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MANUAL
PLANE GEOMETRY,
ON THE HEUKISTIC PLAN,
WITH NUMEROUS EXTRA EXERCISES, BOTH
THEOREMS AND PROBLEMS, FOR
ADVANCE WORK.
BY
G. IRVING HOPKINS,
INSTRUCTOR IN MATHEMATICS AND PHYSICS IN HIGH
SCHOOL, MANCHESTER, N.H.
BOSTON, U.S.A.:
D. C. HEATH & COMPANY.
1891.
COPYRIGHT, 1891,
BY G. I. HOPKINS.
TYPOGRAPHY BY J. S. GUSHING & Co., BOSTON, U.S.A.
PRESSWORK BY BERWICK & SMITH, BOSTON, U.S.A.
PREFACE.
THIS book is published primarily for the author's pupils,
and secondarily for that constantly increasing number of
teachers who are getting more and more dissatisfied with
the old methods of teaching geometry, but who have hitherto
found no manual suited to their needs. That the reasoning
faculties of a child or youth are developed by use, goes
without saying; and consequently the method of teaching
geometry whereby the pupil originates the demonstration, or
simply demonstrates, rather than memorizes the demonstrations
of another, needs no defence at my hands. This manual has
stood the test of three years' work in the class-room, and
differs from other geometries that provide for original work,
in that the original demonstrations and constructions are
not side issues, so to speak, but are required of the pupil in
that sequence of theorems and problems which may be called
the regular required work, and of which complete demonstra-
tions are usually given in other manuals for the pupil to
memorize. In this work demonstrations are given only where
the average pupil would be at a loss to know how to proceed,
and generally, as illustrative of methods, while others are
partially given and left for the pupil to complete. In other
cases, wherever the author has found that his own pupils
were not working to advantage, he has introduced sugges-
tions, of which pupils may or may not avail themselves.
iii
183668
iv PREFACE.
The old method of division into books has been abandoned,
as it serves no practical use, since different authors have made
different divisions. It serves to make the subject a continu-
ous one in the mind of the pupil, without the artificial breaks
of the old way.
The arrangement of the theorems whereby the essential
ones, together with several simple additional ones for giving
the pupil more drill, followed by a set of non-essential but
more difficult ones for advance work, the author has found
advantageous.
He has also found it more useful to the pupil to be com-
pelled to construct his own diagram, and state the converse
of the theorems he is to prove. Practical problems of com-
putation have also been given, immediately following any
given subject, so that the pupil can immediately see the
practical application of the theorems he has demonstrated.
All the problems of construction have been placed together
after the theorems, for the sake of uniformity, as they form
no part of the logical sequence of geometrical truths as em-
bodied in the theorems. Many of them are of practical, as
well as disciplinary, importance, however, and so it is left
to the skilful and judicious teacher to take them up as the
interests of his pupils demand.
The author has received valuable suggestions from Pro-
fessor E. L. Richards of Yale College, and also from Profes-
sor T. H. Safford of Williams College. He also desires to
publicly express his indebtedness to Hon. J. W. Patterson,
Superintendent of Public Instruction for New Hampshire,
and E. R. Goodwin, Principal of the High School at Lawrence,
Mass., for their hearty and outspoken indorsement of his
PEEFACE. V
work, and encouragement to persevere in elaborating the
method.
Finally, thanks are due the publishers and printers for
the excellence and beauty of the mechanical work.
In conclusion, the author would be glad to receive sugges-
tions from those teachers into whose hands this manual may
chance to fall, with a view to its improvement as a regular
class text-book.
G. I. H.
MANCHESTER, N.H.
May, 1891.
TABLE OF CONTENTS.
PAGE
DEFINITIONS 1
GENERAL AXIOMS ~~&
PARTICULAR AXIOMS 7
SYMBOLS 10
ABBREVIATIONS 11
THEOREMS 12
ANGLE MEASUREMENT 15
TRANSVERSALS 20
TRIANGLES 24
ADVANCE THEOREMS 32, 35, 52, 80, 94, 117
QUADRILATERALS 33
CIRCLES 38
RATIO AND PROPORTION 54
PROPORTIONAL LINES 68
POLYGONS 71
PROBLEMS OF COMPUTATION 72, 82, 95, 119
SIMILAR FIGURES 74
SIMILAR TRIANGLES 74
PROJECTION 78
AREAS 84
REGULAR POLYGONS AND CIRCLES 98
MAXIMA AND MINIMA 124
PROBLEMS OF CONSTRUCTION 133
PLANE PROBLEMS 133
OPTIONAL PROBLEMS FOR ADVANCE WORK 143
REQUIRED PROBLEMS OF CONSTRUCTION 147
vii
Vlil TABLE OF CONTENTS.
MISCELLANEOUS PLANE PROBLEMS FOB ADVANCE WORK: PAQB
I. Triangles 155
II. Quadrilaterals 158
III. Circles 159
IV. Transformation of Figures 160
V. Division of Figures 163
APPENDIX.
THEORY OF LIMITS 170
SYMMETRY , 175
THEOREMS ON SYMMETRY... . 177
UNIVERSITY
INTRODUCTION.
GEOMETRY is highly important, and growing in importance
as a branch of mathematical study. This is not only true for
mathematicians ; but, what is extremely interesting to teachers,
to practical men also. The vast development of machinery, of
steam-power, of the applications of electricity, bring chapters
of mathematical science into every-day use which have long
been employed in physical investigation, but never before in
the arts of life; and in the construction of all kinds of
machinery and industrial devices the geometrical representa-
I tion to the eye is of far more immediate and practical value
than abstract calculations. The men who can draw are rapidly
gaining on those who can merely calculate. In such matters
as statistics the old processes are even reversed. We have not
only an application of arithmetic to geometry, but geometrical
representations of arithmetical results ; there is not merely an
algebraic geometry, but a graphic algebra.
In manual training schools geometry is fully as important a
branch of mathematics as arithmetic, even for the future
mechanic. The great defect in American mathematical train-
ing has been that arithmetic and algebra have been too much
favored as against geometry. Teachers have delayed, and
still delay, presenting even the elements of geometry till a
great deal of algebra has been mastered; the meagre facts
which must be stated before even mensuration can be intelli-
gently treated have been reduced to the smallest compass and
the most mechanical shape ; and it is only because we confuse
the difficulty of the subject with our stupid ways of teaching
it that we tolerate the geometrical ignorance of our pupils.
ix
X INTRODUCTION.
There are skilful mathematicians who are unaware how
much better geometers early training would have made them,
if the geometrical side of things had had fair play in their
education. But a reform is impending; and Mr. Hopkins's
text-book here presented is intended to promote it. I desire
to call the attention of all earnest and progressive teachers to
the heuristic method as here expounded.
The word heuristic is derived from the Greek ; it means the
method of discovery. Inventional is another word which has
been somewhat similarly used; but, I think, in a narrower
meaning.
In mathematics the " heuristic" is the same as the " develop-
ment " method ; that is, the method by which the pupil is led
to see the theorems and their demonstrations for himself.
In attempting to use this book, the ordinary laws of good
teaching must be followed. Consequently the pupils, however
mature, must possess all the prior qualifications ; they must be
intelligent as to the subject-matter. The greatest difficulty in
now teaching geometry by any method lies in that neglect of the
elements to which I have before alluded. If a pupil reaches
the age of fifteen (as the pupils of the Massachusetts grammar
schools are said to do) without thorough and systematic train-
ing in the elements of the subject, nothing remains but to pre-
fix a course of instruction in these elements to the study of
demonstrative geometry. A very skilful and celebrated teacher
was about 1875 mentioned to me by his pupils as using the
heuristic method; and I at once wrote him to inquire how he
presented the elements. His answer was that about six weeks
were spent in working up, orally, the doctrine of form; that
important part of geometry whose results are contained in the
definitions and other preliminary matter. This was in America,
with pupils who had passed the grammar school without learn-
ing these elements ; and represents what seems to the writer
to be the minimum of attention to be given to this part of the
subject.
INTRODUCTION. xi
On the other hand, the Austrian higher schools spend four
years (one and one-half to two hours weekly) in that empirical
form of geometiy in which attention is given rather to acquir-
ing a knowledge and practical use of the subject (both plane
and solid) than to its logic as a strictly scientific branch of
study. These boys are from eleven to fifteen years old, on the
average, and correspond to our grammar school pupils in age
and general maturity. Their instruction, so far as I can judge
from the text-books and the books of direction to the teachers,
is almost entirely heuristic in character ; and a former pupil
of mine of Austrian birth (who is now an eminent American
professor of an ancient language) confirmed this, and assured
me that then (twenty years ago) this method was actually
employed in his training.
In the earlier stages of such a course, immediate inspection
shows the simpler geometrical truths with full conviction and
ready acceptance j precisely as the pupil learns that seven times
nine equals nine times seven, not by algebraic demonstration
(which involves several steps), but by practical experience.
Mathematicians in the higher branches make great use of
intuitional proofs, supplemented if need be by logical demon-
stration; and in the lower the same law holds good on the
geometrical side as well as on the arithmetical.
The beginner in geometry should not at first be required to
perform all the geometrical operations at once, nor with great
rapidity. The careful series of objective illustrations of geo-
metrical form, combined with practical exercises and simple
reasonings, cannot, under our present programme, receive all
the expansion which it has in Austria. Our teachers are com-
pelled to deal with their pupils in a short time, and to teach
them to geometrize in comparatively few lessons ; and so the
time to practically carry out the heuristic method must be
obtained by economizing opportunities.
But the old-fashioned method of memorizing the whole book,
definitions, propositions, corollaries, scholia, even the numbers
Xll IN TE OD UCTION.
which were prefixed to these truths to indicate their order, is
happily on the decline. It is a wasteful method, and commu-
nicates far more information than can be permanent or useful;
and it lays a heavy burden on the memory. When the heu-
ristic method takes its place, the pupil will first of all be
brought, by successive steps of abstraction, from the study of
geometrical models, to the idea of geometrical solids, surfaces,
lines, points ; they will then be taught the generation of lines
by the motion of points, the various classes of lines, the sim-
plest figures bounded by straight lines, triangle, square, rec-
tangle, rhomboid, trapezoid, and trapezium. The analysis
should be directed to the cube, the other four regular solids,
the ordinary solids bounded by straight lines. From these
the idea of plane angles in their various classes should be
obtained; the difference and relationship between plane and
diedral angles should be insisted upon. In what does an angle
011 the blackboard differ from that formed when a book or a
door is opened ?
The doctrine of form should be extended somewhat beyond
the things mentioned in plane geometry; the reciprocal relation
between the cube and the octaedron (each has as many faces
as the other has vertices ; the number of their edges is equal)
should be pointed out on the model and exemplified by draw-
ing one inscribed in the other. Similar relations exist between
the dodecaedron and icosaedron ; and two tetraedrons. Again,
in analyzing all convex solids bounded by plane faces the fact
should be brought out that E + 2 = F+ V', that is, the cube
has 12 edges, 6 faces, 8 vertices; 12 + 2 = 6 + 8.
A skilful object teacher, with a few models, can thus readily
develop a great many interesting truths, and thus prepare the
young people's minds for geometry; a few weeks' time can
be well spent on the definitions.
Euclid's text-book, which is on a different plan from the
modern books, begins rather more gradually; and our best
teachers imitate him in going more slowly over the early prop-
IN TR OD UCTION. xiii
ositions. It seems to me that the models bounded by curved
surfaces (cylinder, cone, sphere, frustum of a cone, and the
parts of a cone bounded by the sections, as well as the ellip-
soid) ought to be exhibited and so far analyzed as to furnish
material for the definitions of straight lines and plane surfaces,
in distinction to curved lines and surfaces. And I would go a
step farther; and show the difference between the ruled sur-
faces (like the cylinder and the cone) and those in which no
straight lines can be drawn.
In a word, I would lay the foundation for solid geometry
(which is the practical form of the science) along with plane,
which is a mere abstraction from the other. The introduction
to demonstration may well be combined with additional study
of form. Let the pupils, for instance, be taught to make tri-
angles; then to measure the sides and angles of those they
make; then to classify their triangles into equilateral, isos-
celes, scalene; equiangular, triangles with two equal angles,
with no equal angles ; then to set down a few theorems they
may expect to prove. This is no doubt the way the Egyptian
priests built up the propositions into which, as the foundations
of a secret society, they initiated Pythagoras. Eight angles,
vertical angles, exterior angles, can now be brought in natu-
rally; then triangles with a short side and two long ones;
then the idea of parallels. Any bright teacher who is pos-
sessed with the desire of training pupils to think geometri-
cally will naturally fall into the line of thinking needed; but
it all takes time.
I estimate that, when the definitions and those matters
which make up the first book of our ordinary geometries have
been fully mastered, half the work of the course in plane
geometry has been accomplished; and the teacher must bear in
mind that if this is well done, the pupil has been trained, not
crammed, to a very high point in the subject.
The modern writers on pedagogy lay down the principle
that interest is the end of teaching, and not merely the means.
XIV INTRODUCTION.
If you have brought your pupil to enter upon mathematics
with an earnest desire to accomplish the work in it, you have
done more than if you merely cram it into him. In the col-
leges it is well known that the great mass of students do com-
paratively little in the science. When they have the choice,
they prefer as a rule to take something easier. Just so most
boys never accomplish much in their athletics. The fashion
has come about that about a quarter of our students play
ball, more or less, and the rest look on ; simply because their
muscles have not been trained to the work. Now what we want
in mathematics is precisely analogous, training of the mathe-
matical muscles. The German use of " gymnasium " as the name
for those high schools which do the work preparatory to the uni-
versities schools which accomplish nearly as much at the age
of nineteen or twenty as our colleges do three years later on
is in the same line of thinking. Let us make our high schools
true gymnasia; and a good subject to begin with is geometry.
What, then, is the true exercise of the mathematical powers ?
(I had almost said of the mathematical muscles.) Grube's
numerical analysis of the numbers from 1 to 100 is the heuristic
method in arithmetic; the pupils are led on by the teacher to
analyze and find out arithmetical truths for themselves. It is
much to be wished, as Pestalozzi taught, that form might be
thoroughly taught in the primary school; Steiner, whose
geometrical ideas lie nearly at the foundation of modern
developments, was a pupil and teacher of Pestalozzi's own
school in Switzerland. But if this is not yet practicable, let
the pupil take up geometry from the beginning at a later age;
and really do fundamental and thorough work.
The teacher, in using the heuristic method, must be very
careful of his foundations; he must develop, and thoroughly
develop, his definitions, instead of giving them out as a task.
The first demonstrations must be slowly and carefully done ;
neither the Egyptian priests nor Pythagoras discovered the
early theorems all at once, or in any short time.
INTRODUCTION. XV
The hill of mathematical knowledge is high and steep, and
few go up on it any great distance. There are three ways
of ascending. The pupil may be carried up in a wagon; this
is tolerably easy, and there is some gain of fresh air and a wide
prospect. It is like the committal to memory of a text-
book. He may be dragged up as tourists sometimes ascend
the Alps; tied to a rope with a strong guide at either end.
This is still more beneficial ; but the best way of all is to be
trained to walk up for himself. At first, the trainer must
content himself with gradual development of the muscles, and
little apparent ascent; but the continual exercise finally enables
him to reach a great height with comparative ease.
In preparing pupils to pass college examinations the heuristic
method has a great practical value. Once the method of mak-
ing a demonstration is known and the few data upon which
the questionsxiepend are learned, the student who can demon-
strate originally has a great advantage over him who relies
more largely on his memory. If the latter forgets, he is lost ;
but the former can originate when memory fails, and if he has
been well trained is far less liable to lose his presence of mind.
I congratulate Professor Hopkins on the spirit with which
he has undertaken to fill a gap in our geometrical literature ;
and I hope the book will have the success which it deserves.
TRUMAN HENRY SAFFORD.
DEPARTMENT OF ASTRONOMY,
WILLIAMS COLLEGE,
August, 1891.
"HE
iVERSJTY ' \
,
y
BOOK I.
PLANE GEOMETRY.
1. Space is indefinite extension in every direction.
2. A material substance is anything, large or small, solid,
liquid, or aeriform, visible or invisible, that occupies a portion
of space.
3. It therefore follows that material substances have limited
extension in every direction.
4. For purposes of measurement, extension in three direc-
tions only are considered, called, respectively, length, breadth
(or width), and thickness; they are also called collectively
dimensions.
5. Magnitude, in general, means size, and is applied to any-
thing of which greater or less can be predicated, as time,
weight, distance, etc.; a geometrical magnitude is that which
has one or more of the three dimensions.
6. A geometrical point has position merely; i.e. it has no
magnitude.
The dots made by pencil and crayon are called points, but
they are really small substances used to indicate to the eye
the location of the geometrical point.
7. A geometrical line has only one dimension ; i.e. length.
The lines made by pencil and crayon are substances, and
1
2 PLANE GEOMETRY.
may be called physical lines which serve to show the position
of the geometrical lines.
8. A straight line is one that lies evenly between its ex-
treme points.
This is the definition as given by Euclid. The majority
of modern geometers, however, have substituted the following
as stated by Newcomb ; viz. :
" A straight line is one which has the same direction through-
out its whole length."
Each is designed to express the idea of straightness, and not
to convey it, for it is assumed that the idea already exists in
the pupil's mind prior to the beginning of this study.
9. A curved line, or simply curve, is one no part of which
is straight.
10. Material substances have one . or more faces which
separate them from the rest of space. These faces are called
surfaces, and have, obviously, only two dimensions ; i.e. length
and breadth.
11. The surface considered apart from the substance is called
a geometrical surface.
12. A plane is a geometrical surface such that if any two
points in it be selected at random, the straight line joining
them will lie wholly in that surface.
13. A curved surface is a geometrical surface no portion of
which is a plane.
14. A physical solid is the material composing it, and which
we perceive through the medium of the senses ; while the
geometrical solid is the space, simply, which the physical solid
occupies.
15. A geometrical figure is the term applied to combinations
of points, lines, and surfaces, when reference is had to their
form or outline simply.
PLANE GEOMETRY. 8
16. A plane figure is one whose points and lines all lie in
the same plane.
17. "A plane rectilineal angle is the inclination of two
straight lines to one another, which meet together, but are
not in the same straight line." EUCLID.
" An angle is a figure formed by two straight lines drawn
from the same point." CHAUVENET.
"When two straight liaes meet together, their mutual in-
clination, or degree of opening, is called an angle" LOOMIS.
18. The lines which form an angle are called the sides of
the angle, and the point from which they are drawn is called
the vertex of the angle.
19. When two plane angles have the same vertex and a
common side, neither angle being a part of the other, they are
said to be adjacent angles.
20. When two angles have the same vertex and the sides
of one are the extensions of the sides of the other, they are
called vertical angles.
21. An angle is named by a letter or number placed at its
vertex. If, however, there are two or more angles with the
same vertex, other letters are placed at the extremities of
their sides, and the three letters are used to name the angle,
the letter at the vertex always coming between the other two.
22. Let us consider the point B, in the straight line AC,
a pivot, and BD another starting from the position BC, and
j-v D T
A B C A B C
Fig. I. Fig. H.
turning about B, keeping always in
the same plane. It is evident that, A
as soon as it has started, it forms
two angles with the line AC, of which DBC, Fig. I., is the
4 PLANE GEOMETRY.
smaller. If it continue to revolve, however, it will finally
reach a position as DB, Fig. II., in which the angle DEC is
the larger. Hence, in passing from the first position to the
second, it must have reached a position, Fig. III., where the
two angles DEC and DBA were equal.
23. Hence, when one straight line meets another so as to
form equal adjacent angles, each of the angles is called a right
angle, and the lines are said to be perpendicular to each other.
24. It is also evident that the sum of the angles formed
by any one position of the line ED is equal to the sum of the
angles formed by any other position ; for what is taken from
one angle by the revolution of the line ED is added to the
other.
25. Hence, when one straight line meets another so as to
form two angles, the sum of these two angles equals two
right angles.
26. An angle that is less than a right angle is called an
acute angle.
27. An angle that is greater than one right angle and less
than two is called an obtuse angle.
28. Both acute and obtuse angles are designated as oblique
angles as contrasted with right, angles,
29. A straight angle is a term recently adopted by promi-
nent English and German mathematicians to express, by a
single unit, the sum of two right, angles.
30. When the sum of two angles is equal to a straight
angle, they are said to be supplementary > i.e. each is the
supplement of the other.
31. When the sum of two angles is equal to one right
angle, they are said to be compl&nientq/ry ; i.e. each is the
complement of the other.
UN'
PLANE GEOMETRY. 5
32. It is evident from 22 and 23 that when one straight
line meets another so as to form two angles, these angles are
supplementary.
33. Two straight lines are said to be parallel when, lying in
the same plane, and extended indefinitely both ways, they
do not meet each other.
34. As we have before conceived a line (22) to move, so
we may conceive one geometrical magnitude to be applied to
another for the purpose of comparison. If they coincide,
point for point, they are said to be equal.
35. Thus, if two angles can be so placed that their vertices
coincide in position and their sides in direction, two and two,
the angles must be equal.
36. Conversely, if two equal angles be conceived to be so
placed, one upon the other, that their vertices and one pair
of sides coincide respectively, then the other pair of sides
must also coincide, otherwise one angle would be greater than
the other.
37. Geometrical magnitudes are geometrical lines, angles, sur-
faces, and solids.
38. We shall have occasion to express the addition and
subtraction of geometrical magnitudes, as well as the multi-
plication and division of these magnitudes by numbers.
39. For example, the sum of the two lines AB and CD is
obtained by conceiving them to be placed so as to form one
continuous straight line as HK. . R
Similarly, the difference of two
lines is obtained by cutting off
from the larger a line equal to -r- z
the smaller. Similarly, HKN
represents the sum of the two angles A and B. To multiply
a line by a number is to add it to itself the required number
of times. (See above.)
6 PLANE GEOMETRY.
To divide a line by a number is to conceive the line to be
divided into the required number of equal parts.
HI
The same is true of other geometrical magnitudes. (Illus-
trations of each should be given.)
40. An axiom is a truth that needs no argument ; i.e. the
mere statement of it makes it apparent, e.g. :
GENERAL AXIOMS.
(I. The whole of anything is greater than any one of its parts.
II. The whole of anything is equal to the sum of all its parts.
/Til. Quantities which are respectively equal to the same or
equal quantities are equal to each other.
7"lV. Quantities which are respectively halves of the same or
equal quantities are equal to each other.
V. Quantities which are respectively doubles of the same
or equal quantities are equal to each other.
VI. If equal quantities be added to equal quantities, the
sums are equal.
VII. If equal quantities be subtracted from equal quantities,
the remaining quantities are equal,
VIII. If equal quantities be multiplied by the same or
equal quantities, the products are equal.
PLANE GEOMETRY. 7
/ IX. If equal quantities be divided by the same or equal
quantities, the quotients are equal.
X. If equal quantities be either added to or subtracted from
Lunequal quantities, the results will be unequal.
XI. If equal quantities be either multiplied or divided by
unequals, the results will be unequal.
41. The results obtained by the addition to, subtraction
from, multiplication or division of, unequals by unequals are
indeterminate with one exception. The pupil should ascertain
for himself this exception.
42. Particular axioms.
XII. Between two points only one straight line can be
^ drawn 5 or if others are drawn, they must coincide.
TXIII. A straight line is the shortest of all possible lines
Connecting two points,
XIY. Conversely, the shortest line between two points is
a straight line.
'""XV. If two straight lines have two points in common, they
will coincide however far extended.
XVI. Two straight lines can intersect in only one point.
XVII. In one direction from a point only one straight line
can be drawn ; or if more be drawn, they must coincide,
XVIII. Through a given ,
point (as P) only one line A P B
(as AB) can be drawn par- -
allel to another line (as CD) ;
r if others are drawn, they must coincide,
XIX. If a line makes
an angle with one of two
parallel lines, it will in-
tersect the other if sufficiently extended.
XX. The extension or shortening of the sides of an angle
does not change the magnitude of the angle.
8 PLANE GEOMETRY.
43. A theorem, is a truth which is made apparent by a
course of reasoning or argument. This argument is called a
demonstration.
Every theorem consists of two distinct parts, either ex-
pressed or implied; viz. the hypothesis and conclusion. The
conclusion is the part to be proven, and the demonstration is
undertaken only upon the ready granting of the conditions
expressed in hypothesis ; e.g. :
Hyp. If two parallel lines be crossed by a transversal,
Con. the alternate interior angles are equal.
44. In demonstrating the theorems in this book the pupil
should first analyze the theorem and write it after the above
model. For instance, let us analyze the following theorem;
viz. :
A perpendicular measures the shortest distance from a point
to a straight line.
Now this theorem, analyzed and written according to our
model, would read as follows :
Hyp. If from a given point to a given straight line a per-
pendicular and other lines be drawn,
Con. the perpendicular will be the shortest one of those lines.
45. The converse of a theorem is another theorem in which
the hypothesis becomes the conclusion, and conclusion the
hypothesis. For example, the converse of the theorem men-
tioned in Sect. 43 would read as follows ; viz. :
Hyp. If two straight lines in the same plane be crossed by
a transversal so as to make the alternate interior angles equal,
Con. these two straight lines will be parallel.
The converse of most of the theorems in this work"will be
left for the pupil to state.
46. A problem, in geometry, is the required construction of
a geometrical figure from stated conditions or data ; e.g. :
It is required to construct the triangle which has for two
PLANE GEOMETRY. 9
of its sides AB and CD, and the angle H included between
these two sides.
ZL J3
A postulate is a self-evident problem,
or a construction to the possibility
of which assent may be demanded or
challenged without argument or evi-
dence.
(Both theorems and problems are
commonly designated as propositions.)
47. Before we can accomplish the demonstration of a
theorem in geometry, the following postulates must be
granted ; viz. :
I. A straight line can be drawn from one point to any
other point.
II. A straight line can be extended to any length, or ter-
minated at any point.
III. A circle may be described about any point as a centre
and with any radius.
IV. Geometrical magnitudes of the same kind may be
added, subtracted, multiplied, and divided.
V. A geometrical figure may be conceived as moved at
pleasure without changing its size or shape.
48. A postulate is also used to designate the first step in the
demonstration whereby certain conditions or data are demanded
as fulfilled, or admitted as true, for the basis of the argument,
and which begins with "Let," etc., or "Let it be granted
that," etc. It really demands assent ~
to the general conditions implied in ^ %S X.
the hypothesis, with special reference
to a particular representative diagram
or figure ; e.g. in beginning the demon-
stration of the theorem given in Sect. ^x
43, we should say : "Let AB and HK
be two parallel straight lines crossed by the transversal C7Z>."
10
PLANE GEOMETRY.
This is the postulate ; i.e. it matters not whether AB and
HK are actually straight or actually parallel ; they stand for
straight and parallel lines,, and the argument is just as con-
clusive when based upon their supposed parallelism, as it
would be if we had positive knowledge that those two iden-
tical lines were parallel..
49. The demonstration of the following theorems, should be
written by the pupils, with an occasional oral exercise partici-
pated in by the entire class, each member in turn contributing
a single link in the chain of argument.
50. In order to save time for the pupil in writing and the
instructor in correcting, the author, having used them, recom-
mends the use of the following list of symbols and abbrevia-
tions, as well as such others as the instructor and pupils may
agree upon :
51. SYMBOLS,
4- . . . plus.
... minus.
X ... multiplied by.
= . . . equals, or is equal to.
.'. . . . therefore, or hence.
> ... is greater than.
< ... is less than.
= or X, equivalent to.
O . . . circle.
... circles.
11 ... parallel.
IPs . . parallels.
/. angle.
A . . . . . angles,
rt. Z or [R_ . . right angle,
rt. A or [R_'S . right angles.
A triangle.
A triangles.
rt. A right triangle.
rt. A . . . . right triangles.
_L perpendicular.
Js perpendiculars.
O parallelogram.
17 parallelograms.
PLANE GEOMETRY.
11
52.
ABBKEVIATIONS,
Adj adjacent.
Alt alternate.
Ax axiom.
Comp complementary,
Con conclusion.
Cons construction.
Def definition.
Dem demonstration.
Dist distance.
Ext exterior.
Hyp. .... hypothesis.
Iden identical.
Q.E.D. . . . Quod
Q.E.F. . . . Quod
Int interior.
Line straight line.
Opp opposite.
Post postulate.
Prob problem.
Pt point.
Quad quadrilateral.
St straight.
Sug suggestion.
Sup supplementary.
Trans transversal.
