its opposite sides be equal.
40. If two opposite sides of a quadrilateral figure about or in a
circle be at equal distances from the centre, the other sides will be
parallel.
41. Of all quadrilateral figures contained by four given straight
lines, the greatest is that which is inscriptible in a circle.
42. If any number of quadrilaterals inscribed in a circle, have a
common side, and the sides adjacent to this be produced to meet, the
lines joining the point of concourse with the intersection of the diagonals
of the quadrilateral, shall all meet in the same point.
43. Four circles are drawn, of which each touches one side of a
quadrilateral figure, and the adjacent sides produced; shew that the
centres of these four circles will all lie in the circumference of a circle.
44. Let AB, AC, be the bounding radii of a quadrant; complete
the square ABDC and draw the diagonal AD; then the part of the
diagonal without the quadrant will be equal to the radius of a circle
inscribed in the quadrant.
45. If the sides of a quadrilateral figure circumscribing a circle,
touch the circle at the angular points of an inscribed quadrilateral
figure; all the diagonals will intersect at the same point.
46. If a circle be inscribed in a square, an equilateral triangle in
the circle, and a circle in the triangle, the lines drawn from the vertices
of the triangle to meet in any point of the inner circle, and the radius
of that circle, have their squares together equal to the given square.
47. If any number of parallelograms be inscribed in a given
parallelogram, the diameters of all the figures shall cut one another in
the same point.
48. If equilateral parallelograms successively decreasing in area
ON BOOK IV. THEOREMS.
341
by one half be inscribed one in the other, the sides of the circumscribing
figures must throughout be bisected.
49. Shew that if two circles be inscribed in a third to touch one
another, the tangents of the points of contact will all meet in the same
point.
50. Shew that nine equal circles may be placed in contact, so that
a square whose side is three times the diameter of one of them will cir-
cumscribe them.
51. ABCP, and A'B'C'P' are two concentric circles, ABC,
A'B'C' are any two equilateral triangles inscribed in them. If P, P' be
any two points in the circumferences of these circles, shew that
52.
A'P² + B'P² + C'P² = AP¹² + BP'² + CP/2.
In the fig. Prop. 10, Book Iv, shew that the base BD is the
side of a regular decagon inscribed in the larger circle, and the side of
a regular pentagon inscribed in the smaller circle.
53. In the fig. Prop. 10, Book IV, produce DC to meet the circle
in F, and draw BF; then the angle ABF shall be equal to three times
the angle BFD.
54. If the alternate angles of a regular pentagon be joined, the
figure formed by the intersection of the joining lines will itself be a
regular pentagon.
55. If ABCDE be any pentagon inscribed in a circle, and AC,
BD, CE, DA, EB be joined, then are the angles ABE, BCA, CDB,
DEC, EAD, together equal to two right angles.
56. If the sides of an equilateral and equiangular pentagon be
produced to intersect; the straight lines joining the points of intersection
will form another equilateral and equiangular pentagon.
57. AB, AC are the sides of a regular pentagon and a regular
decagon inscribed in a circle, and drawn on different sides of the point
A, shew that if BC be joined, the difference between BC and AC is
equal to the radius of the circle.
58. If the sides of an equilateral pentagon be produced, the stel-
lated figure which results is equilateral and equiangular likewise. What
is the sum of its re-entrant angles? What is the sum of the exterior
angles which are not in contact with the primitive pentagon?
59. If the sides of the pentagon ABCDE be produced both ways,
so as to meet in the points a, ß, y, d, e; prove that the sum of the angles
at these points is equal to two right angles.
60. A watch-ribbon is folded up into a flat knot of five edges,
shew that the sides of the knot form an equilateral pentagon.
61. Shew that perpendiculars from each of the angles of a regular
pentagon upon the opposite side, pass through the same point. Find
also the value of the angle contained by any two of them, and compare
it with an exterior angle of the pentagon.
62. The difference between the side and the diagonal of a regular
pentagon is equal to the side of another regular pentagon, whose diagonal
is equal to the side of the first.
63. If AB, CD are two diameters drawn at right angles to each
other in the circle whose centre is O, if the radius OB is bisected in E
and on EA there be taken EF equal to EC, shew that CF will be the
side of the inscribed pentagon.
64. If from the extremities of the side of a regular pentagon in-
scribed in a circle, straight lines be drawn to the middle of the arc
342
GEOMETRICAL EXERCISES
subtended by the adjacent side, their difference is equal to the radius;
the sum of their squares to three times the square of the radius; and
the rectangle contained by them is equal to the square of the radius.
65. If all the sides of a regular hexagon be produced both ways
till they meet, the angles thus formed are together equal to four right
angles.
66. If any two consecutive sides of a hexagon inscribed in a circle
be respectively parallel to their opposite sides, the remaining sides are
parallel to each other.
67. Prove that the area of a regular hexagon is greater than that
of an equilateral triangle of the same perimeter.
68. If the alternate sides of a regular hexagon be produced to
meet, the figure so formed will be regular, and have an area equal to
twice the area of the hexagon.
69. If two equilateral triangles be inscribed in a circle so as to
have the sides of one parallel to the sides of the other, the figure com-
mon to both will be a regular hexagon, whose area and perimeter will
be equal to the remainder of the area and perimeter of the two triangles.
70. If from the centre of an equilateral hexagon and each of its
angles, perpendiculars be drawn to any line, the sum of those drawn
from the angles will be equal to six times that which is drawn from the
centre.
71. If a regular hexagon be inscribed in a circle, six circles equal
to it may be described, every one touching the original circle, and two
of the others. If the centres of these circles be joined successively, a
regular hexagon will be formed whose area is four times the area of the
former: and if the outermost points of the six circles be joined, another
hexagon will be formed whose area is seven times the area of the first.
72. If the alternate sides of a regular hexagon be produced to
meet one another, and the angular points of the triangles thus formed
be joined, a regular hexagon will be formed, the area of which is equal
to three times the area of the original hexagon.
73. If in any circle the side of an inscribed hexagon be produced
till it becomes equal to the side of an inscribed square, a tangent drawn
from the extremity, without the circle, shall be equal to the side of an
inscribed octagon.
74. A regular octagon inscribed in a circle is equal to the rect-
angle contained by the sides of the squares inscribed in, and circumscribed
about the circle.
75. If a straight line bisect at right angles any side AB of a regular
polygon of an odd number of sides, shew that it will pass through the
point of intersection of the two sides of the polygon which are most
remote from AB.
76. Shew that every equilateral figure inscribed in a circle is also
equiangular, and that the converse of this is true when the number of
sides is odd, but not necessarily, when the number of sides is even.
77. Prove that the opposite sides of any equiangular rectilineal
figure are parallel when the number of sides is even.
78. If equal straight lines be placed similarly round a circle just
without it, the loci of their extremities will be concentric circles.
79. If any number of equal straight lines be placed in the circum-
ference of a circle, the points of bisection will be in the circumference
of a circle having the same centre.

ON BOOK IV. THEOREMS.
343
80. If a polygon has (n+4) sides, prove that the angles formed
at the points of concourse of these sides produced, are together equal to
(2n) right angles.
81. The interior angles of all polygons are equal, provided the
number of sides be the same in all.
82. Prove that of the polygons of a given number of sides, which
can be inscribed in a given circle, the greatest is that which is equi-
lateral.
83. Of all polygons formed with given sides, the greatest is that
which may be inscribed in a circle.
84. Of all polygons having equal perimeters, and the same number
of sides, the equilateral polygon has the greatest area.
85. In any polygon of an even number of sides, inscribed in a
circle, the sum of the 1st, 3rd, 5th, &c. angles is equal to the sum of the
2nd, 4th, 6th, &c.
86. In any plane equilateral and equiangular polygon of (n) sides,
where (n) is any odd number, if lines be drawn from one of the angles
to the extremities of the opposite side, the angle contained by these
lines is equal to one nth of two right angles.
87. If a polygon be inscribed in a circle, and from each angle be
drawn lines to the next angle but one, the lines so drawn will include a
polygon of the same number of sides, and the sum of the angles of the
included polygon at the extremities of any one of its sides equals the
sum of the angles of the original polygon through which that side pro-
duced, passes: if the original polygon be regular, the included one will
also be regular.
88. If any two sides of a regular polygon taken alternately, be
produced to meet, the figure contained by the straight line made up of
one side and the part produced, the part produced of the other, and the
distances of their extremities to the centre, will be a quadrilateral about
which a circle may be described.
89. What is the difficulty of inscribing geometrically an equilateral
heptagon or undecagon in a given circle? Assuming the demonstrations
of Euclid in his 4th book, shew that any equilateral figure of 2+2, 3×2″,
5 × 2″, or 15 × 2″ sides, (where n is any of the numbers 0, 1, 2, 3, &c.)
may be inscribed in a circle.
90. If a point be assumed in a regular polygon of ʼn sides, from
which perpendiculars are drawn to each of the sides, or the sides pro-
duced, the sum of these perpendiculars is n times the radius of the
inscribed circle.
91. If from any point within a polygon, straight lines be drawn
perpendicular to the several sides, the sum of these perpendiculars shall
be equal to the sum of the perpendiculars drawn from the centre of the
polygon.
92. If from a point whose distance is d from the centre of a circle
whose radius is r, perpendiculars be drawn on all the sides of a regular
polygon of n sides circumscribing the circle, the sum of the squares of
d
the perpendiculars = n(+7²).
93. If two circles have the same centre, and a regular polygon be
described about the outer circle, and from any point in the circumference
of the inner circle perpendiculars be drawn to the sides of the polygon,
the figure formed by joining the bottom of the perpendiculars will have
an invariable magnitude.
344
GEOMETRICAL EXERCISES ·
GEOMETRICAL EXERCISES ON BOOK VI.
PROBLEM I.
To inscribe a square in a given triangle.
Analysis. Let ABC be the given triangle, the base BC, and the
perpendicular AD of which are given.
A
E

G
I
F
B
H
D K
C
Let FGHK be the required inscribed square.
Then BHG, BDA are similar triangles,
and GH is to GB, as AD is to AB,
but GF is equal to GH;
therefore GF is to GB as AD is to AB.
Let BF be joined and produced to meet a line drawn from A
parallel to the base BC in the point E.
Synthesis.
Then the triangles BGF, BAE are similar,
and AE is to AB as GF is to GB,
but GF is to GB as AD is to AB.
Wherefore AE is to AB as AD is to AB;
hence AE is equal to AD.
Through the vertex A, draw AE parallel to BC the
base of the triangle,
make AE equal to AD,
join EB cutting AC in F,

through F, draw FG parallel to BC, and FK parallel to AD;
also through G draw GH parallel to AD.
Then GHKF is the square required.
The different cases may be considered when the triangle is equila-
teral, scalene, or isosceles, and when each side is taken as the base.
THEOREM I.
If from the extremities of any diameter of a given circle, perpen-
diculars be drawn to any chord of the circle, they shall meet the chord,
or the chord produced in two points which are equidistant from the
centre. (Archimedis Lemm. Prop. 13.)
First, let the chord CD intersect the diameter AB in L, but not at
right angles; and from A, B, let AE, BF be drawn perpendicular to CD:
Then the points F, E are equidistant from the centre of the chord CD:
ON BOOK · 345
-
VI.
1
A

D
H
E
B
Join EB, and from I the centre of the circle, draw IG perpen-
dicular to CD, and produce it to meet EB in H.
Then IG bisects CD in G; (III. 2.)
and IG, AE being both perpendicular to CD, are parallel. (1. 29.)
Therefore BI is to BH as IA is to HE; (vi. 2.)
and BH is to FG as HE is to GE;
therefore BI is to FG as IA is to GE;
but BI is equal to IA;
therefore FG is equal to GE.
It is also manifest that DE is equal to CF.
When the chord does not intersect the diameter, the perpendiculars
intersect the chord produced.
ן'
THEOREM II.
If two diagonals of a regular pentagon be drawn to cut one another, the
greater segments will be equal to the side of the pentagon, and the
diagonals will cut one another in extreme and mean ratio. (Euclid's
Elements, XIII. 8.)
Let the diagonals AC, BE be drawn from the extremities of the side
AB of the regular pentagon ABCDE, and intersect each other in the
point H.
D

E
A
H
B
Then BE and AC are cut in extreme and mean ratio in H, and the
greater segment of each is equal to the side of the pentagon.
Let the circle ABCDE be described about the pentagon. (Iv. 14.)
Because EA, AB are equal to AB, BC, and they contain equal angles;
therefore the base EB is equal to the base AC, (1. 4.)
and the triangle EAB equal to the triangle CBA,
and the remaining angles will be equal to the remaining angles,
each to each, to which the equal sides are opposite.
Therefore the angle BAC is equal to the angle ABE;
and the angle AHE is double of the angle BAH, (1. 32.)
but the angle EAC is also double of the angle BAC,
therefore the angle HAE is equal to AHE,
and consequently HE is equal to EA (1. 6.) or to AB.
And because BA is equal to AE,
(vi. 33.)
346
GEOMETRICAL EXERCISES
the angle ABE is equal to the angle AEB;
but the angle ABE has been proved equal to BAH;
therefore the angle BEA is equal to the angle BAH;
and ABE is common to the two triangles ABE, ABH;
therefore the remaining angle BAE is equal to the remaining angle AHB:
and consequently the triangles ABE, ABH are equiangular;
therefore EB is to BA as AB to BH: but BA is equal to EH,
therefore EB is to EH as EH is to BH,
but BE is greater than EH; therefore EH is greater than HB;
therefore BE has been cut in extreme and mean ratio in H.
Similarly, it may be shewn, that AC has also been cut in extreme and
mean ratio in H, and that the greater segment of it CH is equal to the
side of the pentagon.
PROBLEM II.
Divide a given arc of a circle into two parts which shall have their
chords in a given ratio. (Pappi Math. Coll. vII. 155.)
Analysis. Let A, B be the two given points in the circumference of
the circle, and C the point required to be found, such that when the
chords AC and BC are joined, the lines AC and BC shall have to one
another the ratio of E to F.

