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ELEMENTS
OF
GEOMETRY
UPON


THE INDUCTIVE METHOD.
TO WHICH IS ADDED
AN INTRODUCTION
TO
DESCRIPTIVE GEOMETRY.


BY JAMES HAYWARD, A. M.
Lately College Professor of Mathematics and Natural Philosophy
in Harvard University.
CAMBRIDGE:
HILLIARD AND BROWN, BOOKSELLERS TO THE UNIVERSITY.
PHILADELPHIA:
URIAH HUNT, NO. 147 MARKET STREET.
1829.




DISTRICT OF MASSACHUSETTS, to wit:
District Clerk's Office.
BE it remembered, that on the second day of November, 1829,
in the fifty-fourth year of the Independence of the United States
of America, Hilliard & Brown, of the said district, have deposited in
this office the title of a book, the right whereof they claim as proprietors, in the words following, to wit;
" Elemcnts of Geometry upon the Inductive Method. To which is
added an Introduction to Descriptive Geometry. By James Hayward,
A. M. Lately College Professor of Mathematics and Natural Philosophy
in Harvard University."
In conformity to an act of the Congress of the United States, entitled " An act for the encouragement of learning, by securing the copies
of maps, charts, and books, to the authors and proprietors of such copies
during the times therein mentioned;" and also to an act, entitled " An
act supplementary to an act, entitled ' An act for the encouragement of
learning, by securing the copies of maps, charts, and books, to the authors and proprietors of such copies during the times therein mentioned;' and extending the benefits thereof to the arts of designing, engraving, and etching historical and other prints."
JNO. W. DAVIS,
Clerk of the District of JMassachusetts.


CAMBRIDGE:
E. WV. METCALF AND COMPANY,
Printers to ths University.




PREFACE.
IN the volume now offered to the public, the author has
aimed to bring the science of Elementary Geometry within
limits which should adapt it to the convenience of a
greater number than can afford time to acquire a competent knowledge of the subject from the treatises in common
use; while, so far as this branch of elementary preparation
is concerned, he should omit nothing essentially introductory to the higher geometry, or to the physico-mathematical
sciences. He has endeavoured also to state the first principles of the science in a manner better suited to the ap.
prehension of the young student, than that in which they
are usually presented in elementary treatises.
Clairaut, one of the first mathematical geniuses of the
eighteenth century, has said, "Though Geometry be in itself abstract, it must nevertheless be acknowledged, that
the difficulties which the student encounters in the coinmencement of the subject, most frequently arise from the
manner in which it is taught in common Elementary
Treatises. These begin with a great number of definitions, postulates, axioms, and preliminary principles, which
seem to promise little to interest the reader. The propositions which follow, do not fix the mind upon engaging objects; and as they are moreover difficult to be understood,
it commonly happens that beginners are fatigued and dis



Preface.


gusted before they have any distinct idea of what is designed to be taught them."
In the present treatise, the author has endeavoured, as far
as he was able, to remove the difficulties here complained
of; by avoiding the abstract phraseology and technical
forms of the books in common use; by presenting to the
mind of the learner, in a manner as simple as he could, the
elementary truths of the science; and developing by a process of plain and, at the same time, exact examination,
those which are more involved; by following in a great degree the natural order of discovery; and aiming so to conduct the investigation of any particular fact or general principle, as that other collateral truths should unfold themselves
incidentally, and thus, by the very circumstance of their
being unexpected, afford pleasure to the learner, and excite in his mind a curiosity to know what will next discover itself on either side of the path he is pursuing.
This desire to know with what interesting and useful
truths the further pursuit of the subject will bring him acquainted, will enable the student to perform with pleasure
what is commonly (and frequently with too much truth)
called his task, and really to do more with less fatigue. It
is, moreover, a state of the mind in which the truths actually presented to it make the greatest impression; and are
most likely to become a permanent part of the student's
knowledge.
If the result is kept back till the learner has gone
through the process of induction by which any truth is to
be established, and by which also it must have been at first
discovered, his mind will be prepared to perceive the precise meaning of the proposition in which this truth is
enunciated; whereas the slightest misapprehension of the
nature or the extent of the truth stated beforehand in the
proposition, may seriously embarrass him in applying the
process of proof, to say nothing of the difficulty of ridding




Preface.


v


himself entirely of the first impression made by a proposition endowed with the oracular name of theorem.
Another important advantage results from thus withholding the proposition till its truth is apparent. The student will early begin to anticipate, before he arrives at the
statement of the result, the truths as they unfold themselves;
and finding that he has discovered them from the relations
presented, he is gratified by this evidence of his own power, and is encouraged to continue his exertions. He will
be constantly on the watch for new discoveries; and his
studies will be a salutary discipline, not only to his understanding, but to the inventive faculty of his mind.
In preparing these Elements, the author has consulted
various editions of Euclid, and several modern treatises,
among which may be mentioned those of Legendre, Bezout, Reynaud, Lacroix, and Clairaut; besides the meta
physical writings of Lacroix, Carnot, and Laplace. He
has endeavoured to combine, especially in the earlier part
of the work, the simplicity of Clairaut with the certainty of
what are usually called the "severer methods." In the
latter part, the general arrangement of the subject is more
nearly that of Lacroix than of any other. But most of
what may be considered as peculiar, whether in the arrangement or the manner of conducting the inquiries by
which the truths of the science should unfold themselves to
the student, has been suggested by a careful observation of
the difficulties which (in the course of no inconsiderable
experience as an instructer in elementary mathematics) the
author saw his pupils continually encountering; and by the
results of constant efforts to investigate the causes of these
difficulties, and of multiplied endeavours to state the elementary principles of the subject in so simple a manner as to
be apprehended by every mind. Though he may not always have succeeded, yet the results were such as to cona*




Preface.


vince him that the mistakes and misgivings which the young
student experiences in the early part of his geometrical
studies, are very uncertain indications of any peculiar want
of adaptation of mind to this branch of science. So far
from it, these very mistakes should be expected. The subject is new; the leading ideas are many of them expressed
in terms almost exclusively appropriated to this science,
and must therefore be new to the beginner;  though
these terms seem very definite to the metaphysical geometer, there is much uncertainty whether they will convey to
the mind of the student the precise idea which the writer
intended; and it will frequently happen, that the ideas
which many, even of the better scholars, receive from some
of the principal definitions and statements in our elementary books, require to be considerably modified or entirely
changed, to adapt them to the use which is to be made of
theim in the subsequent part of their studies.
How much perplexity and discouragement might have been
spared the learner in such cases, if the mistake could have
been discovered at first. In consequence of a slight misapprehension of some leading principle in the beginning, to
which every mind is liable, the learner tasks himself in vain
to reconcile the subsequent reasoning and results with the
first notions which he received, and which, coming with
all the freshness of novelty, fixed themselves in his mind.
Such mistakes might be, in a great measure, prevented, if'
the instructer would explain and illustrate every sentence
of one or two of the first lessons before they are read by his
pupils; and if, wherever any new element is introduced,
he would be sure that they understand it before they proceed to apply it.
There is always danger in giving the beginner two or
three pages, at first, of elementary matter in the form of
distinct principles and definitions, attended by little or no
explanation. Such an array of detached and unconnected
truths tends only to fatigue and distract him.




Preface.


vii


To prevent, in some degree, this liability to misconception and embarrassment, the definitions and principles, in
the present treatise, are given only as they are to be used
in the course of the investigation, either as principal or
auxiliary truths.
For some of the elements new definitions have been
adopted, which, it is believed, convey to the mind of the
learner more accurate notions of the things defined, and
very much simplify the processes into which they enter.
"A good definition should contain an enumeration of
certain simple characteristic attributes of the thing defined,
by which it may be clearly distinguished from all other
things of a like kind. In this respect the definitions of
mathematics are, in general, peculiarly happy. They usually contain some one simple but characteristic property of
the thing in question, from which all its other properties
may be readily and legitimately deduced."*
In conformity with the spirit of the remark here quoted,
the three following definitions are adopted.
A  straight line is one which has the same direction
throughout its whole extent.
A plane angle is the inclination of two straight lines to
each other.
Parallel lines are straight lines which have the same direction in space.
The simple characteristic property of a straight line,
that is, its straightness, by which it is distinguished from all
other lines, is its identity of direction in every part. The
definition given in several elementary books, namely, that
a straight line is the shortest way from one point to another,
is a proposition which carries to the mind the fullest conviction of its truth; and whether a simple principle or a


* Prof. Elliot of Aberdeen.




Preface.


deducible truth, it is equally a suitable element of geometrical reasoning.  But this identity of direction in all its
parts, is that peculiar property of the straight line, which
enters into every consideration of angles and parallels;
and the neglect of which has been the cause of most of the
embarrassment that has been felt in discussing the doctrine of parallel lines.
The definition of an angle here adopted, is that which
was given by Euclid. It was also adopted by Clairaut and
Laplace; and seems not to be susceptible of improvement.
The inclination of the two lines is precisely the property
which is used in all discussions in which an angle is an
element.
The above definition of parallel lines is adopted because
it is believed to characterize the relation of the parallels to
each other, more precisely than those definitions which make
the parallelism of the lines consist in their not meeting or
their being throughout at the same distance from each other.
Parallel lines are throughout at the same distance from
each other, and cannot meet. These are truths which result from the property or rather the relation of parallelism;
that is, from their having the same direction. This identity of direction is what constitutes the parallelism of the
lines. And this notion of parallels should be, at first, presented to the contemplation of the learner; for this is the
simple principle from which result all those propositions
that make up what is called the doctrine of parallel
lines.
The fundamental truth from which are deduced the various propositions respecting parallels, whatever may be the
form in which it is stated, is essentially this:-Parallel
lines are equally inclined to a straight line meeting them.
It seems to be pretty generally acknowledged that this
proposition has never yet been geometrically proved, in any
of the elementary treatises of geometry, though it is a fun



Preface.


ix


damental principle of a great part of the system. Yet
Laplace said of this proposition, that "its mere enunciation produces the fullest conviction; we ought not,
therefore, in elementary teaching, to insist upon what may
be wanting to the perfect rigor of the proofs which are
given of it; this discussion should be left to metaphysical
geometers, at least until it has been so far elucidated as to
leave no obscurity in the minds of beginners."
It is gratifying to be able to give so high an authority for
such an opinion. The student should at first have presented to him those truths only which he can clearly comprehend, and concerning which he will have no doubt. He
may go back and discuss the difficulties of the subject, when
he has acquired knowledge and power which will enable him
to surmount them with ease.
That the mere enunciation of the above proposition produces the fullest conviction, is no doubt true; but this conviction does not result from the definition of parallels given
in common books; it results rather from the common notion formed of parallel lines before we read of them in
books.  It is hardly credible that the authors themselves, in using parallel lines in the various demonstrations
in which they occur, usually think of them as not Pteeting.
They contemplate them merely as having the same direction, and mentally derive their results from this property.
This is certainly true of those who read their books.
It may be well to consider the three definitions given
above, in their connexion with each other.
The straight line has the same direction in every part.
An angle is the inclination of one straight line to another;
that is, the inclination to each other of these two directions.
Two parallel straight lines have the same direction.
Therefore, a straight line (which has but one direction in
every part), meeting two straight lines which have but one




Preface.


direction in all their parts, must have the same inclination
to both. That is, When a straight line meets two parallel
straight lines, the angles which it makes with the one are
equal to those which it makes with the other. Clearer evidence of the truth of this proposition cannot be desired.
For those propositions which are generally proved by the
method of indivisibles, or the reductio ad absurdum, or both,
the method here employed is essentially the method of limits or ultimate ratios, used by NEWTON in his Principia,
" to avoid," as he says, "the tediousness of deducing perplexed demonstrations ad absurdum." Carnot says of the
method of limits, that " it is of very great importance, as
it relieves us from.the necessity of using the reductio ad
absurdum, the most troublesome of those operations which
constitute the method of exhaustion." It may be added
that, after the student has gone through the labor of committing perfectly to memory one of these troublesome arguments, there is frequently reason to doubt whether he
comprehends the entire force of the process.
A second reason for preferring the method of limits to
the argument ad absurdum, is, that throughout the whole
of the process, it approaches directly the truth sought; and
is substantially that process of the mind by which the truth
may be supposed to have been discovered.
A third reason for adopting this method in preference to
the others will be given in the words of Laplace. " The
method of limits is the basis of the infinitesimal calculus.
To facilitate the student's understanding of this calculus, it
is useful to point out its earliest germs in elementary truths,
which should always be demonstrated by methods the most
general. The student thus gains, at the same time, knowledge and the means of increasing it. In his subsequent studies he merely follows the path which has been traced for
him, and in which he has become accustomed to walk;




Preface.


xi


and thus his advancement in it will grow much less difficult. Moreover a system of knowledge is best preserved
and extended by connecting its parts by a uniform method.
In teaching, therefore, prefer the most general methods,
endeavour to present them in the most simple manner,
and you will find, at the same time, that they are almost
always the most easy."
In discussing those subjects in which the three dimensions of space are concerned, the terms solid and solid angle are not used. The term solid is calculated to mislead
the learner; as, in its common signification, it expresses a
property of which geometry takes no cognizance.
Instead of the term solid angle, the recent method of
designating the different angles formed by any number of
planes, is here adopted. This nomenclature is useful, not
only on account of its definiteness, as presenting to the
mind distinct and precise objects of contemplation, instead
of the more general notion; but also as an introduction
to modern treatises on geometry of three dimensions.
It may be proper to remark, that in the whole course of
the Elements, the author has left something for the learner
to do. He has asked questions which he has left unanswered, proposed problems to be solved and truths to be
proved, as an exercise of the learner's ingenuity; and has
stated certain things as evident, which may require a moment's reflection from the young reader: but by this it is
believed, his progress will be rather facilitated than impeded, as it secures his attention, and gives him the habit of
deducing truths from premises without the assistance of
another.
"If I must apologize," says Bezout, " for neglecting the
use of the words Axiom, Theorem, Lemma, Corollary, Scho



*.
XII


Preface.


lium, &c., two reasons have determined me; the first is,
that the use of these words adds nothing to the clearness
of a demonstration; the second is, that this apparatus of
terms may frequently divert the attention of beginners from
the truth in question, by leading them to suppose that a
proposition invested with the name of Theorem must be a
proposition as remote from their knowledge, as the name is
from terms with which they are familiar. But, to the end
that those of my readers who shall open other books of Geometry may not imagine that they have fallen upon an unknown region, I think it proper to inform them that,
"Axiom signifies a self-evident proposition.
" Theorem, a proposition which makes a part of the science in question, but whose truth, in order to be perceived,
requires a course of reasoning called a demonstration.
" Lemma is a proposition which is not necessarily a part
of the theory in question, but which serves to facilitate the
transition from one proposition to another.
" Corollary is a consequence which is drawn from a
proposition that has just been established.
" Scholium is a remark upon what precedes, or a recapitulation of what precedes."
The latter part of the volume contains an Introduction
to Descriptive Geometry, a science whose practical connexion with so many liberal pursuits gives it a claim to be
reckoned one of the regular branches of academical instruction; aside from the peculiar kind of discipline which
it affords to the mind, the interesting nature of many of the
topics which it discusses, and their effect upon the taste of
the student.
By the study of Descriptive Geometry, the mind sees
bodies and their parts in all their relations of position, mag



Preface.


xii


nitude, and figure; it becomes accustomed to the contemplation of forms, and acquires a certainty and readiness of
the imagination which enables it to make with variety and
skill, new combinations of the elements of form.
Nor is Descriptive Geometry confined to the mere representation of visible forms; many of the severer subjects belonging to mathematical science, are thoroughly discussed
by processes which this science has taught. " In a word,
with the aid of a small number of the elementary propositions of geometry, this science possesses an almost infinite variety of means, by which we may arrive at the solu,
tion of the most difficult problems. It requires but a few
weeks' study to be sufficiently understood; it advantageously replaces the common modes of practice, the long and laborious study of which is rendered unnecessary; at the
same time, it gives us the immense advantage of treating
with equal ease new combinations or unforseen cases."*
In offering to the public this Introduction to Descriptive
Geometry, the Author makes no pretensions to originality.
His object has been to comprise, within convenient limits,
the fundamental principles of the science first given to the
world by Monge, with a few examples of its application to
linear perspective and some other important projections.
It is, however, but an introduction; and though it gives
the general principles of construction, is not meant to be
considered as a treatise. If it should serve the purpose of
calling the attention of teachers and the guardians of our
academical institutions, to the importance of the subject,
the object of this part of the work will have been obtained.
In preparing the Introduction to Descriptive Geometry,
the writings of Monge, Hachette, and Lacroix have been
Crozet.
b




xiv


Preface.


consulted, and also the treatises upon this subject of Cloquet and Professor Davies of West Point. To the latter
treatise the reader is referred for a fuller discussion of Tangent and Cutting Planes and Spherical Projections; also
for a discussion of Spherical Trigonometry and the subject of Warped Surfaces.
In Cloquet's treatise Descriptive Geometry is applied
to the projection of a great variety of bodies, to the laws of
Optics, to the determination of shadows, and to the science
of perspective drawing. In the present treatise, the chapter upon Perspective is mostly taken from Lacroix.
There will be found, both in the Elements and in the Introduction to Descriptive Geometry, errors and slight omissions which can hardly be avoided in a first impression of a
work of this kind; but none, it is hoped, which essentially
affect the work.


Cambridge, October 28, 1829.




CONTENTS.
Page
PREFACE        --                              v
ELEMENTS OF GEOMETRY.
General Notions of Extension  -      -     -   1
PART FIRST.
SECTION I. THE PROPERTIES OF STRAIGHT LINES AND THE
CIRCULAR LINE    -      -     2-52
Of Parallel Lines             ---              4
Of Plane Figures            -     -      -     9
Of Ratios and Proportions -       -        - 22
Of Similar Rectilinear Figures -  -      -    28
Of the Straight Line and Circle      -      - 35
Of Polygons inscribed in a Circle and circumscribed
about a Circle                             46
SECTION II. OF TIIE MEASURE AND COMPARISON OF
PLANE SURFACES   -      - 53-70
Of the JMeasure of Rectilinear Figures  -  -  56
Of the Measure of the Circle  -  -            65
Of the Rectification of the Circumference of the Circle 66
Problems  -      -     -      -     -      -  70
PART SECOND.
SECTION I. OF PLANES AND BODIES TERMINATED BY
PLANES -                     72-96
Of Diedral and Polyedral Angles  -     -      78
Of Polyedral Bodies     -     -            - 80
Of the Volume of Bodies and its Measure  -    89




xvi


Contents.


SECTION II. OF THE ROUND BODIES             97-116
Of the Cone and its Measure  -     -      -      97
Of the Cylinder and its Measure  -     -      - 101
Of the Sphere and its Measure -    -      -     104
Of the Five Regular Bodies -           -         116
Practical Questions  -      -                   118
DESCRIPTIVE GEOMETRY.
Page
SECT. I.  Preliminary Explanations  -      -    121
Orthographic Projections of Points and
Lines   -      -      -      -      123
SECT. II. Of the Straight Line and the Plane    126
SECT. III. Of the Generation of Geometrical Magnitudes        -      -     -       140
Of Curves of Single and of Double Curvature -      -      -      -      - 141
Of Single Curve Surfaces -     -      143
Of the Cone and Cylinder     -      - 144
Of Warped Surfaces      -      -      145
Of Developable Surfaces         -     -146
Of Double Curve Surfaces -     -      147
Of Surfaces of Revolution    -      - 147
SECT. IV. Of the Projection of Curve Lines and
Curve Surfaces        -     -       148
Of Tangent Planes to Curve Surfaces  - 151
Plane Sections of Curve Surfaces  -   153
Of the Intersections of Curve Surfaces - 154
SECT. V. Of Linear Perspective    -      -      158
Of Spherical Projections    -      - 167
The Orthographic Projection of the Sphere 167
The Stereographic Projection of the Sphere 168
The Globular Projection of the Sphere - 172




ELEMENTS OF GEOMETRY.


GENERAL NOTIONS OF EXTENSION.
1. GEOMETRY is that science which teaches us to investigate the magnitudes and forms of extended things, or
extended space, and the relations of their parts.
2. A body, or the space which it occupies, is extended
in three directions; it has length, breadth, and thickness
or depth. If it were destitute of either of these dimensions, it would cease to be.
3. The space which a body occupies, is separated from
other space by what we call the surface or the outside of
the body. A surface is extended in two directions; it
has length and breadth, but is destitute of thickness, and
therefore makes no part of the body itself. A body then
and even definite space, is bounded by surface.
4. If the body have several faces, like a square block
of wood, for instance, these faces may be considered as
so many distinct surfaces, each of which is bounded by
the edges formed by the meeting of this face with the
other faces of the body. These limits are no part of the
surface; they have neither breadth nor thickness; they
have only length, and are called lines. The limit of a
surface, therefore, is a line.
5. The line itself is limited by a point; which has no
extension. A point may also be taken in a line, a surface, a body, or in extended space; it, however, makes
no part of either of these magnitudes. It has position,
but no extent. It may, by moving in space, be considered as generating a line.
1




2


Elements of Geometry.


PART FIRST.
SECTION I.- The properties of straight lines and circular
lines.
6. Between two assumed points (as A and B in
Fig. 1. fig. 1), several lines may be drawn. Among these there
may be one which tends the same way in all its parts;
this is the straight line. We therefore define a straight
line to be, one which has the same direction throughout its
whole extent. It is evident, that the straight line is the
shortest which can be drawn between two points. It is
also evident, that between two points only one straight line
can be drawn.
7. A line which continually changes its direction is
called a curve line.
Curve lines admit of great variety in the character of
their curvature, whereas among straight lines, the only
variety is that of length.
Of the various curves, the only one whose properties
are made the subject of elementary geometry, is the circular curve. This is the most simple of all curves, its
curvature being the same in every part. It is described
by one extremity of a straight line, revolving about the
Fig. 2  other extremity which remains fixed. In Jig. 2, if the
straight line AC be made to revolve about the point C
which does not change its position, the point A will describe the circular curve ABDA. This line, in reference to the circle ABDA, is called its circumference; and
it is manifest that every point in this curve is equally distant from the point C, called the centre of the circle. The
lines AC, BC, DC, &c, which measure this distance, are
called radii of the circle.  A  straight line passing
through the centre and terminating in the circumference, is called a diameter. Any portion of this curve,
as BD, is called an arc of the circle.
In the same circle are the radii equal? and why?
How does the diameter compare with the radius?
8. A surface to which a straight line can be applied
in every direction, so as to touch the surface through the
whole extent of the line, is called a plane surface, or simply
a plane. A surface which is neither a plane surface nor
composed of planes, is called a curve surface. Such is
the surface of a ball, a roll of paper, &c.




Elements of Geometry.


3


9. Two straight lines which have the same direction in
space, are called parallel lines (fig. 3). As they have Fig. 3.
the same direction, they can neither approach towards,
nor diverge from each other; parallel lines, therefore,
must be throughout at the same distance from each other;
and, howeverfar produced, can never meet.
10. Where two straight lines taken in the same plane
are not parallel, they must incline the one to the other,
and are said to make an angle with each other (fig. 4). Fig. 4.
An angle, then, is the inclination of one line to another.
Tile magnitude of the angle depends upon the degree
of inclination, and not at all upon the distance or length
of the lines.
11. The nearer two lines are to being parallel, the
smaller the angle, which becomes nothing when the
lines become parallel. It increases as the lines diverge;
and when this inclination is the same on both sides (as
AB and CD,fig. 5), the angles are called right angles. Fig. 5.
In this case, the line AB is said to be perpendicular
to the line CD. CD is also perpendicular to AB.
When two lines are neither parallel nor perpendicular
to each other, they are said to be oblique to each other.
Remark. We have, then, in geometry, four kinds of
magnitudes. 1. That of a line. 2. That of a surface.
3. That of a body. 4. That of an (inclination or) angle.
Angles may be added, subtracted, multiplied, and divided,
like other magnitudes or quantities.
12. Two lines which form an angle in a plane, either
meet, or being produced, will meet each other in some
point, the point E, for instance, as infig. 6. This point Fig. 6.
of meeting is called the vertex of the angle. We express
the angle, usually, by three letters, thus, the angle AEC.
When there is only one angle at the same vertex, we express it by a single letter; as the angle E (fig. 7), that Fig. 7.
is, the angle whose vertex as at E.
13. Suppose the line AB perpendicular to CD (fig. Fig. S.
8) to move about the point B, into the position A'B, the
angle A'BC will be greater than a right angle, and the
angle A'BD, will be less than a right angle; but the
amount of angular space is not changed by this movement; the angle ABA' is taken from one of the angles
and added to the other; the sum of the two angles is the
same as before. Hence we say, that, when one straight
line meets another straight line, the sum of the two angles
is equal to two right angles.




4


Elements of Geometry.


Fig. 8.  The sum of the angles being equal to two right angles, the exterior lines CB, BD, must be in the same
straight line; otherwise something must be added to or
subtracted from  the angular space to make CBD    a
straight line; which would make the sum of the two angles greater or less than two right angles.
Remark. An angle which is less than a right angle
is called an acute angle; as A'BD. An angle greater
than a right angle is called an obtuse angle, as the angle
A'BC. Both obtuse and acute angles are called oblique
angles.
14. If other lines were drawn'to the same point B, on
Fig. 9. the same side of CD, as in figure 9; the whole amount
of angular space, would evidently be neither increased
nor diminished by it; but would still be equal to two
right angles. And we hence derive this general truth,
that, —l 'Ven a straight line is met by several other
straight lines, in the same point, the sum of all the angles
on the same side of it, is equal to two right angles.,
15. If all these lines be produced through B, it is
manifest that the angular space belzo the line CD, will
be equal to the angular space above; therefore,-The
sum of all the angles made by several straight lines diverging fiom the same point, is equal to fiur right angles.
16. As a straight line has the same direction in all its
parts (6), two straight lines must have the same inclina-.  tion in every part (g. 10); therefore if they intersect each
i 10other, the angles which are opposite, at the vertex, (called
vertical angles,) are equal. Each of the two angles
AEC and DEB, expresses the inclination of the two
straight lines AB and CD, to each other; they are therefore equal angles.
If the two lines are perpendicular to each other, the
two angles on the same side of either of them, are
equal; but these are respectively equal to their vertical
angles (16); it is therefore evident that all right angles
are equal.
Of Parallel Lines.
17. The properties of parallel lines are deduced from
this fundamental proposition.  Two straight lines which




Elements of Geometry.


5


have the same direction in space, must be equally inclined
to the same straight line which meets them.
This proposition results directly from the definition of
a straight line. If two straight lines have different inclinations to the same straight line, that is, to the same
direction, they must be inclined to each other, and therefore cannot be parallel.
18. Suppose the straight line AB (Jig. 11) to intersect Fig. 11.
the two parallel straight lines CD and FG; it must have
the same inclination to both. This inclination is expressed by either of the angles AEC, AHIF; they are therefore equal. As these angles have correspondent situations in relation to these two intersections, they are called corresponding angles.*  And as nothing in this analysis depends upon any particular inclination of AB to
the parallels, we derive this general truth:  Wh7en a
straight line intersects parallel straight lines, the corresponding angles are equal.
19. The sum of the two angles AEC and CEH, is
equal to two right angles (13); if instead of the angle
AEC, we substitute its equal E-IF (18), we shall have
the sum of the angles CEH and EIIF, equal to two right
angles. These angles are called interior on the same
side of the cutting line; and we say that-WhVen a
straight line cuts two parallels, the sum of the two interior angles, on the same side, is equal to two rightangles. And as the sum of the four angles on the same
side of the cutting line, or secant, must be equal to four
right angles.  The sum of the exterior angles on the same
side, is equal to two right angles.
20. The angle EHF is equal to its corresponding angle AEC (18), and the angle DEH is equal to the angle
AEC, being vertical with it (16). Therefore the two
angles, DEH and EHF, (each equal to AEC,) are equal
to one another. These angles are internal with respect
to the parallels, but on alternate sides of the cutting line.
Whence we derive the general truth,-A straight line
cutting parallels, makes the alternate-internal angles
equal.
21. The angle GHB is equal to the angle DEH (18),
and therefore equal to AEC, vertical with DEH. These
* They are sometimes called internal-external angles.
1I




6


Elements of Geometry.


Fig. 11. two angles AEC and GIHB, are exterior with respect to
the parallels, and on alternate sides of the secant; they
are therefore called  alternate-external; and we have
this general proposition, — Then a straighlt line intersects
two parollel straight lines, the alternate-external angles
are equal.
22. These five propositions may be summarily enunciated as follows:  When a straight line meets two parallel straight lines,
(1.) The corresponding angles are equal.
(2.) The sum of the two interior angles upon the
same side, is equal to two right-angles.
(3.) The sum of the two exterior angles on the same
side is equal to two riglt-angles.
(4.) The alternate-internal angles are equal to each
other; and
(5.) The alternate-external angles are equal to each
other.
23. If we incline one of the parallels in question, FG
for instance, the other two lines remaining the same, it
will change the magnitude of each of the angles at H;
and neither of the five propositions will be true. We have
therefore the converse of these propositions; and we say
that, when a straight line meets two straight lines, and
the corresponding angles are equal; or the sum of the
two interior ang4es on the same sidle, or the sumn of the
two exterior, is equal to two righat angles; or the alternateinternal alngles equal to each other, or the alternate-cxternal angles equal to each other, the two lines must be parallel.
Cor. If the two interior angles are less than two rightangles, the lines incline towards each other, and will
meet on that side, if produced sufficiently far. If the
sum of the interior angles on one side is greater than
two right angles, the sum of those on the other side, is
less than two right-angles (13), and therefore the lines
will meet on that side, it produced sufficiently far.
Fig. 12.  24. Let the two angles ABC, DEF, (fig. 12) have the
side AB, parallel to DE, and BC parallel to EF; produce BC to I, and also produce DE till'it meet BI in H.
Then on account of the parallels EF and BI, the corresponding angles DEF and DHI, are equal (22); and
with reference to the parallels DH and AB, the corresponding angles DHI and ABC are equal; that is, ABC




Elements of Geometry.


7


is equal to DEF; and universally-Two angles which
havc their sides parallel and directed the same way, arc
equal.
25. To measure a straight line, we apply to it a scrle or
rule, or some other straigrht line of kl:own and standard
measure, and tlus ascertain its length. This corresponds with general practice in analogous cases; one
quantity is usually measured by another quantity of the
same kind.
In measuring an angle, geometers have adopted a
method somewhat different. To ascertain the magnitude of an angle, they measure thle portion of a circular
arc embraced by the two sides of tile angle, the vertex of
the angle being at the centre ofthe circle.
26. To obtain a clear idea of the magnitude of a an angle, and the connexion wlhich it has with a circular arc,
let us suppose (fig. 13) that at first the two lines AC Fig. 13.
and BD, coincide in tile part BC, and that the part A, of
the line AC, be raised, so that this line, departing from
BC, may revolve about the point C; it becomes immediately inclined to BC; and this inclination increases as
the arc described by the point a increases, and in the
same degree; tlat is, for the same amount of angular
motion in any part of the revolution of the line AC, the
arc described by the point a will be the same; so that
when two angles are equal, the arcs drawn with the same
radius, from the vertices of the angles, as centres, will be
also equal. We ihence see how the magnitude of an angle may be designated by a circular arc.
For this purpose, the ancients divided the circumference of the circle into 360 equal parts, called degrees;
each degree into 60 minutes; and each minute into 60
seconds.  And the magnitude of an angle they expressed by the degrees, (0) minutes (') and seconds (",)
which express the value, or magnitude, of the arc comprehended between the sides, the vertex being at the
centre of the circle; thus, an angle of 35 degrees, 27
minutes, 15 seconds; usually written 350 27' 15'.
27. We have seen (15) that the sum of all the plane
angles made at the same point, is equal to four right
angles; and it is manifest that the sum of all the arcs,
which they would embrace in a circle described from
their common vertex as a centre, would be the entire
circumference. A circumference then, or 360~, is the




8


Elements of Geometry.


measure of four right angles. If, therefore, through the
rig. 14. centre of the circle (fig. 14) we draw two diameters, perpendicular to each other, we shall divide the circle into
four equal parts, called quadrants. ABC is a quadrant;
the arc AB, is an arc of 90~, and the measure of the angle ACB, which is a right-angle. A right angle then is
an angle of 90 degrees. If we divide the arc AB into
90 equal parts, and from C through the 55th division,
draw the line CD, we shall have the angle ACD, an angle of 55~; and the angle BCD, an angle of 35~. The
obtuse angle ACD' is measured by the arc AD', greater
than a quadrant.
28. Of the two acute angles DCA and DCB, each
is called the complement of the other, because each
is just what the other wants to make it a right-angle.
The acute angle DCA, and the obtuse angle DCA', are
called supplements of each other, because each being subtracted from the sum of two right-angles, or 1800, will
give the other. Angles complemental to each other, are
both acute. Of two angles supplemental to each other, if
one be acute the other will be obtuse, and vice versti.
Two right-angles are supplements of each other.
29. A method practised by surveyors for measuring
angles in the field, is by means of an instrument called
Fig. 15. a semicircle, (Jig. 15). At the two extremities of the
diameter AB, sights are fixed, through which you look
directly along the diameter. To the centre C, a moveable index is attached, with sights at its extremities,
D, E, so that the line DCE is a straight line. When an
angle of a field is to be measured, this instrument is
placed horizontally with its centre C at the vertex of
the angle, or, in other words, exactly at the corner of the
feld. The instrument is so placed that, by looking
through the sights A, B, we look along one side of the
field; the index DE is then turned about the central
point C, till the line of sight is directly along the other
side of the field. The magnitude of the angle will be
expressed by the number of degrees, minutes, &c. in the
graduated arc BE, comprehended between the stationary
and moveable diameters.
30. When these angles are to be transferred to paper,
for the purpose of giving a plan of the ground; an instrument called a protractor, is used. This is usually a
Fig. 16. semicircular piece of brass (fig. 16), graduated in de



Elements of Geometry.


9


grees and parts; and the centre C accurately determined. When an angle is to be set off upon the plan, one
of the lines embracing the angle is first drawn; the protractor is then placed with its centre at the point where
the vertex of the angle is to be, and its semidiameter
lying along this line. The number of degrees, &c, is
then counted off upon the arc, or limb (as it is called) of
the instrument, and the point carefully indicated; then
through Ihe centre and this point a line is drawn, and
we have the angle required.
31. Wlen tie magnitude of an angle already constructed, is required, the protractor is placed with its
centre at the vertex of the angle, and its diameter lying
along one of the sides; the number of degrees &c. will
be read upon tile graduated limb, at the point where it
is cut by the other side of the angle.
32. To construct an angle equal to a given angle
without the ]protractor. Suppose ACB (fig. 17) to be Fig. 17.
the given angle; and suppose at c, in the straight line
cb, it is required to make the angle acb, equal to ACB.
With a convenient radius CD, from C as a centre, describe the arc DE; and from c, as a centre with the
same radius, describe the arc d c; then from d as a centre with a radils equal to DE, describe another arc cutting the arc de, in the point e; and through the points
c and c, draw the line c e a, and you have the angle a c b,
equal to the angle ACB.
It is evident that, in the angle ACB, if the sides were
more inclined, that is, if the angle were increased, the
distance of the points D and E would be increased; and
if the angle were diminished this distance would be diminished; therefore, while the distance of tlese points is
the same, and their distance from the vertex of the angle is not changed, the angle must be the same; but
these conditions are the same in the two angles ACIB,
acb; they are therefore equal.
Of Plane Figures.
33. A plane has been defined,-a surface, to which
a straight line, being applied in every direction, will
touch the surface in its whole extent. Plane figures are
portions of plane surface bounded or enclosed by lines.




10


Elements of Geometry.


Those bounded by straight lines, or right lines (as they
are frequently called), are denominated rectilinearfigures.
Those bounded by curve lines are curvilinearfigures.
34. Among rectilinear figures, as two straight lines
cannot enclose a space, the simplest is that of three sides,
called the triangle. Triangles are differently denominated, according to the different relations of their parts.
When the three sides are of equal length it is called an
Fig. 18. equilateral triangle (Jig. 18).  When two of its sides
are equal, it is called an isosceles triangle.  When
no two sides are equal, the triangle is called scalene
Fig. 19. (fig. 19). When it has one right angle, it is called a
Fig. 20. right-angled triangle (fig. 20). In the right-angled triangle, the side opposite the right-angle is called the hypothenuse.
35. The first inquiry concerning the properties of triangles, respects the sum of the three angles; is it always
Fig. 21. the same?  Let us take the triangle ABC (fig. 21), and
through the vertex A draw the straight line DE parallel to
the base BC; we then have AB, a straight line meeting the
two parallel lines DE and BC; the alternate-internal angles DAB and ABC, are equal; and AC being a straight
line meeting the same parallels, the angles EAC and
ACB, are equal, for the same reason; then the angle
DAB being equal to the angle B, and the angle EAC,
equal to the other angle C, the three angles DAB,
BAC and CAE, are equal to the three angles of the triangle; but the sum of these is equal to two rightangles (14); therefore, the sum of the three angles
of the triangle is equal to two right-angles. It is evident
that, however the sides and angles of the triangle may
be changed, they will always admit of a straight line
(as DE) being drawn through the vertex of one of the
angles, and parallel to the opposite side; and that the
two outer angles at this vertex must therefore be equal to
the other angles of the triangle. We say, therefore, that,
the sum of the three angles of every triangle, is equal to
two right-angles, or 180~.
36. (1.) If we know two angles of a triangle, how
can we find the remaining angle?
(2.) How many obtuse angles can a triangle have?
Why?
(3.) How many right angles can any triangle have?
Explain.




Elements of Geometry.


11


(4.) If one angle of any triangle, be a right-angle, Fig. 21.
what will be the sum of the other two?
(5.) If one of the oblique angles of a right-angled triangle be given, how can we find the other?
37. If we produce the base of tle triangle BC to F,
the angle which it makes with AC on the outside is
called the exterior an gle; and because AC is a straight
line meeting the two parallels DE and BF, this angle is
equal to the angle DAC; but DAC is composed of the
two angles DAB and BAC; and DAB is equal to the
angle B; therefore this exterior angle ACF, is equal to
the sum of the two angles ABC and BAC, of the trian.
gle; these two angles with respect to the exterior angle,
are called interior and opposite. We say then —The exterior angle, nmale by producing one of the sides of the
triangle, is equal to the sum of the two interior andl o)pJosite cangles.
38. Problem. The three sides of a triangle being
given, to construct the triangle.
Let the three given sides be the lines A, B. C, (fig. 2-). Fig. 22.
Draw the line DE equal to the given line A; then from D
as a centre with a radius equal to the given line B, describe
an arc; and from E as a centre, with a radius equal to
the other given line C, describe an arc cutting the other
are in F; draw DF and EF, and you have the triangle
required.
39. It is plain that no different triangle can be formed
witll these three lines. The only different construction
which the case admits, is to make the triangle on the
lower side of the base, as the triangle D'E'F; but this
triangle is not really diftcrent from the first. To show
this, turn the last triangle over by lifting up the part F'
and making the whole turn about the base DIE'; then
place it upon the first so that the point D' will be upon
D, and E' upon E; this may be done, as each of the
bases is equal to the given line A. The triangles will
then coincide in all their parts, and must therefore be
equal. Tlhe point D' being upon D, the point F' must be
at the same distance from D, as the point F is, it must
therefore be in the first described arc; and as E' coincides with E, F' must be at the same distance from E, that
F is; it must therefore be somewhere in the other
arc which crosses the first in F; if F' is in each of these
two arcs, it can only be at their intersection, and there



12             Elements of Geometry.
Fig. 22. fore falls upon the point F, and the two triangles coincide in all their parts; they are not different, therefore,
but equal; and eve say, universally, When two triangles
have the three sides of the one equal to the three sides of
the other respectively, the angles will also be equal, resplectively, and the two triangles will be equal in all their
parts. This is the frst case of equal triangles.
40. Problem. Two sides of a triangle being given,
and the angle contained by these sides, to construct the
triangle.
Fig. 23.  Let A and B (fig. 23) be the two giveni sides, and C the
given angle. Draw DE equal to the given line A; at D
make the angle EDG, equal to the give angle C, (32)
and produce the side DG to F, making DF equal to the
given line B; then join FE and you have the triangle
required.
41. It is evident that no different triangle can be constructed with these two sides containing an angle equal
to C. The only different construction with these things
given, would be to make the triangle on the other side
of the base DE; but it is plain that the triangle thus
formed would only be the triangle DEF turned over; the
two triangles would therefore be equal in all their parts.
And we say universally, —Vhen two triangles have two
sides of the one equal to two sides of the other, each to
each, and the angle contained by these two sides of the
one, equal to the angle contained by the two sides of the
other, the two triangles arc equal in all their parts. This
is the second case of equal triangles.
42. Problem. One side of a triangle, and the two angles adjacent t tthat side, being given, to construct the
triangle.
Fig. 24.  Let the line A (jig. 24) be the given side, and the angles B and C, be the angles adjacent to that side. Draw
the line DE equal to A, for the base of the triangle; then
at the point D, make an angle equal to the given angle B;
and at the point E, make another angle on the same side
of DE, equal to the other given angle C; produce these
two sides till they meet in F, and the triangle is constructed.
43. It is manifest that no different triangle can be
formed with these three things given. We may, as in
the other cases, construct the triangle on the lower side
of the base DE; but the triangle thus formed might be




Elements of Geometry.            13
turned over, an'd iaced upon the triangle DEF, and because the bases a e equal, and the angles at the bases the
same, the side DF' would fall upon DF, and the side
EF' would fall upon EF; the point F' would therefore
fall directly upon the point F; and the two triangles
would entirely coincide in all their parts.  We say,
therefore-If two triangles hatce a side andl the two
adjacent angles of the one, equal to a side and the two
adjacent angles of the other, each to each, the two triangleIs will be equal in all their 1parts; that is, the third
angle of the one will be equal to the third angle of tile
other, and the two remaining sides of tile one will be
equal respectively to the two remaining sides of the other. This is the third case of eq/al triangles.
44. Problem. Tlhe hypothenuse and one other side of
a right-angled triangle being given, to construct the triangle.
Let thle line A (fig. 25) equal the hypothenuse, and the Fig. 25.
line B, the other given side of the triangle. Draw the
line CD equal to the given side B: and at D draw the
indefinite line DE, perpendicular to CD; tlen from C as
a centre, with a radius equal to the given hypothenuse,
draw an arc cutting the perpendicular in E; draw CE,
and you have the right-angled triangle required.
[Let the learner show that no different triangle could be constructed with these things given.]
We say then-If two right-angled triangles have
'/he ihypothenuse and a side of the one, eqiual respectively
io the qpot    e npothense   a  side of the other, the two triangrls will be equal in all their parts.
This is the fourth case of equal triangles.
45. We have seen tlat tle sumI of the tliree angles of
every triangle, is equal to two right-angles; yet there is
an infinite variety in the magnitude of the individual an-,ies: and this evidently depends upon the relative magiitudes of the sides. We have learned that when the
sides are the same, the angles must be the same (39).
Let us examine this subject a little farther. We will
take the isosceles triangle ABC (fi. 26), of which AB Fig. 26.
and AC are the two equal sides. From the vertex A, to
the middle of the base BC, draw the straight line AD.
This will divide the triangle ABC, into two triangles ADB, ADC. If we examine them, we shall see
that AB of the one is equal to AC of the other, that DB
2




14


Elements of Geometry.