Vert vertical.
erat demonstrandum,
erat faciendum.
53. The last two expressions are in Latin, and mean respec-
tively, "which was to be demonstrated" or "proven," and
"which was to be performed" or "done." The former is
placed at the close of the demonstration of every theorem to
indicate that the required proof has been completed ;. while the
latter is placed at the close of the work of every problem to
indicate that the required construction has been completed.
54. The pupil must remember that every statement in geo-
metrical demonstration must be " backed up " or substantiated
by giving as authority, definitions, axioms, and previously
established truths, and every unsupported statement, whether
from instructor or fellow-pupil, should be promptly challenged.
55. Complete demonstrations are given of only a few of
the following theorems, and those solely for the. purpose of
giving the pupil a start when encountering a new- subject,
or in places where the pupil's time would not be spent to
advantage in his helpless work ; i.e. where the average pupil
would be at a loss to know how to go at the subject. In
12 PLANE GEOMETRY.
other places occasional suggestions are made for the purpose
of giving the pupil a start without loss of time. In only a
few cases have any diagrams been given, the author thinking
it best to leave them as a part of the pupil's exercise. And
right here he wishes to emphasize this caution to both in-
structor and pupil, based on several years' experience in the
class-room ; viz. : Always employ the most unfavorable diagram.
Care in this particular will save many an inadvertent error,
and prevent the assumption of conditions unwarranted by the
hypothesis.
THEOREMS.
56. Hyp. If one straight line intersects another,
Con. the vertical angles are equal.
B
Post. Let AH and BD be two straight lines intersecting
each other at point C.
We are to prove Z ACD = Z BCH (I.)
and ZACB = Z.DCH. (II.)
Dem. Z ACD + Z ACS = 2 rt. A. (Sect. 32.)
ZBCH + ZACB = 2rt. A (Same reason.)
.-. /.ACD + /.ACB = /.BCH+/.ACB. (Axiom III.)
ZACB= ^ACB. (Iden.)
/. Z ACD = Z BCH. (Axiom VII.)
Q.E.D.
The pupil should now employ a similar method and demon-
strate (II.) and the following easy ones :
PLANE GEOMETRY. 13
57. If two straight lines intersect each other, the sum of
the four angles thus formed equals four right angles.
58. The sum of all the angles formed at one point on the
same side of a straight line equals two right angles.
59. The sum of all the angles formed by any number of
straight lines meeting at one point equals four right angles.
Sug. Extend one of the lines.
Remark. In the demonstration of most theorems it be-
comes necessary to draw one or more auxiliary lines in addi-
tion to those involved in the original figure. These are called,
technically, construction lines, and should always be dotted, in
order that they may readily be distinguished from the given
lines, or those comprising the original figure.
60. If one of the four angles formed by the intersection
of two straight lines is a right angle, the other three angles
are also right angles.
61. All right angles are equal.
62. If two angles be equal, the complements of those angles
are also equal.
63. If two angles be equal, the supplements of those angles
are also equal.
64. If two supplementary adjacent angles be bisected, the
bisecting lines are perpendicular to each other.
65. From one point in a straight line only one perpendicular
to that line can be drawn in the same plane ; or if more be
drawn, they must coincide.
Sug. Consult Sects. 22, 23, and 36 ; also Theorem 61.
66. Hyp. If two adjacent angles be supplementary,
Con. their exterior sides form one and the same straight
line.
14 PLANE GEOMETET.
Post. Let ABH and HBC be 2 sup. adj. A We are to
prove that AB and BC form one and the same st. line.
A. B C
Dem. BC is, by Hyp., a st. line, therefore if it be extended
in either direction, it will still be a st. line. If now we can
prove that the extension of BG must coincide with AB, then
AB and BC must form one line.
There are only two possible relations, as regards position,
between AB and the extension of BC-, i.e. they either coincide
or they do not. Now, if we can prove the impossibility of all
the possible relations except one, whether two or more, knowing
that one of them must be true, that is equivalent to a direct
proof of the existence of that one relation, and is just as con-
clusive.
For example, we know that of the two quantities x and y,
one of three relations must be true; viz. : (1) x > y, (2) x = y,
(3) x<y. There is no other possible relation in respect to
size ; hence, if we can prove the impossibility of the first two
relations in any given case, then that is conclusive that the
third is true, and that consequently x < y.
Let us apply this method to the demonstration of this
theorem.
It is evident that AB is either the extension of BC or it is
not. Let us suppose that the latter relation is true and then
trace the consequences. If AB is not the extension of BC,
then some other line, as BD, must be. Now, if BD is the
extension of BC, CBD must be a st. line, whether it seems to
the eye to be so or not.
.-. HBC + HBD = 2 rt. A, (Sect. 32.)
but Z. HBC + /.HBA = 2rt.A-, (By Hyp.)
PLANE GEOMETRY. 15
.-, Z EEC + Z # = Z /ZBC + Z JIBJL (Axiom III.)
= Z#B(7j (Iden.)
= ZHBA. (Axiom VII.)
Let us examine this last equation. Can the equation be a
true one ? Can the Z.HBD be equal to the ZHBA ? Evidently
not, for the former is a part of the latter (Axiom I.). We
have, therefore, arrived at an impassible result. -It is plain,
therefore, that there must be error somewhere ; either in the
argument or in the supposition upon which the argument is
based. A careful and critical review of the argument dis-
closes no flaw, no error, but, on the contrary, it is in perfect
accord with previously established principles j consequently
the conclusion is inevitable that the error is in our supposition.
If, then, our supposition that BA is not the extension of BG
is erroneous, the only other possible relation must be true;
viz. that BA rs the extension of BC, and therefore they must
form one and the same straight line. Q.E.D.
67. If two vertical angles be bisected, the bisecting lines
form one and the same line.
Sug. Use Theorems 64 and 65.
68. If two pairs of vertical angles be bisected, the bisecting
lines are perpendicular to each other.
Sug. Consult Theorems 64 and 67.
69. The line which bisects one of two vertical angles will,
if extended, bisect the other.
ANGLE MEASUREMENT.
70. The most common unit for measuring the magnitude
of angles is the ninetieth part of a right angle, called a degree.
The degree is subdivided into sixty equal parts called minutes,
and the latter also into sixty equal parts called seconds. By
16 P^ANE GEOMETRY.
this means small fractions of a degree may be expressed in
minutes and seconds. These units are indicated by the fol-
lowing symbols ; viz. :
for degrees, ' for minutes, and " for seconds ;
e.g. 75 20' 34".
This is called the sexagesimal method. Another method,
called the centesimal) has been proposed in France, having
obvious advantages. But owing to the fact that all mathe-
matical tables and instruments had been arranged and con-
structed with reference to the sexagesimal method, it has not
yet come into extensive practical use.
In this method the right angle is divided into one hundred
equal parts called grades, the grade into the same number of
equal parts called minutes, and the latter into one hundred equal
parts called seconds. These are designated by symbols as
follows ; viz. :
45* 15' 28*.
How many degrees in a right angle ?
How many grades in a right angle ?
What is the complement of a right angle ?
What is the supplement of a right angle ?
How many degrees in all the angles formed at one point on
one side of a straight line ? How many grades ?
How many grades in all the angles formed by any number
of straight lines meeting at a common point ? How many
degrees ?
What is the complement of 45; 30; 1; 89 59' 59" 5
45*; 30*; 1*; 89*59'59 n ?
What is the supplement of each of the above angles ?
How many degrees in an angle that is double its comple-
ment ? How many grades ?
How many degrees in an angle that is four times its sup-
plement ? How many grades ?
PLANE GEOMET&Y. 17
What angle, in degrees and grades, do the hour and minute
hands of a clock form at 2 o'clock? At 3 o'clock? At
5 o'clock?
If one of the angles, formed by two straight lines crossing
each other, be 120, find the values of each of the other three
angles in degrees.
If two complementary adjacent angles be bisected, what
is the value of the angle formed by the bisectors in degrees ?
In grades ?
What is the complement of 37 44' 51"?
What is the supplement of 104 33' 21" ?
What is the complement of 41 28' 32"?
What is the supplement of 125^ 76' 84 V ?
An angle of 60 g is how many degrees ?
An angle of 99 is how many grades ?
One-half a right angle is what part of three right angles?
Of two right angles ?
One-fifth of two right angles is what part of one right angle ?
Of four right angles ?
How many degrees in an angle that is one-third its comple-
ment ? How many grades ?
If one of two complementary angles is acute, what kind of
an angle must the other be ?
If one of two supplementary angles is acute, what kind
must the other one be ?
71. A perpendicular is the shortest
line that can be drawn from a point
to a straight line.
(The analysis of this theorem has A. B
been given in the Introduction.) \
Post. Let AD be any st. line, and \
HC a line drawn from pt. H _L to
AD.
K
Also let HB be any other line that can be drawn from pt. H
to AD.
18 PLANE GEOMETEY.
We are to prove that HG < HB.
Cons. Extend HG from pt. G, making GK ' = CH, and join
BK.
Dem. Conceive the figure BHG to be revolved on BO as an
axis until it conies into the same plane with figure BGK.
Then the line CH will fall upon GK.
(Sect. 36 and Theorem 61.)
Pt. H will fall upon pt. K. Why ?
Line BH will coincide with line BK. Why ?
What is the relation, then, between BH and BK?
KG+GH<BH + BK, Why?
or GH+GH<BH + BH. How obtained?
Whence 2CH<2 BH, Why ?
or GH<BH. Why?
But BH represents any line that can be drawn from pt. H
to line AD other than the perpendicular CH. Hence, from
our last inequality, the perpendicular is in every case the
shorter. Q.E.D.
72. The converse of the previous theorem ; viz. :
Hyp. If several lines be drawn from a point to a straight
line, one of which is the shortest that can be drawn,
Con. the latter only will be perpendicular to the said
straight line.
Sug. Let AB be the shortest line that can be drawn from
pt. A to line CD, and from A draw a perpendicular to CD.
By Theorem 15, what must this _L be ?
How must it and line AB be situated relatively, then ?
Remarks, (a) The perpendicular distance from a point to
a line is usually termed simply the distance.
(b) When a _L is drawn to another line, this other line is
called the base of the _L ; and the pt. where the J_ meets the
base is called the foot of the _L.
PLANE GEOMETRY. 19
73. If two oblique lines be drawn from the same point in
a perpendicular to its base, so as to cut off equal distances
from the foot of the perpendicular,
I. the two oblique lines will be equal ;
II. the two angles which the oblique lines make with the
perpendicular are equal ;
III. the two angles formed by the two oblique lines and
the base are equal.
Sug. Use method similar to that in 71.
74. If a perpendicular be drawn to a line at its middle point,
and any point in the perpendicular be selected at random, this
point is the same distance from one extremity of the given
line as from the other.
75. If a perpendicular bisect a given line, and any point out-
side the perpendicular be selected at random, the distances of
this point from the extremities of the given line are unequal.
76. If two lines be drawn from a point to the extremities
of a given line, and two other lines be drawn from a point
within the first two to the same extremities, the sum of the
first two lines will exceed the sum of the other two.
Sug. Extend one of the second pair until it meets one of
the first pair.
77. If from the same point in a perpendicular two oblique
lines be drawn to the base so as to cut off unequal distances
from the foot of the perpendicular, the one cutting the base
the farther from the foot will be the longer.
Sug. In constructing diagram for this Dem. be sure that
both oblique lines are not on the same side of the J_. Consult
73 and 76.
78. Converse of 73, I.
Sug. Use method similar to that illustrated in 66 ; i.e. desig-
nating the distances cut off by x and y, then x > y, or x < y,
or x = y. Prove that the first two relations are impossible.
20 PLANE GEOMETRY.
79. Converse of 77.
JSng. Use method similar to that in 78.
80. Only two equal straight lines can be drawn from the
same point in a perpendicular to its base, or from a point to a
straight line.
81. From a point without a straight line only one perpen-
dicular to that line can be drawn j or if more be drawn, they
must coincide.
82. If two points, each of which is equally distant from the
extremities of a given line, be joined by a line, this line or its
extension will bisect the given line and be perpendicular to it.
83. If two lines in the same plane be perpendicular to the
same straight line, they are parallel.
Sug. They must either be parallel or meet. Show the
impossibility of the latter condition.
84. If a line is perpendicular to one of two parallel lines,
it will be perpendicular to the other also.
85. If two straight lines be parallel to a third, they are
parallel to each other.
Sug. Construct a perpendicular to one, and then consult 83.
TKANSVEKSALS.
86. When two straight lines in the same plane are crossed
by another, the latter is called
a transversal.
In the figure, which line is
the transversal ?
How many angles are formed?
Certain ones of the above
angles are termed interior an-
gles. Name them.
Why are they called interior
angles ?
PLANE GEOMETRY. 21
What name would you give to the others to distinguish
them from those already named ? Why ?
How many pairs of vertical angles ? Name them.
How many pairs of adjacent angles ? Name them.
Two angles situated on opposite sides of the transversal,
either both interior or both exterior, and not adjacent, are
called alternate angles.
How many pairs of alternate interior angles ? Name them.
How many pairs of alternate exterior angles ? Name them.
Two non-adjacent, non-vertical angles, of which one is inte-
rior and the other exterior, and both on the same side of
transversal, are called technically exterior interior angles.
How many pairs of exterior interior angles ? Name them.
87. Hyp. If two parallel lines be crossed by a transversal,
Con. the alternate interior
angles are equal.
Post. Let, etc. (The pupil
should give it with precision.)
We are to prove that any
pair of Alt. Int. angles, selected
at random, are equal ; e.g. :
that ZKBC = ZBCN.
\ T D
Cons. Through Q, the mid-
dle point of BC, construct R W
perpendicular to one of the two parallels, as HK.
Dem. What is the position of R W relative to NP ? Why ?
Conceive the entire figure AHBQRK to revolve about Q as
a pivot, keeping always in the same plane, until the line QA
coincides with QD.
What relation in regard to magnitude exists between the
angles BQR and WQC ? Why ?
When QA has reached the position of QD, where must the
point B be ? Why ?
What position must the line QR take ? Why ?
OF THE
UNIVER
22 PLANE GEOMETRY.
What must be the relative position, then, of the lines CW
and BE? Why? (See 81.)
What must be the relative positions, then, of the points
R and W? Why ?
What is true, then, of the sides and vertices of the two
angles KBC and BCN?
What must be the relation, then, in respect to magnitude,
between these two angles ?
Hence, etc. Q.E.D.
88. If two parallel lines be crossed by a transversal, prove
I. that the exterior interior angles are equal ; (Sug. Consult
87 and 56.)
II. that the alternate exterior angles are equal ;
III. the two interior angles on the same side of the trans-
versal are supplementary ;
IV. that the two exterior angles on the same side of the
transversal are supplementary.
89. Hyp. If two lines in the same plane be crossed by a
transversal so as to make the exterior interior angles equal,
Con. these two lines will be parallel.
Post. Let AC and JVP be two lines in the same plane
crossed by the transversal DH, making the angles DBA and
BKN equal.
We are to prove that AC and NP are parallel.
Dem. It is evident that they must occupy one of the two
relative positions ; viz. parallel or non-parallel.
PLANE GEOMETRY. 23
Let us adopt the supposition that they are non-parallel.
Then a line may be drawn through point K, parallel to AC.
Let the dotted line ST represent this line.
Then Z DBA = Z BKS. (Theorem 88 I. )
But Z DBA = Z BKN. (By Hyp.)
.-. Z BKS = Z BKN. (Axiom III.)
The pupil should finish the argument.
Bug. Consult the argument in the demonstration of Theorem
66.
90. Hyp. If two lines in the same plane be crossed by a
transversal so as to make the alternate interior angles equal,
Con. these two lines will be parallel.
-A
Post. Let AB and CD be two lines in the same plane
crossed by the transversal HQ, making Z CNK~ Z NKB.
We are to prove that AB and CD are parallel.
Dem. Z.CNK=^QND. Why?
Z CNK = Z NKB. Why ?
.-. Z QND = Z #KB. Why ?
.-. CD and .45 are parallel. (Theorem 89.)
91. Converse of 38, II., III., and IV.
91 (a). If two lines be crossed by a transversal making
I. the alternate interior or exterior angles unequal, or
II. the exterior interior angles unequal, or
III. the two interior, or two exterior, angles, on the same side
of the transversal, together less than a straight angle, the two
24 PLANE GEOMETRY.
lines will meet if sufficiently extended, and on that side of
the transversal on which the sum of the angles is less than a
straight angle.
92. If two parallel lines be crossed by a transversal, the
bisectors of the alternate interior, or alternate exterior, angles
are parallel.
93. If two parallel lines be crossed by a transversal, the
bisectors of the two interior, or the two exterior angles, on the
same side of the transversal, are perpendicular to each other.
Sug. Consult 64, 92, and 84.
94. If two angles have their sides parallel, two and two,
they are either equal or supplementary.
95. If two angles have their sides respectively perpendicular
to each other, these angles are either equal or supplementary.
Which of the above results will be true if the angles are
both acute ? If both obtuse ? If one is acute and the other
obtuse ? If both are right angles ?
TKIANGLES,
96. A triangle is a plane figure bounded by three straight
lines, and consequently has three angles. It is sometimes
called a trigon.
Triangles are named,
I. with reference to the character of their angles ;
II. with reference to the relations between their sides.
I. If a triangle has all its angles acute, it is called an acute
triangle.
If it has one obtuse angle, it is called an obtuse triangle.
If it has one right angle, it is called a right triangle.
II. If all its sides are equal, it is called an equilateral triangle.
If two of its sides are equal, it is called an isosceles triangle.
If no two of its sides are equal, it is called a scalene triangle.
PLANE GEOMETRY. 25
The term isosceles is used to designate the fact that two
sides are equal without any reference to the third side, even
though it may then be known, or afterwards ascertained, that
the sides are all equal.
Construct triangles to illustrate each of the above varieties.
Construct three triangles that shall illustrate all the six
varieties.
The sum of the three sides of a triangle is called the perim-
eter.
In a right triangle, the side opposite the right angle is called
the hypothenuse, and the other two sides the legs.
The side on which a triangle is supposed to rest is called the
base. In general, any side of a triangle may be considered
the base ; but in an isosceles triangle the third side, and in a
right triangle one of the legs, is generally the base.
The angle opposite the base is called its vertical angle ; and
its vertex, the vertex of the triangle.
The perpendicular distance from vertex to base, or base
extended, is called the altitude of a triangle.
The line drawn from the vertex of a triangle to the middle
point of the base is called a median.
97. Any side of a triangle is less than the sum of the other
two, and greater than their difference.
98. The sum of the three angles of a triangle is equal to
two right angles.
Sug. Consult Theorems 58 and 87.
99. If one side of a triangle be extended, the exterior angle
thus formed is equal to the sum of the two interior non-
adjacent angles.
100. If two angles of a triangle be known, how can the third
be found ?
Two angles of a triangle are 34 28' 42" and 29 44' 56",
respectively ; find the value of the third angle.
26 PLANE GEOMETRY.
Two angles of a triangle are 64*75W and 86 g 45 r 75 M ,
respectively ; find value of the third angle.
If one angle of a triangle be a right angle, what relation
exists between the other two ?
How many right angles in a triangle ? How many obtuse
angles ?
If the three angles of a triangle are all equal, what is the
value of each angle in terms of a right angle ? In degrees ?
In grades ?
If two angles of one triangle be equal respectively to two
angles of another, what relation exists between their third
angles ? Prove.
If one acute angle of a right triangle is 27 38' 50", what is
the value of the other acute angle ?
Find values of the three angles of a triangle, if the second
is three times the first and the third is two times the second.
In a right triangle one of the acute angles is 17 40' greater
than the other ; find the values of both acute angles.
The sum of two angles is 30 g , and their difference is 9;
find each angle.
101. Hyp. If two triangles have two sides and their in-
cluded angle, of one, equal respectively to two sides and their
included angle, of the other,
Con. the two triangles will be equal in all respects.
Post. Let ABC and DHK be two triangles having side
AB=DH, AC=DK, and their included angles A and D
equal.
We are to prove that the triangle ABC equals the triangle
PLANE GEOMETRY. 27
DHK\ i.e. that all their remaining corresponding parts are
respectively equal.
Dem. Conceive the triangle ABC to be so applied to the
triangle DHK that side AB shall fall upon DH, the point A
falling on point D.
Where must the pt. B fall ? Why ?
Where must the line AC fall ? Why ? (Sect. 36.)
Where must pt. C fall ? Why ?
What, then, must be the relative position of BC and HK?
Why?
Hence, the two triangles are coincident, and are therefore
equal. (Sect. 34.)
Q.E.D.
102. If two triangles have two angles and the side con-
necting their vertices, of one, equal respectively to two angles
and side connecting their vertices, of the other, the two
triangles will be equal in every respect.
Sug. Use method similar to the preceding.
103. If the three sides of one triangle are equal respectively
to the three sides of another, the two triangles are equal in
every respect.
Sug. Apply the triangles as indicated below, then consult
82, 73, IL, and 101.
\
\
~JC
H
104. If two right triangles have their legs respectively
equal, they are equal in all respects.
28 PLANE GEOMETRY.
105. If two right triangles have a leg and an acute angle
of one equal respectively to a leg and an acute angle of the
other, they are equal in all respects.
106. If two right triangles have the hypothenuse and an
acute angle of one equal respectively to the hypothenuse
and an acute angle of the other, they are equal in all
respects.
Query. Would two right triangles be equal if they had
d side and an acute angle of one equal to a side and an acute
angle of the other ?
107. If two right triangles have the hypothenuse and a leg
of one equal respectively to the hypothenuse and a leg of the
other, they are equal in all respects.
Sug. Apply one triangle to the other so that the equal legs
shall coincide ; then consult 78 and 103.
Remark. The corresponding sides and angles of two tri-
angles are those that are similarly situated, and are called
homologous. If the triangles are equal, the homologous sides
are those that are opposite the equal angles ; and conversely,
the homologous angles are those that are opposite the equal
sides.
108. Hyp. If a triangle have two of its sides equal,
Con. the angles opposite those sides are also equal.
Post. Let ACD be a triangle
having the sides AD and CD
equal.
We are to prove that
Cons. Draw DB J_ to AC.
Dem. The two triangles CBD
and ABD are rt. A. Why ?
Name the hypothenuse of each.
What line is a leg of both triangles ?
PLANE GEOMETRY. 29
.-. The two A are equal. (Theorem 107.)
.*. Z G = Z A. (Being homologous angles of equal A.)
Q.E.D.
Query. In the above demonstration how else could the line
BD have been drawn ? Give the corresponding variation in
demonstration.
The pupil should also understand now, at the outset, that
only one condition can be imposed, in drawing an auxiliary
line, which fixes its direction. For instance, he cannot say, in
the previous construction, "Draw DB perpendicular to CA
through its middle point," since either condition gives the
line a definite direction; and, previous to demonstration, he
does not know but that one condition will conflict with the
other.
109. Converse of 108.
Sug. Employ method similar to that in 50.
110. If a triangle be equilateral, it will also be equiangular.
Sug. Consult 108.
. 111. Converse of 110.
Sug. Consult 109.
112. If a perpendicular bisect the base of an isosceles
triangle, it will pass through the vertex and bisect the vertical
angle.
113. Converse of 112.
114. If a line be drawn from the vertex of an isosceles
triangle to the middle point of the base, this line will bisect
the vertical angle and be perpendicular to the base.
115. If one of the equal sides of an isosceles triangle be
extended at the vertex, and the exterior angle thus formed
be bisected, the bisecting line will be parallel to the base.
116. Converse of 115.
30 PLANE GEOMETRY.
117. If a line bisect an angle, any point selected at random
in this bisector will be equally distant from the sides of the
angle.
JSug. Consult 72, Remark (a), and 106.
118. If a line bisect an angle, any point selected at random
outside this bisector is unequally distant from the sides of
the angle.
Sug. If K be the point, then what is the relation between
HS and HB ? Why ?
What is the relation between BK and KH + HS? Why ?
What is the relation of KS to KH + HS? Why ?
What is its relation, then, to BK?
What is the relation between KS and KP ?
What, then, must be the relation between KP and BK?
119. If a point be equally distant from the sides of an
angle, the line joining it with the vertex will bisect the
angle.
120. If the angles at the base of an isosceles triangle be
bisected, the bisectors will form, with the base, an isosceles
triangle.
121. If the angles at the base of an isosceles triangle be
double the vertical angle, a line bisecting either of the former
will divide the triangle into two isosceles triangles.
PLANE GEOMETRY.
31
122. If two angles of a triangle be unequal, the side opposite
the greater of these two angles is longer than the side opposite
the lesser.
Post. Let ABC be any tri-
angle with Z B > Z A.
We are to prove ' that the
side AC, opposite the greater
angle B, is longer than the
side CB, opposite the lesser
angle A.
Cons. Let BD be drawn so
as to cut off a portion of the larger angle B, making it equal
to angle A, so that they will both be in the same triangle
as DAB.
What relation between AD and DB ? Why ?
What relation between BC and CD-}-DB? Why ?
(The pupil should finish it without difficulty.)
123. Converse of 122.
Sug. Three possible relations between the two angles.
Prove the impossibility of two of them ; or, a method similar
to that in 122 may be employed.
123 (a). If two triangles have the two sides of one equal
respectively to two sides of the other, but the included angles
unequal, then the third side of that triangle having the greater
included angle is longer than the third side of the other.
Sug. Apply the two triangles so that two of the equal sides
shall coincide, as is represented in the above diagram.
THE
32 PLANE GEOMETRY.
Will the other pair of equal sides coincide ? Why ?
Join the other two vertices.
What kind of a triangle is HKB'?
What relation, then, between angles HKB 1 and HB'K?
What must be the relation, therefore, between the angles
DKB' and C'B'K?
Consult Theorem 122.
Or, instead of joining B 1 and K, the angle KA'B' may be
bisected, and the point where this bisector cuts DK joined
with B f . Then consult 101 and 97.
124. Converse of 123.
Sug. There are only three possible relations between those
angles. Prove the impossibility of two of them.
ADVANCE THEOREMS.
125. The bisectors of the three angles of a triangle meet in
one point.
Sug. From the point of intersection of two of the bisectors
draw a line to the other vertex, and also perpendiculars to the
three sides. Prove the former a bisector by means of equality
of triangles.
126. The perpendiculars which bisect the three sides of a
triangle meet in a point.
Sug. From the point of intersection of two of the perpen-
diculars draw a line to the middle point of the other side.
Prove this line perpendicular, by consulting 74 and 103.
127. The perpendiculars from the three vertices of a triangle
to the opposite sides meet in one point,
Sug. Through each vertex draw a line parallel to the
opposite side. Then consult 102 and 127.
128. If two angles of an equilateral triangle be bisected,
and lines be drawn through the point of intersection parallel
to the sides, the sides will be trisected.
PLANE GEOMETRY. 33
QUADEILATEKALS,
129. If two lines in the same plane be crossed by two trans-
versals, a figure of four sides may be formed, which is called
a quadrilateral. Hence a quadrilateral may be defined as a
plane figure bounded by four straight lines. A quadrilateral
is also called a tetragon.
If each of the two pairs of lines is parallel) the quadrilateral
thus formed is called a parallelogram.
Define, then, a parallelogram.
If a parallelogram have all its sides equal, it is called a
rhombus.
If its angles are right angles, it is called a rectangle.
If it is both a rhombus and a rectangle, it is called a
square.
Give all the names applicable to a square.
If a quadrilateral have only two of its sides parallel, it is
called a trapezoid.
If no two of its sides are parallel, it is called a trapezium.
A parallelogram whose angles are oblique and adjacent sides
unequal is sometimes called a rhomboid.
A rectangle whose adjacent sides are unequal is sometimes
called an oblong.
The line which joins two opposite vertices of a quadrilateral
is called its diagonal.
The side upon which a parallelogram is conceived to rest,
and the side opposite the latter, are termed, respectively, the
lower and upper bases.
The parallel sides of a trapezoid are always considered as
its bases, the other two sides its legs, while the line bisecting
the legs is called the median.
The altitude of either a parallelogram or trapezoid is the
perpendicular distance between its bases.
The pupil should construct figures to represent all the above-
mentioned quantities.
34 PLANE GEOMETRY.
The sum of the four sides of a quadrilateral is called its
perimeter.
If the legs of a trapezoid are equal, it is called an isosceles
trapezoid.
130. Either diagonal divides the parallelogram into two
equal triangles.
Sug. Consult 87 and 102.