C
A
D
E
B
F
Draw CD touching the circle in C ;
join AB and produce it to meet CD in D.
Since the angle BAC is equal to the angle BCD,(111. 32.)
and the angle CDB is common to the two triangles DBC, DÁC;
therefore the third angle CBD in one, is equal to the third angle
DCA in the other, and the triangles are similar,
therefore AD is to DC as DC is to DB; (v1.4.)
hence also the square of AD is to the square of DC´as AD is to
BD. (VI. 20. Cor.)
But AD is to AC as DC is to CB, (vi. 4.)
and AD is to DC as AC to CB, (v. 16.)
also the square of AD is to the square of DC as the square of AC
is to the square of CB;
but the square of AD is to the square of DC as AD is to DB;
wherefore the square of AC is to the square of CB as AD is to BD;
but AC is to CB as E is to F, (constr.)
therefore AD is to DB as the square of E is to the square of F.
Hence the ratio of AD to DB is given,
and AB is given in magnitude, because the points A, B in the
circumference of the circle are given.
Wherefore also the ratio of AD to ÄB is given, and also the magni-
tude of AD.
Synthesis. Join AB and produce it to D, so that AD shall be to
BD as the square of E to the square of F.
ON BOOK VI.
347
From D draw DC to touch the circle in C, and join CB, CA.
Since AD is to DB as the square of E is to the square of F, (constr.)
and AD is to DB as the square of AC is to the square of BC;
therefore the square of AC is to the square of BC as the square of
E is to the square of F,
and AC is to BC as E is to F.
PROBLEM III.
A, B, C are given points. It is required to draw through any other
point in the same plane with A, B, and C, a straight line, such that the
sum of its distances from two of the given points, may be equal to its
distance from the third. (Phil. Trans. Edin. 1792.)
Analysis. Suppose F the point required, such that the line XFH
being drawn through any other point X, and AD, BE, CH perpendicu-
lars on XFH, the sum of BE and CH is equal to AD.
A

K
H
E
B
G
C
Join AB, BC, CA, then ABC is a triangle.
Draw AG to bisect the base BC in G, and draw GK perpendicular to EF.
Then since BC is bisected in G,
the sum of the perpendiculars CH, BE is double of GK;
but CH and BE are equal to AD, (hyp.)
therefore AD must be double of GK;
but since AD is parallel to GK,
the triangles ADF, GKF are similar,
therefore AD is to AF as GK is to GF
?;
but AD is double of GK, therefore AF is double of GF;
and consequently, GF is one third of AG the line drawn from the
vertex of the triangle to the bisection of the base.
But AG is a line given in magnitude and position,
therefore the point F is determined.
Synthesis. Join AB, AC, BC, and bisect the base BC of the
triangle ABC in G; join AG and take GF equal to one third
of GA;
the line drawn through X and F will be the line required.
It is also obvious, that while the relative position of the points A, B,
C, remains the same, the point F remains the same, wherever the point
X may be. The point X may therefore coincide with the point F, and
when this is the case, the position of the line FX is left undetermined.
Hence the following porism.
A triangle being given in position, a point in it may be found, such,
that any straight line whatever being drawn through that point, the
perpendiculars drawn to this straight line from the two angles of the
triangle, which are on one side of it, will be together equal to the per-
pendicular that is drawn to the same line from the angle on the other
side of it.
348
GEOMETRICAL EXERCISES
PROBLEMS.
4. Given the perimeter of a triangle: describe it so that it may
be similar to a given triangle.
5. On a given base to describe a triangle equal to a given tri-
angle.
6. Given the perimeter of a right-angled triangle, it is required
to construct it,
(1) If the sides are in arithmetical progression.
(2) If the sides are in geometrical progression.
7. Given the vertical angle, the perpendicular drawn from it to
the base, and the ratio of the segments of the base made by it to con-
struct the triangle.
8. Find the locus of intersection of two lines drawn from two
given points, so that their lengths are in a given ratio.
9. In the triangle ABC, let D be the middle point of AB, and E
the foot of the perpendicular let fall from C on AB; then prove that, if
CB-CA=DE, the three sides CA, AB, BC are in arithmetical pro-
gression. Also shew how such a triangle may be constructed.
10. Apply (vI. c.) to construct a triangle; having given the ver-
tical angle, the radius of the inscribed circle, and the rectangle con-
tained by the straight lines drawn from the centre of the circle to the
angles at the base.
11. Having given the difference of the angles at the base, the
perpendicular altitude, and the ratio of the sides, to construct the tri-
angle.
12. Given the vertical angle and the base of a triangle, and also
a line drawn from either of the angles at the base cutting the opposite
side in a given ratio, to construct the triangle.
13. Upon a given straight line as an hypothenuse, describe a right-
angled triangle which shall have its three sides in continued proportion.
14. In any triangle inscribe a triangle similar to a given triangle.
15. Given the base, the ratio of the sides containing the vertical
angle, and the distance of the vertex from a given point in the base; to
construct the triangle.
16. Given the base, the vertical angle, and the ratio of the parts
that the line bisecting the base is divided into, by the line joining the
centres of the circumscribed and inscribed circles, to construct the
plane triangle.
17. The vertical angle and the base of a triangle are given, and
the ratio of the segments into which one of the sides adjacent to the
vertical angle is divided by a perpendicular drawn from the opposite
angle, construct the triangle.
18. Upon the given base AB construct a triangle having its
sides in a given ratio and its vertex situated in the given indefinite
line CD.
19. Describe an equilateral triangle equal to a given triangle.
20. Given the hypothenuse of a right-angled triangle, and the
side of an inscribed square. Required the two sides of the triangle.
21. Given the perimeter of a right-angled triangle, and the per-
pendicular drawn from the right angle upon the opposite side; to find
the three sides (1) geometrically, (2) algebraically.
ON BOOK VI. PROBLEMS.
349
22. Given one of the angles and the perimeter of a plane triangle,
to find the sides, when the area is the greatest possible.
23. To make a triangle, which shall be equal to a given triangle,
and have two of its sides equal to two given straight lines; and shew
that if the rectangle contained by the two straight lines be less than twice
the given triangle, the problem is impossible.
24. From the obtuse angle of a triangle, it is required to draw a
line to the base, which shall be a mean proportional between the seg-
ments of the base. How many answers does this question admit of?
25. To find, by a geometrical construction, an arithmetic, geo-
metric, and harmonic mean between two given lines.
26.
lines.
27.
Find two arithmetic means between two given straight
•
Draw a straight line such that the perpendiculars let fall from
any point in it on two given lines may be in a given ratio.
28. In the figure to Proposition 47, Book 1. draw through Ga
right line, so that the sum of the perpendiculars falling on it from B
and C may be equal to BK.
29. Find a point, such that the perpendiculars let fall from it on
three straight lines given in position may be in a given ratio.
30. If a perpendicular be drawn from a given point of a given
line; find a point in it, such that lines drawn from it to the extremities
of the given line, may be in a given ratio.
31. To find a point P in the base BC of a triangle produced, so
that PD being drawn parallel to AC, and meeting AC produced to D,
AC: CP:: CP : PD.
32. Determine the point in the produced side of a triangle, from
which a straight line being drawn to a given point in the base, shall be
intersected by the other side of the triangle in a given ratio.
33. A point and a straight line being given in position, to draw a
line parallel to the given line, so that all the lines drawn through the
given point may be divided at it in a given ratio.
34. It is required to cut off a part of a given line so that the part
cut off may be a mean proportional between the remainder and another
given line,
35. Find a straight line which shall have to a given straight line
the ratio of 1: √5.
36. It is required to divide a given finite straight line into two
parts, the squares of which shall have a given ratio to each other.
37. The sum of two straight lines and another straight line, a
mean proportional between them being given, to find the two lines.
38. To divide a given line in harmonical proportion.
39. Prove both geometrically and algebraically, that an arithmetic
mean between two quantities is greater than a geometric mean. Also
having given the sum of two lines, and the excess of their arithmetic
above their geometric mean, find by a construction the lines themselves.
40. Determine that point in the base produced of a right-angled
triangle from which the line drawn to the angle opposite the base, shall
have the same ratio to the base produced, which the perpendicular has
to the base itself.
41. On two given straight lines similar triangles are described.
Required to find a third, on which, if a triangle similar to them be
described, its area shall equal the difference of their areas.
350
GEOMETRICAL EXERCISES
42. C is a given point in a given line AB, it is required to deter-
mine a point P, so that AP and BP together may be equal to a given
line, and the angle APB be bisected by PC.
43. From the vertex of a triangle to the base, to draw a straight
line which shall be an arithmetic mean between the sides containing the
vertical angle.
44. To draw a line from the vertex of a triangle to the base,
which shall be a mean proportional between the whole base and one
segment.
45. The distances of three of the angles of a square from a given
point are given, to construct the square. Shew that the same data will
afford two solutions of the problem, and that the assumed distance of
the interjacent angle must have to the sum of the other two, a less ratio
than that of the side of a square to its diagonal.
46. From two straight lines cut off two parts having a given ratio,
so that the sum of the squares of the remainders may be equal to a
given square.
47. If a given straight line AB be divided into any two parts in
the point C, it is required to produce it, so that the whole line produced
may be harmonically divided in C and B.
48. Two straight lines and a point between them being given in
position, to draw two lines from the given point to terminate in the
lines given in position, so that they shall contain a given angle, and
have a given ratio.
49. If a straight line be divided in two given points, determine a
third point such that its distances from the extremities may be propor-
tional to its distances from the given points.
50. Through a given point between two indefinite straight lines
not parallel to each other, to draw a straight line which shall be termi-
nated by them, so that the rectangle contained by its segments shall be
less than the rectangle contained by the segments of any other straight
line drawn through the same point and terminated by the same straight
lines.
51. A rectangular leaf ABCD of which the side AB = 34 a, and
the side BC= 13a has its two corners, A and B, equally bent up at right
angles to the rest of the leaf, and the crease is inclined at the same angle
to each side. From the points A and B straight lines are drawn to the
middle point of CD, these are found to be at right angles to each other,
find how much of the leaf is bent up.
52. ABC is a right-angled triangle; find the point P in AC, so
that the sum of the distances from A and AC is the least possible.
53. ABC is a right-angled triangle, having a right angle at C,
find a point P in the hypothenuse, so situated, that PA may be half
of the perpendicular dropped from P upon BC the base.
54. Find three points in the sides of a triangle, such that they
being joined, the triangle shall be divided into four equal triangles.
55. Determine two straight lines, such that the sum of their
squares may equal a given square, and their rectangle equal a given
rectangle.
56. From the vertex of any triangle ABC, draw a straight line
meeting the base produced in D, so that the rectangle DB . DC= AD².
57. Describe a circle passing through two given points and touch-
ing a given circle.
ON BOOK VI.
351
PROBLEMS.
58. Describe a circle which shall pass through a given point and
touch a given straight line and a given circle.
59. Through a given point draw a circle touching two given
circles.
60. With a given radius to describe a circle, touching two given
circles.
61. Describe a circle which shall have its centre in a given line,
and shall touch a circle and a straight line given in position.
62. In a given circle place a straight line parallel to a given
straight line, and having a given ratio to it; the ratio not being greater
than that of the diameter to the given line in the circle.
63. If any rectangle be inscribed in a given triangle, required the
locus of the point of intersection of its diagonals.
64. In the diameter BB' of a circle produced, a point A is taken
such that AB = 1 of the radius: from A a tangent AC is drawn to the
circle; find a point D in the circle such that AD shall be a mean pro-
portional between AB and AC.
65. Through a given point within a given circle, to draw a
straight line such that the parts of it intercepted between that point and
the circumference, may have a given ratio.
66. From a given point without a circle, it is required to draw a
straight line to the concave circumference, which shall be divided
in a given ratio at the point where it intersects the convex circum-
ference.
67. Find a point external to two circles in the same plane that do
not meet, such that if straight lines be drawn through it cutting both
circles, the portions of all such straight lines intercepted within the
circles, shall be proportional to their radii.
Hence draw a pair of common tangents to two circles, and determine
within what limits a point must be situated, so that a straight line may
be drawn from it cutting both.
68. Given the magnitude and position of two circles; to draw a
chord AB of the greater to touch the less at C, so that CA shall have to
CB a given ratio.
69. From what point in a circle must a tangent be drawn, so that
a perpendicular on it from a given point in the circumference may be
cut by the circle in a given ratio?
70. If a straight line be divided into any two parts, to find the
locus of the point in which these parts subtend equal angles.
71. A given rectangular field is to be laid out as a grass-plot
(having the form of a double square,) surrounded by a walk of uniform
breadth. Find the breadth of the walk, and give a construction for it.
72. Given the sides of a quadrilateral figure inscribed in a circle,
to find the ratio of its diagonals.
73. If in a given equilateral and equiangular hexagon another be
inscribed, to determine its ratio to the given one.
74. From a given point in the side of a triangle, to draw lines to
the sides which shall divide the triangle into any number of equal parts.
75. Any two triangles being given, to draw a straight line parallel
to a side of the greater, which shall cut off a triangle equal to the less.
76. To divide a given triangle into two parts, having a given..
ratio to one another, by a straight line drawn parallel to one of its sides..
77. Inscribe a square in a given right-angled isosceles triangle.
352
GEOMETRICAL EXERCISES
78. Of the two squares which can be inscribed in a right-angled
triangle, which is the greater?
79. The area of a rectangle inscribed in a given triangle being
given, it is required to find its four sides.
80. Inscribe a square in a sector of a circle, so that the angular
points shall be one on each radius, and the other two in the circum-
ference.
81. Inscribe in a square a triangle of which one side and the area
are given.
82. Determine which is the greatest, and which the least, of the
three squares which may be inscribed in a given triangle.
·
83. Inscribe a square in a given equilateral and equiangular pen-
tagon.
84. Inscribe a parallelogram in a given triangle similar to a given
parallelogram.
85. Inscribe the greatest parallelogram in a given triangle.
86. Apply (v1. 25.) to describe a triangle whose sides are as 3, 4,
5, and its area equal to a square of side 6; what are the sides of the
triangle?
87. Inscribe a square in a segment of a circle.
88. Inscribe the greatest parallelogram in a given semicircle.
89. In a given rectangle inscribe a parallelogram of given area
and containing an angle equal to a given angle; and point out the
limitations of the problem.
90. Describe a rectangular parallelogram which shall be equal to
a given square, and have its sides in a given ratio.
91. Through a given point, either within or without a given
triangle, to draw a straight line, which shall cut off from the triangle
any part required.
92. Two similar rectilineal figures being given, to find a third
figure, which shall be similar to them, and also have its area a mean
proportional between them.
93. In a given rectangle inscribe another, whose sides shall bear
to each other a given ratio.
94. To describe a figure similar to each of two other figures, and
equal to their sum or difference.
95. It is required to cut off from a rectangle a similar rectangle
which shall be any required part of it.
96. Inscribe in a given circle an isosceles triangle whose base shall
be equal to its altitude.
97. Inscribe in a circle two isosceles triangles, the area of one of
which shall be one-fourth the area of the other.
98. In a given segment of a circle to inscribe a similar segment.
99. An equilateral triangle and an equilateral and equiangular
pentagon have the same perimeter: compare the radii of the circles
inscribed in them.
100. A series of circles is described all touching two given straight
lines, and also touching each other externally in succession; find the
law of their diameters.
ON BOOK VI. PROBLEMS.
353
THEOREMS.
3. If two triangles are to each other as their bases: prove that
they have the same altitude.
4. Triangles and parallelograms of unequal altitudes are to each
other in the ratio compounded of the ratios of their bases and altitudes.
5. By the construction in Props. 2, a, Book v1, the line made up
of the base and the part produced is divided harmonically.
6. If, in similar triangles, from any two equal angles to the
opposite sides, two straight lines be drawn making equal angles with
the homologous sides, these straight lines will have the same ratio as the
sides on which they fall, and will also divide those sides proportionally.
7. From Prop. 2, Book vi, shew how a line may be drawn on
the ground through a given point parallel to a given straight line by
means of a piece of string.
8. If two systems of parallel straight lines at equal distances cut
one another, the figures between them will be all similar parallelograms.
9. Let AB, AC be two given straight lines, take AD = −th of
AB, and AE=
DC
m+1°
1
m
1
m
th of AC; join BE, CD to meet in F; shew that DF
10. ABC is a triangle; Ab, Ac are taken
AB AC
n
n
respectively,
Cb, Bc meet in D, and AD produced meets BC in E: shew that
2.AE= (n+1) AD; and E is the middle point of AB.
11. Any three lines being drawn making equal angles with the
three sides of any triangles towards the same parts, and meeting one
another, will form a triangle similar to the original triangle.
12. If the original triangle, Prop. 8, Book vi, be isosceles, and
perpendiculars be continually drawn from the foot of the last to the
opposite side, the sum of all the triangles so formed is equal to the area
of the original triangle.
13. If a, b, c represent respectively the sides and hypothenuse of
a right-angled triangle, and h be the length of the line drawn from the
right angle perpendicular to the opposite side: prove that the triangle
whose sides are a+b, h, and c+h, is also right-angled.
14. Shew geometrically that the diagonal and side of a square are
incommensurable.
15. If from any point in the circumference of a circle perpen-
diculars be drawn to the sides, or sides produced, of an inscribed
triangle: shew that the three points of intersection will be in the same
straight line.
16. If the sides AB, AC of a given triangle be cut proportionally
in any two points D and F, and perpendiculars to the sides from those
points intersect in L: prove that all the points L corresponding to
different positions of D and F will be in a straight line passing
through A.
17. Shew that the locus of the vertices of all the triangles con-
structed upon a given base, and having their sides in a given ratio, is a
circle.
23
354
GEOMETRICAL EXERCISES
18. BD, CD are perpendicular to the sides AB, AC of a triangle
ABC, and CE is drawn perpendicular to AD, meeting AB in E: shew
that the triangles ABC, ACE are similar.
19. If one side of a triangle be produced, and the other shortened
by equal quantities, the line joining the points of section will be divided
by the base in the inverse ratio of the sides.
20. CAB, CEB are two triangles having a common angle CBA,
and the sides opposite to it CA, CE, equal: if BAE be produced to D,
and ED be taken a third proportional to BA, AC, then shall the triangle
BDC be similar to the triangle BAC.
21. In any right-angled triangle, one side is to the other, as the
excess of the hypothenuse above the second, to the line cut off from the
first between the right angle and the line bisecting the opposite angle.
22. If from the extremities of the base of a triangle, two straight
lines be drawn, each of which is parallel to one of the sides, and equal
to the other, the straight lines joining their other extremities with the
other extremities of the base, will cut off equal segments from the sides,
and each of these will be a mean proportional between the other two