Fig. 26. of the one is equal to DC of the other, and that AD is a
side common to both; they are therefore equal by the
first case of equal triangles; and if ABD were turned
over upon ADC, we should find that the two would coincide in all their parts. The angle B is therefore equal
to the angle C; the two angles whose vertices are at
A, equal to each other; and the two angles whose
vertices are at D, equal to each other; the two last
are right angles of course (11). Whence we derive
the following important truths-1. In an isosceles, triangle, the angles opposite to the equal sides, are equal.
2. A straight line drawn from  the summit of the isosceks triangle to the middle of the base, is perpendicular to
the base, and bisects the angle whose vertex is at the siumm it.
Fig. 27.  46. If the triangle had been equilateral (fig. 27), that
is, ha-' had its three sides all equal, either side might
have been taken for the base. This would have proved
the angle A equal to the angle B, and the angle C equal
to the angle B; hence-An equilateral triangle is also
equiangular; that is —has all its angles equal.
As the sum of the three angles of every triangle,
is equal to two right-angles, or 180 degrees;  in an
equilateral triangle, each angle is an angle of 60 degrees.
If you have the angle whose vertex is at the summit of
an isosceles triangle, how can you find the other angles?
If you have an angle at the base of an isosceles triangle, how will you find the other angles?
Fig. 28.  47. Take next the scalene triangle ABC (fig. 23). Let
AB be the greater side; take upon AB, the part AD
equal to AC, and draw DC. The angle ADC may be
considered as an exterior angle to the triangle DBC, and
therefore greater than the angle B; the triangle ADC is
isosceles, and consequently has the angle ACD equal to
the angle ADC; but the angle ACB is greater than
ACD; for a still stronger reason is the angle ACB,
greater than ABC. But the side AB is greater than AD,
or than its equal AC. We see then that the greater side
is opposite to the greater angle. If we had compared
either of these angles with the angle A, and also their
opposite sides, the result would have been similar;
whence we have this general rule-In any triangle, the
greater side is opposite to the greater angle. The side
opposite to an angle is frequently said to subtend the angle.




Elements of Geometry.


15


48. If two sides of a triangle are unequal, the opposite
angles will be unequal (47); therefore —If in any triangle
two of the angles are equal, the sides opposite those angles will be equal, and the triangle will be isosceles.
49. In a right-angled triangle, as the two oblique angles must be each less than a right-angle, —the hypothecnuse (the side opposite the right angle) must be greater
than either of the other sides. The two sides which contain the right-angle are frequently called the legs of the
triangle.
50. Either side of a triangle, being a straight line and
therefore the shortest which can be drawn between the
vertices at its extremities, must be less than the sumt of the
other two sides. It follows from this, that cither side of a
triangle must be greater than the differenc between tlhe
other two sides; if this were not the case, the greatest
would exceed or at least equal the sum of tile two others.
51. Suppose AB (Jig. 29) perpendicular to DC, and Fig. 29.
AE and AF two lines drawn fromn the same point A, and
falling obliquely upon DC at equal distances from the
point B; the two triangles ABE and ABF, will have
AB a common side, EB of the one equal to BF of the
other, and the angle ABE equal to the angle ABF (11);
the triangles will be equal by the second case, and will
have the hypothenuses AE, AF equal, and each greater
than the perpendicular AB (49). Draw AG at a greater
distance from the perpendicular than AE is; you will
have the triangle AEG obtuse-angled at E, because this
angle is the supplement of the acute angle AEB (28);
and in the triangle AEG, the side AG being opposite to
the obtuse angle is greater than AE opposite an acute
angle; and we have tlese three propositions. If jfrom a
point w   ut i t rghot a st t line, a perpendicular be drawn to
that line, and also sevcral oblique lines, —
(I.) 7/heperpendicular is less than either of the oblique
lines.
(2.) Oblique lines equally distant from the perpendicular are cqual; and
(3.) Of two oblique lines, that is the greater which is at
the greater distance from the perlpendicular.
52. It follows from the second of these propositions,
that —If a percpndicular be drawn through the middle of
a line, AB, every point in the perpendicular will be
equally distant from the two extremities of this line; for




16


Elements of Geometry.


Fig. 29. straight lines drawn from any point in the perpendicular
to the two extremities, will be oblique lines drawn at
equal distances from the perpendicular, and therefbre
lqutal.
53. As two points are sufficient to fix the position of
a straight line; If one straight line pass through the
middle of another straight line [as AB through the middle of EF]; and if any other point in the first of these
straight lines, be at equal distances from the two extrem'ities of the second, these lines will be perpendicular to each
other. And for the same reason —If each of two points
in a straight line, be equally distant from the two extremities of another straight line, the first will bisect the
second, and be perpendicular to it. To bisect is to divide
into two equal parts.
54. Problem. From any point in a given straight line,
to draw a perpendicular to that line.
Fig. 30.  Let A be a given point in the line BC (fig. 30), from
which it is proposed to draw a perpendicular. Take two
points B and C in the given line, equally distant from
the given point A; from each of these points B and C,
with any radius greater than AB, draw arcs cutting each
other in some point D, above or below the given line;
and from A, draw a line through D, which will be the perpendicular required (53).
It is evident that there can be only one line perpendicular to BC, and passing through the point A; for
if there could be two, as they make the same angle with
the line BC, they will both have the same direction; and
having the same direction, and passing through the same
point, they must coincide entirely, and be one and the
same straight line.  There cannot, therefore, be more
than one eperpendicular drawn through the same point in
the same straight line.
55. Problem. From a given point without a straight
line, to draw a perpendicular to that line.
Fig. 31.  Let A (fig. 31) be the given point, and BC the given
straight line.  From A as a centre, with any radius
greater than the distance of the point A, from the given
line, draw an arc cutting the given line in the points
D and E; then from each of these two points D and E,
with a radius greater than half of DE, draw arcs cutting each other in F; and from the point A, in the direction AF, draw the line AG; this line will be the per.




Elements of Geometry.


17


pendicular required; as it has two points each equally
distant from the extremities of the line DE (53).
56. Problem. From a given point without a straight
line, to draw a perpendicular to that line, when the perpendicular will come near one extremity of the line.
Let A (fig. 32) be the point from which it is required Fig. 32.
to draw the perpendicular to the line BD. From B as
a centre with a radius equal to the distance BA, describe
an arc below BD; and from E as a centre with a radius
equal to the distance EA, describe an arc cutting the first
arc in A'; then from A in the direction AA', draw the
line AG, which will be the perpendicular required. As
B and E are two points in the straight line BG, each
equally distant from A and A'; BG must be perpendicular to AA' (5'3); that is, BGA is a right-angle.
Any other line drawn through A, must be different from AG, and therefore differently inclined to BG;
but AG is perpendicular to BG; any other line must
consequently be oblique to BG. We say thereforeFrom the same point without a straight line, only one
perpend(icular can be.drawn to that line.
57. Problem. To bisect an angle, as the angle C (fg. Fig. 3::.
33). Upon the sides, take CA and CB equal to each
other; draw AB, and from C draw CD perpendicular
to AB; CD will bisect the angle ACB (45).
58. Problem. Through a given point, A (fig. 34), to Fig.;M.
draw a straight line parallel to a given straight line, BC.
From A as a centre, with a radius sufficiently great, describe the arc ED; and from E as a centre with the same
radius describe the arc AF; take EG equal to AF, and
through A and G draw a straight line, which will be parallel to BC: because the alternate-internal angles GAE
and AEF, being measured by equal arcs, are equal (26),
and consequently the two lines AG and BC, to which
they refer are parallel.
59. Suppose two parallel lines, AB and CD (fig. 35), Fig. 35.
to be cut by two other parallel lines, EF, GiH, the space
1KML, comprehended by these parallels, is called a parallelogram. We say then-A parallelogram is a quadrilateral (or four-sided) Jigure of which the opposite sides
areparallel. The interior angles BIK and IKM are equal
to two right angles (22); and the interior angles IKM and
KML, are equal to two right angles; if, therefore, IKM
added to either of the two angles BIK or KML, gives the
2*




18


Elements of Geometry.


Fig. 35. same sum, BIK must be equal to KML. In the same
manner we may show that the two angles IKM and ILM
are equal to each other. We perceive that, in both cases,
the equal angles are opposite to each other; we therefore
say —li every parallelogram, the opposite angles are equal.
60. If we draw through the vertices K and L, the
straight line KL, called a diagonal of the parallelogram;
it will be a straight line meeting the two parallels AB
and CD; the alternate angles ILK, LKM are equal
(22); and because the straight line KL meets the two
parallels EF and GIH, the alternate angles IKL, KLM,
are equal. We have then, the two triangles IKL, KLM,
of which KL is a common side, and in which the angles
IKL, KL[, of the one are respectively equal to the angles KLM, LKM, of the other; the triangles are therefore equal by the third case of equal triangles; and the
sides opposite the equal angles will consequently be
equal, viz. IL to KM, and IK to LM. But these equal
sides are parallels comprehended between  parallels.
From whicl we derive these general truths:
(1.) The diagonal of' a parallelogram divides it into
two equal triangles.
(2.) The opposite sides of a parallelogram are equal.
(3.) Parallel lines comprehended between parallel lines.'c equal.
Fig. 3(.  61. Suppose AB and CD (Jig. 36) to be two parallel.;,id equal lines. By joining their extremities by the.;-.liglt lines AC and BD, we have the quadrilateral
A!^D)C; and by drawing the diagonal CB, we have two
triarb:les, ABC  and CBD.   Since AB and CD    are
parallel, and CB is a straight line meeting these parallels,
the alternate angles ABC, BCD, are equal; and as AB
and CD are equal, and CB a side common to the two triangles, these triangles are equal by the second case, and
therefore equal in their corresponding parts; namely,
the angle ACB is equal to the angle CBD; but these angles are alternate-internal with respect to the two lines
AC and BD, and the secant BC; these lines are therefore parallel (23), and the figure ABDC is a parallelogram. We therefore say —When, in any quadrilateral,
two opposite sides are equal and parallel, the other two
sides will be equal and parallel; and the figure will be a
parallelogram.
62. As two adjacent angles of a parallelogram are




Elements of Geometry.


19


equal to two right angles, the sum of all the angles is
equal to four right angles. And if one of the angles of
a parallelogram is a right angle, they will all be right angles; and the figure will be what is called a rectangle.
If two adjacent sides of a parallelogram are equal; as
the sides opposite are equal to them, the sides will all be
equal. In this case, if the angles are right-angles, the
figure is called a square. If the sides of a parallelogram
are equal, and the angles oblique, the figure is called a
rhombus. If the sides are unequal and the angles oblique
the figure is called a rhomboid.
63. If in the rhomboidal parallelogram ABDC (fig. 37) Fig. 37.
we draw the diagonals AD and CB, and compare the
triangles AEB, CED, we shall have the two angles BAE,
EDC, alternate-internal with reference to the parallels
AB and CD; they are therefore equal; the two angles
AEB and CED, are vertical angles (16), and therefore
equal; consequently the third angle ABE, of the one, is
equal to the third angle ECD, of the other (35); and
the side AB is equal to the side CD, being opposite sides
of the parallelogram. We have, therefore, the side AB,
and the adjacent angles A and B, in the one triangle,
equal respectively to the side CD and the adjacent angles D and C of the other; the triangles are consequently equal by the third case; and their corresponding
parts are equal; namely, AE is equal to ED, and CE is
equal to EB; that is, these diagonals bisect each other
at E. It is evident that this fact results from the equality and parallelism  of the opposite sides; whence we
have the general truth-In every parallelogram, the two
dicagonals bisect each other.
64. If the parallelogram ABDC were a square or a
rhombus, ABD would be an isosceles triangle (34), and
BE would be perpendicular to AD (45). Hence-In a
square or in a rhombus, the two diagonals bisect each other
at right angles.
65. A trapezoid is a quadrilateral which has two of its
sides parallel, and the other two oblique, to each other.
If in a quadrilateral no two of its sides are parallel, the
figure is called a trapezium.
Remark. Rectilinear figures of five sides are called pentagons; of six, hexagons; of seven, heptagons; of
eight, octagons, &c. But all rectilinear figures are comprehended under the general name of polygons. The




20


Elements of Geometry.


sum of all the sides of a polygon is called its perimeter.
66. Problems. (1.) Iaving two contiguous sides and
the contained angle, in a parallelogram, to construct the
parallelogram.
Fig. 38.  Let A and B (fig. 38) be the given sides, and C the
given angle. Draw DE equal to A; and at D, make the
angle EDF, equal to the given angle C; and make DF
equal to B; then through F draw FG parallel to DE,
and through E draw EG parallel to DF, and the parallelogram is completed.
(2.) Construct a square upon an assumed line, as one
of its sides.
(3.) Construct a rhombus upon an assumed line, and
having an assumed angle.
These examples are sufficient to enable us to construct
any rectilinear figure, when we have the sides given, and
the angles which they make with each other. In this
way the plan of a town or an estate is constructed.
67. We have seen that the sum of all the angles in a
parallelogram is equal to four rightt angles (62). It is
desirable to know whether there is any general law respecting the angles of polygons.
Fig. 39.  Let us take the polygon ABCDE (fig. 39); produce
each of its sides outward in succession; and from some
point in the interior of the polygon, draw lines parallel
to each of the sides. It is evident that each of the angles,
a, b, c, &c. must be equal to the corresponding exterior
angles of the polygon (24); but the sum of all the angles
made by lines diverging from the same point in a plane,
is equal to four right angles (15). It is manifest that how
many sides soever the figure may have, from the same
point lines may be drawn respectively parallel to each of
the sides. We say therefore —The sum of all the exterior
angles of every polygon, made by producing the sides
outward in succession, is equal to four right angles.
68. It will be perceived, that at each vertex of the
polygon, the sum of the interior and exterior angles is
equal to two right angles (13); consequently the sum of
all the interior and exterior angles is equal to as many
times two right angles as the polygon has sides. But the
sum of all the exterior angles is equal to four right-angles
(67); therefore-The sum of all the interior angles of
any convex polygon, is equal to as many times two rightangles as the figure has sides, wanting four right-angles.




Elements of Geometry.


21


Remark. These two propositions are limited to convex polygons; or such as have all their angles salient,
that is, having the vertices of the angles outward. In
polygons which have re-etecring angles, that is, angles
whose vertices are towards the interior of the polygon,
as the angle C (fig. 40), these propositions will not be Fig. 40,
true.
69. It is evident tlat any polygon, ABCDEF
(Jig. 41), may be divided into triangles, by drawing from Fig. 41.
one of the vertices, as A, diagonals to each of the other
vertices not adjacent. And it is manifest that if, in two
equal polygons (that is, in two polygons which would coincide if placed the one upon the other, which is called
super-position), diagonals be drawn from corresponding
vertices in the two polygons, to the opposite vertices,
the triangles in the one polygon, would be equal respectively to the triangles in the other; and the triangles
being equal, and similarly disposed, the polygons which
they compose must be equal. That is-Two equal]polygons may be divided into the same number of triangles,
respectively equal. And —If to polygons are composed
of the same number of triangles, equal and similarly disposed, the p)olyrons are equal.
70. A more important method of determining a polygon by triangles, is that in which the triangles have for a
common base, one of the sides of the polygon, as AB
(fig. 42), and their summits respectively in the opposite Fig. 42.
vertices of the polygon. It is manifest that the triangles
ABC, ABD, ABE, ABF, having AB for a common
base, determine the points C, D, E, F; all the vertices
of the polygon are therefore determined, and consequently the polygon itself.
71. In this way surveyors sometime determine the most
important points in a field or an estate, whether on its
border or in the interior. A straight line is measured,
from each extremity of which all the points to be determined can be seen. This line is called the base line,
because it is made the base of a series of triangles whose
summits are to determine the exact position of the other
points. No point can in this way be determined, which
is in the base line produced. This method of determining the points in a survey, is called triang-ulation.
72. We have discussed the general properties of individual triangles, and their relative parts. We have also




22


Elements of Geometry.


compared together such as are equal; that is, such as
may be placed, the one upon the other, so as to coincide
entirely. And we remark that this coincidence is an indispensable condition of equal figures.
Preparatory to a more extended comparison of triangles and other rectilinear figures; and a discussion of
some other properties of individual polygons, it is necessary to consider the doctrine ofproportion.
Of Ratios and Proportions.
73. Before entering upon an examination of the laws
of proportion, it will be proper to explain the meaning of
certain signs which are found very convenient in expressing mathematical truths. When we wish to indicate the addition of two quantities expressed by A and B,
we place between them this sign —, which is equivalent
to the Latin word plus, signifying more; thus, A + B;
which is usually read, A plus B, and means A added to
B. It expresses, of course, the sum of the magnitudes A
and B.
When we would indicate that B is subtracted from A,
we use the sign -, which is equivalent to the Latin
minus, i. e. less; thus, A - B, usually read, A minus B,
signifies the remainder after subtracting the magnitude B
from tie magnitude A, and expresses the difference of
the two magnitudes.
To express the multiplication of two magnitudes, we
use this sign X; thus, A X B; whicl is read, A multiplied by B; and signifies the product of these magnitudes.
A period (.) placed between two magnitudes, also
denotes their multiplication; thus, A * B signifies A multiplied by B.
To denote the division of one magnitude by another,
A
we write them in the form of a fraction; thus, B; which
is read A divided by B, and expresses the quotient arising from the division of A by B.
Instead of the expression A X A, we write, for the
sake of brevity, A2; which signifies the product of A
multiplied by itself, and is read, A second power. If we
would indicate the second power of the sum of two magnitudes, we should write (A + B)2; the second power of




Elements of Geometry.


23


the product, (A X B)2 *. To indicate the third power of a
magnitude, we write A3; and powers generally are designated in a similar manner. The figures 2, 3, &c. are
called exponents.
To indicate any root of a magnitude, we write the
letter indicating the magnitude with a fractional exponent expressing the root, as A", which we read, the
second root of A. So also the third root of A -- B, is
written (A — B).
To show that two magnitudes are equal, we write them
with the sign - between them; as A = B; this is read,
A equals B. The expression 5 + 4 = 9, is read 5 plus
4 equals 9. When we wish to state that A is greater
than B, we write them thus A > B. If we would say
that A is less than B, we write them A < B. The expressions 2A, 3A, &c. indicate double, triple, &c. the
magnitude represented by A.
We now proceed to the consideration of ratios.
74. To find a common measure of two lines, and hence
their ratio, we employ a method similar to that in arithmetic for finding the greatest common divisor of two
numbers.
Let AB and CD (fig. 43) be two straight lines of Fig 43.
which a common measure is required. Apply the less
CD to AB the greater, as many times as CD is
contained in AB; suppose this to be three times with
a remainder EB; so that we shall have AB equal to three
times CD added to EB, or AB = 3CD + EB. Now
let us apply the remainder EB to CD; as it will be
contained four times with the remainder FD, we shall
have CD = 4EB + FD.      Apply now the second remainder FD to EB; it will be contained once, with
the remainder GB; this gives EB = FD + GB. Then
apply GB to FD; it will be contained twice, exactly,
which gives FD   2GB. By retracing the steps of the
operation, we shall perceive that EB = 3GB; that
CD    14 GB; and that AB = 45 GB; from which it is
evident, that-The last remainder GB is a common measure of the straight lines AB and CD. As this measure
---— 2 -     2
* Sometimes we find them  written A  - B, A X B,
&c.




24


Elements of Geometry.


Fig. 43. GB is contained forty-five times in the first, and fourteen
times in the second, we say that the line AB is to the
line CD in the ratio of 45 to 14. This ratio we write in
a fraction, 47-.  The ratio of two lines, therefore, is the
number of times the first contains the second; as the ratio
of two numbers is the quotient arising from the division
of the first by the second.
75. It will be easy to apply this process to any other
example. We should continue the comparison of the
successive remainders, till the last remainder is contained an exact number of times in the next preceding. We
shall soon arrive at this result; or at least so near that
the error will be too small to be recognised by the senses; in this case, we say, we have the aljproximante ratio
of the two lines. This explanation, notwithstanding the
imperfection of the mechanical process of comparing
lines, is sufficient to give us a clear idea of what is meant
by the ratio of one line to another.
76. Remark. It is not at all necessary, in expressing
a ratio, that the greater magnitude should stand as the
numerator of the fraction. In the example above, we
said, that the ratio of AB to CI), is 45; but the ratio of
CD to AB is 4. We say that - 3; that is, 6 contains 2 three times, or the ratio of 6 to 2, is equal to 3;
we may say also, - -; that is, the ratio of 2 to 6 is
equal to one third.
77. When four magnitudes are such that the ratio of
the first to the second, is equal to the ratio of the third to
the fourt., these four magnittelcs are said to be in proportion.  Proportion is equality of ratios. "A PROPORTION," mathematically speaking, is a fornula, expressing the eavality of two ratios. Let A represent a magnitude which contains another magnitude a, as many
times as the magnitude B contains b; these four terms
will form a geometrical proportion. This proportion we
A    B*
write thus: -=. We read this proportion, —The
a    b
* A proportion has generally been written in this form;
A: a"::B: b; and read, A is to a, as B is to b. The
terms A, and B, are called anticcedents; and a, b, are
called consequents. The terms A and b are called the
extreme termns; a and B are called tile?rman terms. A




Elements of Geometry.


ratio of A large to a small, equals the ratio of B large to
b small; or, A divided by a equals B divided by b. That
is, A contains a as many times as B contains b.
78. Suppose that A contains a three times; then B
must contain b three times. Then the ratio of A to a is
A               B
equal to 3; that is- = 3.  So also -  3. We therea               b
A    B
fore see the propriety of the expression A  b.  Upon
a               b
this supposition we have - ', and also -; thereA   3'         B       there — 3
a b6
fore, on account of the common ratio -, we have  -- 
3,        A - B'
If we compare this equation with the one above, we
shall see that the terms of each ratio are the same, but
the numerators have taken the places of the denoininators, and vice versa. We therefore have-I. In any
proportion, the terms of each ratio may be inverted, and
the expression will still be a roportion.  This change we
call inversion.
A   B
79. If, in the original proportion,  —, we multiply
aXB
the equal ratios by a, we shall have A =; if now
A    a
we divide both sides by B, we shall have -= ~. That is
— II. I1 any proportion the ratio of the numerators, equals
that of the denominators. This change we call alternation.
80. If A contain a three times, A 4- a will contain a
four times; so also, if B contain b three times, B - b
will contain b four times; we shall therefore have the
fundamental principle of such a proportion is-The
product of the two means is always equal to the product of
the two extremes. The proportion, therefore, will give
A X b = a X B. If we divide both sides of this equaA B
tion, by a, and also by b, we shall have -  —, the exa    b
pression in the text. It is believed that the principles of
a proportion are more readily apprehended from  this
fractional method of writing them.
3




26


Elements of Geometr-y.


proportion  + -    +b     that is,-III. In a proportion, the sum of the terms of the first ratio, contains its
denominator as many times as the sum of the terms of the
second ratio contains its denominator.
81. How many times soever A may contain a, A -a
will contain a one time less; and how many times soever
B may contain b, B - b will contain b one time less.
A-a B-b
This gives ---         b      that is-IV. The difference of the terms of the first ratio, contains the denominator as many times as the difference of the terms of the
second ratio contains its denominator.
82. From the two last articles, we have by alternation
A - a      a    A- a        a       a.
B --- b-; and B      --    -; but as 7 is common to the
Bb-b'           B-bb                b
A
two proportions, and also equal to    we have
B
A - a     A -- a   A    a
AB + a    BA-a- BA        ba  That is-V. The sum of
B +b-B-b- B             b
the terms of the first ratio of a proportion contains the
sum of the terms of the second ratio as many times as the
first numerator contains the second, or as the first denominator contains the second denominator.  Also-VI. The
difference of the terms of the first ratio, contains the difference of the terms of the second ratio, as many times as
the first numerator contains the second, or as the first denominator contains the second, and, consequently —VII.
The sum of the terms of the first ratio contains the sum
of the terms of the second as many times as the diferencce
of the terms of the first contains the difference of the termn.,-if the second.
S3. If we apply the propositions of the preceding arti
A    a                     A + B
cle to the formula -    (79), we shall have a    +
B    b                     a.+ b
A-B A B
a-       - a b-     That is-VIII. In any proportion,
the sum of the numerators contains the sum of the denominators as many times as one numerator contains its denominator.  Also-IX. The difference of the numerators
contains the difference of the denominators as many times
as one numerator contains its denominator. And, there.




Elements of Geometry.


27


fore-X. The sum of the numerators contains the sum of
the denominators as many times as the difference of the
numerators, contains the difference of the denominators.
A B C
84. If we have a series of equal ratios, -- -  -
a    b   c
D E                                      A B
d   -, &c.; by applying law VIII, we obtain     =
d    e'                                  a + b
B       B    C                           A+B; or as —, we may write instead of this    - + 
b       b    e        -                   a-{-b
C                                       A~ B~ C,  C
-; and applying the same law to this, we have - +  -+
c                                       a + b +i"c
D      consequently A+B+C+D+E            E
d - and consequently  a  -     - +c  - +e  e
wherefore-XI. In any series of equal ratios, the sum of
the numerators contains the sum of the denominators as
many times as one numerator contains its denominator.
A    B     C
85. If we have the two proportions -   anda b c
D
-, we can multiply them together; that is, the first ratio of the first proportion by the first ratio of the second,
and the second ratio of the first, by the second ratio of the
second; and as they are equal quantities, multiplied by
equal quantities, they will give equal products. But to
multiply two fractions together, we multiply their numerators for a new numerator, and their denominators for a
AxC BxD
new denominator. This will give       ---
axe      bX d'
This process is called multiplying the proportions in order; the result is a compoundproportion, and each of the
ratios is a compound ratio. We say then-XII. Proportions multiplied in order, will give a proportion.
86. It is evident from the last article, that a proportio"n
may be multiplied by itself, term by term, and give:a
A B
proportion; thus, the proportion -- B, multiplied by
a    b
A A B B
itself, term by term, gives AX      X    butA X A
a X a     b X 6
is written A2 (73), B X B is B2, &c.  This proportion
A2   B2
may therefore take the form - =  --. If we multiply
a2   u2             r




28


Elements of Geometry.


A2   B2                         A
the proportion -— = b  by the original proportionB         ive A3   B3
-, it will give  -. As we may take the powers, so
it wl     a3    b3
also we may take the roots (of the same degree) of tile
several terms of a proportion and form another propor1     1       1     1
Ai thus       A     B 
tion; thus, -  =   -  and A -        We therefore
as-   by     a3     bhave this general law —XIII. The same powers or the
same roots, of the terms of a proportion, are also in proportion.
A2
87. Remark. The ratio -   is called the duplicate raa-2
A3
tio of A to a; the ratio -  is called the triplicate ratio
a3
A2
of A to a. The ratio-  is called a sub-duplicate ratio:
a2
and -, a sub-triplicate ratio.
a3
We now return to the examination of geometrical
figures.
Of Similar Figures.
88. In constructing the triangles &c. in the preceding
operations, we have considered ourselves as taking the
absolute magnitudes of the sides given in the problems 
but in most cases in practical geometry this is inconvenient and undesirable. The methods used in surveying
fields and townships, and constructing maps, afford a
good illustration of this part of the subject. The lines
by which the field &c. is bounded, are measured, and
also the several angles; and from the notes of the survey
a perfect representation of the outline of the field, township, &c. is constructed upon paper. This is called a
plan of the field, township, RSc. Suppose, for instance,
that we have a triangular field whose sides are eight, ten,
and twelve chains, respectively; the method of construction would be similar to that of the triangle with three
given sides (38). But instead of taking tile absolute




Elements of Geometry.            29
lengths of these known sides, we take lines embracing
eight, ten, and twelve divisions upon a scale of equal
parts, of any magnitude, J- of an inch, for instance,
which will give us a plan of a convenient size, as in the
figure ABC (fig. 44). This is a complete representation Fig. 44.
of the field surveyed; so also it would be, if the sides of
the field had been eight, ten, and twelve rods, or yards,
instead of chains. We should also have respresented the
field as perfectly, if instead of the scale of tenths of an
inch, we had used a scale divided into twentieths of an
inch, as in the figure a b c. If then all these different
triangles are faithfully represented by each other, they
must have a common character; they must be similar.
Let us see in what this similarity consists.
89. 1st. The angles of the one are respectively equal
to the angles of the other. It is evident that this is necessary; for if these artificial triangles have angles different from those of the field, they cannot represent its
form, which is essential to a map or plan. If we had
measured one side and the two adjacent angles, the side
represented by AB, and the angles A and B; in drawing these two plans of the field, we should have made
the angles A and a, each equal to that angle of the field
which they represent; and the angles B and b, each
equal to the other measured angle. And the two sides
a c, b c, having the same inclinations to a b, as AC, BC
have to AB, must have the same inclination to each other, as AC and BC have; that is, the angle c must be equal
to the angle C; and consequently the three angles of the
one must be respectively equal to the three angles of the
other. This then is an essential property of similar triangles, their angles are respectively equal.
90. 2dly. Recurring to the construction of the two
triangles, ABC, a bc; the divisions of the scale with
which the larger was constructed, are just double the divisions of the scale with which the other was constructed; so that each side of the triangle ABC, is just double
the corresponding side of the triangle a b c. If the divisions in the two scales had borne any other ratio to
each other, the corresponding sides of the two triangles
would have borne the same ratio. But the ratio of AB
to a b being equal to the ratio of the two scales; and
also the ratio of AC to a c being equal to the ratio of the
two scales; it follows that AB contains a b as many
3*




30


Elements of Geometry.


AB AC
times as AC contains a c, that is -  -. In the same
ab    ac
manner we may show that AB contains a b as many
times as BC contains b c; that is, -  —. These two
ab    be
AB    BC
proportions may be written together, thus, - B -
-.    We see therefore, that-Int similar triangles, the
ac
corresponding or homologous sides are proportional.
91. If, in making a survey of the field, we had measured one of the angles and the two sides containing that
angle, the construction of the plans would have been
analogous to the solution of the problem in Art. 40; and
plans upon two different scales will accurately represent the field, if we construct, in each, an angle equal to
the measured angle of the field, and take for the lines
which represent the two sides in the larger plan, as many
divisions of the larger scale as there are chains or rods in
these lines respectively, and for the corresponding sides in
the smaller plan the same number of divisions respectively, of the smaller scale. The corresponding sides will
have the ratio of the two scales, and will therefore be in
proportion. We see then, that —Two triangles will be
similar, when an angle of the one is equal to an aLngrl of
the other, and the sides containing the equal angles are
proportional.
If we draw from the vertices C, c, of these triangles, perpendiculars to the opposite sides, as they measure the distance of that corner of the field from the opposite side, they must have the same ratio to each other
as the corresponding sides of the two plans; and so of
any other similar lines in the two plans respectively; whence we derive the proposition-In similar triangles, the homologous angles are equal, and all homologous
lines are proportional.
Remark. Either of these conditions, however, is sufficient to determine their similarity; for by altering the
angles, we alter the ratio of the sides and vice versa.
92. If the field were a rectangle, a parallelogram, or
any other rectilinear figure, it might also be represented
by plans upon different scales; and these plans will be
similar to each other, and similar to the field, when, in




Elements of Geometry.


31


each, the angles are respectively equal to the angles of
the field, and their dimensions also proportional to the
corresponding dimensions of the field. Hence we have
the general truth- Two similar rectilinear figures have the
angles of the one equal to the angles of the other, each to
each, and all homologous lines proportional.
93. Remark. If diagonals were drawn through corresponding vertices in the two polygons; being homologous dimensions, they would be proportional; and being
proportional, the triangles into which these diagonals divided the two figures would be similar respectively. We
say therefore, that-Similar polygons are composed of the
*wame number of similar triangles, similarly placed..4. If the several sides of a polygon were represented
by A, B, C, D, E, and the corresponding sides of a similar
polygon, by a, b, c, d, e; we should have (92) -=
C      D  E
--       -; from which we obtain (81),
A+B+ C +D+E_ A
a+b+cJ   + d-c-+ e    a'
that is-In similar polygons, the perimeters are propo,'tional to their homologous sides.
95. In the base of the triangle ABC (fig. 45), take A c Fig. 45.
of any convenient magnitude, and draw c b parallel to
C B; we shall lhave the triangle A c b, equiangular with
ACB (24), and therefore similar to it (91). Ience-If a
triangle is cut by a straight line parallel to one of the
sides, the small triangle cut off is similar to the whole.
These similar triangles give the proportion
AC    AB                   AC —Ac       AB — A b
A-   Ab;     therefore (81), -Ac            --
Cc     Bb
that is, -A  --; and the line c b divides the two sides
Ac     Ab
AB and AC, proportionally. Thus we have the general
rule, —A straight line drawn through any triangle parallel to one of the sides, divides the other two sides proportionally.
The converse of these two propositions may be proved
fiom Art. 91.
96. If another parallel, b' c', were drawn between b c
A c    C   an
and BC, it would give the proportion A'     B ' and
A b'   B b'




32


Elements of Geometry.


A c'  c c~           CC/   C Cl
A b= bb; therefore, b      Bb'; that is — Vhen two
straight lines are cut by any number of parallel lines they
are divided into proportional parts; and if one of the
lines is divided into equal parts, the parts of the other
wrill be equal.
97. To find a fourth proportional to three given lines
Fig. 46. (fig. 46). Draw the line AB equal to the sum of the
first and second lines in the proportion, take AD equal to
the first; and draw AC equal to the third, and making any
angle with AB; join DC, and through B draw BE parallel to DC; produce AC to E; because DC is parallel to
BE, we have AD-   C   (96); therefore CE is the fourth
proportional required.
98. To find a third proportional to two given lines.
Take a third line equal to the second; and find the
fourth as above; this will be the third proportional required. In this case the second of the given lines is
called a mean proportional between the first and third.
Fig. 47.  99. Let ABC (fig. 47) be a right-angled triangle; from
the vertex A of the right angle draw the perpendicular
AD to the hypothenuse BC. If we compare the triangle
ABD with the original triangle, we shall see that they
have each a right-angle, and the common angle B; the
other angles must therefore be equal (35), viz. the angle C to the angle BAD; the triangle ABD is therefore
similar to the triangle ABC; in the same manner it may
be shown that the other triangle ADC is similar to the
triangle ABC (91); whence we say-If from the vertex
of the right-angle of a right-angled triangle, a straight
line be drawn perpendicular to the hypothenuse, it will divide the triangle into two triangles, each similar to the
original triangle.
100. The similar triangles ABD, ACD, give the proBD AD
portion AD= -DC; that is-The perpendicular is a
AD    DC     t
mean proportional between the two segments of the hypothenuse. The similar triangle ABC, ABD, give the proBC    BA
portion BA = B; that is, BA is a mean proportional
VA = oU
between BC and BD. In the same manner we could
show that CA is a mean proportional between BC and




Elements of Geometry.


33


DC; therefore —Each side of the original triangle is a
mean proportional between the hypothenuse and the segment of the hypothenuse adjacent to that side.
We shall hereafter show the method of finding r i, "*n
proportional between any two given lines.
BD    BA            - (Bt)2
101. The equation BA - B. gives BD         B-C)
DC     CA               (CA)2
and the equation  -    BC   gives DC   - B —; adding these two equations we obtain BD  - DC = BC 
(BA)2 + (CA)2
(A) +C (CA); multiplying both sides by BC, we have
(BC)2 = (BA)2 + (AC)2. From this it is evident that
if we refer to a common measure the three sides of a
right-angled triangle, the second power of the number
expressing the length of the hypothenuse, will be equal
to the sum of the second powers of the numbers which
express the lengths of the other two sides. We can
therefore find the hypothenuse when we have the other
sides. Suppose in this triangle that AB contains the
common measure six times, and AC eight times; then
(BC)2 = 36 -+ 64 = 100; and taking the second root of
both sides of this equation, we have BC  (100) 2 = 10.
So also if we know the hypothenuse and one of the sides,
we can find the other. Suppose we have the hypothenuse equal to 10, and the side AB equal to 6, to find the
side AC. The equation (BC)2= (BA)2 + (AC)2, by
subtracting (BA)2 from both sides, will give (AC)2 =
(BC)2 - (BA)2     100- 36 - 64; and taking the second root of each side, we obtain AC- (64)2- 8.
102. In the triangle ABC (Jig. 48), bisect the angle Fig. 4-.
B by the line BD; and through the point C, draw CE
parallel to DB till it meets AB produced. The angle
ABC, exterior to the triangle CBE, is equal to the sum
of the two interior opposite angles BCE and BEC (37);
but on account of the parallels DB and CE, the angle
ABD (half of ABC) is equal to the angle E; the other
half must be equal to the angle BCE, consequently
BCE and BEC will be equal; therefore, the opposite
sides BC and BE are equal (45). Now in the triangle AEC, DB is drawn parallel to the base, which




34


Elements of Geometry.


gives the proportion, BE- DC (98); but BC is equal
to BE, and may be substituted for it in the proportion;
AB       AD
which  gives, ---    -; that is —A straight line biBC     DC
secting an angle of a triangle, divides the opposite side
into parts proportional to the adjacent sides.
103. Let us now take an obtuse-angled triangle, as
Fig. 49. ABC (fig. 49) and draw perpendiculars from the vertices of each of the angles to the opposite sides. The
lines AF and BG, respectively perpendicular to BC and
CA, will meet in the point D; we wish to ascertain the
point M, in which CE perpendicular to BA, meets BG.
First, it is manifest that, as the two triangles ABG
and EBM, have each a right angle, and the angle EBM
common, they are equiangular and similar; and give the
BA    BM
proportion BG =BE'
If we now compare the triangles EBC, and ABF, having each a right angle and the common angle ABC, they
BC    BA
give the proportion BE- -BA   And a comparison of
the two triangles BGC, BFD, give the proportion
BD BC
BF    BG. Multiplying in order, the two last proportions,
BC X BD         BC X BA
gives BF   BE    BFX BG; if now we divide each
BF x BE     BF x BG'
BC
ratio of this proportion by the common multiplier BFp, we
BA    BD
have BG-   B-E; and comparing this proportion with
BA BM
the first, B-G-BE; we perceive that they have the
BA
common ratio T-; the other two ratios are therefore equal,
BM     BD
and give the proportion BE - BE; that is, BM is equal
to BD, and the three perpendiculars meet at the same
point D.
If the original triangle were acute-angled, as ABD
in the same figure, it is manifest that the three perpen



Elements of Geometry.


35


diculars drawn from the vertices to the opposide sides,
would meet in the point C.   We say therefore-In
any triangle, the perpendiculars drawn from the vertices
of the angles to the opposite sides, will meet in the same
point.
Remark. If the triangle be acute-angled, the point of
meeting will be within the triangle; if the triangle be
obtuse-angled, the point of meeting will be without; and
in the right-angled triangle, the point of coincidence is
the vertex of the right-angle.
Of the Straight Line and Circle.
104. The circle is the plane surface embraced by
the circular curve already described (7), called the
circumference of the circle.  A straight line which
touches the circumference only in one point, is called a tangent, as AB (fig. 50). When it is said that Fig. 50.
the straight line touches the circumference, it is meant
that its direction is such that if it were produced ever so
far each way, it would have only one point in common
with the circumference. This point is called thepoint
of contact.
105. If the straight line cuts the circumference in two
points, it is called a secant. A straight line which has
its two extremities in the circumference, as DE (fig. 51), Fig. 51.
is called a chord; it subtends both the arc and the angle
at the centre measured by this arc. The portion of surface
contained between the two radii DO, EO, and the arc
DGE, is called a sector. The portion contained between
the chord and the arc is called a segment of the circle.
106. If the two radii DO, EO, the chord DE and the
arc DGE, be made to revolve together about the centre
0, of the circle, the points D and E will be continually
in the circumference; and the arc DEG will always coincide with the part of the circumference then embraced
by the two radii (7); if this were not the case, some
parts of the circumference must be nearer to the centre
than other parts; moreover, the straight line DE would
be constantly the chord of this arc. Now as the angle
DOE would not be changed by this revolution of the
sector DOE, we say that this angle would have the same
arc in every part of the circumference, and this arc
would also have the same chord. We have, therefore,




36


Elements of Geometry.


these propositions:-(l). In the same circle or in equal
circles, equal angles at the centre are measured by equal
arcs; and equal arcs subtend equal angles.  (2). In the
same circle, or in equal circles, equal arcs are subtende'
by equal chords; and when, in the same circle, the chordi
are equal, the arcs are equal.
As the triangle would not be changed by this revolution, the distance of the vertex O from the opposite side
DE, would not vary; that is-In the same circle, equal
chords arc at the same distance from the centre.
107. If the arc were increased, the angle at the centre
would be increased (26); but the two sides of the triangle OD and OE remaining the same, if their angle is increased, the third side Dig must be increased (32); therefore — The greater arc is subtended by the greater chord.
By increasing the side DE of the triangle, we increase
the angle 0, and the arc which measures this angle;
therefore —The greater chord subtends the greater arc.
108. We said that the chord increases as the arc is increased; this is true to a certain extent; that is, wlile the
arc is less than 180~, or a semi-circumference.  At this
point the chord is confounded with the two radii, and becomes a diameter (7); it subtends no angle, and is no
longer called a chord.  But as the chord continues to
increase till it is confounded with the diameter, we see
that —Of two unequal chords, the greater is at the less
distance from the centre.
If the arc be increased beyond 1800, the two radii
will make an angle with each other on the other side,
and the chord which subtends this arc, will make the
third side of the triangle, and is therefore less than the
sum of the other two (50), and consequently less than
the di;m-leter. Therefore-In the same circle a chord is
alwrays less than the diameter.
It will be perceived that every chord correspIonds to two
arcs, which taken together constitute the circumference.
Fig. 52.  W!,9. In fig. 52, suppose the radius CD to be drawn
pe:'cj,dicular to the chord AB; if we draw the other
two radii CA and CB, we shall have the isosceles triangle ACB; CE perpendicular to the base AB, bisects the
base and the angle at C (45); as the two angles at C
are equal, the arcs AD and DB are also equal (107);
whence we say-Radius perpendicular to a chord, bisects
the chord and its arc.