Query. Is the converse of this theorem true ?
131. The sum of the four angles of a quadrilateral equals
four right angles.
132. The opposite sides of a parallelogram are equal.
132 (a). Two parallel lines are everywhere equally distant.
133. Converse of 132.
tSug. Draw one diagonal, then consult 45 and 34.
134. The opposite angles of a parallelogram are equal.
135. Converse of 134.
Sug. Consult 131 and 91.
136. If two sides of a quadrilateral are equal and parallel,
the quadrilateral is a parallelogram.
137. The two diagonals of a parallelogram bisect each other.
138. Converse of 137.
139. If one angle of a parallelogram be a right angle, the
other three angles are right angles, and the parallelogram is
therefore a rectangle.
140. The diagonals of a rectangle are equal.
141. Converse of 140.
142. The diagonals of a rhombus are perpendicular to each
other.
143. Converse of 142.
144. The diagonals of a rhombus bisect the angles.
145. Converse of 144.
PLANE GEOMETRY. 35
146. If two parallel lines be crossed by a transversal, the
bisectors of the interior angles form a rectangle.
Sug. Consult 88, III., 92, 93, and 64.
147. The lines which bisect the angles of a rhomboid form
a rectangle.
148. If two parallelograms have two sides and their in-
cluded angle of one equal respectively to two sides and their
included angle of the other, the two parallelograms are equal
in every respect.
149. If a line be drawn parallel to one side of a triangle,
and bisecting another side, this line will bisect the third side
of the triangle, and be equal to one-half the side parallel to it.
150. Converse of 149.
151. The median of a trapezoid is parallel to its bases and
equal to one-half their sum.
Sug. Through one extremity of the median draw a line
parallel to one leg, and extend the shorter base to meet it.
ADVANCE THEOREMS.
152. If from any point in the base of an isosceles triangle
lines are drawn parallel to the equal sides, the perimeter of
the parallelogram thus formed will be equal to the sum of
the two equal sides of the triangle.
153. If the legs of a trapezoid are equal, the angles which
they make with either base are equal.
154. Converse of 153.
155. If the angles at one base of a trapezoid are equal, the
angles at the other base are also equal.
156. The line which is parallel to the bases of a trapezoid
and bisects one leg is a median.
36 PLANE GEOMETRY.
157. The line joining the vertex of the right angle to the
middle point of the hypothenuse in a right triangle is equal
to one-half the hypothenuse.
158. The lines which join the middle points of the sides
of a triangle divide the triangle into four equal triangles.
159. The three medians of a triangle meet in one point.
Sug. From the point of intersection of the two medians
draw a line to the third vertex. From the middle point of
this line draw another to the point midway between one of
the other vertices and the point of intersection of the two
medians. By extending the first line and connecting certain
points a parallelogram may be formed.
160. The lines which join the middle points of the sides
of any quadrilateral form a parallelogram.
Sug. Draw the diagonals, then consult 150.
161. The lines which join the middle points of the sides of
a rhombus form a rectangle.
162. The lines which join the middle points of the sides of
a square form a square.
163. The lines which join the middle points of the sides
of a rectangle form a rhombus.
164. The lines which join the middle points of the sides
of an isosceles trapezoid form a rhombus.
165. The median and diagonals of a trapezoid intersect at
the same point.
Sug. Consult 151 and 149.
166. The diagonals of an isosceles trapezoid are equal.
167. Converse of 166.
168. If a trapezoid be isosceles, the opposite angles are
supplementary.
PLANE GEOMETRY. 37
169. The line which joins the middle points of the diagonals
of a trapezoid equals one-half the difference of the bases.
170. The two perpendiculars from the extremities of the
base to the equal sides of an isosceles triangle are equal.
171. The medians drawn to the equal sides of an isosceles
triangle are equal.
172. The bisectors of the angles at the base of an isosceles
triangle are equal.
173. The two perpendiculars from the middle point of the
base of an isosceles triangle to the equal sides are equal.
174. State and prove the converse of each of 170, 171, 172,
173.
175. If one of the equal sides of an isosceles triangle be
extended at the vertex, making the extension equal to the
side, the line joining the end of the extension with the nearer
extremity of the base is perpendicular to the base.
176. If one angle of an isosceles triangle be 60, the triangle
is equilateral.
177. If from the middle points of two opposite sides of a
parallelogram lines be drawn to the vertices of the angles
opposite, these lines will trisect the diagonal that joins the
other two vertices.
178. If the two base angles of a triangle are bisected, and
through the point of intersection of these bisectors a line be
drawn parallel to the base and terminating in the sides, this
line is equal to the sum of the two segments of the sides
between this parallel and the base.
179. The bisectors of the vertical angle of a triangle and
the angles formed by extending the sides below the base meet
in a point which is equally distant from the base and the
extensions of the sides.
38 PLANE GEOMETRY.
180. If one of the acute angles of a right triangle is double
the other, the hypothenuse is double the shorter leg.
181. If from any two points selected at random in the base
of an isosceles triangle perpendiculars be drawn to the equal
sides, the sum of the perpendiculars from one point equals the
sum of the perpendiculars from the other point.
182. If from any point selected at random in an equilateral
triangle perpendiculars be drawn to the sides, the sum of these
perpendiculars is constant, and equal to the altitude of the
triangle.
CIRCLES.
183. A circle is a portion of a plane bounded by a curved line,
all points of which are equally distant from a point within
called the centre.
The bounding line is called the circumference.
Any portion of the circumference is called an arc.
A radius (plural radii) is any line from centre to circum-
ference.
A diameter is any line passing through the centre and ter-
minating both ways in tho circumference.
How do the radius and diameter of the same circle compare
in magnitude ?
How do the radii of a circle compare in magnitude ? The
diameters ?
A semicircumference is one-half the circumference.
A sector is that part of a circle included between an arc and
the radii drawn to its extremities.
A quadrant is a sector which is one-fourth the circle.
A quadrant arc is one-fourth the circumference.
A chord is any straight line whose extremities are in the
circumference.
A segment is that portion of the circle included between an
arc and the chord which joins its extremities.
PLANE GEOMETRY. 39
Every chord, therefore, must divide the circumference into
two arcs and the circle into two segments.
If the arcs are unequal, they are designated as major and
minor arcs, and the segments as major and minor segments.
The chord is said to subtend the arc, and the arc is said to
be subtended by the chord.
Whenever a chord and its subtended arc are mentioned, the
minor arc is meant unless it is otherwise specified.
If two circles have the same centre, they are said to be
concentric.
A central angle is an angle formed by two radii.
An inscribed angle is an angle formed by two chords, with
its vertex in the circumference.
When would an angle be said to be inscribed in a segment ?
A tangent is a straight line that touches the circumference of
a circle, but on being extended does not intersect it ; i.e. the
tangent and the circumference have one point, and only one,
in common. This point is called the point of contact, or point
of tangency.
Two circumferences are tangent to each other when they
have one point in common but do not intersect ; i.e. when they
touch each other.
If one of two tangent circumferences lies within the other,
they are said to be tangent internally; if it lies without, they
are said to be tangent externally.
A secant is a straight line that intersects a circumference in
two points lying partly within and partly without the circle ;
e.g. a chord extended in either direction becomes a secant.
The term circle is also sometimes used to designate a cir-
cumference.
Construct diagrams illustrating the above magnitudes and
their relations.
184. A diameter is greater than any other chord.
Sug. Draw the diameter AD, and let HB be any other
chord. Join CH and CB. Consult 97.
40 PLANE GEOMETRY.
185. A diameter bisects both the circle and circumference.
Sug. Fold over one part on the
other, using the diameter as an axis
of revolution.
185 (a). Converse of 185.
186. A straight line cannot intersect
the circumference of a circle in more
than two points.
Sug. Suppose it could intersect in three points, and draw
radii to the three points ; then consult 80.
187. If two circles have equal radii, they are equal.
Bug. Apply one to the other.
188. If the diameters of two circles are equal, the circles
are equal.
189. Converse of 187 and 188.
190. If two equal circles be concentric, their circumferences
will coincide.
191. If, in the same or equal circles, two central angles be
equal, the arcs which their sides intercept will also be equal.
Remark. In this and the following theorems where the
expression " same or equal circles " occurs, the demonstration
will be more satisfactory if two circles are used.
Sug. Apply one circle to the other.
192. Converse of 191.
193. If, in the same or equal circles, two chords be equal,
the arcs which those chords subtend are also equal.
Sug. Draw radii to the extremities of the chords, then con-
sult 103 and 191.
194. Converse of 193.
Sug. Draw radii as above. Consult 192 and 101.
PLANE GEOMETRY. 41
195. If, in the same or equal circles, two central angles be
unequal, the arc intercepted by the sides of the greater angle
is greater than the arc intercepted by the sides of the lesser
angle.
Sug. Apply one circle to the other.
196. Converse of 195.
197. If, in the same or equal circles, two chords be unequal,
the arc subtended by the greater chord is greater than that
subtended by the lesser.
Sug. Draw radii to extremities, then consult 124 and 195.
198. Converse of 197.
Sug. Draw radii as above, then consult 196 and 123. Or,
three possible relations.
199. If a diameter (or radius) be perpendicular to a chord,
it will bisect the chord, and also the arcs into which the chord
divides the circumference.
Sug. Draw radii to extremities of the chord, then consult
78 or 107 and 191.
200. A straight line that is perpendicular to a radius at its
extremity is a tangent to the circle.
Post. Let BHKbe a circle, DB a radius, and AC a straight
line perpendicular to DB and passing through the point B.
We are to prove that AC is a tangent to the circle BHK.
42 PLANE GEOMETRY.
Now, if we can prove that every point in AC except B is
out vide the circumference, then AC must be a tangent. (See
Del of Tangent.)
Cons. From D draw any other line to AC, as DP.
Then P, the extremity of DP, must represent any point in
AC except B.
Now DP > DB. Why ?
But DB is a radius, and if DP is longer than a radius,
where must its extremity be ?
Hence, etc.
201. If a radius be drawn to the point of contact of a
tangent to a circle, it will be perpendicular to the tangent.
Sug. If it can be proved that the radius is the shortest
distance from the centre to the tangent, then it must be per-
pendicular. (See Theorem 72.) Use previous diagram.
What must be the situation of all the points in the tangent,
except the point of contact, with reference to the circum-
ference of the circle ?
What, then, must be the relation to the radius of a line
drawn from the centre to any point in the tangent except the
point of contact ?
Hence, etc.
202. If a perpendicular be erected to a tangent at the point
of contact, this perpendicular, if extended, will pass through
the centre of the circle.
Sug. Draw a radius to point of contact, then consult 201
and 65.
203. If a line be drawn from the centre of a circle perpen-
dicular to a tangent, this line will pass through the point of
contact.
204. If a chord and tangent be parallel, the arcs which they
intercept are equal.
Sug. Draw a radius to point of contact, then consult 201,
84, and 199.
PLANE GEOMETRY. 43
205. If two chords be parallel, the arcs which they inter-
cept are equal.
206. If two tangents be parallel, the arcs which they in-
tercept are equal.
207. The line joining the points of contact of two parallel
tangents is a diameter.
208. If two tangents have their points of contact the oppo-
site extremities of a diameter, the tangents are parallel.
209. If a radius bisect a chord, it will bisect the arc and be
perpendicular to the chord.
210. If a perpendicular bisect a chord, it will, if extended,
pass through the centre of the circle.
211. If a radius bisect an arc, it will also bisect its sub-
tending chord and be perpendicular to it.
211 (a) . If a line bisect an arc and its subtending chord, this
line will, if extended, pass through the centre of the circle.
212. If two non-intersecting chords intercept equal arcs,
they are parallel.
213. Converse of 204.
214. If, in the same or equal circles, two chords be equal,
they are at equal distances from the centres.
Sug. Consult 72, Remark (a).
215. Converse of 214.
216. If, in the same or equal circles, two chords be unequal,
the less chord is the farther from the centre.
Post. Let ABD and HPK be two equal circles, and chord
AB > chord HK
We are to prove that HK is farther from the centre N than
AB is from the centre Q.
44 PLANE GEOMETRY.
Cons. Draw the radius C V perpendicular to AB and radius
NZ perpendicular to HK. Draw also the radii CA, CB, NH,
and NK.
Then CO is the distance AB is from C, and NS is the dis-
tance HKis from N. (72, Eemark (a).)
So we are, in reality, to prove NS > CO.
Dem. AB > HK Why ?
05 is what part of AB ? Why ?
ITS is what part of HK? Why?
Then what is the relation of OB and HS in respect to
magnitude ?
Express that relation by symbols.
What is the relation between the two arcs A VB and HZKt
Why?
What relation, then, between the A ACB and HNK? Why ?
How, then, does the sum of the angles A and B compare
with the sum of the angles IT and N? Explain.
How does Z A compare with Z B ? ZHwithZK? Why?
What relation, then, in respect to magnitude, between the
AH and B? Why?
Now, on BO mark off a distance from B equal to HS, as
BT, and join CT.
Compare CT and NS. (See Theorem 123 (a).)
Now compare them both with CO.
Hence, etc.
PLANE GEOMETRY.
45
The following variation of the above method was given by
one of the author's pupils.
Make AE = CD. Then, since HR is less than PD, ZAia
less than Z C. (Theorem 124.)
The rest is similar to the one above.
217. Converse of 216.
Sug. Three possible relations.
218. Through any three points selected at random, provided,
however, they are not in the same straight line, one circum-
ference can be made to pass, and only one; or, if more be
drawn, they must coincide.
Sug. If the points be joined, these lines must be chords
of the circle. Then consult 210.
219. If two unequal circles are concentric, the chords of the
greater which are tangents of the less are equal.
Sug. Consult 201, 215, and 72, Remark (a).
220. If two circles be tangent to each other externally, the
radii drawn to the point of contact form one and the same
straight line.
Post. Let the two circles PCN and CDK be tangent to
each other externally at C, and AC and BC the radii drawn
to the point of contact C.
We are to prove that ACB is a straight line. If we can
prove that ACB is the shortest distance from A to B, then
ACB must be a straight line. For if a straight line be drawn
46 PLANE GEOMETRY.
from A to B, it will be the shortest distance between A and B.
(Sect. 42, XIV.) And since there can be but one shortest
distance, if ACB is proved also to be the shortest distance,
then ACB must coincide with that line, and consequently be
a straight line.
Jf
So we are to prove that ACB is the shortest distance from
-4 to B.
Cons. Draw any other radius than BC in circle CDK, as
BD, and join AD.
Dem. Since every point in the circumference CDK except
C is outside the circumference CPN (see Def.), AD must
extend beyond the circumference, and
AD > AC. Why ?
DB=CB. Why?
DB>AC+CB. Why?
But D is any point in the circumference CDK except C.
Hence, the distance from A to B by way of D is always greater
than by way of C. Or ACB is the shortest distance between
A and B, and is therefore a straight line. Q.E.D.
221. If radii be drawn to the point of contact of two circles
that are tangent to each other externally, and a straight line
be drawn through the point of contact perpendicular to one of
the radii, this line will be a common tangent to the two circles.
PLANE GEOMETRY. 47
222. If two circles be tangent to each other externally, the
straight line joining their centres will pass through the point
of contact.
223. If two circles be tangent to each other internally, and a
line be drawn through the point of contact tangent to the
outer circle, it will also be tangent to the inner circle.
224. If two circles be tangent to each other internally, and
radii be drawn to point of contact, these radii will lie in the
same straight line.
225. If two circles be tangent to each other internally, the
line joining their centres will, if extended, pass through the
point of contact.
226. If two circles be tangent to each other internally, and
a common tangent be drawn through the point of contact, a
perpendicular to this tangent at the point of contact will, if
extended, pass through the centres of both circles.
227. If two circles be tangent to each other internally, the
radius of the smaller to the point of contact will, if extended,
pass through the centre of the larger, and the radius of the
larger to point of contact will also pass through the centre of
the smaller.
228. If two circles be tangent to each other internally, and
a line be drawn from the centre of either circle perpendicular
to their common tangent, this perpendicular will pass through
the centre of the other circle and also through the point of
contact.
229. Two unequal circles may have five positions relative to
each other, viz. :
I. One may lie wholly within the other without contact.
II. They may be tangent to each other internally.
III. They may intersect each other.
IV. They may be tangent to each other externally.
V. One may lie wholly without the other without contact.
48 PLANE GEOMETRY.
If the circles are equal, how many positions relative to each
other may they have ?
In Case I. what relation exists between the line joining
their centres and the sum of their radii ? Between the same
line and the difference of their radii ? Prove.
Prove both the above relations in each of the other four
cases.
What must be the position of two circles, then, if the dis-
tance between their centres is
Jf. ft?
II. Less than the sum, and also less than the difference, of
their radii ?
III. Equal to the difference of their radii ?
IV. Less than the sum, and greater than the difference, of
their radii ?
V. Equal to the sum of their radii ?
VI. Greater than the sum of their radii ?
230. If two circles intersect each other, the line joining
their centres will bisect their common chord and be perpen-
dicular to it.
231. If two circles intersect each other, and radii be drawn
to the middle point of their common chord, these radii will
form one and the same straight line.
232. If two circles intersect each other, and either radius
be drawn bisecting their common chord, this radius will, if
extended, pass through the centre of the other circle.
233. If two circles intersect each other, and a perpendicular
bisect their common chord, this perpendicular will, if extended,
pass through the centres of both circles.
234.* A central angle is measured by the arc which its sides
intercept on the circumference.
* See Appendix.
PLANE GEOMETRY.
49
Dem. In the circle ANQ let us conceive the radius CA
to move about C as a pivot, remaining always in the same
plane. In one complete revolution it is evident that the
extremity A will have traced the entire circumference. Like-
wise, in one-half a revolution,
as when it has reached the
position CP } making PGA a
diameter, the extremity A will
have traced one-half the circum-
ference. So when it has reached
the position CN perpendicular
to AP, it will have traced one-
fourth the circumference (see
199, and 59 and 60). In like
manner, when it has reached
the position CK, making the
angles ACK and KCN equal,
then tne arcs AK and KN are also equal (see 191) ; that is,
the arc AK is one-eighth of the circumference, and the angle
ACK is one-eighth of the angular space about the point C.
In like manner, the arc AH is one-sixteenth, AD one thirty-
second, AB one sixty-fourth, etc., of the entire circumference,
while their corresponding central angles are the same parts of
four right angles, or the angular space about the centre (7.
So that, whatever the position of the radius CA, the arc
described will be the same part of the entire circumference
that its corresponding central angle is of the angular space
about the centre. If, now, the entire circumference be divided
into 360 equal parts, each part would correspond to a central
angle of 1. Consequently, if an arc contain 27 of the 360
equal parts, it may be called an arc of 27, and its correspond-
ing central angle would contain -ff^ of four right angles, or
would be an angle of 27. Likewise, if the arc contains a
whole number of degrees and a fraction, the latter can be
expressed in the smaller units. For example, suppose the arc
50
PLANE GEOMETRY.
to contain 48 37' 24.92" ; then the corresponding central angle
would contain the same number of angular units ; i.e. it would
be an angle of 48 37' 24.92", to the same degree of approxima-
tion. Therefore the arc is said to measure the corresponding
central angle, because, as above stated, it is the same part of
the entire circumference that its corresponding central angle is
of the angular space about the centre. Q.E.D.
235. An inscribed angle is measured by one-half the arc
intercepted by its sides.
Remark. In demonstrating this theorem it will simplify it
somewhat to make three cases of it ; viz. :
I. When one of the sides of the angle is a diameter.
II. When the centre is between the sides of the angle.
III. When the centre is without the angle.
K
D
Fig. I.
CASE I. Post. Let B be an inscribed angle with one of
its sides, as AB, a diameter.
We are to prove that the angle B is measured by one-half
the arc AD; that is, one-half the arc AD is the same part
of the entire circumference ABD that the angle B is of the
angular space about B.
Cons. Draw the radius CD (Fig. I.).
Dem. What relation exists between CD and CB ? Why ?
What relation, then, between angles B and D ? Why ?
PLANE GEOMETRY. 51
What relation exists between the sum of the angles B and
D and angle DCA? Why?
What relation, then, exists between the angle B and the
angle DCA? Whj?
What measures the angle DCA ? Why ? (Consult 234.)
What, then, must measure the angle B ?
Again, Fig. II., Case I.
Cons. Di v aw a diameter, as UK, parallel to BD.
Dem. What relation exists between the angles BCK and
HCA ? Why ?
What relation, then, .exists between the two arcs HA and
BK? Why? (See 191.)
What relation between the two arcs DH and BK? Why?
What relation, then, between the two arcs DH and AH?
Why?
What relation, then, between the two arcs AH and AD ?
Why?
What relation between the two angles HCA and B ? Why ?
What measures the angle HCA ? Why ?
What, then, must measure the angle B ?
After answering correctly the foregoing questions the pupil
should have no difficulty in writing "out a complete demonstra-
tion of each of the three cases, using Case I. in demonstrating
II. and III.
236. If two angles be inscribed in the same or equal circles,
and their sides intercept the same or equal arcs, the two angles
are equal.
236 (a). Converse of 236.
236 (6). What part of the circumference measures a right
angle ? Prove.
What relation exists between angles inscribed in the same
segment ? Why ?
What measures an angle inscribed in a semicircle ?
What kind of an angle, then, must it be ?
52 PLANE GEOMETRY.
If an angle be inscribed in a segment less than a semicircle,
what kind of an angle must it be ? Why ?
If an angle be inscribed in a segment greater than a semi-
circle, what kind of an angle must it be ? Why ?
237. If an angle be formed by two chords whose vertex is
between the centre and circumference, it will be measured by
one-half the sum of its two intercepted arcs.
238. If the vertex of an angle formed by two secants is
without the circle, this angle is measured by one-half the
difference of the two intercepted arcs.
239. If an angle be formed by a tangent and chord, this
angle is measured by one-half the arc intercepted by its sides.
240. If the vertex of an angle formed by a tangent and
secant is without the circle, this angle is measured by one-half
the difference of the two intercepted arcs.
241. If an angle be formed by two tangents to the same
circle, this angle is measured by one-half the difference of the
two intercepted arcs.
242. If from the same- point without a circle two tangents
be drawn, these two tangents are equal; i.e. the distances
from the common point to the points of contact are equal.
Sug. Consult either 201 and 107, or 239 and 109.
ADVANCE THEOREMS.
243. The line which joins the vertex of an angle formed
by two tangents to the centre of the circle bisects the angle,
and also the chord which joins the points of contact.
244. If, from the same point, two tangents to a circle be
drawn whose points of contact are the extremities of a chord,
the angle formed by the two tangents is double the angle
formed by the chord and diameter drawn from either ex-
tremity of the chord,
PLANE GEOMETRY. 53
245. If two circles which are tangent to each other externally
have three common tangents, the one that passes through the
point of contact of the two circles bisects the other two.
Sug. Consult 242.
246. If four tangents form a quadrilateral, the sum of two
opposite sides equals the sum of the other two opposite sides.
247. If two mutually equiangular triangles be inscribed in
equal circles, the triangles are equal in all respects.
248. If two opposite sides of an inscribed quadrilateral are
equal, the other two sides are parallel.
248 (a). Converse of 248.
249. The opposite angles of an inscribed quadrilateral are
supplementary.
250. If two equal chords be drawn from opposite extremities
of a diameter and 011 opposite sides of it, they will be parallel.
251. If two chords be drawn through the same point in a
diameter making equal angles with it, they are equal.
252. If one of the equal sides of an isosceles triangle be the
diameter of a circle, the circumference will bisect the base.
253. If an equilateral triangle be inscribed in a circle, and
a diameter be drawn from one vertex, the triangle, formed by
joining the other extremity of the diameter and the centre of
the circle with one of the other vertices of the inscribed
triangle, will also be equilateral.
254. The bisectors of the angles formed by extending the
sides of an inscribed quadrilateral are perpendicular to each
other.
255. If an equilateral triangle be inscribed in a circle, and
any point in the circumference be selected at random, one of
the lines which join this point to the three vertices will equal
the sum of the other two.
54 PLANE GEOMETRY.
256. If an inscribed isosceles triangle have its base angles
each double the vertical angle, and its vertices be the points
of contact of three tangents, these tangents will form an
isosceles triangle each of whose base angles is one-third its
vertical angle.
257. If through any point selected at random in a circle a
diameter and other chords be drawn, the least chord will be
the one that is perpendicular to the diameter.
258. ADB is a semicircle of which the centre is (7, and
AEG is another semicircle on the diameter AC, and AT is a
common tangent to the two semicircles at A. Prove that, if
from any point F in the circumference of the first a straight
line be drawn to (7, the part FK, cut off by the second semi-
circle, is equal to FH, perpendicular to the tangent AT.
259. If a triangle ABO be formed by the intersection of
three tangents to a circumference whose centre is 0, two of
which, AM and AN, are fixed, while the third, BGj touches
the circumference at a variable point P, prove
I. that the perimeter of the triangle is constant.
II. that the angle BOO is constant.
260. If the sides of any quadrilateral be the diameters of
circles, the common chord of any adjacent two is parallel to
the common chord of the other two.
EATIO AND PBOPOKTION,
(It is assumed that the pupil is familiar with the algebraic
processes involved in the manipulation of equations of the
first and second degree. For definition of Geometrical Magni-
tudes, see 37.)
261. If of two unequal magnitudes the less is contained an
exact number of times in the greater, the latter is said to be a
multiple of the former, and the former an aliquot part of the
latter.
PLANE GEOMETRY. 55
262. If of three unequal magnitudes the smallest is con-
tained an exact number of times in each of the two larger, the
latter are said to be commensurable, and the former is said to
be their common measure.
263. When no magnitude can be found which is contained
an exact number of times in each of two magnitudes, the latter
are said to be incommensurable.
264. If two commensurable magnitudes contain their com-
mon measure, one n times and the other m times, they are said
to be to each other as n to m, or in the ratio of n to m.
x
A
B
For example, if the line x is contained in the line A 3 times,
and in line B 5 times, then A is to B as 3 to 5, and the line x
is a common measure of the two lines A and B.
265. Equimultiples of two or more magnitudes are the results
obtained by multiplying these magnitudes by the same number.
(See 39.)
Thus, if x and y be any two lines, and a and b two other
lines such that the former contains the line x n times (in this
particular case n is 3), and the latter contains the line y
n times, then a and b are equimultiples of x and y respectively.
266. If two magnitudes are to each other as m to n (see 264),
it is usually expressed thus, m:n, called the ratio form;
or thus, ~, called the fractional form,
n
267. Hence a ratio may be defined as an expression of com-
parison, in respect to size, of two magnitudes of the same
kind.
56 PLANE GEOMETRY.
268. If x and y are two magnitudes, and y a multiple of x
(see 261), the latter being contained in the former n times,
then the ratio of y to x is as n to unity, or n : 1.
269. If x and t/ are commensurable (262), and their common
measure is contained n times in a?, and m times in y, then the
ratio of x to y is as n to ra, or, technically expressed, n : m.
270. If x and ?/ are incommensurable (see 263), the ratio
cannot be exactly expressed. If, however, x and y are num-
bers, the ratio may be approximately expressed by placing 1
for the first term, and the quotient of the greater divided by
the less to any number of decimal places for the second term ;
thus, l:n.pgr-K In such a case the computation can be
carried to any required degree of approximation.
271. In a similar way can be obtained an approximate ratio
of two incommensurable magnitudes, expressed numerically;
If x and y be two incommensurable magnitudes, suppose x
to be divided into a certain number, M, of equal parts, and that,
on applying one of these equal parts to y, it is found to be
contained m times with a remainder less than this part. Then
it is evident that the ratio of y to x is neither m : w,, nor
m + l:n, but is greater than the former and less than the
latter; i.e. greater than the value of , and less than the
value of m "*" , or -| It is thus seen that the smaller
n n n
the unit into which x is divided, the greater becomes the value
of n, and consequently the smaller becomes the value of -.
And hence, neglecting this value, will express the approxi-
PLANE GEOMETRY. 57
mate ratio of y to x, the degree of approximation depending
upon the size of the unit of measure.
272. The first term of a ratio is called the antecedent, and
the second term the consequent.
273. A proportion is an expression of equality between two
equal ratios, usually indicated by four dots between the two
ratios ; thus, a : b : : x : y, and read, a is to b as x is to y.
274. The four magnitudes forming a proportion are called
proportionals.
275. If a proportion contains only two ratios, it is called a
simple proportion ; if more than two and all equal, it is called
a continued proportion.
