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INTRODUCTORY
MODERN GEOMETRY
OF
POINT, RAY, AND CIRCLE
PART I
BY
W. B. SMITH, Ph.D.
^ -~ ■■■ o
Ij^
IN MEMORIAM
FLORIAN CAJORI
f/a^ C?/y-^^-^
The price of Part I. is 75 cents, and of the co7?iplete
hook $1.10. Copies of this part, if returried to the pub-
lishers in good condition, will be exchanged for the complete
book, on payment of the difference in price.
INTRODUCTORY MODERN GEOMETRY
.^>f)m
INTRODUCTORY
MODERN GEOMETRY
OF
POINT, RAY, AND CIRCLE
Part I
BY
WILLIAiVrBENJAMIN SMITH, A.M., Ph.D. (Goett.)
Professor of Mathematics and Astronomy
University of the State of Missouri
aeltefleS berool^rt mit trcue,
^reunblid(> aufgefagted 3leue.
— Goethe.
Ncto gorft
MACMILLAN & CO.
AND LONDON
1892
All rights reserved
Copyright, 1892,
By MACMILLAN AND CO.
Typography by J. S. Cushing & Co., Boston, U.S.A.
Presswork by Berwick & Smith, Boston, U.S.A.
GEOMETRY.
INTRODUCTION.
1 . Geometry is the Doctrine of Space.
What is Space? On opening our eyes we see objects
around us in endless number and variety : the book here,
the table there, the tree yonder. This vision of a world
outside of us is quite involuntary — we cannot prevent it,
nor modify it in any way ; it is called the Intuition (or Per-
ception or Envisagement) of Space. Two objects precisely
alike, as two copies of this book, so as to be indistinguishable
in every other respect, yet are not the same, because they
differ in place, in their positions in Space : the one is here,
the other is not here, but there. In between and all about
these objects that thus differ in place, there lies before us
an apparently unoccupied region, where it seems that noth-
ing isy but where anything might be. We may imagine or
suppose all these objects to vanish or to fade away, but we
cannot imagine this region, either where they were or where
they were not, to vanish or to change in any way. This
region, whether occupied or unoccupied, where all these
objects are and where countless others might be, is called
Space.
2. There are certain elementary facts ^ that is, facts that
cannot be resolved into any simpler facts, about this Space,
and these deserve special notice.
K7i«nf».*ii
2 GEOMETRY.
A. Space is fixed, permanent, unchangeable. The objects
in Space, called bodies, change place, or may be imagined
to change place, in all sorts of ways, without in the least
affecting Space itself. Animals move, that is, change their
places, hither and thither ; clouds form and transform them-
selves, drifting before the wind, or dissolve, disappearing
altogether ; the stars circle eternally about the pole of the
heavens ; sun, moon, and planets wander round among the
stars ; but the blue dome of the sky,* the immeasurable
expanse in which all these motions go on, remains unmoved
and immovable, as a whole and in all its parts, absolutely
the same yesterday, to-day, and forever.
B. Space is homoBoidal ; i.e. it is precisely alike through-
out its whole extent. Any body may just as well be here,
there, or yonder, so far as Space is concerned. A mere
change of place in nowise affects the Space in which the
change, or motion, occurs.
C. Space is boundless. It has no beginning and no end.
We may imagine a piece of Space cut out and colored (to
distinguish it from the rest of Space) ; the piece will be
bounded, but Space itself will remain unbounded.
N. B. When we say that Space is unbounded, we do not
mean that it is infinite. Suppose an earthquake to sink all
the land beneath the level of the sea, and suppose this latter
at rest ; then its outside would be unbounded, without begin-
ning and without end, — a fish might swim about on it in
any way forever, without stop or stay of any kind. But it
would 7iot be infinite ; there would be exactly so many
square feet of it, a finite number, neither more nor less.
Likewise, the fact that bodies may and do move about in
space every way without let or hindrance of any kind implies
* Appearing blue because of the refraction of light in the air.
INTRODUCTION. 3
that Space is boundless, but by no means that it is infinite.
For all we know there may be just so many cubic feet of
Space ; it may be just so many times as large as the sun,
neither more nor less. This distinction between unbounded
and infinite, first clearly drawn by Riemann, is fundamental.
D. Space is continuous. There are no gaps nor holes in
it, where it would be impossible for a body to be. A body
may move about in Space anywhere and everywhere, ever
so much or ever so litUe. Space is itself simply where a
body may be, and a body may be anywhere.
E. Space is triply extended, or has three dimensions.
This important fact needs careful explication.
In telling the size of a box or a beam we find it necessary
and sufficient to tell three things about it : its length, its
breadth, and its thickness. These are called its dimensions ;
knowing them, we know the size completely. But to tell the
size of a ball it is enough to tell one thing about it, namely,
its diameter; while to tell the size of a chair we should
have to tell many things about it, and we should be puzzled
to say what was its length, or breadth, or thickness. Never-
theless, it remains true that Space and all bodies in Space
have just three dimensions, but in the sense now to be made
clear.
We learn in Geography that, in order to tell accurately
where a place is on the outside of the earth, which may
conveniently be thought as a level sheet of water, it is
necessary and sufficient to tell two things about it ; namely,
its latitude and its longitude. Many places have the same
latitude, and many the same longitude ; but no two have
the same latitude and the same longitude. It is not suffi-
cient, however, if we wish to tell exactly where a thing is in
Space, to tell two things about it. Thus, at this moment
the bright star Jupiter is shining exacdy in the south ; we
4 GEOMETRY.
also know its altitude, how high it is above the horizon (this
altitude is measured angularly — a term to be explained
hereafter, but with which we have no present concern) . But
the knowledge of these two facts merely enables us to poiiit
towards Jupiter ; they do not fix his place definitely, they
do not say how far away he is : we should point towards
him the same way whether he were a mile or a million of
miles distant. Accordingly, a third thing must be known
about him, in order to know precisely where he is ; namely,
his distance from us. But when this third thing is known,
no further knowledge about his place is either necessary or
possible. Once more, here is the point of a pin. Where
is it in this room ? It is five feet above the floor. This is
not enough, however, for there are many places five feet
above the floor. It is also ten feet from the south wall, but
there are yet many positions five feet from the floor and ten
feet from the south wall, as we may see by sHpping a cane
five feet long sharpened to a point, upright on the floor,
keeping the point always ten feet from the south wall. But
as it is thus slipped along, the point of the cane will come
to the point of the pin and then will be exactly twelve feet
from the west wall. If it now move ever so little either way
east or west, it will no longer be at the pin-point and no
longer twelve feet from the west wall. So there is one, and
only one, point that is five feet from the floor, ten feet from
the south wall, and twelve feet from the west wall. Hence
it is seen that these three facts fix the position of the pin-
point exactly. A fourth statement, as that the point is nine
from the ceiling, will either be superfluous, if the ceiling is
fourteen feet high, being implied in what is already said, or
else incorrect, if the ceiling is not fourteen feet high, contra-
dicting what is already said. In general, with respect to any
position in Space it is necessary to know three independent
JXTROnUCTION. 5
facts (or data), and it is impossible to know any more. All
other knowledge about the position is involved in this knowl-
edge, which is necessary and sufficient to enable us to an-
swer any rational question that can be put with respect to
the position. Accordingly, since any position in Space is
known completely when, and only when, three independent
data are known about it, we say that Space is triply or three-
fold extended^ or has three dimensions. The dimensions are
any three independent things that it is necessary and suffi-
cient to know about any position in Space, as of the pin-
point or of Jupiter, in order to know exactly where it is.
3. But with respect to the outside of the earth, viewed
as a level sheet of water, we have seen that only two data,
as of latitude and longitude, are necessary and sufficient
to fix any position on it ; neither are more than two inde-
pendent data possible ; all other knowledge about the posi-
tion is involved in the knowledge of these two data about
it. Accordingly we say of such outside of the earth that it
is doubly or two-fold extended y is bi-dimensional, or has two
dimensions ; and we name every such outside^ every such
bi-dimensional region, a surface. Such is the top of the
table : to know where a spot is on it we need know two, and
only two, independent facts about it, as how far it is from
the one edge and how far from the other. (Which other?
and why?)
We see at once that a surface is no part of Space, but is
only a border (doubly extended) between two parts of Space.