segments.
23. If a line AB be divided in extreme and mean ratio in C, and
produced to D so that BD is equal to BC the larger segment, then AD
is similarly divided in B. Also if a point E be taken in BC, such that
BE is equal to AC, AE is similarly divided in C.
24. In any triangle, if a perpendicular be let fall upon the base
from the vertical angle, the base will be to the sum of the sides, as the
difference of the sides to the difference or sum of the segments of the
base made by the perpendicular, according as it falls within or without
the triangle.
25. From the vertex of an isosceles triangle two straight lines
drawn to the opposite angles of the square described on the base, cut
the diagonals of the square in E and F: prove that the line EF is
parallel to the base.
26. If an nth part be cut off from any straight line, the rectangle
contained by the whole line and the remaining part, shall be equal to
n times the rectangle contained by the two parts.
27. If two triangles are equal, and the sides about one angle of
the one are reciprocally proportional to the sides about one angle of the
other, then two angles are either equal or supplements to each other.
28. If the vertical angle CAB of a triangle ABC be bisected by
AD, to which the perpendiculars CE, BF, are drawn from the remain-
ing angles: bisect the base BC in G, join GE, GF, and prove these lines
equal to each other.
29. If perpendiculars be drawn from the extremities of the base
of a triangle on a straight line which bisects the angle opposite to the
base, the area of the triangle is equal to the rectangle contained by
either of the perpendiculars, and the segment of the bisecting line
between the angle and the other perpendicular.
30. If lines be drawn from the angles of a triangle bisecting the
opposite sides, shew that the line joining the points of bisection is
parallel to the third side of the triangle.
31. If the three sides of a triangle be bisected, the lines which
join the points of bisection will divide the triangle into four equal
triangles, each of them similar to the whole triangle.
ON BOOK VI. THEOREMS.
355
32. If in the figure, Prop. 10, Book IV, the straight lines DC, BA
be produced to meet the circle again in E, F, and EF be joined: shew
that the triangle CEF is to the triangle ABD as 3+ √5: 2; and that
the triangle ABD is a mean proportional between CEF and BCD.
33. The square of the line bisecting the vertical angle of any
triangle is a mean proportional between the differences of the squares
of each side containing that angle, and the square of the adjacent seg-
ment of the base.
34. If the triangle ABC has the angle at C a right angle, and
from C a perpendicular be dropped on the opposite side intersecting it
in D, then AD : DB :: AC² :˜¨˜CB².
35. If the triangle ACB has a right angle C, and AD be drawn
bisecting the angle A, and meeting CB in D, prove that
AC² : AD² :: BC : BC + BD.
36. If through the point of bisection of the base of a triangle any
line be drawn, intersecting one side of the triangle, the other produced,
and a line drawn parallel to the base from the vertex, this line shall be
cut harmonically.
37. If the interior angle BAC, and the exterior angle DAC of any
triangle ABC be bisected by lines AE, AF, which also cut BC in E, F:
shew that BF, BC, BE are in harmonic progression.
38. ACB is a triangle whose base AB is divided in E, and pro-
duced to F, so that AE: EB, and also AF: FB, as AC: CB. Join
CE, CF, and shew that the angle ECF is a right angle.
39. The square described upon the side of a regular pentagon in
a circle, is equal to the square of the side of a regular hexagon, together
with the square upon the side of a regular decagon in the same circle.
40. From B the right angle of a right-angled triangle ABC, Bp
is let fall perpendicular to AC, from p, pq is let fall perpendicular to
BA, &c.: prove that
·
Bp+pq+&c.: AB :: AB+ AC: BC.
41. If CD be drawn from the vertex C to any point D in the base
of the triangle ABC, then
AC². BD+BC², AD = CD². AB+ AB. AD.BD.
42. If on the two sides of a right-angled triangle squares be
described, the lines joining the acute angles of the triangle and the
opposite angles of the squares, will cut off equal segments from the
sides; and each of these equal segments will be a mean proportional
between the remaining segments.
43. ABC is an equilateral triangle; E any point in AC; in BC
produced take CD=CÂ, CF=CE, AF, DE intersect in H;
HC AC
EC AC+ EC'
44. ABC is an isosceles triangle; draw CE perpendicular to the
base AB; draw ADF intersecting CE in D, and CB in F;
DE CA - CF
CE
CA+ CF
45. If in a triangle ABC, CA be produced to D, and from CB,
BE be cut off equal to AD; then the line ED joining the points of
section will divide the base in the inverse ratio of CD: CE.
23-2
356
GEOMETRICAL EXERCISES
:
46. In any right-angled triangle ABC, (whose hypothenuse is
AB) bisect the angle A by AD meeting CB in D, and prove that
2AC²: AC² - CD2 :: BC : CD.
47. ABC is a right-angled triangle, ADE is an isosceles triangle,
and AF is perpendicular to DE: shew that AF is a mean proportional
between AC and 1 (AB+ AC).
48. If from the angles of a triangle straight lines be drawn bisect-
ing the opposite sides, the sum of the squares of the sides: sum of thẹ
squares of the bisecting lines :: 4 : 3.
49. From any two angles of a triangle, straight lines are drawn to
the points of bisection of the opposite sides: prove that the point of their
intersection is distant from either angle, two-thirds of the line drawn
from it.
50. If the exterior angle CAD of a triangle BAC be bisected by a
straight line AE which likewise cuts the base BC produced in E: shew
that the rectangle BE, EC is equal to the rectangle BA, AC together
with the square of AE.
51. If any point be taken in the plane of a parallelogram from
which perpendiculars are let fall on the diagonal, and on the sides
which include it, the rectangle of the diagonal and the perpendicular on
it, is equal to the sum or difference of the rectangles of the sides and
the perpendiculars on them.
52. BE, AC are parallel lines; F, G, H, &c. a series of equidistant
points in AC; draw Bfh cutting BE in B, and EF, EG, EH, &c. in
f, g, h, &c. Bf, Bg, Bh, &c. are in harmonic progression.
53. If diverging lines cut a straight line, so that the whole is to
one extreme, as the other extreme is to the middle part, they will inter-
sect every other intercepted line in the same ratio.
54. A straight line CD is drawn bisecting the vertical angle C of
a triangle ACB, and cutting the base AB in D; also on AB produced
a point E is taken equidistant from C and D: prove that AE.BË
DE2.
55. From an angle of a triangle a line is drawn to the middle
point of the opposite side, and through the point of bisection of this line
another is drawn from either angle to the side subtending it. Prove
that the latter line divides this side into segments which are as 2 to 1.
56. No triangle can be inscribed in a square, greater than half
the square.
57. If the opposite sides of a hexagon inscribed in a circle be
produced till they meet, the three points of intersection will be in the
same straight line.
58. If a hexagon be described about a circle, the three lines joining
the opposite angular points, intersect in a point.
59. The square inscribed in a circle is to the square inscribed in
the semicircle :: 5 : 2.
60. If a square be inscribed in a right-angled triangle of which
one side coincides with the hypothenuse of the triangle, the extremi-
ties of that side divide the base into three segments that are continued
proportionals.
61. The square inscribed in a semicircle is to the square inscribed
in a quadrant of the same circle :: 8: 5.
62. Shew that if a triangle inscribed in a circle be isosceles,
having each of its sides double the base, the squares described upon
ON BOOK VI. THEOREMS.
357
the radius of the circle and one of the sides of the triangle shall be
to each other in the ratio of 4: 15.
63. If on the diameter of a semicircle, two equal semicircles be
described, and in the curvilinear space a circle be inscribed touching
the three semicircles, the radius of this circle shall be to the radius of
either less semicircle :: 2 : 3.
64. If a straight line be drawn through the points of bisection
of any two sides of a triangle, it will divide the triangle into two
parts which are to each other as 1 to 3.
65. If A₁, A be the areas of the squares inscribed in, and cir-
cumscribed about a given circle, A that of a regular octagon inscribed
in the same circle, shew that A₁ : A :: A : A2.
66. The diagonals of a trapezium, two of whose sides are parallel,
cut one another in the same ratio.
67. If from any two points within or without a parallelogram,
straight lines be drawn perpendicular to each of two adjacent sides and
intersecting each other, they form a parallelogram similar to the former.
68. ABCD is a trapezium of which the opposite sides AD, BC
are parallel, shew that
AC² ~ BD² : AB²~ DC² :: BC+AD : BC ~ AD.
69. The rectangle contained by two lines is a mean proportional
between their squares.
70. If AD is drawn bisecting the angle BAC, and AE is drawn
at right angles to AD; if BDCE be any straight line cutting AB,
AD, AC, AE respectively in the points B, C, D, E: shew that
1
+
1
2
BE¯ BD¯ BC'
71. If from a point to a straight line there be drawn four lines in
harmonical proportion, two of which terminate in its extremities; the
whole line and the three parts into which it is divided, will be in har-
monical proportion also.
72. If through a given point within a triangle lines be drawn from
the angles to the opposite sides, and the points of section be joined, the
three first drawn lines will be harmonically divided.
73. HKL, PQR are two triangles; prove that if the straight
lines HP, KQ, LR meet in one point, the intersections of KL, QR;
LH, RP; HK, PQ lie in a straight line.
74. Four lines AB, CD, EF, GH, drawn in any direction, inter-
sect in the same point P; then if from any point m in one of these
lines, another be drawn parallel to the next in order, cutting the remain-
ing two in p and q ; the ratio of mp: pq is the same in whichever line
the point m is taken.
75. If a straight line be harmonically divided, and the angle con-
tained by lines drawn from the extremity and one point, be bisected by
a line drawn from the other point; that line shall be at right angles to
the line which is drawn from the other extremity.
76. Through the point of intersection Ď of two straight lines
drawn from two angles of a triangle to bisect the opposite sides, a
straight line EDFG is made to pass, meeting two sides in E and F and
the remaining side produced in G: prove that
DE
DFDG'
1
1
1
358
GEOMETRICAL EXERCISES
77. From A, the lines AB, AC, AD, AE, are drawn intersecting
in B, C, D and E; and F, G, H and K; the lines BE and FK which are
inclined to each other at any angle, shew that if
BE: ED :: BC : CD,
then FK: KH :: FG : GH.
78. If two lines intersecting each other, are each terminated by
two unlimited lines given in position, the ratio of the rectangles con-
tained by their respective segments will be the same with the ratio
of the rectangles made by the segments of any other two lines inter-
secting each other which are similarly terminated, and are respectively
parallel to the former.
79. If from any point four straight lines be drawn, so that a
line parallel to one of them falling upon the other three may be bi-
sected by the middle one of these, then will the four lines cut any
line in harmonic proportion.
80. Let DACEB be points in a straight line such that
AC: CB: AD: DB, and AE= EB;
shew that rectangle DCE equals rectangle ACB.
81. APB is a quadrant, SPT a straight line touching it at P, PM
perpendicular to CA; prove that triangle SCT: triangle ACB:: triangle
ACB triangle CMP.
82. ABC is a triangle, right-angled at A, and having the angle
ABC twice the angle ACB, draw BD, bisecting the angle ABC, and
meeting AC in D; AE, DF are perpendiculars on BC, meeting it
respectively in the points E and F. Shew that
1
1
1
BE .DF˜ AE.BF˜ BE. AE
83. In the triangle ABC, AC=2.BC. If CD, CE respectively
bisect the angle C, and the exterior angle formed by producing AC;
prove that the triangles CBD, ACD, ABC, CDE, have their areas as
1, 2, 3, 4.
84. D, E, are the middle points of the sides CA, CB of a triangle;
join D and E, and draw AE, BD, intersecting in 0; then shall the
areas of the triangles DOE, EOB, BOA, be in geometrical progression.
85. If triangles AEF, ABC have a common angle A, triangle
ABC : triangle AĔF :: AB. AC: AE.AF.
86. If ABC be a right-angled triangle having the right angle at
B, and squares ABDE, CBFG, be described on the sides AB, CB; and
DE, FG be produced to meet AC produced in the points H and K,
the triangle ABC will be a mean proportional between the triangles
AEH, CGK.
87. If there be three equal rectangles contained by AB, AG;
CD, CH; EF, EK. Then if EK –CH=CH – AG, AB – CD will be
to CD– EF :: AB: EF. Also if this latter proportion hold, EK – CH
=CH-AG. Prove this, and explain how this is really a geometrical
proof of the common propositions of harmonic progression.
88. If about an equilateral triangle another be described, and
about this latter another, and so on. Prove that the areas of the tri-
angles are in geometrical progression.
ON BOOK VI. THEOREMS.
359
89. If a tangent to two circles be drawn cutting the straight line
which joins their centres, the chords are parallel which join the points
of contact and the points where the line through the centres cuts the
circumferences.
90. Given three unequal circles which do not intersect, and let
pairs of double tangents be drawn internally to each pair of them, the
three intersections will be in one right line.
91. If one chord in a circle bisect another, and tangents drawn
from the extremities of each be produced to meet, the line joining
their points of intersection will be parallel to the bisected chord.
92. If corresponding points at which the tangents are parallel, be
taken in the circumferences of two circles which cut each other, the
straight lines joining them will cut the line passing through both cen-
tres in the same point; and if ♪ be the distance of that point from
the centre of the circle whose radius is (r), d= where (a) is
the distance between the centres of the circles, and R the radius of the
other circle.
γα
Rr
93. If two lines AP, BP, drawn from two fixed points A, B, have
a constant ratio, the locus of P is a circle: prove this and find the radius
of the circle.
94. In any isosceles triangle having the angles at the base each
double of the vertical angle, if r, r' be the radii of the inscribed and
circumscribed circles, (a) either side, shew that
r.r' : §.a² :: √√5-1 : √5+3.
95. Let ACB be the diameter of any circle whose centre is C, P
any point in it, Q a point so taken that CP: CA :: CA: CQ; take M any
point in the circumference, join QM, AM, PM. Prove that AM always
bisects the angle QMP.
96. The rectangle contained by the sides of any triangle is to
the rectangle by the radii of the inscribed and circumscribed circles, as
twice the perimeter is to the base.
97. In any triangle, the rectangle contained by the excess of half
the perimeter above each of the two sides including an angle, is equal
to the rectangle contained by the radius of the inscribed circle, and
the radius of the circle which touches the base and the two sides pro-
duced.
98. If a straight line touch a circle, and a perpendicular be drawn
from the point of contact upon any diameter; and if from the extremi-
ties of the same diameter, and from the centre, perpendiculars be raised
upon the diameter and produced to meet the line which touches the circle,
these four perpendiculars are proportional.
99. A point is taken in the diameter of a circle, and another point
in the diameter produced, so that the radius is a mean proportional
between their distances from the centre: prove that the lines drawn
from these points to any, the same point in the circumference, are
always in the same ratio.
100. If from the extremity of a diameter of a circle tangents be
drawn, any other tangent to the circle terminated by them is so divided
at its point of contact that the radius of the circle is a mean proportional
between its segments.
101. If through any point in the arc of a quadrant whose radius is R,