276. The first and last terms of a simple proportion are
called the extremes, and the other two the means.
277. In a simple proportion, where the terms are all of
different values, each term is said to be a fourth proportional
to the other three.
278. In a simple proportion, where the two means or the two
extremes are alike, the repeated quantity is said to be a mean
proportional to the other two, and the proportion is called a
mean proportion.
279. In a mean proportion, either of the quantities not
repeated is said to be a third proportional to the other two.
280. When a line is so divided that the larger part is a
mean proportional between the whole line and the smaller
part, it is said to be divided in extreme and mean ratio.
281. It will be found on investigation that the order of four
proportionals, when all of the same kind, can be varied, as well
as certain other transformations effected, without destroying
the proportion. The principal of these are as follows :
58 PLANE GEOMETRY.
282. Magnitudes are said to be in proportion by alternation
when either the two means or the two extremes are made to
exchange places.
283. Magnitudes are said to be in proportion by inversion
when the means are made to exchange places with the extremes.
284. Magnitudes are said to be in proportion by composition
when the sum of the terms of the first ratio is to either term
of the first ratio as the sum of the terms of the second ratio
is to the corresponding term of that ratio.
285. Magnitudes are said to be in proportion by division
when the difference of the terms of the first ratio is to either
term of that ratio as the like difference of the terms of the
second ratio is to the corresponding term of that ratio.
286. Magnitudes are said to be in proportion by composition
and division when the sum of the terms of the first ratio is to
their difference as the sum of the terms of the last ratio is to
their like difference.
THEOREMS.
287. Equimultiples of two magnitudes are in the same ratio
as the magnitudes themselves.
CASE I. When the magnitudes are commensurable.
a \ i i .'
b I i 1 5
Let x and y be two commensurable magnitudes, and a and b
be their respective equimultiples.
We are to prove x : y : a : b.
Dem. Since x and y are commensurable (262), they must
have a common measure, as the line c. Suppose it to be con-
tained in x n times, and in y m times ; i.e. that on dividing
PLANE GEOMETRY. 59
the two lines x and y into n and m equal parts respectively,
the parts will all individually be equal to the line c. Con-
sequently if the line c be multiplied first by n and then by m
(see 39), the resulting lines will be exactly equal in length to
the lines x and y respectively.
These relations may be symbolically expressed as follows:
(1) - = c and ^ = c.
n m
(2) n x c = x and m x c = y.
n X c x n x
Whence = -, or = -.
m x c y my
Let q be the number by which x and y are multiplied to
obtain the equimultiples a and b ; then
(3) q x x = a and q xy = b.
But from (2) g x x = q x ft X c,
and qxy = qxmxc.
Whence a = q xnx c,
and b qxmxc.
a a x n x c n
Therefore - = i =
b q xmx c m
Whence - = -, or a:b::x:y. Q.E.D.
b y
CASE II. When the magnitudes are incommensurable.
Let x and y be two incommensurable magnitudes, and a and
b two equimultiples of x and y.
60 PLANE GEOMETRY.
We are to prove y : x : : b : a.
Let x be divided into n equal parts, and designate the value
of one of these parts by p. Let q denote the number of times
x and y are contained in a and b respectively. Since x and y
are incommensurable, p will be contained in y m times with a
remainder r.
Then x = np (1) and qx = qnp ;
but a = qx, .*. a = qnp. (2)
Again, y = mp + r (3) and qy = qmp -f qr ;
but bqy, .'. b = qmp + qr. (4)
Dividing (4) by (2), and (3) by (1),
b _ qmp qr y_mpr
a~~ qnp qnp x ~ np np'
Simplifying,
& = + Jl and ? = + .
a n np x n np
Whence - = -, or y : x : : b : a. Q.E.D.
x a
288. If two commensurable ratios can be expressed by the
same numerical value, then these two ratios are equal, and
consequently the four magnitudes composing those ratios are
proportionals.
Let a, b, x, and y represent four magnitudes, a and b being
commensurable, also x and y, so that, on applying their re-
,. . a m x m
spective units of measure, =- = and - =
b n y n
n o*
Then r = - (Axiom III.) or a: b:: x: y. Q.E.D.
PLANE GEOMETRY. 61
289. If two incommensurable ratios can be expressed by the
same approximate numerical value, however small the unit of
measure, then these two ratios are equal, and the four magni-
tudes comprising those ratios are proportionals.
Post. Let a : b and x : y be two ratios, each of which is
incommensurable and whose true values lie between the ap-
proximate values and u . (See 271.)
n n n
We are to prove that a:b::x:y; i.e. that there is no dif-
ference between the values of - and - ; i.e. that they are equal.
b y
Dem. It is evident that - and - are either equal or unequal.
b y
Let us assume that they are unequal, and designate their
difference by d. This difference being fixed and definite
cannot be changed by any legitimate manipulation or trans-
formation of the ratios.
By the hypothesis,
(1) ^> and ?>.
b n y n
(2) ?<!? + l and 2<5 + i.
b n n y n n
Consequently the difference between - and - must be less
-, by
than .
n
Now by decreasing the value of the common unit of measure,
the value of n increases and that of - decreases accordingly.
Hence we may conceive n to be so great that - will be smaller
n
than d ; i.e. the difference between ? and ~ is equal to d and
b y
less than d at the same time. This is of course absurd or
impossible. Hence they cannot differ in value ; i.e. they are
equal, and a:b::x:y. Q.E.D.
62 PLANE GEOMETRY.
290. In the demonstration of the following theorems, the
expression " four quantities " means the numerical measures
of four geometrical magnitudes of the same kind. It would
be well, however, to represent the magnitudes either by angles
or lines, as in 287.
291. If four quantities form a proportion, the product of the
means equals the product of the extremes.
292. Remark. Our ability to demonstrate the theorems in
proportion depends on our knowledge of the principles govern-
ing the manipulation of the equation. Hence the first thing to
be done is to change the proportion form to the equivalent
equation form. The author is of the firm belief that mathe-
maticians have no right to amalgamate these two forms of
expression, and that pupils should be taught to rigidly dis-
criminate in their use.
Post. Let the four quantities a, 6, c, and d form a propor-
tion so that
a : b : : c : d.
We are to prove that ad = be.
Dem. a - = *
b d
(Changing to equation form.)
(Multiplying both members of the equation by d.)
ad = be.
(Multiplying both members of the previous equation by b.)
Q.E.D.
293. If three quantities form a proportion, the product of
the extremes is equal to the square of the mean.
PLANE GEOMETRY. 63
294. If the product of two quantities equals the product
of two others, the two factors of either product may be made
the means, and the other two the extremes of a proportion.
Post. Let ex = an.
We are required to form a proportion from the four factors
c, x, a, and n, in which c and x shall be the extremes.
Dem. c = . Why?
= 1 Why?
a x
.*. c : a : : n : x.
(Changing from equation form to the proportion form.)
Form proportions from the following equations in which the
factors of the product marked Ex. shall form the extremes,
avoiding the factor 1.
Ex.
I.
2 x = ac.
Ex.
II.
6 = ax.
Ex.
III.
3 Va = 14.
Ex.
IV.
a(x + y) = n 2 .
Ex.
V.
2/ 2 = an -f- ac.
Ex.
VI.
ab + bx = en +
yc.
Ex.
VII.
aic + aj = 2/6 2 +
b 2 .
Ex.
VIII.
a 2 1 = a 2 6 2 .
Ex.
IX.
ax + xb + cx =
nd + 7m + wfc.
Ex.
X.
x*+2cx + c?=
an + ny.
64 PLANE GEOMETRY.
295. If four quantities form a proportion, they will be in
proportion by inversion.
. Divide 1 by each member of the equation.
296. If four quantities form a proportion, they will be in
proportion by alternation.
Post. Let the four quantities a, a, c, and n form a propor-
tion so that
a : x : : c : n.
We are to prove that
a : c : : x : n.
Dem. - = -. Why ?
x n
- = -. Why?
xc n
Why?
.*. a : c : : x : n. Q.E.D.
Query. If the corresponding terms of two proportions be
added, will the sums form a proportion ? Numerical example.
297. If four quantities form a proportion, they will be in
proportion by composition.
Post. Let the four quantities x, y, a, and b form a pro-
portion so that
x : y : : a : b.
We are to prove,
I. x -\-y :y : : a -\-b :b.
II. x -\- y : x : : a + b : a.
Dem. - = -. Why?
y b
PLANE GEOMETRY. 65
+1 = + 1. Why?
~ + y - = l + l Why?
y y b^ b
-. Why?
y b
.: x + y:y::a + b:b. Why?
Q.E.D.
The pupil should demonstrate Part II.
Sug. Consult 295.
298. If four quantities form a proportion, they will be in
proportion by division.
Sug. There are four cases. In Case I. subtract 1 from each
member of the equation. The other cases may be similarly
demonstrated by reversing the above process, and consulting
295.
299. If two proportions have a ratio in each equal, the other
two ratios will form a proportion.
300. If two proportions have the two antecedents of one
equal respectively to the two antecedents of the other, the
consequents will form a proportion.
301. If two proportions have the two consequents of one
equal respectively to the two consequents of the other, the
antecedents will form a proportion.
302. If four quantities form a proportion, they will be in
proportion by composition and division.
Sug. Consult 298, 297, 296, and 299.
303. If any number of magnitudes of the same kind form a
proportion, the sum of the antecedents is to the sum of the
consequents as any antecedent is to its consequent.
66 PLANE GEOMETRY.
Post. Let the quantities a, x, c, w, d, and r form a continued
proportion, so that
a : x: : c:n: :d:r.
We are to prove,
or II. : : c : n ;
or III. : : d : r.
Dem. Now, if this theorem can be demonstrated, then
Case I. must be a true proportion. Let us temporarily assume
it to be true, and trace the results.
First. x(a + c + d) = a(x + n+r), Why?
or ax -\- ex + dx = ax -f an + ar ; Why ?
but (1) ax = ax. Why?
.. ex + dx = an + ar. Why ?
(2) ca; = an. Why ?
.-. (3) dx = ar. Why?
Now it is evident that if we could obtain the equations (1),
(2), (3) from our given proportions, we could reverse the
above process, and thus demonstrate the theorem.
Let us bear in mind that the ratio a : x in Case I. is the
one that is to form the proportion with that composed of the
sums of the antecedents and consequents, and that equation (1)
is formed from the product of the terms of that ratio.
Hence (1) ax = ax ; Why ?
(2) ex = an ; Why ?
(3) dx = ar. Why?
Hence ax -f- ex + dx = ax -f an -f- ar, Why ?
or x(a + c + d) = a(x + n + r). Why?
The pupil should be able to finish this case, and also demon-
strate the other two without difficulty. Consult 204.
PLANE GEOMETRY. 67
304. If four quantities form a proportion, the terms of
either ratio may be either multiplied or divided by the same
quantity, and the results still form a proportion.
304 (a). If the antecedents or consequents of a proportion
be either multiplied or divided by the same quantity, the
results will still form a proportion.
305. If four quantities form a proportion, and the terms of
one ratio be either multiplied or divided by the same quantity,
while both terms of the other ratio be either multiplied or
divided, either by the same or different quantity from that
used in the first ratio, the results will still form a proportion.
Sug. The pupil should take each case involved in the state-
ment of the above theorem separately.
306. If two proportions be given, the products of the corre-
sponding terms will also form a proportion.
307. If two proportions be given in which two corresponding
ratios have the antecedent of one equal to the consequent of
the other, the remaining antecedent and consequent, together
with the products of the corresponding terms of the other two
ratios, will form a proportion.
308. If the antecedents of a proportion are equal, the con-
sequents are equal.
308 (a). Converse of 308.
309. If four quantities form a proportion, their like powers
and like roots will also form a proportion.
310. If three quantities form a proportion, the first is to
the third as the square of the first is to the square of the
second.
311. If four quantities form a proportion, the sum of the
squares of the first two terms is to their product as the sum
of the squares of the last two is to their product.
68 PLANE GEOMETRY.
311 (a). Substitute "difference" for "sum" in 311.
312. If two quantities be either increased or diminished by
like parts of each, the results will be in the same ratio as the
quantities themselves.
312 (a). If three terms of a simple proportion are equal
respectively to the three corresponding terms of another pro-
portion, the fourth terms of the two proportions ^re equal.
PKOPOBTIOffAL LINES.
313.* If a line be drawn parallel to one side of a triangle,
the four parts into which it divides the other sides will form
a proportion.
Post. Let ABC be any tri-
angle, and S T a line drawn
parallel to the side CB.
We are to prove that the
four parts AS, SB, AT, and
TC form a P r P rtion -
Dem. Let us conceive the
line DH passing through the
vertex A and parallel to BC,
to move toward BC and remaining always parallel to it. The
instant it starts there will, of course, be two points of inter-
section, as and P. It is evident that when the point has
reached the point B, the point P will have reached the point C.
Why?
Again, if Q and R be the middle points of the sides AB and
AC respectively, it is evident that when the point reaches
the point Q, the point P will reach the point R. Why ?
Hence when the point O has moved over one-half of AB,
the point P has moved over one-half of AC. Similarly, when
has moved over one-fourth, one-tenth, one-thousandth, or
* See Appendix.
PLANE GEOMETRY. 69
one-7ith of AB, P has moved over the same part of AC.
Hence, no matter what the position of the moving line, AO is
always the same part of AB that AP is of AC; that is, the
ratio of AO to AB is the same as the ratio of AP to AC', or
AO : AB : : AP : AC.
Now, since by Hyp. ST is parallel to BC when point
reaches $, point P reaches T.
Hence AS : AB : : AT : AC.
But AB-AS:AS::AC-AT:AT. Why?
But AB -AS = SB and AC-AT=TC.
.'. SB:AS::TC: AT. Why ?
Q.E.D.
314. If a line be draw'n parallel to one side of a triangle,
the other two sides and either pair of corresponding parts
form a proportion.
Sug. Use result obtained in previous theorem, and consult 297.
315. Converse of 313. 4
Post. Let ACH be a tri-
angle, with line BD drawn
so that
AB:BC::AD: DH.
We are to prove BD par-
allel to CH.
Dem. First, suppose BKto be drawn through B parallel to CH.
Then AB : BC :: AK: KH. Why?
/. AD:DH:: AK : KH. Why ?
.-. AD +DH:DH: : AK+KH : KH; Why ?
or AH:DH::AH:KH.
.-. DH=KH. Why?
.-. points D and K must coincide. Why ?
70 PLANE GEOMETRY.
What, then, must be the relative position of the lines BK
and BD ? Why ?
But BK was drawn parallel to CH. Consequently _BZ>,
which coincides with BK, must also be parallel to CPL Q.E.D.
316. If a series of parallel transversals, intersecting any
two straight lines, intercept equal distances on one of these
lines, they will also intercept equal distances on the other.
317. If a series of parallel transversals, intersecting any
number of straight lines, intercept equal distances on one of
these lines, they will also intercept equal distances on each of
the others.
318. If two lines be drawn parallel to one side of a triangle
intersecting the other two sides, the parts thus intercepted
form a proportion.
318 (a) . If a line be drawn parallel to one side of a triangle,
this side, its parallel, and either of the other two sides, together
with the segment of the latter joining the third side, will form
a proportion.
319. If any number of straight lines be intersected by a
series of parallel transversals, the parts thus intercepted by
the latter form a proportion.
320. If a line be drawn parallel to the base of a triangle,
and another from vertex to base, the parts of both the base
and its parallel will form a proportion.
321. If any number of straight lines which pass through a
common point be intersected by a series of parallel transversals,
the parts intercepted by both parallels and non-parallels form
a proportion.
322. Converse of 321.
323. The bisector of an angle of a triangle divides the oppo-
site side into segments proportional to the other two sides.
Sug. From the vertex of the bisected angle extend one of
PLANE GEOMETRY. 71
the sides, and from one of the other vertices draw a line
parallel to the bisector meeting the side extended.
324. If the bisector of an exterior angle of a triangle meet one
of the sides extended, this side plus its extension, the extension,
and the other two sides of the triangle, form a proportion.
Sug. From extremity of the side extended, draw a parallel
to the bisector.
POLYGONS.
325. A polygon is a portion of a plane bounded by straight
lines.
What is the least number of sides that a polygon can have ?
The greatest number ?
326. The word "polygon " is from two Greek words meaning
many angles.
What is a polygon of three sides usually called ? It is also
called a trlgon. A polygon of four sides is usually termed
what ? It is also called a tetragon. A polygon of five sides
is called a pentagon, one of six sides a hexagon, one of seven
sides a heptagon, one of eight sides an octagon, one of nine sides
a nonagon or enneagon, one of ten sides a decagon, one of
eleven sides an undecagon, one of twelve sides a dodecagon, and
one of fifteen sides a pentedecagon.
327. A convex polygon is one each of whose angles is less
than 180.
A concave polygon is one that has one or more re-entrant
angles ; i.e. where, at one or more vertices, the angular space
within the polygon is more than 180.
Whenever polygons are mentioned in this work, convex
polygons are meant unless otherwise specified.
The diagonal of a polygon is a line joining any two vertices
not contiguous.
328. A regular polygon is one that is both equilateral and
equiangular.
72 PLANE GEOMETRY.
The perimeter of a polygon is the sum of its sides. How
many angles has each of the above-named polygons ? How
does the number of sides compare, in each case, with the num-
ber of angles ? How many angles, then, has a polygon of n
sides ? How many diagonals can be drawn from a single
vertex in each of the above-named polygons ? Compare the
number of diagonals in each case with the number of sides.
If the polygon has n sides, how many diagonals can be drawn
from a single vertex? How many triangles are formed by
drawing diagonals from a single vertex in each of the above-
named polygons ? How does the number of triangles compare
with the number of sides ? If the polygon has n sides, how
many triangles would be formed by diagonals similarly
drawn ?
If from any point selected at random in a polygon lines be
drawn to the vertices, into how many triangles will it be thus
divided ?
329. The sum of the interior angles of a polygon is equal to
two right angles, taken as many times, less two, as the polygon
has sides ; or twice as many right angles as the polygon has
sides, less four right angles.
Sug. Divide the polygon up into triangles, and let n equal
the number of sides. Having found an expression in terms of
n, and a right angle for the sum of the angles, what would be
the value of one angle if the polygon were equiangular?
330. The sum of the exterior angles of any polygon formed
by extending each of its sides in succession similarly, is equal
to four right angles.
Sug. Consult Sect. 25 and Theorem 325.
PROBLEMS OF COMPUTATION.
331. Find the value in right angles of the sum of the interior
angles of a pentagon, hexagon, octagon, decagon, dodecagon, pen-
tedecagon, and a polygon of fifty-two sides.
PLANE GEOMETRY. 73
332. Find the value, in terms of a right angle as the unit,
of one of the angles of an equiangular pentagon, octagon, dodec-
agon, a polygon of twenty sides, one hundred sides.
333. Find the values of the above angles in degrees, also in
grades.
334. How many sides has an equiangular polygon, the sum
of four angles of which is equal to seven right angles ?
335. How many sides has an equiangular polygon, the sum
of three angles of which is equal to five right angles ?
336. How many sides has an equiangular polygon, the sum
of nine angles of which is equal to sixteen right angles ?
337. How many sides has the polygon, the sum of whose
interior angles is equal to the sum of its exterior angles ?
338. How many sides has the polygon, the sum of whose
interior angles is double that of its exterior angles ?
339. How many sides has the polygon, the sum of whose
exterior angles is double that of its interior angles ?
340. How many sides has the polygon, the sum of whose
interior angles is equal to nine times the sum of its exterior
angles ?
341. How many sides has the equiangular polygon, when
the sum of nine of its interior angles is four times the sum of
its exterior angles ?
342. How many sides has the equiangular polygon, when
the sum of five of its interior angles is equal to two and one-
fourth times the sum of its exterior angles ?
343. How many sides has the equiangular polygon when
(a) 5^==9frt.z? (e) 8 A = 3J times its ext. A ?
(b) 4 = 7J " (/) 6 " = 3f " " "
(c) 16 " = 31 (g) 7 " = 6|rt. A ?
(d) 5 = 8 (h) 3 " = 5|
74 PLANE GEOMETRY.
SIMILAR FIGUKES.
344. Similarity in general means likeness of form; i.e. two
figures are said to be similar when they have the same shape,
although they may differ in size.
Similar geometrical figures are those whose homologous angles
are equal, and whose homologous sides form a proportion.
345. Homologous sides of similar polygons are those which
join the vertices of equal angles in the respective polygons.
In similar triangles the homologous sides are those that are
opposite the equal angles.
The pupil should be careful to observe at the outset that
similarity involves two things ; viz. equality of angles and pro-
portionality of sides.
346. Two polygons are said to be mutually equiangular when
the angles of one are equal respectively to the corresponding
angles of the other, and mutually equilateral when each side of
one has an equal side in the other.
SIMILAE TEIAttaLES.
347. If two triangles be mutually equiangular, they are
similar.
A
p
Post. Let ABH and KDN be two triangles having
and
We are to prove the triangles similar ; i.e. that their homol-
ogous sides form a proportion.
PLANE GEOMETRY. 75
Cons. Mark off on HB a distance HP equal to KN, and on
HA a distance HQ equal to DN, and join PQ.
Dem. What relation exists between the two A RDN and
PQH? Why?
What, then, must be the relation between the A PQH and
D ? What between A QPH and K?
What, then, must be the relation between the A PQH and
A ? What between A QPH and B ? Why ?
What, then, must be the relative position of the lines AB
and QP ? Why ?
Apply Theorem 314, and the remainder of the demonstration
should be easy. Q.E.D.
348. If two triangles have two angles of one equal respec-
tively to two angles of the other, the two triangles are similar.
349. If two right triangles have an acute angle of one equal
to an acute angle of the other, the two triangles are similar.
350. If two triangles have an angle in each equal, and the
sides including those angles form a proportion, the two tri-
angles are similar.
Sug. Apply one to the other so that the equal angles shall
coincide ; then consult Theorem 315. Or proceed in a manner
similar to that in Theorem 347.
351. If two triangles have their sides respectively parallel,
they are similar.
Sug. Consult 94 and 98, then three possible hypotheses
regarding the relations of the respective pairs of angles.
352. If two triangles have their sides respectively perpen-
dicular to each other, they are similar.
Sug. Consult 95 and 98, and see suggestions to 351. Consult
the caution in Sect. 55.
353. If two triangles be similar, their homologous altitudes
are in the same ratio as either pair of homologous sides
including the vertical angle.
76 PLANE GEOMETRY.
354. If two triangles be similar, their homologous altitudes
are in the same ratio as their homologous bases.
355. If the homologous sides of two triangles form a pro-
portion, the triangles are similar.
Post. Let ABC and DHK be two triangles, their homol-
ogous sides forming the continued proportion.
AB : DHi: AC : DK : : BC : HK.
We are to prove that the two triangles are similar ; i.e. that
they are mutually equiangular.
Cons. Make NC equal to DK, PC equal to HK, and join
NP.
Dem. In the given proportion substitute for DK and HK
their equals NC and PC. Then
AC:NC::BC: PC.
What is true, then, of the two triangles ABC and NPC ?
See Theorem 350.
.-. AB:NP::BC: PC. Why ?
But by Hyp.,
AB-.DH'.'.BC'.HK.
What is true of the first terms of these two proportions ?
Of the third terms ? Of the last terms ? What must be true,
then, of the second terms ?
The pupil should finish the demonstration without difficulty.
PLANE GEOMETRY. 77
356. If two polygons be similar, the diagonals drawn
from homologous vertices will divide them into the same
number of triangles, similar two and two, and similarly
placed.
357. Converse of 356.
358. The perimeters of two similar polygons are in the same
ratio as any two homologous sides or any two homologous
diagonals.
359. The perimeters of two similar polygons are in the same
ratio as the bisectors of any two homologous angles or any
two homologous lines howsoever drawn.
360. If, in a right triangle, a perpendicular be drawn to the
hypothenuse from the vertex of the right angle,
I. the two triangles thus formed are similar to the original
triangle and to each other.
II. the perpendicular and the two segments of the hypothe-
nuse form a proportion.
(Where three quantities form a proportion, what relation
must one of these quantities sustain to the other two ?)
III. either leg, the hypothenuse, and that segment of the
latter which joins the former, form a proportion.
(The pupil should preserve these proportions for use further
on.)
IV. the squares of the two legs and the two segments of
the hypothenuse form a proportion.
Sug. Use the two proportions in III. by Theorem 293, and
divide.
V. the square of the hypothenuse bears the same ratio to
the square on either leg, as the hypothenuse bears to that seg-
ment of the latter joining that leg.
VI. the two legs, the hypothenuse, and the perpendicular
form a proportion.
78
PLANE GEOMETRY.
360 (a). The square of the hypothemise of a right triangle
is equal to the sum of the squares of the two legs. See
Theorem 426.
Sug. Use result in 360, IV., by composition, and then com-
pare with 360, V.
361. The square of the diagonal of a square is equal to twice
the square of one of its sides.
PKOJEOTION.
362. The projection of one line upon another is that portion
of the latter included between two perpendiculars to the latter
drawn from the extremities of the former.
K' B
Fig. I.
For example, HKis the projection of the line CD upon AB,
and RS is the projection of RQ on NP; it is also the projec-
tion of RQ on RF. Mention all the cases of projection in
Figure III., the dotted lines being perpendicular to the respec-
tive sides.
363. In any triangle the square of the side opposite an acute
angle is equal to the sum of the squares of the other two sides,
minus twice the product of one of those sides and the projec-
tion of the other upon that side.
If C be the acute angle, then by a glance at the following
diagram it will be readily seen that there may be two cases
depending upon whether the projection involves an extensiot
PLANE GEOMETRY.
79
of one side, which will evidently be the case if the side to be
projected is opposite an obtuse angle.
C V
Fig. I.
We are to prove
In Case I.
In Case II.
Fig. H.
=BU 2 +AU* - 2BC - DO.
DB =BC -DC.
DB=DC-BC.
The square of either of these two equations gives the same
result ; hence to the square add the squares of the perpendicu-
lar, and then combine the terms by using Theorem 360 (a).
364. In an obtuse triangle the square of the side opposite
the obtuse angle is equal to the sum of the squares of the other
two sides, plus twice the product of one of those sides and the
projection of the other upon that side.
Sug. Form an equation by placing the projection of the side
opposite the obtuse angle equal to the sum of its two parts,
then proceed as in 363.
365. If any median of a triangle be drawn,
I. the sum of the squares of the other two sides is equal to
twice the square of one-half the bisected side, plus twice the
square of the median ; and
II. the difference of the squares of the other two sides is
equal to twice the product of the bisected side and the pro-
jection of the median upon that side.
Sug. Use 363 and 364, then combine the resulting equations.
80 PLANE GEOMETRY.
366. In any quadrilateral the sum of the squares of the four
sides is equal to the sum of the squares of the diagonals, plus
four times the square of the line joining the middle points of
the diagonals.
Sug. Use 365. What modification of the above theorem
would result if the quadrilateral were a parallelogram ?
367. If from any point, selected at random, in the circum-
ference of a circle, a perpendicular be drawn to a diameter,
this perpendicular will be a mean proportional between the
two segments of the diameter.
368. If two chords of a circle intersect each other, the four
parts form a proportion.
369. If two secants be drawn from the same point without
the circle, the entire secants and the parts that are without
the circle form a proportion.
370. If, from the same point, a tangent and secant to a cir-
cle be drawn, the tangent, the secant, and that part of the
latter that is outside the circle, will form a proportion.
371. The two tangents to two intersecting circles from any
point in their common secant are equal.
Sug. Consult 370.
372. If two circles intersect each other, their common chord
extended bisects their common tangents.
ADVANCE THEOREMS.
373. In any triangle the product of any two sides is equal
to the diameter of the circumscribed circle multiplied by the
perpendicular drawn to the third side from the vertex of the
angle opposite.
Sug. Construct the diameter from the same vertex as the
perpendicular, and join its extremity with one of the other
vertices, making a right triangle similar to the one formed by
the perpendicular, whence the necessary proportion.
PLANE GEOMETRY. 81
374. In any triangle the product of any two sides is equal
to the product of the segments of the third side, formed by
the bisector of the opposite angle, plus the square of the
bisector.
Sug. Circumscribe a circle, and extend the bisector to the
circumference, and connect its extremity with one of the other
vertices of the triangle. Then if the vertices of the triangle
be lettered A, B, and (7, and the bisector be drawn from A,
and D the point where the bisector crosses the side BC, and
E the extremity of the bisector extended, then the triangles
BAD and ACE can be proved similar.