Thus, the whole earth-surface is no part either of the earth-
space or of the air-space around the earth, but is the boun-
dary between them. A soap-bubble floating in the air is
not a surface ; though exceedingly thin, it has some thick-
ness and occupies a part of space ; the outside of the film
6 GEOMETRY.
is a surface, and so is the inside, and these are kept apart
by the film itself. If the film had no thickness, the outside
and the inside would fall together, and the film would be a
surface ; namely, the outside of the Space within and the
inside of the Space without.
4. Consider now once more this earth- surface, still viewed
as a smooth level sheet of water. From Geography we
learn that there are two extreme positions on this surface
that are called poles and that do not move at all as the
earth spins round on her axis. We also learn that there is
a certain region of positions just midway between these
poles and called the Equator. This Equator is no part of
the surface ; it is only a border or boundary between two
parts of the globe-surface, which are called hemispheres.
To know where any position is on this border, it is neces-
sary and sufficient to know o?ie thing about it, namely, its
longitude ; neither is any other independent knowledge
about the position possible ; all other knowledge is involved
in this one knowledge. Accordingly we say of this border,
the Equator, that it is simply extended, or has one dimension
only. Every such one-dimensional border is called a line,
and its one dimension is named length. A line, then, has
length, but neither breadth nor thickness.
5. Lastly, consider a part of a line, as of the Equator,
say between longitudes 40° and 50°. The ends of this part
bound it off from the rest of the equator, but they them-
selves form no part of the Equator. They are called points ;
they have position merely, but no extent of any kind, neither
length nor breadth nor thickness, — they are wholly non-
dimensional.
INTRODUCTION. 7
*6. It is noteworthy that extents, or regions, are bounded
by extents of fewer dimensions, and themselves bound
extents of more dimensions. Thus, lines are bounded by
points, and themselves bound surfaces ; surfaces are bounded
by lines, and themselves bound spaces ; spaces are bounded
by surfaces, and themselves bound — what? If anything
at all, it must be some extent of still higher order, oi four
dimensions. But here it is that our intuition fails us ; our
vision of the world knows nothing of any fourth dimension,
but is confined to three dimensions. If there be any such
fourth dimension, we can know nothing of it by intuition ;
we cannot imagine it. In music, however, we do recognize
four dimensions : in order to know a note completely, to
distinguish it from every other note, we must know four
things about it : its pitch, its intensity, its length, its timbre,
— how high it is, how loud it is, how long it is, how rich
it is. While, then, extents of higher dimensions may be
unimaginable, they are not at all unreasonable.
This doctrine of dimensions is of prime importance, but
rather subtile ; let not the student be disheartened, if at first
he fail to master it.
6. We may see and handle bodies, which occupy portions
of Space ; but not so surfaces, lines, points, which occupy
no Space, but are merely regions in Space. Here we must
invoke the help of the logical process called abstraction,
i.e. withdrawing attention from certain matters, disregard-
ing them, while regarding others. A sheet of paper is not
a surface, but a body occupying Space. However thin, it
yet has some thickness. But in thinking about it we may
leave its thickness out of our thoughts, disregard its thick-
ness altogether ; so it becomes for our thought, though not
for our senses or imagination, a surface. The like may be
8
GEOMETRY.
said of the film of the soap-bubble. Again, consider the
pointer. It is a body or solid, not only long, but wide and
thick ; it occupies Space. It is neither line nor surface.
But we may, and do often, disregard wholly two of its
dimensions, and attend solely to the fact that it is long.
Thus it becomes for our thought a line, though not for our
senses or imagination. So the mark made with chalk or
ink or pencil is a body, triply extended ; but we disregard
all but its length, and it becomes for our reason a line.
Lastly, we make a dot with pen or chalk or pencil ; it is a
body, tri-dimensional, occupying Space. But we may dis-
regard all its dimensions, and attend solely to the fact that
it has position, that it is here, and not there. So it becomes
in our thought a point. By such abstraction the earth, the
sun, the stars, the planets, may all be treated as points.
Fig. I.
7. Inasmuch as Space is continuous, there may also be
continuous surfaces and lines ; and the only surfaces and
lines treated in this book are continuous, without holes, gaps,
rents, breaks, or interruptions of any kind in their extent. ,
It is important to note that in passing from any position A
INTRODUCTION. 9
to another B on a continuous line, a moving point P must
pass through a complete series of intermediate positions ;
/>. there is no position on the line between A and B that
the point P would not assume in going from A to B.
(Fig. I.)
*8. Starting from the notion of Space, we have attained
the notions of surface, line, and point, in two ways : by treat-
ing them as borders, and by the process of abstraction. But
we may reverse this order and attain the notions of line,
surface, and solid or space from the notion of point, with
the help of the notion of motion, thus : Let a point be
defined as \izs\x\g position without parts or inagnitude of any
kind. Let it move continuously through Space from the
position A to the position B. To know where it is at any
stage of its motion along any definite path, it is necessary
Fig. 2.
and sufficient to know one thing ; namely, how far it is
from A. Hence its path is a 6? w^- dimensional extent, or
what we call a line.
10 GEOMETRY.
Now let a line move in any definite way from any position
Q to any other position R. To know the position of any
point of its path, it is necessary and sufficient to know two
things ; namely, the position of the point on the moving line
and the position of the moving line itself: hence the path
of the line is a /z^/^-dimensional extent, which we have
already named a surface. (Fig. 2.)
Now let a surface move in any definite way from any
position U to any other position V. To know the position
of any point on its path it is necessary and sufficient to
know three things about it ; namely, its position on the
moving surface (which, we know, counts as two things) and
the position of the moving surface itself. Hence the path
of the surface is a //jr^^f-dimensional extent, which we have
already named a solid or a part of space.
Now, if we let a solid move, what will its path be?
Naturally we should expect it to be a /^z^r- dimensional
extent, but no such extent is yielded in our experience by
any motion of a solid — the path of a solid is nothing but
a solid. The explanation of the apparent inconsistency is
very simple, to-wit : A piece of a line traces out a surface
only when it moves out from the line itself, — if one part
were to slip round on another part of the same line, it would
trace out no surface at all as its path ; likewise, a piece of
a surface traces out a solid as its path only by moving out
from the surface itself, — if one part were to slip round on
another part of the surface, it would trace out no solid at
all as its path. So, if a piece of our space could move out
from space itself, it would trace out a four-fold extent as its
path ; in fact, however, no part of space can move out from
space ; on the contrary, it can only slip along in space, from
one part of space to another, and hence does not trace out
any four-fold extended path.
INTROD UCTION. 1 1
9. Space, we have seen, is homceoidal, everywhere alike.
We naturally inquire : Is there any homueoidal surface ? In
general, surfaces are certainly not homceoidal. Consider an
egg-shell, and by abstraction treat it as a surface. It is not
alike throughout ; the ends are not like each other, and
neither is like the middle region. Suppose a piece cut out
anywhere ; if slipped about over the rest of the shell, this
piece will not fit. But now consider a smooth round ball
covered with a thin rigid film, and treat this film as a sur-
face, by disregarding its thickness. Suppose a piece of the
film cut out and slipped round over the rest of the film : the
piece will fit everywhere perfectly, the surface is homceoidal ;
it is called a sphere-surface.
N.B. The precise definition of this surface is that all
its points are equidistant from a point within, called the
centre. Suppose a rigid bar of any shape, pointed at both
ends, and movable about one end fixed at a point ; then
the other end will move always on a sphere-surface, which
is the whole region where the moving end may be. Since
Space is homceoidal around the fixed point, the surface
everywhere equidistant from the point is also homceoidal.
Now turn over the piece cut out of this spherical film
and slip it about the film : it no longer fits anywhere at all
— the surface is homceoidal, but not reversible.
10. But now consider a fine mirror covered with a deli-
cate film, which by abstraction we treat as a surface. Sup-
pose a piece cut out of the film and slipped about over it :
the piece fits everywhere ; turn it over, re-apply it, and slip
it alK)Ut : it still fits everywhere — the surface is both homa-
oidalzxA reversible ; it is called a plane-surface.
* N.B. A precise definition of this surface is the following :
Take two points A and B and suppose two equal spherical
12 GEOMETRY.
bubbles formed about A and B as centres. Let them ex-
pand, always equal to each other, until they meet, and still
, ^ keep on expanding. The line where
^ S the equal (Fig. 3) spherical bubbles,
^^' ^" regarded as surfaces, meet, has all
its points just as far from A as from B. As the bubbles
still expand, this line, with all its points equidistant from A
and B, itself expands and traces out a plane as its path
through Space.