360
GEOMETRICAL EXERCISES
two circles be drawn touching the bounding radii of the quadrant, and
r' be the radii of these circles: shew that rr' = R².
r₁
102. If a circle be inscribed in an equilateral triangle, another
similar triangle in the circle, another circle in the last triangle, and so
on for ever; the radius of the first circle shall be equal to the sum of
the radii of all the others.
103. Let the two circles, radii R, r, which touch (1) the three sides
of a triangle ABC, and (2) one side, BC, and the other two produced,
touch AB in D1, D2, AC in E₁, E2; shew that
BD₁. BD₂ = CE, . CE₂ = R. r.
104. ACB is the quadrant of a circle, AD, BD, lines drawn from
the extremities A and B, so that AD is always equal to CD. Prove
that the point D is in the circumference of a semicircle whose diameter
is to the diameter of the circle to which the quadrant belongs, as
1 to √2.
105. Let the two diameters AB, CD, of the circle ADBC be at
right angles to each other, draw any chord EF, join CE, CF, meeting
AB in G and H; prove that the triangles CGH and CEF are
similar.
106. Let P be a given point within a circle upon the radius AC,
and let a point Q be taken externally upon the same radius produced,
so that CP: CA :: CA : CQ; if from any point M in the circumference,
straight lines MP, MQ be drawn to the two points P and Q, these
straight lines will every where have the same ratio, or MP: MQ::
AP: AQ.
107. If any triangle ABC be inscribed in a circle, and at the points
A, B, C, tangents be drawn, and produced to meet the opposite sides
of the triangle produced in the points D, E, F ; these points will be in
a straight line.
108. Prove that there may be two, but not more than two, similar
triangles in the same segment of a circle.
109. If the rectangles under the segments of the diagonals of a
quadrilateral figure be equal, a circle may be described about it.
110. By help of VI. D. it may be shewn, that if three of the angular
points of a quadrilateral figure in a circle, be equally distant from each
other, then the distance of the fourth from the middle one equals the
sum of the distances from the other two.
111. ABCD is any trapezium inscribed in a circle. Its sides are
produced to meet in P and Q. Prove the following proportion ;
PD.DQ:AD.DC::PB.BQ: AB.BC.
112. If a straight line drawn from the vertex of an isosceles
triangle cutting the base, be produced to the circumference of a circle
described about the triangle, the rectangle contained by the whole line
produced and the part of it between the vertex and the base, is equal to
the square of either of the sides of the triangle.
113. A circle may be described about any quadrilateral figure, if,
when any two of its sides are produced to meet, the rectangle by one
side produced and the part produced, shall be equal to the rectangle by
the other side produced and the part produced.
114. In the figure to Prop. 47, Book 1, the points A, D, L, E, are
in the circumference of a circle.
115. If the line bisecting the vertical angle of a triangle be divided
ON BOOK VI. THEOREMS.
361
into parts which are to one another as the base to the sum of the sides,
the point of division is the centre of the inscribed circle.
116. In a plane triangle, if the line joining the centre of the cir-
cumscribed circle and the point of intersection of the perpendiculars, be
bisected; the point so determined is distant from the vertical angle,
two thirds of the line drawn from that angle to the bisection of the base.
117. If two circles touch each other externally and also touch a
straight line, the part of the line between the points of contact is a
mean proportional between the diameters of the circles.
118. If a perpendicular be drawn from the right angle to the hypo-
thenuse of a right-angled triangle, and circles be inscribed within the
two smaller triangles into which the given triangle is divided, their
diameters will be to each other as the sides containing the right angle.
119. If from a point without a circle there be drawn three straight
lines, two of which touch the circle, and the other cuts it, the line
which cuts the circle will be divided harmonically by the convex cir-
cumference, and the chord which joins the points of contact.
120. The perpendicular drawn from the centre of a circle on the
chord of any arc, is a mean proportional between half the radius and the
line made up of the radius and the perpendicular drawn from the centre
on the chord of double the arc.
121. About a triangle ABC describe a circle, and through the
points ABC draw tangents DAF, FBE, ECD, forming a triangle
DFE; shew that the rectangle DE, AF, is to the rectangle EC, DF,
in the duplicate ratio of AB to CB.
122. If two straight lines be drawn from the same point, and
divided into parts reciprocally proportional to the whole lines, a circle
may be described through the extremities and the points of division.
123. Let S be any point in the radius of a circle whose centre is
O and radius OA. In OA produced, take OL a third proportional to
the two OS, OA, draw any straight line LE to the circumference, and
join SE, AE; AE bisects the angle LES. Required a demonstration.
124. If a straight line AB be bisected in C, and upon AB and AC
semicircles be described, and from B a straight line be drawn touching
the inner semicircle and terminated by the other in D;
then BD: AC:: 4√2:3.
125. If through the vertex, and the extremities of the base of a tri-
angle, two circles be described, intersecting one another in the base or its
continuation, their diameters are proportional to the sides of the triangle.
126. A straight line ABCD is terminated and trisected at the
points where it meets the circumferences of two circles that cut each
other, and it is intersected in E by FG which joins their centres: shew
that AE: ED :: 2 EF + EG: 2EG + EF.
127. If a triangle be inscribed in a circle, and from its vertex, lines
be drawn parallel to the tangents at the extremities of its base, they will
cut off similar triangles.
128. If a semicircumference be divided into any number of equal
parts AB, BC, CD, DE, &c., and the straight lines AB, AC, AD, AE,
&c., be drawn, then AB: AC:: AC: AB+ AD and AC: AD + AB ::
AD: AC + AE.
129. If BEHN be a line cutting a circle whose centre is H, and points
in the circumference E and N; also if BFIK be a line touching it where
EF, NK, are perpendicular on BI; then BI²
EH²
=
BF.BK.
FI2
362
GEOMETRICAL EXERCISES
130. BEF is a circle inscribed in the triangle ABC touching the
sides BC, AC, AB, in D, E, F; HIK is a circle touching AB in K
and CB, CA, produced in H, I; in CH take CL = CA, and in CI take
CM = CB; then FK = AM and 4AF. FB = AL. MB.
131. If the sides of a triangle be in arithmetic progression, and a
and a₁ be the longest and shortest sides, and r and r, the radii of the
inscribed and circumscribed circles, 6rr₁ = ɑɑ₁.
132. ABC is a triangle inscribed in a circle; AB, AC, BC, pro-
duced, cut a line given in position, in points m, n, p, respectively. If
tmt tp, be the lengths of the lines drawn from m, n, p, touching the
circle, shew that tm. t. tp = An. Bm..Cp Am. Bp. Cn.
=
133. If two circles cut each other so that the circumference of one
may pass through the centre of the other, and from that centre, lines be
drawn cutting both circumferences, the segments of the lines inter-
cepted between the circumferences, will be to each other as the per-
pendiculars let fall from the intersections with the inner circumference
upon the line which joins the points of intersection of the circles.
134. If one point be taken in the radius of a circle and another in
the radius produced, such that the radius is a mean proportional between
their distances from the centre, and from the nearer point, a perpen-
dicular be erected from any point in which tangents are drawn to the
circle, the line joining the points of contact will always pass through
the farther point.
135. If A be the area of any triangle, prove that the area of a
triangle whose angular points divide the sides of the former in the ratio
of n to 1, is equal to
n² − n + 1
(n + 1)2
A.
136. If two chords in a circle intersect each other, and the two
segments of one chord have the same ratio to each other as the two
segments of the other chord; a straight line bisecting the angle con-
tained by the two segments which form corresponding terms of the
ratios, will pass through the centre of the circle.
137. The product of all the sides of any triangle is equal to four
times its area multiplied into the radius of its circumscribed circle.
138. If as in Euclid v1. 3, the vertical angle BAC of the triangle
BAC be bisected by AD, and BA be produced to meet CE drawn
parallel to AD in E; shew that AD will be a tangent to the circle
described about the triangle EAC.
139. If a triangle ABC be inscribed in a circle, and a straight line
BD be drawn parallel to the tangent at A, meeting AC or AC produced
in D, then will AD: AB:: AB: ÁC,
140. If a circle be inscribed in a right-angled triangle and another
be described touching the side opposite to the right angle and the pro-
duced parts of the other sides, shew that the rectangle under the radii
is equal to the triangle, and the sum of the radii equal to the sum of the
sides which contain the right angle.
141. ACB is an isosceles triangle inscribed in a semicircle, of
which the diameter is AB; from any point in AC produced, DE is
drawn perpendicular to AB, and cutting the semicircle in F, and CB
in G. Prove that EF is a mean proportional between EG and ED.
142. If two circles cut each other, and a common chord ABCDE
be drawn, meeting one circle in A and D, the other in B and E, and
ON BOOK VI. THEOREMS.
363
the line joining these points of intersection in C; then the square of
BD shall be to the square of AE as the rectangle BC, CD, is to the
rectangle AC, CE.
143. If a circle be inscribed in a triangle, and another be described
about it; the distance between their centres is a mean proportional
between the radius of the circumscribed circle, and its excess above
twice the radius of the inscribed circle.
144. The diagonals AC, BD, of a trapezium inscribed in a circle
cut each other at right angles in the point E; the rectangle AB.BC:
the rectangle AD.DC:: BE: ED.
145. If two circles touch each other internally in a point A, from
which a straight line ABC is drawn cutting the circles in B, C ; then if
another straight line drawn through either of these points perpendicular
to the line which joins the centres, cut the other circle in D, AD shall
be a mean proportional between AB and AC.
146. Two circles whose radii are as 2:3 touch each other in-
ternally; through the centre of the smaller circle a straight line is
drawn perpendicular to their common diameter; and from the points
where this straight line meets the circumference of the larger circle,
tangents are drawn to the smaller circle; shew that these tangents will
be perpendicular to each other.
147. If two circles cut each other in any two points, and from
either point diameters be drawn, the extremities of these diameters and
the other point of intersection are in the same straight line. Also, if
on the line joining the points of intersection, as a diameter, a circle be
described, and any line be drawn from one extremity of this diameter
cutting the two circles, the parts of this line intercepted between the
circumference of the circle, and the circumferences of the other two
circles shall be proportional to the perpendiculars drawn in these circles
from the other extremity of the diameter.
148. If two circles can be described so that each shall touch the
other and three sides of a quadrilateral figure, one fourth of the differ-
ence of the sums of the opposite sides is a mean proportional between
the radii. Express the area of the quadrilateral figure in terms of the
sides and the radii of the circles.
149. Two circles intersect in A and B; AD, AD', are diameters;
AC, AC', are chords, each of which touches the circle of which it is not
a chord; the line AEE' bisects the angle DAD' and cuts the circles in
E and E'; then the common tangent to the circles is a mean propor-
tional between the chords DE, D'E'; and their common chord AB'is
a mean proportional between the chords BC, BC'.
150. The sides containing a given angle are in a given ratio and
the vertex is fixed; shew that if the extremity of one of the sides
moves in a given line, so does the extremity of the other.
151. Two equal circles are drawn intersecting in the points A and
B, a third circle is drawn with centre A and any radius not greater than
AB intersecting the former circles in D and C. Shew that the three
points B, C, D, lie in one and the same straight line.
152. Two equal circles are drawn intersecting in the points A and
B, a third circle is drawn with centre A and any radius not greater than
AB intersecting the former circles in D and C; join BA and upon it as
a diameter, describe a semicircle BEA; then if from the point B any
line BFEG be drawn cutting ABD in F, AEB in E, and ACB in Ğ,
FE equals EG.
364
GEOMETRICAL EXERCISES
153. S is a fixed point in OA the radius of a circle, centre 0; in
OA produced, OL is taken a third proportional to OS and OA, and
from the point L, any line LPQ is drawn cutting the circle in P and Q,
and if the line LPQ cut the line drawn from S at right angles to AO in
the point E, prove that LQ is divided harmonically in the points E
and P, that is, LQ : LP :: QE : EP.
154. S is a fixed point in OA the radius of a circle, centre 0; in
OA produced, OL is taken a third proportional to OS and OA, and
from the point L any line LPQ is drawn cutting the circle in P and Q;
it is required to demonstrate the following properties:
(1) If PS be joined, the ratio of LP to PS is a constant ratio.
(2) If QS be also joined, the rectangle contained by PS, SQ is a
constant rectangle.
(3) If through S a line be drawn at right angles to AO, tangents
to the circle at P and Q will cut this line in one and the same point.
N.B. By means of the first property, it may be shewn that the
following proposition inserted in Simson's edition of the "Data" is not
true. (Prop. 80.) "If the sides about an angle of a triangle have a
given ratio to each other, and if the perpendicular drawn from that
angle to the base has a given ratio to the base; the triangle is given int
species."
155. A circle, a straight line, and a point being given in position,
required a point in the line, such that a line drawn from it to the
given point may be equal to a line drawn from it touching the circle.
What must be the relation among the data, that the problem may
become porismatic, i. e. admit of innumerable solutions?"
156. If a point (C′) be taken in any one (as AB) of three indefinite
straight lines that intersect in A, B, and C; and lines (as C'B'A') be
drawn from C' cutting AC, BC (as in B', A'); then all the intersections
of each pair of lines (as BB', AA') drawn from B and A to the points
of section (B', A'), lie in a line that passes through C.
157. In the sides AB, AC of a given triangle ABC, take two
points M, N, and in the line joining them take a point P, such that
MB AN MP
AM¯NC PN
:
prove that if PB, PC, be joined, the triangle BPC is twice the triangle
AMN; and that the circle described about the triangle AMN will
always pass through a fixed point.
158. Lines are drawn from the angular points A, B, C of a
triangle through any point, meeting the opposite sides in a, b, c; a
circle is described through these three points cutting the same sides
in a', b', c'. Shew that Aa', Bb', Cc' meet in one point.
Assuming that Ab. Bc. Ca - Ba. Cb. Ac, and conversely, that
when this relation holds, the lines pass through one point.
159. In the figure annexed to the 47th Proposition, Book 1, if
the points G, H; F, K; be joined, the lines GÌ, FK, and BC, if
produced, meet in one and the same point.
160. If any point D be taken in the interior of a triangle, and AD,
BD, and CD be respectively produced to meet BC, AC, and AB in E,
F, G: then AF. BĞ.CE = AG.BE.CF.
161. If a straight line intersect the two sides AC, BC of a plane
triangle in the points b, a, and the base AB produced in c; prove that
Ab × Bc × Ca = Ac× Ba × Cb.
ON BOOK VI, THEOREMS.
365