375. If one side of a right triangle is double the other, the
perpendicular, from the vertex of the right angle to the hypoth-
enuse, divides it into segments which are to each other as 1
to 4.
376. A line parallel to the bases of a trapezoid, passing
through the intersection of the diagonals, and terminating in
the non-parallel sides, is bisected by the diagonals.
377. In any triangle the product of any two sides is equal
to the product of the segments of the third side, formed by the
bisector of the exterior angle at the opposite vertex, minus the
square of the bisector.
Sug. Consult Theorem 374.
378. The perpendicular, from the intersection of the medians
of a triangle, upon any straight line in the plane of the tri-
angle, is one-third the sum of the perpendiculars from the ver-
tices of the triangle upon the same line.
379. If two circles are tangent to each other, their common
tangent and their diameters form a proportion.
380. If two circles are tangent internally, all chords of the
greater circle drawn from the point of contact are divided pro-
portionally by the circumference of the smaller.
82 PLANE GEOMETRY.
381. In any quadrilateral inscribed in a circle, the product
of the diagonals is equal to the sum of the products of the
opposite sides.
Sug. From one vertex draw a line to the opposite diagonal,
making the angle formed by it and one side equal to the angle
formed by the other diagonal and side which meets the former.
382. If three circles whose centres are not in the same
straight line intersect one another, the common chords will
intersect each other at one point.
383. If two chords be perpendicular to each other, the sum
of the squares of the four segments is equal to the square of
the diameter.
384. The sum of the squares of the diagonals of a quadri-
lateral is equal to twice the sum of the squares of the lines
joining the middle points of the opposite sides.
PROBLEMS OF COMPUTATION.
385. The chord of one-half a certain arc is 9 inches, and the
distance from the middle point of this arc to the middle of
its subtending chord is 3 inches. Find the diameter of the
circle.
386. The external segments of two secants to a circle from
the same point are 10 inches and 6 inches, while the internal
segment of the former is 5 inches. What is the internal seg-
ment of the latter ?
387. The hypothenuse of a right triangle is 16 feet, and the
perpendicular to it from the vertex of the opposite angle is
5 feet. Find the values of the legs and the segments of the
hypothenuse.
388. The sides of a certain triangle are 6, 7, and 8 feet
respectively. In a similar triangle the side corresponding to
8 is 40. Find the other two sides.
PLANE GEOMETRY. 83
389. The sides of a certain triangle are 9, 12, and 15 feet
respectively. Find the segments of the sides made by the
bisectors of the several angles.
390. If a vertical rod 6 feet high cast a shadow 4 feet long,
how high is the tree which, at the same time and place, casts
a shadow 90 feet long ?
391. The perimeters of two similar polygons are 200 and
300 feet respectively, and one side of the former is 24 feet.
What is the corresponding side of the latter ?
392. How long must a ladder be to reach a window 24 feet
high, if the lower end of the ladder is 10 feet from the side of
the house ?
393. Find the lengths of the longest and the shortest chord
that can be drawn through a point 6 inches from the centre of
a circle whose radius is 10 inches.
394. The distance from the centre of a circle to a chord 10
inches long is 12 inches. Find the distance from the centre
to a chord 24 inches long.
395. The radius of a circle is 5 inches. Through a point
3 inches from the centre a diameter is drawn, and also a chord
perpendicular to the diameter. Find the length of this chord,
and the distance from one end of the chord to the ends of the
diameter.
396. Through a point 10 feet from the centre of a circle
whose radius is 6 feet tangents are drawn. Find the lengths
of the tangents and of the chord joining the points of contact.
*
397. If a chord 8 feet long be 3 feet from the centre of the
circle, find the radius and the distances from the end of the
chord to the ends of the diameter which bisects the chord.
398. Through a point 5 inches from the centre of a circle
whose radius is 13 inches any chord is drawn. What is the
84 PLANE GEOMETRY.
product of the two segments of the chord ? What is the length
of the shortest chord that can be drawn through that point ?
399. From the end of a tangent 20 inches long a secant is
drawn through the centre of a circle. If the exterior segment
of this secant be 8 inches, what is the radius of the circle ?
400. A tangent 12 feet long is drawn to a circle whose
radius is 9 feet. Find the external segment of a secant through
the centre from the extremity of the tangent.
401. The span of a roof is 28 feet, and each of its slopes
measures 17 feet from the ridge to the eaves. Find the height
of the ridge above the eaves.
402. A ladder 40 feet long is placed so as to reach a window
24 feet high on one side of the street, and on turning the
ladder over to the other side of the street, it just reaches a
window 32 feet high. What is the width of the street ?
403. The bottom of a ladder is placed at a point 14 feet
from a house, while its top rests against the house 48 feet
from the ground. On turning the ladder over to the other
side of the street its top rests 40 feet from the ground. Find
the width of the street.
404. One leg of a right triangle is 3925 feet, and the differ-
ence between the hypothenuse and other leg is 625 feet. Find
the hypothenuse and the other leg.
AEEAS,
405. The area of a surface is its numerical measure ; i.e.
the numerical expression for the number of times it contains
another surface arbitrarily assumed as a unit of measure. For
example, the area of the floor of a room is the number of
times it contains some one of the common units of surface,
square foot, square yard, etc. This unit of surface is called
the superficial unit. The most convenient superficial unit is
PLANE GEOMETRY.
85
the square of the linear unit. The square of a linear unit (or
line) is the area of the square constructed with that linear
unit for its sides. Surfaces that are equal in area are said to
be equivalent.
406.* If two rectangles have equal altitudes, their areas
will be in the same ratio as their bases.
CASE I. When the bases are commensurable (262).
D
w w w"
-JV"
r
Q
s
s'
1
J
J
?
K
t 1
V t"
a a' a"
Post. Let ABCD and HKNP be two rectangles, having their
altitudes AD and HP equal, and their bases AB and HK com-
mensurable. Designate their areas by Q and R respectively.
We are to prove Q:R::AB: HK.
Dem. Since AB and HK are commensurable, they must
have a common measure (262).
Suppose this common measure to be contained in AB n times,
and in HK m times.
At the several points of division, a, a', a", etc., t, t 1 , t", etc.,
construct ab, a'b', a"b", etc., perpendicular to AB, and tw }
t'w', t"w", etc., perpendicular to HK.
Then the rectangle ABCD will be divided into n equal
rectangles r, r', r", etc., and the rectangle HKNP will be
divided into m equal rectangles s, s', s", etc. These smaller
rectangles are also equal to one another.
(The pupil should demonstrate the above propositions.)
Hence AB : HK : : n : m, (264.)
and Q : R :: n : m. (264.)
.-. Q : R ::AB:HK. (288.) Q.E.D.
* See Appendix.
86
PLANE GEOMETRY.
CASE II. When the bases are incommensurable.
Suppose AB to be divided into any number of equal parts,
as , and that one of these parts be applied to HK and found
to be contained m times with a remainder less than that part.
From the points of division construct lines as in Case I.
Then the rectangle ABCD will be divided into n equal rec-
tangles, and rectangle KNPH into m equal rectangles, with a
remaining rectangle less than one of the m rectangles.
Then Q : R and AB : HK are two incommensurable ratios.
T- TJ -J
(1) 5:>S and - <- + -.
Q n Q n n
and
AB n AB n n
Hence the two ratios R : Q and HK : AB have the same
approximate numerical value, whatever the magnitude of n.
.'. R:Q::HK:AB, (289.)
or Q'.R'.-.AB: HK. Q.E.D.
407. If two rectangles have equal bases, their areas are in
the same ratio as their altitudes.
Sug. Consider their altitudes as bases, and bases as alti-
tudes, and proceed as in 406.
408. The areas of any two rectangles have the same ratio
as the products of their respective bases and altitudes.
A rt A B n
A.
G D
K
C
"*,
N
P
H
K
TT
a
Kg. I.
Fig. H.
PLANE GEOMETRY. 87
(By the product of two lines is meant the product of the
numbers which represent them when both are measured by
the same linear unit.)
Post. Let ABCD and HKNP be any two rectangles, with
DC and PN their respective bases. Let us designate their
areas by A and a respectively, their altitudes by H and h,
and their bases by B and b.
We are to prove that A : a : : B x H : b X h.
Dem. Consider the two rectangles so placed that the ver-
tices C and K shall coincide, and the four angles all equal.
Then BCH is one straight line, as is also DKN. Why ?
Complete the rectangle BQNC. Then considering CN the
base of rectangle BQNC, it has the same altitude as rectangle
ABCD.
.-. Area ABCD : Area BQNC : : DC : KN. (1) Why ?
Again, considering BC the base of the rectangle BQNC,
Area HKNP : Area BQNC : : HK : BC, (2)
or Area BQNC : Area HKNP : : BC : HK (3) Why ?
Hence
Area ABCD : AreafflOTP : : BC X DC : HKxKN,
(See Theorem 307.)
or A: a::BxH :b X h. Q.E.D.
408 (a). Hence the above demonstration shows that if the
bases and altitudes of any two rectangles be measured by the
same linear unit, the ratio of their respective products (see
408) will be the ratio of their respective surfaces, and conse-
quently either may be assumed as the measure of the other.
For example, suppose the linear unit to be contained 5 times
in AD, 18 times in DO, 3 times in HP, and 6 times in PN.
Then A : a : : 5 X 18 : 3 X 6,
A 90
or
a
88
PLANE GEOMETRY.
This simply means that, using a as the unit, the area A is
5 times as large as the area a ; or that the area a, using A as
the unit surface, is ^ as large as A.
Practically it is more convenient to compare the areas of
both rectangles with the square of the linear unit (405) as a
unit of surface. This comparison is formally shown in the
enunciation and demonstration of the following theorem.
409. The area of a rectangle is measured by the product of
its base and altitude.
Post. Let ABCD be any rectangle ; and in whatever linear
unit the base and altitude be expressed, let r be a square whose
sides are the same unit. Designate its area by R, and its base
and altitude by 6 and h respectively.
We are to prove
Dem.
1x1, (408.)
or
R b x h .
r lxl ?
or, since r is the unit of area,
R = bxh.
Q.E.D.
When the base and altitude are commensurable, this is ren-
dered evident by dividing the rectangle
into squares, each equal to the superficial
unit, as shown in annexed figure. The
above demonstration, however, includes the
case when the base and altitude are in-
commensurable.
PLANE GEOMETRY.
89
410. If a parallelogram and rectangle have the same or equal
bases and the same or equal altitudes, they are equivalent.
K
Sug. Place them so that their bases shall coincide, as repre-
sented in above diagram. Then AB and HK are in the same
straight line. Why ?
Prove equality of the triangles ADH and BCK, and apply
Axiom VII.
411. The area of a parallelogram is measured by the prod-
uct of its base and altitude.
412. If two parallelograms have the same or equal bases
and the same or equal altitudes, they are equivalent.
413. If two parallelograms have the same or equal bases,
their areas are in the same ratio as their altitudes.
414. If two parallelograms have the same or equal altitudes,
their areas are in the same ratio as their bases.
415. The areas of any two parallelograms are in the same
ratio as the products of their respective bases and altitudes.
416. If a triangle and parallelogram have the same or equal
bases and the same or equal altitudes, the area of the latter
is double that of the former.
Sug. Place them so that their bases will coincide.
417. The area of a triangle is measured by one-half the
product of its base and altitude.
418. If two triangles have the same or equal bases and the
same or equal altitudes, they are equivalent.
419. If two triangles have the same or equal bases, their
areas are in the same ratio as their altitudes.
90 PLANE GEOMETRY.
420. If two triangles have the same or equal altitudes, their
areas are in the same ratio as their bases.
421. The areas of any two triangles are in the same ratio as
the products of their respective bases and altitudes.
422. The area of a trapezoid is equal to the product of its
altitude and one-half the sum of its parallel sides.
423. The square described upon the sum of two lines is
equivalent to the sum of the squares upon the two lines, plus
twice the rectangle formed by the two lines.
424. The square described on the difference of two lines is
equivalent to the sum of the squares upon the two lines, minus
twice the rectangle formed by the two lines.
425. The rectangle formed by the sum and difference of two
lines is equivalent to the difference of the squares upon the
two lines.
426. The square described upon the hypothenuse of a right
triangle is equivalent to the sum of the squares described
upon the legs.
This theorem was first demonstrated by Pythagoras, about
450 B.C., and hence is called the Pythagorean theorem. It has
been a favorite one with mathematicians, and consequently
about fifty different demonstrations of it have been recorded.
Among the many diagrams used, the following are a few ; and
it is hoped that the pupil will endeavor to demonstrate it for
himself, using either one of these, or, preferably, one of his
own. No. I. is the diagram used by Euclid in the demonstra-
tion of this theorem, constituting his famous "47th." No.
VII. is the diagram used by the late President Garfield, who
was said to have utilized his leisure hours in Congress in
mathematical investigation. In each figure ABC is the given
triangle, and in No. VII. ADC is one-half the square upon
the hypothenuse. DH is drawn parallel to BC to meet BA
extended. The method is algebraic, and involves an equation
between the sum of the areas of the three triangles and the
trapezoid BCDH.
PLANE GEOMETRY.
91
Fig.VH.
Fig. VI.
92
PLANE GEOMETRY.
Fig. VIH.
See also Theorem 360 (a).
Fig. IX.
427. The following demonstration is original with the
author :
A
Post. Let AC be any chord in the circle DAC, and DK a
diameter perpendicular to AC. Also let DH be drawn inter-
secting AC. Join HK-j then
but
hence
but
DN : DK : : DB : DH,
.'. DNx DH=DKx DB-,
DK=DB+BK;
DN x DH = DB (DB + BK).
DN x DH = DB~ + DB x BK-,
DB x BK= AB x BO =
.-. DN X Z>#=
PLANE GEOMETRY.
93
Now conceive chord DH to revolve about D as a centre
until the point H coincides with point A ;
then DN = DH = DA.
.-. DA*= DB 2 + Aff. Q.E.D.
For other methods of demonstrating this theorem, see Jour-
nal of Education, June 23 and July 7, 1887 ; and May 24 and
June 28, 1888.
427 (a). Converse of 426.
428. The diagonal of a square is incommensurable with its
side.
Sug. Find value of diagonal in terms of its side by Theorem
426.
429. If two triangles have an angle in each equal, their
areas are in the same ratio as the products of the sides which
include the equal angles.
Post. Let ABC and DHKke two triangles with
Designate their areas by S and s, and the sides opposite the
several angles by the corresponding small letters.
We are to prove that
S : s::cxb:7cx d.
Apply the triangles so that the equal angles shall coincide,
or extend the sides of one until they equal the corresponding
sides of the other, as in the above diagram.
94 PLANE GEOMETRY.
Dem. I. Area DHK : Area NHK: : DH : NH. Why ?
II. Area NHK : Area NHP : : HK : HP. Why ?
Whence
AreaDffiT : Area NHP : : DHx HK : NH* HP.
Why?
or S :s::kxd:cxb. Q.E.D.
430. The areas of two similar triangles have the same ratio
as the squares of any two homologous sides.
Sug. Consult 421 and 354.
431. The areas of two similar polygons are in the same ratio
as the squares of any two homologous sides.
Sug. Consult 356, 43Q, and 303.
432. The areas of two similar polygons are in the same ratio
as the squares of any two homologous diagonals.
433. Any two homologous sides or diagonals of two similar
polygons are in the same ratio as the square roots of their
areas.
434. If similar polygons are described upon the sides of a
right triangle as homologous sides, the polygon described upon
the hypothenuse is equivalent to the sum of the polygons
upon the other two sides.
ADVANCE THEOREMS.
435. If two triangles have two sides of one equal respec-
tively to two sides of the other, and their included angles
supplementary, the triangles are equivalent.
436. If a straight line be drawn through the centre of a
parallelogram, the two parts are equivalent.
PLANE GEOMETRY. 95
437. If through the middle point of the median of a trape-
zoid a line be drawn, cutting the bases, the two parts are
equivalent.
438. In every trapezoid the triangle which has for its base
one leg, and for its vertex the middle point of the other leg, is
equivalent to one-half the trapezoid.
439. If any point within a parallelogram, selected at ran-
dom, be joined to the four vertices, the sum of the areas of
either pair of opposite triangles is equivalent to one-half the
parallelogram.
440. The area of a trapezoid is equal to the product of one
of its legs and the distance of this leg from the middle point
of the other.
441. The triangle whose vertices are the middle points of
the sides of a given triangle is equivalent to one-fourth the
latter.
442. The parallelogram formed in 101 is equivalent to one-
half the quadrilateral.
443. If two parallelograms have two contiguous sides respec-
tively equal, and their included angles supplementary, the par-
allelograms are equivalent.
444. The lines joining the middle point of either diagonal
of a quadrilateral to the opposite vertices, divide the quadri-
lateral into two equivalent parts.
445. The line which joins the middle points of the bases of
a trapezoid divides the trapezoid into two equivalent parts.
PROBLEMS OF COMPUTATION.
446. Compute the area of a right isosceles triangle if the
hypothenuse is 100 rods.
96 PLANE GEOMETRY.
447. Compute the area of a rhombus if the sum of its diago-
nals is 12 inches and their ratio is 3 : 5.
448. Compute the area of a right triangle whose hypothe-
nuse is 13 feet and one of whose legs is 5 feet.
449. Compute the area of an equilateral triangle, one of
whose sides is 40 feet.
450. The area of a trapezoid is 3^ acres ; the sum of the two
parallel sides is 242 yards. Find the perpendicular distance
between them.
451. The diagonals of a rhombus are 24 feet and 40 feet
respectively. Compute its area.
452. The diagonals of a rhombus are 88 feet and 234 feet
respectively. Compute its area, and find length of one of its
sides.
453. The area of a rhombus is 354,144 square feet, and one
diagonal is 672 feet. Compute the other diagonal and one
side.
454. The sides of a right triangle are in the ratio of 3, 4,
and 5, and the altitude upon the hypothenuse is 20 yards.
Compute the area.
455. Compute the area of a quadrilateral circumscribed about
a circle whose radius is 25 feet and the perimeter of the quad-
rilateral 400 feet.
456. Compute the area of a hexagon having the same length
of perimeter and circumscribed about the same circle.
457. The base of a triangle is 75 rods and its altitude 60
rods. Find the perimeter of an equivalent rhombus if its
altitude is 45 rods.
PLANE GEOMETRY. 97
458. Upon the diagonal of a rectangle 40 yards by 25 yards
an equivalent triangle is constructed. Compute its altitude.
459. Compute the side of a square equivalent to a trapezoid
whose bases are 56 feet and 44 feet respectively, and each of
whose legs is 10 feet.
460. Find what part of the entire area of a parallelogram
will be the area of the triangle formed by drawing a line from
one vertex to the middle point of one of the opposite sides.
461. In two similar polygons two homologous sides are 15
feet and 25 respectively. The area of the first polygon is 450
square feet. Compute the area of the other polygon.
462. The base of a triangle is 32 feet and its altitude 20
feet. Compute the area of the triangle formed by drawing a
line parallel to the base at a distance of 15 feet from the
base.
463. The sides of two equilateral triangles are 20 and 30
yards respectively. Compute the side of an equilateral tri-
angle equivalent to their sum.
464. If the side of one equilateral triangle is equivalent to
the altitude of another, what is the ratio of their areas ?
465. The radius of a circle is 15 feet, and through a point
9 inches from the centre any chord is drawn. What is the
product of the two segments of this chord ?
466. A square field contains 5f acres. Find the length of
fence that incloses it.
467. A square field 210 yards long has a path round the
inside of its perimeter which occupies just one-seventh of the
whole field. Compute the width of the path.
468. A street 1J miles long contains 5 acres. How wide is
the street ?
98 PLANE GEOMETRY.
469. The perimeter of a rectangle is 72 feet, and its length
is twice its breadth. What is its area ?
470. A chain 80 feet long incloses a rectangle 15 feet wide.
How much more area would it inclose if the figure were a
square ?
471. The perimeter of a square, and also of a rectangle
whose length is four times its breadth, is 400 yards. Compute
the difference in their areas.
472. A rectangle whose length is 25 m is equivalent to a
square whose side is 15 m . Which has the greater perimeter,
and how much ?
473. The perimeters of two rectangular lots are 102 yards
and 108 yards respectively. The first lot is -f as wide as it is
long, and the second lot is twice as long as it is wide. Com-
pute the difference in the value of the two lots at $1 per
square foot.
474. A rhombus and a rectangle have equal bases and equal
areas. Compute their perimeters if one side of the rhombus
is 15 feet and the altitude of the rectangle is 12 feet.
475. The altitudes of two triangles are equal, and their
bases are 20 feet and 30 feet respectively. Compute the base
of a triangle equivalent to their sum and having an altitude
J as great.
REGULAR POLYGONS AND OIEOLES.
See 328.
476. An equilateral polygon inscribed in a circle is regular.
Sug. See Theorem 193.
477. An equiangular polygon circumscribed about a circle
is regular.
PLANE GEOMETRY.
99
478. If a polygon be regular, a circle can be circumscribed
about it ; i.e. one circumference can be constructed which shall
pass through all its vertices.
Post. Let ABCDH be a regular
polygon of n sides.
We are to prove that a circum-
ference can be constructed which
will pass through all its vertices.
Dem. A circumference may be
passed through any three vertices,
as A, B, and C. (See Theorem
218.)
From the centre, K, of this
circumference draw lines to all the vertices.
Z ABC = Z BCD. Why ?
What relation exists between the two triangles AKB and
BKC? Why?
What relation, then, exists between the two angles ABK
and BCK? Why ?
What relation must consequently exist between the two
angles KBC and KCD ? Why ?
How, then, do the two triangles KBC and KCD compare ?
Why?
What must therefore be the relation between KD and KC ?
Why?
What must be true, then, of the circumference passing
through the vertices A, B, and C, as regards the vertex D ?
In a similar manner, fix the position of the circumference
with regard to each of the other vertices.
479. If a polygon be regular, a circle may be inscribed in it ;
i.e. a circle may be constructed which shall have the sides of
the polygon as tangents.
Sug. First circumscribe a circle by Theorem 478, then con-
sult 214.
100 PLANE GEOMETRY.
480. One side of a regular hexagon is equal to the radius
of the circumscribed circle.
Post. Let ABCDHKbe a regular
hexagon, and let Nloe the centre of
the circumscribed circle (Theorem
478), and let NA and NKbe drawn.
the are to prove AK= AN.
Dem. The arc AK is what part of
We entire circumference ? Why ?
What is the value of the angle N
then?
What, then, must be the value of
H
/.NAK+Z-NKAt Why?
What relation exists between the A NAK and NKA ?
Why?
What, then, must be the value of each one of them ?
From this relation of the three angles what must be the
relation of the sides of the A AKN ?
Hence AK = AN. Q.E.D.
481. If the circumference of a circle be divided into any
number of equal arcs, the chords joining the successive points
of division will form a regular polygon.
482. If the circumference of a circle be divided into any
number of equal arcs, the tangents at the points of division
will form a regular polygon.
483. Tangents to a circumference at the vertices of a regular
inscribed polygon form a regular circumscribed polygon of the
same number of sides as the inscribed polygon.
484. Def. The radius of the circumscribed circle is called
the radius of the polygon.
The radius of the inscribed circle is called the apothegm of the
PLANE GEOMETRY. 101
polygon. The common centre of the circumscribed and in-
scribed circles is also the centre of the polygon.
The angle formed by two polygonal radii is called the
polygonal central angle, and that formed by two sides simply
the polygonal angles.
485. If all the radii of a regular polygon of n sides be
drawn, how many triangles will thus be formed ?
What kind of triangles will they be ?
What will be their relation to one another as regards magni-
tude?
Find an expression for the central polygonal angle in terms
of n and a right angle.
Find an expression for the polygonal angle in terms of the
Su-me quantities. (See 329.)
What relation exists between the polygonal angle and the
polgonal central angle ? Prove.
= p i. an gie, and - = Cent, angle.
n n
Sug. See above and compare the two expressions.
How does the radius of the polygon divide the polygonal
angle ?
486. If radii be drawn to a regular circumscribed polygon,
and chords of the circle be drawn joining the successive
points of intersection, these chords will form a regular poly-
gon of the same number of sides as the circumscribed polygon,
and the sides of the two polygons will be parallel, two and
two.
487. If tangents be drawn at the middle points of the sev-
eral arcs subtended by the sides of a regular polygon, these
tangents will form a regular polygon whose vertices lie on the
radii extended of the inscribed polygon, and whose sides are
respectively parallel to those of the latter.
102 PLANE GEOMETRY.
488. If the vertices of a regular inscribed polygon of n
sides are joined to the middle points of the arcs subtended by
the sides of the polygon, the lines thus drawn will form a
regular inscribed polygon of 2n sides.
489. If tangents are drawn at the middle points of the arcs
between adjacent points of contact of the sides of a regular
circumscribed polygon of n sides, a regular circumscribed poly-
gon of 2 n sides will thus be formed.
490. The perimeter of a regular inscribed polygon of n
sides is less than the perimeter of the regular polygon of 2n
sides inscribed in the same circle.
491. The perimeter of the regular circumscribed polygon of
n sides is greater than the perimeter of the regular polygon of
2 n sides circumscribed about the same circle.
492. Two regular polygons of the same number of sides are
similar.
H"
Post. Let P and P be two regular polygons, each having n
sides. We are to prove that they are similar ; i.e. that their
homologous sides form a proportion, and that they are mutu-
ally equiangular.
Dem. What is the relation between all the angles of poly-
gon P ? Of polygon P f ? Why ?
I. What is the value of one of the polygonal angles, as -4,
of polygon P ? See 329 and 485.
PLANE GEOMETRY. 103
What is the value, expressed in same terms, of one of the
polygonal angles, as JVJ of the polygon P' ?
What, then, is the relation between the angles A and JV?
Hence, what must be true regarding the corresponding
angles of the two polygons ?
II. What is the relation between AB and EG ? Why ?
Hence
What is the relation between NQ and QR ? Why ?
Therefore = . Why?
Or AB : BC : : NQ : QR, (proportion form)
or ABiNQiiBC: QR. Why ?
The pupil should finish the demonstration.
493. The areas of two regular polygons of the same number
of sides are in the same ratio as the squares of any two
homologous sides.
Sug. Consult 431.
494. The perimeters of two regular polygons of the same
number of sides are in the same ratio as the radii of their cir-
cumscribed circles.
Sug. Consult 359 and 492.
495. The perimeters of two regular polygons of the same
number of sides are in the same ratio as their apothegms.
496. The areas of two regular polygons of the same number
of sides are in the same ratio as the squares of the radii of
their circumscribed circles.
497. The areas of two regular polygons of the same num-
ber of sides are in the same ratio as the squares of their
apothegms.
104 PLANE GEOMETRY.
498. The area of a regular polygon is equal to one-half the
product of its perimeter and apothegm.
Sug. Draw radii, then consult 485.
499. Any curved or polygonal line which envelops an arc of
a circle from one extremity to another is longer than the
enveloped arc.
Post. Let AMB be the arc of a circle. Then, of all envelop-
ing lines, there must be one shorter than any of the others.
Left ACDEB be that line. (Of course, if there should be
others equal in length to ACDEB, the argument would not
be vitiated. The essential condition is, that there shall be
none shorter.)
Dem. There are three possible relations between the arc
AMB and the enveloping line ACDEB.
I. ACDEB = AMB.
II. ACDEB < AMB.
III. ACDEB > AMB.
Let us suppose Case I. to be true. Then join any two points
in ACDEB by a line that shall not cut the arc AMB, as PQ.
(The possibility of constructing this line cannot be challenged,
as ACDEB envelops the arc AMB.)
Then, since PQ < PC + CD + DQ, APQED > ACDEB.
But APQED is a line that envelops the arc AMB. Hence
we have arrived at the result that this enveloping line is
shorter than ACDEB. This is in direct conflict with the
PLANE GEOMETRY. 105
hypothesis, which was that the latter was the shortest en-
veloping line. Hence Case I. is inadmissible. Case II. leads
to the same result. Hence Case III. is alone true.
In the same manner it can be proved that any convex line
which returns into itself is shorter than any line enveloping
it on all sides, whether the enveloping line touches the given
convex line in one or several places or surrounds without
touching it.
500. The circumference of a circle is greater than the perim-
eter of any polygon inscribed in it.
501. The circumference of a circle is less than the perimeter
of any polygon circumscribed about it.
Sug. Consult 499.