Hence we may define \he plane as the region (or surface)
where a point may be that is equidistant from two fixed
points. Instead of region it is common to say locus, i.e.
place. Briefly, then, a plane is the locus of a point equidis-
tant from two fixed points. It is evident that the plane, as
thus defined, is reversible ; for since the bubbles about A
and B are all the time precisely equal, to exchange A and
B^ or to exchange the sides of the plane, will make no dif-
ference whatever. Thus the plane cuts the Space evenly
half in two ; and since Space itself is homoeoidal, so also is
this section or surface that halves it exactly. The superiority
of this definition consists in its not only telling what surface
the plane is, but also making clear that there actually is such
a surface.
1 1 . The mirror is the nearest approach that we can make
to a perfect plane surface ; the blackboard is not plane, it
is rough and warped ; but we shall disregard all its uneven-
ness and treat it as a plane extended through Space without
end. Any surface may be dealt with as a plane by abstrac-
tion, being thought as ho?nceoidal and reversible.
12. On this board, regarded as a plane, we. draw a chalk-
mark, abstract from all its dimensions but its length, and
INTRODUCTION.
13
treat it as a line. This line is plainly not alike throughout ;
a piece cut out and slipped along it will not fit (Fig. 4).
FUJ. 4.
But here is a line homceoidal, alike in all its parts ; it is
drawn with a pair of compasses and is called a circle
(Fig. 5). One point of the compasses is held fast at the
centre Oy while the other traces out the circle as its path in
B
Fig. s.
• W
Fig. 6.
the plane. The circle is the locus of a point in the plane
equidistant from a fixed point. Since the plane is homce-
oidal, so too is this circle (see Art. 10) ; a piece, called an
arc, cut out and slipped round will everywhere fit on the
circle. But turn it over and slip it round, — it fits nowhere ;
the circle is not reversible. It divides the plane into two
14 GEOMETRY,
parts, not halves, that are not ahke along the dividing line.
But now suppose a perfectly flexible string fastened at S and
stretched by a weight W. Its length only being regarded,
it is a line homoeoidal, alike throughout, and also reversible ;
any part AB will not only fit perfectly anywhere on it, but
will also fit when reversed, turned end for end. Such a
line is called right, or straight, or direct, or a ray. Extended
indefinitely, it cuts the whole plane into two halves pre-
cisely alike along the ray itself.
*N.B. The common line where the two spherical bub-
bles of Art. lo meet is a circle, for it is plainly precisely
alike all around ; it is homoeoidal, being the intersection of
two homoeoidal surfaces, namely, the two equal sphere-
surfaces ; it is also in a plane, and in fact traces out the
plane by its expansion as the bubbles expand.
To get accurately the notion of the ray or straight line,
we need another point C, and a third expanding bubble
always equal to those about A and B. The circular inter-
section of the bubbles about A and B will trace out one
plane ; of those about B and C will trace out another plane ;
of those about C and A will trace a third plane. All the
points where the first two planes intersect will be equidistant
from A and B and C, and no other points will be; the
same may be said of all points where the second and third
planes meet, and of all points where the third and first meet ;
hence all three of the planes meet together, and they meet
only together. Also, the line where they meet has every
one of its points equidistant from all the thre5 points, A, B,
C ; hence it is the locus 0/ a point equidistant from three
fixed points. Moreover, it is homoeoidal and reversible,
since it is the intersection of two planes, which are homoe-
oidal and reversible ; hence it is what we call a straight
line, or right line, or ray.
INTRODUCTION. IS
13. We may now define :
A sphere- surface is the locus of a point at a fixed
distance from a fixed point. It is homceoidal, but not
reversible.
A plane is the locus of a point equidistant from two fixed
points. It is both homceoidal and reversible.
A ray is the locus of a point equidistant from three fixed
points. It is both homceoidal and reversible ; it is also the
intersection of two planes.
A circle is the locus (or path) of a point in a plane at a
fixed distance from a fixed point. It is homceoidal, but not
reversible. It is also the locus (or path) of a point in space
at a fixed distance from two points ; it is also the intersec-
tion of two equal sphere-surfaces.*
14. It is only with the foregoing figures and combinations
of them that we have to deal in this book. Circles and rays
may be drawn with exceeding accuracy, but any lines, how-
ever roughly drawn, may answer our logical purposes as
well as the most accurately drawn ; we have only, by
abstraction, to treat them as having the character of the
lines in question.
Circles and sphere-surfaces are unbounded, without be-
ginning or end, but both are finite : we shall learn how to
measure them.
* In the foregoing free use has been made of the notion of equidistance without
formal definition, because of its familiarity. We may, however, say precisely: \{ A
and B be two points, the ends of a rigid bar of any shape, and if /I be held fast, then
all the points on which B can fall are equidistant from A , and no other points arc
equidistant with them. They all lie on a closed surface, called a sphere-surface.
All points within this surface are said to be less distant, and all points without are
said to be more distant, from A than B is. Herewith, then, we tell exactly what
we mean by equidistant, less distant, and more distant, but we make no attempt to
define distatue in general, which is difficult and unnecessary to our purpose.
16 GEOMETRY,
15. Any geometric element or combination of geometric
elements, as points, lines, surfaces, is called a geometric
figure. It is a fundamental assumption, justified by experi-
ence, that space is homoeoidal, that figures or bodies are not
affected in size or shape by change of place. Two figures
that may be fitted exactly on each other, or may be thought
so fitted, are called congruent. Any two points, lines, or
parts of the two figures, that fall upon each other in this
superposition are said to correspond. It is manifest that all
planes are congruent and all rays are congruent. Rays and
planes are unbounded, but whether or not they are finite is
a question that we are unable to answer.
16. Any part of a circle or ray, as AB, is bounded by
two end-points, A and B, and is finite ; the one is named
an arc (Fig. 5), the other a tract, sect, or line-segment.
Each is denoted by the two letters denoting the ends, as
the tract AB, the arc AB. Sometimes it is important to
distinguish these end-points as beginning and end proper ;
we do this by writing the letter at the beginning first.
A'
B'
A
B
C
D
D'
A'
B'
I-T-
E
Fig. 7.
17. Two tracts, AB and AB\ are called equal when the
end-points of the one may be (Fig. 7) simultaneously fitted
on the end-points of the other.
If we have a number of tracts, AB, CD, EF, etc., and
we lay off successively on a ray tracts A^B\ C D\ E^P, etc.,
INTKODUCriON. 17
respectively equal to AB, CDy EF, etc., the end of the first
being the beginning of the second, and so on, while no part
of one falls on any part of another, we are said to add or
sum the tracts ABj etc. Each is called an addend or sum-
mand, and the whole tract from first beginning to last end
is called the sum.
Equality is denoted by the bars ( = ) between the equals,
as AB = CD.
i8. If, when the beginning A is placed on the beginning
C, the end B does not fall on the end Z>, the tracts are
unequal, and we write AB ^ CD. If B falls between C
and A then AB is called less than CD, AB<CD\ but if
D falls between A and B, then AB is called greater than
CD, AB > CD. In either case, the tract BD or DB,
between the two ends of the tracts, whose beginnings coin-
cide, is called the difference of the two tracts, and we are
said to subtract the one from the other. Ordinarily we
mention the greater tract first in speaking of difference.
19. The symbols of addition and subtraction are + and
— (plus and minus), thus :
AB-^- CD^AD and AB - CD=BD.
It is important to note here the order of the letters. In
summing a number of tracts, as AB, CD, EF, etc., to A'Z,
Fig. 8.
we have AB + CD -\- EF-- + KL = AL (Fig. 8). The
order of the summands is indifferent, and this important
fact is called the Commutative Law of Addition. Thus
AB+ CD+EF=AB^-EF-\- CD=EF-\-AB-\- CD, etc.
18 GEOMETRY.
' 20. When beginning and end of a tract or of any mag-
nitude are exchanged, the tract or magnitude is said to be
reversed, and the reverse is denoted by the sign — . Thus
the reverse of AB is BA, or AB = — BA. If we add a
magnitude and its reverse, the sum is o, or
AB -h{-AB) = AB-\-BA = o.