162. If perpendiculars be raised upon the middle points of the
sides of a triangle, and respectively equal to half of those sides, and the
extremities of those perpendiculars be joined; the sum of the squares
of these last lines is equal to the sum of the squares of the sides of the
triangle together with six times its area.
163. If R be the radius of the circle inscribed in a right-angled
triangle ABC, right-angled at 4; and a perpendicular be let fall from
A on the hypothenuse BC, and if r, r' be the radii of the circles inscribed
in the triangles ADB, ACD: prove that r² + r²² = R².
12
164. If from the extremities of the base of a triangle, perpen-
diculars be let fall on the opposite sides, and likewise straight lines
drawn to bisect the same, the intersection of the perpendiculars, that of
the bisecting lines, and the centre of the circumscribing circle will be
in the same straight line.
165. If a circle be inscribed in any plane triangle, and a right line
drawn from the vertex to the point where the circle touches its base:
prove that the middle point of that line, the centre of the circle, and
the middle point of the base of the triangle, are in the same straight line.
166. A circle is inscribed in a triangle; and a second triangle is
formed, whose sides are equal to the distances of the points of contact
from the angles of the triangle. If r be the radius of the circle inscribed
in the first triangle; and p, p' the radii of the inscribed and circum-
scribed circles of the second triangle, then will r² = 2pp'.
167. The side of an equilateral triangle inscribed in a circle: side
of a square inscribed in the same :: √3: √2; and the area of the
triangle: area of a square :: 3√3 : 4√4.
168. If two equilateral polygons be drawn, one within, and the
other about a circle, and another of half the number of sides be in-
scribed in the circle, this shall be a mean proportional between the
other two.
169. One of the sides of a rectilinear figure is divided into any
number of parts, and figures described upon the parts, similar and
similarly situated to the rectilinear figure, shew that the sum of the
peripheries of the smaller figures is equal to the periphery of the larger;
and if the parts be equal and ʼn in number, that the areas of the
smaller figures are together equal to th of the area of the larger.
1
n
170. Area of inscribed pentagon: area of circumscribing pentagon
:: the square of the radius of the circle inscribed within the greater
pentagon the square of the radius of circle circumscribing it.
171. If the sides of any regular figure be bisected and each two
adjacent points of bisection be joined, a similar figure will be formed,
and its area will be to that portion of the area of the original figure
which is without it as the square on the diameter of the circle circum-
scribing it is to the square on one of its sides.
172. A regular hexagon inscribed in a circle is a mean proportional
between an inscribed and circumscribed equilateral triangle.
173. If two regular polygons P and Q of the same number of
sides be, the one inscribed in a circle, and the other circumscribed about
it, and also a polygon of twice the number of sides R be inscribed
in the same circle, then R shall be a geometrical mean between P
and Q.
366
GEOMETRICAL EXERCISES
GEOMETRICAL EXERCISES ON BOOK XI.
THEOREM I.
Define the projection of a straight line on a plane; and prove that if
a straight line be perpendicular to a plane, its projection on any other
plane, produced if necessary, will cut the common intersection of the two
planes at right angles.
Let AB be any plane and CEF another plane intersecting the
former at any angle in the line EF; and let the line GH be perpen-
dicular to the plane CEF.
H C

A
E
F
L
G
B
Draw GK, HL perpendicular on the plane AB, and join LK,
then LK is the projection of the line GH on the plane AB;
produce EF, to meet KL in the point L;
•
then EF, the intersection of the two planes, is perpendicular to
LK, the projection of the line GH on the plane AB.
Because the line GH is perpendicular to the plane CEF,
every plane passing through GH, and therefore the projecting
plane GHKL is perpendicular to the plane CEF;
but the projecting plane GHLK is perpendicular to the plane
AB; (constr.)
hence the planes CEF, and AB are each perpendicular to the third
plane GHLK;
therefore EF, the intersection of the planes AB, CEF, is perpen-
dicular to that plane;
and consequently, EF is perpendicular to every straight line which
meets it in that plane;
but EF meets LK in that plane.
Wherefore, EF is perpendicular to KL, the projection of GH on
the plane AB.
ON BOOK XI.
367
THEOREM II.
Prove that four times the square described upon the diagonal of a
rectangular parallelopiped, is equal to the sum of the squares described
on the diagonals of the parallelograms containing the parallelopiped.
Let AD be any rectangular parallelopiped; and AD, BG two diago-
nals intersecting one another; also AG, BD, the diagonals of the two
opposite faces HF, CE.
H
B
A
G
F

E
C
D
Then it may be shewn that the diagonals AD, BG, are equal; as
also the diagonals which join CF and HE: and that the four diagonals
of the parallelopiped are equal to one another.
The diagonals AG, BD of the two opposite faces HF, CE are equal
to one another: also the diagonals of the remaining pairs of the opposite
faces are respectively equal.
And since AB is perpendicular to the plane CE, it is perpendicular
to every straight line which meets it in that plane,
therefore AB is perpendicular to BD,
and consequently ABD is a right-angled triangle.
Similarly, GDB is a right-angled triangle.
And the square of AD is equal to the squares of AB, BD, (1. 47.)
also the square of BD is equal to the squares of BC, CD,
therefore the square of AD is equal to the squares of AB, BC, CD;
similarly the square of BG or of AD is equal to the squares of AB,
BC, CD.
Wherefore the squares of AD and BG, or twice the square of AD,
is equal to the squares of AB, BC, CD, AB, BC, CD;
but the squares of BC, CD are equal to the square of BD, the
diagonal of the face CE;
similarly, the squares of AB, BC are equal to the square of the
diagonal of the face HB;
also the squares of AB, CD, are equal to the square of the diagonal
of the face BF; for CD is equal to BE.
Hence, double the square of AD is equal to the sum of the squares of
the diagonals of the three faces HF, HB, BC.
In a similar manner, it may be shewn, that double the square of the
diagonal is equal to the sums of the squares of the diagonals of the three
faces opposite to HF, HB, BC.
Wherefore, four times the square of the diagonal of the parallelopi-
ped is equal to the sum of the squares of the diagonals of the six faces.

368
GEOMETRICAL EXERCISES
•
PROBLEMS.
1. Two of the three plane angles which form a solid angle, and
also the inclination of their planes being given, to find the third plane
angle.
2. Having three points given in a plane, find a point above the
plane equidistant from them.
3. To describe a circle which shall touch two given planes, and
pass through a given point.
4. Two triangles have a common base, and their vertices are in a
straight line perpendicular to the plane of the one; there are given the
vertical angle of the other, the angles made by each of its sides with the
plane of the first and the distance of the vertices of the two triangles,
to find the common base.
5. Find the distance of a given point from a given line in space.
6. Find a point in a given straight line such that the sums of its
distances from two given points (not in the same plane with the given
straight line) may be the least possible.
7. Draw a line perpendicular to two lines which are not in the
same plane.
8.
Two planes being given perpendicular to each other draw a
third perpendicular to both.
9. Required the perpendicular from the vertex upon the base of
a triangular pyramid, all the sides of which are equilateral triangles of
a given area.
10. Given the lengths and positions of two straight lines which
do not meet when produced and are not parallel; form a parallelopiped
of which these two lines shall be two of the edges.
11. How many triangular pyramids may be formed whose edges
are six given straight lines, of which the sum of any three will form a
triangle?
12. Bisect a triangular pyramid by a plane passing through one
of its angles, and cutting one of its sides in a given direction.
13. Define a cube. When this solid is cut by a plane obliquely
to any of its sides, the section will be an oblong, always greater than the
side, if made by cutting opposite sides. Draw a plane cutting two
adjacent sides, so that the section shall be equal and similar to the side.
THEOREMS.
3. If two straight lines are parallel, the common section of any
two planes passing through them is parallel to either.
4. If three limited straight lines be parallel, and planes pass
through each two of them, and the extremities be joined, a prism will
be formed, the ends of which will be parallel if the straight lines be
equal.
ON BOOK XI. THEOREMS.
369
5. If two straight lines be parallel, and one of them inclined at
any angle to a plane, the other also shall be inclined at the same angle
to the same plane.
6. Parallel planes are cut by parallel straight lines at the same
angle.
7. If from a point A above a plane, straight lines AB, AC be
drawn, meeting it in B and C, of which AB is perpendicular to the
plane, and AC perpendicular to a straight line DC in that plane, and
CB be joined, CB shall be perpendicular to DC.
8. Three straight lines not in the same plane, but parallel to and
equidistant from each other, are intersected by a plane, and the points
of intersection joined; shew when the triangle thus formed will be
equilateral and when isosceles.
9. If two straight lines in space be parallel, their projections on
any plane will be parallel.
10. Of all the angles, which a straight line makes with any
straight lines drawn in a given plane to meet it, the least is that which
measures the inclination of the line to the plane.
11. Three parallel straight lines are cut by parallel planes, and the
points of intersection joined, the figures so formed are all similar and
equal.
12. If a straight line be at right angles to a plane, the intersection
of the perpendiculars let fall from the several points of that line, on
another plane, is a straight line which makes right angles with the com-
mon section of the two planes.
13. If two straight lines be cut by four parallel planes, the two
segments, intercepted by the first and second planes, have the same ratio
to each other as the two segments intercepted by the third and fourth
planes.
14. If there be two straight lines which are not parallel, but which
do not meet, though produced ever so far both ways, shew, that two
parallel planes may be determined so as to pass, the one through the
one line, the other through the other; and that the perpendicular dis-
tance of these planes is the shortest distance of any point that can be
taken in the one line from any point taken in the other.
15. If from a point above the plane of a circle straight lines be
drawn to the circumference there will be only two of them equal in
length, and they will be equidistant from the shortest and longest, and
on opposite sides. What is the exception to this proposition?
16. Two planes intersect each other, and from any point in one
of them a line is drawn perpendicular to the other, and also another line
perpendicular to the line of intersection of both; shew that the plane
which passes through these two lines is perpendicular to the line of
intersection of the plane.
17. Two points are taken on a wall and joined by a line which
passes round a corner of the wall. This line is the shortest when its
parts make equal angles with the edge at which the parts of the wall
meet.
18. If four straight lines in two parallel planes be drawn, two
from one point and two from another, and making equal angles with
another plane perpendicular to these two, then if the first and third be
parallel, the second and fourth will be likewise.
19. If, round a line which is drawn from a point in the common
24
370
GEOMETRICAL EXERCISES
section of two planes at right angles to one of them, a third plane be
made to revolve, shew that the plane angle made by the three planes is
then the greatest, when the revolving plane is perpendicular to each of
the two fixed planes.
20. The content of a rectangular parallelopipedon whose length is
any multiple of the breadth and breadth the same multiple of the depth
is the same as that of the cube whose edge is the breadth.
21. A rectangular parallelopiped is bisected by all the planes
drawn through the axis of it.
22. If a straight line be divided into two parts, the cube of the
whole line is equal to the cubes of the two parts together with thrice the
right parallelopiped contained by their rectangle and the whole line.
23. If a right-angled triangular prism be cut by a plane, the volume
of the truncated part is equal to a prism of the same base and of height
equal to one third of the sum of the three edges.
24. If a four-sided solid be cut off from a given cube, by a plane
passing through the three sides which contain any one of its solid angles,
the square of the number of standard units in the base of this solid, shall
be equal to the sum of the squares of the numbers of similar units, con-
tained in each of its sides.
25. In an oblique parallelopiped the sum of the squares of the
four diagonals equals the sum of the squares of the twelve edges.
26. If any point be taken within a given cube, the square de-
scribed on its distance from the summit of any of the solid angles of the
cube, is equal to the sum of the squares described on its several perpen-
dicular distances from the three sides containing that angle.
27. Shew that a cube may be cut by a plane, so that the section
shall be a square greater in area than the face of the cube in the propor-
tion of 9 to 8.
28. Why cannot a sheet of paper be made to represent the vertex
of a pyramid, without folding?
29. A, B are two fixed points in space, and CD a constant length
of a given straight line; prove that the pyramid formed by joining the
four points A, B, C, D is always of the same magnitude, on whatever
part of the given line CD be measured.
30. Define similar rectilineal figures; and prove, that if a pyramid
with a polygon for its base be cut by a plane parallel to the base, the
section will be a polygon similar to the base.
ON BOOK XII.
371
GEOMETRICAL EXERCISES ON BOOK XII
THEOREM I.
If semicircles ADB, BEC be described on the sides AB, BC of a
right-angled triangle, and on the hypothenuse another semicircle AFBGC
be described, passing through the vertex B; the lunes AFBD and BGCE
are together equal to the triangle ABC.
F

B
G
D
F
A
C
It has been demonstrated (XII. 2.) that the areas of circles are to one
another as the squares of their diameters; it follows also that semicircles
will be to each other in the same proportion.
Therefore the semicircle ADB is to the semicircle ABC, as the square
of AB is to the square of AC,
and the semicircle CEB is to the semicircle ABC as the square of
BC is to the square of AC,
hence the semicircles ADB, CEB, are to the semicircle ABC as
the squares of AB, BC are to the square of AC;
but the squares of AB, BC are equal to the square of AC: (1.47.)
therefore the semicircles ADB, CEB are equal to the semicircle
ABC. (v. 14.)
From these equals take the segments AFB, BGC of the semicircle
on AC, and the remainders are equal,
that is, the lunes AFBD, BGCE are equal to the triangle BAC.
THEOREM II.
If on any two segments of the diameter of a semicircle semicircles be
described, all towards the same parts, the area included between the three
circumferences (called äpßnλos) will be equal to the area of a circle, the
diameter of which is a mean proportional between the segments. (Archi-
medis Lemm. Prop. 4.)
Let ABC be a semicircle whose diameter is AB,
and let AB be divided into any two parts in D,
and on AD, DC let two semicircles be described on the same side;
also let DB be drawn perpendicular to AC.
;
24-2
372
GEOMETRICAL EXERCISES
Then the area contained between the three semicircles, is equal to
the area of the circle whose diameter is BD.
B

A
D
C
Since AC is divided into two parts in C,
the square of AC is equal to the squares of AD, DC, and twice the
rectangle AD, DC; (II. 4.)
and since BD is a mean proportional between AD, DC;
the rectangle AD, DC is equal to the square of DB, (vi. 17.)
therefore the square of AC is equal to the squares of AD, DC, and
twice the square of DB.
But circles are to one another as the squares of their diameters or
radii, (XII. 2.)
therefore the circle whose diameter is AC, is equal to the circles
whose diameters are AD, DC, and double the circle whose diameter
is BD;
wherefore the semicircle whose diameter is AC is equal to the circle
whose diameter is BD, together with the two semicircles whose
diameters are AD and DC:
if the two semicircles whose diameters are AD and DC be taken
from these equals,
therefore the figure comprised between the three semi-circumferences
is equal to the circle whose diameter is DB.
THEOREM III.
Every section of a sphere by a plane is a circle.
If the plane pass through the centre of the sphere, it is manifest that
the section is a circle, having the same diameter as the generating semi-
circle.
But if the cutting plane does not pass through the centre,
let AEB be any other section of the sphere made by a plane not
passing through the centre of the sphere.
H