Elements of Geometry.


37


As one point and the direction of a straight line, determine its position (53)-A straight line drawn perpendicularly through the middle of a chord, bisects the arc,
and passes through the centre of the circle.
110. This enables us to find the centre of a circle to
which any arc belongs (fig. 53). We have only to as- Fig. 53.
sume two chords, AD, DB, in the arc, and draw perpen- *
diculars through the middle of each; the point of meeting of these two lines is the centre of the circle.
The operation would be the same if we were required
to draw a circular curve through the three given points
A, D, B. If, however, these three points were in the
same straight line, it is manifest that the perpendiculars
would be parallel (23); they would therefore never
meet; consequently —No    three points in the same
straight line, can be in the circumference of a circle.
Can you make two or more circumferences pass
through three given points not in the same straight line?
Can you make a circumference pass through anyfour
points not in the same straight line?
111. As the tangent has but one point in common
with the circumference, namely, the point of contact
(fig. 54), every other point must be without, and conse- Fig. 54.
quently farther from the centre of the circle; and if radius
be drawn to this point, it will be less than any other
straight line drawn from the centre to the tangent; it
must consequently be perpendicular to the tangent (51);
therefore —To draw a tangent to any point in a circumference, we have only to draw a straight line perpendicular to radius at that point.
As many circles, therefore, as have the point A, a
point in the circumference, and BD, a common tangent,
must have their centres in the same straight line CA,
produced if necessary. In this case the circumferences
are said to be tangent to each other.
When the circles have their centres on opposite sides
of the point of contact, they are said to be tangent externally. When they have their centres on the same
side, they are said to be tangent internally.
Can a straight line pass through the point of contact
between a circumference and a rectilinear tangent, which
shall cut neither of them?
Can you draw a circular line between a right-line tangent and the circumference 1
4




38             Elements of Geometry.
Can you draw a circular line between two circular
curves tangent to each other, and having their centres
on the same side of the point of contact?
Fig. 55.;  112. If we draw the secant, MN (fig. 55) parallel to
the tangent, it will be perpendicular to the radius CA;
and as the interior portion of the secant, will be a chord,
*  the arc which it subtends will be bisected by this radius
(109) at the point of contact. So if another parallel secant OP be drawn, the arc which this cuts off will also
be bisected by the same radius; so that we shall hlave the
arc AO equal to the arc AP; and AM    equal to AN;
iicerefore MO and NP are equal.   Hence —Parallels
i7ctercept equal arcs in the circunference.
113. We have seen (106) that equal angles at the cenl1e of a circle, are subtended by equal arcs; if, therefore,
v  find a common measure of tile two arcs AB, A'B,
(< '. 56) drawn with equal radii; and divide each of the
Fig. 56. arcs by this common measure; and to the several points
of division draw radii; the two angles ACB, A'C' i, will
be divided into equal angles. If the arc a b is a common measure of the arcs, and the angle a c b, a common
measure of the two angles; as there will be in each of
the sectors as many of the small arcs as angles; we
ACD     AB      A'C'B'   A'B'
shall have -       -   and                  as the
a cb             a acb     ab
denominators are the same in both proportions the
numerators must be in proportion; which gives (79)
A/'C'B   A/B/
-ACB -- AB; that is, in circles drawn with equal radii,
ACB     AB )
the angles at the centre are to each other in the ratio of
the arcs which subtend them, if these arcs are commensurable.
Suppose the arcs have no common measure; we may
take a measure of one of them, A'B' for instance (fig.
57), so small that by applying it to AB as many times as
Fig. 57. it is contained in AB, the remainder b B which we will
designate by m, will be less than any assignable arc;
and the angle b CB, which it subtends, less than any assignable angle, which indefinitely small angle we designate by n. The arcs A'B' and A b are commensurable,
ACB- n AB-m
and therefore give the proportion A'C'B-   ABA'C'IBI'     A'W'




Elements of Geometry.


39


Now suppose that, to AB produced we apply this small
measure of A'B', once more than before, it will extend to
b'; and suppose also that the excess of this arc A b' over
AB, is less than any assignable arc, which excess we designate by m', and the corresponding indefinitely small
ACB-+ n    AB_ +m
angle by n'; we shall haveAC +      AB  ~' It is
A'C'B'     AB       ' ti
manifest, therefore, fiom these proportions, that, one of
the arcs, with its angle, remaining the same, if any magnitude however small be subtracted from the other arc,
a corresponding magnitude must be subtracted from
its angle, to preserve the proportion; and if any magnitude however small be added to this arc, a corresponding
magnitude must be added to its angle to preserve the
proportion; from which it follows that if nothing be
either added to or subtracted from the one, nothing must
be either added to or subtracted friom the other. We
ACB      AB
shall therefore have the proportion, A    = C _  Beven when the arcs are incommensurable. We therefore say —lIf f.omlt the vertices of two ainglcs, arcs are
described with the same radius, the portions of these
arcs embraced by the sides, will be in the ratio of the an'[cs.
-    114. We see from the preceding article, what is to be
ilnderstood when it is said that-Anl a.ngle has for its
u:!''.tre'c the circular arc com)prchended( between its sides,
bti vertex bein.g at the centre of the circle. By this we
moan tiat the angle is the same part of four right-angles,
as the arc is of an entire circumference or 3600. When
therefore, we speak of an angle of 30~, we mean an angle whicl wvould embrace between its sides, 30 parts out
of 360, into which any circumference described from the
vertex of this angle as a centre, should be divided.*
115. Suppose that the vertex of the angle, instead of being at the centre of the circle, is at some point between
the centre and the circumferance, as at C (fig. 58). If Fig. 58.
* This is according to the ancient division of the circumference, mentioned in article 26; a different system
has been adopted in France. This system divides the
circumference into 400 degrees; each degree into 100
minutes, and each minute into 100 seconds.




40


Elements of Geometry.


we draw through the centre, the two lines a n and
b m, parallel to the two lines AN and BM, they will make
an angle at the centre equal to the angle to be measured (24); and the arc m n will be as much greater than
MN as the arc a b is less than AB; so that the sum of
the arcs intercepted by the lines passing through the
centre, is equal to the sum of the arcs intercepted by the
lines which meet at C; therefore the arc a b is equal to
half the sum of the two arcs AB and MN. But a b
measures the angle a 0 b equal to ACB. Whence we
have this truth —An angle whose vertex is in the interior of a circle, made by two chords, has for its measure
half the sum of the intercepted arcs.
116. Suppose the vertex of the angle to be in the circumference (in which case the angle is called an inscribed angle). By drawing parallels to the chords through
Fig. 59. the centre (fig. 59) we shall have the two arcs C n and
C m, equal to A a and B b; therefore the arc m n equals
A a - B b; consequently the arc a b equals A a + B b;
that is, the arc a b is half the arc AB; but a b measures
the angle whose vertex is at 0, equal to the inscribed
angle.
When both the chords which form the angle are
on the same side of the centre, as the angle ACB';
this angle may be considered as the difference of the two
angles BCB', BCA; BCB' has for its measure half the
arc BB'; BCA has for its measure half of AB; consequently, the remaining angle ACB' must have for its
measure half the remaining arc AB'; and thereforeAny inscribed angle has for its measure half the arc
etobraced by its sides.
117. This result includes the case of the angle made
by the chord and tangent; for if AC be made to revolvo about the point C till it come into the position
A'C   as the angle ACB increases, the arc AB increases,
half of it still measuring the angle; and this reasoning
holds till AC come infinitely near to the position of the
tangent. We infer, therefore, that the error, if there be
one, must be infinitely small.
But to ascertain whether there is any error in this result, draw through the centre parallels to the chord and
Fig. 60. tangent (fig. 60); the angle ACB will be equal to a c b,
measured by the arc a b. On account of the parallels,
the arcs C a and C n are equal; and we have the arc mn




Elements of Geometry.


41


equal to C a added to B b; but the arc m n is equal to
a b; therefore the arc a b is equal to a C added to B b,
that is, equal to half the arc C a B. Hence we have rigorously demonstrated that- The angle made by a tangent
and chord, hasfor its measure half the included arc.
118. If the meeting of the two lines is beyond the circumference (fig. 61), draw M b parallel to CB; the arc Fig. 61
A b will be equal to the difference between the two arcs
AB and MN; and the angle AM b, will be equal to the
angle C (24); but the angle AM b has for its measure
half the arc A b (116); therefore —The ancgle made by
the two secants meeting without the circumference, hasfor
its measure half the difference of the two arcs embraced by
the sides.
119. This includes the case of the angle made by two
tangents; for drawing the chord A b (fig. 62) from one Fig. 62.
point of contact parallel to the other tangent; the arcs
AB and B b, will be equal (112), and the arc A b will be
the difference between the two arcs A b B and AB. But
half this arc A b measures the angle DAb (116), and conquently the angle ACB, equal to D)Ab. Hence we say —
The angle made by two tangents, is measured by half the
di'ferencce f' the arcs between their points of contact.
120. Froln the reasoning in article 116, it follows that
— A/l inscribed angl1es subtended by the srame arc, are
equal (fi.!. 63), as each has for its measure half this arc. Fig. 63
121. If the arc embraced by the angle is a semi-circumference, that is, if the two sides of the inscribed angle are in the extremities of a diameter (fig. 64), the Fig.GL.
semi-arc, which measures the angle, will be an arc of
900; tllherefore (27)-An angle inscribed in a semicircle
is a r~ight-angle.
122. Two angles, inscribed in opposite segments of
the same circle, as ACB, ADB (fig. 63), will have, each Fig. 63.
for its measure, half the opposite arc; therefore the two
will have for their measure, half the circumference;
consequently —The sum of two angles inscribed in opposite segments of the same circle, is equal to two rightangles.
12:3. As every tangent to the circumference of a circle,
is perpendicular to the radius whose extremity is the point
of contact (111); if we were required to draw a tangent
to the circumference BFD (fig. 65) from the point A Fig. 65.
without the circle; we should draw a straight line from
4*




42


Elements of Geometry.


the point A to the centre C, and make AC the diameter of a circle whose circumference cuts that of the
given circle in two points B, D, to which we should
draw the straight lines AB, AD; these will be tangents
to the circumference BFD. For, drawing the radii CB,
CD, the angles ABC, ADC, being each inscribed in a
semicircle, will be right angles; that is, the straight
lines AB, AD, are perpendicular to the radii CB, CD, at
their extremities (121). It appears that, from any point
without a circle, two tangents may be drawn to that circle. 
124. If we have the straight line FG given in posiFig. 66. tion (fig. 66), and the point B also given, we can describe a circumference passing through B, to which GF
shall be tangent at the point A. Draw AB; bisect this
line by the perpendicular DC; and from A draw a line
perpendicular to FG, meeting DC in the point C; this
point will be the centre, and AC the radius of the required circumference.
If BA be produced to E, the acute angles at A, made
by the tangent and chord produced, will be equal to the
angles inscribed in the grcater segment of this circle;
the obtuse angles at A, made by the tangent and chord
produced, will be equal to the angles inscribed in the
less segment (116, 117). It is therefore easy to solve
this general problem,-Uponz a given straight line (as
AB in this figure) to describe the segment 'f a circle, in
which a given angle may be inscribed.. 125. We have seen how we are to measure the angles
which the chords, tangents, and secants of any circle
make with each other. Let us inquire into the laws
which regulate the proportion of their parts.
Let AB and CD be two chords which clt each other
Fig. 67. at E fig. 67). By drawing the chords CA and BD, we
have two triangles CICA and BED; the two angles at C
and B are equal, because they have each for a measure
half the arc AD; the angles at E are vertical angles,
and are therefore equal; consequently the other two angles A and D, are equal, and the two triangles are thereAE    ElD
fore similar, and give the proportion, -EC  — EB  That
is, the greater part of each chord divided by the less
part of the other, gives the same ratio. This is what is
called an inverse or reciprocal proportion; and we say



Elements of Geometry.


43


Two chords which intersect, divide each other into parts
reciprocally proportional.
126. If these chords are perpendicular to each other,
and one of them  (as AB fJi. 68) is a diameter, it will Fig. 6S.
bisect the other in E (109); and therefore in the proportion, instead of ED, we may use its equal EC. This
AE       EC
will give EC -  EB_ that is, EC is a mean proportional
EC      EB'
between AE and EB.
Suppose we draw also the chord BC; since AB is a
diameter, the angle ACB inscribed in a semicircle, is a
right angle; and CE is a perpendicular drawn from the
vertex of the right-angle of the right-angled triangle
ABC, to the hypotheniise, dividing the triangle into two
smaller triangles.
If we examine the triangles into which the triangle
ACB is divided, we shall perceive that the triangle ACE
has a right angle and the angle A in common with the
larger triangle; the triangle BCE has a right angle, and
the angle EBC   in common with the larger triangle.
Each of these partial triangles is therefore equiangular
with the larger; they are consequently similar to it and
similar to each other. If we compare each of them with
AE     AC
the larger we have the proportions AC -AB'     and
EB     CB
-CB — AB; tthat is, AC is a mean proportional between
CB     AB'
AB and AE; and CB is a mean proportional between
AB and EB.
If, therefore, we consider these lines AC, AB, CB,
and CE, in their connexion with the circle, we should
enunciate in this manner the truths just proved;
(1). A perpendicular drawn from any point in the
circunmference of a circle to the diameter, is a mean
proportional between the two segments of the diameter;
(2). If, from the rertex of an angle inscribed in a semicircle a perpendicular be draiwn to the diameter, each
of the chords which form this angle, is a mean proportional between the whole diameter and the segment adjacent to that chord.
The above result affords us the means of giving the
solution which we promised (101) of the problem-To
find a line which shall be a mean proportional between two
given lines.




44


Elements of Geometry.


[The learner will perceive that there are here two
methods of solving this problem; he can give examples
of each.]
127. If, however, we regard the triangle ACB, independently of its connexion with the circle, these truths
will be enunciated thus:
If from the vertex of the right-angle in a right-angled
triangle, a perpendicular be drawn to the hypothenuse,(1). This perpendicular will be a mean proportional
between the two segments of the hypothenuse; and will divide the triangle into two triangles similar to each other
and similar to the whole.
(2). Each of the sides wvill be a mean proportional between the entire hypothenuse and the segment at'jacent to
that side.
[These propositions were differently obtained in articles 99 and 100.]
128. If these lines instead of meeting within the circumference, were to meet beyond the circumference, as
Fig. 69. in figure 69; by drawing the chords CA and BD, we
have also, two equiangular triangles AEC and DEB;
the angle E is common to the two triangles, and the two
angles at A and D, are measured by half the same are;
consequently they are equal, and the other two angles
are equal, and the triangles similar; they will give us
EA     ED
this proportion, EC E-; that is-T- hcn two secants
meet beyond the circumference, the entire secants will be in
the inverse ratio of the parts without the circle.
129. As the truth of this last proplosition does not
depend upon any particular position of tile secant, we
Fig. 70. may suppose the secant DE (jif. 70) to revolve about
the point E till the interior part CD vanishes and the sescant becomes a tangent; the proposition will hold true
to the vanishing point of the secant; but at that point
the exterior part becomes equal to the whole; and we
AE     EF
shall have F - EB; that is-If, from the same point
without a circle, a tangent be drawn and also a secant, the
tangent will be a mean proportional between the entire secant and the exterior part.
By drawing the two chords AF and BF, the two similar triangles AEF and BEF, will give the same proportion. So that we are sure of the accuracy of this result.




Elements of Geometry.


45


Why are the triangles AEF and BEF similar?
130. If we make the tangent AB equal to the diameter of the circle (fig. 71), draw the secant through the Fig. 71.
centre, and take upon the tangent the part AF equal to
the exterior pait of the secant, the last article will give
the proportion
AE     AB        AE-AB        AB-AD
AB -AD; (81.)        AB          AD
But AE —AB is AD, and AB - AD is BF; the proAF     BF
portion therefore becomes B —  AF; and by inverAB    AF
sion (7S) we have AF - BF   that is, the line AB is divided at F, so that the part AF is a mean proportional
between the whole line and the other part BF. This
is called dividing a line into mean and extreme ratio.
To divide the line AB into mean and extreme ratio,
draw through one extremity a perpendicular equal to half
the line; make this the radius of a circle, through the centre of which draw a secant meeting the given line in the
other extremity; the exterior part of the secant transferred to the given line, gives us the point of division F.
131. If we make the perpendicular BC equal to the
given line AB (fig. 72), the angle at A will be an angle Fig. 72.
of 45~, that is, half of a right-angle (35), and the portion
AC will be the diagonal of a sqnare whose side is AB.
This suggests a method of comparing the diagonal with
the side of the square.
The method of making this comparison was given in
article 74. If the side be applied to the diagonal, it will
be contained in it once with a remainder AD; if then
we apply AD to AB, it will be contained in it twice with
some remainder. Now we know that AB contains AF
as many times as AE contains AB; so that instead of
comparing AF with AB, we have only to compare AB
with AE; but AE contains AB twice, with the remainder AD. This remainder is to be compared with AB,
which is the same thing as comparing AB with AE, but
this is the same as the last. It appears, then, that this.
is the commencement of a series of operations which can
never terminate; therefore- The side of a square and
its diagonal are incommensurable magnitudes-a result
which we shall obtain hereafter by a different method.




46


Elements of Geometry.


Of Polygons inscribed in a Circle, and circumscribed
about a Circle.
132. IN figure 68, we have the right-angled triangle
ACB, whose vertices are in the circumference of a circle of which the diameter forms the hypothenuse of the
triangle; and as the vertices of any triangle are three
points not in the same straight line (110), it is manifest
that a circumference may be described passing through
these vertices. Under such circumstances the triangle is
said to be inscribed in the circle.
133. If we examine the inscribed triangle ABC (fig.
Fig. 73. 73), we shall perceive, that as each angle is an inscribed
angle, it will have for its measure half the opposite arc,
of which the side opposite to this angle is the chord;
and as the greater chord subtends the greater arc (107),
we have an additional proof that the greater angle in a
triangle is opposite to the greater side (47). And as
equal chords have equal arcs, and equal arcs subtend
equal angles (106), an equilateral triangle must have its
angles equal (46); and the equal sides in an isosceles
triangle, must be opposite to equal angles (45).
We have also a new proof that the sum of the angles
in every triangle, is equal to two right angles (35); for
as each of the angles is measured by half of the opposite
arc (116), the sum of three angles will have for their
measure half the sum of the three arcs, or half the circumference; which is the measure of two right angles
(27).
134. To inscribe a circle in a triangle; bisect two of
Fig. 74. the angles, as A and B (fig. 74); from the point of meeting of the two bisecting lines, draw perpendiculars OD,
OE, OF, to the sides of the triangle; the point O will
be the centre, and these perpendiculars will be the radii
of the circle. The two triangles AOE, AOF, have each
a right-angle, and their angles at A equal; their other
angles will therefore be equal; and having the common side AO, they will be equal by the third case of
equal triangles; OE and OF are therefore equal. In
the same way.it may be shown that OE and OD are
equal. The circle, therefore, which has O for its centre
and OE for its radius, will have each side of the triangle
ABC tangent to its circumference. The circle is said




Elements of Geometry.


47


to be inscribed in the triangle; and the triangle is circumscribed about the circle.
135. It will be readily seen that no circumference
passing through the three points A, B, C, can also pass
through the point E (fig. 73). A quadrilateral, there- Fig. 73.
fore, cannot always be inscribed in a circle. It may be
well to inquire under what circumstances a quadrilateral may be inscribed in a circle. If we consider the
inscribed quadrilateral ABCD, we sliall perceive that the
angle ADC has for its measure half the arc ABC; if the
angle ADC were less than it is, the others remaining the
same, its vertex would be beyond the circumference; if
it were greater, the vertex would be within the circumference; it must therefore be such as to be measured by
the arc ABC. So also the three points A, D, C, remaining the same, the angle at B must be such as to have for
its measure half the arc ADC. But these two angles
have for their measure halfof an entire circumference;
their sum, tlerefore, must be equal to two right angles.
In the same manner we may show that the sum of the other
two angles is equal to two right angles; whence we infer, that-A quadrilateral may be inscribed in a circle,
when its opposite angles are slip)lements of each other.
A square, therefore, and any other rectangle, may be
inscribed in a circle; but no other parallelogram.
136. In polygons, generally, of a greater number of
sides, it would be more difficult to ascertain whether
they could be inscribed.
Suppose, however, we take a certain number of equal
lines, as AB, BC, CD and DE (fig. 75), and dispose Fig. 75.
them as in the figure, so as to make equal angles whose
vertices are at B, C, D. The circumference of only one
circle can be drawn through the three points A, B, C;
the relative position of these points, therefore, determines
the curvature. But on account of the equality of the
lines and angles, the relative position of any three contiguous vertices will be the same; the same curvature,
therefore, would be required throughout, and the circular are which passes through the three vertices A, B, C,
if produced will pass through the other vertices. We
say, therefore, that-Any polygon which has its sides
equal and its angles equal, may be inscribed in a circle.
Remark. As the circumference of a circle may be divided into unequal parts, and chords may be drawn to




48


Elements of Geometry.


to the points of division, forming inscribed polygons; it
will be perceived that there is an endless variety of polygons which may be inscribed in a circle.
137. If we construct the isosceles triangle ABC (fig.
Fig. 76. 76), the angle ACB being any measure of four rightangles; the perpendicular C a will bisect the side AB
and measure the distance of the point C from this side.
If this triangle be made to revolve about the side BC,
till the point A comes again into the same plane, it
will occupy the place of the isosceles triangle BCA',
equal to ABC; the perpendicular C a will fall upon C a',
which is therefore equal to C a; we have also the angle
ABA' double the angle ABC, the angle at the base of
the triangle. If we now revolve this triangle aboutthe
line CA' till B meets the plane of the paper in B', we
shall have C a falling upon C a" perpendicular to A' B'
and therefore measuring the distance of the point C from
this line; we have a polygonal angle BA'B' equal to twice
the angle at the base of the isosceles triangle, and therefore equal to ABA'. If we continue this process, as the
angle ACB is a measure of four right-angles, the triangle
after a certain number of turns, will have occupied all
the angular space about C, bringing one of its sides to
the line AC, whence the triangle first departed. We
shall therefore have the polygon ABA'B'A"B" of which
all the sides are at the same distance from C; if, therefore, from the point C, with a radius equal to this distance C a, we describe a circumference, it will touch
each of the sides of the polygon; the polygon is therefore circumscribed about the circle. If we examine this
polygon we shall see that the sides are all equal, and all
the angles at the vertices are equal, being each equal to
twice the angle at the base of the isosceles triangle used
in generating this polygon. We say therefore-A polygon, whose sides are equal and whose angles are equal,
may be circumscribed about a circle. Because CA, CB,
CA', &c. are sides of the same isosceles triangle, they
are all equal; and if from C as a centre with the radius CA, we describe a circumference, it will pass through
the vertices of the polygon. The polygon will then be
inscribed.
Which of the several classes of parallelograms (62)
can be inscribed in a circle?
Which of the several kinds of parallelograms can be
circumscribed about a circle?




Elements of Geometry.


49


138. In the polygons whose sides are equal and angles
equal, whichever side be taken for the base, the base will
be of the same magnitude, the angles at the base will be
the same, and the opposite vertices will be determined
by equal triangles similarly situated. On account of this
regularity of their parts, these figures are called regular
polygons. We have seen (136, 137) that polygons whose
sides are equal and angles equal, may be iuscribed in a
circle, and also circumscribed about a circle. We say,
therefore —Regular poolgons ma/ be inscribed in a circle,
and also circumscribed about a circle.
139. In the regular hexagon ABA'B'A"/B" (fig. 76),Fi. 76.
the point C is called the centre 'o the polygon; the equal
angles ACB, BCA', &c. are called angles at the centre.
As regular polygons of the same number of sides must
have equal angles at the centre (15), they will be composed of the same number of similar isosceles triangles
(46), and will tlerefore be similar polygons (('S, and as
the sides of the isosceles triangles of wlhich they are conposed, are the radii of tlil circumscribed circles, and the
heights of tlese triangles, the radii of the inscribed circles; it follows (92) that-hi, reg'utlar );olyg;,ons of the
same number of sides, the pJ:e'ileters are to each other as
the radii of' the circumC7scribcedl circles, and also as the radii
of thec inscribed circecs.
140. Each angle of an equilateral triangle is an angle
of 60~ (40); the chord of 609 is equal to radius.
Therefore (fg. 77) —To draw t an eqilateral triangle, Fir. 77.
describe a circular arc with a radiuts equal to the given
side, take a chord equal to this radi!s, and draw radii
fromn the centre to the extreiCcties of this chord.
141. It will be perceived that, as radius is equal to the
chord of 60~, it may be applied six times to the circumference.  This will give us a regular, inscribed hexagon,
as ABDEFG. If we draw chords to the alternate angles
of the hexagon, we shall have an inscribed equilateral
triangle.  And if we bisect each of the arcs subtended
by the sides of the hexagon, and draw chords to the same
arcs, we shall have a regular polygon of twelve sides,
called a dodecagon. By continuing the process, we obtain regular inscribed polygons of 24, 48, 96, &c. sides.
142. If we draw two diameters perpendicular to each
other (fig. 78) and draw the chords (AB, BC, &c.) to Fig. 78.
their extremities, we shall have an inscribed quadrilat5




50


Elements of Geometry.


Fig. 78. eral, the sides of which are equal. We perceive that
this figure is composed of four right-angled triangles,
which are also isosceles; the acute angles therefore are
each equal to half a right-angle; but each angle of the
inscribed quadrilateral is composed of two of these equal,angles, and is therefore a right-angle; consequently the
figure is an inscribed square.
If we bisect the arcs AB, BC, &c. and draw the
chords of these semi-arcs, we shall have an inscribed
octagon. And by continuing the process of subdividing
the arcs and drawing chords to them, we shall have inscribed figures of 16, 32, 64, &c. sides.
143. In the right-angled triangle AOB, we have (101)
(AB)2 = (AO)2 + (OB)2, or (AB)2 =2 (AO)2. as AO
is equal to OB; and if we extract the second root of both
sides of this equation, we shall have AB - AO X (2);
or considering the radius AO as unity or one, it gives
AB = (2)2. This is the side of the inscribed square.
It is always the diagonal of a square whose side is AO,
equal to unity. Now it is known that there is no numerical expression for this value (2)2 independent of the
radicalform; and as this quantity is incommensurable
with unity, it follows that the diagonal is incommensurable with the side of a square (131). We see, however,
that notwithstanding this incommensurableness, the geometrical process gives us their ratio exactly.  The ratio
of the diagonal to the side of the square, is (2)2.
The side of the inscribed equilateral triangle is obtained from figure 77, thus: (EB)2 = (EA)2 - (AB)2;
but as AB is equal to radius or one, and EA equals twice
radius or two, we have (EB)2  4-1 -= 3; therefore
EB - (3)
The side of the regular inscribed hexagon equals radius, or one. Having the side of an inscribed polygon, we
have its perimeter, and radius entering into the expression, we have the ratio of the perimeter to radius.
Fig. 79.  144. If in the circle ABFGH, &c. (fig. 79) we divide the radius AC into extreme and mean ratio at E,
EC being the larger part, take the chord AB equal to
EC, and join EB and BC; the result of this division of
AC    EC
radius gives -C  AE     but substituting for EC its
I    EC    AEL




Elements of Geometry.


51


AC AB
equal AB, we have AB-AE-.      We have, therefore,
in the two triangles ABC, ABE, the angle A common
and the sides about this angle proportional; they are
therefore similar (91); but as ABC is an isosceles triangle, ABE must be an isosceles triangle, and EB is equal
to AB or to EC; the triangle EBC is therefore isosceles, and the angle ECB, which, on account of the similar
triangles, is equal to ABE, is also equal to EBC; consequently each of the angles at the base of the isosceles
triangle ABC is double the angle at the vertex; whence
the angle ACB is I of two rigllt angles or -1 of four right
angles  the arc AB which subtends this angle, must
therefore be y- of the circumference; and the chord AB
is the side of a regular inscribed deccraon.
By joining the alternate vertices BG, GI, &c. of the
regular inscribed decigon, we have the regular inscribed pent^ao"n. And by bisecting the arcs and drawing
chords, we obtain regular inscribed polygons of 20, 40,
80, &c. sides.
145. The arc subtended by the side of a regular inscribed decagon, is one tenth of the circumference; and
the arc subtended by the side of the regular inscribed
hexagon, is one sixth of the circumference. Let the arc
BP be one sixth of the circumference; if we subtract
from this the arc BA, which is 1 of the circumference,
we shall have the arc AP -  -  I -- -' of the circumference; and the chord AP will be the side of a regular
inscribed polygon of 15 sides. By the process of bisecting the arcs, we obtain polygons of 30, 60, &c. sides.
146. If we could divide any arc into three equal parts,
we could obtain successively the arcs subtended by the
sides of regular inscribed polygons of 9, 18, 27, 36, 45,
&c. sides; but this division refers itself to the famous
problem of the trisection of the angle, whose solution was
so much sought by the elder geometers; "being in
point of difficulty, or rather perhaps of impossibility, on a
footing with the other two celebrated problems, viz. the
duplication of the cube, and the quadrature of the circle." Elementary geometry has not yet furnished us
with the means of solving this problem.
147. Having a regular inscribed polygon of any number of sides, we may circumscribe about the circle a regular polygon of the same number of sides, by drawing tan



52


Elements of G2coincft.y,


gents to the circle through the vertices of tile inscribed
polygon. If we draw the radii CA, CB, CD, &c. (fig.
Fig. SO. 80) to the vertices of the inscribed polygon ABDEFG,
and draw through these vertices, perpendicular to radius,
the straight lines nb, bd, rc, f;feJ;,,g hi we shall have the
regular circunmscribed p!ygon abcefg, of the same
number of sides.
tlcRemark. The I-tiperpldicular CA, CB, &c. drawn from
the centre of the regular polygon to one of its sides,
(which always bisects that side) is called the apothegm
of the polygon. C mn is the apothegm of the inscribed
polygUon.
14t. It is manifest that by increasing the number of
sides in the inscribed po'lygon, we increase its perimeter,
whichl can never ie greater than thcircumference; and
by increasing the number of sidcs of the circumscribed
pologon, we diminish its periimeter, which can never be
less tban the circumference. 'Ihe circumference of the
circle being then the limit of this approximation, or the
value to which these tw\'o magnitudes approach; we say
that- The circmn'-erncc of the circle is Ceuivallet to the
perimelter oj the 'reular' insclrib  (o:' ci)rcumscribec) lpo1 -ygon of an ibfnite tnumler' of sides.
Two circles may therefore be considered two similar
polygons; whence (139) —    he cit rc(umflrences of two
circles care to each otherl in the ratio of theilr radii, or of
their diamecters.
Remnark. The apothegm of the inscribed or circumscribed polygon of an infinite number of sides must be
infinitely near to radius.
The consideration of the ratio of the circumference to
radius, or to the diameter, we shall defer for the present.




Elements of   eometry.


53


PART FIRST.
SECTION II.-Of the Measure and Comparison of Plane
KSurfaces.
149. In discussing the properties of plane figures thus
far, we have considered only the lines and the angles,
and their several relations to each other, without regarding the quantity of surface embraced by the outline or
perimeter of the figure. The whole amount of surface
or superficies in any geometrical figure is called its area.
The word suzface, in general, we use to signify superficial extent without regard to quantity.
150. In speaking of figures as equal, we have said
that it is an indispensable condition of geometrical equality that the figures compared should coincide by superposition. But it is evident that two figures may have the
same amount of surface, and still be very different i9
form. Sucl figures we call equivalent. A field of a
circular form may have the same superficial extent with
another field whose form is quadrangular; we should say
that the two fields are equivalent or equal in area.
In comparing parallelograms we consider one side as
the base of the figure, and the perpendicular distance of
this side from  the opposite, the height of the figure.
Sometimes we call one side the inferior base, and the
opposite, the superior base. In triangles one side is
taken for tile base, and the perpendicular distance of this
side from the vertex of the opposite angle, is the height
of the triangle.
151. To compare two parallelograms ABCD, ABEF,
(fig. 81) whose bases are equal, and heights equal, place Fig. 81
their bases together as in figure 82. On account of their Fig. 82.
equal heights and the parallelism of their opposite sides,
their superior bases CD, EF, will be in the same straight
line CF. We shall have two triangles CAE, DBF, of
which the sides CA and DB are equal and parallel, being opposite sides of the same parallelogram (59); AE
and BF are equal and parallel for the same reason. The
two angles CAE, DBF, having their sides parallel and
directed the same way, are equal; and the twvo triangles
CAE, DBF, are equal by the second case of equal trian
gles. If from the whole quadrilateral CABF, we take
5*




54


Elements of Geometry.


the triangle DBF, there will remain the parallelogram
ABDC; and if from the same quadrilateral we take the
equal triangle CAE, there will remain the parallelogram
ABFE; the two parallelograms must therefore be equivalent. We therefore say-Parallelograms of equal bases
anId Cqual heights, 'are equivalent.
152. As every rectangle is a parallelogram, it follows
that-Every parallelogram is equivalent to a rectangle of
an equal base and equal height.
153. We have seen that the diagonal of a parallelogram divides it into two equal triangles (GO); the trianFig. 83. gle ABC (fig. 83) is then one half of the parallelogram ABCD; and therefore (152)-Evezrc triangle is
equivalent to one half of' a rectangle of an equial base and
equal height. And consequently-All trian les of equal
bases and equal heqights, are equl(ialent.
154. We see tlat the triangle and parallelogram  are
both referred to tle rectangzle for the measure of their
area. If then we can firnd the exact measure of the
rectangle, we have the means of measuring the area of
any parallelogram or triangle, and, hence. of any rectilinear figure, as all rectilinear figures may be divided into
triangles ly drawing diagonals through ol)posite vertices.
155. To compare two rectangles of the same height
and unequal bases, place the less upon the greater so
that two equal sides may coincide, and the base CF lie
Fig. 4..along the base CD (fig. 84).
If the bases are commensurable, divide them by a common measure; and through the points of division, draw
lines parallel to the side CA. Suppose this common
measure to be contained eight times in AB, and five times
in AE; these lines parallel to CA will divide the larger
rectangle into eirh.t equal rectangles, of which the smaller rectangle AECF will contain five. This will give us the
ABCD     AB
proportion AECF-   AE; that is, the two rectangles are
to each other in the ratio of tleir bases.
Fig. 85.  If the bases AB, AE are incommensurable, (Jig. 85) divide the larger AB by a measure of the smaller. Suppose it to be contained a certain number of times with the
indefinitely small remainder B b, less than any assignable
magnitude, which denote byd; and draw through b a
straight line parallel to BD. Let us denote the difference
between the rectangle whose base is AB and the rectangle




Elements of Geometry.


55


whose base is A b by m; as A b and AE are commenABCD-m
surable, we shall have the proportion  AECF
AECF
AB -d
AE- -  If we apply the measure of AE one time more
than it is contained in AB, it will reach beyond B to b';
and suppose this excess B b', which we denote by d', to be
less than aly assignable magnitude, (since we can divide
AE into parts so small that this excess shall be as little
as we please), if we draw through b' a line parallel to BD,
meeting CD produced, we shall have a rectangle which.
exceeds the rectangle ABCD by an indefinitely small
excess which we designate by im'. According to what is
proved in the preceding part of this article, we have
ABCD +- mi',AB -- d1'
AB  - CD +  -   AB -'    We see, therefore, that in
comparing these two rectangles and their bases, if any
mag:litude however small be added to tile rectangle
ABCD, a corresponding magnitude must be added to its
base AB to preserve tie proportion; and if any magnitude however small be subtracted from tile rectangle
ABCD, a corresponding magnitude must be subtracted
from its base to preserve tlie proportion; it therefore follows necessarily, that if nothing be added to, or subtracted from the rectangle, notlinTg must be added to, or subtracted from its base, to preserve the proportion; and we,ABCD     AB
consequently have A-C-     AF; that is, the rectangles are in the ratio of their bases though the bases are
incommensurable.  We therefore say- Two rectangles
of the same height, are to each other in the ratio of their
bases.
-156. As the side AC might have been taken as the
common base of the two rectangles whose heights
are AB and AE; we have from the same proportion
ABCD      AB
AECF    - AF' this general rule also-Two rectangles
of equal bases are to each other in the ratio of their
heights.
157. As a triangle is equivalent to half the rectangle
of the same base and height, it follows from the last two
articles, that-Two triangles, of equal heights, are to




56


Elements of Geometry.


each other in the ratio of their bases; and tiwo triangles
of equal bases, are to each other in the ratio of their
heights.
Fig. 86. 1158. If we place two rectangles (fig. 86) so as to have
the point C a vertex in each, and their sides parallel; and
produce BA and FE to meet in H; we shall have another rectangle ACEH, with which each of the others may
be compared; these will give the following proportions
ABCD      CD   ACEH      AC
(15) ACEH      CE' CEFG -CG          If we multiply
ABCD X ACEHI
these ratios in order, it will give ACEH   X CEFG
CE -X   C; but in the first ratio of the proportion, the
CE X CG
same factor ACEH, occurs both in the numerator and
denominator; it may therefore be stricken out, which
ABCD CD X AC
will give   C         X   C   But CD X AC is the
CEFG - CE x CG
product of the base of the rectangle ABCD by its height,
and CE X CG is the product of the base of the rectangle CEFG by its height. Therefore —Two rectangles
are to each other as the products of their bases by their
heights.
159. It is usual to estimate areas by square feet, square
yards, square rods, &c. By a square foot, we mean a
square whose side is one foot; by a square yard, a
square whose side is one yard, &c. If we have a rectangular floor whose length is six yards, and breadth five
yards, in order to ascertain its area let us divide the end
Fig. 87. of the room AC (jig. 87) into yards, and through each
division draw lines parallel to AB; this will divide the
floor into five equal rectangles, each a yard in width and
six yards long; if then we divide the line AB into portions of one yard each, and through these points of division draw lines parallel to A a, we shall divide the rectangle AB a b into six equal squares whose side is one
yard; each of the other rectangles a b c d, &c. will contain six square yards; so that the whole rectangle will
contain five times six square yards, or thirty square
yards. This might have been at once obtained by multiplying the number expressing the linear yards in AB,
by the number of linear yards in AC.
If we would have the area of a rectangle in square




Elements of Geometry.


57


feet, we must multiply the number of linear feet in one
side by the number of feet in the contiguous side. This
shows us what is meant by the product of two lines; and
we therefore say- The arca  a f  rectang le is expressed by
the prorhuct of its base mittiijlied b' its hceight.
1()6. As any parallelo gram is equivalent to a rectangle
of tlie same base and heirht  e      f ean/ p, f arallCloyeravtl is th bse )rwodlt Of t  b'sec mn ltiplicd bI/ the height.
And therefore (153) — 7Th ac61'c o!f a trianglC is half the
Io('odicEt of the base  ul:tipeliheda b] the height. Consequen'tllj aniy lto tr.iangles are to cach otheri as the products. of their bases by thelir h ig, 'hts.
161. If we designate the h]eight of a rectangle by a,
and tlhe base by b, tlhe area will be expressed by a X b;
or a. b; and the rectangle is said to be containe(l by the
two lines a and b. If tie base of the rectangle is equal
to its heiglht, the figure is a square ((62), and its area is
expressed by a. ca, or a2 (73). I[ence the second power
of a quantity is fiequently called its square.*  A triangle,
whose height is a, and wlose base is b, will be reprea.b.sented by - a. b, or —.
162. Prloblem.  To construct upon a given line c
(fig. 88) a rectangle equivalent to the given rectangle Fig. 88.
a. b. Find a fourth proportional (97) to the three given
lines c, a, b; let d be this fourth proportional; we have
c    b
a - =d. Multiply both sides by a and also by d, we have
c X d= a X b; the rectangle contained by the two lines
c and d is tlle rectangle required.
163. Problem. To find a square equivalent to a
given rectangle. Let a and b be the two sides of the
given rectangle; find a mean proportional between these
two lines (98); let c be this mean proportional; we have
- =-; multiply both by c and b, we have a X b = c2;
c is the side of the required square.
* The expression L( a square," or l" the square of a,"
may be used when, as in the present case, we wish to
designate the area of a rectangle, each of whose sides is
expressed by a; but when the product a X a, or a2, is
not considered as representing such in area, it should
never be read " a square."