The same result o is obtained by subtracting, from a magni-
tude, itself or an equal magnitude ; and, in general, it is
plain that to subtract CB> yields the same result as to add
(Fig. 9) the reverse B>C. The reverse of a magnitude is
A B
C D C
H 1 J^
A B
Fig. 9.
often called its negative, the magnitude itself being called
its positive.
Similar rules hold for adding and subtracting arcs of a
circle or of equal circles.
ANGLES.
21. The indefinite extent of a ray on one side of a point
O, as OA, is called a half-ray : it has a beginning O, but no
end. Two half-rays, OA and OA\ which together make up
a whole ray, are called opposite or counter (Fig. 10).
Now let two half-rays, OA and OB, have the same be-
ginning O ; the opening or spread between them is a mag-
nitude : it may be greater or less. Suppose OA and OB to
be two very fine needles pivoted at O ; then OB may fall
exactly on OA, or it may be turned round from OA ; and
JXTKODUCriON. 19
the amount of turning from OA to OB, or the spread
between the half- rays, is called the angle between them.
We may denote it by a Greek letter, as «, ^vritten in it ; or
by a large Roman letter, as O, at its vertex (where the half-
rays meet) ; or by three such letters, as AOB, the middle
Fig. io.
one being at the vertex, the other two anywhere on the half-
rays. The symbol for angle is ^.
22. The angle is perfectly definite in size, it has two
ends or boundaries ; namely, the two half-rays, sometimes
called arms. When we would distinguish these arms as
beginning and end, we mention the letter on the beginning-
arm first, and the letter on the end-arm last; thus, A OB;
here OA is the beginning and OB the end of the angle.
Exchanging beginning and end reverses the angle ; thus,
BOA = -AOB.
23. Two angles whose ends or arms may be made to fit
on each other simultaneously are named equal ; they are also
congruent. Two angles whose arms will not fit on each other
simultaneously are unequal; and that is the less angle whose
end-arm falls within the other angle when their beginnings
20
GEOMETRY.
coincide; the other is the greater; thus, AOB> AOC
(Fig. II).
24. We sum angles precisely as we sum tracts ; we lay
off a, 13, etc., around O, making the end of each the begin-
ning of the next : the angle from first beginning to last end
Fig. II.
is the stmt. So, too, in order to subtract /? from «, lay off
/8 from the beginning towards the end of a ; the angle from
the end of (3 to the end of a is the difference, a — fi. Or
we may add to a the reverse (or negative) of ^ : the sum
will be « + (-/5) or a-jS (Fig. 12).^
^ It is important to note the close correspondence of tract and angle: the former is
related to points as the latter is to rays (or half-rays). The tract is the simplest
magnitude that lies between points, that distinguishes them and keeps them apart;
likewise the angle is the simplest magnitude that lies between rays (in a plane), that
distinguishes them and keeps them apart. So, too, we define equality and inequality
among tracts and among angles, quite similarly, and without being compelled before-
hand to form the notion of the size either of a tract or of an angle. We may now
define the distance between two points to be the tract between them, and the dts-
tatice between two (half-) rays to be the angle between them, leaving for future
decision which tract and which angle if there should prove to be several.
INTRODUCTION.
'D
21
Fig. 12.
AXIOMS.
25. At this stage we must recognize and use certain dic-
tates or irresoluble facts of experience, called axioms.
{'kiniiiuimQ^ns somethifig worthy, like the Latin dignitas ; in
fact, older writers use dignity in the sense of axiom. But
Euclid's phrase is kqlvqx twouu = common notions.^ Some
have no special reference to Geometry, but pervade all of
our thinking about magnitudes ; such are
(i) Things equal to the same thing are equal to each
other.
(2) If equals be added to, subtracted from, multiplied by,
or divided by, ccjuals, the results will be equal.
22 GEOMETRY.
(3) If equals be added to or subtracted from unequals,
the latter will remain unequal as before.
(4) The whole equals, or is the sum of, all its distinct
parts, and is greater than any of its parts.
(5) If a necessary consequence of any supposition is
false, the supposition itself is false.
Others concern Geometry especially, as :
(6) All planes are congruent.
(7) Two rays can meet in only one point.
The extremely important axiom (7) may be stated in
other equivalent ways, thus : Two rays cannot meet in two
or more points ; or, Two rays cannot have two or more
points in common ; or. Only one ray can go through two
fixed points ; or, A ray is fixed by two points.
26. A statement or declaration in words is called -a, p7'op-
osition. The propositions with which we have to deal state
geometric facts and are also called Theorems (Oeiopyjfia, from
OewpeLv, to look at, means the product of metital contempla-
tioTi). Propositions are often incorrect; theorems, never.
Subordinate facts, special cases of general facts, and facts
immediately evidenced from some preceding facts, are
called Corollaries or Porisms {■Tropi(Tixa = dedi/ctio?i).
We may now proceed to investigate lines and angles, and
find out what we can about them. The first and simplest
things we can learn concern
Tn. I.] CONGRUENCE, 23
CONGRUENCE.
27. Theorem I. — All rays are congruent.
Proof. Let L and Z' be any two rays (Fig. 13). On
L take any two points, A and B ; on L' take any two
points, A' and B', so that the tract AB shall equal the
tract A'B'. Think of Z and Z' as extremely fine rigid
spider-threads, and in thought place the ends of the tract
AB on the ends of the tract A'B', A on A\ and B on B'.
B
A B
Fig. 13.
Then A and A' become one and the same point, and B and
B' become one and the same point ; through these two
points only one ray can pass (by Axiom 7) : hence Z and
Z', which go through these two points, now become one
and the same ray ; that is, they fit precisely, they are con-
gruent. Quod erat demonstrandum = which was to be
/>rotred = oTTtp iStL Sci^ai, — the solemn Greek formula;
whereas the Hindu, appealing directly to intuition, merely
said Pafya — Behold !
28. In the foregoing proof we assumed that on any ray
we could lay off a tract equal to a given tract, or that on
any ray we could find two points, A and B, as far apart as
two other points. A' and B'. This assumption that something
can be done, is called a Postulate (alr-qfia), I.e. a demand,
which must be granted before we can proceed further.
Actually to carry out the construction, we need a pair of
compasses.
24 GEOMETRY. [Th. 11.
29. Theorem II. — If two points of a ray lie in a certain
plane, all points of the ray lie in that plane.
Proof. Regard the surface of paper or of the blackboard
as a plane, and suppose it covered with a fine rigid film,
itself a plane. Let L be any ray having two points, A and
B, in this plane. Through these two points suppose a
second plane drawn or passed; by definition (Art. 13) it
will intersect our first plane, or film, along a ray /; this ray
/ goes through the two points, A and B, and lies wholly
(with all its points) in the first plane ; also the ray L goes
through A and B, and only one ray can go through the same
two points, A and B, by Axiom 7 ; hence L and / are
the same ray ; but / has all its points in the first plane ;
hence L has all its points in the first plane, q. e. d.
Query : What postulate is assumed in this proof ?
Corollary. If a ray turn about a fixed point P, and glide
along a fixed ray Z, it will trace out a plane (Fig. 14).
Fig. 14.
For it will always have two points — namely, the fixed
point and a point on the fixed ray — in the plane drawn
through the fixed point and the fixed ray.
Th. II.] CONGRUENCE. 25
Query: What postulate is here implied? — Henceforth it
is understood that all our points, lines, etc., are complanar,
/>. lie in one and the same plane.
30. In the foregoing Theorem and Corollary we observe
clauses introduced by the word if. Such a clause is called
an Hypothesis, i.e. a supposition. The result reached by
reasoning from the hypothesis and stated immediately after
the hypothesis, is called the Conclusion.
31. All logical processes consist in one or both of two
things : the formation of concepts, as of lines, surfaces,
angles, etc., and the combination of these concepts into
propositions. Geometric concepts are remarkable for their
perfect clearness and precision — we know exactly what we
mean by them ; this cannot be said of many other concepts,
about which diverse opinions prevail, as in Political Economy.
Hence it is that Geometry offers an unequalled gymnasium
for the reason or logical faculty. We shall now generate
some new concepts. Let the student note their definiteness
as well as the mode of their formation.