A
B
E
K
Take the centre C, and draw the diameter HCK perpendicular to
the section AEB, and meeting it in D;
draw AB passing through D, and join AC;
take E, F, any other points in the line AEB, and join CE, DE;
CF, DF.
ON BOOK XII, PROBLEMS.
373
Then since CD is perpendicular to the plane AEB, it is perpen-
dicular to every straight line which meets it in that plane,
therefore the angles CDA, CDE, CDF are right angles,
and CA, CF, CE, being lines drawn from C, the centre of the
sphere to points in the surface, are therefore equal to one
another.
Hence, in the right-angled triangles, CDA, CDF, CDE;
the square of DA is equal to the difference of the squares of CA
and CD;
and the square of DF is equal to the difference of the squares of
CF, and CD;
also the square of DE is equal to the difference of the squares of
CE and CD:
therefore the squares of DA, DF, DE are equal to one another;
and therefore the lines DA, DF, DE are equal to one another.
But DA, DF, DE are three equal lines drawn from the same point
D, in the same plane,
hence the points A, F, E lie in the circumference of a circle of
which D is the centre.
THEOREM IV.
There can be only five regular solids.
If the faces be equilateral triangles. The angle of an equilateral tri-
angle is one third of two right angles; and six angles, each equal to the
angle of an equilateral triangle, are equal to four right angles; and
therefore a number of such angles less than six, but not less than
three are necessary to form a solid angle. Hence there cannot be more
than three regular figures whose faces are equal and equilateral tri-
angles.
If the faces be squares. Since four angles, each equal to a right
angle, can fill up space round a point in a plane. A solid angle may be
formed with three right angles, but not with a number greater or less
than three. Hence, there cannot be more than one regular solid figure
whose faces are equal squares.
If the faces be equal and regular pentagons. Since each angle of a
regular pentagon is a right angle and a fifth of a right angle: the
magnitude of three such angles being less than four right angles, may
form a solid angle, but four, or more than four, cannot form a solid angle.
Hence, there cannot be more than one regular figure whose faces are
equal and regular pentagons.
If the faces be equal and regular hexagons, heptagons, octagons, or
any other regular figures; it may be shewn that no number of them
can form a solid angle.
Wherefore there cannot be more than five regular solid figures,
of which, there are three, whose faces are equal and equilateral trian-
gles; one, whose faces are equal squares; and one, whose faces are
equal and regular pentagons.
374
GEOMETRICAL EXERCISES
PROBLEMS.
1. In what sense is it said that the circle does not admit of
quadrature? Describe generally the process by which Archimedes ob-
tained a first approximation to the ratio of the circumference of a circle
to its diameter.
2. Given a circle traced upon a plane, describe another whose
area is exactly twice as great as that of the former.
3. ABC is a circle of given radius, describe another concentric
circle abc whose area shall be equal to 3 times the area of ABC.
4. Divide a circle into any number of equal parts by means of
concentric circles.
5. To divide a circle into any number of equal parts, the peri-
meters of which shall be equal to the circumference of the circle.
6. Two circles touch each other internally, and the area of the
lune cut out of the larger is equal to twice the area of the smaller circle.
Required the ratio of the diameters of these circles.
7. If on one of the radii of a quadrant a semicircle be described;
and on the other, another semicircle so described as to touch the former
and the quadrantal arc; compare the area of the quadrant with the
area of the circle described in the figure bounded by the three curves.
8. The centres of three circles A, B, and C are in the same right
line, B and C touch A internally, and each other externally; P, Q being
the points where A is touched by B, C respectively: to find a point R
on A such that the portion of the lune PR intercepted between B and
A may be equal to the portion of QR between C and A.
9. The diameter of a circle is divided into two parts, upon each
of which as diameters circles are described; when the remaining area
of the great circle is equal to that of one of these two circles, find the
ratio which the parts of the diameter bear to one another.
10. Describe a sphere about a given regular tetrahedron ; and the
edge of the tetrahedron being 1, find the radius of the sphere.
11. Having given an irregular fragment, which contains a portion
of spherical surface: shew how the radius of the sphere, to which the
fragment belongs, may be practically determined.
12. Let three given spheres be placed on a horizontal plane in
mutual contact with each other; find the sides of the triangle formed
by joining the points in which the spheres touch the plane.
13. Construct the five regular solids.
14. Having given six straight lines, of which each is less than the
sum of any two, determine how many tetrahedrons can be formed, of
which these straight lines are the edges.
15. Inscribe a sphere within a tetrahedron.
16. Find the dihedral angle contained by two adjacent faces of a
regular octahedron: and find its solidity.
17. Shew how to find the content of a pyramid, whatever be the
figure of its base, the altitude and area of the base being given.
18. A pyramid of triangular base is composed of ten spherical
balls of given radius; the base is composed of six, the next layer of
three, and the remaining one is placed upon them. Find the distance
of the upper ball from the ground.
ON BOOK XII. THEOREMS.
375
THEOREMS.
5. The area of a circle is equal to half the rectangle contained by
two straight lines which are equal to its circumference and radius.
6. Compare the angle in a segment, which is one fourth part of
a circle, with a right angle.
7. In different circles the radii which bound equal sectors con-
tain angles reciprocally proportional to their circles.
8. Prove that the sectors of two different circles are equal, when
their angles are inversely as the squares of the radii.
9. If the diagonals of a quadrilateral inscribed in a circle cut each
other at right angles, and circles be described on the sides; prove that
the sum of two opposite circles will be equal to the sum of the other two.
10. If circles be inscribed in the triangles formed by drawing
the altitude of a triangle right-angled at the vertex, the circles and the
triangles are proportional.
11. Two straight lines are inclined to each other at a given angle,
find the area of all the circles which can be described touching each
other and the two given lines, the position of the centre of the last circle
being given.
12. If two chords of a circle intersect each other either within or
without the circle at right angles; and if on these segments as diame-
ters, circles be described, the areas of these four circles are together
equal to that of the original circle.
13. Shew that the semicircles described on the diagonals of a
right-angled parallelogram together equal the sum of the semicircles
described on the sides.
14. The diameter of a semicircle is divided into any number of
parts, and on these parts semicircles are described. Shew that their
circumferences are together equal to that of the given semicircle.
15. If two circles be so placed that their planes are parallel, and
the straight line which joins their centres perpendicular to the plane of
each; the straight lines which join the opposite extremities of any pair
of parallel diameters will all intersect the straight line which joins the
centres in the same point: if the circles be equal, that point will be the
bisection of the aforesaid straight line.
16. If through a point without the plane of a circle straight lines
be drawn to its circumference and a plane be drawn parallel to the
circle on either side of the point, the points of intersection of the lines
with the plane will be in a circle, and the area of this and of the first
circle will be as the squares of their distances from the point given.
17. If the arc of a semicircle be trisected, and from the points of
section lines be drawn to either extremity of the diameter, the difference
of the two segments thus made will be equal to the sector which stands
on either of the arcs.
18. The centres of three circles A, B, and C are in the same
right line, B and C touch A internally, and each other externally.
Shew that the portion of the area of A, which is outside B and C, is
equal to the area of the semicircle described on the chord of A which
touches B and C at their point of contact.
19. If three equal circles intersect, so that each of the circum-
ferences pass through the centres of the other two, the spaces bounded
by the circumferences intercepted will be all equal.