58


Elements of Geometry.


164. Problem. To find a square equivalent to a given
triangle. Let a be the height of the triangle and b the
base. Find c a mean proportional between a and  b.
a    C
We shall then have --   -; and by multiplying by
both the denominators we obtain I a X b  c2; that is,
c is the side of the square sought.
165. Any polygon may be changed into an equivalent
polygon of a less number of sides, by using the principle
that triangles of the same base and height are equivalent.
Suppose we have the pentagon ABCDE. Draw the diFig 89 agonal AD (fig. 89); through the vertex E draw EF
parallel to this diagonal and meeting CD produced in F;
and draw AF; the pentagon ABCDE will be transformed into the equivalent quadrilateral ABCF. In the same
manner all polygons may be reduced to equivalent triangles.
166. If we would ascertain the measure of a trapezoid,
Fig. 90. as ABCD (fig. 90), AB and CD being the parallel sides;
draw the diagonal AD; this will divide the trapezoid into two triangles whose bases are respectively the two
parallel sides of the figure, and whose common height is
the perpendicular distance AF of these two sides. Each
triangle will have for its measure the height multiplied
by half the base (160). Therefore the sum of the two
triangles will have for its measure the product of the
common height into half the sum of their bases. We
therefore say- The measure of a trapezoid is the product
of half the sum of the parallel sides, multiplied by their
perpendicular distance.
If we take a for the height of a trapezoid, and b and b'
for the two parallel sides, the area will be expressed by
+        or a.(b + b')
2a. (b +- b'), or  ( 
167. Let us now compare two similar triangles, as
Fig. 91. M and A', (fig. 91). Let their heights be a and a', and
their bases b and b'; this will give  —,  b' );  or
Al _ a.b
leaving out the common factor 1, we have   -.
But because the triangles are similar, we have (91)
a    b                              b
_. -; we might therefore substitute  for its equal




Elements of Geometry.


59


a..                  b.b
- in the preceding equation; which gives   -Ib =b' -;
M     b2
or which is the same thing, - l -- -"  We see, therefore, that the two triangles are to each other in the ratio
of the second powers of their bases. Any other homologous sides being taken for bases, would give a similar
result. And as all homologous lines are proportional in
similar triangles, we have this general rule-Similar triangles are to each other as tie second powers of their homologous dimensions; or, as the squares described upon
their homologous lines.
168. As two similar polygons of any number of sides
(93) are composed of the same number of similar triangles of proportional dimensions, this result may be readily applied to them.
The two similar pentagons (ji. 92) are divided by the Fig. 92.
diagonals, each into three triangles similar to the corresponding triangles in the other (93). The homologous diaga    b
onals a, b, and a', b', give the proportion,-  -  (92).
The similar triangles, in the two figures, give the proporL     a'  a1            N     b2    N     a'2
tions        21l      ' and    -   b, or N        -
L + M+ N        a"
(167); therefore (84) L +   l; that is,
L+     l 3I/ +N// Nthat
-Tlhe areas of similar polygons are to each other as the
second powers of their homologous dimensions; or as the
squares described upon their corresponding lines.
169. Suppose we have the two triangles ABC, AEF,
(fig. 93) having the common angle A; draw the perpen- Fig. 93.
diculars BD and EG to the side AC; the similar triangles
BD AB
ABD, AEG, give the proportion EG - A  E.  But the two
ABC AC X BD
triangles ABC, AEF, give AEF = AF X     G (160). If
we multiply these two proportions in order, and suppress
BD               ABC ABXAC
the common factor  -, we shall have AE-AEX — A
That isIf two trianles have an angle of the one equal
That is —If two triangles have an angle of the one eqfual




60


Elements of Geometry.


to an angle of the other, their areas will be as the products
of the sides containing the equal angles.
Fig. 94.  170. If we take CE (fig. 94) equal to the sum of the
two lines a and b, and construct upon it tie square
CEFG; and upon the part CD, equal to a, construct the
square CDIH, and produce the sides HI and DI to L and
K; we shall have the square of CE (the sum of the two
lines a and b) divided intofour parts, viz. CDIHI, IKGL,
DELI, and HFKI. Because CEGF is a square, CF is
equal to CE; and because CDIH is a square, CH is
equal to CD; therefore F-I is equal to DE, or b; and
HI being one side of the square CD or a, the rectangle
HFKI is a rectangle contained by two lines, one equal to
a, and the other equal to b; DELI is is a rectangle of
the same dimlensions; and 1LGK is a square whose side
is equal to DE, or b; so that the whole figure, which is
the square of CE (the sum of tle two lines, a and b),
contains two squares, which are the squares of tlhe parts
a and b; and two equal rectangles each of which is contained by these two parts. This gives the equation
(a -- b)2     a~ — ~ b —  (a. b), which may be enunciated as follows-If a line is cqual to the sutm of t.c)o other
lines, the square describlcd,1ptl thiis linle is equral to the
sum of the squares of the t:vo lines, jplus twice the rectacilgle contained by the two lines.
Fig. 95.  171. Take (fig. 95) CD the difference between two
lines a and b; construct upon BD, equal to a, the square
ABDI; and upon CD, equal to the difference of tie two
lines, another square CDEF; produce EF to G, maling
FG equal to b; and complete the square FGHI, which
will be the square of b; so that the whole figure is equal
to the square of a plus the square of; but ABCK is a
rectangle contained by the two lines a and b; EGIUK is
an equal rectangle; if we take away these two rectangles, we shall have left the square CDFE, (the square of
a —b). This gives the equation (a - b)2 = a2 -- b2
-2 (a. b), which gives this general rule-The square
of the difference of two lines, is equal to the square (f the
greater, plus the square of the less, minus twice the rectangle contained by the two.
172. Let a represent the hypothenuse of a right-angled
triangle whose two sides are b and c. From the vertex
Fig. 96 f the right-angle draw AD (fig. 96) perpendicular to the
hypothenuse. We have (101) the two equations, CD



Elements of Geometry.


61


(CA)2    b2            (AB)-     C2
(CA) —  b —, and DB -(AB)       -.   Adding these
CB      a              CB       a
equations, we have CD + DB =CB =a =           +;
a
and multiplying both sides by a, we have a  b3 + c2,
that is- -'he square (lescribed upon the hypothenuse of a
right-anglcd triangle, is equivalent to the sum of the
squares described (upon the other two sides.
173. We may demonstrate this truth from the areas
immediately, without referring the lines to numbers, as
in the above proof. Upon the hypothenuse CB (fig. 97) Fig. 97.
construct the sqluare CF; and upon the inside of the
other sides CA, AB, construct the two squares CII and
AL; fiom the vertex A of tlhe right-angle draw AD
perpendicular to the hylpothenuse and produce it to G;
and produce  aL aInd I1N till tlley meet tlie peripl)ery of
the square CIFB: tile points of meeting will el the vertices F and E; for tlih triangle LBF has its si(des respectively l)erl)endicular to tliose of the original triangle ABC,
its angles are therefore equal; and having LB cequal to
AB, being a side of the same square, the two triangles
are therelbre equal by tle tlhird case, their hypothenuses
are consequently equal, and ML produced must meet
BF in the point F; in tlhe same manner we may show
that UN producedl will fall upon the point E.
Tlhe square AL and the parallelogram AKFB, are
upon the same base AB, and lhave the same heighlt BL,
they are therefore equivalent (15>1); BK and BG are parallelograms upon the same base BF, and have the same
height BD, they are therefore equivalent; tllat is, the
square upon the side AB is equivalent to tle rectangle
BG: in thle same manner we may show that the square
upon CA is equivalent to the rectangle CG: therefore
the sum of the squares b2 + c2 is equivalent to the sum
of the rectangles CG and BG; but these two rectangles
taken togetler make the square of the hypothenuse a2.
This gives us the equation a2 = b2 + c2, the same result as tlat obtained by the other process.
174. Remark. This is one of the most important propositions in elementary geometry. A method of proving it
geometrically was long a desideratum with the ancient
geometers. It was at length discovered by Pythagoras.
It is known by the name of the Pythagorean proposition;
and is the 47th of the 1st book of Euclid.
6




62


Elements of Geometry.


Next to this in importance, as a geometrical truth, and
nearly connected with it, is the proposition-Similar
figures have their homologous measures proportional. In
the application of the elementary principles, these two
propositions are continually used.
175. If b2 +c2= a2, then b = a2 - c2, and c2
-_a2 - b2; that is, The square of one side of a rightangled triangle, is equivalent to the square of the hypothenuse, minus the square of the other side.
Fig. 98.  176. In figure 98, let AB -a, and CB = b. Produce AB to E, making BE    b, then AE will be equal
to a- b, and AC will be equal to a - b. Upon AB
make the square AD, which will be the square of a or a2;
and upon AC construct the square AF, which will be the
square of a —b, that is (a —b)2. If we complete the
rectangle A H, it will be the rectangle contained by
a + b and a -b; and as the part M' is equal to the part
of the figure, instead of the rectangle All, we may
take its equivalent AIKFLB; but this part of the figure
is equal to the square IB with the square KL taken from
it, that is, equal to a2 - b2; so that the rectangle AH is
equivalent to the difference between the square described
upon a and the square described upon b, which gives the
equation (a + b) X (a - b)   a2 - b2. The rectangle
contained by the sum and difference of two lines, is equivalent to the difference of the squares described upon the
two lines.
Fig. 97. '177. The two rectangles CEGD, DBFG, (fig. 97) of
the same height are in the ratio of their bases CD and DB
(155); but they are equivalent respectively to the squares
of the two sides, CA and AB, to which they are adjacent;
consequently- The squares of the two sides of the rightangled triangle, are to each other in the ratio of the adjacent segments of the hypothenuse.
178. Suppose that the angle A, instead of being a rightFig. 99. angle, is an obtuse angle (fig. 99); produce the side CA
till it meet in D, the perpendicular BD, drawn from the opposite vertex; denote the perpendicular by y, and the distance AD by x. The right-angled triangle ABD (172)
gives the equation c2 -y2 + x2; and the triangle CBD
gives a2 = y2 + (x + b)2 =y2 +a2 + -     + 2 (6 X x)
(170). And substituting c2 for its value, we have a2 -
c2 + b2 + 2 (b X  ). We therefore say-In an obtuseangled triangle, the square of the side opposite the obtuse




Elements of Geometry.


63


angle, is equivalent to the sum of the squares of the other
two sides, plus twice the rectangle contained by one of
those sides and the distance of the vertex of the obtuse angle from the foot of the perpendicular drawn from the opposite vertex to this side produced.
179. If the angle A be acute, the perpendicular BD
(fig. 100), will fall within the triangle. Designating AD
by x, and BD by y, we have (CD)2        - (b -  )2 
b2 +2-2 (b X:), (171);c2 -y2        z"; a = -y2 +
(CD)2 =y?/2 +       -  23  - z,-  (b X;); and substituting cs
for its value, we have a2  c2 4- &  - 2 (b X z); that is
-In an obliqlue-anl rled triatngle, the squarie of the side opposite an acute a ngle, is eqitivalent to the sum o'fthe squares
of the other two sides, minus itwice the rectangle contained by
one of the sides ad!jaclnt to this anugrc, and the distance of
this angle from the foot of the perpendicular drawn to
this adjacent side from the ppop)site vertex.
180. In figure 99, (170) we have (CD)2 -2 2 + b2
+ 2 (b X   ); and subtracting x2 from each side, we have
(CD) — x2- b +-2 (b x x).      We had also on the
same figure, a2 =- c2 + b2 - 2 (b X.); subtracting c2
from  each side we have a2 -c2 - b2 + 2 (b X x);
therefore a2 -  c2- (CD)2 - x2.
In figure 100 (179) we have (CD)2 = x2 + b2 -2
(b X x); therefore (CD)2 - x2 = b2 -2 (b X x); but
we had in tie same article a2 = c2 - b~ - 2 (b X x );
which gives a2 - C- b2  - 2 (b X x); therefore a2 -c2
= (CD) 2-x. We therefore say-If Jiom         the vertex of cither of the angles of any triangle, a perpendicular be lrawnzon  the oplisite side (produced if necessary)
the cifference of the squares of the sides containing this
angle, will be equivaleit to the difference of the squares
of the two distances of the foot of this perpendicular from
the other vertices of the triangle.  >
181. If we would find the side of a square equivalent
to two given squares; we have only to make a rightangled triangle, the two sides of which are respectively
equal to the sides of the given squares; the hypothenuse
will be the side of the square sought (172).
If we would find the side of a square equivalent to the
sum of several squares, as a2 +- b2 + c + d2, &c. (fig.
101); construct a right-angled triangle ABC, whose two Fig.101.
sides are a and b, the square of the hypothenuse (m2) will
be equivalent to the two squares a2 and b2; then make




64


Elements of Geometry.


another right-angled triangle with the two sides m and c;
the square of the hypothenuse (Wn) will be equivalent to
the sum of tlle first three squares; that is, u2 -= mn 1- 2;
therefore n l - a2- 2 - -2; or n   (a +   2 b+ (2);
proceed in the same way with any number of squares.
182. If we would find a square equivalent to the difFig.102. ference between two given squares, (a and b6 (fi(. 102);
draw a line AC, equal to the side of tlle smaller of the
two squares, 62; at A draw the indefinite line AB, perpendicular to AC  then from C as a centre with a radius
equal to the side of the larger square, draw an arc cutting
the perpendicular in the point B; BA will be tle side of
the square souglht; for we have c2 -  2 -  Ib; c 
(a2- 62f.
183. If fiom the summit of any triangle, as ACB
Fig.103. (fig. 103), we draw a straight line, as CE, to the middle
of the base, it will divide the triangle into two triangles,
one of which (if the sides CA and CB are unequal) will be
an obtuse-angled triangle, and the other an acute-anlled
triangle. If we draw the perpendicul;ir CD, and designate
the line CB by a, AC by b, AE and EB each by  E c, ED by
x, and CE by m?; the triangle ACE will give tlhe equation
62-_ m2 + (,c)2 + -2 ( c X x), (178)  the other triangle gives the equation a2 = — n2 - ( c)2 -  2 (2 c X x),
(179); adding these two equations, we have a + b" - 
2 m2 +- 2 (- c)2; t!at is-i-f a li,i e be (rancl fot tlhe
'vrtex of 1any angle of a triangrle to the middl(c of the opposite side, twice the square of this line, alcd to tw ice
the square of half this ojosite sid, wlill be equivalent to
the sum of the squares of the two sides which contain this
angle.
Fig.104.  184. If in the parallelogram  ACBD (fig. 104) we
draw the two diagonals, they will bisect each otler at E;
and according to the last article, the triangle ACB gives
6-2 +- 2 - 2 (A E)2 + 2 (CE)2;
c2 + d-2- 2 (AE)2- + 2(DE)2;
adding now the two equations, and considering that CE
equals DE, we have a2 + 7b + c2 - d2 - 4 (AE)2 + 4
(CE)2. But 4 (AE)2 is the square of 2AE; tliat is, the
square of AB; and 4 (CE)2 is the square of 2CE, the
square of CD.     The equation, therefore, becomes
a2 +- 62 + c2 + d2 - (AB)2 + (CD)2.   That is-The
sum of the squares of the four sides of a parallelogram,




Elements of Geometry.


65


is equivalent to the sum of the squares of the two diagonals.
185. A regular polygon (fig. 76) is composed of a Fig. 76.
certain number of equal isosceles triangles, whose bases
make the perimeter of the polygon, and whose common height C a is the apothegm      of the polygon.
The area of each of these triangles is the product of its
base into half the height; therefore —The area of a regular polygon is expressed by the product of its perimeter
multiplied by ha/J its apothegm; that is, by half the radius of the inscribed circle.
186. As two regular inscribed polygons of the same
number of sides are composed each of the same number
of equal triangles; and as the triangles in the one polygon are similar to those of the other, these similar trianBOC     (BO)2
gles (fig. 105) will give the proportion  -  (b ) = Fig05.
(ON)2      6 (BOC)     (BO)2    (ON)2
(onn)2; or -— (o      --- = (ot —2 -   But six times
(o n)2     6 (b o C)   (b o"2  (o 71)2
BOC is the larger polygon, and six times b oc is the
smaller polygon; BO, b o, are radii of the circumscribed
circles; and ON, on, are radii of the inscribed circles.
We therefore say- The areas of regular polygons of the
same number of sides, are to each other as the squares of
the radii of the circumscribed circles, and also of the inscribed circles.
187. It is manifest that as we increase the number of
sides in the inscribed regular polygon, we increase its
area, which may approach indefinitely near to that of the
circle, but can never exceed it. And as we increase the
number of sides in the regular circumscribed polygon,
we diminish its area, which, though it may approach indefinitely near to that of the circle, can never be less.
We have also seen (148) that as the number of sides of
the inscribed and circumscribed polygons are increased,
their perimeters approach in value to that of the circumference.
If we suppose that the number of sides of the circumscribed polygon is so great that the excess of its perimeter over the circumference is less than any assignable magnitude, which excess we denote by d; the area
of the polygon will exceed the area of the circle, which
we express by A, by an exceedingly small quantity, which
6*




66              Elements of Geometry.
we denote by m. If we denote the circumference by C,
the perimeter of the polygon will be expressed by C + d;
the radius of tlis circle, which we denote by RI, will evidently be the apothegm of this circumscribed polygon,
and its area will be expressed by 1A q- m -  R X ( C  d).
Tile perimeter of the inscribed polygon of the same
number of sides will be less than the circumference of
the circle by an indefinitely sinall nmagnitude 1d'; and
the area of this polygon will be less tllhn tha t of the circle b)y an indefinitely small magnitude nl'. ''The apothegm of tllis inscribed polygon will be less than the radius of the circle by an indleinitely small maIgnitude x. To
express the area of this inscribed polygon, we laive the
equation A - m'l --    (I -  x) X (C — d).  The area
A -i- m is greater thanl thle area of tlhe circle by an indefinitely small quantity, and tlle product -1 RI X (C- (1d)
which expresses thlis area, is greater tlan I  1' X C.
The area A - m', is less than tile area of tie circle by
an indefinitely small quantity, and the product I (1?- z)
X (C —d') which expresses this area, is less than
R X C(. If, then, any area,rntcatcr than the area of
tlhe circle requires for its expression a product grcater
than I P X C; and if any area less than the area of
the circle requires for its expression a product less than
-   X C; it manifestly follows tliat the product.1? X C
cannot be the measure of an area greater or less than
that of the circle whose radius is 1? and circumference
C; it must therefore be the measure of an area equal to
the area of this circle.  Ience we say- 'i'ec area of a
circle has for its measure half the p'roduct of radius multiplied by the circumference. 'That is-A circle is cyquivalcnt to a rectangle whose base is /quiva(lent to the entire
circulmference, and lwhose cheigt is eq ual to htalf radiu's.
188. Having a rectangle equivalent to a given circle,
a mean proportional between its two adjacent sides, will
be the side of an equivalent square.
To find the side of a square equivalent to a circle of a
given diameter, constitutes the hitherto unsolved problem of the quadrature of the circle. This problem, it will
be seen, depends upon another, that is, to find the circumference of a circle, the radius or diameter being
given, or in other words, to find the definite ratio of the
circumference of a circle to its diameter. This is called-The definite rectification of the circumference of a
circle.




Elcments of Geometry.


67


Although the length of the circumference of a given
circle, or its ratio to the diameter, has never been exactly obtained; yet it may be approximated to any required degree of exactness.
" 189. Problem. To find the approximate ratio of the
circumference to the diameter of the circle. Designatingthis ratio by ar, the radius of any circle by R, the circumference of a circle whose radius is R, by C, and the
area of the same circle, by A, we have r = - -;
whence C -2r X R, and A           2Xr X RX  I R
7r. 1, 2.
If R = 1, then A -=; that is-Thie area of a circle
whiose radlitus is one, equals the valule of; that is, the
area of the circle whose radius is 1, bears the same ratio
to the square wlhose side is one, as the circumference bears
to the diameter. Tie question is therefore reduced to
finding the area of a circle whose radius is unity or one.
The area of a circle being less than that of any circumscribed polygon and greater than that of any inscribed;
and as tle values of two similar polygons, the one inscri bed and the other circumscribed, approach more
nearly to an equality, in proportion as the number of their
sides is greater; and as the number of sides of these polygons may be increased till the difference in their areas
shall be less than any assignable quantity l: If then
we take the arithmetical mean between the areas of the
circumscribed and inscribed polygons as the area of the
circle, the error will be much less than half d.
A square circumscribed about a circle whose radius is
1, is tile square of the diameter, or the square of 2, equal
to 4;  the inscribed square (143) is the square of
()2", or 2. The proposed approximation, therefore, will
be readily made if we can find an easy solution of the
following problem.
190. Problem. The area of a regular inscribed polygon
and that of a similar circumscribed polygon being given,
to find the areas of regular inscribed and circumscribed
polygons of double the number of sides. Let EF and GH
(fig. 106) be the sides of regular polygons of t sides, the Fig.106.
one circumscribed and the other inscribed; by drawing
the chords BG and BH, and the tangents GI and HK,
the straight lines BG and IK will be the sides of regular




6S


Elements of Geometry.


polygons of 2n sides, the one inscribed and the other
circumscribed. The triangles ECF and GCH will be
contained n times in the respective circumscribed and
inscribed polygous of n sides, of which they are parts;
and the triangles ICK and BCG, wvill be contained 2n
times, in the regular circumscribed and inscribed polygons of 2 a sides. Let A represent the area of the regular circumscribed polygon of n sides; a the area of the
similar inscribed polygon; A' and a' the areas of the
regular circumscribed and inscribed polygons of 2n
sides. We shall have the areas of the several polygons as
follows: circumscribed, of n sides =- A -- 2  X EBC,
inscribed,    of n sides  -- a -  n X GDC,
circumscribed, of 2 sides: A'/  4 n X BCI,
inscribed,   of 2 n sides  a'  2 n X GBC.
A    EBC       a'   GBC
Then a      B    and     GDC; and triangles of
a    GBC'     a    GDCI
the same altitude being to each other as their bases, we
EBC     C      CB
have GBC= - CG -; for the same reason, we
(jot/   CG     CDJ
GBC     CB    EBC               A    a'
have         C     GBC; consequently -   -,which
gives a' =(A X a).     -    -    -    (1).
Also, because CI bisecting the angle BCE, divides the
base into parts BI and IE, proportional to the sides CB
and CE, we have
BCI    BI    BC     a         BCI       BCI
ICE-    E    EC    a an BCI+ ICE        BCE
a      I-A'   A'                  A'      a
— a  -   a-  AA; but the equation  - =
a+a' - a  A                         2 A   a + a'
2AXa
gives A'   -     -.         - -  -   (2).
a —. a'
The problem then is solved by means of these two
formulas; for knowing the areas of the inscribed and
circumscribed polygons of n sides, we have the quantities A and a; from these, the formulas give us A' and a',
the areas of the inscribed and circumscribed polygons of
2 n sides.
191. We are now prepared to solve the problem enunciated in article 189, viz.-To find the area of a circle
whose radius is one; or, in other words, to find the value
of rt. The area of the circumscribed square, radius being




Elements of Geometry.


69


1, is equal to 4; the area of the inscribed square is equal
to 2. Making A = 4, and a = 2, the formulas (1), (2),
16
wvill give a' = (8)2 = 2,8284271 = the area of an inscribed octagon; and A' - +    -         3,3137085
2 --- 2,8!4`217 I
the area of a circumscribed octagon.
If we substitute these values of A' and a', for A and a
in tile formulas, we shall have i' -3,06146741 -the
area of a regular inscribed polygon of 16 sides, and
A' = 3,1'25979 -= the area of a regular circumscribed
polygon of 16 sides.
The values of these polygons will enable us to find the
areas of inscribed and circumscribed polygons of 32
sides; and, as the farther we proceed in the calculation
the nearer wvill the two polygons approach to an equality,
we may continue the process till the values of thle inscribed and circumscribed polygons do not differ for any
number of lecimals to which it may be thought best that
the expressions should extend. Having carried the process thus far, and knowing that the circle cannot be
greater tlan the circumscribed polygon, nor less than
the inscribed polygon, we take this value for the area
of the circle, as far as the expression extends.
The following table gives the result of this calculation
pursued till the expressed values of the circumscribed
and inscribed polygons in a circle whose radius is 1, do
not differ for the first seven decimals.


Numbler of sides in
tihe polyrgon.
4
8
16
64
128
256
512
10'24
2048
4096
8192
16384
32768


Value of circumscribed polygon.
4,0000000
3,3137085
3,18.25979
3,1517249
3,1441148
3,1422236
3,1417504
3,1416 321
3,141 6025
3,1415951
3,1415928
3,1415927
3,1415926


Valtue of inscribctd 
2,0000000
2,8284271
3,0614674
3,1214451
3,1365485
3,1 403311
3,1412- 772
3,1415138
3,1415 729
3,1415377
3,1415914
3,1415923
3,1415925
3,1415926




70


Elements of Geometry.


We therefore take 3,1415926 for the approximate area
of a circle whose radius is 1. It is also the approximate
value of the ratio of the circumference of any circle to
its diameter, that is, 7 == 3,1415926 nearly; and multiplying the diameter of a circle by this quantity, will
always give us the circumference to a sufficient degree of
exactness.
192. If we denote the radius of a circle by R, the diameter will be 2R, and the circumference 2 r.R, and
the area, A - - R X 27.R -.R2.   That is-The
square of radius multiplied by 3,1415926, will give the
area of any circle.
193. Let A denote the area of a circle whose radius
is R, and a the area of a circle whose radius is r; we
have the two equations, A - ~r. R", a =  r.r2. These two
A    7i.R'2
equations will give the proportion  7.2; and divida    7r. r2
ing each term of the second ratio by the common factor
A    R2
nr, we have -_ -     that is —The areas of circles are
a    r22 
to each other as the squares of their radii.  The radii of
circles are in the ratio of their diameters, and therefore
in the ratio of their circumferences (148). WVe see,
therefore, that it is with curvilinear as with rectilinear
figures (168); and we have this general rule-Similar
plane figures are to each other as the squares of their homologous lines.
Fig.107.  194. The sector AEBC (fig. 107) is evidently the
same part of the circle that the arc AEB is of the entire
circumference; if the circumference multiplied by half
radius gives the area of the entire circle, it follows that
— The arc of the sector multiplied by half radius, gives
the area of the sector.
To obtain the area of the segment AEBD, subtract
the area of the triangle ABC from the area of the sector.
PROBLEMS.
(1). Suppose a castle, whose walls are 48 feet high,
surrounded by a ditch 64 feet wide; what is the length
of a ladder which will reach from the outside of the ditch
to the top of the castle wall?
(2). A triangular field has one right-angle; the side
opposite the right-angle measures 75 chains; one of the




Elements of Geometry.


71


sides adjacent to the right-angle, measures 45 chains;
what is the length of the third side?
(3). How many square chains in a rectangular field
whose length is 12- chains, and whose width is 8
chains?
(4). What is the side of a square field of an equal
area with the above?
(5). One side of a triangular field is 20 rods; and
the perpendicular distance of the vertex of the opposite
angle from this side, is 20 rods; what is the side of a
square field whose area is twice as great?
(6). What length of carpeting 7 of a yard wide, will
cover the floor of a room whose length is 30 feet And
width 21 feet?
(7). The length of a rectangular lawn is 25 rods,
and its width S rods; what is the length of an equivalent
lawn whose width is 12- rods?
(8). What is tie circumference of a circle whose
radius is 4? and what is the area of the same circle?
(9). What is the approximate value of the side of a
square equivalent to.a circle whose radius is 9?.
(10). What is the radius of a circle whose circurnference is 31,416?
(11). What is the value of the apothegm of a regular
hexagon, inscribed in a circle whose radius is 20? and
what is the area of this hexagon?
(12). What is the area of an equilateral triangle inscribed in a circle whose radius is 20?
(13). What is the area of a square inscribed in a circle whose radius is 10? What is the area of the circumscribed square?
(14). What is the area of a semi-circular pond whose
straight side is 200 yards?
(15). What is the area of a sector embracing 60~ in
a circle whose radius is 20 yards?




PART SECOND.
SECTION I. —Of Planes, and of Bodies ternincted
by Planes.
195. IN the preceding sections, we have considered all
the lines of any one of the figures which have leen discussed, as in a plane surface; that is, in a surface to
which a straight line may be applied in every direction, so
as to lie entirely in this surface. This has enabled us to
unfold the elementary principles of linear geometry, and
to measure the areas of surfaces bounded by straight
lines or circular curves. We proceed now to investigate
the geometrical relations in bodies.
As the surfaces of many bodies consist, in whole or in
part, of planes; and as all bodies involve the three dimensions of space, length, breadth and thickness; in discussing their geometrical character, it is necessary to consider not only the forms and magnitudes of planes, but their
relative positions, their inclinations, and their relation to
lines and points wnithout them.@
In discussing the general relations of lines to planes,
and of planes to each other, we consider them as indefinitely extended, except where some limit is expressly
stated.
196. If two straight lines are in the same plane, a
second plane may be conceived to pass through one of
these lines without coinciding with the other line; it
will therefore cut the first plane. But if any two points
* In what follows, the figures though represented on
a plane, are considered as embracing the three dimension of space. And where any line is supposed to pass
behind any part of a body or of a plane, it is indicated
in the figure by a dotted line.




Elements of Geometry.


73


common to the two planes, be taken, a straight line
drawn between them will lie wholly in each of the
planes (8); therefore-The intersection of two planes is
a straight line.
197. If the second plane he turned about the straight
line through which it passes, till it coincides with the
other line, it must also coincide with the other plane:
Two straight lines, therefore, are sufficient to determine
the position of a plane:  And th'ee points, not in the
same straight line, will also determine the position of a
plane; for the plane may be conceived to revolve about
a straight line passing through two of the points; and if
it be placed so as to coincide with the other point, it can
revolve no farther without leaving it; it is therefore
fixed.
It follows from this that any two straight lines which
cut each other, are in the same plane; for a plane may
be passed through one of them, as AB (fig. 108), and Fig.108.
turned till it coincide with the other CD.
It is evident, therefore, that-any three points, not in
the same straight line, being joined, two and two, by
three straight lines, a plane triangle will be formed; if
four points are so joined, the quadrilateral, thus formed,
may be a plane figure; but if the plane passing through
three of the points, do not at the same time pass through
the fourth, the quadrilateral cannot be considered as a
single surface.
19S. A straight line is said to be parallel to a plane,
when it dues not incline towards the plane, in either direction. A line parallel to a plane will therefore be, in
every part, at the same distance firom  the plane; and
consequently can never meet it, however far produced.
A straight line is perpendicular to a plane when its inclination to the plane is the same on all sides. It is evident that the straight line will, in this case, be perpendicular to every straight line drawn in this plane, through
the foot of the perpendicular.
199. Let us suppose a straight line PD (fig. 109) Fig.109.
to revolve about the straight line AB, always perpendicular to it, AB being considered as fixed. It is
evident that PD, by this revolution, will generate a plane
surface; for in whatever position the revolving line be
placed, in the position PE, for instance, a plane may be
conceived to pass through EPA, which being extended
7




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Elements of Geometry.


Fig.109. will also pass through the opposite position of the revolving line, as PC. But as this line must be always perpendicular to AP, the sum of the two angles APE, APC,
will be equal to two right-angles, and CPF will always
be a straight line, that is. the surface will be a plane, to
which AB will be perpendicular (8).
It follows, therefore, that-If two straight lines are
perpendicular to the same straight line, at the same point,
they are in the same plane perpendicular to this line.
200. Remark. It is evident that-There can be only
one line passing through P, perpendicular to this plane;
for if there could be another it must diverge from PA,
and therefore its inclination to the plane must be greater
on one side than the other, which cannot be the case
with a perpendicular (198).
201. It is also evident that-Through any point,
without a plane, as A for instance above the plane MN,
only one perpendicular can be drawn to that plane. For
a straight line passing through A in the same direction
with AP must be the same line; another line, therefore,
which passes through A must be inclined to AP, and
consequently will be differently inclined to the plane
MN N; it cannot, therefore, be perpendicular to this
plane.
It is equally evident that-Only one plane can cut a
straight line at the same point, perpendicular to it. For
two planes passing through the same point, must be inclined to each other; they cannot, therefore, be both
perpendicular to the same straight line.
202. As any two of the lines, CE, FD, passing through
the point P in this plane, are sufficient to fix the position
of this plane with respect to the line AB perpendicular
to these two lines (197); we say-If a straight line is
perpendicular to each of two straight lines drawn in a
plane through its foot, it is perpendicular to every other
straight line drawn through its foot in this plane, and is
therefore perpendicular to the plane.
203. Suppose AP to be perpendicular to the plane
Fig.lIl. MN (fig. 110); draw in this plane, from the foot of this
perpendicular, the equal straight lines PC, PD; and
draw AC, AD. The two triangles APC, APD, are both
right-angled at P; the sides containing the equal angles
are equal respectively; the triangles are therefore equal
by the second case, and give AC = AD. We therefore




Elements of Geometry.


75


say-Oblique lines drawn to a plane friom any point
without it and equally distant from a perpendicular to
the plane, drawn through the same point, are equal.
204. It follows from the last result, that-If from any
point without a plane, a straight line be drawn perpendicular
to that plane, and also several equal oblique lines be drawn
to the plane from this point; these oblique lines will meet
the plane in the circumference of a circle whose centre is
the foot of the perpendicular.  Consequently-If a
straight line be (lrawn perpendicular to a circle through
its centre, any point in this straight line is equally distantfrom every point in the circunlference qf the circle.
205. If we produce PD to G, and draw AG, it will be
easy to show that AG must be greater than AD (51),
and thence to show that-Of two oblique lines falling
upon a plane at unequal distances from a perptdiicular,
that is the greater which falls at the greater distance
from the perpendicular.
206. Suppose the line AG (jig. 111) to be oblique to Fig.1l.
the plane MN; draw from A to the plane the perpendicular AP; join PG, and through G draw BC perpendicular to PG. Then taking GC equal to GB, draw PB,
PC, AB, AC; PB and PC will be equal (41), and AB
and AC will therefore be equal; and AG will be a straight
line (rawn from the vertex of an isosceles triangle to the
middle of the base; that is-BC is perpendicular to the
oblique line AG, when it is perpendicular to the straighf
line which joins the foot qf the oblique line with the foot
of the perpendicular A P.
In this case AP and BC are said to be perpendicular
to each other, though they cannot meet.
207. Suppose the plane OP (fig. 112) to cut the plane Fig.112
MN, in the line AB; draw through the point C, CD in the
plane MN, and CE in the plane OP, each perpendicular
to the intersection AB of these planes; then turning the
plane OP about the intersection till the two planes coircide, CE will coincide with CD. If the part E of the plane
OP be now raised from the plane MN, turning this plane
about the intersection, this point will describe the arc
DE, which in every stage of the process will measure
the inclination of the two planes; the centre of this arc
is C; therefore- To n/measure the angle which two planes
make with each other, draw in each plane, through the
same point, a straight line perpendicular to the intersec



76


Elements of Geometry.


tion, and the angle which these two lines make with each
other is the angle of the planes.
208. Remark. The angle made by two planes is called a diedral angle, that is, an angle of two faces; and is
designated by four letters, of which the two middle ones
are in the intersection, and the two others are in the
different planes out of the intersection; as the diedral
angole PABN. The intersection AB of the two planes is
called the edge of the diedral angle. When this angle
is 90~, the inclination of the plane PO to the plane MN,
is the same on each side of the intersection, and the
planes are said to be perpendicular to each other.
It is evident that two diedral angles are to each other
as the arcs of the plane angles which are their measures.
It is also manifest that two diedral angles which are opposite at the edge, as MABO and PABN, are equal.
Fig.113.  209. Suppose the two planes CD and EF (fig. 113)
to be each perpendicular to the plane AB; from the
point G where the three planes meet, draw EH perpendicular to the plane AB; it will be in each of the planes
CD and EF (207); therefore-The intersection of two
planes each perpendicular to a third plane, is also perpendicular to this third plane.
210. If two straight lines are perpendicular to the
same plane, they have the same direction in space and
are therefore parallel to each other. Let CE be perpenFig.114. dicular to the plane MN (fig. 114), and BD be parallel to CE; let the plane AD pass through the parallels; this plane will be perpendicular to the plane
MN (207), and DB parallel to EC, will be perpendicular
to the intersection AB; it will therefore have the same
inclination to the plane MN, as the plane AD has to the
plane MN; and as this inclination is a right angle, DB
is perpendicular to the plane MN.    Therefore.-A
straight line parallel to a second which is perpendicular
to any plane, is also perpendicular to this plane.
Fig.115.  211. Take the two planes AB and CD (fig. 115) perpendicular to the same straight line GH; draw GR and
GL in the plane AB, and from the point H draw HM
and HS parallel respectively to GL and GR; GL and
GR will be both perpendicular to the line GH, and con.
sequently will lie in the plane AB (201). Now the two
lines HM and HS determine the position of the plane
CD (198), and GL and GR determine the position of the




Elements of Geometry.


77


plane AB; but HM and GL are parallel, therefore the
planes in which they are situated are not inclined to
each other in that direction; for the same reason they
are not inclined to each other in the direction of the parallels HS and GR. The same may be shown with respect to lines in the two planes in every direction from
the points G and H; consequently the two planes are
not inclined to each other in any direction; and we say
that- Two planes are parallel when they are not inclined
to each other, in any direction.  We say, therefore, that
- Two planes perpendiculvr to the same straight line, are
parallel. And- Two parallel planes have common perpendiculars.
212. Remark 1st. As two parallel planes are not inclined to each other, thy must be throughout at the same distancefrom each other; and, however far produced, can
never meet; therefore-All the perpendiculars between
parallel planes, are equal.
213. Remark 2d. If two straight lines which cut each
other, are parallel respectively to two other straight lines
which cut each other, the plane determined by the first
two lines, will be parallel to the plane determined by the
other two (211).
214. Let the two parallel planes AB and CD (Jig.
116) be cut by a third plane FH. The two intersec- Fig.116.
tions EF and GH, being in the same plane FH, will
meet unless they are parallel (12); and if these lines
will meet, the planes AB and CD in which they are situated, will also meet; but these planes being parallel to
each other, can never meet; consequently the two intersections EF and GIH cannot meet; they are then parallel; and we say- -Vhen two parallel planes are cut by
a third plane, the two intersections are parallel.
Remark. If the lines FG and EH are parallel, being
between the parallels EF and GH they must be equal
(60); we say therefore-Parallel lines between parallel
planes, are equal.
215. Let the two straight lines GH and 1K (fig.
117) be cut by the three parallel planes, AB, CD, EF. Fig.117.
Draw HI piercing the plane CD in M; draw also GI and
LM, MN and HK. Then GI and LM are the intersections of the plane HGI with the parallel planes AB
and CD, they are therefore parallel; for a similar reason
MN is parallel to HK. In the triangle GHI, LM being par7*




78


Elements of Geometry.


GL     IM 
allel to the base GI, we have L =      and in the
I — MH;
triangle IHK, MN   being parallel to HK, we have
IM     IN
MH -     NK; but these two equations having the comIM
mon member M-,' the other two members are equal;
GL     IN
that is, L  =  K-; and we say-If two straight lines
LH     NKI
meet three-parallel planes, they will be cut into proportional parts by these planes.
216. When several planes, as ASB, BSC, CSD, DSE,
Fig 118. ESF, FSG, and GSA, (fig. 118) all passing through the
point S, are joined, two and two, by their sides diverging
from S, the space which they comprehend, unlimited in
the direction opposite to S, is called a polyedral angle.*
Angles of this kind are distinguished by the number of
planes which meet at the summit or vertex of the angle.
A triedral angle is an angle of three faces, as the angle
SABC (fig. 119); a tetraedral angle is one which has
four faces. The angle SABCDEFG (fig. 118) has seven
faces and is called a heptaedral angle, &c.
The things to be considered in a polyedral angle, are
the plane angles formed by the edges of the polyedral
angle; and the respective inclinations of these planes,
or the diedral angles formed by contiguous faces.
217. Problem. Three plain angles being given to construct a triedral angle. Upon the side AS of one of the
Fig.119. angles, ASB (fig. 119), place the side AS of another of
them, so that their vertices may coincide at S; then
raising the part C of the second, make it turn about
the intersection AS; it will be perceived that, as the
inclination of the plane CAS to the plane BAS increases, the angular space CSB will increase; there is
therefore a certain inclination of the first two planes,
which allows the third angle to be introduced to complete the triedral angle; therefore the magnitude of each
of the plane angles, determines the inclination of the
other two. We say therefore-If two triedral angles
are formed of plane angles respectively equal, the inclination of any two planes in the one, will be equal to the
inclination of the homologous planes in the other.
* This was formerly called a solid angle.




Elements of Geometry.