32. Let OA and OB be any two co-initial half- rays,
forming the angle A OB. Think of OA as held fast and of
OB as turning about the pivot (9, starting from the position
OA. As it turns (counter-clockwise), the (Fig. 15) angle
Fig. 15.
AOB increases. Finally, let it return to its original posi-
tion, OA ; then the whole amount of turning from the upper
26 GEOMETRY. [Th. III.
side of OA back to the under side of OA, or the full spread
around the point O, is called a ///// angle (or round angle,
or c'lrcum-dcci^t, or perigon) , Think of a fan opened until
the first rib falls on the last. — Note that the upper and
under sides of OA are exactly the same in position, and are
distinguished only in thought. (Think of a circular piece
of paper slit straight through from the edge to the centre.)
The like may be said of the two sides of any line or surface.
We can now prove
33. Theorem III. — All round angles are congruent.
Proof. Let AOB and AO'B' (Fig. 16) be any two
round angles. Slip the half-ray OA down, and turn it till
OA falls on O'A^ \ they will fit perfectly (why?); the
Fig. 16.
whole round angle about O will fit perfectly on the whole
round angle about O' (why?); hence the two full angles
are congruent, q. e. d.
N.B. In this slipping of figures about in the plane, it is
well to imagine the plane to consist of two very thin, per-
fectly rigid, smooth and transparent films ; also, to imagine
one figure drawn in the lower film and one in the upper ;
and to imagine the upper slipped about at will over the lower.
Query : On what cardinal property of the plane do these
considerations hinge ?
Th. IV.]
CONGRUENCE.
27
34. From O draw any half-ray OA ; then any second
half- ray from O, as OB, will (Fig. 17) cut the round angle
AOA into two angles, A OB and BOA. The end OB of
the first falls on the beginning, OB, of the second ; while
A
B
Fig. 17.
the end, OA, of the second falls on the beginning, OA, of
the first. Hence the round angle AOA is their sum, by
Art. 24.
If we draw any number of half- rays, OB, OC, etc., -"OL,
the round angle will still be the sum of the consecutive
angles AOB, BOC, etc., --^LO A ; hence we discover and
enounce this
Theorem IV. — T/ie sum of the consecutive angles about a
point in a plane is a round angle.
N.B. We cannot apply Axiom i immediately, because
we do not know, except by Art. 24, what is meant by a sum
of angles.
35. In the foregoing article we have exemplified the
erotetic, questioning, investigative method, in which the result
28 GEOMETRY. [Th. IV.
is not announced until it is actually discovered and estab-
lished. In Theorems I., II., III., on the other hand, the
dogmatic procedure was illustrated, the fact or proposition
being announced beforehand, while the demonstration fol-
lowed after. Each method has its merits, and we shall
employ both.
36. As OB turns round from the upper to the under
side of OA, the angle AOB begins by being less than BOA
and ends by (Fig. 19) being greater than BOA. The plane
Fig. 19.
is continuous, the turning is continuous, the change m size
is continuous ; hence, in passing from the stage of being
less to the stage of being greater, the angle has passed
through the intermediate stage of being equal; let OA^ be
the position of the rotating half-ray at this stage of equality,
then AOA^ = A'OA. Two equal parts making up a whole
are called halves; hence AOA' and A'OA are halves of
the full angle AOA ; they are named straight (or flat)
angles.
37. Now, — Halves of equals are equal ;
All straight angles are halves of equals
(namely, equal round angles) ;
Th. VII.J CONGRUENCE. 29
Hence
Theorem V. — All straight angles are equal.
This argument here given /*// extenso is a specimen of a
syllogism ((rvAAoyto-/tAo« = computation = thinking together).
The first two propositions are called premisses, the third
and last, in which the other two are thought together, is
called conclusion. All reasoning may be syllogized, but
this is rarely done, as being too formal and tedious.
38. Theorem VI. — Two counter half- rays bound a
straight angle.
O
Fig. 20.
For, let OA and OA' be two such counter half- rays (Fig.
20) forming the whole ray AA\ Turn the upper half of
the plane film round O as pivot until the upper OA' falls on
the lower OA-, then, since the ray is reversible, the ray
A A' will fit' exactly on the ray A^A; i.e. the two angles
AOA' and A' OA are congruent and equal; and the two
compose the round angle A OA ; hence each is half of
A OA ; i.e. each is a straight angle, q. e. d.
39. Theorem VII. — Conversely, The half-rays bounding
a straight angle are counter.
P
Fic;. 21.
Let OA and OA' bound a straight angle (Fig. 21) AOA' ;
also let TB and J*B' be two counter half-rays ; then they
30 GEOMETRY. [Th. VII.
bound a straight angle BPB\ by Theorem VI. Since all
straight angles are congruent, we may fit these two on each
other ; i.e. we may fit OA and OA^ on PB and PB^ ; but
B B^ is a ray ; so then is AA^ ; i.e. OA and OA^ are
counter, q. e. d.
40. We may define a straight angle as an angle bounded by
counter half-rays. Then we may prove Theorem V. thus :
The ends of all straight angles are pairs of counter half-
rays (or form whole rays) ;
But all such pairs (or whole rays) are congruent (by
Theorem I.) ;
Therefore, all ends of straight angles are simultaneously
congruent.
But when the ends of angles are (simultaneously) congru-
ent, so are the angles themselves.
Hence all straight angles are congruent, q. e. d.
Here the first conclusion, introduced by " therefore," is
deduced from two premisses ; but the second, introduced by
"hence," is apparently deduced from only one. Only
apparently, however ; for one premiss was understood but
not expressed ; namely, all straight angles are angles whose
ends are congruent. Without some such implied additional
premiss, it would be impossible to draw the conclusion.
Such a maimed syllogism, with only one expressed premiss,
is called an enthymeme. The great body of our reasoning
is enthymematic. We shall frequently call for the suppressed
premiss or reason by a parenthetic question (Why?).
41. Now draw two rays, LV and MM\ meeting at O.
Each divides the round angle about O into two equal
straight angles, and together they (Fig. 22) form four angles
«, ^, «', y8'. Two angles, as « and /8, that have a common
arm, are called adjacent. Accordingly we see at once :
Th. IX.] CONGRUENCE. 31
Theorem VIII. — Where tiuo rays iniersect^ the sum of two
difjaccnt angles is a straight angle.
M
Fig. 22.
*
Two angles whose sum is a straight angle are called
supplemental ; two angles whose sum is a round angle we
may call explemental. Two angles as « and «', the arms of
the one being counter to the arms of the other, are called
opposite^ or vertical^ or counter.
Theorem IX. — When two rays meety the opposite angles
formed are equal.
For a-{-P = S (a straight angle) (why?) ; and «'-f ^ = 5
(why?).
Hence u + p = u' -{- /3 (why?) ; therefore a = a'. Simi-
larly let the student show that ft = )S'. q. e. d.
An important special case is when the adjacentSy a and y8,
are equal. Each then is half of a straight angle, and there-
fore one fourth of a round angle ; and each is called a right
angle. Now let the student show that if « = ^, then «' = ^
and a — /?', or
Corollary, When two intersecting rays make two equal
adjacent angles y they make all four of the angles equal (Fig.
23).
Def Rays that make right angles with one another are
called normal (or perpendicular) to each other. N.B.
The normal relation is mutual. How?
32
GEOMETRY.
[Th. IX.
Def. Two angles whose sum is a right angle are called
complemental.
Fig. 23.
42. Are we sure that through any point on a ray we can
draw a normal to the ray? Let O be any point on the
ray LV (Fig 24). Let any half-ray, pivoted at O, start
R
f
•^ o^
Fig. 24.
from the position 01. and turn counter-clockwise into the
position 0L\ At first the angle on the right is less than the
angle on the left, at last it is greate?' ; the plane, the turning,
and the angle are all continuous ; hence in passing from the
stage of being less to the stage of being greater, it passes
Til. X.] CONGRUENCE. 33
through the stage of equality. Let OR be its position in
this stage ; then ^LOR^^ ROV ; i.e. OR is normal to
LV. Moreover, in no other position, as OSy is the ray
normal to LV \ for LOS is not = LOR unless OS falls on
ORy but is less than L OR when OS falls within the angle
LOR, while SOV is greater than LOR ; hence LOS and
SOV are not equal ; />. 6?^ is not normal to LL when (7.S
falls not on OR. Similarly, when Z (96" is greater than LOR.