376
GEOMETRICAL EXERCISES
20. Take any three points, A, B, C in the circumference of a
circle. Join AB, BC, AC, and draw AD. AE parallel to the tangents
at B and C, and meeting BC produced if necessary in D and E; and
prove that the segments BD and EC are to each other in the duplicate
ratio of AB to AC.
21. If the diameter AB of a circle be divided into an odd number
th
(n) of equal parts, and C and D be the
n
th
("-1) and ("+1)" divisions;
2
and AEC, AFD, CGB, DHB be semicircles: shew that the perimeter
of the figure AECGBHDF is equal to that of the circle, and its area an
part of the area of the circle.
Nth
22. The circle inscribed in a square is equal to four equal circles
touching one another and the sides of that square internally.
23. If AB be a circular arc, centre O, and AD be drawn perpen-
dicular to BO, and the arc AC taken equal to AD, then the sector BOC
equals the segment ACB.
24. Let AB and DC be two diameters of a given circle, at right
angles to each other; AEB a circular arc described with radius DB or
DĂ; prove that the area of the lune AEBC = area of triangle ADB.
25. If two points B, D, be taken at equal distances from the ends
of the arc of a quadrant, and perpendiculars BG, DH be drawn to the
extreme radius; the space BGHD shall be equal to the sector BOD.
26. ABC is an isosceles right-angled triangle. On BC is described
a semicircle BDEC, and BFC is a circle whose radius is AB and centre
A. The segment BCF is equal to the segments BAD, ACE.
27. If a semicircle be described on the hypothenuse AB of a
right-angled triangle ABC, and from the centre E, the radius ED be
drawn at right angles to AB, shew that the difference of the segments
on the two sides equals twice the sector CED.
28. If semicircles be described upon the sides of a right-angled
triangle on the interior, the difference between the sum of the circular
segments thus standing upon the exterior of the sides and segments of
the base equals the space intercepted by the circumferences described
on the sides.
29. Semicircles are described upon the radii CA, CB of a qua-
drant, and intersect each other in a point D, shew (a) That the points
B, D, A are in one straight line. (b) That the area common to both semi-
circles is equal to the area without them. (c) That the remaining areas
of the two semicircles are equal, each is one fourth of the square on AC.
30. If two parallel planes cut a sphere so that the sections are
equal, they are equidistant from the centre.
31. Shew that all lines drawn from an external point to touch a
sphere are equal to one another; and thence prove that if a tetrahedron
can have a sphere inscribed in it touching its six edges, the sum of
every two opposite edges is the same.
32. If two equal circles cut one another in the diameter, and
a plane cut them perpendicularly to the same diameter, the points of
section of this plane with the circumferences, are in a circle.
33. In any polyhedron having different faces, some with an even,
some with an odd number of sides: shew that the number of those
faces which have an odd number of sides is necessarily an even number.
34. The angles of inclination of the faces of a regular tetrahedron
and of a regular octahedron are supplementary to each other.
|
INDEX
TO THE
PROBLEMS AND THEOREMS
IN THE
GEOMETRICAL EXERCISES.
ABBREVIATIONS.
Senate House Examination for Honors. S. H.
Smith's Mathematical Prizes. S. P.
Bell's University Scholarships. B. S.
St Peter's College. Pet.
Clare Hall. Cla.
Pembroke College. Pem.
Gonville and Caius College. Cai.
Trinity Hall. T. H.
Corpus Christi College. C. C.
King's College. Ki.
Queens' College. Qu.
St Catharine's Hall. Cath.
Jesus College. Jes.
Christ's College. Chr.
St John's College. Joh.
Magdalene College. Mag.
Trinity College. Trin.
Emmanuel College. Emm.
Sidney Sussex College. Sid.
Downing College. Down.
In the years the centuries are omitted and
the places are supplied by a comma prefixed,
thus,45 means 1845.
PROBLEMS ON BOOK I, p. 297, &c.
1 Sid.,30.,43. Qu.,34.
Trin.,40.
2 Emm.,22.,35. Sid.,30.
Trin. 37.
3 S. H. 17. Trin.,24.
,37. Qu.,25. Emm.,27.
Cath.,29. Pem. 39.
Sid. ,40. Chr. ,45.
4 Emm.,21. Qu. ,23.
,40.,42. Trin. ,26.,27.
,29. C. C.,30.
Pem.,32.,38.
5 Qu.,26.
6 Qu.,28.
7 Qu.,30.
8 Ki.,45. Cath.,30.
9 Trin.,32.,37.
10 Qu.,31. Cath. 35.
Emm.,35. Sid.,38.
B. S.,40. Trin. ,27.
11 Trin.,34.
12 Trin.,44.
.13 Qu.,32. Jes. ,36.
14 Chr. 42.
15 Qu.,19.
16 Trin.,38.
17 Trin.,36.
18 Trin.,43.
Jes.,36.
19 Trin.,28.
20 Mag.,35.
21 Cai.,44.
22 Cai.,33.
23 Emm.,24.
24 Trin.,29.
25 Emm.,22.
26 Jes.,28. Sid.,36.
27 Sid.,45.
28 S. H.,33.
29 Pet.,25.
30 Qu.,19.
31 Chr.,39.
32 Cai.,33. Qu.,33.
33 Sid. 30. Qu.,30.,35.
,40. Chr.,34. Emm.,44.
34 Cai.,31.
35 Qu.,33. S. H.,43.
36 Qu.,24.
37 Qu.,19.,37. Emm.,25.
Mag. ,29.,32. Cai.,34.
Trin.,37.,38. Pet. ,44.
38 Joh.,16. Qu.,30.
Pem. ,33.
39 Qu.,29.,37.,26.
Trin. ,27.,33.,36.,40.
Chr.,44. Pem. ,45.
40 Qu.,39.
41 Pet.,33.
42 Pet.,27.
43 Emm.,33. Mag.,45.
44 Emm.,32.
45 Joh.,31.
46 Pet. ,45.
47 Cla.,36.
48 Pet.,36.
49 Joh. ,15. Jes.,19.
50 Joh.,13.
51 Trin. ,40.
52 Qu.,31. Sid.,36.
53 Qu. 19.
54 Qu.,24.
55 Joh. ,26.
56 Sid.,35.
57 Joh.,26.
58 Trin. ,33.
59 Trin.,42.
60 Trin.,35.,45.
61 Emm.,24. Cai.,38.
Mag.,42.
62 Trin.,25. Chr.,35.,39.
63 S. H.,20. Joh.,38.
64 Qu.,34. Emm.,39.
65 Qu.,22.
66 Trin.,23.
67 Trin.,28.
378
INDEX.
#
1 Trin.,40.
2 Mag.,38.
3 Emm.,34.
4 Cath. 31.
5 Cai.,41.
6 Trin.,19. Qu.,19.,31.
Emm.,33.,42.
Mag. 34. Pem.,36.
Pet.,44.
7 Trin.,31.
Sid.,43.
8 Chr.,28. Pem. ,42.
9 Emm.,25.
10 Trin. ,26.
11 S. H.,36.
12 Pem.,29.
13 Mag.,39.
14 Qu. ,32.
15 Joh.,26. Qu.,31.
16 Trin.,26.
17 Jes.,42.
18 Trin. ,40.
19 Mag.,40.
20 Trin.,29.
23 Emm.,32. Qu.,39.
21 Trin.,25.
22 Joh.,30.
24 Cai.,29.
25 Cath.,33.
27 Jes. ,34.
28 Mag.,37.
26 Mag.,34.
THEOREMS, p. 300, &c.
31 Trin. 37.
29 Mag.,36.
30 Trin.,37.
32 Qu.,37.
33 Qu.,37.
34 Jes.,39.
35 Qu.,37.
36 Trin.,41.
37 Emm.,31. Chr.,38.
38 Cai.,42.,45.
39 S. H..02.,10. Joh.,15.
Jes.,32. Chr.,34.
Mag. 34. Trin. ,37.
Joh.,37.
40 S. H.,35.
41 Jes.,41.
42 C. C.,45.
43 Pet.,24.
44 Pet. 38. Chr.,39.
45 S. H.,19. Trin. ,29.
Qu,,35.
46 S. H.,36.
47 Cla.,36.
Pem. 44.
48 Emm.,21. Trin. ,44.
49 Qu.,30.
50 Cai.,36.
51 Trin.,37.
52 Trin.,44.
53 Emm.,30. Cath.,34.
54 Mag.,44.
55 Chr.,43.
56 S. H.,14.
57 Cath.,31.
58 Qu.,38.
59 Chr.,38.
60 Joh.,13.
61 Pem.,31.
62 Pet.,45.
63 S. H.,04.
Chr. 29.
C. C.,23.
Cath.,35.
Pet. 36. Trin. 37.
Qu.,39. Cai. ,40.
64 S. H.,18.,03. Trin.,25.
,44. Cla.,31.,36.
65 Qu.,25. Trin. ,25.,38.
Pet. 39. Pem.,42.
66 Chr.,26.,41. Joh. ,31.
Pet. 38. Trin. ,39.
67 Cai.,40.
68 Cai.,31.
69 S. H. ,04.
H.,04.
Emm.,39.
70 Trin.,21.
Cai. 34.
Cai.,34.
71 Joh.,15. Trin.,19.
Emm. ,32. Qu.,35.
C. C.,36. Mag. ,39.
72 Cai.,37.
73 Jes.,20.
Cath. ,35.
74 Trin. ,40.
Qu.,32.
75 Pet. ,32.,35.
76 Jes.,35.
1 Emm. .22.
PROBLEMS ON BOOK II, p. 307, &c.
2 S. H.,14. Joh.,18.
Trin.,35.
3 C. C.,34.
4 Qu.,22.
5 Joh. ,19. Trin.,25.
Qu.,26.
Joh.,17.
2 Emm.,34.
3 Jes.,32.
4 Emm.,23. Chr.",38.
5 S. H.,20
6 Trin. ,42.
7 Trin. 31.
8 Pet.,37.
9 T. H.,40.
10 Cai.,44.
11 Trin.,37.
12. Pet.,25.
13 S. H.,38.
14 Joh.,18.
15 S. H03. Joh.,18.
Qu. 21. Trin.,37.
Sid.,42. Chr.,45,
16 Trin.,38.
6 Qu.,19. C. C.,26.
7 Chr.,40.
8 Trin. 41.
9 Joh. ,34.
10 Pem.,44.
11 Pet.,33.
12 Pem.,45.
THEOREMS, p. 308.
17 Joh.,41.
18 Joh. ,44.
19 Cath.,31.
20 S. H.,40.
21 S. H.,10.,04. Trin. 29.
22 S. H.,16. Trin. ,27.
,30.,37. Mag. ,31.,43.
Pet. 29.,38. Sid. 34.
Emm.,21.,27.,37.,44.
Cai.,43. Qu.,37.
23 Pet. .43.
24 Sid.,38.
25 Joh.,13.
26 Qu.,24.
27 Joh. ,13. Emm.,25.
,36. Trin. ,32.
Mag. ,33.,40.
28 Joh.,21.
13 Qu.,31. Pem.,42.
14 Qu.,36.
15 Sid.,31.
16 C. C.,37.
17 Pet.,26.
18 Qu.,38.
29 Joh.,25.
30 Cai.,42.
31 Joh.,20. Emm.,26.
32 Joh. 26. Jes. ,37.
Mag.,42.
33 Pet.,44.
34 Emm.,23.,26.,28.,43.
Trin. ,27.,44. Pet.,30.
Mag.,33.,43. B.S.,38.
Chr.,41.
35 Emm.,28. Sid.,33.
C. C.,39.
36 Joh.,19. Qu. ,29.,30.
37 Chr.,30.
38 S. H.,07. T. H. ,44.
39 Emm.,28. Trin.,32.
40 Cai.,28.
INDEX.
379
1 Qu.,23.
2 S. H.,04.
3 S. H.,11.
4 Emm.,22.
5 Trin. 42.
6 Pet.,29.
Sid.,41.
Pem.,36.
Chr. 44.
PROBLEMS ON BOOK III, p. 315, &c.
32 Qu.,33.
33 Joh.,19. Qu.,26.
34 Qu.,33.
35 Joh. ,17.
63 Pem.,44.
64 Trin.,20.
65 Joh.,28.
36 Chr.,44.
37 Qu.,36.
39 Qu.,34.,44.
7 Trin.,11.
8 Trin.,30.,39. C. C.,35.
,45. Emm.,37.
Chr.,39.
9 Cai.,40.
10 Joh. ,30.
11 Joh.,21.
12 Sid.,35.
13 C. C.,32.
14 C. C.,34.
Pem. ,40.
38 Qu.,25.
40 Trin. 34.
41 Joh.,17.
42 Sid. 35. Pem. ,42.
43 Joh.,21.
44 S. H.,13.
66 Joh.,16.
67 Emm.,21. Pem.,32.
Cla.,36.
68 Trin.,34.
69 Qu, ,40.
70 Trin.,43.
71 Qu.,43.
72 Joh. ,30.
73 Trin.,24.
Cai.,45.
74 Trin. ,26.,27.
75 S. H.,03.
76 Qu.,32.
45 Pet.,26.
46 Mag.,35.
47 Joh.,28.
77 Pet.,44.
48 S. H.,20. Trin.,22.,25.
78 Pem.,37.
Mag.
37. Qu.,39.
79 Sid.,41.
49 Qu.,30.
80 Pet.,25.
50. Trin. ,41.
15 Joh.,20.
16 Joh.,13.
17 Mag. 27. Trin.,29.
18 Cai.,32.,41.
19 S. H.,03. Qu.,22.
Emm.,27. Sid.,30.
Cath.,30.,35. Mag.,34.
,37.,45. B.S.,39.,43.
20 B. S.,26. Chr.,40.
21 Emm.,24. Qu.,44.
22 Cai.,28.
23 Joh. ,22.
24 Trin.,26.
25 Trin.,45.
26 Cai. ,43. S. H.,45.
27 Chr.,27.
28 Joh.,17.
29 Qu. 39. Pem.,43.
30 Qu.,41.
31 Pet.,39. Qu.,39.
1 Chr.,28. S. H.,36.
Cai.,44.
2 Trin.,27.
3 Emm. 28.
4 Trin.,19.,23. Qu.,21.
,22. Pem.,30.,39.
Sid.,36. Pet. ,31.
Emm.,34.,42.,44.
5 Trin.,19. Sid.,33.
Cai.,34. Emm. ,34.
Qu.,36.
6 Cath.,33.
7 Cai. .44.
8 S. H.,43.
9 Sid.,44.
10 S. H.,10.
11 S. H.,14. Qu.,20.,32.
Joh. 25. Emm.,32.
Cai.,44.
Chr. 45.
12 Cai.,39. Jes.,26.
C. C.,38.
13 Pet.,39. Pem.,45.
14 Chr. ,40.
15 T. H.,40.
16 Cai.,40.
17 Pet.,44.
18 Qu.,35.,36. Pem.,37.
19 Trin.,39.
20 Cai.,39.
51 S. H.,45.
52 Trin.,21.
53 Pet..43.
54 Joh.,25.
55 Emm.,21. Mag.,40.
56 Joh.,22.
57 Trin.,42.
58 Qu.,37.
59 S. H.,04. Joh.,16.
Qu. ,20.,35.,29.
Trin. ,22.,23. Pet.,31.
B. S.,30.,34.
60 Pet.,43.
61 Trin.,33.
62 Qu.,37.,29.
THEOREMS, p.. 320, &c.
21 Joh.,18.
22 Sid.,29.
23 Joh. ,30.
24 Cai.,41.
25 Cai.,45.
26 Joh. 22. Mag.,29.
27 Jes.,20. Trin.,25.
28 Pet.,30.
29 Trin.,29.,32.,38.
S. H.,08. Pet.,19.,20.
,21. Qu.,20.,28.
Mag.,30. B. S.,39.
30 Pet.,42.
31 Joh.,31.
32 Cai.,38.
33 Trin.,30. Pem.,35.,40.
34 Cath.,33. Mag. ,37.
Pem.,40. Cai.,43.
35
36 Pem.,45.
37 Joh.,21.
38 Emm.,24.
41 Qu.,33.
42 Cai.,39.
81 Pet.,32.
82 Trin.,44.
83 Trin.,20.
84 Joh.,36.
85 Chr.,44.
86 Cai.,40.
87 Sid.,38.
88 Joh. ,43.
89 Cai.,31. Pet.,34.
90 S. H.,19.
91 C. C.,25. B. S.,28.
Mag. ,45.
92 Joh.,20.
93 Qu.,32. Sid.,44.
94 Trin. ,31.
95 Joh.,34.
46 Joh.,28. Qu.,35. ·
47 Qu.,42.
48 Trin.,41.
49 S. H.,29.
50 Cath.,30.
51 Qu.,36.
52 Joh. ,18.
53 Mag. ,42. (or ,44.)
54 Pem.,31.
55 Joh.,29.
56 Cath. ,35.
57 C. C.,42.
58 Jes.,34.
59 Emm.,35. Joh.,38.
60 Jes.,33.
61 Qu.,42.
62 Joh. ,21.
63 Trin.,39.
64 S. H. 36.
65 Pet.,37.
66 Joh.,26.
67 Joh.,22.
68 S. H.,42.
69 S. H.,25.
70 S. H.,28.
39 S. H.,14.
40 Mag. ,37.