79


218. Remark 1. If the plane angles be unequal, but are
placed in the same order in both, the two triedral angles
will coincide, when placed together; in this case they
are called equal. If the order of the planes be different
(fg. 120), the two triedral angles cannot coincide; for, Fig.120.
placing two equal faces together, so that they may coincide, two other equal faces will be inclined to this plane
on opposite sides. In this case they are said to be symmetrical.
219. Remark 2d. If the three plane angles which form
a triedral angle are all equal, the three inclinations, that
is, the three diedral angles will be equal. If each of the
plane angles is a right-angle, the diedral angles will be
right-angles; and the triedral angle may be called
right-angled.
220. If the magnitude of the third angle is just equal
to the difference between the other two, this third plane
can be introduced without inclining the other two, that
is, the diedral angle will be nothing; the three planes
will then be in the same plane, and will therefore comprehend no space. If the third angle be less than the
difference between the other two, it will not fill the remaining space, the inclination of the two being nothing.
We say, therefore, that- The sum of any two of the plane
angles which form a triedral angle, is always greater
than the third.
221. Remark. Though the magnitude of the plane
angles determines their inclination, in a triedral angle;
this is not the case with any other polyedral angle. If
we have four plane angles given, we can form an infinite
variety of tetraeclral angles with them, as they will admit
of an infinite variety of inclinations; but if, in a tetraedral angle, one of the inclinations be fixed (the plane
angles and their order of arrangement being given) this
fact will determine all the others.
222. If the sum of the three plane angles given to
form a triedral angle (fig. 121) were equal to four right Fig.121.
angles, the angular space left for the introduction of the
third plane, would not be sufficient to receive it until the
second plane was turned quite over into the same plane
with the first (15). In this case the three plane angles
would be in the same plane. The same would be true,
of any number of plane angles given to form a polyedral angle; if their sum were equal to four right-angles




80


Elements of Geometry.


their sides would coincide when they were all in the same
plane, their vertices being placed together. And it is
evident that, if the sum of three angles were greater than
Fig.122. four right angles (jig. 122) they could not be brought
together at their vertices, their sides coinciding two and
two; they could not therefore form a triedral angle.
Whence we say- The sum (f all the plane angles which
form a convex polyedral angle, is always less thanfour
right-angles.
223. Remark. By a convex polyedral angle, we mean
one, all of whose diedral angles are salient, that is, have
their edges standing out, as in figure 118. When there
are, in the polyedral angles, re-entering diedral angles,
Fig.123. (as the diedral angle DESF (fig. 123), there is no general limit to the sum of the plane angles which may form
a polyedral angle.
It is evident that the sum of the diedral angles in triedral angles, varies with the plane angles of which the
triedral angle is composed. Are there any limits to this
variety? and if so, what are these limits?
Of Polyedral Bodies.
224. From the discussion of polyedral angles, we proceed naturally to that of polyedral bodies, or polyedrons,*
that is-bodies whose surfaces are composed entirely of
planes. The word polyedron signifies a body of many
faces. When we speak of geometrical bodies we do not
include necessarily the idea of matter or substance which
resists the approach of other bodies or excludes them
from the places which these occupy, as a block of
wood or of marble. Geometry, as we said at first, is the
science of magnitude and form, and takes no cognizance
of even the existence of matter. It discusses, it is true,
the form and extent of material substances; but it is the
form and extent merely; or rather, it is the form and meas* These bodies have been called solid polyedrons;
but it has been very properly objected to the term solid,
that it has in common language an appropriate use, totally different from that which has been assigned to it in
geometry. This new application of a familiar term, is
calculated to mislead the learner, and should never be
adopted.




Elements of Geometry.


81


ure of those portions of space in which these bodies are
extended, that makes the subject of geometry. When
we speak of a geometrical body, therefore, we mean a
portion of space of that particular form and extent; and
when we speak of the fice of a polyedral body, %ve mean
that ideal division between tip portion of space occupied
by the body, and the portion without.
225. In a polyedral angle it is observed, that the
figure is open and unlimited in the direction opposite to
the summit (216); it follows, therefore, that with only
one polyedral angle, of how many planes soever it be
composed, no portion of space can be entirely enclosed.
In orde- then to form a polyeclral body, it will be necessary to have more than one polyedral angle.
The simplest polyedral angle we have seen (fig. 124 ) Fig.124.
is one of three faces (216). If these three planes were
cut by a fourth, a certain portion of space would be enclosed; and if the contiguous sides of the plane faces,
forming a triedral angle, are equal, the sides opposite to
the common vertex S in these triangles, as AB, BC, and
CA, will be in the same plane (197); the plane triangle
ABC may therefore be considered as one face of the polyedron SABC. Without this last plane ABC, we had
but one polyedral angle, that whose vertex is at S; we
have now four polyedral angles; the summits of which
are S, A, B and C. Each of these is a triedral angle,
the most simple of polyedral angles; and each of the
four faces is a triangle, the simplest of rectilinear figures.
It seems quite evident, therefore, that if merely these
circumstances are considered, this is the simplest of polyedral bodies. It is called a tetraedron, a body having
four faces.
226. We consider this body as contained by the three
triangles ASB, BSC, CSA, (having a common vertex in
S,) and the triangle ABC; which last we will call the
base of the polyedron.
If we had four triangles, having the vertices in S (fig.
125) and their bases in the same plane ABCD, the body Fig.125.
contained by these five planes we should call apentaedron,
and in general, a body contained by six plane faces is
called a hexaedron; of eight, an octaedron, &c. But all
bodies of which one of the faces is a polygon, and all
the others triangles having their summit in the same point,
are called pyramids. The body SABCDE (fig. 126) is Fig. 126.




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Elements of Geometry.


a pentagonal pyramid, because its base ABCDE is a pentagon; the point S is called the summit of the pyramid.
Fig.127. The tetraedron SABC (fig. 127) is a triangular pyramid. And in general, a pyramid is called triangular,
quadrangular, pentagonal, hexagonal, &c., according as
its base is a triangle, quadrangle, pentagon, hexagon, &c.
In any pyramid, each of the polyedral angles adjacent
to the base, has only three faces. In a tetraedron, therefore, we may take either of the faces for the base; but
in a pyramid of a greater number of faces, we can never
take a triangle for the base.
2-27. If the three faces which form the tetraedral anFig.124. gle whose summit is at S (fig. 124) were equal issoceles
triangles, their respective inclinations to each other
would be the same (217); and they would have the same
inclination to the base ABC, which must be an equilateral triangle as it is contained by the bases of three
equal isosceles triangles. The body would then be called
a regular pyramid. But it would not be a regular tetraedron, or a regular polyedron. To be a regular tetraedron, all the faces must be equal. They will, therefore, be equilateral and equally inclined to each other,
and the triedral angles will be all equal. And in general, a regular polyedron has all its faces equal regular
polygons, all its diedral angles equal, and all its polyedral angles equal.
228. Of a regular pyramid it is requisite that the base
should be a regular polygon of any number of sides, and
that the other faces be equal isosceles triangles; these
triangles will be therefore equally inclined to the base
(217), and a perpendicular drawn from the summit or
apex (as it is sometimes called) will meet the base in the
centre. This perpendicular measures the height of the
pyramid.
229. We have seen that the plane angles in a tetraedron determine the inclinations of the planes respectively, and consequently the triedral angles; therefore-If
two triangular pyramids are contained by planes respectively equal and similarly arranged, they are equal
to each other; and if placed together will coincide in all
their parts.
230. Now let the two triangular pyramids S-ABC,
Fig.128. S-DEF, (fig. 128) be contained by similar triangles, arranged in the same order in both. Similar triangles have




Elements of Geometry.


83


their angles equal; therefore the two pyramids will have
their triedral angles respectively equal (217), and the
inclination of any two faces in the one, equal to that of
the corresponding faces in the other. These bodies are
therefore called similar pyramids; they are also called
similar tetraedrons. And generally, we call similarpolyedrons, those which have the same number of faces respectively similar and similarly arranged, and their diedral angles respectively equal. We have seen that, in the
tetraedron and the pyramids, this last circumstance is a
necessary result of the others (226). We have, therefore, this general truth-Any two pyramids are similar
when the faces of the one are similar respectively to the
faces of the other, and similarl!b disposed.
231. In the tetraedron SDEF, let us suppose a section to be made by passing a plane through ABC, parallel to the base DEF; AB will be parallel to DE, being
the intersection of the plane SDE with the two parallel
planes ABC, DEF, (209); for a similar reason, BC is
parallel to EF, and AC to DF. -But in any triangle a
line drawn parallel to the base, cuts off a partial triangle
similar to the whole; therefore the triangle SAB is similar to SDE, SBC similar to SEF, and SAC to SDF.
Now in the triangle SDE, AB being parallel to DE, we
SE    DE
have -    -B; and BC being parallel to EF, we have
SE    EF            DE     EF 
SB -     BC' therefore AB    - B; in a similar manner
DE     DF
we show that      =AC; that is, the two triangles
AB     AC
ABC and DEF have their sides proportional, and are
similar, and the tetraedron SABC is similar to the tetraedron SDEF (229). And we say generally —f a tetraedron be cut by a plane parallel to one of its faces, the
portion containing the triedral angle opposite to this face,
will be a tetraedron similar to the whole.
232. Let the pentagonal pyramid S-FGHIK (fig.129) Fig.129.
be cut by a plane ABCDE, parallel to its base. Suppose
SQ to be a perpendicular drawn from the summit to the
base produced, the cutting plane will pass through the
point P. The pentagon ABCDE will be the base of the
smaller pyramid whose height is SP, as SQ is the height
of the larger. The sides AB, BC, CD, DE, and EA,




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Elements of Geometry.


Fig. 129. are respectively parallel to the sides FG, GH, HII 1K,
and KF, of the base of the larger pyramid; and being
parallel and directed the same way, the angles made by
these sides are respectively equal to the angles of the
base FGHIK. Now because each triangular face of the
larger pyramid is cut parallel to its base, we have the
SF     FG     SG     SG      GH       SH
proportions       K1 -              -   w_ 
proportions SA -=  A B       '   SB     B      SC 
SF     SG
&c.; from which we derive the following; SA - SBSH           PFG     GHl
SC   &c.; and _B     B' &c., that is, the triangular
faces of the two pyramids, have their sides proportional,
and are therefore similar triangles; and the bases of the
two pyramids have their sides proportional and their angles equal, they are consequently similar pentagons.
These two pyramids are contained by similar planes
similarly arranged, and the inclination of any two planes
in the less pyramid, is equal to the inclination of the corresponding planes in the larger. The diedral angles
made by similar triangular faces, are equal; and the
corresponding diedral angles at the bases are equal, because the corresponding triangular faces are in the same
plane, and the bases parallel. From the above analysis
we derive the following truths:
(1.) If any pyramid be cut by a plane parallel to its
base, the smaller pyramid thus separated, is similar to the
whole.
(2.) Any section of a pyramid parallel to its base, is
similar to the base. All parallel sections in a pyramid,
are similar polygons.
233. We can find the height of a pyramid when we
know the dimensions of a trunk or frustum of the pyramid, as FGHIK-ABCDE, which remains after cutting
off the superior part S-ABCDE, and whose bases are parFG      SQ
allel to each other; for taking the proportion AB-  SP,
FG-AB        SQ-SP       PQ
we have (81) -AB        =       P ---   SP    which
AB x PQ
gives, SP = FG - AB; and adding SP to PQ, we have
the height, and therefore all the dimensions of the entire
pyramid.




Elements of Geometry.


85


SF     FG
234. We had S —; but FG           and AB are ho- Fig.129.
f9A      AB '
mologous sides of the similar sections FGHIK, ABCDE,
of the pyramid. These sections being similar, are to
each other as the squares of their homologous sides,
at is, FGHIK      (FG)2       (FG)      a (SF)n
t s ABCDE - (AB)2      (AB)2    (SA)";
cause QF and PA are parallel, we have S-    SP; and
SA     SP'
(SF)2    (SQ)2                           FGIIIK
(SA)   (S); consequently we have ABCDE=
(SA)2    (SP)A
(S-S)2. We say, therefore, that-In any pyramid, sec(SP)2
tions made by parallel planes, are to each oiler as the
squares of the distances of these planes from the summit.
SF    SG     SGI
235. We had S      -- SC-       &c. We have also
SA SB SC'
SF SQt
SA-   -SP; we say, therefore,-Similar pyramids have
SA     ^)r
their homologrous edges proportional; and these proportional to their heights. The same may be proved of
all other homologous measures; whence we have the
more general truth —In similar pyramids homologous dimensions are proportional.
236. Remark. As the homologous faces of similar pyramids, are similar polygons; it follows that-In similar
pyramids the homologous faces are to each other, as the
squares of the homologous edges, as the squares of their
heights, or as the squares of their homologous dimensions
generally.
237. By drawing the diagonals FH, FI, in the base of
the pyramid, and passing planes through these diagonals
and the summit, we cut the pentagonal pyramid into
three tetraedrons. And a slight inspection of any polyedral body whatever, will be sufficient to show, that it may
be divided into tetraedrons, in a similar manner. And
consequently-If we have two similar polyedrons, by
passing planes through them both in the same manner, we
may divide them into the same number of similar tetraedrons. The tetraedrons must be similar; for the bodies
being similar, and the cutting planes passing through
corresponding parts, these planes will have the same inclinations to the other faces and to each other, in the one
8




86


Elements of Geometry.


case as in the other.  We say also-When two polyedrons are composed of the same number of similar tetraedrons, similarly disposed, they are similar.
From the preceding article, we infer that-In similar
polyedral bodies, homologous dimensions are proportional;
and homologous faces and corresponding sections, are to
each other as the squares of homologous dimensions in the
bodies.
238. Let A, B, C, D, E, F, be the faces of any polyedron; and a, b, c, d, e,f, be the corresponding faces of
a similar polyedron; let Albe any diagonal in the first,
and m the corresponding diagonal in the second:  We
A    B     C    D    E     J M
have    -       -    -       -  -     But in a series
a     b    c    d    e     m '2
of equal fractions, the sum of the dividends contains
the sum of the divisors as many times as one dividend
contains its divisor; that is A  B  C+D+E
a+ b +c + d + e     f
M'2
--.  But the sum  of all the faces of a polyedron,
m~2
constitutes its area, or whole amount of surface; therefore-The areas of similar polyedrons are to each other
as the squares of their homologous dimensions.
239. Among polyedral bodies, that which demands a
particular examination is the prism.  A prism is a body
which has two opposite faces equal and parallel, and all
Fig.130. its other faces parallelograms.  In figure 130, the body
ABCDEFGHIK, (frequently designated by AG) is a
prism; the pentagon ABCDE is its base. The perpendicular distance of the base from its opposite face, is called
the height of the prism. The body is formed by drawing
from each of the angles of the base, the parallel lines
AF, BG, CH, &c. above the plane of this base and terminated by the plane FGHIK, parallel to ABCDE. The
lines AF, BG, &c. taken two and two, determine the
lateral faces of the prism; and being parallel lines comprehended between parallel planes, they determine these
faces to be parallelograms.  The pentagon FGHIK,
which forms what is sometimes called the superior base
of the prism, is contained by lines respectively equal
to those of the inferior base ABCDE (60); they are
also parallel to them respectively; consequently the angles are equal, and the two polygons, having the homolo



Elements of Geometry.


87


gous sides equal and their homologous angles equal, are
equal to each other. It is evident, from  similar considerations, that-Any plane section   LMNOP      of the
prism, parallel to its base, will be equal to its base; consequently, plane sections of a prism, parallel to each other,
are equal.
240. In any prism, each of the polyedral angles is
composed of three plane angles. A prism in which the
lateral faces are rectangles, is called a right prism; all
others are called oblique prisms.
241. Suppose, in the prism Al (jig. 130), we have the Fig 130.
base given, and the two faces, AG and BfI; if we apply the two faces to the base, in the same manner as they
are disposed in the figure, to form the triedral angle
whose vertex is at B, tlle magnitude of these angles will
determine the inclinations of the planes. These determine the positions of the points F, G and II; the three
points II, C, D, determine the plane Cl; and this being a parallelogram fixes tlle point 1. In a similar manner it may be shown thatall the other parts of the prism
are fixed by the above conditions. We say, thereforeAll the parts of any prism are determined by the three
planes Jorminng cither of the triedlral angles, and the disposition of those planes; therefore —f two prisms have
the three planes forming a triedral angle in one, equal
respeetivel to the three planes which form a triedral alzgle in the other, and dlisposed in the same order, the
prisms are equal. If the planes arc respectively equal,
but the arrai'uement dif/erent, the prisms will be symmetrical. Of rirght prisms, those which are symmetrical are
also equal.
'  212. The prism ABCDEFGH       (usually designated by
AG) (Jig 131), whose bases are parallelograms, is called Fig.131.
a parallelopiped; it has its opposite faces equal and parallel. Its base ABCD is a parallelogram, therefore AD
is equal and parallel to BC; ABFE is a parallelogram,
consequently AE is equal and parallel to BF; but when
two lines wlich cut each other are parallel to two other
lines which cut each other, the plane determined by the
first two is parallel to the plane determined by the second
two (211);   therefore the plane AEHD, is parallel
to BFGC; and they are evidently equal, their sides and
angles being equal. In the same manner we may show
that the two other opposite faces are equal and parallel.




88


Elements of Geometry.


We say, therefore-In a parallelopiped any two opposite
faces arc equal and parallel; and —A prism contained
by six planes, of which each is parallel to one of the others,
is a parallelopiped.
243. By comparing the plane angles at two opposite
vertices in any parallelopiped, the learner will perceive
that they are equal respectively, being the corresponding
angles in equal parallelograms; and as each is a triedral
angle, the inclination of two faces in the one, must be
equal to tlat of the corresponding faces in the other; but
the plane angles in one are in the inverse order of those
of the other; it will therefore follow, that-In every inclined parallelopiled the olpposite triedral angles are symmetrical.
244. A parallelopiped is called rectangular when each
of its faces is a rectangle. This will be the case, if, at
one of the triedral angles, each of the edges is perpendicular to the plane of the other two. The rectangular parallelopiped of which each of the faces is a square, is
called a cube. It is also called a regular hexacledron, being contained by six equal regular polygons, and having
its diedral angles all equal, and its triedral angles equal.
- 245. The bases of a parallelopiped are parallelograms;
consequently the sum of the squares of their four diagonals, is equal to the sum of all the squares described
upon the eight edges at the bases (184). If through opposite diagonals in the bases, and therefore through opposite
Fig. 131. vertices of the body (fig. 131), two planes are passed, the
two sections thus made will be two parallelograms, four
sides of which are the four diagonals of the bases, and their
other sides the four lateral edges of the body. In these
two sections there are four diagonals, which are evident.
ly the diagonals,(f the parallelopicped; and the sum of
the squares of these tliagonals, must be cqual to the sium
of the squares of the sides (184), that is, to the sum of the
squares of the twelve edges of the parallelopiped. Therefore-In a r ectangular parullelopiped, the square of one
of the diagonals is equal to the suml of the squares of the
three efdges which meet at tthe same tirtex, And-In the
cube, the square of a diagonal is equal to three times
the square of one of the edges.
246. Tlie cube is distinguished among bodies, as the
square among surfaces. All the plane angles in its faces
are right-angles; all its diedral angles are right angles;




Elements of Geometry.            89
each of its triedral angles is composed of three rightangles; -and all its edges are equal. This character of
its angles, and the equality of its sides, give to its form a
remarkable simplicity; on which account it has been
selected as a unit in the measure of bodies.
Of the Volume of Bodies, and its Measure.
247. By the volume of a body, we mean the quantity
of sjace which the body occupies, or the whole amount
of space included by its surface. The volume of a body
is very distinct from its form. Two bodies may be equal
in volume and very different in form. A cubic cistern
may be constructed which shall contain the same quantity of water as a hogshead. If the water be drawn from
the hogshead into tlh cistern, the quantity of water remaining the same, the whole amount of space which the
water occupies will be the same; the volume is not
changed, though the form is very different. We see,
therefore, that two very different geometrical bodies may
be equal in volume. If two bodies are equal in volume,
they are called equivalent bodies.
24S.' Suppose we have two parallelopipeds, AG and
AL, standing upon the same base ABCD, and having
their superior bases in the same plane parallel to ABC Fig.132.
(fig. 132). We observe first, that the whole figure consists of the parallelopiped AG, and the triangular prism
BFK-CGL, and also of the parallelopiped AL, and the
triangular prism AEI-DHM. If, therefore, we can
show that these two triangular prisms are equal, as one of
them added to one of the parallelopipeds, makes the same
body as the other prism added to the other parallelopiped, the two parallelopipeds must be equal in volume.
The triangles AEI, BFK, have their sides AE and
BF equal, being opposite sides of the same parallelogram;
for the same reason Al = BK; as these equal lines are
also parallel the angles EAI and FBK are equal; the
two triangles are therefore equal by the second case; the
faces EC and FD are also equal being opposite faces of
the same parallelopiped (242); EF is equal to IK, being
each equal to AB; if from each of these we take IF, we
have El =FK; EH is also equal and parallel to FG,
therefore the angle IEH is equal to the angle KFG;
8*




90


Elements of Geometry.


and the two parallelograms EM and FL, having two adjacent sides and the contained angle the same, are
equal; we therefore have the three faces containing the
triedral angle whose vertex is at E, respectively equal to
the three faces which contain the corresponding triedral
angle in the other prism; the two prisms are therefore
equal, and the parallelopipeds AG and AL are equivalent.
In a similar manner we can show that tile perallelopiped ABCD-IKLM, is equivalent to ABCD-EFGH
Fig.133. (fig. 133), they being upon the same base ABCD, and
having their superior bases in the same plane. So also
it may be shown that the parallelopiped ABCD-NOPQ,
whose edges are perpendicular to the base is equivalent
to the parallelopiped ABCD-EFGH; therefore the parallelopiped ABCD-IKLM is equivalent to ABCD-NOPQ.
But this parallelopiped AP, may be changed into the
equivalent parallelopiped ABRS-NOTU, whose bases
Fig.134. are rectangles (fig. 134) as the triangular prism BCROPT is equal to the prism ADS-NQU.
It is evident that all the faces of this last parallelopied, are rectangles; and each of these inclined parallelopipeds is equivalent to this rectangular )parallelopiped.
We therefore say —Any parallelopiped may be changed
into an equivalent rectangular parallelopiped, having the
same height and an equivalent base.
249. Any parallelopiped may be divided into two
equivalent triangular prisms by a plane passing through
two opposite lateral edges; for each base of the triangular
prism is half the lase of the parallelopiped, and their
heights are the same. If, therefore, an inclined parallelopiped may be changed into an equivalent rectangular
parallelopiped of the same base and height; the triangular
prisms which compose the inclined parallelopiped, must
be respectively equivalent to the triangular prisms which
compose the right parallelopiped, as in either case each
prism is one half of the parallelopiped. Whence it appears that-Any inclined triangular prism   may be
changed info an equivalent right prisrf of the same height
and an equivalent base.
- 250. Any prism of more than three lateral faces, may
be cut into triangular prisms, by passing planes through
opposite lateral edges. These prisms will have the same
volume when their bases are the same and their height




Elements of Geometry.


91


the same, whether their edges are inclined to the base or
not (249). We say therefore-Any two prisms having
equivalent bases and the same height are equivalent, or
equal in volume.
251. In a comparison of pyramids, we will take the
triangular pyramid, or tetraedron.
Let us suppose that the two pyramids S-ABC, s-a b c,
(fig. 135) have the same height and equivalent bases in Fig.135.
the same plane; let their height be represented by AH,
perpendicular to the plane of their bases. Divide AH into
any number of equal parts, at the points w, x, y, z, and
through these points, suppose planes to be passed parallel
to the plane of the bases; the corresponding sections of
the two pyramids, made by these planes, will beequivalent (234), namely DEF to d ef, GHI to g h i, &c.
Upon the upper side of the triangles ABC, DEF, GHI,
&c. as bases, construct prisms, which will be exterior to
the corresponding segment of the pyramid; and upon
the lower side of d e  g h i, k lm, &c., construct prisms
which will be interior to the corresponding segments. It
is manifest that the sum of all the exterior prisms of the
pyramid S-ABC, is greater than the pyramid; and
that the sum of all the interior prisms in the pyramid
s-a b c is less than the pyramid; therefore the pyramid
S-ABC minus the pyramid s-a b c, must be less than the
difference between the sum of the interior and the sum
of the exterior prisms. But the prisms all have the same
height; and therefore, those upon equivalent bases are
equivalent; consequently the sum of all the interior
prisms in the pyramid s-a b c, is equal to the sum of all
the exterior prisms about the other pyramid, excepting
the lower prism  whose base is ABC.    This prism,
therefore, is the difference between the sum of the exterior prisms about the pyramid S-ABC, and the sum
of the interior prisms in the pyramid s-a b c.
If we suppose the pyramids to be cut by planes only
half the distance asunder before supposed, and prisms
constructed as above; the prism whose base is ABC
will still be the difference between the sums of the two
series of prisms;. but having the same base and a less
height, its volume must be less. It is evident that as we
increase the number of prisms in the series, the volume
of that prism which is equal to the difference of the -two
series, must be continually diminished; whence we may




Elements of Geometry.


continue the process till this difference is as small as we
please; but this difference will always be greater than
the difference of the two pyramids, which must consequently be infinitely small, that is, nothing. We say,
therefore —Two triangular pyramids which have the
same heigh/t and equivalent bases, are equal il volume.
252. As every pyramid of more than three lateral
faces, may be cut into triangular pyramids of the same
height, by planes passing through the summit and the
diagonals of the base, this law of triangular pyramids
will be true of all pyramids composed of triangular pyramids. We say, therefore, that-Pyramids of the same
heitght and equivalent bases, are equal in volume.
253. We have seen that polyedral bodies may be cut
into tetraedrons (or triangular pyramids) by planes passing through them diagonally (237). Suppose we have
Fig.136. the triangular prism  ABC-DEF  (fig. 136).  First,
pass a plane through the vertex E and the edge AC of
the base; it will cut off the tetraedron E-ABC, whose
base and altitude are the same as those of the prism.
Suppose the remainder ACDEF, to be cut through the
vertices C, D, E, it will be divided into two tetraedrons,
one of which, CDEF, may be considered as having C
for its summit, and DEF, one of the bases of the prism,
for its base; this tetraedron has also its base and
height equal to those of the prism. The remaining segment EACD, may be considered as a tetraedron whose
summit is at E, and whose base is ACD; but as the
edge BE is parallel to the opposite face of the prism, the
point B is at the same distance from the plane of the opposite face, as the point F is; we may therefore substitute for the tetraedron EACD, the equivalent tetraedron BACD, having the same base ACD, and the same
height. But in the tetraedron BACD, we may take ABC
for the base, and the point D for the summit; this tetraedron has also the same base and the same height with
the prism. We have, therefore, three equivalent tetraedrons composing this prism. We, therefore say-A tetraedron is equivalent to one third of a triangular prism
of the same base and height..  254. By reviewing the analysis of the last article, it
will be perceived that the body ABC-DEF is equivalent
to three tetraedrons which have ABC for a common base,
and their vertices respectively in the three points D, E,




Elements of Geometry.


93


and F. If, therefore, the body were a truncated prism,
that is, if the plane DEF were inclined to the base, we
should say —A truncated triangular prism is equivalent
to three pyramids whose common base is the base of the
prism, and whose heights arc respectivel i equal to the distances of the opposite vertices from the base.
255. As all pyramids of more than three lateral faces
may be cut into tetraedrons of the same height, the sum
of whose bases will be the base of the pyramid; and as
the prisms constructed upon the bases of these tetraedrons taken together, will constitute a prism of the
same height, constructed upon the base of the pyramid
in question; and as each of the tetraedrons is one third
of the corresponding triangular prism, the sum of all the
tetraedrons will be one third of the sum of all the triangular prisms. We therefore say (249)-Any pyramid is
one third of a prism of the same base and height, or one
third of a rectang-ular parallelolipcd of the same height
and air equivalent base.
256. Let us now compare together bodies of different
heights. Suppose the two rectangular parallelopipeds
AG and AL (fg. 137) to have the same base ABCD; Fig.137.
and AE and AI, for their heights. Supppose these
heights to be to each other in any definite ratio, for instance in the ratio of 9 to 5. Divide AE into 9 equal parts,
AI will contain 5 of them; through these points of division pass planes parallel to the base; they will divide
the parallelopiped AG into 9 rectangular parallelopipeds
of equal bases and equal heights; and as the parallelopiped AL contains 5 of these, the two parallelopipeds
will be to each other as 9 to 5; that is, as AE to AI,
We should have the same result if their heights were in
any other commensurable ratio; the bodies would be in
the ratio of their heights.
Suppose their heights to be incommensurable; we may
divide one of the lines into equal parts so small that the
difference between one of these parts and the remainder
of the division of the other line by such part, is less than
any given difference; we thus find an approximate ratio
of the two bodies, which is indefinitely near to the ratio
of the two lines; but which ratio we never pass, on which
side soever we commence the approximation; we therefore infer that the ratio of the heights is, in this case also,
the ratio of the two bodies. We therefore say-Two




94


Elements of Geometry.


rectangular parallelopipeds of the same base, are to each
other in their ratio of the heights.
257. Let us now compare the two rectangular paralFig.138. lelopipeds AG and IP (Jig. 138) of different bases and
different heights. Upon the edge IN take II'   AE,
and through this point suppose a plane section I'L' parallel to the base; and upon BC take BC' = IM, and
through the point C' suppose the plane section C'III parallel to the face AF. We shall then have the parallelopipeds AG' and IL' whose bases are the equal rectangles
AH' and IM/, and whose heights are the edges AB and
vol. AG'    AB
IK; we shall therefore have (256) vol. IL~~   IK;
and by comparing the parallelopipeds AG and AG', considered as having for their common base the rectangle
vol. AG    AD
AF, we obtainvo. AG  - AD' By multiplying these
vol. AG'   AD/
two proportions in order, omitting the factor AG' common to the numerator and denominator, and substituting
for AD' its equal IM, we make a comparison of the parvol. AG   AB X AD
allelopeds AG and IL'; this gives ol.IL' G   K  X AM'
Comparing now the parallelopipeds IL' and IP, having the same base IKLM, we obtain the proportion
vol. IL'  II'
vol IP   -j I-N  Multiplying together these two equations, omitting the common factor IL' in the two terms
of the first ratio, and substituting in the second AE for
vol. AG
its equal II', we obtain the following: vol. IPAB X AD X AE
1K X   - X IN  We therefore say-Any two rectanIK X IM X IN '
gular parallelopipeds are to each other as the products of
the three edges which meet at one of the vertices; that,is, as the products of their three dimensions..-  258. To illustrate this result, we take as the unit of
the measure of volume, the cube whose side or edge is a
unit in the measure of lines. Let ag be this cube (fig.
Fig.139. 139). If we compare this small cube whose side is 1,
with the parallelopiped AG, we shall have
vol. AG     AB >< AD X AE; that is, the par
vol, ua b         1




Elements of Geometry.


95


allelopiped AG contains the cube whose side is a b as Fig.139
many times as the product of the lines AB, AD, AE, referred to the common measure a b, contains unity. And
this is what is meant, when it is said that-The measure of volume in a rectangular parallelopiped, is the product of its three contiguous edges.
259. We observe that the product AB X AD expresses
the number of squares, whose side is unity [a b], contained in the base AC; that is, expresses the area of the
base; we say therefore, that-The volume of a rectangular parallelopiped has for its measure the product of
its base by its height.
260. As any parallelopiped may be changed into a
rectangular parallelopiped of the same height and an
equivalent base (248), it follows that-Any parallelopiped hasfor the measure of its volume, the product of
its base by its height.
261. As a rectangular parallelopiped may be divided
into two equivalent triangular prisms of the same height
with the parallelopiped, whose bases are each half
the base of the parallelopiped (249); it follows, thatThe volume of a triangular prism has for its measure the
product of its base by its height.
262. As every prism of more than three lateral faces
may be considered as composed of triangular prisms of
the same height with itself, the sum of whose bases
make the base of the prism (250), we say-The volume of every prism has for its measure the product of its
base by its heYighi'. And consequently (255)-The volume
of any lpyramid has for its measure, the product of one
third of the base by the height.
Remark. To obtain the volume of any frustum of a
pyramid, find the volume of the entire pyramid, and subtract from it the volume of the partial pyramid cut off.
263. Let P represent any prism, B its base and Hits
height; we shall have the formula P =B X H; let P'
denote a similar prism, B' its base, and H' its height;
this gives P' =  B' X H'. If we compare these, we
have P-   B — X.   But in similar polyedrons, hol -DI X rlT
mologous faces are proportional to the squares of their
homologous measures (237); the bases are then as the




96


Elements of Geometry.


B     fZ2
squares of their heights; this gives B  =.it; we may
therefore substitute for  in the proportion above, the
12                      P     HIa X H
equal ratio  72; this will give us p/ i,    X H=
H3
-H; that is —Similar prisms are to each othler as the
third powers of their heights, or as the cubes of their homologous measures generally.
264. Letp represent a pyramid, b its base and h its
height; p' another similar pyramid, b' its base, and h'
its height; for the first of these pyramids, we have the
expression p =   b X h;   and for the second, p'
b' X Ah.  By comparing these, we obtain the proporP     I b X It
tion P =      XXh; and by a process of reasoning simp     h y
ilar to the above, we obtain ---  3; that is-Similar
P'    h13
pyramids are to each other as the cubes of their heights;
or, as the third powers of their homologous dimensions
generally.
265. As all polyedral bodies may be considered as
composed of pyramids; we have from the above rule, a
method of ascertaining the measure of their volume.
And as similar polyedral bodies are composed of the
same number of similar tetraedrons or triangular pyramids, we infer that-Similar polyedrons are to each
other as the cubes described upon their homologous measures.




Elements of Geometry.


97


PART SECOND.
SECTION II.-Of the Round Bodies.
266. ROUND bodies are those which are produced by
the revolution of any plane figure about a straight line;
they are called bodies of revolution. Those usually discussed in the elementary treatises, are the right cone,
the ri ght cylinder and the sphere.
The right cone is generated by the revolution of a
right-angled triangle about one of the sides containing
the right-angle, as the triangle SCA (fig. 140) about the Fig.140.
side SC. The hypothenuse SA by this motion generates
the conical suface.  'Jach point in the hypothenuse, describes the circumference of a circle whose centre is in
the line SC; the circle generated by the line CA is called the base of the cone. The line CS, upon which the
generating triangle turns, is called the axis of the cone.
The point S is called the summit or apex of the cone.
It is evident that a plane passing through the axis will
cut the conical surface in two straight lines. A plane
perpendicular to the axis, will have the circumference of
a circle for its section of the conical surface.
267. The cone just described is the right cone whose
base is a circle. The inclined cone with a circular base
(fig. 141) may be considered as generated by the motion Fig.141.
of a straight line, as AS, one point of which, as S, being
fixed, the other part being carried round the circumference of a circle, as ADB, situated in a plane which
does not pass through S. The straight line CS joining
the apex with the centre of the base, is here also called
the axis. This cone is also called an oblique or scalene
cone.
268. In the right cone (fig. 140) the similar triangles Fig.140.
AC     CS
ACS, A'C'S, which give the proportion AC  - CS 
AS        h
7, show that the radii of the circles ADB, A'D'B', are
proportional to their distances from the apex of the cone;
9




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Elements oJ Geometry.


Fig.140. but the circumferences of circles being as their radii
(148), and their areas being as the squares of their radii
circ. ADB     AC     CS     AS
192), we have  c            AD-B -A'C' =C  A'S' and
circ. ADBCS- 
area ADB   -   (AC)2    (CS)2    (A S)2 
area A'D'B'    (A'C')2- (C'S)2 (- A'S)2'
In figure 141, supposing A'D'B' to be parallel to the base,
circ. ADB     AC     CS     AS     OS
we have also           -      -
circ. A'DB' -A'C'    C'S    A'S    O'S'
area ADB        (AC)2     (CS)2   _  (AS)2
area A'D'B' -   (A'C')2    (C'S)2  -  (A'S)2
(OS)2. We see, therefore, that it is with cones as with
pyramids (232).
(1). If any cone be cut by a plane parallel to its base,
the smaller cone cut of is similar to the whole.
(2). Similar cones have their homologous dimensions
proportional, and their bases proportional to the squares
of their homologous lines; and
(3). In any cone, parallel sections are to each other as
the squares of their distances from the apex.
269. Remark. When we have the dimensions of any
truncated cone whose bases are parallel, as ADB, A'D'B',
we can find by a process analogous to that of article 233,
the dimensions of the entire cone, and also of the superior cone, S-A'D'B'.
Fig.142.  270. If we inscribe in the base of the cone (fig. 142)
a regular polygon, and circumscribe also about, the base
a similar polygon; by drawing straight lines from the
apex of the cone to the vertices of these polygons, we
shall have a pyramid inscribed in the cone, and also a
pyramid circumscribed about the cone. The figure, not
to make confusion, exhibits but one of the lateral faces
of each pyramid. The circumscribed pyramid exceeds
the inscribed pyramid by a certain magnitude; but by
increasing the number of sides in the polygons which
are their bases, we increase the number of lateral faces
in the pyramids. The inscribed pyramid is increased,
and the circumscribed pyramid is diminished by this
process, which may be continued till the difference between the two pyramids is less than any assignable magnitude. By this process, also, the lateral surfaces of the
two pyramids are made to approach indefinitely near to




Elements of Geometry.


99


each other. The lateral surface of the inscribed pyra- Fig.142.
mid car} never be greater than the conical surface, and
the lateral or convex surface of the circumscribed pyramid can never be less than the conical surface. The
difference between the convex surface of the circumscribed pyramid and the conical surface. must always
be less than the difference between the two pyramidal surfaces which we suppose less than any assignable magnitude.
271. The convex surface of tlhe pyramid circumscribed about a right cone, is composed of a certain number
of triangles, whici, on account of the regularity of the
base, and the summit being in a straiglit line perpendicular to the middle of the base, are isosceles and equal.
The common height of all these triangles is SG the side
of the cone; one half the product of SG multiplied by
the sum of their bases, will give their area. The sum of
the bases of these triangles is the perimeter of the polygon; therefore, if we denote this perimeter by P, we
shall have for the area of the pyramid  P X SG. This
perimeter exceeds the circumference of the circle, and
therefore this product exceeds the product 4circ. X SG
by an indefinitely small magnitude which we designate
by d; and denoting the area of the cone by A, and
the excess of the pyramidal surface over the conical
surface by m; we shall have the equation A +m =
-circ. X (SG) 4-d. The area of the inscribed pyramid, calling p the perimeter of its base, will be ~p X sg;
and denoting the excess of the conical surface over the lateral surface of this pyramid by nm'; and also denoting by
d the excess of the product  circ. X (SG) over the product Ip X (sg), which may be less than any assignable
quantity, as by increasing the sides of the polygon, the
perimeter becomes sensibly confounded with the circumference, and sg, with SI;, the side of the cone: we shall
have A - m' -= cirC. X (SG() - d; we see, therefore,
that if any magnitude, however small, be added to the
area of the cone, something must be added to the product I circ. X (SG) to make the equation; and if any
magnitude, however small, be subtracted from this area,
a corresponding quantity must be subtracted from this
product to balance the equation; fromn which it follows
that circ. X (SG) expresses neither more nor less than
the area of the conical surface. Wherefore-The con.




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Elements of Geometry.


Fig.142. vex surface of a right cone, is measured by half the product of the circunference of the base multiplied by the
side.
272. To find the area a truncated cone.   Subtract
from the area of the entire cone the area of the smaller cone cut off. Calling A' the area of the frustum
Fig.143. ADB-A'D'B' (fig. J43;, A the area of the entire cone,
and a the area of the less, we shall have A' - A - a
-   (circ. OA) X (SB) -     (circ. O'A') X (SB'), or A'
(circ'. OA) X (SB) —     (circ. OA) X (SB') +  (circ.
O'A') X (SB)-    '(circ. O'A') X (SB)     (circ.OA)
X (SB - SB') + I (circ. O'A') X (SB-SB') - I (circ.
OA + circ. O'A') X (BB'). Hence- The area ofthe frustum of a right cone has for its measure the product of its si(e
multiplied by half the sum of the circumferences of its bases.
Or supposing a section A'"D'B" E"', at equal distances
from the two bases of the frustum; the similar triangles
SO      SO"      SO'
SOA, SO"A", SO'A', give OA       -O'A";
SO    SO"       SO" -- SO'
whence (82)    A-   -A      O"        A'' that is,
0"0             0e0"
b-A-_ O"A" -= O        '-O7A. These numerators are
OA     ()A        O"A" - O'A'
equal by construction; the denominators are therefore
equal; and OA is as much greater than O"A", as 0)"A"l
is greater than O'A'; but these radii are as their circumferences. The circumference A"D"B"E"' is therefore an arithmetical mean between the other two; and is
consequently equal to half their sum. Hence-The area
of the frustum of a cone, has for its measure the product
of its side multiplied by the circumference of the plane
section made at equal distances from its bases.
Remark. By substituting the apex of the cone for the
superior base, the first formula gives the area of the
entire cone.
273. We have seen (270) that two pyramids may be
constructed the one inscribed in a cone and the other
circumscribed about it, such that the difference of their
volumes shall be less than any assignable magnitude.
The cone, therefore, may be considered s s a pyranid of
an infinite number of lateral faces; it will therefore have
for the measure of its volume, one third of the product of




Elements of Geometry.


101


its height by its base. We shall however give this a
different proof.
The common height of these pyramids will be SO the
height of the cone (jig. 142). If we call B the base of the Fig.142.
cone, and d the difference between the base of the cone
and the base of the circumscribed pyramid, and d' the difference between the base of the cone and the base of the
inscribed pyramid, V the volume of the cone, m the excess of the volume of the circunlscribed pyramid over
the volume of tie cone, and rm' the excess of the volume
of the cone over the volume of the inscribed pyramid.
This will give V+ m - - (SO) X (B +- d); V- m'
- (SO) x (B - d'). That is, if any thing be added to
the volume of th i ne, somethillg must be added to the
product (SO) X,- to balance the equation; and if any
magnitude however small be subtracted from the volume
of the cone, a correspondilg quantity must be subtracted
front this product, to preserve the equation; from which
it follows that (SO) X 1, expresses neither more nor
less than the volume of the cone; and we say —The volume of a cone has for its measure, one third of the product of the base multiplied by the height.
174. If the radius of tie base of a cone be designated
by R, and the height of the cone by H; we shall have
for the base of the cone 7r. R2. Denoting its volume by
V, we have V = I 7. R2 X iI.
JrY7ar7'k. To find the volume of a truncated cone, subtract the volume of the less from the volume of the entire
cone.
275. If we conceive the rectangle AC C'A' (fig. 144) Fig.144.
to revolve -about one of its sides CC', it will generate the
body ADB-A'D'B', called a right cylinder.    The
straight line AA' will, by this revolution, generate the
cylindrical su face.
Any point, as A", in this straight line, will describe
the circumference of a circle, as A"D"B", equal and
parallel to the circle ADB called the base of the cylinder and generated by AC the base of the rectangle.
For A"C" perpendicular to CC', and AC also perpendicular to CC', will, by this motion, describe plane circles
perpendicular to CC' (199); they are therefore parallel;
A"C" being equal to AC, and they being the radii of
these circles, the circles are e4ial. From which it follows that —Any section of a right cylinder parallel to its
9*




102


Elements of Geometry.