502. The area of a regular inscribed polygon of 2n sides
is greater than the area of a regular polygon of n sides
inscribed in the same circle.
503. The area of a regular circumscribed polygon of 2n
sides is less than the area of a regular polygon of n sides
circumscribed about the same circle.
504. If the number of sides of a regular inscribed polygon
be continuously doubled, its perimeter will as continuously
increase in length, and consequently approach nearer and
nearer the length of the circumference.
505. If the number of sides of a regular inscribed polygon
be continuously doubled, its apothegm will as continuously
increase in length, and consequently approach nearer and
nearer the length of the radius.
506. If the number of sides of a regular circumscribed
polygon be continuously doubled, its perimeter will as con-
tinuously decrease in length, and consequently approach nearer
and nearer the length of the circumference.
106
PLANE GEOMETRY.
507. If the number of sides* of a regular circumscribed
polygon be continuously doubled, its radius will as continu-
ously decrease in length, and consequently approach nearer
and nearer the length of the radius of the circle.
508. If the number of sides of a regular inscribed polygon
be continuously doubled, its area will as continuously increase,
and consequently approach nearer and nearer the area of the
circle.
509. If the number of sides of a regular circumscribed
polygon be continuously doubled, its area will as continuously
decrease, and consequently approach nearer and nearer the
area of the circle.
510. If, in 504 to 509 inclusive, the number of sides be
doubled an/infinite number of times, the approach will be
infinitely/near; iiia. thay^C^BBafli*&: hence a circle may
be regarded as a regular polygon with an infinite /number of
sides, with the circumference its perimeter, ana radius its
apothegm.
511. The area of a regular inscribed polygon of 2n sides
is a mean proportional between the areas of two polygons, each
of n sides, one inscribed within, and the other circumscribed
about, the same circle.
H G I F Post. Let AB be one side of the
regular inscribed polygon of n sides,
and EF one side of the regular cir-
cumscribed polygon of n sides and
parallel to AB. Let C be the centre
of the circle. Draw the radii CA,
CG, and CB, G being the point of
contact of the tangent EF. Then
CA and CB, if extended, will pass
through points E and F respectively. At A and B construct
the tangents AH and BI } and join CH. Then HI will be
PLANE GEOMETRY. 107
one side of the regular circumscribed polygon of 2n sides.
Designate the area of this polygon by P 9 .
Again, let p represent the area of the inscribed polygon
whose side is AB, or the polygon of n sides ; P the area of
the circumscribed polygon of n sides, or whose side is EF-,
and p' the area of the inscribed polygon of 2n sides, and
whose side is AG.
We are to prove p : p r : : p' : P.
Dem. It is evident that the areas of the A ACD, ACG, and
EGG are - - part of the respective polygons p, p' } and P.
71
Area A ACD : Area A ACG ::CD:CG',
and Area A ACD : Area A ACG : : p : p'.
.-. p:p'::CD:CG.
Again, Area A CAG : Area A CEG ::CA: CE,
and Area A CAG : Area A CEG :: p r : P.
.\p':P::CA:CE.
But, since AD is parallel to EG, the two A CAD and CGE
are similar.
Hence CD:CG::CA: CE.
.-. p:p' : ip'i P. Q.E.D.
512. The area of a regular circumscribed polygon of 2n
sides is equal to the quotient obtained by dividing twice the
product of the areas of two regular polygons, each of n sides,
one inscribed within, and the other circumscribed about, the
same circle, by the sum of the areas of two regular polygons,
one of n and the other of 2n sides, both inscribed in the same
circle. (See diagram and postulate of previous theorem.)
n -p
We are to prove P' = . ,.
p+p'
108
PLANE GEOMETRY.
Dem. Since CH bisects the /. EGG,
OH :HE::CG: CE. (See Theorem 323.)
But Area A CGH : Area A CHE ::GH: HE.
.-. Area A CGH: Area A CHE ::CG : CE.
Again, CG : CE : : CD : CA } or CG : CE : : CD : CG.
But p:p'::CD: CG. (Theorem 511.)
.-. p:p'::CG:CE.
.-. Area A CGH : Area A CHE ::p:p'.
Hence
Area A CGH : Area A CGH + Area A CHE ::p:p +p',
or Area A CGH : Area A CGE : : p :p+p',
or 2 Area A CGH : Area A CGE ::2p:p+p',
or Area A CHI : Area A CGE ::2p:p +p'.
But areas of A CHI and CGE are part respectively of
the polygons P' and P.
Hence Area A CHI : Area A CGE : : P'
.'. P': P :: 2p:p+p'.
o ^ r>
Whence J
P.
Q.E.D.
p+p'
513. The chords which join the extremities of two per-
pendicular diameters form a
square.
Post. Let CB, BD, DA, and
AC be chords of the circle
ACBD connecting the extremi-
ties of the two perpendicular
diameters CD and AB.
We are to prove that ACBD
is a square.
Dem. What measures the
ZACB?
PLANE GEOMETRY.
109
What kind of an angle, then, must it be ?
Determine the kind of angle at B, D, and A.
Compare the lengths of AC, CB, BD, and DA.
Hence the figure ACBD is a square.
Q.E.D.
H
D
514. The diagonals of an inscribed square will be diameters
of the circle and perpendicular to each other.
515. The tangents to a circle whose points of contact are
the vertices of an inscribed square
will form a square.
Post. Let ABCD be a square
inscribed in the circle whose cen-
tre is P; BD and AC its diago-
nals; and HK, KN, NQ, and QH,
tangents whose points of contact
are the vertices of the inscribed
square.
We are to prove that HKNQ is
a square.
Dem. What measures each of the A K, N, Q, and H?
What kind of angles, then, must they be ?
What is the relative position of HK and QN? Of KN
and HK? (See Theorem 208.)
What is the relation between .RET and KC? Between BK
and BH?
What must be the relation, then, between HK and KN ?
Hence HKNQ is a square. Q.E.D.
516. The area of an inscribed square is equivalent to twice
the square of the radius. (Use diagram of previous theorem.)
It is evident that the area of the right triangle DPC is
one-half that of the square upon PC, or the radius of the
circle.
The area of the square ABCD is equal to four times the
area of the triangle DPC,
110 PLANE GEOMETRY.
or Area AZ>PO = (representing radius by r),
2
and 4 x Area A DPC =2r*= Area ABCD. Q.E.D.
517. The area of a circumscribed square is equal to four
times the square of the radius of the circle. (Use same
diagram as before.)
It can easily be shown that the square HKNQ is composed
of four smaller squares, each of which is the square of the
radius; hence
Area HKNQ = 4 r 8 . Q.E.D.
518. Problem. From 511 and 512 we have
I. j>'=VpxP and II. P =
P+p'
Letting p and P represent inscribed and circumscribed
squares respectively, we have from 516 and 517,
p = 2r* and P=4?- 2 .
By substituting these values in I., we have
p'= 2.82843 r 2 .
Then by substituting values of p } p\ and P in II., we have
P= 3.31371 r 2 .
Thus we have computed the areas of the regular inscribed
and circumscribed octagons in terms of the radius of the circle.
Again, calling these p and P, p' and P' will be polygons of
sixteen sides; and using the formulae as before, the areas
of the latter may be computed. Kepeating the process, those
of thirty-two, sixty-four, etc., sides may be, in like manner,
computed.
Below are the tabulated results to seven decimal places for
thirteen doublings of the number of sides of the polygons.
PLANE GEOMETRY.
Ill
No. Sides.
Area of Inscribed
Polygon.
Area of Circumscribed
Polygon.
4
2.0000000 r 2
4.0000000^
8
2.8284271 r 2
3.3137085^
16
3. 0614675 r 2
3.1825979^
32
3.1214452 r 3
3.1517240^
64
3.1365485 r 2
3.1441184 r 2
128
3.1403312^
3.1422236 r 2
256
3.1412773^
3.1417504 r 2
512
3.1415138 r 2
3.1416321 v*
1024
3.1415729^
3.1416025 r 2
2048
3.1415877^
3.1415951^
4096
3.1415914r*
3.1415933 r 2
8192
3.1415923 r 2
3. 1415928 r 2
16384
3.1415925^
3.1415927^
32768
3 1415926 r 2
3.1415926^
Hence, since the area of the circle is greater than that of
the inscribed polygon and less than that of the circumscribed
polygon, 3.1415926 r 2 must be the area of the circle correct
to within less than one tenth-millionth part of r 2 . But by
continuing the process, the areas of the two polygons may be
made to agree to any desired number of decimal places, and
therefore such result may be taken as the area of the circle
without sensible error. If r be taken as unity, it would,
of course, vanish from the expression, and consequently
3.1415926 may be taken as the area of a circle ivhose radius
is unity.
519. It can easily be shown that the two formulae
and II. P'=
P+P'
will be true if these quantities represent perimeters instead
112
PITANE GEOMETRY.
of areas. Regarding them as such, and using the diagram
in 515, we will compute the perimeters in terms of the
diameter, calling the latter D.
Each side of the circumscribed square is, of course, equal to
the diameter, and hence its perimeter is 4 J9.
In the right triangle APD,
,
V2
V2
Hence the perimeter of the inscribed square is 2.8284271 D.
Using the formulae as suggested, and tabulating the results,
we have the following :
No. Sides.
Perimeter of Inscribed
Polygon.
Perimeter of Circumscribed
Polygon.
4
2 8284271 D
4 0000000 D
8
3 0614675 D
3 3137085 D
16
3 1214452 D
3 1825979 D
32
3. 1365485 D
3 1517249 D
64
3. 1403312 D
3 1441184 D
128
3 1412773 D
3 1422236 D
256
512
3.1415138 D
3 1415729 D
3.1417504 D
3 1416321 D
1024 .... . .
3 1415877 D
3 1416025 D
2048
3.1415914 D
3 1415951 D
4096
3. 1415923 D
3 1415933 D
8192
3.1415924 D
3 1415928 D
16384
3 1415925 D
3 1415927 D
32768
3.1415926 D
3 1415926 D
Hence, since the circumference of the circle is greater than
the perimeter of the inscribed polygon, and less than that of
PLANE GEOMETRY.
113
the circumscribed polygon, 3. 1415926 D must be the circum-
ference of the circle correct to within less than one ten-
millionth part of D. But by continuing the process the
perimeters of the two polygons may be made to agree to
any desired number of decimal places, and therefore such
result may be taken as the circumference of the circle without
sensible error. If D be taken as unity, it would, of course,
vanish from the expression, and consequently 3.1415926 may
be taken as the circumference of a circle whose diameter is unity.
520. The circumferences of two circles are in the same ratio
as their radii, and also their diameters.
Post. Let ABHK, etc., and STVY, etc., be two circles
whose centres are JVand W, and designate the circumferences
by C and c, and their radii by R and r, and their diameters
by D and d, the capital letter referring to the larger circle.
We are to prove
I. C:c::E:r.
II. C:c::D:d.
Cons. Inscribe in each a regular polygon of n sides, and
construct the radii NB and WT, and the apothegms NQ and
WE.
Dem. Designating the perimeters by P and p,
P:p::NQ: WR.
Why?
114 PLANE GEOMETRY.
If now we inscribe polygons with double the number of
sides, and continue this process indefinitely, the perimeters
will coincide with the circumferences, and the apothegms with
the radii.
Hence I. C:c:: R : r,
and C:c::2jft:2r; Why?
or II. C : c : : D : d. Q.E.D.
521. The areas of two circles are in the same ratio as the
squares of their radii and of their diameters.
Sug. Use method similar to the above, and consult 496
and 497.
521 (a). Def. Similar arcs, sectors, and segments are those
that correspond to equal central angles.
522. Similar arcs are in the same ratio as the radii of the
circumferences of which they are a part, and also as the
diameters.
C f
Post. Let CB and KH be two similar arcs, and A and D
the centres of the circumferences of which they are a part.
Cons. Draw the radii AC, AB, DK, and DH.
(Designate circumferences, diameters, and radii, as before.)
We are to prove
I. ATcCB:ATcKH::K:r.
II. ATcCB:AicKH::D:d.
Dem. ZA = Z.D. Why?
Hence arc CB is the same part of the circumference C, as
arc KH is of the circumference c.
.-. Arc CB : Arc KH : : C : c.
PLANE GEOMETRY. 115
But R:r::C:c. (See Theorem 520.)
.-.I. ATcCB:A.TcKH::R:r', Why?
II. Arc 05: AicKH::D:d. Why?
Q.E.P.
523. The areas of similar sectors are in the same ratio as
the squares of the radii, and also of the diameters, of the
circles of which they are a part.
524. The area of a circle is equal to one-half the product of
its circumference and radius.
Sug. Consult 498 and 510.
525. The area of a sector is equal to one-half the product of
its arc and radius.
526. The areas of similar segments are in the same ratio
as the squares of their radii, the squares of their diameters,
and as the squares of their chords.
527. Let us designate the circumference of a circle whose
diameter is unity by v, and the circumference of any other
circle by C ; its diameter by D ; its radius by R ; and its area
by A. -'
Then 0:ir::D:l. Why?
Hence I. O = 7rZ>;
C 1
whence II. = TT,
or III. C = 7rx2#.
TD
Multiplying both members of this equation by , we have
CR
But, by 524, = the area of the circle ; hence
2
IV. A =
116 PLANE GEOMETRY.
528. Hence the area of any circle is equal to the square of its
radius multiplied by the constant quantity TT, and the circumfer-
ence of every circle is equal to the product of its diameter (or
twice its radius) by the same quantity IT.
From II. above, it is readily seen that TT is the ratio of the
circumference of any circle to its diameter, or of a circumfer-
ence to its radius.
The exact numerical value of TT can be only approximately
expressed. As computed in 352 it is 3.1415926, but for prac-
tical purposes in computing its value is usually taken as
3.1416.
The symbol TT is the first letter of the Greek word meaning
perimeter or circumference.
529. The quadrature of the circle is the problem which re-
quires the finding of a square which shall be equal in area to
that of a circle with a given radius. Now since the area of a
circle is equal to its circumference multiplied by one-half its
radius, if a straight line of same length as circumference be
taken as the base of a rectangle, and one-half the radius as its
altitude, their product will be the area of the rectangle, also
of the circle. It is also evident that, if a line which is a mean
proportional between these two be taken as the side of a
square, the area of this square will be equal to that of the
rectangle, and consequently to that of the circle. It will be
shown in the series of problems of construction (Prob. 736)
how this mean proportional can be found ; hence, to " square
the circle " we^ must be able to find the circumference when
the radius is known, or vice versa. For accomplishing this, we
must know the ratio of the circumference to its diameter or
radius. But this raticf, as has been remarked before, can be
only approximately expressed ; for, as the higher mathematics
prove, the circumference and diameter are incommensurable,
but the approximation has been carried so far that the error is
infinitesimal. Archimedes, about 250 B.C., was the first to
assign an approximate value to TT. He found that it must be
PLANE GEOMETRY. 117
between 3.1428 and 3.1408. In 1640 Metius computed it cor-
rectly to 6 places. Later, in 1579, Vieta carried the approx-
imation to 10 places, Van Ceulen to 36 places, Sharp to 72
places, Machin to 100 places, De Lagny to 128 places, Ruther-
ford to 208 places, and Dr. Clausen to 250 places. In 1853
Rutherford carried it to 440 places, and in 1873 Shanks com-
puted it to 707 places, but these latter results do not appear
to have been verified. The following is its value to 208 places,
as computed by Rutherford :
v = 3.141592653589793238462643383279-
502884197169399375105820974944-
592307816406286208998628034825-
342717067982148086513282306647-
093844609550582231725359408128-
484737813920386338302157473996-
0082593125912940183280651744 +.
530. Some idea of the accuracy of the above value may be
formed from the following statement taken from Peacock's
Calculus : "If the diameter of the universe be 100,000,000,000
times the distance of the sun from the earth (about 93,000,000
miles), and if a distance which is 100,000,000,000 times this
diameter be divided into parts, each of which is one 100,000,-
000,000th part of an inch ; then if a circle be described whose
diameter is 100,000,000,000 times that distance, repeated 100,-
000,000,000 times as often as each of those parts of an inch
is contained in it ; then the error in the circumference of this
circle, as computed from this approximation, will be less than
one 100,000,000,000th part of the one 100,000,000,000th part
of an inch."
ADVANCE THEOREMS.
531. In two circles of different radii, angles at the centres
subtending arcs of equal length are to each other inversely as
the radii.
118 PLANE GEOMETRY.
532. If, from any point within a regular polygon of n sides,
perpendiculars be drawn to all the sides, the sum of these per-
pendiculars is equal to n times the apothegm.
533. If perpendiculars be drawn from the vertices of a regu-
lar polygon to any diameter of the circumscribed circle, the
sum of the perpendiculars upon one side of the diameter is
equal to the sum of those on the other side.
534. An equiangular polygon inscribed in a circle is regular
if the number of its sides be odd.
535. An equilateral polygon circumscribed about a circle is
regular if the number of its sides be odd.
536. The sum of the squares of the lines joining any point
in the circumference of a circle with the vertices of an inscribed
square is equal to twice the square of the diameter of the
circle.
537. The area of the ring included between two concentric
circles is equal to the area of the circle whose diameter is that
chord of the outer circle which is tangent to the inner.
538. If the sides of a right triangle be the homologous sides
of similar polygons, the area of the polygon on the hypothe-
nuse is equal to the sum of the areas of the other two.
539. If three circles be described upon the sides of a right
triangle as diameters, the area of that described upon the
hypothenuse is equal to the sum of the areas of the other two.
540. If upon the legs of a right triangle semi-circumferences
are described outwardly, the sum of the areas contained be-
tween these semi-circumferences and the semi-circumference
passing through the three vertices is equal to the area of the
triangle.
541. If the diameter of a circle be divided into two parts,
and upon these parts semi-circumferences are described on
PLANE GEOMETRY. 119
opposite sides of the diameter, these semi-circumferences will
divide the circle into two parts which have the same ratio as
the two parts of the diameter.
542. If two chords of a circle be perpendicular to each other,
the sum of the areas of the four circles described upon the four
segments as diameters will be equal to the area of the given
circle.
543. If squares be constructed outwardly upon the sides of
a regular hexagon, their exterior vertices will be the vertices
of a regular dodecagon.
544. The radius of a regular inscribed polygon is a mean
proportional between its apothegm and the radius of a regular
polygon of double the number of sides circumscribed about
the same circle.
545. The area of a regular dodecagon is equal to three times
the square of its radius.
546. If the radius of a circle be divided in extreme and
mean ratio, the larger part will be the side of a regular decagon
inscribed in the same circle.
547. The apothegm of a regular inscribed pentagon is equal
to one-half the sum of the radius of the circle and the side of
the regular decagon inscribed in the same circle.
548. The square of the side of a regular inscribed pentagon
is equivalent to the sum of the squares of the side of the regu-
lar inscribed decagon and the radius of the circle.
PROBLEMS OF COMPUTATION.
549. Compute the side of a regular inscribed trigon in terms
of the regular trigon circumscribed about the same circle.
Compare their areas.
550. Compute the side of an inscribed square in terms of
the square circumscribed about the same circle.
120 PLANE GEOMETRY.
551. Compute the apothegm of a regular inscribed trigon in
terms of the regular hexagon inscribed in the same circle.
552. Compute the apothegm of a regular inscribed hexagon in
terms of the regular trigon inscribed in the same circle.
553. Regular trigons and hexagons are both inscribed and
circumscribed about the same circle. Compare their areas.
554. Compute the area of a regular polygon of 24 sides
inscribed in a circle whose radius is 10 inches.
555. Compute the perimeter of a regular pentagon inscribed
in a circle whose radius is 12 feet.
556. The perimeter of a regular hexagon is 480 feet, and
that of a regular octagon is the same. Which has the greater
area, and how much ?
557. If paving blocks are in the shape of regular polygons
(i.e. their cross-sections), how many shapes can be employed
in order to completely fill the space ?
558. Compute the diameter of a circle whose circumference
is 12 feet and 10 inches.
559. The diameter of a carriage wheel is 4 feet and 3 inches.
How many revolutions does it make in traversing one-fourth
of a mile ?
560. What is the width orf the ring between two concentric
circumferences whose lengths are 480 feet and 360 feet ?
561. Find the length of an arc of 36 in a circle whose diam-
eter is 36 inches.
562. In raising water from the bottom of a well by means
of a wheel and axle, it was found that the axle, whose diameter
was 8 inches, made 20 revolutions in raising the bucket. Com-
pute the depth of the well.
563. Find the central angle subtending an arc 6 feet and 4
inches long, if the radius of the circle be 8 feet and 2 inches.
PLANE GEOMETRY. 121
564. If the radius of a circle is 5 feet and 3 inches, find the
perimeter of a sector whose angle is 45.
565. If the central angle subtending an arc 10 feet and 6
inches long is 72, what is the length of the radius of the circle ?
566. If the length of a meridian of the earth be 40,000,000
metres, what is the length of an arc of 1" ?
567. Two arcs have the same angular measure, but the
length of one is twice that of the other. Compare the radii
of those arcs.
568. Compute the area of a circle whose circumference is
100 yards.
569. Two arcs have the same length, but their angular
measurements are 20 and 30 respectively. If the radius of
the first arc is 6 feet, compute the radius of the other.
570. Find the circumference of a circle whose area is 2
acres and 176 square yards.
571. The diameter of a circle is 40 feet. Find the side of a
square which is double the area of the circle.
572. The area of a square is 196 square rods. Find the area
of the circle inscribed in the square.
573. A circular fish-pond which covers an area of 5 acres
and 100 square rods is surrounded by a walk 5 yards wide.
Compute the cost of gravelling the walk at 6J cents per square
yard.
574. What must be the width of a walk around a circular
garden containing If acres, in order that the walk may contain
exactly one-fourth of an acre ?
575. A carpenter has a rectangular piece of board 15 inches
wide and 20 inches long, from which he wishes to cut the
largest possible circle. How many square inches of the board
must he cut away ?
122 PLANE GEOMETRY.
576. The perimeters of a circle, a square, and a regular
trigon, are each equal to 144 feet. Compare their areas.
577. If the radius of a circle be 12 inches, what is the radius
of a circle 10 times as large ?
578. What will it cost to pave a circular court 30 feet in
diameter, at 54 cents per square foot, leaving in the centre a
hexagonal space, each side of which measures 4 feet ?
579. A circle 18 feet in diameter is divided into three equiv-
alent parts by two concentric circumferences. Find the radii
of these circumferences.
580. If the chord of an arc be 720 feet, and the chord of its
half be 369 feet, compute the diameter of the circle.
581. The chord of half an arc is 17 feet, and the height of
the arc 7 feet. Compute the diameter of the circle.
582. The radius of a circle is 12 feet ; the chords which sub-
tend two contiguous arcs are 6 feet and 9 feet respectively.
Compute the chord subtending the arc equal to the sum of the
other two.
583. The lengths of two chords, drawn from the same point
in the circumference of a circle to the extremities of a diam-
eter, are 6 feet and 8 feet respectively. Compute the area of
the circle.
584. The chord of an arc is 32 inches, and the radius of the
circle is 34 inches. Compute the length of the arc.
585. The diameter of a circle is 106 feet. Compute the
lengths of the two arcs into which a chord 90 feet long divides
the circumference.
586. The area of a sector is 385 square feet, and the angle
of the sector is 36. Compute the radius of the circle and
perimeter of the sector.
PLANE GEOMETRY. 123
587. Compute the area of a segment, if the chord of the arc
is 56 feet and the radius of the circle is 35 feet,
588. A room 20 feet long and 15 feet wide has a recess at
one end in the shape of the segment of a circle, the chord being
15 feet, and its greatest width 4 feet. Compute the area of
the entire room.
589. Compute the number of square feet of brick that would
be required in blocking up one of the arches of a railway via-
duct, if the span of the arch is 60 feet, height above the piers
20 feet, and distance from the ground to the spring of the
arch 20 feet.
590. Compute the area of a circle in which the chord, 3 feet
long, subtends an arc of 120.
591. Compute the area of a segment whose arc is 300, the
radius of the circle being 20 inches.
592. The areas of two concentric circles are as 5 to 8. The
area of that part of the ring which is contained between two
radii making the angle 45 is 300 square feet. Compute the
radii of the two circles.
593. What is the altitude of a rectangle equivalent to a
sector whose radius is 15 feet, if the base of the rectangle is
equal to the arc of the sector ?
594. Compute the radius of a circle, if its area is doubled by
increasing its radius one foot.
595. The radius of a circle is 10 inches. Through a point
exterior to the circle two tangents are drawn, making an angle
of 60. Compute the area of the figure bounded by the tan-
gents and the intercepted arc.
596. Three equal circles are drawn tangent to each other,
with a radius of 12 feet. Compute the area contained between
the circles.
124 PLANE GEOMETRY.
597. Upon each side of a square, as a diameter, a semi-
circumference is described within the square. If the side of
the square is 10 inches, find the sum of the areas of the four
leaves.
598. In a circle whose radius is 100 feet two parallel chords
are drawn on the same side of the centre, one equal to the side
of a regular hexagon, and the other to the side of a regular
trigon, both inscribed in the given circle. Compute the area
of the circle comprised between the two parallel chords.
MAXIMA AND MINIMA,
599. Def. Among quantities of the same kind, that which
is greatest is called the maximum (plural, maxima), and
that which is the smallest is called the minimum (plural,
minima).
For example : of all lines inscribed in the same circle, the
diameter is the greatest, and therefore is a maximum; and of
all lines drawn from the same point to a given straight line,
that which is perpendicular is the shortest, and is therefore a
minimum.
Def. If two or more plane figures have equal perimeters,
they are said to be isoperimetric.
Def. A plane figure is said to be a maximum or a minimum
when its area is a maximum or a minimum.
600. If any number of triangles have the same or equal
bases and equal areas, that which is
isosceles has the minimum perim-
eter.
Post. Let ABC and ADB be two
triangles having the same base
AB and equal areas, and let ACB
be isosceles, having CA equal to
CB.
PLANE GEOMETRY. 125
We are to prove
AC + CB + AB < AD + DB + AB.
Or, since AB = AB,
we are to prove
AC + CB <AD + DB.
Cons. From B construct a perpendicular to AB, and extend
it to meet AC extended in K. Join DK and draw CH
through point D.
Dem. How must the altitudes of the two A ACB and ADB
compare ?
What must be the position, then, of the line CH relative to
AB?
How, then, do the two A HCB and CBA compare ? Why ?
The two A KCH and CAB ? Why ?
How, then, must the two A KCH and HCB compare ?
Why?
What is the position of CH relative to KB ? Why ?
How, then, must OfiT and CB compare ? DK and DB ?
Now compare AC + CK with AD 4- DK, and the pupil
should be able to write, or give orally, a complete and accurate
demonstration of this theorem.
601. If any number of triangles have the same area, that
which is equilateral has the minimum perimeter.
602. If any number of triangles have the same or equal
bases and equal perimeters, that which is isosceles is the
maximum.
/Sug. Through the vertex of the isosceles triangle draw a
line parallel to the base. Then prove that the vertex of the
other triangle cannot fall on that line. Then compare their
altitudes, and consequently their areas.
603. If any number of triangles be isoperimetric, that which
is equilateral is the maximum.
126
PLANE GEOMETRY.
604. If any number of triangles have two sides in each
respectively equal, that in
which these sides are per-
pendicular to each other is
the maximum.
Sug. Place them so that
one set of equal sides shall
coincide, as AB. Then com-
pare their altitudes DK, CA, and HN, and consequently their
areas.
605. If any number of equivalent parallelograms have the
same or equal bases, the perimeter of that which is rectangu-
lar is the minimum.
606. Of all rectangles of given area, the perimeter of the
square is a minimum.
607. If any number of triangles have the same or equal
bases and the same or equal altitudes, the perimeter of that
which is isoceles is a minimum.
608. Every closed plane figure of given perimeter whose
area is a maximum, must be convex.
Post. Let ACBN be a plane concave
figure, with the straight line AB which
'' joins two of its points in its perimeter
lying without.
We are to prove that ACBN cannot be
a maximum.
Dem. Conceive the figure CAB to be revolved about AB as
an axis till it comes to the position AC'B. Then the figures
ACBN and ^IC'jBJVhave equal perimeters, but the area of the
latter will exceed that of the former. Hence ACBN cannot
be a maximum among isoperimetrical figures. But ACBN is
any concave, i.e. non-convex, plane figure. Therefore, of all
isoperimetrical plane figures, the maximum must be convex.
PLANE GEOMETRY.