71 S. H.,16.
43 Joh. 20. Emm.,26.
72 Trin.,39.
73 Joh.,42.
74 Cai. 43. Emm.,44.
44 Jes.,43.
45 Cath, ,31.
380
INDEX.
75 Cai.,37.
76 Trin.,29. Sid.,45.
77 Chr.,26..
78
79 Pem.,32.
80 Sid.,29.
81 Pet.,39.
82 S. H.,02. Pem.,32.
T. H.,44.
83 Trin.,45.
84 Pet.,45.
85 S. H.,04.
86 Qu.,38.
87 Cai.,31.
88 Cai.,37.
89 Trin. .41.
90 Cai.,42.
91 S. H.,38.
92 Cai.,40.
93 Cai.,36.
94 S. H.,13.,42.
95 Trin. .43.
96 B. S.,45.
97 Chr.,44. Trin.,23.
Jes.,34.
98 Joh. 18.,19. Qu.,26.
,39. Mag.,29.
Emm. ,30. Pem.,44.
99 Cath.,30.
100 C. C.,42.
101 Trin. 42.
102 Qu.,33.
103 Joh. ,34.
104 Trin.,24.
105 Trin. ,43.
106 Qu.,34.
107 Joh. ,34.
108 Trin.,26. Pem.,34.
109 Joh.,19. Qu.,26.
Trin. 34.
110 Joh. .35.
111 Trin. ,38. Joh. ,19.
Chr.,39. Jes.,43.
S. H.,42.
112 Joh.,29.
113 C. C.,29.
114 Jes.,38. C. C.,38.
115 Joh.,33.
116 C. C.,39.
117 Mag. ,42.
118 Cath.,33.
119 Jes.,44.
120 Joh. ,21.
121 C. C.,33.
122 Joh.,25.
123 Joh.,20.
124 Joh.,19.
125 Trin.,21.
126 Joh..30. Cai.,36.
127 Cai.,39.
128 Cai.,31.
129 S. H.,39.
130 Pet.,29.
PROBLEMS ON BOOK IV, p. 333, &c.
1 Qu.,20. Emm.,27.
Cath. 34. Trin. ,44.
2 Qu.,20.,30.,34.
Trin. ,29. Emm.,30.
C. C.,35. B. S.,36.
Pem.,40.
3 Joh.,16. Pet.,36.
4 Trin.,31.
5 Qu.,41.
6 Qu.,19. S. H. ,05.
Sid.,40. Ki.,45.
7 Emm.,24. Qu.,32.
8 Chr.,41.
9 Pet.,40.
10 Joh. 28. Trin.,23.
11 Qu.,44.
12 Cai.,33.
13 C. C.,44.
14 Trin. ,40.
15 Trin. 41.
16 Joh.,33.
17 Trin. 31.
18 Cai. 43. Pet.,42.
19 Emm.,21. Trin. ,36.
Trin.,36.
Pem. 42.
20 Sid.,40. C. C.,41.
21 Joh. ,33.
22 Sid.,29. Qu.,43.
23 Pet.,24.
24 Chr. 26.,42.
25 Joh.,28.
26 Pet.,30.
27 Emm.,24.
28 Emm.,29.
29 Pem.,29.,35.
30 Trin. 11.
31 Chr.,32.
32 Pet.,43.
33 Joh.,25.
34 Trin. ,26. Qu.,32.
Chr. ,40.
35 S. H.,13.
36 Joh. ,42.
37 Pem.,29.
38 Trin. 28. Joh. ,34.
Pet.,37.
39 S. H.,02.
40 Trin.,20.
41 Cai.,32.
42 Mag.,35.
43 Trin.,44.
44 Cath. ,29. Qu. ,29.
S. P.,43. Emm.,39.
Trin. ,37.,43. 44.
Mag. 39. Sid. ,44.
Cath.,29.
45 Trin.,37.
46 Trin. ,45.
47 Trin. 23. Qu.,37.
48 Qu.,21.,26.,36.
49
50 Qu.,23.
51 Qu.,26.
52 Cai. 33. B. S. ,40.
53 Emm.,25. Mag. ,42.
54 Joh.,14.,16.,37.
S. H.,44.
55 Qu.,33.
56 Jes.,31.
57 Joh. 18.
58 Joh.,17. Trin. ,36.
59 Pet.,25.
60 Pem.,45.
61 Qu.,31.
62 Chr.,29.
63 Qu.,31.,40. Trin. ,42.
64 S. H. 04.
65 Qu.,31.
66 Trin.,27. Mag.,43.
67 Emm.,23. Trin.,26.
Pem.,34. Mag.,34.
68 Joh. ,20. Emm.,31.
Trin. 35.
69 Qu.,34. C. C.,41.
70 Trin.,22. B. S.,27.
Pem.,36.
71 Trin.,36.
72 S. H.,10.
73 Cai.,34.
74 Joh. 13. S. H.,16.
75 Joh.,23.
76 Trin.,38.
77 Trin.,21. Chr.,30.,34.
78 Cai.,35.
79 C. C.,33.
80 Trin. ,23.
,32.,36.
81 S. H.,04.
Sid.,33.,34.
82 S. H. .03.
83 Qu.,28.
84 Emm.,32.
85 Chr.,34.
Emm.,23.
Emm. 26.
86 Qu.,35. Trin.,44.
87 S. H.,11.
88 Qu.,29.
89 Qu.,37.
90 Emm.,37.
91 Chr.,32.,33.,38.
Emm. 32.
92 Qu.,33.
93 Trin.,36.
94 Joh.,33.
95 Trin.,24.
96 Trin.,19.
97 Emm.,24. Mag.,33.
,37.,39.,41. Trin.,24.
Chr.,43.
98 Trin.,37.
99 Trin. ,41.
100 Joh. ,44.
101 Mag.,41.
102 Joh.,13.
103 Trin.,21.
104 S. H.,41.
INDEX.
381
1 S. H.,08.,12. Chr.,33.
Pet.,34.,38.
2 Trin.,37.
3 Mag.,30. Pem.,30.
4 S. H.,16. Qu.,20. ‚27.
C. C.,28.
5 Jes.,33.
6 Chr. 41.
7 Emm.,37.
Joh. ,39.
8 Trin. ,23. Sid.,39.
Qu.,41.
9 Joh. ,39.
10 S. H.,13.
II Cai.,39.
12 Joh.,20.
13 Chr.,27.
14 S. H.,13.
C. C.,45.
Trin. ,22.
Qu.,19.
Emm.,21.,33. B.S.,26.
Cai.,35. Pem.,36.
15 Chr.,38. Pet.,26.
Emm.,44.
16 Trin.,40.
THEOREMS, p. 338, &c.
Sid.,42.
30 Pem.,29. C. C.,41.
31 Cai.,37.
32 Cai.,41.
33 Emm.,21. ‚25.,40. ‚45.
Chr.,39. Pet. ,35.
B. S.,41.
34 Chr.,39.,45.
35 Cai.,38.
36 Jes. 30. Cai.,35.
37 Trin.,19.,29.
Emm.,36.,45.
38 Joh.,13. S. H.,34.
Pem.,39. Emm. ,40.
,43.,45. Chr.,44.
39 S. H.,39. Trin. ,41.
Emm. 42.
40 Cai.,38.
41 Trin.,20.
42 Joh.,41.
43 Joh. 41.
44 C. C. 38.
45 Joh.,20.
46 Trin.,41.
60 Trin.,33.
61 Emm.,34.
62 Chr.,39.
63 Qu.,42.
64 C. C.,24.
65 Joh.,14.
66 S. H.,38.
67 Chr.,45.
68 Cai.,42.
69 Cai.,38.
70 Joh. ,20.,23.
71 Emm.,35.
72 Trin.,45.
73 Pet.,25.
74 Joh.,18.
75 S. H.,32.
76 Joh.,21. Pet.,30.
Emm.,36. Trin. ,37.
77 Trin.,31.
78 Cai.,34.,39.
79 Cai. 37.
80 S. H. 15.
81 Qu.,36.
82 Trin.,34. Joh.,34.
83 Joh.,28.
84 S. H.,03. Trin., 24.,30.
Qu.,31.,35. Cai.,35.
17 Trin.,38.
18 Joh. 42.
47 Trin.,29.
19 Cath.,30. Mag. ,30.
Qu.,44.
48 Cai.,34.
49 Jes.,38.
50 Cai.,38.
51 Pet.,45.
52 Cath.,30. Mag.,33.,37.
87 Cai.,41.
88 Cai.,34.
85 Joh. ,25.
86 Pet.,32.
20 Joh. 30.
21 Qu.,20.
22 Pem.,31.
23 Trin. ,43,
24 Cai.,42.
25 Joh.,22.
26 Chr.,36. S. H.,36.
27 Joh.,31.
28 Pet.,27.
29 Trin.,30. Qu.,32.
53 Cai.,40.
54 Sid.,38. Trin. ,39.
55 Cai.,41.
56 Pet.,36.
57 Trin. ,40.
58 Trin. 37.
59 Trin.,45. Pem.,45.
89 Trin. 19.
90 Joh.,18.
91 Qu.,32. Pem.,35.
92 Cai.,41.
93 Qu.,43.
1 Qu.,38. Trin
PROBLEMS ON BOOK VI, p. 348, &c.
2 Qu.,23. Sid.,34.
C. C.,40.
3 Ki.,45.
4 Qu.,33.
5 Qu.,29.,35.,37.
B. S.,39.
6 Joh.,13. Trin.,20.
Emm.,24. Chr.,37.
45. Qu.,36.,22.‚44.
7 Trin.,44.
8 Trin. ,23.,33.,41.
9 Pem.,43.
10 Trin.,32.
11 Trin. 30. Pem.,34.
12 Joh.,18.
13 S. H.,25.,17. Qu.,23.
14 Pem.,34. C. C.,30.
15 Joh.,29. Qu.,43.
16 Qu.,31.
17 Jes.,35.
18 Qu.,21.
19 Trin.,36.
20 S. H.,25,
21 Trin. 34.
22 S. H.,26.
23 Pet.,33.
24 Trin.,11.,28.,43.
Jes.,19. Qu. ‚21.,23.
,26. C. C.,26.
Pem.,32.,34. ,43.
Cai.,33. Emm.,21.
25 Qu.,38. Chr.,43.
Trin.,44.,33.
26 Cai.,31.
27 Joh. ,28.
28 Joh. 15. Trin. ,22.
Qu.,36. Chr.,35.
Cath.,35.
29 Joh.,29. Cai.,45.
30 Joh.,28.
31 Jah.,19.
32 Pet.,27.
33 Joh.,23.
34 Qu.,34.
35 Joh. ,34.
36 Qu.,24.
37 Emm.,25.
38 Trin.
39 Emm., 23,, 30. B.S.,29.
40 T. H.,44.
41 Joh.,17.
42 Trin. ,30.
43 Pem.,33.
44 Joh. ,26.
45 Trin. ,24.
46 Joh.,21.
47 Joh.,20.
48 Joh.,20..,21. Trin.,45.
49 Joh.,19.
50 S. H.,02.
Trin.,20.,22.,33.
51 Pem.,45.
52 C. C.,26.
53 Pem.,35.
54 Trin.,38.
55 Qu.,30. C. C.,40.
56 Pet.,28.,35.
57 Trin.,22. Qu.,39.
Chr.,42.
58 Qu.,22.,38.
Trin. ,42. ,44.
59 Qu.,29.,35.,41.
S. P.,43.
60 Trin. ,35.
382
INDEX.
61 Qu.,36.,38.
62 Joh. ,15.
63 Qu.,41.,42.
64 Cai.,41.
65 Qu.,22.
66 Pet.,25.
67 Joh. ,43.
68 Qu.,41.
69 Joh.,17.
70 Trin.,21.
71 Trin.,39.
72 Joh.,19.
73 Joh.,20.
74 Joh.,21.
1 C. C.,31.
2 Jes.,19. Trin.,32.,44.
3 S. H. .18.
4 Pet.,38.
5 Trin. ,32.,35.
6 Joh.,23.
7 Chr.,43.
8 Cai.,36.
75 Pet.,32.
76 Pet.,25.
77 Sid.,34.
78 Joh. ,38.
79 Qu.,22.
80 Joh.,13.,21. Trin.,29.
,34. Qu.,43.,38.
81 Joh.,28.
82 Trin.,24.
83 C. C.,28.
Pem.,42.
84 C. C.,35. S. H.,11.
85 Trin.,23.
86 Trin.,42.
87 Trin.
THEOREMS, p. 353, &c.
43 Joh.,29.
44 Joh.,29.
45 Qu.,34.
88 S. H.,09. B. S.,30.,31.
89 Qu.,27.
90 Joh. 14.
91 Qu.,24.
92 Qu.,24.
93 S. H.,36.
94 Chr.,38.
95
Qu.,25.
96
Trin.,25. Mag.,37.
97 Qu.,29. Mag.,31.
98 Sid.,29.
99 Trin.,29.
100 Sid.,33.
93 Trin.,35.
94 Pet.,24.
95 Pet.,24.
96 Pet.,26.
97 Joh. ,23.
46 Trin.,25.
47
48 Joh. 18. S. H.,34.
49 Cath.,31.
98 Pet.,39.
99 Emm.
100 Mag.,41.
101 Joh.,31. Qu.,44.
102
103 Joh.,37.
50 Joh.,19.
9 Jes.,37.
10 Jes.,39.
11 Cath.,30. Emm.,34.
Sid.,44.
12 Emm.,34.
51 Qu.,37.
52 Joh.,29.
53 Qu.,38.
54 Cath.,33.
104 Pet. 31.
55 Joh.,23. Emm.,29.
Chr.,30.,32.
105 Qu.,21.
106 Qu.,31.
Trin. 27.
56 Joh.,28.
107 Joh. ,15.
57 Pem.,39.
67 Qu.,36.
68 Joh. ,25.
69 Joh.,20.
70 Jes.,44.
71 Qu.,40.
27 Joh. ,28.
72 Qu.,43.
28 Joh.,19. Qu.,25.
73 S. H.,38.
74 Joh.,14.
75 Joh.,20.
76 Qu.,29.
77 Qu.,29.
78 Trin.,24.
79 Joh.,21.
80 Trin. 36.
13 Cath.,33.
14 Joh. 14.
,28.,32.,34.,41.,44.
Cath.,34. Chr.,44.
15 Pem. 30. S. P.,42.
16 Cath.,30.
17 S. H.,18.
18 Trin. 23.
Mag. 37.
19 Joh.,26.
20 Trin. ,33.
21 Joh.,26.
22 Joh. 18.
>
Qu.,20.
Cai. 35.
23 Trin.,42.,44.
24 Trin.,30. S. H. ,04.
Mag. 44.
25 Cath.,31.
26 Pet.,27.
29 Trin.,28,
30 Qu.,34.
31 Pem.,29. Cath.,34.
Emm.,34. Pet.,35.
Chr.,39.
32 Joh. ,39.
33 Trin. ,25.
58 Joh. ,42.
59 Pet.,36.
60 Cai.,39.
61 Trin. ‚11.,20.,32.,33.
Chr.,35.
62 Pet.,37.
63 Chr.,37. Pem.,31.
64 Cai.,40.
65 Pet. 24.
66 Cai.,44.
108 Pet.,35.
109 Joh.,15. Jes. 36.
110 Emm.,35.
111 Joh.,16.
112 Emm.,21.
113 Trin,,29.
114 Joh. ,15.
115 Joh.,35.
116 S. H.,33.
117 Pem. 31. ,43. Qu.,19.
,25.,43. Trin. ,22. ,37.
Cai.,43. Mag.,32.
118 Qu.,41.
119 Joh. ,15.
120 Trin.,43.
121 Trin. 43.
122 Joh.,23.
124 Pet.,33.
125 Trin. 35.
126 Pet.,33.
127 Sid.,30.
128 Chr.,29.
129 Qu.,34.
123 Joh.,13.
130 Joh.,29.
34 Cai.,36.
81 Cai.,31.
131 Qu.,29.
82 Jes.,42.
83 Joh. 42.
84 Sid.,44.
85 Qu.,20.,26.,32.
35 Qu.,32.
36 Chr.,36.
37 S. H.,15. Qu.,28.
Chr.,39. Emm.,26.
Cai.,35.
38 Joh.,34.
39 Jes.,19. Trin.,22.,25.
,27. Qu.,35. Pem.,37.
Mag. ,45.
40 Joh.,25.
41 C. C.,24.
42 Joh.,15. C. C.,37.
86 Sid.,39.
87 Qu.,33.
88 Emm.,24.
89 Cai.,45.
138 Joh.,19.
90 Trin.,19.,41. Sid.,45. 139 B. S.,42. Ki.,44.
C. C.,44.
91 Cath.,36.
92 S. H.,37.
140 S. H.,39. Pem.,43.
141 Pem.,32.
142 Pet.,44.
132 Joh. ,33.
133 Joh.,21.
134 Joh.,21.
135 Cai.,37.
136 Pet.,45.
137 Joh.,19. Qu.,26.
Emm.,31.
INDEX.
383
143 Joh.,20.
144 Joh. 22. Emm.,26.
145 Pet.,35.
146 Mag.,35
147 Joh.,15.
148 S. H.,45,
149 S. H.,40.
150 Joh.,44.
151 Joh.,14.
152 Joh.,14.
153 Joh.,14.
154 Joh.,14.
155 Trin.,26.
156 Joh. ,40.
157 Joh. 38.
158 Joh.,43. S. H.,33.
159 Joh. 14.
160 S. Hí,03. Cai. 43.
Pet.,41.
161 Qu.,31. Trin. ‚42.
162 Cai.,29.
163 C. C.,30.
164 Joh.,18. Cath.,31.
165 Qu.,40.
166 Joh.,42.
167 Joh.,18.
168 Chr.,36.,38.
169 Pet. ,40.
170 Trin. ,20.
171 Cai.,39.
172 Emm.,37.
173 Chr.,37.
1 Trin.,30.,41.
2 Cai.,42.
3 Trin.
4 Trin.
5 Trin.,32.
PROBLEMS ON BOOK XI, p. 368,
1 Trin.,29. Chr. ,43.
2 Chr.,44.
3 S. H.,12.
4 Cai.,38.
6 Joh.,29.
7 S.H.,25. Qu.,30.
Trin. ,25.,35.
8.Emm.,32.
9 S. H.,14. Qu.,20.,28.
THEOREMS, p. 368, &c.
12 Trin.,31.
13 Trin.,43.
14 Trin.
15 Cai.,37.
16 Pet.,40.
17 Trin.,28.
5 Pet.,25.
6 Cai.,37.
7 Trin.,31.,36.
18 Emm.,34.
8 Cai,,43.
19 Joh. ,14.
9 S. H.,20.
20 Cai.,45.
10 Joh.,31.
11 Cai.,34.
21 Cai.,44.
&c.
10 Trin. 29.
11 S. H.,37.
12 Qu.,30.
13 Cai,,36.
22 Joh.,17.
23 Trin.,42.
24 Trin.
25 Trin.,22.',25. Mag.,29.
Joh.,34. Chr. ,45.
26 Trin.
27 S. H.,33.
28 Emm.,35.
29 Cath.,33.
30 Joh.,22.
PROBLEMS ON BOOK XII, p. 374, &c.
1 S. H.,03. Joh.,41.
2 Joh.,33.
3 C. C.,34.
4 Qu.,25.,29. Trin. ,33.
,35.,44. Chr. ‚34. ‚41.
,45. Emm. ,39.
Mag. ,34.
5 Trin.,21.
6 S. H.,18. Chr.,33.
7 Joh.,21. Chr.,31.
8 S. H.,44.
9 Pem.,45.
10 Trin.,41.
11 S. H.,37.
12 Sid.,33.
THEOREMS, p. 375, &c.
13 Trin. ,34.,32.
14 S. H.,26.
15 Mag. 29. Trin. ,32.
16 Trin.,37.,29. S. H.,20.
Mag. ,29.
17 S. H.,01.
18 Qu.,33.
1 Trin.,23.,27.
Pem.,30. Sid.,31.,44.
Cai.,31.,34.,41.
Emm.,36.,40. Chr.,42.
2 Jes.,19. Pem.,32.
3 Chr.,32. B. S.,35.,37.
4 Cath.,30. Trin. ,32.
S. H.,03.,43. Chr.,35.
5 Qu.,39.
6 Joh.,13.
7 Emm.,33.
8 S. H.,16.
9 Cai.,32.
10 Cai.,45.
23 Chr.,37.
11 S. H.,28.
24 S. H..01.
12 Trin. ,21. Joh.,15.
25 Joh.,15.
13 Qu.,36.
26 Joh.,17.
14 Qu. ,37.
27 Joh.,31.
15 Cai.,39.
16 Cai.,38.
17 Qu.,24.
18 S. H.,44.
28 Cai.,37.
29 Joh.,22. Cai.,38.
S. H.,43.
30 Cai.,45.
19 Cai.,39.
20 Pet.,37.
31 Joh.,37.
32 Cai.,44.
21 Pem.,37.
33 Pet.,28.
22 Joh.,17.
34 S. P.,39.

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