Hence
Theorem X. — Through a point on a ray one^ and only one^
ray can be drawn normal to the ray,
43. Def. A ray through the vertex of an angle, and
forming equal angles with the arms of the angle, is called
the inner Bisector or mid-ray of the angle. The inner
bisector of an adjacent supplemental angle is called the outer
bisector of the angle itself. Thus OI bisects innerly and
OE bisects outerly the angle AOB (Fig. 25).
Exercise. Prove that there is one and only one such inner
mid-ray.
34
GEOMETRY.
[Th. XI.
44. Theorem XI. — The inner Bisector of an angle
bisects also its cxp lenient innerly.
Proof. Let 6>/ bisect ^ AOB innerly ; then '^AOI=
^ lOB; call each a; then a^-BOI'=:a + AOr (why?) ;
take away «; then BOV^AOV (why?) ; i.e. the ray //'
bisects innerly the angle BOA, the explement of A OB.
Show that the angles marked «' are equal.
45. Theorem XII. — The inner and outer Bisectors of
an angle are tiormal to each other.
Proof. Let 01 and OE bisect (Fig. 25) innerly and
outerly the angle AOB. Then, by definition, the angles
marked a are equal, and the angles marked /? are equal ;
also the sum of + a + « + /8 + y8 = *S ; hence a + y8 = J 6* ;
or, lOE = a right angle, q. e. d.
TRIANGLES.
46. Thus far we have treated only of rays intersecting in
a single point. But, in general, three rays Z, J/, N (Fig.
Fig. 26.
26) will meet in three points, since each pair will meet in
one point, and there are three pairs : {MN) , (iVZ) , {LM) .
Th. XIII.]
TRIANGLES.
35
Denote these points by A^ B^ C. Then the figure formed
by these three rays is called a triangle, trigon, or three-side.
Af By C are its vertices ; <f, )8, y its inner angles ; BC, CA,
AB, its inn^r sides, or simply its sides. Its angles and sides
are called its parts. It is the simplest closed rectilinear
figure, and most important. If instead of taking three
rays we take three points A^ B, C, then we may join them
in pairs by rays; and since there are three pairs, BC, CA,
AB, then there are three rays, which we may name Z, M, N.
Thus we see that three points determine three rays, just as
three rays determine three points. This equivalent deter-
mination of the figure by the same number of points as of
rays makes the figure unique and especially important. We
denote it by the symbol A. We now ask, When are two
triangles congruent?
47. Theorem XIII. — Two A having two sides and the
included angle of the one equal respectively to two sides and
the included angle of the other are congruent.
The data are: Two A, ABC and AB'C, having the
three equalities, AB = A'B\ AC==A'C', a = u' (Fig. 27).
Fig. 27.
Proof. Fit the angle « on the angle a' ; this is possible,
because the angles are eqwal and congruent. Then A falls
36
GEOMETRY.
[Th. XIV.
on A^ ', also the point B falls on B^ (why? Because AB
= A'B^), and C falls on C (why?). Hence the three ver-
tices of the two A coincide in pairs ; therefore the three
sides of the two A coincide in pairs (why? Because through
two points, as A (A') and B {B'), only one ray can pass).
Q. E. D.
Corollary i. The other parts of the two A are equal
or congruent in pairs of correspondents : /? = /?', 7 = y',
BC=:B'C'.
Corollary 2. Pairs of equal parts lie opposite to pairs of
equal parts.
48. Theorem XIV. — Two A having two angles and the
included side of the one equal respectively to two angles and
the included side of the other are congruent (Fig. 28).
Fig. 28.
Data: Two A ABC, A'B'C\ having «=«', ^ = ^\
AB=A'B\
Proof. Fit AB on A'B'; this is possible (M'hy?). Then
a will fit on a' (why?), and ^ on ft' (why?) ; i.e. the ray
^C will fit on A'C, and the ray BC on B'C Then the
point C will fall on C (why ? Because two rays meet in only
one point) ; i.e. the two A fit exactly, q. e. d.
Th. XVI.]
TRIANGLES.
37
49. We m:iy now use the conditions of congruence thus
far established to generate new notions that may be used in
estabHshing other Theorems.
Dff. The ray normal to a tract at its mid-point is called
the mid-normal of the tract.
Theorem XV. — Any point on the mid-normal of a tract is
equidistant from its ends (Fig. 29).
P>..C
Data : AB a tract, M its mid-point, Z the mid-normal,
PsLny point on it.
Proof. Compare the A APM and BPM. We have
AAf=BM (why?). PM=PM, ^ AMP=:^ BMP
(why?) ; hence the A are congruent (why?) ; and PA =
PB. Q. E. D.
Dff. A A with two equal sides, like APB, is called
isosceles ; the third side is called the base, and its opposite
angle the vertical angle.
50. Theorem XVI. — The angles at the base of an isosceles
A are equal; and conversely.
38 GEOMETRY. [Th. XVI.
Data: ABC an isosceles A, AB its base, AC and BC its
equal sides (Fig. 30).
Proof. Take up the A ABC, turn it over, and replace it
in the position BCA, Then the two A ACB and BCA
have the equal vertical angles, C and C, also the side AC =
BC (why?) and BC = AC (why?) ; hence they are con-
gruent (why?), and the '^A = :^B. Q. e. d.
Conversely, A A whose basal angles are equal is isosceles.
Let the student conduct a proof quite similar to the fore-
going.
Def. The ray through a vertex and the mid-point of the
opposite side is called the medial of that side.
Corollary i . In an isosceles A the medial of the base is
normal to it, and is the mid-ray of the vertical angle.
Corollary 2. When the medial of a side of a A is normal
to the side, the A is isosceles. Prove it.
Corollary 3. When the medial of a side bisects the
opposite angle, the A is isosceles. Can you prove it ?
Th. XVI.] LOGICAL DIGRESSION. 39
LOGICAL DIGRESSION.
51. When the subject and predicate of a proposition are
merely exchanged, the proposition is said to be converted^
and the new proposition is called the converse. Thus X is
K; coiiicrsch\ Kis X. In general, converses of true prop-
ositions are not true, but false. Thus, The horse is an
animal is always correct, but The animal is a horse is
generally false. A proposition remains true after simple
conversion only when subject and predicate are properly
quantified, thus : All horses are some animals ; conversely,
Some animals are all horses. Both propositions are correct
and mean the same thing. But they are awkward in ex-
pression, and such forms are rarely or never used. When
the quantifying word is all or its equivalent, the term is
said to be taken universally ; when it is some or its equiva-
lent, the term is said to be taken particularly. Thus in the
foregoing example horse is taken universally, but animal
particularly. The only useful conversions are of proposi-
tions in which both subject and predicate are ufiiversal. In
the great body of propositions only the subject is quantified
universally, the quantifier is omitted from the predicate, but
a particular one is understood. To show that a universal
quantifier is admissible requires in general a distinct proof.
52. In order to convert an hypothetic proposition, we
exchange hypothesis and conclusion. Thus, \{X is K, U is F;
the converse is, if 6^" is V^ X is K All such hypothetic
propositions may be stated categorically ^ thus : All cases of
A' being Fare cases of 6^" being V \ conversely, All cases of
U being V are cases of X being Y. This converse is plainly
false except when the quantifier all is admissible in the first
predicate.
40 GEOMETRY. [Th. XVII.
53. But while the converse of a true hypothetic propo-
sition is generally false, the contrapositive is always true.
This latter is formed by exchanging hypothesis and conclu-
sion and denying both. Thus : If Jf is K, then 17 is F;
contrapositive y If U is not V, then X is not K Or, if a
point is on the mid-normal of a tract, then it is equidistant
from the ends of the tract ; contrapositive. If a point is not
equidistant from the ends of a tract, then it is not on the
mid-normal of the tract.
54. Theorem XVII. — An outer angle of a A is greater
than either inner non-adjacent angle.
Data: Let ABC be any A, a' an outer angle, ^' a non-
adjacent inner one (Fig. 31).
Fig. 31.
Proof. Draw the medial CM and lay off MD — MC ;
also draw AD. Then in the A AMD and BMC we have
AM=BM {vfhy}), MD = MC (why?), and '^AMD =
'^.BMC (why?) ; hence the A are congruent (why?), and
^MBC=^MAD (why?). But -^^ MAD is only part of
the ^ «' ; hence «'> ^ MAD (why?) ; i.e. «'>/?'. q. e. d.