127
E 1
609. Of all plane figures. that are isoperimetric, that which
is a circle is the maximum.
Dem. I. It is evident that,
with a given perimeter, an in-
definite number of figures of
different shapes and areas may
be constructed. It is also evi-
dent that we can diminish the
area indefinitely, but cannot
thus increase it. Consequently,
there must be among all these
figures having the same perim-
eter either one maximum figure,
or several maximum figures of different forms and equal
areas.
II. Let ACBFG be a maximum figure with a given perim-
eter ; then by 608 it must be convex. Let also the line AB
divide the perimeter into halves ; then it must also divide the
area into halves. For suppose one of the parts, as AFB, to be
greater than the other, and conceive this part to be revolved
on AB as an axis until it comes into the same plane with
ACB, and let AF'B be its position after revolution. Hence
the perimeter of the figure AF'BEGA is equal to that of the
figure ACBFG A, but the area of the former is greater than
that of the latter. Therefore the figure ACBFG cannot be a
maximum. But by hypothesis it is a maximum. Hence AB
must bisect the area of ACBFG.
Since ACBFG is a maximum, and AB bisects the area, it
follows that the figure AF'BFG is also a maximum.
Again, let F be any point in BEG A selected at random, and
F 1 its position after revolution. Join FF 1 , FB, FA, F'B, and
F'A. Then AF=AF', and FQ = F*Q. Hence the two tri-
angles AQF and AQF' are equal.
Therefore FF' is perpendicular to AB.
Similarly the two triangles AF'B and AFB are equal.
128 PLANE GEOMETRY.
The triangle AFB must be a maximum, otherwise its area
could be increased without increasing its perimeter ; i.e. with-
out increasing the lengths of the two chords .AF and FB, which
would consequently leave the areas of the two segments AGF
and FEE unchanged, and therefore make up an area greater
than ABEG, by which it is evident that ACBFG could not
be a maximum; but this also conflicts with the hypothesis
which grants that ACBFG is a maximum. Consequently the
triangle AFB must be a maximum, and therefore the angle
AFB must be a right angle. (See Theorem 604.) But F is
any point in the curve BEFGA. Hence BFA must be a semi-
circle, as also ACB. Hence the whole figure ACBFG must
be a circle. Q.E.D.
610. Of all plane figures having equal areas, the perimeter
of that which is a circle is the minimum.
Post. Let C be a circle, and A any other plane figure having
the same area.
We are to prove Perimeter C < Perimeter A.
Dem. For let B be a circle having a perimeter equal to that
of A. Hence by 609 area B is greater than area A, and hence
greater than area C. Hence if area C is less than area B,
what must be true of their perimeters, i.e. their circumfer-
ences ? But by construction
Perimeter B = Perimeter A.
Hence perimeter of C is a minimum. Q.E.D.
PLANE GEOMETRY.
129
611. Of all mutually equilateral polygons, that which, can
be inscribed in a circle is the maximum.
Post. Let ABCDH = P and A'B'C'D'H' = P' be two mutu-
ally equilateral polygons, having AB equal to A'B', BC equal
to B'C', etc., of which ABCDH can be inscribed in a circle,
and let N be the centre of such circle.
Cons. With the radius NB construct the arcs A'B 1 , B'C',
C'D', etc.
Dem. The Arc AB = Arc A'B', and Arc BC= Arc B'C', etc.
Hence Circumference ABCDH = sum of the Arcs A'B', B'C 1 ,
etc.
Hence Perimeter of S = Perimeter of S 1 .
Therefore Area S > Area S'. (Theorem 609.)
But the corresponding segments are equal. Hence, sub-
tracting their respective sums from the above inequality leaves
Area P > Area P 1 . Q.E.D.
612. Of all isoperimetric poly-
gons of the same number of
sides, that which is equilateral
is the maximum.
Post. Let ABCDHK be the
maximum of all isoperimetrical
polygons of any given number
of sides.
130 PLANE GEOMETRY.
We are to prove that it is equilateral ; i.e. that
KH=HD=DC, etc.
Cons. Connect any two alternate vertices, as AH.
Dem. The AAKH must be a maximum of all isoperimet-
rical triangles having the common base AH\ otherwise another
triangle, as ANH, could be constructed, having the same perim-
eter and a greater area, in which case the area of the polygon
ABCDHN would be greater than that of ABCDHK. Hence
the latter would not be a maximum. This result conflicts with
our hypothesis, which grants that it is a maximum. Therefore
the triangle AKH must be a maximum, and consequently
isosceles. (See Theorem 602.)
Hence AK= KH.
Similarly, by joining KD,
KH= HD, etc.
Hence the polygon is equilateral. Q.E.D.
613. The maximum of all equilateral polygons of the same
number of sides is that which is regular.
614. Of all polygons having the same number of sides and
equal areas, the perimeter of that one
which is regular is a minimum.
Post. Let P be a regular polygon,
and M any irregular polygon having
the same number of sides and same
area as P; and let N be a regular
polygon having the same number of
sides and isoperimetrical with M.
PLANE GEOMETRY.
131
(See 612 and 613.)
Why?
Dem. Area M<N.
Area M Area P.
.-. Area P < N.
But of two regular polygons of the same number of sides,
that which has the less area must have the less perimeter.
Why?
Hence Perimeter P< Perimeter N,
and . . Perimeter P < Perimeter M.
Hence perimeter of P is a minimum. Q.E.D.
615. Of all isoperimetric regular polygons, that which has
the greatest number of sides is the maximum.
-A J) B
Post. Let ABC be a regular trigon, and P a regular tetragon,
having equal perimeters.
We are to prove Area P > Area ABC.
- Cons. Draw from C any line CD to AB. At C make
Z DCH equal to the Z CD.B, and <7ZT equal to DB, and
join .HZ).
Dem. A <mff = A CDB. Why ?
Hence Area ABC = Area AMTC. Why ?
But Area P > Area ADHC. Why ?
Hence Area P > Area AB(7.
Similarly, P could be shown to be less than an isoperimetric
regular pentagon, etc. Q.E.D.
132 PLANE GEOMETRY.
616. The area of a circle is greater than the area of any
polygon of equal perimeter.
617. Of all regular polygons having a given area, the perim-
eter of that which has the greatest number of sides is a
minimum.
Post. Let Q and P be two regular polygons having equal
areas, and Q having the greater number of sides.
We are to prove that the perimeter of P is greater than
that of Q.
Dem. Let R be a regular polygon whose perimeter is equal
to that of Q, but the number of sides the same as P.
Then Q > R. Why ?
But Area Q = Area P.
.: Area P > Area R.
.. Perimeter P> Perimeter R. Why?
But Perimeter R = Perimeter Q.
.-. Perimeter P > Perimeter Q.
Hence perimeter Q is a minimum. Q.E.D.
618. The circumference of a circle is less than the perimeter
of any polygon of equal area.
619. The rectangle formed by the two segments of a line
is maximum when the segments are equal.
PLANE GEOMETRY.
PROBLEMS OF CONSTRUCTION.
620. In the demonstration of the foregoing theorems it has
been assumed that certain constructions were possible; i.e.
that perpendiculars and parallels could be drawn, that lines
and angles could be bisected, etc. It is now proposed to
show that those and many other problems can be performed,
so our previous demonstrations are not vitiated that are in
any way dependent upon such constructions.
621. The solution of a geometrical problem of construction
involves in general three steps; viz.:
I. The construction proper, by the use of the compass and
ruler.
II. Demonstration to prove the correctness of the construc-
tion.
III. Discussion of its limitations and applications, including
the number of possible constructions.
If numerical or algebraical results are also required, then
there is, of necessity,
IV. Computation, by which numerical values are ascertained,
involving also the use of general symbols in obtaining algebraic
formulae.
622. Each pupil should be provided with a good pair of
compasses, for the use of either ink or pencil, a good ruler
with straight edges, besides one hard and one soft pencil. The
lead of the pencil should be sharpened flat, so that a fine line
can be made.
PLANE PKOBLEMS.
623. It is required to find a point which is a given distance
from a given point.
Post. Let C be the given point, and AB the given dis-
tance.
134 PLANE GEOMETEY.
We are required to find a point which shall be at the dis-
tance AB from C.
A B
Cons. First, with the ruler, from C draw an indefinite
straight line, as CD, in any direction. Then with the com-
passes, using C as a centre and AB as a radius, draw an arc
cutting the line CD, as at H.
Then H is the required point ; for,
Dem. If, in applying the compasses to AB a circle had
been constructed, and the circle HKN also completed, then
the circles would have equal radii.
.-. His the same distance from C that A is from B.
Discussion. Since, from the definition of a circle, all points
in the circumference are equally distant from the centre, it
follows that every point in the circumference IIKN is the
same distance from C that A is from B. Consequently any
point in that circumference answers the conditions of the
problem. Q.E.F.
623 (a). Whenever a line is found such that any point in it
selected at random will fulfil certain specified conditions, or
such that all points in it have a common property, that line
is called the LOCUS of that point or points. Hence we may say,
in the above case, that the circumference HKN is the locus
of the point which is the distance AB from point (7, or, as
some prefer to put it, it is the locus of all points which are
at the distance AB from point C.
PLANE GEOMETRY.
135
624. It is required to find the point, having given its
distances from two given points.
A B
Post. Let the two given distances be A and B, and the two
given points C and D.
We are required to find a point which is A distant from C,
and B distant from D, or vice versa.
Sug. Find locus of the point which is A distant from point
(7, and locus of the point which is B distant from point D, and
vice versa.
Dis. How many points, then, satisfy the condition of the
problem ?
Determine the result if
I. The distance between C and D had been greater than the
sum of A and B.
II. If it had been equal to the sum of A and B.
III. If it had been equal to the difference of A and B.
IV. If it had been less than the difference of A and B.
(See Theorem 229.)
625. It is required to find the point which is equally distant
from two given points.
Post. Let A and B be the two given points.
We are to find a point which is the same distance from A
as from B.
136
PLANE GEOMETRY.
It is evident that whether there be more or not, there must
at least be one midway between A and B ; so join AB.
With A and B as centres, construct, with the same radius,
two circles which shall intersect. How can you tell whether or
not they will intersect ?
K
\
\
/
N
Connect the points of intersection, as D and H. Then DH
sustains what relation to the two circles ?
The line AB sustains what relation to the two circles ?
Then what relation, as regards their position, between AB
How is the point D situated with reference to points A and B ?
How is the point .ff situated with reference to the same points?
Then what must be the relation of AC and CB as regards
magnitude ? (Consult Theorem 82 or 230.)
Suppose, now, DH be indefinitely extended, and any point
in it selected at random. How will this point be situated with
reference to the points A and B ? Why ?
What name, then, shall we give to the line NK? Why ? Q.E.F.
626. It is required to bisect a given straight line.
Sug. Employ method similar to the previous one.
PLANE GEOMETRY. 137
627. It is required to construct a perpendicular to a given
straight line which shall pass through a given point in that line.
Post. Let AB be the given line, and C the given point.
We are required to construct a perpendicular to AB passing
through point C.
Sug. Lay off equal distances each side of (7, as CD and
CB ; then use Problem 625.
627 (a). It is required to construct a perpendicular to a
line at one extremity.
Sug. Extend the line ; then use previous problem. In case
the extension should not be possible or convenient, use the
following.
Select any point at random, as (7,
making sure that it lies between A ^ ^ N
and B. Then, with CB as a radius,
construct the circle or arc KBN.
Through D, point where this circle
intersects the given line, and (7, draw
the straight line DH, the point H
being where this line intersects the "^^s, ^ B
arc KBN. Join HB.
Then HB is the perpendicular required. Why ? Q.E.F.
628. It is required to construct a perpendicular to a given
line from a given point without the line.
C'
JD s
Let ^15 be the given line, and C the given point.
138 PLANE GEOMETRY.
Sug. With. C as a centre, construct an arc which shall inter-
sect AB.
Point C is how situated with reference to the two points of
intersection D and H ?
Can you find another point the same distance as C from
those two points ?
If that point and point C be joined by a straight line, how
will that line be situated with reference to DH ? (See
Theorem 82.)
628 (a). It is required to bisect a given arc.
Sug. Connect the extremities of the given arc; then use
Problem 626, and for proof consult Theorem 199.
629. It is required to bisect a given angle.
Sug. Use Problem 628 (a) and Theorem 192.
630. At a given point in a given line, it is required to con-
struct an angle equal to a given angle.
Sug. Consult 191 and 192.
631. It is required to draw through a given point a line
parallel to a given line.
Sug. Consult 87 and 84, and Problem 630.
632. Two angles of a triangle being given, it is required to
construct the third angle.
Sug. Consult Theorems 58 and 98, and Problem 630.
633. Having given two sides of a triangle and their included
angle, it is required to construct the triangle.
634. Having given two angles of a triangle and the side
joining their vertices, it is required to construct the triangle.
635. Having given the three sides of a triangle, it is re-
quired to construct the triangle.
636. It is required to construct the locus of the point which
is a given distance from a given straight line.
PLANE GEOMETRY. 139
637. It is required to construct the locus of the point which
is a given distance from a given circumference.
638. It is required to construct the locus of a point which
is equally distant from two given parallel lines.
639. It is required to construct the locus of the point which
is equally distant from two non-parallel lines in the same
plane.
640. It is required to construct the locus of the point which
is equally distant from the circumferences of two equal circles.
641. Having given the hypothenuse of a right triangle, it
is required to construct the locus of the vertex of the right
angle.
Bug. Consult Theorem 236 (6).
642. It is required to find in a given line, as AB, a point
which is equally distant from two given points, as C and D.
D
Sug. Consult Problem 625.
643. It is required to find a point which is equally distant
from three given points.
Sug. Join the points, then consult Problem 625.
644. Through a given point without a given line, it is
required to draw a line which shall make an angle with the
given line equal to a given angle.
645. It is required to construct the triangle, having given
the base, the vertical angle, and one of the other angles.
646. It is required to construct the triangle, having given
two sides and an angle opposite one of them.
647. It is required to find the centre of a given circumfer-
ence or arc.
140 PLANE GEOMETRY.
648. It is required to construct the circumference, having
given three points in it.
649. It is required to construct a circumference which shall
pass through the vertices of a given triangle.
650. It is required to find the locus of the centre of the
circumference which shall pass through two given points.
651. With a given radius, it is required to construct the
circle which shall pass through two given points.
652. It is required to construct the isosceles triangle, having
given the base and the vertical angle.
653. It is required to construct a circumference which shall
be a given distance from three given points.
654. It is required to construct a circle which shall have
its centre in a given straight line and circumference passing
through two given points.
655. It is required to find a point which shall be equally
distant from two given points, and at a given distance from a
third given point.
656. It is required to construct the equilateral triangle, hav-
ing given one side.
657. It is required to trisect a right angle.
658. It is required to find a point which shall be equally
distant from two given points, and also equally distant from
two given parallel lines.
659. It is required to find a point which shall be equally
distant from two given points, and also equally distant from
two given non-parallel lines in the same plane.
660. It is required to find a point which shall be equally
distant from two given parallel lines, and also equally distant
from two non-parallel lines in the same plane.
PLANE GEOMETRY. 141
661. It is required to construct
I. an angle of 45 ; VI. an angle of 75 ;
II. an angle of 60 ; VII. an angle of 22 30' ;
III. an angle of 30 ; VIII. an angle of 52 30' ;
IV. an angle of 15 ; IX. an angle of 135 ;
V. an angle of 105 ; X. an angle of 165.
662. It is required to find a point in one side of a
triangle which shall be equally distant from the other two
sides.
663. It is required to find a point which shall be equally
distant from two non-parallel lines in the same plane, and at
a given distance from a given point.
664. It is required to construct the right triangle, having
given the two legs.
665. It is required to construct the right triangle, having
given the hypothenuse and one acute angle.
666. It is required to construct the right triangle, having
given one leg and adjacent acute angle.
667. It is required to construct the right triangle, having
given one leg and the acute angle opposite.
668. It is required to construct the right triangle, having
given the hypothenuse and one leg.
669. It is required to construct a tangent to a given circle,
having given the point of contact.
670. It is required to construct a tangent to a given circle,
which shall pass through a given point outside the circle.
Sug. Connect the centre of the given circle with the given
exterior point. On this line as a diameter construct a circle,
and join the points of its intersection of the given circle with
extremities of the diameter.
142
PLANE GEOMETRY.
671. It is required to construct the parallelogram, having
given two sides and their included angle.
672. It is required to construct a circle within a given tri-
angle, so that the sides of the triangle shall be tangents of the
circle.
Sug. Consult Theorems 117 and 125.
673. It is required to construct the circle, having given a
chord and angle made by the chord and a tangent.
Let AB be the given chord, and C the angle made by the
chord and tangent. Then one extremity of the chord must be
the point of contact.
Take D.H=AB. Construct angle DHN equal to angle C.
Then H is the point of contact, and NH the tangent.
Find the locus of the centre of the circumference passing
through V and H. (Problem 650.)
Then consult Theorem 202.
Hence T must be the centre of the circle required.
If from any point in the arc DOH (call the point Q) lines
be drawn to D and ff, how will the angle Q compare in magni-
tude with the angle DHN?
PLANE GEOMETRY. 143
What is formed by the lines DH, DQ, and HQ ?
If we call DH the base, what would you call the angle Q ?
What the point Q ?
What might the arc DQH be named, then ? Q.E.F.
674. Having given the base and vertical angle of a triangle,
it is required to construct the locus of its vertex.
Sug. Consult the previous problem.
675. It is required to construct the triangle, having given
the base, the vertical angle, and the altitude.
Sug. Use previous problem.
676. It is required to construct the triangle, having given
the base, the vertical angle, and the median.
OPTIONAL PKOBLEMS FOB ADVANCE WOEK.
677. It is required to construct the triangle, having given
the base, vertical angle, and perpendicular from one extremity
of the base to the opposite side.
678. It is required to construct the isosceles triangle, having
given the altitude and one of the equal angles.
679. It is required to construct the chord in a given circle,
having given the middle point of the chord.
680. It is required to construct a circle whose circumference
shall pass through the vertices of a given rectangle.
681. It is required to construct a tangent to a given circle
which shall be parallel to a given straight line.
682. It is required to construct a tangent to a given circle
which shall be perpendicular to a given straight line.
683. It is required to construct a tangent to a given circle
which shall make a given angle with a given straight line.
684. It is required to construct the isosceles triangle, having
given the vertical angle and a point in the base, in position.
144 PLANE GEOMETRY.
685. It is required to construct the triangle, having given
the altitude, the base, and an adjacent angle.
686. It is required to construct the triangle, having given
the altitude, the base, and an adjacent side.
687. It is required to construct a rhombus, having given its
base and altitude.
688. It is required to construct the triangle, having given
the altitude and the sides which include the vertical angle.
689. It is required to construct the triangle, having given
the altitude and angles adjacent to the base.
690. It is required to construct an isosceles triangle which
shall have its vertical angle twice the sum of its other two
angles.
691. It is required to construct the square, having given its
diagonal.
692. It is required to construct the right triangle, having
given the hypothenuse and perpendicular from the vertex of
the right angle to the hypothenuse.
693. It is required to construct the locus of the centre of
the circle of given radius tangent to a given straight line.
694. It is required to construct the circle of given radius
which shall be tangent to two given non-parallel lines.
695. It is required to construct the circle of given radius
which shall be tangent to a given straight line, and whose cen-
tre shall be in a given line not parallel to the former.
696. It is required to construct the circle which shall be
tangent to a given line, and whose circumference shall pass
through a given point.
697. It is required to find the locus of the centre of the
circle of given radius which shall be tangent externally to a
given circle.
PLANE GEOMETRY 145
It is required to construct the locus of the centre of
the circle of given radius which shall , be tangent internally to
a given circle.
699. It is required to construct a circle which shall be tan-
gent to a given line and a given circle.
700. It is required to construct a circle which shall be tan-
gent to two given circles.
701. It is required to construct a circle which shall cut
three equal chords of given length from three given non-
parallel lines.
702. It is required to construct in a given circle a chord of
given length passing through a given point.
703. It is required to construct in a given circle a chord of
given length and parallel to a given straight line.
704. It is required to construct a line of given length pass-
ing through a given point between two given parallel lines.
705. It is required to construct a line of given length between
two given non-parallel lines, and which shall be parallel to a
given line.
706. It is required to construct a line of given length between
two non-parallel lines, and which shall pass through a given
point.
707. It is required to construct the right triangle, having
given the hypothenuse and radius of the inscribed circle.
708. It is required to construct the right triangle, having
given the radius of the inscribed circle and one acute angle.
709. It is required to construct the triangle, having given
in position the middle points of its sides.
710. It is required to construct the triangle, having given
the base, vertical angle, and radius of the circumscribing circle.
146 PLANE GEOMETRY.
711. It is required to construct the isosceles triangle, having
given the base and radius of the inscribed circle.
712. It is required to construct a straight line which shall
pass through a given point and make equal angles with two
given lines.
713. It is required to find a point in a given secant to a
given circle such that the tangent to the circle from that
point shall be of given length.
714. It is required to construct the right triangle, having
given one leg and radius of the inscribed circle.
715. It is required to construct the right triangle, having
given the median and altitude from the vertex of the right
angle to the hypothenuse.
716. It is required to find the locus of the centre of the
chord which passes through a given point in a given circle.
717. It is required to construct the triangle, having given
the base, vertical angle, and sum of the other two sides.
718. It is required to construct a circle which shall be
tangent to two given lines and at given point of contact in one.
719. It is required to inscribe a circle in a given sector.
720. It is required to construct a common tangent to two
given circles.
Sug. Make five cases according to relative position of the
circles. (See Theorem 229.)
721. It is required to inscribe a square in a given rhombus.
722. Having given two intersecting circles, it is required to
draw a line through one of the points of intersection, so that
the two intercepted chords shall be equal.
Sug. Join centres. Draw from point of intersection to
middle of that line. From centres draw radii parallel to latter
line. Through point of intersection draw line perpendicular to
these radii.
PLANE GEOMETRY. 147
723. It is required to construct an equilateral triangle having
its vertices in three given parallel lines.
724. It is required to construct a tangent to a given circle
with two given parallel secants so that the point of contact
shall bisect the part between the secants.
725. It is required to construct three equal circles which
shall be tangent to each other and also to a given circle
externally.
726. It is required to construct three equal circles which
shall be tangent to each other and also to a given circle
internally.
727. It is required to construct three equal circles which
shall be tangent to each other and also to the sides of an
equilateral triangle.
728. It is required to construct a semicircle having its
diameter in one of the sides of a given triangle and tangent
to the other two sides.
729. It is required to construct a triangle, having given the
radius of the inscribed circle and two sides.
730. It is required to find a point in a given line such that
lines to that point from two given points without the line
make equal angles with the line.
731. It is required to construct the triangle, having given
the perimeter, altitude, and vertical angle.
EEQUIRED PROBLEMS OF CONSTRUCTION.
It is required,
732. To divide a given line into any number of equal parts.
Sug. From one extremity of the given line construct a line
of indefinite length, making any convenient angle with the
given line. Then, with any convenient unit of length assumed
as a unit of measure, beginning at the vertex, lay off on the
148 PLANE GEOMETRY.
indefinite line this unit as many times as it is required to
divide the given line into parts. Then consult 316.
733. To divide a given line into parts proportional to any
number of given lines.
Sug. Place the given parts so as to form one straight line,
making any convenient angle with the line to be divided.
Then consult 313 and 319.
734. To construct a line that shall be a fourth proportional
to three given lines.
Sug. Consult Ratio and Proportion, 277, Theorem 313, and
previous problem.
735. To construct a line that shall be a third proportional
to two given lines.
Sug. Consult Katio and Proportion, 279.
What must one of the given lines be if those two and one
other are to form a proportion? Consider that in the con-
struction.
736. To construct a mean proportional between two given
lines.
Sug. Consult Theorem 367.
737. Given a polygon and an homologous side of another
similar polygon, to construct the latter.
Sug. Consult Theorems 356 and 357.
738. To inscribe in a given circle a triangle similar to a
given triangle.
739. To circumscribe about a given circle a triangle similar
to a given triangle.
740. To construct a square which shall be equivalent to the
sum of two given squares.
Sug. Consult Theorem 360.
741. To construct a square which ktiall be equivalent to the
sum of three or more given squar.es..
PLANE GEOMETRY. 149
Sug. Construct a square equivalent to two of the given
squares, then one equivalent to that and one other of the
given squares, and so on.
742. To construct a square which shall be equivalent to
the difference of two given squares.
743. To construct a square equivalent to a given rectangle.
JSug. If x and y are the base and altitude of a rectangle,
and n one side of an equivalent square, then
xy = n 2 .
Whence x : n : : n : y ; (See Theorem 294.)
or n is a mean proportional between x and y.
Hence consult Problem 736.
744. To construct a square which shall be equivalent to a
given parallelogram.
745. To construct a square which shall be equivalent to a
given triangle.
746. To construct a triangle which shall be equivalent to a
given polygon of more than three sides.
Let ABCDH be a polygon
of n sides.
Extend one of the sides, as
AB, and construct the diag-
onal DB. Through vertex C
draw CK parallel to DB, and
join DK.
Considering DB the com-
mon base of the two triangles DCB and DKB, what is the
relation between the areas of those two triangles ? Why ?
How does the area of the polygon DKAH, then, compare
with that of DCBAH? Why?
How many sides has the former as compared with the
latter ?
Proceed in the same way with the polygon DKAB.
150
PLANE GEOMETRY.
746 (a). To construct a square equivalent to any given polygon.
Sug. Use Problem 746, then 745.
746 (6). To construct a square equivalent to the sum of any
number of given polygons.
746 (c). To construct a rectangle which shall be equivalent
to a given square and the sum of whose base and altitude
shall be equal to a given line.
Sug. Upon the given line as a diameter construct a circle.
Construct a line parallel to the diameter, distant from it one
side of the given square. Then consult Theorem 367.
746 (d) . To construct a rectangle which shall be equivalent
to a given square and the difference of whose base and altitude
shall be equal to a given line.
Sug. Proceed as in 746 (c), then at extremity of the diam-
eter construct a tangent equal to one side of given square, and
from other extremity of this tangent construct a secant
through centre of circle. (Consult Theorem 370.)
747. To construct a right triangle which shall be equivalent
to a given triangle and its hypothenuse equal to a given line.
748. To divide a given line into extreme and mean ratio.
(See 280.)
Post. Let AB be the given line. At one extremity erect
a perpendicular CB equal to one-half AB. With C as centre
PLANE GEOMETRY.
151
Dem. Since CjB= DK=AB.
and CB as a radius construct the circle DBN. Construct the
secant AK passing through the centre C. With A as a centre
and radius AD (point D being point of intersection of secant
and circumference) construct the arc DH. Then the line AB
will be divided in extreme and mean ratio at H.
2
AD : AB : : AB : AK. Why ?
AD : AB - AD : : AB : AK- AB. Why ?
AH : AB - AH: : AB : AK-DK. Why ?
AH:HB::AB:AD. Why?
AH:HB::AB:AH', Why?
or HB:AH::AH: AB. Why ?
Hence the line AB is divided in extreme and mean ratio.
Q.E.F.
749. To inscribe a regular decagon in a given circle.
Post. Let ABH be the given
circle, and A C one of its radii.
Consult Theorem 546, then use
Problem 748.
Cons. With A as a centre and
D C as a radius, construct chord
AB. Join BC and BD.
Dem. AC:AB::AB:AD.
Why?
Hence the two AABD and
ABO are how related ? (See Theorem 350.)
What kind of a triangle is ^OB ?
What kind of a triangle must ABD be, then ?
What relation, then, between AB and DB ? What between
What relation, then, between the A CAB and <3R4 ? Be-
tween the A DAB and AD? Between the A BDA and
? Why ? Between the A DBC and DCB ?
152 PLANE GEOMETRY.
What relation does the Z BDA bear to the sum of DCB and
DBG? Why?
What relation does Z DAB bear to Z C, then ?
Hence, what relation does Z AB(7 bear to Z ?
Compare now Z DAB + Z ABC with the Z (7, and finally
compare Z 1MB + Z ABC + Z with the Z (7.
If the pupil has answered the above questions correctly, he
will now have the equation,
Z DAB + Z JLBC + Z <7 = 5 Z G.
What is the value of the first member of the above equa-
tion ? Why ?
Then 5ZC=2rt.Zs;
or 10 Z C = 4 rt. A.
Whence Z <7 = T V of 4 rt. A
Hence the arc ^..B is what part of the circumference ?