Fig.144. base, is a circle equal to its base.  Any plane section
through the axis or parallel to the axis, is a rectangle.
The straight line A'C', also perpendicular to CC', generates the circle A'D'B', also equal and parallel to ADB;
this we call the superior base. The line CC' about
which the rectangle revolves, is called the axis of the
cylinder, and is perpendicular to the bases.
276.   If we conceive a circle, as ADB, around
Fig.145. which a straight line, as AA' (fig. 145) moves always
parallel to itself, but oblique to the plane of the circle;
this line will generate the surface of an oblique or inclined cylinder.  Or we may conceive this cylinder
to be generated by the motion of the circle ADB parallel to itself, along the straight line AA', every point in
the circle describing a straight line by this motion. The
circumference of the circle will generate the cylindrical
surface; the centre of the circle will generate the axis
CC'; and as the generating circle is always parallel to
itself, every section of this cylinder parallel to the base,
will be a circle equal to the base. Any section of this
cylinder through the axis or parallel to the axis, is a parallelogram. Sections parallel to the axis and having the
same inclination to the base as the axis has, are rectangles.
277. We have here given two methods of generating
the inclined cylinder. In the first, we call the straight
line AA' which generated the cylindrical surface, the
generatrix, and the circle by whose circumference its
motion was guided, we call the (irectrix. In the latter
method we call the circle by which the cylinder was
generated, the generatrix; and the straight line by
which its motion was directed, the directrix. If in either
of these two cases the straight line had been perpendicular to the plane of the circle, the body generated would
have been a right cylinder.*
278. A slight inspection will satisfy us that the cyl* These methods of defining a cylinder are preferable
to the description given in Art. 274; for they will apply
to all cylinders, if, instead of the term circle, we substitute a plane surface embraced by a curve line; as it is
not an essential property of a cylinder, that any section
of it should be a circle. Nor is it an essential property
of a cone, that any section of it should be a circle.




Elements of Geometry.


103


inder belongs to the family of prisms; andl therefore to
find the area and volume of a cylinder we should proceed as we would with the prism in the same cases.
That is, the area of any right cylinder lhas for its
measure the height of the cylinder multiplied by the periphery of the base; and the volume of,ni/. cylinder is
measured by tlhe the product of its base multiplied by its
height. This may be readily demonstrated.
279. [fwe inscribe a regular polygon il the circular
base of the right cylinder (/fr. 146) and circumscribe about Fig.146.
the circle a similar polygon, and make these two polygons the bases of two prismns the oiie inscribed in the
cylinder and the other circumscribed about it; as the
number of sides of these polygons may be so great that
the difference between the polygons shall be as little as
we please, it is manifest that we may erect upon the
polygons two prismls, the one inscribed in the cylinder
and the other circumscribed about it, wlhose ditierence
shall be less than any assignable magnitude.
The area of the cylinder is less than the area of the
circumscribed prism by an indefinitely small magrnitude
m, and greater than that of the inscribed prism, by an
indefinitely small quantity mi'. The perimeter of tlhe base
of the circumscribed prism exceeds tle circumference
of the base of tlhe cylinder, by a quantity less than any assignable magnitude, which excess we denolede by d. This
circumference exceeds the perirmeter of the base of the
inscribed prism by an indefinitely small excess which we
designate by d'. Denoting the area of the cylinder by
A andl the heilght by I, we shall have for the area of the
exterior prisnm A + -  nm- II X (circ. +- d); and tor the
area of the interior prism, A - m' = II X ((ie. - d);
that is, if any rmagnitude, however small, be added
to the area of the cylinder, something must be added
to the product of its height multiplied by the circumference of the base, to express this area; and if any magnitude however small be subtracted from the area of this
cylindrical surfiace, a corresponding magnitude mnust be
subtracted from the above mentioned product, to express
this area; that is-The product of the height of the right
cylinder multiplied by the circumnerence of the base, (being the measure of an area neither greater nor less) is
the measure of the area of the cylindrical surface.
280. The learner is requested to give an analogous




104


Elements of Geometry.


proof that-The volumne of a cylinder has for its measure
the product of the area o 'its base multi)lied by its height.
Remark. It is manifc'st that this last proposition applies as well to oblique as to right cyiillders, there being
nothing in the reasoning fwhich supposes the axis of the
cylinder perpendicular to the base.
Fig.147.  2S1. Iftlhe semicircle ACB (fig. 147) revolve about
its diameter AB, it will generate a sphere; the semicircurnfmlrence generates the spherical surface. Every
point in this semi-circumference, as D, generates the
circumference of a circle whose centre is in thle diameter AB (here called the axi:) andl whose l)lane is perpendicular to ttlis axis. 'iTle extremities A, B, of tile axis,
are, in reference to this revolution, called poles. They
are particularly the poles of all tile circles whose circumferences are generated by the several points in the arc
ACB of the revolving semicircle.
As every point of tile arc of the semicircle is at the
same distance from its centre, and as tilis centre does not
move during tile process of revolution; it follows that
every point of tile sphlerical surface is equally distant
from the point 0, the centre of the generating semicircle
and also the centre of the sphere.  It is also evident that
any radius of the semicircle, in any part of its revolution,
is also radius of the sphere; and that these radii are all
equal.
282. If we suppose tile sphere to be cut by a plane
passing through its centre, it is perfectly manifest that
the section will be a circle, wlose radius will be the radius of tle sphere.
But suppose the plane to pass through the spliere on
one side of the centre; and let the section be DGFH;
from () the centre of the sl)here, draw OE perpendicular
to this section, and draw radii to the several points in the
periphery of this section; they will be equal oblique lines,
and will therefore meet tile plane DGFI1, at equal distances fiom the perpendicular OE (206). Every point
in this periphery is therefore at the same distance from
E; the section is consequently a circle whose centre is
E. And as this reasoning applies to any plane section
on either side of the centre, we infer that —L'ey plane
section of a sphere is a circle; and also that —Radius




Elements of Geometry.


105


rprpendicular to any plane section of the sphere, passes Fig.147.
through its centre.
2-3. Remark. As DE is always one side of a rightangled triangle whose hypothenuse is radius of the
sphere, DE rmust be less th;an radius; the plane section
whose radius is radius of the sphere, is called a great
circle; other plane sections are called less circles.
284. As two great circles are planes, their intersection
must be a straiglit line; and as each passes through the
centre of the sphere, the line of intersection must pass
through the centre of each of them, that is —Two great
circles muttually bisect each other.
285. Any portion of a spherical surface embraced by
the arcs of three great circles, is called a spherical triangle. As two circumferences of great circles bisect each
other, if each anigle of a spherical triangle is salient, each
of the arcs must be less than a semni-circutnference.
286. Let ('M  be a spherical triang(le, and draw the
radii OC, OM, 01; it is evident that they determine a
triedral angle whose vertex is at 0, and whose plane angles are measured by the arcs CI, CM, IM. And as the
sum of any two of these plane angles is greater than the
third (220) the sum of the arcs whiich measure them must
be greater tlian the third arc (113), that is — 'lhe sum of
any two siles of a spherical trianglle, is greater than the
third. Whence we inlfer that-The shortest way from one
point to another, on the suiface (f a sphere, is in the arc of
a great circle passing through these points. For if we suppose it shorter to pass through any point out of tills arc,
that point and the two l)oints proposed may be made the
vertices of a spherical triangle, one side of whicl will be
the arc first proposed and less than the sum of the other
two; the inference is therefore manifestly true.
287. Suppose the two semi-circumflterences ACB and
AIB which meet each other in A and B, to be cut by the
arc IM of another great circle; we shall have MI < MB
+ BI; and consequently AM + AI -+ IM <      AMB +
AIB; that is —The sum of the three si(les of any spherical triangle, is less than the circumference of a great
circle.
Remark. We might have deduced this truth from the
limit of the nagnitude of the plane angles which form
the triedral angle at the centre of the sphere (222); this
sum being less than four right angles, the sum of the




106


Elements of Geometry.


Fig.147. arcs which measure these angles. must be less than an
entire circumtference ('c7).
283. As any straight line drawn through any point in
a curve line (in the plane of the curve) and oblique to a
right-line tangent at that point, must cut the curve line
(111), the tan;gent may be considered as representing tle
direction of the curve at that point. It follows from this,
that-The a(nle which/ any two curves make with each
other at their intersection is neither more nor less than the
angle which their respective tangents at this point make
with C(Ich other.
The angle CIM of the spherical triangle will therefore be tile same as the angle which the tangents to the
two arcs CI, MI, at the point I, make with each other.
These tangents are perpendicular to radius 10 (111),
and are in the planes of the two circles respectively;
but being in these planes and perpendicular to their intersection, their angle is the inclination of these planes,
or the diedral angle made by the two faces of the triedral angle whose edge is 1O.    We therefore say-A
spherical triangle determines a triedral angle whose vertex is at the centre of the sphere, whose plane angles are
measured by the sides of the spherical triangle, and whose
diedral angles are respectively the angles of the spherical
triangle.
2S9. It will be readily seen that as the sum of the
plane angles in a triedral angle, approaches to four rightangles, the value of each of the diedral angles approaches
to two right-angles; and their sum of course approaches
to six right-angles; and tile less the plane angles of the
triedral angle the nearer does tile sum of the diedral angles approach to two right-angles. Therefore —The sum
of the three angles of a spherical triangle is not constant;
and the limits of this variation are two right-ungles and
six right-angles.
Remark. The other properties of spherical triangles
analogous to those of plane triangles, discussed in articles 38 to 52 inclusive, the learner is requested to investigate by processes analogous to those adopted in the
articles referred to.
29)0. If through the point E the centre of the less circle DGFII, a straight line be drawn perpendicular to this
circle, it will pass through the centre of the sphere (282);
it is called the axis of this circle, and is also the axis of




Elements of Geometry.


107


every circle of the sphere, parallel to this. It will meet Fig.147.
the spherical surface in two points which are the poles
of this circle, and also the poles of all parallel circles in the sphere.  Each of these poles is equally
distant from every point in the circumference of this circle (204); tlat is, the chords drawn fiomn tle pole of
any circle to the several points in its circumrference, are
all equall; consequently the arcs are equal. This enables us to describe the circumference of a circle upon
the surface ofa sphere from one of its poles, without finding the centre of the circle.
291. A plane is tanLgent to a sphere when it touches it
in only one point, as C (fig. 148). This point is called Fig.148.
the point of contact, and as every other point of this
plane is without the sphere, it must be at a greater distance from its centre (281). Therefore a straight line
drawn from the centre of the sphere to the point of contact, must be perpendicular to the tangent plane (2o)5).
Hence-A planle tanglent to a sphere is perpendicular to
radius at the point of contact.
292. Let us inscribe in any arc of a semi-circle (ig. 
149) a portion of a regular polygon, and circumscribe Fig.149.
about the same arc the corresponding portion of a regular polygon of the same number of sides. The area generated by the portion abcd of the inscribed polygon, in
a revolution of the figure about the diameter ap, is composed of several conical surfaces generated by the sides.
The side a b generates a complete conical surface; the
others, surfaces of truncated cones. The area of one of
these surfaces, that generated by c d for instance, is
obtained by drawing from I the middle of c d, I q perpendicular to a O; it is expressed by (c d) X circ. I q, (272).
This expression may be changed to one more general;
for this purpose, draw c r perpendicular to dO, and
draw also O1. The triangles d c r, q 1O, having their
sides perpendicular to each other, are equiangular and
therefore similar, and give the proportion -  ' --  But
cr 1q
as c r is equal to f 0, and as the circumferences of circles are to each other as their radii, we havef Ocirc. O
cire. 1; and, multiplying by the denominators, we obtain
circ. I q




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Elements of Geometry.


Fig.149. (c d) X circ. q - (f O) X circ. 10; conseqlently the
area of the conical surface described by c d, may have
for its expression (f O) X circ. I 0; that is, the product
of the height multiplied by the circumlference of the circle inscribed in the polygon of which this line is a side.
As there is nothing in the reasoning which fixes the
part of the arc in which this line c d is situated, it applies
equally to every side of the polygon; and consequently
the sum of the areas of the surfaces described by the several sides of the polygon between a and (d, will be the sum
of tleir heights multiplied by the common factor, the
circumference of the inscribed circle; that is, (a O) X
circ. 10.
So also the area of the body generated by the revolution of the portion of the circunscribed polycon. will be
(AO) X circ. OL. If we had continuel the polygons
quite through the semi-circumference (the arc (a b being
supposed a measure of the semi-circumference), thle surface generated by the larger polygon would be equal to
the height of the polygon multiplied by the circumlference of the circle; and the area of the surface generated by the less polygon, would be equal to its height
multiplied by the circumference of the inscribed circle.
These polygons, by having the number of their sides
increased, may be made to approach each other so nearly that the difference between the surfaces generated by
them respectively, shall be as little as we please. But as
these surfaces approach to an equality, the products expressing them approach also to an equality. The height
of the inscribed polygon is constantly equal to tlie diameter of the revolving semicircle; the other factor, the
circumference of the inscribed circle, continually increases with the increasing number of the sides of the
polygon; and the limit of this increase is the circumference of the circle in which the polygon is inscribed.
The limit, therefore, of increase for the product expressing the area of the surface described by the interior polygon, is that of the diameter multiplied by the circumference of the circle.
With respect to the other product, the height of the
greater polygon is diminished continually as the number of sides increase; and the limit of this diminution is the diameter of the circle; the other factor remains constantly the circumference of the circle; so




Elements of Geometry.


109


so that the limit of decrease for the product express- Fig.149.
ing the area of the surface generated by the exterior polygon, is that of the diameter multiplied by
the circumference of the circle.  But the area of the
surface generated by the semicircular arc, is the
limit to which the areas, generated by the polygons,
approach as the number of sides in the polygons are
increased. It therefore follows, that the diameter multiplied by the circumference expresses an area neither
greater nor less than the area generated by the semicircumference in a revolution about its diameter. This
surface is the surface of a sphere (281), which we designate by S. We therefore say —The area of a sphere
has for its measure, the product of its diameter multiplied
by the circumference of a great circle: That is, S  2 R
X circ.
293. Remark 1st. The area of a great circle has for its
measure, half the radius multiplied by the circumference,
or I R X circ. =. R2. Therefore - The are- of a sphere
is equivalent to that of foulr great circles of the sphere.
And we have S -- 4 wr. R2.
294. Remark 2d. It follows from  the above reasoning, that any portions of the same spherical surface comprehended between parallel planes, and called spherical
zones, are to each other as the perpendicular distances of
the planes by which they are determined; whether these
planes are secants or secants and tangents.
295. The portion of the spherical surface comprehended between the two semi-circumferences ALB, AIB,
(fig. 147) is called a lunary surface, and the angle IAL, Fig.147.
the lunary angle. Suppose the angle IALof this lunary
surface to be commensurable with four right-angles,
(that is, the sum of all the angles made by arcs diverging from the point A); and suppose that their common
measure is contained forty-eight times in four rightangles, and ten times in the angle IAL. If we divide
the angle IAL into ten equal parts, and the remaining
angular space about the point A, into parts equal to 1
of IAL; and produce the arcs till they meet in the point
B; the entire spherical surface will be divided into forty
eight equal lunary surfaces, of which the lunary surface
BIALB contains ten; therefore this lunary surface is to
the entire surface of the sphere, as ten is to forty-eight
or as the lunary angle is to four right-angles.
10




110


Elements of Geometry.


Fg.147.  If the lunary angle IAL is incommensurable with four
riglt-angles, we can take a measure of four right-angles
so small that the remainder after dividing the angle IAL
by this measure, is less than any assignable angle, which
remainder we designate by d, and the lunary surface answering to this angle by mn; the lunary angle minus this
remainder, IAL - d, is commensurable with four rightangles; and they therefore gives us the proportion
spher. surf.  4 right-angles
lun. su-f. - ni-   AL -d. Suppose that we ap
ply this small measure of four right-angles, once more
than it is contained in the lunary angle, so that the angular space occupied by this measure shall exceed the
lunary angle by an indefinitely small excess, which excess we denote by d', and the lunary surface answering
to this excess by i'. The angular space thus occupied
by this small measure, is commensurable with four rightangles, and we shall consequently have the proportion
spher. surf.   4 right-ancgles
lun. s.   --- -IL         We see therefore that,
lun.surf. - -m    IAL -{- d
if any thing be added to the lunary surface, a corresponding quantity must be added to the lunary angle to
balance the proportion; and if any thing be subtracted
from the lunary surface, a corresponding quantity must
be subtracted from the lunary angle to balance the proportion; that is, the sum of four right-angles is to the lunary angle as the entire spherical surface is to a magnitude neither greater nor less than the lunary surface.
And we say —The area of a Ilnary surface is to!f'e area
of a sphere, as the lunary angle is to four right-ingles.
Representing the lunary surface whose angle is IAL
by Lun.A, calling S the area of the spherical surface,
and denoting a right-angle by D;     we shall have
Lun.A -       S X A
4D'
296. Remark. If two great circles cut each other at
right-angles, they will divide the surface of the sphere
into four lunary surfaces whose angles are all rightangles. And a third great circle perpendicular to the common intersection of the first two circles, will divide each
of these lunary surfaces into two triangles, each of whose
angles is a right-angle.  The surface of the sphere,
therefore, is composed of eight spherical triangles whose
angles are all right-angles.




Elements of Geometry.


111


297. It will be perceived that the three great circles Fig.147.
ACBL, CILK and MIFK, which form the spherical triangle CIM, divide the spherical surface into eight triangles, of which CKM and FIL are symmnetrical, which is
readily seen by observing that the triedral angles at the
centre, to which they correspond, are equal in all their
parts. This equality of the parts of the triedral angles,
shows at once the equality of the parts of the corresponding spherical triangles. But this equality of the parts of
one spherical triangle, to the parts of another, does not,
as in plane triangles, prove the equalit/ of the triangles,
unless the equal parts are arranged in the same order;
their parts being arranged in the contrary order, we cannot turn one of them over as we did the plane triangle (39) to make the two coincide; for placing them so
that their vertices may coincide, the two triangles would
be on opposite sides of the plane passing through the
three points of these vertices; they are therefore symmetrical.
298. Let ABC, DEF, (fig. 150) be two symmetrical Fig.150.
triangles. Let P be the nearest pole of the less circle
whose circumference passes through the vertices A, B,
and C; and let Qf be the nearest pole of the less circle
whose circumference passes through the three vertices
of the other triangle. If we drawv the equal arcs PA,
PB, PC, and also the equal arcs QD, QE, QF; it is
evident that the three last arcs will be equal to the three
first; for the relative position of the points D, E, F, being symmetrical with that of the three points A, B, C,
the curvature of the circumference passing through the
first three points, must be the same with the curvature
determined by the other three points (136); the circles
are therefore equal, and consequently at equal distances
from their poles; and the arcs PA, PB, PC, QD, QE,
CF, are all equal, We have then three isosceles spherical
triangles PAB, PBC, PAC, equal respectively to tile three
spherical triangles QDE, QEF, Q)DF; the quadrilateral
PABC, composed of the two triangles PAB, PBC, must
therefore be equivalent to the quadrilateral QDEF, composed of the two equal triangles Q.)E, QEF; if then we
take from the one quadrilateral the triangle PAC, and
from the other, the equal triangle QDF, the remaining
triangle ABC of the one will be equivalent to the remain



112


Elements of Geometry.


Fig.150. ing triangle DEF of the other; and we say-Two symmetrical spherical triangles are equal in area.
Fig. 147.  299. The two opposite triangles CMK, LFI, (fig. 147)
being symmetrical, are equal in area; if we add to each of
these equal areas the triangle CMI, we shall have CMK
+ CMI =_ CMI + LFI; but the triangles CMK, CMI,
constitute the lunary surface whose angle is CIM; the
sum of two triangles CMI, LFI, in the nearer hemisphere,
is therefore equivalent to this lunary surface. We have
therefore this general rule:-If two semi-circumferences
cut each other in a hemisphere, the sum of two opposite
triangles will be equal to a linary surface whose angle
is equal to the angle madle by the semi-circumferences.
Fig.1.51.  300. Let ABC (fig. 151) be a spherical triangle, and
DEFGHI the circumference of a great circle which does
not cut this triangle. Produce the sides of the triangle both
ways till they meet this circumference; and we shall
have IBD + GBF = Lun.B, HAG        - DAE =Lun.A,
FCE + HCI = Lun. C. It is manifest that the sum of
these six triangles is equal to the hemisphere plus twice
the triangle ABC, that is, equal to 2 ABC + Lun. 2 D;
we shall therefore  have 2 ABC + Lun. 2 D = Lun.
(A -+B-   C), or ABC = -Lutn.(A++4-       C-2 D).
We say therefore-The area of a spherical tri(ngle is
equivalent to hclf the lunary surface whose angle is equal
to the sum of the three angles of the triangle minus two
right-angles.
Suppose a, b, c, to designate the arcs which measure
the angles A, B, C, of any spherical triangle; and let C
denote the circumference of a great circle, and R  the
radius. We shall have, for the area of the spherical triangle, R X (a + b - c    C).
301. To ascertain the volume of the body generated
by the revolution of the polygonal sector a b c d  (fig.
Fig.149. 149), it will be necessary to examine the body generated
Fig.152. by the isosceles triangle ABO (fig. 152) revolving about
the straight line OD as an axis. Produce AB till it
meet the axis in D; and from the points A, B, draw
AM and BN, perpendicular to OD.
The volume of the body generated by the triangle
AOD, will have for its measure I nr X (AM)2 X OD,
(174). The body generated by the triangle OBD, has
for its measure 3 n (BN)2 X OD; therefore the difference of these bodies, or the body generated by ABO,




Elements of Geometry.


113


will have for its measure  r r. [(AM2) - (BN2)] X OD. Fig.152.
This expression may take another form. From  I, the
middle of AB, draw IK lerpendicular to OD, and
through B draw   BP parallel to OD; we shall have
AM + BN =2 IK, and AM - BN = AP;             consequently (AM + BN) X     (AM -    BN) =   (AM)2 -
(BN)2, (176), = 2 IK X AP. The volume of the body
generated by ABO, may therefore be expressed by
2  X IK X AP X OD. But if we draw 01, as OAB
is an isosceles triangle, 01 will be perpendicular to AB;
and the two triangles APB, OID, are similar, and give
OI    OD
the proportion  -    AB; hence AP X OD       OI X
AB; and substituting OI X AB for its value in the
above expression we have 2 r X IK X AB X OI. But
the similar triangles ABP, OIK, give the proportion
BP    - I, wherefore AB X IK    OI X BP = OI X
MN; and substituting this value of AB X IK in the expression for the volume of the body in question, we have
2 T4  X OI2 X MN. The area of a circle whose radius
is OI is expressed by w X (OI)2. We therefore sayThe volume of a body generazted by the revolution of an 
isosceles triangle about a straight line which meets it only
in its summit, and is in the same plane, has for its measure
two thirds of the area of a circle whose radius is the height
of the triangle, multiplied by the part of the azis intercepted by perpendiculars drawn to it from the vertices at
the base of the triangle.
302. [If the axis of revolution is parallel to the base of
the triangle, MN  will equal AB, the surface generated
by AB will be a cylindrical surface, and the volume of
the body generated by the triangle will be two thirds of
the vollume of a cylinder the radius of whose base is OI,
the height of the triangle, and whose height is AB, the
base of the triangle. It follows, therefore, that the volume of a body generated by the revolution of the same
triangle about its base as an axis, will be equivalent to
one third of the cylinder of the above dimensions.]
303. Returning to figure 149, it will be readily seen
that the portion of the polygon abcd, inscribed in the
quadrant ad is composed of isosceles triangles whose
common height is the radius of the inscribed circle. The
volume of the body generated by aOb, will have for its
10*




114


Elements of Geometry.


Fig.149. measure 2-r X (Oh)2 X ae; that generated by bOc
will have 2 rT X (Oi)2 X ef; and that generated by
dOc will have 2 GT X (0 1)2 Xf. As O/h, Oi and 01
are equal, the sum of these bodies, or the body generated by the portion of the inscribed polygon abc d, will
have for the meaaure of its volume, 32 r X (O h)2 X (a e +
ef+f 0), or - IT X (O h)2 X a 0; and if the portion of
the inscribed polygon occupied the entire semicircle, the
body generated by it, in a revolution about the diameter
ap, will have for its measure 2 n X (0 /h)2 X adp.
If we apply this result to the body generated by the
circumscribed polygon, we shall have for the measure of
its volume, <r X (a 0)2 X AP.
By increasing the number of sides of the polygons,
the inscribed polygon increases continually, and the limit
of this increase is the semicircle; but with this increase
of the number of sides, the factor 0 h increases, and
the limit of this increase is O a; and as the other factors
remain the same, the limit for the product is  7 r X
(O a)2 X a).
But by increasing the number of sides of the exterior
polygon, it becomes continually less, and the limit of
this diminution is the semicircle; but with this increase
in the nu-mber of sides, the factor AP is continually diminished, and the limit of this diminution is ap; and
as the other factors remain the same, the limit of the
product expressing the volume of the body generated by
the portion of the polygon circumscribed about the semicircle, is 2 7 X (Oa)2 X ap. It appears therefore that
2 -a X (0 (a) X ap, cannot be the measure of a volume
greater or less than that of the body generated by the
semicircle whose radius is aO, revolving about the diameter ap;  but this body is the sphere whose radius
is aO.
If we denote a 0 by R and ap by 2 R, the expression becomes 2 nr X R2 X 2R, or 4 > X R3; and may
take the form 4 r. R2 X  R. But 4 r R2 is the area
of the sphere whose radius is R; we therefore say- The
sphere has for the measure of its volume, the area of its
surface, multiplied by one third of its radius; or the area
of a great circle multiplied by two thirds of the diameter.
304. If we have a cone and a cylinder whose common
Fig.148. height is equal to the diameter of the sphere (fig. 148),




Elements of Geometry.


115


and whose common base is equal to a great circle of the Fig.148.
sphere; we may express these convex surfaces as follows:
that of the cone   = (5) 7r. R2;
sphere  -   4   rr. R2;
cylinder    4   r. R2.
The volumes of these bodies have the following expressions:
that of the cone  =     2 r X R 3  2= T7. R2 X R;
sphere     4   X R3 = s    r..R2 X2R;
cylinder 6-   r X R3 =-    R.2 X 3R.
Remark. The sphere is therefore two thirds of the
circumscribed cylinder. The cone is one third of the
circumscribed cylinder.  The area of the sphere is
equivalent to that of the convex surface of the circumscribed cylinder.
305. The bodies of revolution which have now been
discussed, are said to be similar when the figures which
generate them are similar. Two cones generated by
similar triangles revolving about homologous sides, have
their heights in proportion to the radii of their bases.
Let A represent the area of a cone whose height is H
and the radius of whose base is R, and whose side is S;
we shall have the equation JA_2 7r.R X I S, (272).
For another similar cone we shall have a - 2 r. r X - s.
If we compare these expressions, cancelling the common
A    R X s              R     S
factors, we have      --; or because -      -, we
a     r Xs              r     s
A     R2     S2  JH2
may write - =      =     --       We therefore saya     r2     s2    h2
The areas of similar cones are to each other as the second
powers of their correspondin'.y dimensions.
306. Two similar cylinders are generated by similar
rectangles (Jig. 144). Their areas will have for their ex- Fig.144.
pressions, A - 2 7. R X II, a= 2 r. r X h. If we compare these neglecting the common factors in the terms of
A    R    II  R/2    H2
the second ratio, we shall have --  r        --
a    r X A     7r2   h2
that is- The areas of similar cylinders are to each other
as the second powers of their homologous measures.
307. All spheres are similar, because every sphere
has for its generatriz a semicircle, and all semicircles




116


Elements of Geometry.


are similar. The areas of two spheres have for their
expressions, A = 4 r. R 2, a  4 r.r2; comparing these
and cancelling the common factors in the second ratio,
A    R2
will give  -     2; that is —The areas of two spheres
are to each other as the second powers of their radii.
308. If we compare the volumes of two similar cones,
which are expressed as follows; V=  I rr. R2 X H, and
V  R2 x H      R3
V-   1.r2 X h; we shall have   -    R-  - 
2                         v      r2Xh    -  r3
IP3     R    HI
A-3 as, —    h; that is-The volumes of similar
cones are to each other as the third powers of their homologous lines.
309. The volumes of two similar cylinders have for
their expressions, V' =  r. R2 X H, v' = w r2.  X h. If
V,    R2 X H
we compare these volumes, we have        --      -
v'     r2 Xh 
R3    tI3          R    H
R      h 3, because -   -.    Whence we say-The
r3    Ah           r     A
volumes of similar cylinders are to each other as the third
powers of their homologous men(sures.
310. Two spheres, being always similar bodies, and
having  for the expressions of their volumes, V"-=
-4.R3, and v"= -7.r3, may be readily compared;
V"    R3     D3
and will give  -   -- =.   That is-The vol'v      r3     d3 
umes of two spheres are to each other as the third powers
of their radii, or as the third powers of their diameters.
311. It was observed (227) that a regular polyedron
has all its laces equal regular polygons, all its diedral
angles equal, and all its polyedral angles equal.
Let us inqulire into the number of regular polyedrons
which may be constructed.
in the firsl place, it is manifest that the faces of these
regular bodies must be equilateral triangles, squares or
regular pentagons; for the angles of a regular hexagon
are equal to four right angles; no polyedral angle therefore, can be formed with regular hexagons, or with regular polygons of a greater number of sides.
Of those bodies contained by equilateral triangles. the




Elements of Geometry.


117


polyedral angles must be either triedral, tetraedral or
pentaedral; for six angles of 60~ each are equal to four
right-angles, and cannot therefore form a convex polyedral angle.
The regular tetraedron is contained by four equilateral triangles; and has four triedral angles. Triedral Fig.153.
angles admit only of four faces in the body.
The reqular octaedron is contained by eight equilateral triangles; and it has six tetraedral angles. The te- Fig.154.
traedral angle allows only eight faces in the body.
The regular icosaedron, contained by twenty equilateral triangles, will have twelves pentaedral angles. These Fig.155.
are all the regular polyedrons which can be formed with
triangular faces.
The regular hexaedron or cube is evidently the only Fig.156.
regular polyedron which can be formed by square faces.
The regular dodecaedron contained by twelve pentag- Fig.157.
onal faces, has twenty triedral angles.
These jive bodies are the only regular polyedrons.
They can each of them, be inscribed in a sphere, and
each may have a sphere inscribed in it; and, in each
case, the inscribed and circumscribed sphere will have
the same centre. The right cone whose angle at the
apex is 600, and the right cylinder whose height is equal
to the diameter of its base, may each have an inscribed
and circumscribed sphere whose centres are the same
point.  The right prism and pyramid, whose heights
have the proper ratio to their other dimensions, and
whose bases are regular polygons, may have inscribed
and circumscribed spheres with a common centre.
Besides the five polyedrons mentioned above, the
sphere is the only other regular body.




118


Elements of Geometry.


Practical Questions.
What is the volume of a prism whose height is 20 feet
and whose base is a rectangle the contiguous sides of
which are 3 feet and 8 feet?
Required the cubic feet in a stick of timber 60 feet
long, 18 inches wide and 8 inches thick?
What is the volume of a prism whose height is 7 yards,
and whose base is a right-angled triangle one side of
which is one foot and the other 30 inches?
Required the volume of a prism whose height is 12
feet and whose base is an equilateral triangle whose side
is 6 feet.
A rectangular parallelopiped has its three contiguous
edges, 4 feet, 6 feet, and 3~ feet; what is it its volume?
An obelisk is composed of the fiustum of a square
pyramid, the side of whose inferior base is 12 feet, the
side of the superior base 6 feet and the height 681 feet,
this is surmounted by a pyramidal summit 1.- feet high,
making the entire height 70 feet; what is the entire volume of the obelisk?
Suppose the above obelisk to be of stone, what would
it cost at one dollar per foot of surface, the base of the
obelisk not being included?
What is the area of a rectangular parallelopiped whose
three contiguous edges are 3 feet, 4 feet, and 5 feet?
What is the area of a right pyramid, the base of which
is a square of 6 feet on a side, and whose height is 4
feet?
A house 40 feet wide and 50 feet long, has a cellar
under the whole of it, 8 feet deep below the sills; the
bottom of the sills (which are 10 inches wide) is 2 feet
above the natural surface of the ground; the inside of
the cellar wall, which is 22 inches thick, coincides with
the inside of the sills; the underpinning stones are 18
inchess deep and 8 inches thick; there are 6 cellar windows of 3 feet each in width and 18 inches high. H-ow
many cubic yards of earth were removed from the cellar?
What did the cellar wall cost at $3 per cubic yard? How
many cubic feet of stone in the underpinning; and what
did it cost at 50 cents per foot of surface?
A ditch is to be dug 1 mile in length, 5 feet in width




Elements of Geometry.


119


and 4 feet in depth; what will it cost at 10 cents pi
cubic yard?
An embankment for a road is to be made across a
meadow, 1 of a mile in width; the road is to be 6 feet
above the surface of the meadow, 30 feet wide at top,
with banks sloping 2 feet in 3; what will this embankment cost at 121 cents per cubic yard?
A bridge is to be built of stone; its length 60 feet, its
width 25 feet and its height 12 feet, with two semi circular arches of 8 feet radius and 9 feet high in the intrados;  What will be the cost of this bridge, at $2 per
cubic yard?
What is the area of a right cone the radius of whose
base is 36 feet, and whose height is 48 feet? What is
the volume of this cone?
What is the area of a cylinder whose height 'is 20 feet
and the radius of whose base is 10 feet? What is its
volume?
What is the area of a sphere inscribed in the same cylinder; and what is its volume?
What are the area and volume of a right cone inscribed in this cylinder?
How many cubic feet of water will a hemispherical
vessel contain, its radius being 10 feet?
How many cubic feet of water in a pond of 1000 acres,
10 feet deep? IHow many miles of canal will such a
reservoir supply for three months, at 50 cubic feet per
mile per minute; supposing one foot in depth of the
reservoir to be lost by evaporation?
Supposing this reservoir is at one extremity of the
canal to be supplied by it, what will be the velocity of
the current in this part of the canal, with 5 feet of water
and an average width of 40 feet?
A cubic block measures 6 feet on a side; what is
the edge of a cubic block eight times as large?
Cleopatra's Needle, a celebrated pyramidal obelisk
near Alexandria in Egypt, consists of a single piece of
red granite; its height is 60 feet, and its inferior base is
7 feet square. Supposing its superior base to be 5 feet
square, what will be its area and volume; and what its
weight at 170 Ibs per cubic foot?
The pyramid of Cheops, the largest of the Egyptian
pyramids, has a square base whose side is 693 feet; its
height is 499 feet. How many acres of ground does it




120           Elements of Geometry.
cover; and what is the side of a cube of equal volume?
Suppose 1000 men to be employed at a time, in erecting this structure, and suppose 10 men, on an average,
to complete 29 cubic feet of the work per day: Iow
long would it be in building?




AN INTRODUCTION


TO
DESCRIPTIVE GEOMETRY.
SECTION I.-Preliminary Explanations.
1. IN the Second Part of the Elements of Geometry
we discussed geometrical magnitudes embracing the
three dimensions of space. But it must have been perceived by the learner, that, in the graphical representations of these bodies, there was much indefiniteness respecting the relative position and relative measure of
their several parts. In the application of Geometry to
the arts, we have frequent occasion, not only to represent
the various forms and relative positions of bodies, but
accurately to determine the various dimensions and other
relations of their parts, and to deduce from these determinate relations, a series of results, the accuracy of
which must depend entirely upon the fidelity of the
graphical representation of tie given magnitudes, and of
the graphical construction of the consequent relations.
2. The object of Descriptive Geometry is to teach this
accurate representation of bodies upon a plane, and this
graphical solution of problems. A geometrical magnitude is represented upon a plane by a method called projection.
The simplest projection, and that in most frequent use,
is called orthogonal or othographic projection. It consists in drawing straight lines from the points to be projected, perpendicular to the plane on which the projection is to be made, and which is called the plane of projection; the straight lines by which the projection is
made are called prijecting lines.
3. When the projecting lines are inclined to the plane
of projection, the projection is called oblique. When the
11




122


Descriptive Geometry.


projecting lines converge towards some fixed point, the
projection is called linear perspective.
The orthographic projection may be considered a perspective in which the point of convergence is at an infinite distance from the body.
4. In the orthographic projections used for the general
purposes of Descriptive Geometry, two planes of projection are commonly useld; and these are usually taken
perpendicular to each other, the one horizontal and the
other vertical, as the construction is thus rendered the
most simple.
These planes are called co-ordinateplanes. They are
considered to be indefinitely extended; and will therefore meet. The line of intersection is horizontal, and is
called the ground line.
Fig. 1.  Let AB (fig. 1) represent the ground line, and the
plane of the paper the horizontal plane of projection.
The vertical plane of projection will pass through AB
perpendicular to the plane of the paper. We may conceive this vertical plane to revolve about the ground line
as an axis, till it shall coincide with the other plane, that
is, the plane of the paper, the part of the vertical plane
above the horizontal moving backward to coincide with
that part of the horizontal plane which is behind the vertical. That part of the plane of the paper which lies
above the ground line, will now represent not only that
portion of the horizontal plane which lies behind the vertical, but also that portion of the vertical plane which is
above the horizontal; and that part of the plane of the
paper which lies below the ground line, will represent
that portion of the horizontal plane which is in front of
the vertical, and also that portion of the vertical plane
which is below the horizontal.
5. This rectangular position of the co-ordinate planes
gives usfour diedral angles, each of which is a right angle (El. 208).'* The diedral angle in front of the vertical plane and above the horizontal, we call thefirst angle; that behind the vertical and above the horizontal,
we call the second; the diedral angle behind the vertical
and below the horizontal, the third; and that below the
horizontal and in front of the vertical, thefourth.


* This reference is to the Elements, article 208.




Descriptive Geometry.


123


In the projection of geometrical magnitudes gener- Fig. i.
ally, the planes of projection may be so chosen that
the points to be projected shall be all in the first diedral
angle. The projection consists of two parts.  First, the
several points of the body are projected by perpendiculars drawn from  them to the horizontal plane; the intersections of these perpendiculars with the horizontal
plane, make their projection upon this plane: This is
called their horizontal projection.  Secondly, the several
points are projected by perpendiculars drawn to the vertical plane: the intersections of these perl)endiculars
with the vertical plane, constitute the vertical projection
of these points.
6. This projection is used by architects and engineers
in representing the several parts of any edifice or structure.  A drawing is made upon paper, which is a milniature representation of a horizontal projection of the edifice. This is called the ground plan, or simply the plai of
the edifice; and exhibits the relative situations of all the
remarkable points in the edifice, referred to a horizontal
plane by perpendiculars to this plane.
The plan of an edifice, it will be perceived, shows us
nothing respecting the height of its different parts.  In
order to fix the heights, and therefore to determine the
relative positions of the several points in the edifice, as
well as its various dimensions, a drawing is also made
representing a vertical projection of these points.
When the vertical projection represents the interior of
the building, it is called a section or proJile. When this
projection exhibits the exterior of the building, it is called an elevation.
It is manifest that those lines in the building which
are oblique to each of the planes of projection, will not
be projected in their true magnitudes upon either of the
planes.  We shall see, however, that descriptive geometry accurately determines these magnitudes notwithstanding this apparent defect of the process.
7. Let AM (fig. 1) be a point in the first angle.  Draw
MM' perpendicular to the horizonal plane, and VMM"
perpendicular to the vertical plane; Al' is the horizontal
projection of the point Al, and Ml" is the vertical projection of this point.  If we suppose a plane to pass through




124


Descriptive Geometry.


Fig. 1. the projecting lines MM', MM", it will be perpendicular
to each of the planes of projection BC, BD, (El. 207),
and therefore perpendicular to the ground line (El. 209).
But if JMl " be perpendicular to AB, by revolving the
vertical plane, about the ground line, through the second
diedral angle into the plane of the paper, MM" will be
in the same straight line with IM' also perpendicular to
AB (El. 14). We therefore say-The vertical and horizontal projections of the same point, are in a straight
line perpendicular to the ground line.
8. Remark. The distance of the vertical projection of
a point from the ground line is equal to the distance of
the point itself from the horizontal plane; and the distance of the horizontal projection of a point from the
ground line, is equal to the distance of the point, in
space, from the vertical plane.
It follows, therefore, that all points situated in the
vertical plane, have their horizontal projections in the
ground line; and all points situated in the horizontal
plane, have their vertical projections in the ground line.
All points in either plane are their own projections in that
plane.
9. The two projections of a point determine its position; for M' and M" being the two projections of a point,
if we draw through the projections straight lines perpendicular to each of these planes, these perpendiculars
must pass through the proposed point; it must therefore
be at their intersection M.
When we speak of a point in space given in position
by its projections, we say, the point (M', M"); by which
we mean, the point whose horizontal projection is M'
and vertical projection M".
10. Two lines which are parallel or which cut each
other, determine the position of a plane (El. 197). If,
therefore, we have the intersections of any plane with the
two planes of projection, this plane is given in position.
11. The two intersections of the proposed plane with
the planes of projection, are called its traces upon
these planes. Its intersection with the vertical plane is
called its vertical trace; and its intersection with the
horizontal plane, is its horizontal trace. If the proposed
plane be parallel to one of the planes of projection it can




Descriptive Geometry.