Similarly, prove that a! > y.
Th. XVIII] triangles, 41
55. Theorem XVIII. — If tivo sides 0/ a A are unequal,
then the opposite angles are unequal in the same sense {i.e.
the greater angle opposite the greater side) (Fig. 32).
Fig. 32.
Data: ABC a A, AC > AB, AR the mid-ray of the
angle at A, AB' laid off = AB.
Proof. ABR and AB^R are congruent (why?) ; hence
^ ABR = ^ AB'R (why ?) ; but ^ AB'R > C (why ?) ;
i.e. 7(.ABC>^ACB. q. e. d.
Conversely, If two angles 0/ a A are unequal, the opposite
sides are unequal in the same sense.
Proof. The opposite sides are not equal ; for when the
sides are equal, the opposite angles are equal (Theorem
XVI.), and contrapositively, when the angles are unequal,
the opposite sides are unequal. Then, by the preceding
Theorem, the greater angle lies opposite the greater side.
56. Join BB' ; then AR is the mid-normal of BB' (why ?),
and hence angle CBB'= angle BB^R (why ?) . Hence angle
BB'C>B'BC (why?) ; hence BC>B'C (why?). But
B'C=AC-AB; hence BC> AC - AB : i.e.
42
GEOMETRY.
[Th. XIX.
Theorem XIX. — Any side of a ^ is greater than the
difference of the other two.
Add AB to both sides of this inequahty and there results
AB + BOAC] i.e.
Theorem XX. — Any side of a A is less than the sum of
the other two.
This fundamental Theorem is here proved on the sup-
position that AB <AC; if AB were =ACot> AC, it would
need no formal proof.
57. Theorem XXI. — A point not on the mid-normal of
a tract is not equidistant from the ends of the tract.
Data : AB the tract, MN the mid-normal, Q any point
not oviMN {Y\g. 33).
Fig. 33.
Proof. Draw QA and QB \ one of them, as QA, must
cut MN^X. some point, as P. Then QB< QP-\-PB (why ?) ,
and PB^PA (why?); hence QB<QP-\-PA', i.e.
QB < QA. Q. E. D.
Of what Theorem is this the converse ?
If now we seek for a point equidistant from A and B, we
can find it on the mid-normal of AB and only there ; hence
the locus of a point equidistant from the ends of a tract is
the mid- normal of the tract.
Th. XXIII.]
TRIANGLES,
43
58. Theorem XXII. — Two A with the three sides of the one
equal respectively to the three sides of the other are congruent.
Data: ABC and A'B'C the two A, and AB = AB\
BC==B'C, C^= C'y^' (Fig. 34).
Proof. Turn the A A'B'C over and fit A'B^ on AB so
that C shall fall (say) at D. Draw CD. Then A and B
are on the mid-normal of CD (why?) ; hence the ray AB
is the mid-normal of CD (why?) ; hence the angle CAB =
angle DAB, and angle CBA = angle DBA (why?). Hence
the A are congruent (why?), q. e. d.
N.B. As to when the A must be turned round and when
turned over, see Art. 94.
59. Theorem XXIII. — A. From any point outside of a
ray one normal may be drawn to the ray.
Data: /*the point, LV the ray (Fig. 35).
Proof. From P draw a ray far to the left, as PA^ making
the angle PAL > angle PAL\ Now let the ray turn about
/* as a pivot into some position far to the right, making
angle PA^L < PA^V. The plane, the angle, the motion, all
being continuous, in passing from the stage of being unequal
44
GEOMETRY.
[Tii. XXIII.
in one sense to the stage of being unequal in the opposite
sense, the angles made by the moving ray with the fixed ray
must have passed through the stage of equality. Let PN be
Fig. 35.
the ray in this position so that angle PNL — angle PNL! ;
then each is a right angle by Definition, and /Wis normal to
LL. Q. E. D.
B. There is only one ray through a fixed point and normal
to a fixed ray.
Proof. Any other ray than PN, as PD, is not normal to
LV ; for the outer angle PDL is > the right angle PND
(why?). Q. E. D.
C. The normal tract PN is shorter than any other tract
from P to the ray LV .
Proof. For the right angle at Wis > angle PDN (why?) ;
hence PN < PD (why ?) . q. e:. d.
D. E. Equal tracts from point to ray meet the ray at
equal distances from the foot of the normal; and conversely.
Proof. For, if DPD^ be isosceles, then the normal PN is
the medial of the base (why ?) .
F. Two, and only t7i>o, tracts of given length can be drawn
from a point to a ray.
Th. XXIV.]
TRIANGLES.
45
Proof. For two, and only two, points are on the ray at a
given distance from the foot of the normal.
G. Of tracts drawn to points unequally distant from the
foot of the normaly the one drawn to the remotest is the
longest.
Proof. In the A PDA, angle PDA > PAD (why?) ;
hence PA > PD (why ?) . q. e. d.
Similarly, PA' > PD.
H. Equal tracts from the point to the ray make equal
angles with the normal from the point to the ray and also
equal angles with the ray itself; and conversely.
I. Of unequal tracts from the point to the ray, the longest
makes the greatest angle with the normal and the least with
the ray.
I^t the student conduct the proof of H and /.
60. Theorem XXIV. — Two A having two angles and an
opposite side of one equal respectively to two angles and an
opposite side of the other are congruent.
Data : ABC and A^B'C two A having AB — A'B\ angle
u = angle a', angle y = angle y' (Fig. 36).
Fig. 36.
46
GEOMETRY.
[Th. XXIV.
Proof. Fit «' on a ; then B' falls on B (why?), and AC
falls along A C. Draw the normal BN. Then ^ C and B C
make the same angle, y = y', with the ray AN; hence they
are = and meet the ray in the same point (why ?) ; i.e. C
falls on C ; i.e. the A are congruent. Q. e. d.
6i. We now come to the so-called ambiguous case, of
two A with two sides and an opposite angle in one equal to
the two sides and the corresponding opposite angle in the
Fig. 37.
other. Let ABC and A B' C (Fig. 37) be the two A, with
AB = A'B', BC=B'C, and angle a = angle «'. Fit a' on
Th. XXV.]
TRIANGLES.
47
« ; then A^B^ falls on ABy B^ on B ; but since from a point
B {B') we may draw two equal tracts to the ray AL, the
side B'C may be either of these equals and may or may not
fall on BC, In general, then, we cannot prove congnience
in this case. But if ^C be > AB^ then angle « > angle y
(why?), and there is only one tract on the right of AB drawn
from B to the ray AC and equal io BC ; the other tract
equal to BC must be drawn outside of AB and to the left.
Hence in this case, when the angle lies opposite the greater
side, the A are congruent. Hence
Theorem XXV. — Two A having two sides and an angle
opposite the greater in one equal to two sides and an angle
opposite the greater side in the other are congruent.
Corollary. Two right A having a side and any other
part of one equal to a side and the corresponding part of
the other are congruent.
Fig. 38.
62. We have seen (Art. 47) that when two A have two
sides and included angle in one equal to two sides and
48 GEOMETRY. [Tii. XXVI.
included angle in the other, they are congruent. But what
if the included angles are not equal ? Let ABC and A'B' C
be the two A, having AB^AB\ BC= B'C, but y8 > /8'.
Slip the upper film of the plane along until A^B^ fits on AB
and let C fall on D. Draw the mid-ray BM of the angle
CBD, let it cut AC Tui M, and draw DM. Then the A
CBM SiXid DBM 3ixe congruent (why?) ; hence AM-{- MD
= AC (why?), and^C>^Z>, or^O^'C. Hence
Theorem XXVI. — Two A having two sides of one equal
to two sides of the other, but the included angles unequal^
have also the third sides unequal, the greater side lying
opposite the greater angle.
Conversely, Two A having two sides in one equal to two
sides in the other, but the third sides unequal, have the
included angles also unequal, the greater angle being opposite
the greater side.
Proof. The included angles are not equal; for if they
were equal, the A would be congruent (why?) and the
three sides would be equal. Hence the included angles are
unequal, and the relation just estabUshed holds ; namely, the
greater angle Hes opposite a greater side. q. e. d.
63. Theorem XXVII. — Every point on a mid-ray of an
angle is equidistant from its sides.
Data : O the angle, MJVV the mid-ray, P any point on it.