.-. AB is the side of a regular inscribed decagon. Q.E.F.
750. To inscribe a square in a given circle.
tSug. Consult Theorem 513.
751. To inscribe a regular hexagon in a given circle.
Sug. Consult Theorem 480.
752. To inscribe a regular pentedecagon in a given circle.
Sug. Find the difference between a central angle of the
regular decagon and that of a regular hexagon.
753. To inscribe in a given circle
I. a regular trigon ;
II. a regular pentagon ;
III. a regular octagon ;
IV. a regular dodecagon ;
V. a regular polygon of sixteen sides ;
VI. a regular polygon of twenty sides.
754. To circumscribe around a given circle all the above-
mentioned regular polygons.
PLANE GEOMETRY.
153
755. To inscribe in a given circle a regular polygon similar
to a given regular polygon.
Sug. Construct a central angle equal to that of the given
polygon, etc.
756. To circumscribe a circle about any given regular
polygon.
757. Upon a given line as a base, to construct a rectangle
equivalent to a given rectangle.
758. To construct a square whose ratio to a given square
shall be the same as that of two given lines.
Post. Let Q be the given square, and I and p the two given
lines.
We are required to construct a square (Q') so that
Area Q : Area Q' : : I : p.
Cons. Place the two given lines so as to form one straight
line, &s\ADB.
On this as a diameter construct a semicircle, and at Z> erect
the perpendicular DK. Join AK and BK.
Make KN equal to one side of the given square, as ST.
Draw NH parallel to AB. Then KH is the side of the
required square.
Dem. ZK 2 : BK 2 ::AD: DB. (Theorem 360, IV.)
Or, AK 2 :BK 2 ::l:p.
Again, NK : HK : : AK : BK. Why ?
Hence NK 2 : HK 2 : : AK 2 : BK 2 .
.'. NK 2 :HK 2 ::l:p. Why?
154 PLANE GEOMETRY.
Hence the square constructed on HK as a side is the re-
quired square. Q.E.F.
759. To construct a polygon similar to a given polygon, the
ratio of whose areas shall be that of two given lines.
C
Post. Let ABCDH be the given polygon (P), and p and q
the two given lines.
We are required to construct a polygon (P') similar to P,
so that Area p . Area P , ::p:q
Cons. Upon any side of the polygon P, as AB, construct a
square.
Then, by the previous problem, find the side of a square
whose ratio to that of the square on AB shall be that of the
two lines p and q.
Upon this line construct a polygon similar to polygon P.
This will be the polygon required.
Dem. This will be left for the pupil.
759 (a). To construct a polygon similar to one of two given
polygons and equivalent to the other.
Q
Post. Let P and Q be the two given polygons. We are
PLANE GEOMETRY. 155
required to construct a polygon similar to P and equivalent
to Q.
Cons. Construct squares equivalent to each of the polygons
P and Q.
Then find a fourth proportional to the sides of these squares
and any side of the polygon P selected at random, as AB.
Upon this fourth proportional as an homologous side construct
a polygon similar to P. Then this polygon will be similar to
P and equivalent to Q, and is therefore the polygon required.
Dem. This is also left for the pupil.
760. To construct upon a given line, as one side,
I. a regular trigon ; V. a regular pentagon ;
II. a regular tetragon ; YI. a regular decagon.
III. a regular hexagon ; VII. a regular dodecagon ;
IV. a regular octagon ; VIII. a regular pentedecagon.
761. To construct a regular hexagon, given one of its shorter
diagonals.
762. To construct a regular pentagon, given one of its
diagonals.
763. To construct a circle equivalent to the sum of two
given circles.
764. To construct a circumference equal to the sum of two
given circumferences.
765. To divide a given circle by a concentric circumference
into two equal parts.
MISCELLANEOUS PLANE PROBLEMS FOR ADVANCE WORK.
I. TRIANGLES.
It is required to construct the triangle, having given,
766. Its base, vertical angle, and difference of the other two
sides.
156 PLANE GEOMETRY.
767. Its base, vertical angle, and a square which is equal to
the sum of the squares upon the other two sides.
768. Its base, vertical angle, and a square which is equiva-
lent to the difference of the squares upon the other two sides.
769. Its base, vertical angle, and sum of its altitude and the
two remaining sides.
770. Its base, vertical angle, and the sum of its altitude
and difference of the other two sides.
771. Its base, vertical angle, and difference between its alti-
tude and sum of its other two sides.
772. Its base, vertical angle, and difference between its alti-
tude and difference of the other two sides.
773. Its base, vertical angle, and ratio of its altitude to the
sum of its other two sides.
774. Its base, vertical angle, and ratio of its altitude to the
difference of its other two sides.
775. Its base, altitude, and sum of its other two sides.
776. Its base, altitude, and difference of its other two sides.
777. Its base, altitude, and ratio of the other two sides.
778. Its base, altitude, and a square equivalent to the rec-
tangle of the other two sides.
779. Its base, altitude, and a square which is equivalent to
the sum of the squares upon the other two sides.
780. Its base, altitude, and a square which is equivalent to
the difference of the squares upon the other two sides.
781. Its vertical angle, sum of base and altitude, and sum
of the other two sides.
782. Its vertical angle, sum of base and altitude, and differ-
ence of its other two sides.
PLANE GEOMETRY. 157
783. Its vertical angle, sum of base and altitude, and ratio
of the other two sides.
784. Its vertical angle, sum of base and altitude, and a
square equivalent to the rectangle of its other two sides.
785. Its vertical angle, sum of base and altitude, and sum
of the three sides.
786. Its vertical angle, sum of base and altitude, and differ-
ence between the base and sum of the other two sides.
787. Its vertical angle, sum of base and altitude, and differ-
ence between the base and difference of its other two sides.
788. Its vertical angle, sum of base and altitude, and the
ratio of the base to the sum of the other two sides.
789. Its vertical angle, sum of base and altitude, and the
ratio of the base to the difference of its other two sides.
790. Its vertical angle, altitude, and the square equivalent
to the sum of the squares of the sides which include the ver-
tical angle.
791. Its vertical angle, altitude, and radius of the circum-
scribing circle.
792. Its vertical angle, radius of the inscribed circle, and
perimeter.
793. Its vertical angle, radius of the inscribed circle, and
ratio of the sides including the vertical angle.
794. Its vertical angle, radius of the inscribed circle, and
a square equivalent to the rectangle of the sum of the two
sides including the vertical angle and the base.
795. Its vertical angle, radius of the inscribed circle, and
a square whose area is equal to the difference between the
sum of the squares of the sides including the vertical angle
and the square of the base.
796. Its base, meclian, and sum of the other two sides.
158 PLANE GEOMETRY.
797. Its base, median, and difference of the other two sides.
798. Its three altitudes.
799. Its three medians.
800. Two sides, and difference of the angles opposite them.
801. Its vertical angle, difference of the angles at the base,
and difference of the other two sides.
802. Difference of the angles at the base, difference of the
segments of the base made bj the altitude, and sum of the
other two sides.
803. It is required to construct the equilateral triangle,
having given the three distances from its vertices to a common
point within the triangle.
804. The same as 803, but the common point without the
triangle.
IL QUADRILATERALS.
805. It is required to construct a square, having given
I. the sum of its diagonal and side.
II. the difference of its diagonal and side.
It is required to construct a rectangle, having given
806. The sum of two adjacent sides and its diagonal.
807. The difference of two adjacent sides and its diagonal
808. One side, and sum of diagonal and adjacent side.
809. One side, and difference of diagonal and adjacent side.
It is required to construct the rhombus, having given
810. Its side and altitude.
811. Its altitude and lesser angle.
812. Its side and sum of its diagonals.
813. Its side and difference of its diagonals.
PLANE GEOMETRY. 159
814. Its lesser angle and sum of its diagonals.
815. Its lesser angle and difference between its longer diago-
nal and altitude.
It is required to construct the rhomboid, having given
816. The longer side, sum of its diagonals, and larger angle
made by the diagonals.
817. Its lesser angle, longer diagonal, and sum of two adja-
cent sides.
818. Its lesser angle, longer side, and sum of its altitude
and lesser side.
819. Its lesser angle, shorter side, and difference of its
longer diagonal and longer side.
It is required to construct an isosceles trapezoid, having
given,
820. One leg, diagonal, and longer base.
821. Its longer base, diagonal, and lesser angle.
822. Its diagonal, altitude, and leg.
823. Its longer base, lesser angle, and sum of altitude and
leg.
It is required to construct the trapezoid, having' given
824. Its longer base, lesser angle (i.e. angle formed by longer
base and a leg), and its diagonals.
825. Its longer base, one leg, lesser angle, and altitude.
826. Sum of its bases, the two legs, and lesser angle.
827. Difference of its bases, the two legs, and angle formed
by its diagonals.
III. CIRCLES.
Having given a circle, it is required to construct
828. Three equal circles, tangent to the given circle exter-
nally, and tangent to each other.
160 2 LANE GEOMETRY.
829. Three equal circles, tangent to the given circle inter-
nally, and tangent to each other.
830. Four equal circles, as in (828) and (829).
831. Five equal circles, as in (828) and (829).
832. Six equal circles, as in (828) and (289).
833. A circle tangent to three given circles.
IV. TRANSFORMATION OF FIGURES.
It is required,
834. To transform a given triangle into an equivalent
isosceles one having the same base.
835. To transform a given isosceles triangle into an equiva-
lent equilateral one.
836. To transform a given triangle into an equivalent
equilateral one.
837. To transform a given triangle into another equivalent
triangle whose base and altitude shall be equal.
838. To transform a given triangle into another equivalen
triangle, and similar to a given triangle.
839. To transform a given triangle into a triangle with
one angle unchanged and its opposite side parallel to a given
line.
840. To transform a given triangle into an equivalent tri-
angle with a given perimeter.
841. To transform a triangle into a trapezoid, one of whose
bases shall be the base of the triangle, and one of its adjacent
angles one of the basal angles of the triangle.
:
PLANE GEOMETRY. 161
842. To transform a given triangle into a right triangle with
given perimeter.
843. To transform a given triangle into a parallelogram with
given base and altitude.
844. To transform a parallelogram into a parallelogram with
a given side.
845. To transform a parallelogram into a parallelogram hav-
ing a given angle.
846. To transform a parallelogram into a parallelogram with
given altitude.
To transform a square into
847. A right triangle.
848. An isosceles triangle.
849. An equilateral triangle.
850. A rectangle with given side.
851. A rectangle with given perimeter.
852. A rectangle with given difference of sides.
853. A rectangle with given diagonal.
To transform a rectangle into
854. A square.
855. An isosceles triangle.
856. An equilateral triangle.
857. A rectangle with given side.
858. A rectangle with given perimeter.
859. A rectangle with given difference of sides.
860. A rectangle with given diameter.
It is required to construct a parallelogram equivalent to the
861. Sum of two given parallelograms of equal altitudes.
162 PLANE GEOMETRY.
862. Difference of two given parallelograms of equal alti-
tudes.
863. Sum of two given parallelograms of equal bases.
864. Difference of two given parallelograms of equal bases.
865. Sum of two given parallelograms.
866. Difference of two given parallelograms.
It is required to transform a given parallelogram into
867. A triangle.
867 (a) . An isosceles triangle.
868. A right triangle.
868 (a) . An equilateral triangle.
869. A square.
870. A rhombus having for a diagonal one side of the par-
allelogram.
871. A rhombus having a given diagonal.
872. A rhombus having a given side.
873. A rhombus having a given altitude.
874. A parallelogram having a given side and diagonal.
875. To transform a rhombus into a square.
876. To inscribe in a given circle a rectangle equivalent to
a given square.
To transform a trapezoid into
877. A triangle.
877 (a). A square.
878. A parallelogram having for one base the longer base of
the trapezoid.
879. An isosceles trapezoid.
PLANE GEOMETRY. 163
To transform a trapezium into
880. A triangle.
881. An isosceles triangle with given base.
882. A parallelogram.
883. A trapezoid with one side and the two adjacent angles
unchanged.
V. DIVISION OF FIGURES.
884. It is required to divide a given triangle into any num-
ber of equivalent parts, by lines drawn from one vertex.
885. It is required to divide a given triangle into any num-
ber of equivalent parts, by lines drawn from any point in its
perimeter selected at random.
886. It is required to divide a given triangle into any num-
ber of equivalent parts, by lines drawn from any point selected
at random in the triangle.
887. It is required to divide a given triangle into any num-
ber of equivalent parts by lines parallel to one side.
888. It is required to divide a given triangle into any num-
ber of parts whose areas shall be in a given ratio, by lines
drawn from one vertex.
889. It is required to divide a given triangle into any
number of parts, whose areas shall be in a given ratio, by lines
drawn from any point selected at random in the perimeter.
890. It is required to divide a given triangle into any num-
ber of parts, whose areas shall be in a given ratio, by lines
drawn from any point selected at random in the triangle.
891. It is required to divide a given triangle into any num-
ber of parts, whose areas shall be in a given ratio, by lines
drawn parallel to one side.
892. It is required to divide a given parallelogram into any
164 PLANE GEOMETRY.
number of equal parts by lines drawn parallel to one pair
of sides.
893. It is required to divide a given parallelogram into
any number of parts, whose areas shall be in a given ratio, by
lines parallel to one pair of sides.
894. It is required to divide a given parallelogram into two
equivalent parts by a line drawn from any point selected at
random in the perimeter.
895. It is required to divide a given parallelogram into two
equivalent parts by a line drawn through any point in the
parallelogram selected at random.
896. It is required to divide a given parallelogram into
two parts, whose areas shall be in a given ratio, by a line
drawn from one vertex.
897. It is required to divide a given parallelogram into two
parts, whose areas shall be in a given ratio, by a line drawn
from any point in the perimeter selected at random.
898. It is required to divide a given parallelogram into two
parts, whose areas shall be in a given ratio, by a line drawn
from any point in the parallelogram selected at random.
899. It is required to divide a parallelogram into two
equivalent parts by a line drawn parallel to a given line.
900. It is required to divide a parallelogram into two parts,
whose areas shall be in a given ratio, by a line drawn parallel
to a given line.
901. It is required to divide a parallelogram into any num-
ber of equivalent parts by lines drawn from either vertex.
902. It is required to divide a parallelogram into any num-
ber of equivalent parts by lines drawn from any point in its
perimeter selected at random.
903. It is required to divide a parallelogram into any num-
PLANE GEOMETRY. 165
her of equivalent parts by lines drawn from any point in the
parallelogram selected at random.
904. It is required to divide a parallelogram into any num-
ber of equivalent parts by lines parallel to a given line.
905. It is required to divide a parallelogram into any num-
ber of parts, whose areas shall be in a given ratio, by lines
parallel to a given line.
It is required to divide a trapezoid into two equivalent
parts by a line drawn
906. Parallel to the bases.
907. Perpendicular to the bases.
908. Parallel to one of the legs.
909. Through one of its vertices.
910. Through a given point in one of its bases.
911. Through any point selected at random in its perimeter.
912. Through any point selected at random in the trapezoid.
913. Parallel to a given line.
It is required to divide a trapezoid into any number of
equivalent parts by lines drawn
914. Parallel to the bases.
915. Perpendicular to the bases.
916. Parallel to one of its legs.
917. Through one of its vertices.
918. Through any point selected at random in one of its
bases.
919. Through any point selected at random in its perimeter.
920. Through any point selected at random in the trapezoid.
921. Parallel to a given line.
166 PLANE GEOMETEY.
It is required to divide a given trapezoid into any number
of parts whose areas shall be in a given ratio by lines drawn
922. Parallel to the bases.
923. Perpendicular to the bases.
924. Parallel to one of the legs.
925. Through either vertex.
926. Through any point selected at random in one of the
bases.
927. Through any point selected at random in its perimeter.
928. Through any point selected at random in the trapezoid.
929. Parallel to a given line.
930. It is required to divide a trapezium into two equivalent
parts by a line drawn from either vertex.
931. It is required to divide a trapezium into two equivalent
parts by a line drawn from any point selected at random in
its perimeter.
932. Given the diameter of a circle, it is required to con-
struct a straight line equal in length to the circumference.
From Theorem 519 it is evident that it can only be approxi-
mated.
\Q
A*
~"~ '
Let AB be the given diameter and C its middle point.
Extend AB indefinitely as AS. Make BD and DE each equal
to AB. At E erect the perpendicular EQ, and on it make
EFaud FG each equal to AB. Join AG, AF, DG, and DF.
Lay off EH and HK each equal to AG, and from K lay off
PLANE GEOMETRY. 167
KL equal to AF. Again, from L make LM equal to DGr, and
MN equal DF. Bisect EN v& P; bisect EP at 72; and then
trisect -EJjR at T arid TF. Then CT will be the required line
approximately equal to the circumference of the circle whose
diameter is AB ; for, calling the diameter unity,
EL = 2 CH- KL = 2VL3 - VlO,
and
ET = T V (2 Vl3 - VlO + V5 + V2), and therefore
CT= 2J + Jg- (2V13 - VlO + V5 + V2) = 3.1415922 + .
Q.E.F.
168
PLANE GEOMETRY.
APPENDIX.
1. The following theorem, and demonstrations of it, are
given as a substitute for 234, for those teachers who may
prefer it. Eatio and proportion (261), however, should be
taken previous to attempting it.
2. In the same or equal circles two central angles are in the
same ratio as the arcs which their sides intercept.
Post. Let NBA and QKH be two equal circles, and C and
I) two central angles.
We are to prove Z C : Z D : : Arc AB : Arc HK
CASE I. When the angles are commensurable.
Dem. If the angles are commensurable, there is some angle,
as W } which will be contained an exact number of times in
each.
Suppose it is contained m times in Z (7, and n times in Z. D.
Then ZC:ZD::m:n.
If, now, lines be drawn from C and D dividing the two
angles into m and n equal parts respectively, each part being
equal to angle W, then arc AB will be divided into m equal
PLANE GEOMETEY. 169
arcs, and HE into n equal arcs, the divisions all being equal,
by Theorem 191.
.*. Arc AB : Arc HK : : ra : n.
.-. Z C : Z D : : Arc AB : Arc HK (Theorem 288.)
CASE II. When the angles are incommensurable.
Post. Let C and D be two incommensurable central angles
in equal circles.
We are to prove Z CiZDnArcAB: Arc HK
Conceive them to be applied as in I.
Then if arc HKis not the fourth term of this proportion,
some other arc greater or less than HK must be. Suppose
it to be greater as A W.
Then Z ACB : Z ACK: : Arc AB : Arc A W. (a)
Now conceive the arc AB to be divided into equal parts by
continued bisection until each part is less than KW. Then
there must be at least one point of division between K and
W, as N.
.-. Z ACB : Z ACN: : Arc AB : Arc AN. (b)
.'. Z ACK : Z ACN: : Arc AW: Arc AN.
(Theorem 300.)
But the arc AN is less than the arc AW. Consequently,
if the proportion be a true one, the angle ACN must be less
170 PLANE GEOMETRY.
than the angle ACK. On. the contrary, it is greater, and
therefore the proportion cannot be a true one. Therefore the
supposition on which the argument was based ; viz. that the
fourth term must be an arc greater than HK, cannot be true.
A similar argument will prove that it cannot be less.
Consequently it must be the arc HK.
.-. ZC:^D::AicAB: Arc HK Q.E.D.
3. This really means the same thing as 234 ; viz. that the
numerical measure of an angle at the centre of a circle is
the same as the numerical measure of its intercepted arc, if
the adopted unit of angle is the angle at the centre which
intercepts the adopted unit of arc.
4. Another favorite method of demonstrating this theorem
is that of the method of limits, hereto appended.
THEOKY OP LIMITS,
5. A constant quantity, or simply a constant, is a quantity
whose value remains unchanged throughout the same dis-
cussion.
6. A variable quantity, or simply a variable, is a quantity
which may assume different values in the same discussion,
according to the conditions imposed.
7. The limit of a variable is a constant quantity, which
the variable is said to approach in value whenever a regular
and definite increase or decrease in value is assigned to the
latter.
8. Whenever it can be shown that the value of a variable,
by such constant increase or decrease in value, can be made to
differ from that of its limit by less than any appreciable or
assignable quantity, however small, this variable is said to
approach indefinitely to its limit.
PLANE GEOMETRY. 171
9. For example :
A-, - 1 - 1 - 1 KB
C D H K
Suppose a point move from A toward B under the condition
that during the first second it shall move over one-half the
distance AB, or AC, and that during each successive second it
shall move over one-half the remaining distance. Then at
the end of the second second it would be at D, at the end of
the third at H, at the end of the fourth at K, and so on. It
is evident that it can never reach the point B, for there will
constantly remain one-half the distance ; but if its motion be
continued indefinitely, it will approach indefinitely near to B.
Consequently, the distance from A to the moving point is
an increasing variable, and AB is its limit; while the distance
from B to the moving point is a decreasing variable, with zero
as its limit.
10. Other illustrations may be given ; e.g.
0.3333 + ... = A + T f + + ^n +..,
Here the sum of the series of fractions is the increasing
variable, and approaches -| as its limit.
Again : let ABC be a right triangle, with
C the right angle, and consider the point B
to move toward (7; the angle A will then
be a decreasing variable approaching zero as
its limit, and the angle B will be an increas-
ing variable approaching a right angle as
its limit.
11. Theorem. If two variables are always equal, and each
approaches a limit, their limits are equal.
D Q
D'
172 PLANE GEOMETRY.
Post. Let AB and A'B' be the limits to which the two equal
variables AD and A'D' indefinitely approach.
We are to prove AB = A'B'.
Dem. AB and A'B' are either equal or unequal* Let us
suppose them unequal, and that AB is the greater. Mark off
AQ equal to A'B'.
Then the variable A'D' cannot exceed A'B', but the variable
AD may exceed AQ, and consequently the variable AD be-
comes greater than the variable A'D'. This, however, is con-
trary to the hypothesis that the two variables must always
be equal. Therefore AB and A'B' cannot be unequal ; i.e.
they are equal. Q.E.D.
12. Theorem. If two variables are in a constant ratio, their
limits are in the same ratio.
Post. Let AD and AH be two unequal variables, and
approaching their respective limits AB and AC, and having
AD
a constant ratio such that - = m.
We are to prove that = 41*,
or that AD : AH ::AB:AC.
Dem. Since = m, AD == m x AH.
AH
And since m X AH will vary as AH varies* AD and m x AH
PLANE GEOMETRY.
173
are two equal variables, and therefore, from the previous
theorem,
Limit of AD = Limit of m X AH, or
Limit of AD = mx Limit of AH.
Limit of AD _
' Limit of AH~
but the limit of AD is AB, and the limit of AH is AC.
AB
and since
or
AC
AD
AH
AB
AC
110 ,
= w;
AD
AH'
AD:AH::AB:AC.
Q.E.D.
13. Let us now apply these principles to the demonstration
of the theorem enunciated at the beginning of the Appendix.
In Case I. the demonstration will remain unchanged.
Case II., when the arcs are incommensurable.
Post. Let AQB and NHK be two equal circles, and C and
D two central angles whose arcs AB and HK are incom-
mensurable.
174 PLANE GEOMETRY.
Z C Arc AB
We are to prove =
or Z C: Z D : : Arc AB : Arc
Ztera. Conceive the arc AB to be divided into any number
of equal arcs, and one of these arcs to be applied as a unit
of measure to the arc HK. It will be contained a certain
number of times with a remainder, SK, less than the unit
of measure.
Construct DS.
Then, since AB and HS are commensurable,
Z C
Z D Arc HS
If, now, the number of equal parts into which the arc AB
is divided be indefinitely increased, the unit of measure of the
arc HK will be correspondingly diminished, and the point
S will get indefinitely near to K. Consequently the arc
HS approaches indefinitely to HK, and the Z HDS to the
Z.HDK.
Consequently the variables
its limit
and m h
Arc
Arc .## Arc
But Arc
Z J^Z)^ Arc
Z ACB Arc
Ai-cHK'
because if two variables are always equal and approaching
their limits, their limits are equal. Q.E.D.
14. Theorems 406 and 313 may be Demonstrated by similar
methods.
PLANE GEOMETRY.
175
SYMMETRY,
15. Two points are said to be symmetrical with respect to
a point when they are equidistant from, and in the same line
with, this point. This point is called the centre of symmetry.
16. Two points are said to be symmetrical with respect to
a line when the line that joins them is perpendicular to, and
bisected by, this line. This line is called an axis of symmetry.
17. Two points are said to be symmetrical with respect to
a plane when the line that joins them is perpendicular to,
and bisected by, this plane. This plane is called a plane of
symmetry.
18. The distance of either of two symmetrical points from
the centre of symmetry is called the radius of symmetry.
19. Two plane figures are symmetrical with respect to a
centre, axis, or a plane, when' any point in either figure selected
at random has a correspondingly symmetrical point in the
other.
K
\
\
JET'
Fig. I.
3
Fig. II.
Fig. III.
Fig. IV.
176
PLANE GEOMETRY.
C
TT
N
M
P'
Fig. V.
Fig. VI.
Thus, in Figs. I., II., and III., the lines AB and QD are
symmetrical, with respect to the centre (7, the axis /$, and
the plane MN, respectively. In Figs. IV., V., and VI., the
same is true of the triangles ABQ and HR W.
20. A plane figure is symmetrical
I. when it can be divided by an axis into two figures sym-
metrical with respect to that axis ;
II. when it has a centre such that, if a line be drawn
through it in any direction at random, the two points at which
it intersects the perimeter are symmetrical with respect to
that centre.
Thus figure ABCDH is symmetrical with respect to the
axis K/S, and ABQDKH with respect to the centre (7. In
the latter case PP 1 or NN' is called a diameter of symmetry.
(See 18.)
PLANE GEOMETRY.
177
21. A geometrical solid is symmetrical
I. when it can be divided by a plane into two solids sym-
metrical with respect to that plane ;
II. when it has a centre such that, if a line be drawn
through it in any direction selected at random, the two points
at which it intersects the surface are symmetrical with respect
to that centre.
z
Fig. II.
Fig. I.
Thus figure ABCDfiFHK, Fig. I., is divided by the plane XZ
at the lines LM, MN, NP, and PL, into two figures, symmet-
rical with respect to the plane XZ. Hence it is symmetrical.
Similarly, in Fig. II., the points P and P being symmetrical
with respect to the point (7, according to above definition, the
figure is symmetrical.
THEOREMS.
22. The centre of a circle is a centre of symmetry.
23. The diameter of a circle is an axis of symmetry.
24. The line which bisects the vertical angle of an isosceles
triangle is an axis of symmetry.
25. Either altitude of an equilateral triangle is an axis of
symmetry.
26. The point of intersection of two altitudes of an equi-
lateral triangle is a centre of symmetry.
178 PLANE GEOMETRY.
27. A segment of a circle is a symmetrical figure.
28. That part of a circle included between two parallel
chords is a symmetrical figure.
29. The common part of two intersecting circles is a sym-
metrical figure.
30. The diagonal of a square is an axis of symmetry.
31. Every equilateral tetragon is a symmetrical figure.
(How many axes of symmetry does a square have ?)
32. The point of intersection of the diagonals of a parallelo-
gram is a centre of symmetry.
33. An isosceles trapezoid is a symmetrical figure.
34. If one diagonal of a tetragon divides it into two isosceles
triangles, the other diagonal is an axis of symmetry.
35. The bisector of an angle of a regular polygon is an axis
of symmetry.
36. The perpendicular bisector of one side of a regular poly-
gon is an axis of symmetry.
37. If the angles at the extremities of one side of an
equilateral pentagon be equal, the pentagon is a symmetrical
figure.
38. If two diametrically opposite angles of an equilateral
hexagon are equal, the hexagon is a symmetrical figure.
39. Every equiangular tetragon is a symmetrical figure.
40. If a figure have two axes of symmetry perpendicular
to each other, their intersection is a centre of symmetry.
Post. Let ABDH, etc., be a figure having the two axes of
symmetry PP' and MM' _L to each other.
We are to prove that their point of intersection C is a
centre of symmetry.
PLANE GEOMETRY.
179
Cons. From any point in the perimeter selected at random,
as Q, construct QOA.MM', QFPP', and join TL, OC, and
FC.
Dem. OT=TQ = LC. Why ?
,TT
Then what kind of a figure is OTLC? What relation, then,
between TL and OC ? Compare in a similar manner TL and
OF. Finally compare OC and CF.
Hence points and F are in the same line with and equi-
distant from C. Hence C is a centre of symmetry. Q.E.D.
YB 1 7298