1~25


have no trace upon that plane. Therefore, if a plane Fig. 1.
has but one trace, it is parallel to that plane of projection
which contains no trace. A plane whose horizontal
trace is M'N', and whose vertical trace M"N", is called
usually, the plane (M'N', M"N").
12. If a plane be inclined to each of the planes of
projection, but parallel to the ground line, it can never
meet the ground line (El. 198); and consequently its traces upon the planes of projection cannot meet the ground
line; they are therefore parallel to it:  And we sayA plane parallel to the ground line, will have its traces
parallel to the ground line. A nd the traces of a plane
inclined to this line, will intersect it.
13. If the two traces of the proposed plane are perpendicular to the ground line, the plane itself must be perpendicular to tle ground line (El. 199); and if the ground
line is perpendicular to the proposed plane, the co-ordinate planes, both passing through this line, must be
perpendicular to the proposed plane (El. 207). Therefore
-If the two traces of the proposed plane are perpendicular to the ground liie, the plane itself is perpendicular to each of the co-ordinate planes.
14. As one point is projected by a perpendicular
drawn to the plane of projection; all the points in a
straight line are projected by perpendiculars drawn to
the plane of projection (fig. 2). These perpendiculars Fig. 2.
being parallel, and passing through the same straight line,
are in the same plane. We therefore say-The horizontal
projection of a straight line, is the horizontal trace f' a
plane passing through the line, and perpendicular to the
horizontal plane.  The vertical projection of a straight
line, is the vertical trace of a plane passing through the
line, and perpendicular to the vertical plane.  These
planes are called projecting planes.*
15. It is here supposed that the proposed line is not
perpendicular to either of the co-ordinate planes.  The
projection of a straight line perpendicular to the plane of
projection, is a point.
16. If a line be parallel to either of the co-ordinate


* It is important to distinguish between projecting
planes and the planes of projection.
11*




126


Descriptive Geometry.


Fig. 2. planes, its prqjection upon the other plane will be parallel
to the ground line. For the line itself and the projecting line of any point in it, are two lines which determine this projecting plane to be parallel to one of the
planes of projection (El. 197); their intersections with
the other plane of projection are therefore parallel
(El. 214).
From this it follows that- The projections, on the same
plane, of parallel lines, are parallel.
If, however, the lines are perpendicular to either of the
co-ordinate planes, their projections will be points, and
cannot be said to be parallel.
17. Two projections of a line determine its position in
space. For drawing a plane through its horizontal projection perpendicular to the horizontal plane, the plane
must pass through the proposed line; and passing a plane
through the vertical projection perpendicular to the vertical plane, this plane will also pass through the line; the
line must therefore be the intersection of these two
planes, which is determined, the planes themselves being
determined: The line is therefore given in position by
its projections.
Remark. It is readily seen that each of the projections
of the proposed line, answers to an infinite number of
lines all in this projecting plane; but the two projections
answer only to a line in each of the planes; this line
must be their intersection.
If the proposed line were in a plane perpendicular to
each of the co-ordinate planes, and inclined to these
planes, the two projections would not determine it.
Some other condition would be necessary; such as its
projection upon a third plane, or the distances of two
points of the line from their projection on one of the
planes.
When we have occasion to speak of a line given by its
horizontal projection M'N', and its vertical projection
M"N", we say, the line (M'N', M"N").
18. [Right lines and planes are considered as indefinite in extent. In the diagrams, those parts of the given
or the required lines which are in the first angle, are
made full; the dotted lines in the figures represent portions of the proposed lines which are situated behind a
plane, and supposed to be seen through the plane. Those




Descriptive Geometry.


127


lines which are composed of short lines and dots are the
auxiliary lines used in constructing the solution.]
SECTION II.-Of the Straight Line and the Plane.
19. PROBLEM. The two projections of a line being
given, to determine the points in which it pierces the coordinate planes.
Let N'N (fig. 3) be the horizontal projection of the Fig. 3.
proposed line, and AIM" its vertical projection. The
point where it pierces the vertical plane must have its
horizontal projection in the ground line (8); N must
therefore be the horizontal projection of this point. If
then we draw, in the vertical plane, NP" perpendicular
to the ground line, and meeting the vertical projection in
P", this point P" will be the point in which the proposed
line meets the vertical plane. And if we draw in the
horizontal plane, the line  IP' perpendicular to the
ground line, the point P' will be the point in which the
proposed line pierces the horizontal plane.
It is evident that this problem admits of four cases;
as the part of the proposed line between the points where
it meets the two planes, may be situated in either of the
four diedral angles. The solution in figure 3, answers
to the case in which this portion of the line is in the
first angle. Figure 4 gives us the second case; namely, Fig. 4.
that in which the portion of the line between the two
points in which it pierces the planes, is in the second
angle. The other cases we shall leave to the ingenuity
of the learner.
20. PROBLEM. The projections of a line being given, to
find the angle which it makes with one of the planes of
projection.
The inclination which a line has to a plane, is measured in a plane passing through the line, perpendicular
to the first plane (El. 207). Suppose that the angle required, is that which the proposed line makes with the
horizontal plane (fig. 5). This angle will be in the pro- Fig. 5.
jecting plane, and will be contained between the line itself and its horizontal projection. If, therefore, the projecting plane be conceived to revolve round its horizontal trace, till it coincide with the horizontal plane, the
required angle will, by this process, be transferred to the
horizontal plane, and can be measured.




128


Descriptive Geometry.


Fig. 5.  For this purpose the horizontal projection P'N is considered to be the axis about which the projecting plane
P'P"N revolves. In such a rotation, it is perfectly manifest that every point in the revolving plane will be,
throughout its revolution, at the same distance from the
axis; and that any line in the revolving plane, perpendicular to the axis, will be perpendicular to the axis at
the end of this revolution. The line NP" will therefore
be perpendicular to the axis when the projecting plane
shall be revolved into the horizontal plane. If, therefore, we draw Np perpendicular to P'N, take Np equal
NP", and draw P'p, the angle p P'N will be the angle
required.
21. PROBLEM. Two points being given by their projections, to Jind their distance asunder in space.
Let the two points be (M', M") (O', O") (fig. 5). It
is obvious from the preceding process, that m o will be
the distance required. This might be found at once by
drawing through the point M', M'm perpendicular to
M'O', and equal l"M; and through the point 0', the
line O'o perpendicular to M'0' and equal to 0"0; this
gives us m o.
22. Remark. The projection of a line upon either
plane, is less than the line itself, unless the line be parallel
to this plane. The length of a line is equal to the hypothenuse of a right-angled triangle, of which one side is
its horizontal projection, and the other, the difference of
the heights of the two extremities of the line above the
horizontal plane. The projection of a line parallel to the
plane of projection is equal to the line.
23. PROBLEM. Through a given point to draw a line
parallel to another line given in position by its projections.
The projections, upon the same plane, of parallel lines
are parallel (16); therefore, if through the projection of
the given point in each plane, we draw lines parallel to
the projection, in that plane, of the given line, these will
be the projections of the required line.
24. PROBLEM. Having the projections of two lines, to
ascertain whether they cut each other.
The horizontal projections of the two lines may inter



Descriptive Geometry.


129


sect, and also their vertical projections, and yet the two
lines may not come near each other. If the lines intersect
in space they will have one point common; and if they
are not in the same plane perpendicular to one of the coordinate planes, their projections on each plane will intersect; and these two intersections must be the two
projections of the point where the lines actually cut each
other. The two projections of the same point are in a
straight line perpendicular to the ground line (7); therefore, to ascertain whether two lines given by their projections, actually cut each other, draw from the point
where their projections intersect, perpendiculars to the
ground line; if these perpendiculars meet the ground
line in the same point, the two points where the projections intersect are the projections of the same identical
point; and this point must be in each of the given lines;
that is, the two given lines cut each other, and are therefore in the same plane.
Figure 6 shows a case in which the projections of the Fig. 6.
two lines in each of the planes, cut each other, when the
lines themselves do not meet. Figure 7 gives us the Fig. 7.
projections of two lines which actually intersect in
space.
25. Remark. If the two proposed lines cut each other
in a plane perpendicular to one of the co-ordinate planes,
their projection upon this plane will be but one straight
line; if they cut each other in a plane perpendicular to
each of the co-ordinate planes, their vertical and horizontal projections will be in one and the same straight
line perpendicular to the ground line.
26. PROBLEM. Two planes being given by their traces,
to find the projections of their line of intersection.
When the traces of the proposed planes upon one of
the co-ordinate planes, cut each other, their point of
meeting is common to the two planes proposed, and is
therefore a point in the required line.
Let (NN', NN"), (MM', MM"), be the two given Fig. 8.
planes; it is manifest that the point P' is common to the
two proposed planes and the horizontal plane ABC;
this is therefore one of the points sought. The point Q"
evidently belongs to the common section of the two proposed planes with the vertical plane; this then is also a
point in the required line. The problem is therefore re



130


Descriptive Geometry.


Fig. 8. duced to finding the projections of a line passing through
the points P', CQ". The horizontal projection of Q'" is
Q  (8); P' is its own horizontal projection; P'Q is
therefore the horizontal projection of the proptsed line.
For similar reasons PQ" is its vertical projection.
Fig. 9.  27. Remark 1st. If the traces NN', MM', (fig. 9) of the
proposed planes upon one of the co-ordinate planes, the
horizontal for example, are parallel to each other, the
intersection of the proposed planes will be parallel to the
horizontal plane; and of this line one point is known,
namely, the point P". As this line is in the same plane
with each of the horizontal traces of the proposed planes,
it will meet them unless it is parallel to them; but it
cannot meet them, unless they meet each other, which is
impossible as they are parallel; this line of intersection
is therefore parallel to each of the horizontal traces of the
proposed planes. This question therefore refers itself to
the problem in article 23.
28. Remark 2d. It may happen that the line of intersection of the proposed planes meets neither of the coordinate planes; in this case it must be parallel to them
Fig. o1. and parallel to the ground line (fig. 10). To find, under
these circumstances, the line of intersection of the proposed planes, it will be necessary to refer to a third
plane; which, for the sake of simplicity, we take perpendicular to each of the other two co-ordinate planes.
We shall not now detain the reader with this process; a
little familiarity with constructions will render the solution perfectly easy.
29. If, however, the proposed planes are parallel to
each other, their traces upon each of the co-ordinate
planes will be parallel; but it is perfectly manifest that
they cannot intersect.
30. PROBLEM. To find the projections of a point, when
we know three planes in each of which it is situated.
As these three planes have only this single point in
common; to find the projections of this point, seek first
the two projections of the common section of any two of
the proposed planes; this line being cut by the third
plane, will thus give the point required.
We shall come to the same result, by finding the intersection of two of the planes, and then seeking the intersection of one of these with the third. These two




Descriptive Geometry.


131


intersections will be two straight lines which cut each
other. This problem will then refer itself to the preceding (26).
31. The most simple method of determining a point
by means of three planes, is to suppose them perpendicular to each other; and then give the distances of the
proposed point from three planes parallel respectively to
the three planes in which this point is situated.
Suppose the three planes BAC, BAD, and DAC, (fig.
11) to be perpendicular to each other, and that we know Fig. 11.
that the point M in space is situated at a distance MM'
from the first, MM" from the second, and MM'" from the
third. As parallel planes are equally distant throughout
their whole extent (El. 212),if we draw the planes MM"M"',
MM' M"', MM' M", at the given    distances respectively from the three planes BAC, BAD, DAC, and respectively parallel to them, the proposed point will be
found at their mutual intersection.
These three planes, with the co-ordinate planes, form
the rectangular parallelopiped whose diagonal is AM.
The square of this diagonal is equal to the sum of the
squares of the three edges which meet at the proposed
point (El. 245); that is (AM)" = (MM')2 + (MM")2 +
(MM"')2. Whence we say-The square of the distance
of any point in space firom the point of meeting of three
rectangular co-ordinate planes, is equal to the sum of the
squares of the distances of the proposed point from each
of these planes.
32. Remark. It will be perceived that three planes
which cut each other form eight triedral angles, in each
of which the proposed point may be situated conformably
with the above conditions; so that to fix the position in
space, of the proposed point, we must not only have its
distance from each of the three co-ordinate planes, but
must know on which side of each it is situated; or we
must in some way particularize the triedral angle in which
it is situated.
When a point is given by a line and a plane, it is the
same thing as if it were given by three planes; for, instead of the given line we must use its two projecting
planes.
33. PROBLEM. A plane being given by its traces, and
a line, inclined to this plane, being given by its projections, to find the point at which the line pierces the plane.




132


Descriptive Geometry.


Find the intersection of the proposed plane with one
of the projectinig planes of the given line; this line of intersection, being in each of these planes, will meet the
given line at the point at which it pierces the proposed
plane.
The several steps of this solution may be performed
from what was said in article 26.  Let (N"O, M'O)
Fig. 12. (fig. 12) be the given plane, and (QQC", NM') the given
line; N"NM' will be the plane which projects this line
upon the horizontal plane: M' and N" are two points in
the common section of this plane with the given plane,
N"M is the vertical projection of this common section,
and P" is the vertical projection of the point of meeting
of this common section with the proposed line; that is,
P" is the vertical projection of the point at which the
given line pierces the given plane. If then we draw
P"P' perpendicular to the ground line, and meeting the
horizontal projection of the given line, in the point P',
this point will be the horizontal projection of the point
sought.
34. PROBLEM. Having the vertical and horizontal
traces of a plane, to find, for any point in the horizontal
plane, the height of the corresponding point in the proFig. 13. posed plane.
Let (GN", GG') (fig. 13) be the given plane, and M'
the given point in the horizontal plane; through M' pass
a vertical plane whose horizontal trace shall be parallel
to that of the given plane; the proposed point will be in
this auxiliary plane, and therefore in its intersection with
the given plane, which intersection will be parallel to the
horizontal plane (27); the point N" will be a point in
this intersection, and consequently the line N"N will
measure the height of the required point above the horizontal plane. M" is the vertical projection of this point,
and M' its horizontal projection (7).
35. To find the angle which the inclined plane (GG',
GN") makes with the horizontal plane BAC; we suppose, from the required point M in the inclined plane,
and from its projection M' in the horizontal plane, perpendiculars drawn to the intersection GG' of these
planes; these will form the right-angled triangle MM'G',
of which we know the two sides MM', M'G', and which
may therefore be readily constructed. The angle MG'M'
will be the inclination required.




Descriptive Geometry.


133


36. Having the horizontal trace of the inclined plane, Fig. 13.
and the angle of its inclination to the horizontal plane,
we may find the height MM' of the point M above the
horizontal plane.  Through the point M' draw NM'
parallel to GG' the horizontal trace of the given plane;
draw M'G' perpendicular to GG', and make the angle
M'G'M equal to the given angle; this will give us MM'
the height of the point M.
37. Having the horizontal trace of the inclined plane,
and its inclination to the horizontal plane, to find the
vertical trace. Find the height of this plane above the
point M', by the last article; and through the point N,
where M'N, parallel to the horizontal trace, meets the
vertical plane, draw NN" perpendicular to the ground
line; and, taking NN" equal M'M, draw GN", which
will be the vertical trace of the proposed plane.
38. PROBLEiM. Through a given point to draw a plane
parallel to a given plane.
We know that the traces of the required plane will be
parallel respectively to those of the given plane (29);
they are therefore easily drawn, if we can find upon
either of the co-ordinate planes one point of the required
plane. Suppose, through the proposed point, a straight
line parallel to the horizontal trace of the required
plane; it will lie wholly in this plane, and will therefore
meet the vertical plane in the vertical trace of the required plane.
Let (JIM', MM") (fig. 14) be the given plane, and Fig. 14.
(P', P") the given point.  Through P' draw P'E parallel to M'1M; draw EE" perpendicular to the ground
line, making EEi" equal to PP"; and draw NE" parallel to M1M", and NN' parallel to  MM'.  These two
lines NN', NN", are the traces of a plane passing
through P (given by its projections, P', P") parallel to
the given plane (IM', JMM").
39. Let (MM', lJM") (fig. 15) be an inclined plane, Fig. 15.
to which the line L'M is supposed perpendicular. Suppose a vertical plane L'EM' to pass through this line
cutting the inclined plane in EM'.  This auxiliary
plane, being perpendicular to the inclined plane and to
the horizontal plane, must be perpendicular to their intersection, that is, to the horizontal trace of the given
plane. The horizontal projection of the line L'M, must
12




134


Descriptive Geometry.


F;i. 15. be the line L'M'; but IMM' being perpendicular to the
plane L' E M', is perpendicular to L'M' (El. 202); that
is, the horizontal projection of the given line is perpendicular to the horizontal trace of the given plane.
By a similar process we may show that the vertical projection of the given line is perpendicular to the vertical
trace of the given plane: We therefore say-If a line
is perpendicular to a plane, the projections of this line will
be respectively perpendicular to the traces of the plane.
40. The converse of this proposition is also evident;
that is —f the projections of a line are perpendicular respectively to the traces of a plane, the line is perpendicular to the plane. For the projecting planes of the line,
being perpendicular respectively to the traces of the
plane, must be perpendicular to the plane itself; therefore the' line, which is the intersection of these projecting planes, is perpendicular to the plane.
41. PROBLEM. To drawt through a given point a line
perpendicular to a given plane; and to find the point in
which it pierces the plane.
Each projection of the given point will of course be a
point in the corresponding projection of the proposed
line; and the projections of the proposed line will be
fc)erpendicular to the corresponding traces of the given
plane.
'ic. 16.  Let (L', L") be the given point (fig. 16), and (JIM',
JIM") the given plane. Draw   L'E' perpendicular to
M1 M', and L" E" perpendicular to MM"; the line (L'E',
L"E/") is the line required.
The other part of this construction refers itself to article 33.
42. PROBLEMI. To draw through a given )point, a plane
perpendicular to agiven straight line.
It is manifest from the constructions which have just
been performed, that the traces of the required plane
will be perpendicular to the projections of the given
line; if, therefore, we assume a plane which answers
this condition, the problem reduces itself to the construction in article 38.
43. Remark. A simple method of constructing the two
preceding problems, is to take the vertical plane of the
Fig. 17. projection in the proposed line (fig. 17), in each case.




Descriptive Geometry.


135


In the first problem, the plane might be given by its Fig. 17.
horizontal trace and its inclination to the horizontal
plane. In the second problem, the line might be given
by its horizontal projection and its inclination to the
horizontal plane.
The details of this construction we shall leave as
an exercise for the learner.
44. PROBILErI. To pass a plane through three points,
given in space, not in the same straight line.
Join the three points by two straight lines; find the
points where these lines pierce the co-ordinate planes
(18); and the straight line joining the two points in
each of the co-ordinate planes will be the trace of the required plane in that plane.
Let the three given points be (M', AM"), (N', N"), and
(P', P"), (fig. 18); the two lines joining these points and Fig. lS'
determining the required plane, will be (M'N', MNIN')
(M'P', M"P"); and E', F', will be the points where
these lines pierce the horizontal plane. The straight
line E'F' joining these two points, will consequently
be the horizontal trace of the required plane.
To find the vertical trace of the proposed plane, we
draw, according to article 38, MIG parallel to E'F'.
and GG" perpendicular to the ground line and equal to
MM"; the straight line ItG" will be the vertical trace
of the required plane.
45. We may construct this problem by passing a vertical plane through one of the given points and each of
the others, and revolving these two planes about their intersection with the horizontal plane, as in article 20, till
they coincide with this plane (fig. 19). If we then draw Fig. 1}
MN meeting MIN' in F', and MP meeting MIP' in E',
the straight line E'F' will be the horizontal trace of the
required plane.
It is manifest that, by drawing M'G' perpendicular to
this horizontal trace, and constructing the triangleG' M' M"
of which the angle at M' is a right-angle and the side
M'M" equal to M'M, the angle MG'M' will be the inclination of the required plane to the horizontal plane
(35).
46. The lines ME', MF', are the distances in space of
the point M from the points E' and FV where these lines
pierce the horizontal plane; we have therefore the three




136


Descriptive Geometry.


Fig. 19. sides of the triangle whose vertices are M, E', and F'.
This triangle may be conceived to turn about the side
E'F' till it meets the horizontal plane on the other side
of this line; its vertex M will then be at the point m.
47. Remark. In this construction we have the developement of the entire surface of a tetraedron whose base
is M'E'F' and whose lateral faces are ME'IF, ME'M'.
and MF'M'.
48. PROBLEM. Tw o planes being given in position, to
find the angle w2hich they make with each other.
The angle made by two planes being the plane angle
made by two lines drawn in these planes through the
same point perpendicular to their line of intersection;
to determine this angle, we may construct a plane perpendicular to the intersection of the given planes. The
intersections of this auxiliary plane with the given planes,
will form an angle with each other equal to the angle of
the proposed planes. This angle may be easily found
from the construction in article 46.
The following very simple solution of this problem
was communicated to Lacroix by Monge.
" Suppose the two given planes to be (EH', EF"),
Fig. 20. (GH', GF"), (fig. 20), and the horizontal projection of
their intersection to be FH'. We construct in the vertical plane passing through this straight line, the intersection of the two given planes, by drawing F F perpendicular to F'II' and equal to F'F"; then through any point M'
taken at pleasure in the line I-I' s construct a plane perpendicular to the line F HI (43); and find also the straight
line PM', which is the intersection of this plane with the
vertical plane FFI1'. But if we revolve the first of these
planes about its horizontal trace L'N', the line PM', being perpendicular to M'N', will necessarily fall upon
H'M' (20), and the point P will be found in P'; the triangle whose vertices are L', P, N' will not be altered by
this revolution; and the angle L'P'N' will be the angle
of the given planes."
49. PROBLEM. A plane being given, and also a line in
that plane, to draw throlgh this line a second plane making with the first a given angle.
This problem is easily solved by retracing the several
steps of the preceding solution. In this case, the data




Descriptive Geometry.


137


are the plane (EH' EF"), and the vertical plane FH'F Fig. 20.
which is revolved into the horizontal plane. We construct L'N' and the point P', as in the preceding problem, and make upon L'P' the angle L'P'N equal to the
given angle; through the point N' thus determined, we
draw H'G, we then draw G F"' and we have the required plane (G H', G F").
50. PROBLEM. Two lines which cut each other cbeinC
given in space, tofind their angle.
The two projections of the point of meeting of these
two lines, will be in a straight line perpendicular to the
ground line (24). Construct the given lines; find the
points P', Q', (fig. 21) where they pierce the horizontal Fig. 21.
plane (18); draw the straight line P'Q'; and this
straight line together with the two given lines will form
a triangle, of which the angle opposite to this side is the
angle required.
To measure this angle, revolve the triangle about the
side P'Q' into the horizontal plane. To ascertain at
what distance from M' the point M will fall, suppose N'
to be the foot of a perpendicular drawn from M to the
base P'Q' of the triangle. The line MN', which measures the height of this triangle, will be the hypothenuse
of a right-angled triangle whose two sides are MM',
M'N'. Revolve this triangle about the side MM' till it
is parallel to the vertical plane, and the vertical projection NM" of its hypothenuse will be the height of the
triangle P'MQ'.  The required angle P'^IQ  is now
easily measured.
51. PROBLEM. The projections of a line and the traces
of a plane being given, to construct their angle.
The angle which the line makes with the plane is the
angle which it makes with its projection upon that plane;
if, therefore, from any point in the given line, a perpendicular be drawn to the plane, it will meet this projection and a right-angled triangle will be formed; consequently the angle which this perpendicular makes with
the given line, is the complement of,the angle which the
given line makes with the plane.  This question is
therefore reduced to finding the angle of the two lines.
The angle P'MQ' (Jig. 22) is the complement of the Fig. 22.
required angle, whose construction will be readily under12*




138


Descriptive Geometry.


stood from the last article, observing that (OK', OK")
represents the proposed plane, and that the other parts
of the construction are the same as in the preceding solution.
52. PROBLEM. Two straight lines which do not cut
each other being given in space, to find the distance between their nearest points.
Suppose first that one of the given straight lines is perFig. 23. pendicular to the horizontal plane (Jig. 23), its projection
in this plane will be the point MI (16), and its vertical
projection will be AIM" perpendicular to the ground
line. Draw M'P' perpendicular to the horizontal projection of the second given line, and this will be the distance sought.
53. The shortest distance between two straight lines,
Fig. 24. M'N' and EF (fig. 24) may also be found, by drawing
through one of them a plane HiG' parallel to the second,
and then drawing from any point in the second a perpendicular EF' to this plane. This perpendicular is the
distance sought, and determines the plane FEE' which
cuts the straight line M'N' in the point P', where this
straight line approaches the nearest to EF.
Fig. 25.  Let the line (EP", E/P') (fig. 25) be the first of these
lines, and (OM, O'lM) the second. E' is the point
wliere the first given line pierces the horizontal plane;
and (E'L', EL") is a line drawn through E' parallel to
the second given line, to determine a plane parallel to
this last. The plane passing through the line drawn as
above, and through the first of the given straight lines,
is constructed by a process analogous to that of article 44;
this plane is G' G G"; and as it is parallel to the second
of the given straight lines, it is only requisite to draw,
through any point of this last, a perpendicular to this
plane; this is done through the point O 0'.
We next seek the point of meeting of this perpendicular with the plane G'G G" (33); this point is (N', N"').
In order to find the nearest points of the proposed
lines, we draw, through the point N', (parallel to M1O'
the horizontal projection of the second straight line) the
line N/P', which is evidently the horizontal projection of
the intersection of the plane G'GG" with a plane drawn
perpendicular to this and through the second given
line, since it belongs to a straight line parallel to this,




Descriptive Geometry.


139


and which passes through the foot of the perpendicular Fig. 25.
drawn from this straight line to the plane in question; in
other words, it is the projection of E' F' in figure 24.
The point P' in which the line N' P' meets the horizontal projection E' P' of the first of the given lines, is
the projection of the point P' of figure 24, and consequently the projection of the point where the first straight
line approaches the nearest possible to the second. The
perpendicular to the plane G'GG", drawn through this
point, is the shortest distance sought; its projections are
P' K', P" K", parallel to N' O', N" 0, respectively; its
length is the hypothenuse of a right-angled triangle, of
which one side is P' K', and the other, the difference of
the distances of the points P" and K" above the ground
line. This will be found by article 21.
54. PROBLEM. Having two of the plane angles in azny
triedral angle, and the inclination of these planes, to construct the developcement of this triedral angle.
Suppose one of the given faces Aff to revolve into the
plane of the other AfE (fig. 26); if through any pointf, Fig. 26.
taken at pleasure in the edge Af, a perpendicularff'
be drawn, it will describe in this developement, a plane perpendicular to this edge. In this plane, the angle of inclination of the given planes will be found; if, therefore, we
make uponfF the angle F"fF equal to the given angle, and takef F" equal toff', we shall thus have the
situation of the point f' in relation to the linef F, when
the planes have their given inclination. But it is evident that the three points fF" and F, determine the
base of a pyramid formed by the proposed triedral angle
and the plane which has been drawn perpendicular to
the edge Af; and the required face of this triedral angle will meet the first face in AF, and the second in Af;
aud will apply itself to the constructed triangle, in the line
FF". The third face will therefore be determined by
the triangle whose three sides are AF, Af' and FF".
Remark. This is one case of the general problem
Any three of the six things in a triedral angle (namely,
the three plane angles and the three diedral angles) being given, to find the others. As every spherical triangle answers to a triedral angle at the centre of the
sphere, (the sides of the spherical triangle measuring
the plane angles in the triedral angle, and the angles of




140


Descriptive Geometry.


Fig. 26. the triangle being the diedral angles in the triedral angle,) it is manifest that the solution of this general
problem is the solution of all problems which can occur
in spherical trigonometry.
As all polyedral bodies are determined by planes and
their intersections, the preceding problems will enable
us to construct all kinds of polyedrons.
SECTION III.-Of t4h Generation of Geometrical
Magnitudes.
55. If we conceive a point to move constantly in the
same direction, it describes a straight line. The straight
line is the path described by the moving point; that is,
it consists of the successive positions occupied by this
point. And these positions, qr points, we call the elements of the line; and we say that the line is generated
by the motion of the point.
It will be readily seen that the number of these elementary positions in a finite line, is illimitable; we
therefore say, that the line is composed of an infinite
number of points.
If a straight line be conceived to move in such a manner that any two points in it shall describe equal and
parallel straight lines, it will be always in the same
plane, and will generate a parallelogram. This surface
will be composed of the successive positions of this
straight line; and these positions or lines constitute the
elements of the surface. And all plane surfaces may be
considered as composed of right-line or rectilinear etetnents.
A circle may be considered as composed of an infifinite number of parallel straight lines. It must then be
considered to have been generated by a straight line (its
diameter) moving in a plane and changing its magnitude
with its distance each way from its first position, according to a certain law.
Or, as a circle is generated by the motion, in a plane,
of a straight line equal to radius, about one extremity
(El. 266); each elementary point in this moveable radius
may be considered as generating a circular curve. These
circular curves compose the entire surface of the circle,
and constitute its curvilinear or circular elements.




Descriptive Geometry.


141


56. The motion of a surface, out of the plane of its
original position, generates a body.
If a square move in such a manner that any three
points in it, not in the same straight line, shall describe
parallel straight lines, each equal to a side of the square
and perpendicular to the surface, the product of this motion will be a cube; and the successive positions of this
square are the elements of the cube.
If the several points in the square should move in straight
lines passing through the same fixed point, the dimensions
of the square will be in a constant ratio to its distance
from that point; and must therefore all be zero (0) at the
instant of passing that point. The body thus generated
by the moving square will be a pyramid whose summit is
the fixed point, whose base is the square, and whose elements are the successive positions and magnitudes of this
square.
A triangular or any other polygonal pyramid may be
conceived to be generated in the same manner. The
analogous method of generating the cone and cylinder,
was noticed in the Elements (El. 266, 275).
A sphere may also be considered as generated by the
direct motion of its great circle always parallel to itself;
its centre describing a straight line perpendicular to its
plane; and its area diminishing with its distance (each
way) from its original position, according to a certain
law.
In these examples, every successive position of the
generating plane is a section of the body thus generated. And as these successive positions constitute the
whole intermediate space, and are all parallel, the body
may be considered as composed of an infinite number of
parallel sections.
Of Curves.
57. If a point in motion continually change its direction, its path is a curve line. If this line lie wholly in
the same plane, it is called a curve of single curvature.
If the points of the curve are not in the same plane,
it is called a curve of double curvature.
58. Mathematical curves are considered to be given,
when the law, according to which the generating point
changes its direction, is given.
A point in motion will generate a circular curve, when




142            Descriptive Geometry.
the conditions of its motion are that it shall continue in
the same plane, and at a constant distance from a given
point in that plane.
If the conditions of its motion are, that it shall continue in the same plane, and that the sum of its distanFig. 27. ces MF, Mf, from two fixed points F,f, (jig. 27) is constantly the same, it will generate a curve called an ellipse.
The longest diameter of the ellipse is called its transverse
axis; the shortest is called its conjugate axis. These axes
are perpendicular to each other; and each divides the
curve into two parts which are equal, and symmetrical
with respect to this axis. The points of the ellipse at the
extremities of the axes, are called the vertices of the ellipse.
If the condition is, that, continuing in the same plane,
Fig. 28. the difference of the two distances MF, Mf (fig. 28) of
the generating point from two fixed points shall not vary,
it will describe a curve called a hyperbola. The axis of
the hyperbola is a straight line dividing the curve into
two symmetrical portions. The point where the axis
meets the curve, is called the vertex of the hyperbola.
If the point move with the conditions, that it shall not
depart from the plane, and that its distance MX from a
Fig. 29. straight line, as XZ (fig. 29) drawn in that plane, and
its distance MF, from a fixed point F in that plane, shall
be constantly equal, the curve generated by the point
will be a parabola. The axis of the parabola is a straight
line dividing the curve into two symmetrical parts, as
in the hyberbola; the point where the axis meets the
curve is called the vertex of the parabola.
These four curves, the circle, the ellipse, the hyperbola, and the parabola, may be obtained by cutting a right
cone with a plane making different angles with the axis
of the cone.
(1). If a right cone whose base is a circle, be cut by
a plane perpendicular to its axis, the section will be a
circle (El. 268).
(2). If the cutting plane make an acute angle with
the axis, greater than that which the side of the cone
makes with the axis, the section will be an ellipse.
(3). If the cutting plane make with the axis of the
cone, an angle less than that which one side makes with
the axis, and do not pass through the vertex, the section
will be a hyperbola. In this case the cutting plane must
be parallel to two right-line elements of the conical surface.




Descriptive Geometry.          143
These elements, therefore, would not be cut by this
plane, though all the other elements would. And if
these parallel elements were projected upon the cutting
plane, their projections could not touch the curve,
though they would continually approach it, as we follow
the curve down from the vertex. These projections are
called asymtotes of the curve.
It is important to observe that the plane, in the case
of the hyperbola, though it cuts all the right-line elements
of the surface excepting two, it does not cut them on
the same side of the apex of the cone. But these elements being produced, the plane will cut the remainder
beyond the apex. The curve, therefore, consists of two
branches: The straight line joining their vertices is
called the transverse axis; and a straight line perpendicular to the middle of this, and in the plane of the
curve, is generally assumed as the coiljgate axis of the
hyperbola.
(4). If the angle which the cutting plane makes with
the axis of the cone, is equal to that which the side
makes with the axis, the section is a parabola. The
parabola has but one axis.
These curves, on account of this method of obtaining
them, are usually called the Conic Sections.
59. If a point be conceived to move il the surface of
a cylinder or a cone, with this condition, that its path
shall, in every part, make the same angle with the
axis of the cone or cylinder, it will, in each case, describe a spiral curve, which is a curve of double curvature. The more important curves of double curvature
are those produced by the intersection of curve surfaces.
Of Curve Sutfaces.
60. The curve surfaces of which geometry takes cognizance, are generated by lines, according to certain mathematical laws which serve to determine their parts and
the character of their curvature.
As a plane surface is considered as consisting of an
infinite number of straight lines, so we may conceive a
curve surface to be composed of an infinite number of
curves; and this surface will be perfectly determined,
when we have these curves and the law which connects




144


Descriptive Geometry.


each one of them with the next. A curve line, therefore,
which moves in space, or which changes, at the same
time, its magnitude and position, according to some determinate law, generates a curve surface. This surface
may be considered as composed of the successive positions and forms which the line takes during its motion.
61. The most simple of curve surfaces are those of
the cone and the cylinder.
A cylindrical surface is generated by a straight line
moving in space, always parallel to itself, and directed
in its motion by a curve line. The straight line is here
called the generatrix, and the curve line the directrix.
If the directrix be a circular curve, the surface will be
that of the cylinder considered in the Elements.
The cylindrical surface may also be generated by the
motion of a curve line along a straight line, each point
in the curve moving in a direction parallel to the straight
line. In this case the curve line is called the generatrix, and the straight line the directrix.
If the cylindrical surface be cut by a plane parallel to
the right-line generatrix, the section will be a straight
line coincident with one position of this generatrix. The
successive positions of this right-line generatrix we call
the right-line elements of the cylindrical surface.
If this surface be cut by parallel planes making any
angle with the right-line elements, these sections will be
all the same curve; and each of these is an element of
the cylindrical surface.
62. If the motion of the right-line generatrix were determined by the conditions, that this line should pass
constantly through the same point, and that it should
move along a curve line, the product of this motion
would be a conical surface. If the curve line were the
circumference of a circle, the conical surface thus produced would be that of the cone considered in the Elements. But it is not essential to a cylindrical or conical
surface, as the terms are used in Descriptive Geometry,
that the curvilinear directrix should be of a nature to return into itself.
Cones and cylinders are distinguished by their bases;
they are called circular, elliptical, hyperbolical, or para



Descriptive Geometry.


145


bolical, according as the base is a circle, an ellipse, a hyperbola or a parabola.
The conical surface may also be conceived to be generated by the motion of the curve which was before
taken for the directrix; the conditions being, that all
the points in the curve shall move in straight lines converging to a fixed point. this point will be the summit of
the cone.
63. As the right-line generatrix is considered as infinite, it is manifest that the conical surface generated
by its motion will consist of two parts perfectly symetrical, but in opposite directions from their common summit. These two parts (called, in French, nappes) are
considered as constituting but one and the same surface.
If a conical surface be cut by a plane passing through
the summit, the section will be in a right-line element of
the surface; if it be cut by planes parallel to each other,
none of which pass through the summit, these several
sections will be similar curves. and are considered as
curvilinear elements of the surface.
64. If the straight-line generatrix, instead of moving
according to the conditions given for the cylinder or the
cone, were to move along two straight lines or a straight
line and a curve line not in the same plane, or along
two curves not in the same plane and so situated that
the straight line shall neither pass constantly through
the same point, nor have its successive positions parallel, the surface generated by its motion will be what is
called a warped surface.
65. It will be at once seen that there may be an infinite variety in the character and curvature of sirfaces
which fall under the general division of warped surfaces.
We shall give one or two examples.
(1.) Let the right-line generatrix have two right-line
directrices; suppose that one of them is perpendicular to
a horizontal plane, and the other inclined to this plane;
and suppose that a condition of the motion of the generatrix, is, that it shall be constantly parallel to the horizontal plane. It will be readily perceived what is the
kind of surface generated by this motion of the straight
line.
'13




146


Descriptive Geometry.


(2.) Let there be one right-line directrix in the vertical plane, and perpendicular to the horizontal plane;
and let the other directrix be a circular arc of 900,
whose plane is perpendicular to each of the co-ordinate
planes, whose centre is in their common intersection,
and whose extremities are in these planes respectively;
and suppose the right-line generatrix to be always parallel
to the horizontal plane. The body embraced by the surface generated by this straight line, and the two co-ordinate planes, is what was called by Wallis, a conoidalwedge.
66. Warped surfaces, although generated by straight
lines, differ from cylindrical and conical surfaces, in a
very important particular. The latter are capable of being developed or unrolled upon a plane; the former are
not capable of such developement.
That any surface may be developed, it is necessary
that any two contiguous positions of the right-line generatrix should be in the same plane.
67. To understand the developement of a cylindrical
surface, suppose a plane applied to it; it is manifest that
the plane will touch the cylindrical surface along one of
its right-line elements. Let the cylinder be rolled upon
the plane, the right line elements of its surface will come
successively in contact with the plane, and when the cylinder has rolled once round, every right line element has
touched the plane; and the portion of the plane thus
brought into contact with the cylindrical surface, in one
entire revolution, is equivalent to the cylindrical surface,
and may be considered its developenment. Or suppose a
piece of plane paper rolled about a cylinder, so as just
to cover it, to be unrolled into a plane; it will represent
the developement of the cylindrical surface.
The developement of a conical surface may be illustrated in a similar manner.
68. A more common method of illustrating the developement of these surfaces, is to begin with the prism and
pyramid. One of the lateral faces of the prism, for instance, is applied to the plane upon which the surface is
to be developed.  The body is then made to revolve
about the edge common to this face and. the next, till this
second face comes into the same plane; the body is then




Descriptive Geometry.


147


revolved about the edge which separates the second face
from the third, till this third face comes into the same
plane. The process is continued, the body revolving
upon each successive edge, as upon a hinge, till each of
the faces has been in contact with the plane. The portion of the plane thus brought in contact with the prismatic or pyramidal surface, constitutes the developement
of this surface.
To apply this process to cylindrical and conical surfaces, we have only to consider the cylinder as a prism,
and the cone as a pyramid, each of an infinite number of
lateral faces.
69. If, instead of the right-line generatrix employed in
the production of cylindrical and conical surfaces, we
employ a curve line, the plane of which, in the progress
of its motion, always cuts the plane of the curvilinear directrix in a line perpendicular to the tangent to the directrix at the point of intersection of the generatrix and
directrix, the surface generated will be a surface of double curvature,
70. The most simple of double curve sutfaces, are
those which are also surfaces of revolution.
Of surfaces of revolution we have, in The Elements,
three examples: those of the right cone, the right cylinder, and the sphere. Of these the surfaces of the cone
and the cylinder, are single-curve surfaces. The spherical surface is both a double curve surface and a surface
of revolution.
71. If, instead of the semicircle used in the production of the sphere, we employ a semi-ellipse revolving
about one of its axes, the surface thus generated will be
a spheroidal surface; the body thus produced is usually
called a spheroid, sometimes an ellipsoid. If the semiellipse revolve about its transverse axis, the product of
its revolution is called a prolate spheroid; if the semiellipse revolve about its conjugate axis, it will generate
an oblate spheroid.
72. If the curve-line generatrix be a hyperbola revolving about either axis of the curve as an axis of revolution,
the surface generated by this motion will be a hyperboloidal surface; the body embraced by this surface begin a hyperboloid.




148            Descriptive Geometry.
73. If the generatrix be a parabola revolving round
the axis of the curve, the surface generated will be a
paraboloidal surface, the body thus generated being a
paraboloid.
These three last bodies have been called conoids, on
account of their being generated by the conic sections.
These conoidal surfaces are surfaces of double curvature as well as surfaces of revolution.
74. There is manifestly an infinite variety of surfaces
of revolution. But they all agree in one particular;
namely, that every plane section perpendicular to the
axis of rotation, gives a circular curve.
A plane section of these surfaces through the axis, will
always give the generating curve. This section is called
the meridian section of the surface'; or its meridian curve.
75. Another class of surfaces are called annular surfaces; these admit of great variety; the simplest is that
of the common ring, which will serve to give a general
idea of the class. This surface is generated by the
motion of one circle along the circumference of a greater
circle; the centre of the generating circle moving in
the circumference of the directrx, with its plane at
each point perpendicular to the tangent of this directing
circumference.
In surfaces of this class, it is not necessary that either
the generatrix or directrix should be circular, or that the
directrix should be a curve of single curvature. We
Fig. 30. have, in figure 30, another example of this kind of surfaces; GH is the generating curve, and XZ is the curvilinear directrix.
SECTION IV. —Of the Projection of Curve Lines
and Curve Surfaces.
76. It is evident that plane curves parallel to the plane
of projection, will have, for their projection, curves equal
to themselves. That is, the horizontal projection of a
circle, parallel to the horizontal plane, is an equal circle:
The horizontal projectiou of an ellipse parallel to the
horizontal plane, is also an equal ellipse; and so of any
other curve.
But if plane curves are parallel to one of the co-ordinate planes they must be perpendicular to the other;




Descriptive Geometry.


149


their projections, therefore, upon this other, will be
straight lines.
77. To project, upon the horizontal plane, a plane
curve parallel to it, perpendiculars are drawn to the
plane from every point in the curve. These perpendiculars will form a cylindrical surface of which the proposed curve is a perpendicular section. And, in general, any curve line not perpendicular to the plane of projection, is projected by a cylindrical surface passing
through the curve perpendicular to the plane of projection. This cylindrical surface is called the projecting
cylinder of the curve. The projection is the intersection
of this cylindrical surface with the plane of projection.
78. If the plane curve to be projected be inclined to
the plane of projection, the projection will be a curve
different from the proposed curve.
To illustrate this principle, let us suppose a circle perpendicular to the vertical plane and inclined to the horizontal. One diameter of the circle will be perpendicular
to the vertical plane and therefore parallel to the horizontal plane. The horizontal projection of this diameter
will be a straight line equal to itself (22); and all the
chords parallel to this diameter, will have their projected
lines equal to themselves. But the diameter, perpendicular to that which is parallel to the plane of projection,
will have the same inclination to this plane that the circle itself has to this plane; it will therefore have for its
projection a line shorter than itself, whose magnitude depends upon the degree of inclination (22).  So
also the chords parallel to this diameter, will have for
their projections straight lines bearing the same ratio to
the chords, themselves as the projection of this diameter
bears to the diameter itself. That is, the right-line elements of the circle taken in the direction of their greatest inclination to the plane of projection, have for their
projections lines which bear a constant ratio to these elements.
Now it is shown, in treatises on Conic Sections, that
the straight lines, parallel to the conjugate axis, in an ellipse, have all the same ratio to the corresponding chords of
the circumscribed circle, that is, the ratio of the conjugate to the transverse axis.  The projection of a circle
13*




150


Descriptive Geometry.


inclined to the plane of projection, is therefore an ellipse
whose transverse axis is equal to the diameter of the circle, and whose conjugate axis is the projection of that diameter which has the greatest inclination to the plane of
prqjection.
This is sufficient to illustrate the principle, which the
reader may apply to the ellipse and to other plane
curves.
79. Curves of double curvature may also be determined by their projections upon two planes, as every point
in the curve may be determined by its two projections.
But those which result from the intersection of curve
surfaces, will be determined by the projections of the intersecting surfaces.
80. Curves of double curvature have their parts frequently determined by the rectilinear tangents to points
in these parts. The relation of a right-line tangent, to a
curve of double curvature, is not easily explained. The
rectilinear tangent to a plane curve, must not cut the curve
and must be in the plane of the curve. Without this last
condition an infinite number of straight lines might be
drawn touching a plane curve in any point, conformable
with the first condition: So also an infinite number of
straight lines may be drawn through any point of a curve
of double-curvature, without cutting the curve; yet
only one of these can be a tangent to the curve. To determine the position of a rectilinear tangent to curves of
this kind, the tangent has been defined to be "a straight
line passing through two contiguous points in the curve."
It is perhaps as philosophical to fix the position of the
tangent by saying that, it must be in the plane determined
by the point of contact and the two contiguous points of
the curve.
81. A cylindrical surface is determined when any
plane section of the surface and any right line element
are determined by their projections; as the whole surface is composed of right-lines of each of which we have
one point (namely, that in which it is cut by the plane
making the given section) and the direction which is
parallel to the given element. Any point in the surface
may therefore be determined by article 23.




Descriptive Geometry.


151


A conical surface is determined by the projection of
its summit and any curvilinear section, upon each of the
co-ordinate planes.
82. A surface of revolution is most easily constructed
by choosing one of the planes of projection perpendicular,
and the other parallel, to the axis of revolution. Its projection upon the plane to which its axis is parallel, will
be a meridian section.  This, together with the projection of this axis upon the other plane, will enable us to
determine every point in the surface.
Of Tangent Planes to Curve Surfaces.
83. The theory of tangent planes to curve surfaces is
important, as it enables us to determine the relative positions of these surfaces, when it could not be done by
the surfaces themselves.
A plane is tangent to a curve surface when it touches
the surface, and lies wholly on the same side of the surface.
Suppose a convex surface to be cut by planes all passing through the same point in that surface; the sections
would be plane curves and the rectilinear tangents to
these curves at their common point, would be all in the
same plane. Any two of these right lines determine this
plane; and this plane is tangent to the surface at this
point. So that the construction of a tangent plane to
any curve surface, reduces itself to the construction of
two tangents to the curves in the surface at that point.
84. Suppose a straight line perpendicular to the tangent plane at the point of contact; this line is called a
normal to the curve surface at that point: as also the
normal to a plane curve at any point, is a line in the
plane of the curve, drawn through the point of contact
and perpendicular to the tangent.
If through the normal to a curve surface, at any point
planes be supposed to pass, cutting the surface, these
planes will be normal to the surface at this point; they
are called normal planes.
85. PROBLEM. To draw a tangent plane to a cylinder.
A plane which touches a cylindrical surface, must, from




152


Descriptive Geometry.


the nature of the surface, meet it in one of its right-line
elements; but it will meet the plane of the base in a
Fig. 31. straight line, as PT (fig. 31,) which will be a tangent
to the periphery of the base.
If therefore we draw through the proposed point M,
the straight line MP parallel to the generatrix AD, and
construct the tangent PT to the point P of the curvilinear directrix, the tangent plane will be determined by the
two lines PM and PT.
Remark 1st. If the proposed point through which the
tangent plane is to be drawn, is without the cylinder, it is
evident that there will be two planes which will answer
to the conditions of the problem. To construct this solution find the point where a straight line passing through
the given point and parallel to a rectilinear element of
the cylinder, meets the plane of projection.  A line
drawn through this point, tangent to the intersection of
the cylindrical surface with this plane, will be the trace
of the tangent plane upon this plane of projection.
Remark 2d. If we would find the rectilinear tangent
to a curve made by the intersection of two curve surfaces, we should construct tangent planes to these surfaces
at the proposed point; and the intersection of the two
planes would be the right-line tangent required.
86. PROBLEM. To construct a tangent plane to a
cone.
The only difference between this construction and
that of the last problem, is, that the line MP must be
drawn through the summit S instead of being parallel to
Fig. 32. the generatrix (fig. 32).
87. PROBLEM. To draw a tangent plane to a surface
of revolution.
The most convenient method of determining a tangent
plane to surfaces of this class, is by means of the rectilinear tangents to the meridian curve and circular section, at the proposed point of contact.
Fig. 33.  The plane tangent to the point M (fig. 33) will therefore be determined by the straight lines MT and Mt;
the first being tangent to the generating curve MX, and
the other to the circle MZ. If therefore we can draw
a tangent to the meridian curve at the proposed point
of contact, it is easy to construct the plane tangent to
the surface at that point.




Descriptive Geometry.


153


The tangent plane to a surface of revolution, is always
perpendicular to the meridian plane passing through the
point of contact. It therefore follows that, a meridian
plane to a surface of revolution is always normal to that
surface.
Plane Sections of Curve Surfaces.
88. Plane sections of a cylindrical surface parallel to
its axis and of a conical surface through its summit, are
all straight lines (61, 63).
Plane sections of these surfaces parallel to the bases
of these bodies are similar to these bases (El. 268, 275).
In the right cylinder and right cone these sections are
perpendicular to the axis; if the bases are circles these
sections will be circles. But an oblique section of a
right cone by a plane which cuts all its rectilinear elements on the same side of the apex, is an ellipse (58).
And it is easy to show to those acquainted with the properties of this curve, that a plane section of a right cylinder, oblique to its axis, is also an ellipse.
These sections are easily determined; for having
these surfaces given, their rectilinear elements are
known; and knowing these right lines and the cutting
plane, the curves themselves are determined by finding
the points where these lines meet this cutting plane,
(33).
89. A plane section of a sphere is always a circle.
The determination of these sections belongs to the solution of the general problem —To find the position and
magnitude of the circle which is the intersection of a given
sphere and a given plane.
By drawing a perpendicular from the centre of the
sphere to the cutting plane, and determining the meeting of this line and the proposed plane, we have the
centre of the circle required.
This solution will be very simple, if we take the plane
of the vertical projections, DAB (fig. 34) perpendicular Fig. 34.
to the horizontal trace of the proposed plane, which may
always be done. Then, 0' and O' being the projections
of the centre of the sphere, if we suppose it cut by a vertical plane drawn through the line H' O' perpendicular to
AC; this plane will pass through the centre of the




154


Descriptive Geometry.


Fig. 34. sphere, and will contain the perpendicular drawn from
this point to the proposed plane DAC; but it is parallel
to the vertical plane DAB: We may therefore suppose
it to be applied to this vertical plane without any
change of the lines which it contains, either in magnitude or in position relative to its horizontal trace H' 0'
which will then coincide with AB. This being premised, M"E"N" will be the great circle which results from
the section of the sphere by the vertical plane just mentioned; the perpendicular /0G// will determine the vertical projection G" of the centre of the section sought.
We thence obtain its horizontal projection G'; and, revolving the plane DAC about it horizontal trace, as this
centre will fall upon the point G in the horizontal plane,
we may describe from this point, with the radius G"M",
the circle MN which will be the required intersection of
the sphere and plane proposed.
Of the Intersections of Curve Surfaces.
90. The most simple method of constructing the intersections of curve surfaces, is to suppose a series of
plane sections of the two surfaces, and to determine the
section made by each plane with each of these surfaces;
the points common to the two curves, in any one of
these plane sections, are of course points in the intersection required.
This method will be the most simple if we draw the
cutting planes parallel to one of the co-ordinate planes.
Suppose them parallel to the vertical plane; the sections
which they make with the proposed surfaces will have
their horizontal projections respectively in the horizontal
traces of these cutting planes; and their vertical projections will not only show the height of any required point
above the horizontal plane, but will be exactly equal to
the auxiliary curves whose respective projections they
are.
A few examples will illustrate these principles, and enable the reader to apply them generally to the construction of analogous problems.
91. PROBLEM. To construct the intersection of a cylinder and a sphere.
We shall take for the plane of the horizontal projec



Descriptive Geometry.


155


tions a plane perpendicular to the rectilinear elements of
the cylinder; the vertical plane will therefore be parallel
to these elements. We suppose cutting planes parallel
to the vertical plane; their intersections with the cylindrical surface will be right-line elements of this surface
and their intersections with the sphere will be circles
whose centres will have their vertical projection in the
same point with that of the centre of the sphere. The
radii of these circles may be easily found from the discussion in article 89.
The details of these constructions will be mostly left
to the reader, as they will be found easy when the principles already discussed are well understood.
In the present case we draw straight lines G' I', gi', (fig.
35) in the horizontal plane, parallel to the ground line. Fig. 35.
These lines are the horizontal traces of the auxiliary planes,
which are supposed parallel to the vertical plane of projection; g' i' is the radius of the circle which is the section made by one of these planes in the sphere whose
centre has its horizontal projection in E' and its vertical
projection in E". Thie points P',, P'3, where the line
g' i' meets the base of the cylinder, are the horizontal
projections of two rectilinear elements of this surface
which are met each in two points by the circumference
of the circle which is the section of the sphere by this
auxiliary plane whose horizontal trace is g' i'. If, therefore, from the point E" as a centre and with a radius
equal to g'i', we describe a circumference, the points
P",,,  3 P"1, p' 3, where it meets the vertical projections of the rectilinear elements just mentioned, will belong to the intersection of the cylinder and sphere.
By a similar process we may determine as many other
points as we wish of this intersection. There is, however, one circumstance in the general problem under discussion, which demands particular attention; this is the
case in which the cylinder entirely penetrates the sphere,
which it will do if the radius of a circular section of the
cylinder is less than the radius of the sphere minus the
distance of the axis of the cylinder from the centre of the
sphere. In this case there will be two intersections of
the surfaces in question, the one where the cylinder enters the sphere, and the other where it emerges from it.
The horizontal projections of these two curves will of
course be the same; and it will be readily seen that the




156


Descriptive Geometry.


Fig. 35. parts P",, P"3, P"4, P"2, belong to the vertical projection
of the curve, where the cylinder enters the sphere; and
that p"1, p"3, "4, P"2, belong to the vertical projection
of the curve where the cylinder emerges from the spherical surface.
The following figure represents the case in which the
Fig. 36. distance of the axis of the cylinder from the centre of the
sphere is greater than the difference of the radii of the
sphere and cylinder. In this case the cylinder does rot
entirely penetrate the sphere; there is therefore but one
intersection of the two surfaces.
92. PROBLEM. To find the projection of the curve
made by the intersection of the sphere and the cone.
Suppose the cone to be cut by planes passing through
its apex, and perpendicular to the horizontal plane. The
sections thus made in the conical surface, will be straight
lines which are easily determined; the sections which
these auxiliary planes make in the sphere, will be circles
whose centres and radii are also easily found (89).
Fig. 37.  Let S' K' (fig. 37) be the horizontal trace of one of these
cutting planes. Instead of revolving about this trace to coincide with the horizontal plane, we suppose it applied
to the vertical plane DAB by bringing the line S'K' to
AB, so that the point S' will fall upon S; then by taking
S k equal to S'K' we can draw the straight lines S"k,
which will be the sections of the conical surface by the
cutting plane.  Making mg equal to S'G', the point g
will be the position of the centre of the circle in which
the sphere is met by the cutting plane, and whose radius
is g h equal to G'H', (89).
The pointsp where the circumference of this circle
meets the straight lines S" k, belong to the intersection
of the proposed cone and sphere.  There are four of
these points; two where the cone enters the sphere, and
two where it emerges from it.
The points p are situated in the cutting plane. To
have their projections, we must take S'P' equal to p i,
and the point P' will be the horizontal projection; by
drawing P'P" perpendicular to AB, the vertical projection P" will be found at the meeting of this straight line
with p i; for the point p being taken in a vertical plane,
is at the same height with its projection upon every other
vertical plane.




Descriptive Geometry.


157


That we might not render the figure two complicated Fig. 37
we have performed this process upon only one of these
points P; it is readily seen, however, that the process is
applicable to the other three.
93. PIROBLENi. To construct the intersection of two
cones.
Suppose that the proposed cones have their bases in
the same plane, or (which is the same thing) that we
know the curvilinear section made in each of these cones
by one of the co-ordinate planes, the horizontal, for example. The problem may always be reduced to this
state.
Now imagine a plane passing through the line which
joins the summits of the proposed cones, and turning
about this line; this plane in each of the positions where
it meets the cones will cut each of them in two rectilinear elements of their surfaces; and as these are all in
one plane, the two which belong to the first cone, will
meet those which belong to the second; and these points
of meeting will be points in the intersection required.
Let (S', S"), (s', s"), (fig. 38) be the summits of the Fig. 3s.
cones, F'F' and ff' the curves which are their intersections with the horizontal plane, E the point where the
line joining their summits meets the horizontal plane; it
is evident that the cutting plane, in all its positions, will
pass through this point.
We now draw the straight line E'F' at pleasure, but
in such a manner that it will meet the bases of the cones,
and consider this line as the horizontal trace of the cutting plane.
We next construct the projections of lines drawn from
the points F' to the summit of the first cone, and from
the points f to the summit of the second; these lines are
respectively the projections of rectilinear elements of the
conical surfaces proposed, situated in a plane which
passes through the straight line which joins the summits
of the cones, and through the straight line F'F'; their
points of meeting marked upon each of the co-ordinate
planes by the figures 1, 2, 3, 4, will be points in the intersection required.
The figure represents a case in which one of the cones
entirely penetrates the other. Two of the four points
found by the preceding construction, belong to that part
14




158


Descriptive Geometry.


of the curve where the penetrating cone enters the other,
and the other two points belong to the part where the
first cone emerges from the second.
94. To find the intersection of a cone and cylinder,
we imagine a straight line passing through the summit
of the cone parallel to the axis of the cylinder; then
planes passing through this line will cut the proposed
conical anti cylindrical surfaces in their recti-linear elements.
95. To find the intersection of two cylinders, it will
be necessary to cut these bodies by planes parallel to
their axes; if the cylinders have their bases upon the
same plane, the construction will be analogous to that
given above for the cones.
We first determine the horizontal traces of the cutting
planes, which we do by passing through any point two
straight lines parallel respectively to the axes of the two
cylinders; the horizontal trace of the plane of these lines
will of course be parallel to the horizontal traces of the
cutting planes. We may draw as many of these traces
as we wish; the points where they meet the periphery
of the base, will be points in the rectilinear elements according to which these surfaces are supposed to be cut
by these auxiliary planes; these right lines being constructed will give, by their mutual intersections, points in
the required intersection.
SECTION V. Linear Perspective.
96. Perspective is a science which teaches us to represent upon any surface whatever, the outline of objects,
such as they appear when viewed from any given point.
Light in passing through a homogeneous medium
micVes in straight lines; and objects become visible by
means of the rays of light which proceed from their surfaces to the eye.  These rays, by their inclinations
among themselves, determine the images of bodies.
Thus we perceive the contour, or apparent outline, of
'ii. 39. the quadrilateral ABCD (Jig. 39) because from each
point of it a ray of light is conveyed to the eye. It is
manifest that these rays, taken together, constitute the




Descriptive Geometry.


159


pyramid formed by the lines drawn from the different in.:',,
points of the object to the eye.*
Let O-ABCD represent this pyramid of rays, the summit 0 being the place of the eye. Each of these rays or
rectilinear elements of this pyramid, must appear to the
eye but a single point. If, therefore, this pyramid be
cut by a plane or by any other surface, this section will
exhibit to the eye at 0 the same outline as the quadrilateral ABCD. It is not necessary, therefore, in order to
give us the sensation produced in us by the organ of vision, that the object itself should be presented to the
eye; it is sufficient for this purpose, to determine an assemblage of rays disposed in the same manner respectively as those which pass to the eye from the different
points in the object.t
Hence we can represent objects upon a plane; for if
we conceive the pyramid formed by the assemblage of
rays transmitted from different parts of the object to our
eye, to be cut by a plane, an image would be formed
which would represent the contour of the body and the
relative position of its differents parts.
It follows from what precedes that the determination
of this image depends entirely upon finding the intersections of the lines proceeding from the eye to different
conspicuous points of the object, with the plane or surface on which it is to be represented.
This surface is called the picture or plane of delineation. The respective positions of the eye, the picture,
and the object, must be determined, in order that the
image may be determined.
The knowledge of the true form and dimensions of
* This supposes the object to be either white or colored, but not black; for in that case it would be perceived only by the absence of light; thus we might say
that the pyramid was determined by the absence of
rays from the space occupied by the quadrilateral.
t It is evident that a perspective of the object would
also be formed, supposing the visual rays produced beyond it and extended until they meet the plane situated
behind it; the image in this case would be greater
than the object.




160


Descriptive Geometry.


the body which we wish to represent, will give us the
projections of the conspicuous points which determine
its contour, and the situation of the parts which compose it.
The problem will then be reduced to finding upon the
plane of delineation, the image of each of these points,
or in other words, the meeting of a given straight line
with a given plane.
We shall discuss some of the different cases which
the problem presents.
97. PROBLEM. To find upon the plane of the picture,
situated in any manner whatever, the appearance or the
perspective of a point given in space.
Take the vertical projection of the proposed point,
on a plane perpendicular to the common intersection of
the plane of the picture with the horizontal plane. Let
Fig. 40 T'AT" (fig. 40) be the plane of the picture; 0' and 0"
the projections of the eye 0; P' and P" those of the
point P which is to be put in perspective; O'P' and O"P"
will be the projections of the visual ray OP.
The meeting p of this line with the plane of the picture, will determine the perspective sought, which may
be found by article 33; but as this point must be constructed upon the plane of the picture, the projections p'
and p" are not sufficient. This meeting of the visual
ray with the plane of delineation is called the perspective of the point from which the ray emanates, and is
really an oblique projection of this point upon the plane
of the picture. As the lines by which the several points
in the object should be projected upon this plane, must
converge to the point in which the eye is situated, this
projection is called a perspective projection.
We therefore draw p'p through the horizontal projection of the required point p, and perpendicular to the horizontal trace of the plane of the picture. We now have
the distances Ap and A p" of the required point from
the two lines AT" and AT' perpendicular to each other
in the plane of the picture.
The line AT', which is ths intersection of the plane
of the picture with the horizontal plane, is called in perspective the base line; it is considered as limiting the
bottom of the picture, and as exhibiting the ground on
which the original object stands.




Descriptive Geometry.


161


93. When the plane of delineation is perpendicnlar to Fig. 40.
the horizontal plane, as the plane T'A t", then the projections O'P' and O"P" themselves determine, by their
intersections with the lines TA and t"A, the distances
Aq' and A q" of the perspective q from each of these
straight lines.
We take for example a pyramid (fig. 41) of which Fi.41..
the four triedral angles have their summits projected at
the extremity of the rays drawn from the points 0' and
0". The construction of the perspective of one of these
summits is designated by the same letters as in figure 40,
In the case where the plane of delineation is vertical,
the construction is very much simplified by taking the
plane of the picture itself for the co-ordinate vertical
plane. The eye, being supposed behind the plane of the
picture, (fig. 42) has its horizontal projection in 0'; that Fig. 42.
of the point in question is in P', and p is the perspective
of this point.
99. Remark. If the object to be represented is terminated by straight lines and planes, we may construct
its image by seeking the perspectives of the vertices of
the polyedral angles by which it is terminated; and in
order to this, it will only be necessary to repeat the process which has just been indicated. Two points will determine a straight line, and the faces of the proposed
object will be formed by a certain number of lines.
When the object is terminated by curve surfaces, no
particular point is presented by which we may determine
its form; we must first find its visible limit.  The
visible limit of a body is the curve which separates
the part which is seen from that which is not seen; it is
evidently formed by the series of points in each of which
the visual ray merely touches the surface of the body. If
we conceive a conical surface having its summit placed
at the eye, enveloping the proposed body, by touching
it, the curve of contacts will be precisely that of the
visible limit.  If we cut this cone by planes drawn
through the eye, in any manner whatever, each of them
will form in the proposed body a section to which two of
the right line elements of the cone will be tangent.
From this results a general method of constructing the
visible limit of a curve surface.
Let us suppose this surface to be cut by a series of ver14*




162            Descriptive Geometry.
Fig. 40. tical planes such as OO'P'P, (Jfg. 40) passing through the
eye; construct upon the vertical plane the projection P"X"
of each of the sections, and from the point 0" draw O"P"
tangent to this curve. Having the projections of the
visual ray, we can find as in the preceding problem, the
perspective of the point P situated upon the visible limit
of the proposed object.
100. If we suppose the eye to be situated at a'n infinite distance from the object, so that the visual rays
may be considered as parallel; having designated by a
straight line the direction in which the body must be
seen, nothing more remains to place the points in perspective but to draw from these points, lines parallel to
the given line, and to find their intersection with the
plane of delineation.
It is readily seen that in this hypothesis, the apparent
contour of the body is determined by tangents to its surface which are parallel to each other, the whole of which,
taken together, form a cylindrical surface.  To determine these tangents we choose cutting planes vertical
and parallel to the given line; and the tangents to the
vertical projections of the sections must be drawn parallel to the given line which marks the direction of the
visual ray.
This perspective has a great analogy to the orthographic projection discussed in the former part of this
introductory treatise; and may be used in constructing
problems of the same kind with those which were solved
in the second section. There is no necessity for making
these projections by perpendicular lines; in many instances the solutions would be as simple if the projections were made by oblique lines.
1 shall proceed to give some propositions which may
serve as the foundation of another method of perspective
which may be applied with great facility to bodies terminated by planes and straight lines.
101. If we draw through the eye a straight line parallel to a straight line situated in any manner whatever
with respect to the plane of delineation, the point where
this parallel meets this plane belongs to the perspective of
the straight line proposed.
In fact, all the lines drawn from the eye to the dif



Descriptive Geometry.


163


ferent points of any proposed straight line, form a plane
which, by its intersection with the plane of the picture,
determines the perspective of this straight line; but the
line 00' (fig. 43) being parallel to the proposed line and Fig. 43.
passing through the eye which we suppose in 0, is situated necessarily in this plane; therefore, the point 0'
where it meets the plane of delineation TA, belongs to
the perspective of the straight line proposed.
It is evident that the point P' where the proposed
straight line meets the picture itself, makes also a part
of its own perspective; therefore, to trace this perspective, it is sufficient to know the points where the proposed line and a line drawn parallel to it through the
eye, will meet the plane of the picture.
102. It follows from the preceding article, that the
perspectives of any number of lines parallel to each other, will all meet in a single point in the plane of the picture; this point is called in treatises on perspective, their
vanishinlg point. Only one line, indeed, can be drawn
through the eye, which will be parallel to all these others; their perspectives must therefore pass through the
point where this line meets the plane of the picture.
103. If the proposed lines were at the same time parallel to the plane of the picture, the straight line 00'
drawn through the eye, would not meet the picture; and
consequently their perspectives would be parallel among
themselves. We may be convinced, a priori, of the
truth of this proposition, by the following reasoning:
The two proposed straight lines being parallel, the planes
formed by the assemblage of rays drawn from the eye to
different points of these lines, and containing their
perspectives, have necessarily their intersection parallel
to these same lines, (El. 214) and consequently, to the
plane of the picture. The perspectives can meet only
in the points common to the intersection and to the plane
of the picture; they will therefore be parallel to each
other.
Hence we derive a very simple method of putting lines
and points in perspective.
We draw a perpendicular 00" from the eye to the
plane of delineation (fig. 44); the point 0" where it meets Fig. 44.
plane is called the centre of the picture, sometimes call



164


Descriptive Geometry.


Fig. 44. ed the point of sight. It follows from what has been
said, that the perspectives of all lines perpendicular to
the plane of the picture, must meet in this point.
We project the proposed point P upon the plane of the
picture, which we suppose to be vertical; the point P"
where this projection falls, will be that in which the perpendicular drawn from the proposed point to the plane of
the picture, meets it; and P"O" will be the perspective
of this line.
Next draw P'1I, making with AB an angle equal to
half a right angle; this will be the projection of a horizontal line drawn from the point P to the plane of the
picture and making the same angle with it; and its meeting with this plane will be at the point M" placed at a
height MM" equal to P'P. But if we take, upon O"D"
parallel to AB, a magnitude O"D" equal to the distance
00" of the eye from the picture, it is evident that the
line OD" will be parallel to all the horizontal lines which
can be drawn to the picture, at an inclination of 450, in
the direction MP'; consequently the perspectives of
these lines must all meet in the point D", whichis called
the point of distance. Having drawn M"D", this straight
line must contain the perspective of the point P; but
this perspective must also be found under O"P"; it is
therefore in R".
From the three preceding articles we derive the following rules.
(1). When an original straight line is parallel to the
base of the picture, the perspective of this line is also parallel to the base of the picture (103).
(2). The perspectives of all straight lines perpendicular to the picture, are directed towards the centre of the
picture (102).
(3). Theperspectives of all horizontal straight lines
which make with the picture an angle of 45~, will meet in
the point of distance (103).
(4). The perspective of a horizontal straight line making any angle whatever with the picture, will have its
vanishing point in the horizon, at the intersection with
the picture of a straight line drawn parallel to the proposed line, through the eye.
(5). The horizon of the picture is the vanishing line of
the perspectives of all horizontal planes.
(6). And  by  analogy-The vertical line passing




Descriptive Geometry.


165


through the centre of the picture, is the vanishing line of
all vertical planes perpendicular to the plane of the picture.
(7). The perspectives of planes parallel to the picture,
can have no vanishing lines, and are always fgures similar to their originals.
104. We can find the perspective of objects by means
of the receding scale, which dispenses with tracing their
ground plan, and elevation; it is constructed in the following manner.
We refer the objects to three rectangular co-ordinate
planes, the first horizontal, and passing through the ground
line AB (fg. 45); the second vertical, perpendicular to Fig. 45.
the plane of delineation, and passing through the edge
BT; the third, the plane of delineation itself, ABT,
which we here suppose to be vertical. A point will then
be given when we know its respective distances from
these three planes (30). The distance of this point from
the plane of the picture will be measured upon BC; its
distance from the vertical plane passing through BC and
BT, will be measured upon AB; and its distance from
the horizontal plane, or the height of the point, will be
measured upon BT. Now, the two lines AB and BT
being in the plane of the picture, it is sufficient to transfer to this plane the divisions of the third line, BC; this
is done by drawing to the centre of the picture the line
BO" which will be the perspective of the line BC, and by
drawing to the point of distance D", the straight lines
a D", 1 D", 2 D" &c. which will cut BO" in the points
c, 1, 2, 3, &c. corresponding to the parts B b, b 1,
1 2, &c. of the line BC.
The line BO", thus divided, is the receding scale which
marks the apparent sinking of objects in the picture; and
if we draw through the points of division, of this scale,
lines parallel to AB, they may be considered as the
ground lines of several planes drawn parallel to the plane
of the picture at the depths marked by the corresponding
divisions of the scale; they would contain the perspectives of horizontal projections, or the bases of objects situated in these planes.
If we then take upon the straight line AB, which is
called the front scale, a part Be equal to the distance of
the proposed point from the vertical plane passing through




166


Descriptive Geometry.


Fig. 45. BC and BT, and draw to the centre of the picture the
straight line e 0", the meeting of this line with g 2 parallel to AB, will give the perspective of the horizontal
projection, or of the base of the proposed object.
Finally, if upon BT, the scale of heights, we take the
part ef equal to the height of the proposed point, and
drawf O", this last straight line will meet g h perpendicnlar to g2 at the point h which will be. the perspective
sought.
We see that this process gives, by operating directly
upon the plane of delineation, the perspective of all the
objects which we may wish to represent, when we have
constructed the receding scale.
105. When the picture is so large as to render the
construction inconvenient, we may calculate the divisions of the receding scale, by considering the similar
triangles 0" c D" and a c B; fiom  which we have
aB     O"D" aB+O"D" _ aB        a B+ O"D"    a B
Be — "c' Bc+O"e -B '               BO"    - Be
The division of this scale gives the distances of straight
lines which represent the base lines of perspective planes
parallel to the picture. The heights khg are also calcuef    e O"
lated by a similar proportion, since -  -0g,, and the
straight lines e 0" and g 0" are evidently to each other
as the distances BO" and 0" 2.
a B   0"D"
The proportion  - -,-  will give the distance
O"D" of the eye from the picture, when we have given
the straight line BO", the space aB, and its perspective B c.
105. General Remark. We have, in what precedes,
the general means of putting into perspective the apparent outlines and the remarkable points of objects; but
these processes, which constitute linear perspective, are
by no means sufficient to give a complete representation
of a body.  Light and shade, and the gradations of
tint, all concur to represent the prominences, depressions, and distances of objects. All these circumstances may be rigorously determined by methods analogous
to those which have been given. To do this, it is only
necessary to make such an analysis of the enunciation of




Descriptive Geometry.


167


the question as will enable us to discover the mathematical conditions upon which it depends.
With respect to shadows, for example, if the luminous
body is reduced to a point, it is evident that they will be
determined by the space comprehended by a conical surface tangent to the opaque body, and having for its summit the luminous point. Consequently, to determine the
shadow thrown upon any surface whatever, is to find the
portion which this cone embraces of the surface in question, that is, the portion circumscribed by the curve
which is the intersection of this cone with the proposed
surface.
We cannot here discuss those considerations which
require knowledge foreign from Geometry; they are mentioned only to show of what use, in the arts, may be a
familiarity with the geometry of space.
Spherical Projections.
107. IN spherical representations for geographical, astronomical, and nautical purposes, a perspective of the
outline with other remarkable lines and points in the
body is made upon a plane supposed to pass through the
centre of the sphere and perpendicular to a straight line
drawn through this centre to the eye. This section of
the sphere by the plane of delineation is called the primitive circle of the perspective projection. The plane itself is called the primitive plane.  For these purposes
two kinds of projections are used, the orthographic and
the stereographic.
108. Orthographic Projection. In the orthographic
projection the eye is supposed to be situated at an
infinite distance from the body to be projected; so
that the rays of light which proceed to the eye from
the various points in the body, may be considered
as making an infinitely small angle with each other;
that is, they are parallel. And the sphere being transparent as well as the plane of delineation, these rays will
project orthographically upon this plane, the various lines
and points belonging to the farther hemisphere.
108. As all great circles perpendicular to the plane of
the pers),ctive, are in planes passing through the eye,




168


Descriptive Geometry.


they will be projected in straight lines drawn through the
centre of the primitive.  And every small circle perpendicular to the primitive, will have for its projection, that
chord of the primitive which results from the intersection
of the plane of this circle with the plane of the primitive, (76).
109. All circles of the sphere, which are parallel to the
primitive, have for their projections, circles equal to themselves; as the visual rays proceeding from the original
circle to the eye, form a cylinder of which the circle and
its perspective are parallel sections, (El. 275).
110. [The orthographic projections of a circle inclined
to the primitive, will be an ellipse (77) whose transverse
axis is equal to the diameter of the circle, and whose
conjugate axis is equal to this diameter multiplied by the
cosine of the inclination.  If the circle be a great circle, the centre of the eliptical projection will be in the
centre of the primitive.
Any ordinate of the ellipse will be equal to the corresponding semi-chord of the circle multiplied by the cosine of inclination.]
111. PROBLEM. To find the conjugate axis of the orthographic projection of an inclined great circle.
Let the arc M m" measure the inclination of the proFig. 46. posed circle to the plane of the primitive (fig. 46); draw
m"m parallel to the horizontal diameter of the primitive
(which is also the transverse axis of the ellipse) till it
meets the vertical diameter in m; take O m' equal to O m,
and m m' will be the conjugate axis required.
To find the ordinate on, corresponding to the semichord o N; make the angle N o n" equal to the inclination of the proposed circle to the plane of the primitive;
take o n" equal to o N, and draw n"n parallel to A o; o n
will be the ordinate required. By finding a sufficient
number of ordinates the curve may be approximately
traced.
112. Stereographic Projection. The stereographic
projection of a sphere is a perspective in which the eye
is situated at the pole of the primitive circle. The sphere
and the plane of delineation being here also supposed to
be transparent, the various lines and points upon the
farther hemisphere, will be projected upon the primitive




Descriptive Geometry.


169


circle by the visual rays passing from the several points Fig. 46.
to the eye; and the lines and points of the nearer hemisphere will be projected upon the same plane beyond
the circumference of the primitive.
113. All great circles perpendicular to the primitive,
as they pass through the axis of the primitive, will have
their planes pass through the eye, and will therefore be
projected in straight lines passing through the centre of
the primitive.
114. Every circle parallel to the primitive will have a
circle for its projection; for the visual rays, in this case,
constitute a right cone whose base is the original circle
and of which the projection is a section parallel to the
base (El. 268).
As the axis of this cone of rays passes through the
centre of the primitive, The centre of the projection of
circles parallel to the primitive, will be in the centre of
the primitive circle.
115. PROBLEM. To find the distancefrom the centre of
the primitive, of the prqjection of any point whose distance
from the farther pole is given.
Let AP'BO' (fig. 47) be the horizontal projection of the Fig. 47.
sphere; AB the horizontal trace of the plane of delineation, or the horizontal projection of the primitive circle;
and M' the horizontal projection of the proposed point
(supposed in the circumference of the horizontal great
circle), 0' being the horizontal projection of the eye.
Draw O'M' meeting AB in the point m; and C' m will
be the distance of the perspective of the proposed point
from the centre of the primitive circle.
116. [The angle P'O'M' is measured by half the arc
P'M' (El. 116); and by drawing the arc C'E it will be
perceived that C' m is the tangent of this angle, that is,
the tangent of half the arc P'M', which measures the distance of the point M from the farther pole and is called
the polar distance of tlle point.  We therefore say,The projection of eme ry pilntl in the surface of a sphere,
is at a distance firwl t he rcentre of the primitive circle,
equal to the tano,'ent /i i half the polar distance of that
point.
117. This gives us tie radius of the projection of
every circle parllid t,,I(!,pri!itive; for every part of
its circumference hc.,irl  H;t th.li same distance from the
1i)




170


Descriptive Geometry.


Fig. 47. pole, The tangent of half the polar distance of a circle
parallel to the primitive, is the radius of the projection
of that circle.]
118. PROBLEM. To find the projection of a circle inclined to the plane of the primitive.
Let the circle in question be a small circle; the visual
rays proceeding to the eye from the several points in this
circle will constitute a scalene cone. The intersection
of this cone with the plane of the primitive will be the
projection required. Suppose the sphere to be so situated that the plane determined by the axis of the cone
and the axis of the primitive circle shall be horizontal;
Fig. 48. and let figure 48 represent the horizontal projection of
the sphere, AB being the horizontal trace of the plane
of the picture, and M'N' the horizontal projection of the
circle proposed.
The angle n m O', made by the two chords AB and
O'M', is measured by half the sum of the arcs AO', BM',
(El. 115) or (as AO' is equal to BO') by half of the arc
O'M'; but this is the measure of the angle O'N'M' (El.
116), therefore the angles n m 0' and M'N'O' are equal.
The angle N'O'M' is common to the two triangles nmO'
and N'M'O'; and having two angles of the one equal to
two angles of the other, the other angles must be equal,
that is, the angle O'M'N' is equal to the angle O' n m;
and this intersection of the cone of rays by the perspective plane, is a sub-contrary section, and is therefore a
circle, the base of the scalene cone being a circle.
To show that this section is a circle, suppose a section
parallel to this, made by a plane whose horizontal trace
is n' m', and whose intersection with the base of the cone
is a vertical line having its horizontal projection in the
point c'. This line is a common chord of the two curvilinear sections of this cone; and as the section whose
horizontal projection is M'N' is a circle, the square of
half this common chord is equal to the product of the
two segments of the diameter; that is, equal to c' N' X
c'M', (El. 126).
The angle d m' M' has its sides parallel to those of the
angle c m O' which was shown to be equal to c'N' n';
therefore the angle c'm' M' is equal to c'N'n'; and as
the two angles at c' are vertical angles, the two triangles
are equiangular and therefore similar, and give the pro



Descriptive Geometry.


171


c'N'   c' m'
portion -  = -; and, by multiplying by the de- Fig. 48.
nominators, we obtain c' N' X c' M'  c' m' X c' n'. This
last product, therefore, equals the square of half the common chord of the curves; this last curve is consequently a circle; and as this is a section parallel to the
perspective in question, the perspective is therefore a
circle. And we say, —The stereographic projection of a
small circle inclined to the plane of the primitive, is also
a circle.
119. If the circle in question were a great circle, the
line M'N' would pass through the point C'; and a process similar to the above would show, that the intersection by the plane of the primitive circle, of a cone of
which this is the base and O the summit, would be a circle. We thence conclude that,- The stereographic projections of all circles of the sphere are circles.
120. As the process in article 118, gives us the two
extremities m and n of the diameter of the projection of
the inclined circle proposed; c the middle of m n will be
the centre of the projection, and c m the radius. We
can therefore find the projection of any circle inclined
to the plane of the primitive.
121. If the small circle be perpendicular to the primitive, so that its horizontal projection may be EF (fig.
47); the polar distance O'E or P'F will give two points of Fig. 47.
the projection, E and F; and by drawing the straight
line O'F we obtain the distance g C' from the centre of
the picture to the nearest point of the curve; which take
from the centre of the primitive, on the horizontal line.
Having now three points in a circular curve, the curve is
readily drawn, (El. 110).
122. We have, in figure 49, an orthographic projec- Fig. 49.
tion of the sphere upon the plane of the equator. The
meridians, being perpendicular to the plane of the projection, are straight lines; the parallels of latitude are
circles concentric with the primitive.
In figure 50 we have a stereographic projection of the Fig. 50.
same. The meridians, as they pass through the eye, are
projected in straight lines; all the other cir. — i i'ines
upon the sphere have circular curves for their?rojections.




172


Descriptive Geometry.


Fig. 51.  123. Figure 51 exhibits an orthographic projection of
the sphere upon the plane of a meridian; this gives all
the other meridians elliptical curves, and all the parallels
Fig. 52. of latitude straight lines. In figure 52 we have a stereographic projection of the sphere upon the plane of a meridian. As the eye is supposed to be at the intersection
of one of the meridians represented and the equator, the
projections of these are straight lines. All the other circles are projected in circular curves.
Remark. It will be perceived that the several portions
of a spherical surface are not represented in their proportional magnitudes by either of these projections; the
orthographic having the parts most crowded near the
circumference of the primitive; and the stereographic
being most crowded near the centre of the primitive.
These projections, notwithstanding these imperfections, serve for the representation of astronomical phenomena, and the construction of astronomical and nautical
problems. But in those geographical representations in
which a hemisphere is to be exhibited at once, a method
called globular projection is generally used. This is not
indeed a projection; it is a construction, made by dividing the diameter which represents the equator, into equal
parts to represent the same number of degrees of longitude, and drawing meridian circles through these divisions and the poles; and by dividing the polar diameter into equal parts, and also the semicircles on each
side of it into the same number of equal parts, and drawing circular curves through the corresponding divisions
Fig. 53. to represent the parallels of latitude, (fig. 53).
There are other methods of representing portions of
the earth's surface, but we cannot go farther in this subject. Our object has been to give merely an introduction to Descriptive Geometry; a sketch or outline of
some of the more practical, and thence more generally interesting subjects, which this science instructs us to discuss.


THE END.




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