Proof. From P draw the normals PC and PD ; they are
(Fig. 39) the distances of P from the ends of the angle.
Then the A POC and POD are congruent (why?) ; hence
PD = PC. Q. E. D.
Th. XXVIII.]
TRIANGLES.
49
Conversely, A poitit equidistant from the ends of an angle
is on a mid-ray of the angle (Fig. 39).
FIG. 39.
Proof. If PC = PD, then the ^ POC and POD are
congruent (why ?) ; hence angle POD = angle POC. q. e. d.
Accordingly we say that the mid-rays of an angle are the
locus of a point equidistant from its ends.
*64. It is just at this stage in the development of the
doctrine of the Triangle that we are compelled to halt and
introduce a new concept before we can proceed any further.
The necessity of this step will appear from what follows
(which may, however, be omitted on first reading, at the
option of teacher or student) .
Def. Two A not congruent are called equivalent when
they may be cut up into parts that are congruent in pairs.
Theorem XXVIII. — Any A is equivalent to another A
having the sum of two of its angles equal to the smallest
angle of the given A.
Data : ABC the A, a the least angle (Fig. 40).
50 GEOMETRY. [Th. XXIX.
Proof. Through M, the mid-point of BC, draw AM SLud
make MD = MA. Then the A A CM and DBM are con-
gruent (why?), the part AMB is common to ABC and
ABD, and the sum of the angles ADB and ^^Z? = angle
BAC. Q.E.D.
Fig. 40.
Corollary. The sum of the angles in the new A is equal
to the sum of the angles in the old A.
* 65. We may now repeat this process, applying it to the
smallest angle, as A, of the A ABD. In the new A ABE
the smallest angle, as A, cannot be greater than \ of the
original angle a in ABC ; after n repetitions of this process
we obtain a A, as ALB, in which the sum of the angles A
and L cannot be > — of the original angle « in the A
^" I
ABC. By making n as large as we please, we make —
as small as we please, and so we make — of angle a
smaller than any assigned magnitude no matter how small.
Meantime the other angle B has indeed grown larger and
larger, but has remained < a straight angle. Hence the
sum of the angles in the A ALB cannot exceed a straight
angle by any amount however small ; but the sum of the
angles in ALB = sum of the angles in ABC ; hence
Theorem XXIX. — The sum of the angles in any A can-
not exceed a straight angle by any finite amount.
Th. XXX.]
TRIANGLES.
51
Corollary i. The outer angle of a A is not less than the
sum of the inner non-adjacent angles.
Corollary 2. From any point outside of a ray there may
be drawn a ray making with the given ray an angle small at
will.
Proof. From P draw any ray PA, and lay off AB = PA
(Fig. 41). Then the angle PBA is not greater than
Fio. 41.
\PAN (why?); now lay off BC=^PB (why?); then
angle PCB is not > \ angle PBA (why?) ; proceeding
this way, we obtain after n constructions an angle PLN
not > — of the angle PAN^ and by making n large enough
2"
we may make this ^ as small as we please, q. e. d.
♦66. Theorem XXX. — If the sum of the angles in any A
equals a straight angle, then it equals a straight angle in
every A (Fig. 42).
52
GEOMETRY,
[Th. XXX.
Hypothesis : ABC a A with the sum of its angles
Proof, (i) Draw any ray through C, as CD. Then if
the sum of the angles in the A ACD and BCD be ^ — ;c
and S—y, x and y being any definite magnitudes however
small, then on adding these sums we get 2S—{x-\-y)',
and on subtracting the sum, S, of the supplemental angles
at Z> we get 6* — (^-f-jv) for the sum of the angles of the
A ABC. Now if this sum be S, then x and y must each
be O ; i.e. the sum of the angles in each of the A A CD
and BCD is 5. Now draw DE and DF\ in each of the four
small A the sum of the angles is still = ^. (2) We may
now make a A as large as we please and of any shape what-
ever, but the sum of the angles will remain = S. For, take
the same A ABC, and draw CD normal to AB. Then the
sum of the (Fig. 43) angles in the A ACD is S, as has
Fig. 43.
been shown above ; also angle Z> is a right angle ; hence
the angles A and ACD are complementary. Now along
^C fit another A ACD^ congruent with ACD) then all
the angles of the quadrilateral AD CD' are right, and the
figure is called a rectangle. Now we can place horizontally
side by side as many of these rectangles, all congruent, as we
please, say p of them ; we can also place as many of them
vertically, one upon another, as we please, say q of them ;
Th. XXX.] TRIANGLES. 53
and we can then fill up the whole figure into a new rectan-
gle, as large as we please. About each inner junction-point
of the sides of the rectangles there will be four right angles
plainly. Now connect the two opposite vertices, as A and Z,
of this rectangle. So we get two congruent right A, in each
of which the sum of the angles is S. Then any A that we
cut off from this right A will, by the foregoing, have the
sum of its angles equal to 6". Since / and q are entirely in
our power, we may make in this way any desired right A
and from it cut off any desired oblique A, with the sum of
its angles = S. q. e. d.
Hence either no A has the sum of its angles = 6", or
every A has the sum of its angles = S.
67. A logical choice between these alternatives is impos-
sible, but the matter may be cleared up by the following
considerations :
Across any ray LM draw a transversal T, cutting LM zX
O, and making the angles «, )3, y, 8. Through any point,
as O', of T'draw a ray (Fig. 44) L'M' making angle «' = a.
Fig. 44.
This is evidently possible (why?). Then plainly /8' = /3,
y' = y, S' = 8, «' = u ; they are called corresponding angles ;
5+ GEOMETRY. [Th. XXX.
also a and y', /? and V are equal, — they are called alternate
angles ; also « and V, as well as ^ and y', are supplemental,
— they are called interadjacent angles.
68. Now let /*be the mid-point oi 00^ ; on it as a pivot
turn the whole right side of the plane round through a
straight angle until O falls on 0\ and 6>' falls on O. Then,
since the angles about O and O^ are equal as above stated,
the half-ray OL will ffill and fit on the half-ray 0'M\ and
the half-ray O^V on the half-ray OM. Accordingly, if the
rays LM and Z'J/' meet on one side of the transversal T,
they also meet on the other side of T.
69. Three possibilities here lie open :
(i) The rays ZJ/and Z'J/' may meet on the left and
also on the right of T, in different points.
(2) They may meet on the left and also on the right of
T, in the same point.
(3) They may not meet at all.
No logical choice among these three is possible. But
in all regions accessible to our experience the rays neither
converge nor show any tendency to converge. Hence we
assume as an
Axiom A. Two rays that make with any third ray a pair
of corresponding angles equal, or a pair of alternate angles
equal, or a pair of interadjacent angles supplemental, are
non-intersectors.
70. But another query now arises. Is it possible to draw
another ray through (9' so close to Z' that it will not meet
'OL however far both may be produced ? Here again it is
impossible to answer from pure logic. An appeal to experi-
ence is all that is left us. This latter testifies that no ray
Th. XXX.] PARALLELS, 55
can be drawn through O^ so close to 0*V as not to approach
and finally meet the ray OL. Hence we assume as another
Axiom B. Through any point in a plane only one non-
intersector can be drawn for a given straight line.
This single non-intersector is commonly called the parallel,
through the point, to the straight line.
71. It cannot be too firmly insisted, nor too distinctly
understood, that the existence of any non-intersector at all,
and the existence of only one for any given point and given
ray, are both assumptions, which cannot be proved to be
facts. The best that can be said of them, and that is quite
good enough, is that they and all their logical consequences
accord completely and perfectly with all our experience as
far as our experience has hitherto gone. Even then, if
there be any error in our assumptions, we have thus far been
utterly unable to find it out.
A geometry that should reject either or both of these
assumptions would have just as much logical right to be as
the geometry that accepts them, and such geometries lack
neither interest nor importance. They may be called Hyper-
Euclidean in contradistinction from this of ours, which from
this point on is Euclidean (so-called from the Greek master,
Euclides, who distinctly enunciated the equivalent of our
Axioms in a Definition and a Postulate).
Note. — Observe the relation of Axioms A and B : the one is the
converse of the other.
Observe also that the necessity of assuming the first lies in our igno-
rance of the indefinitely greats and the occasion of assuming the other
lies in our ignorance of the indefinitely small. See Note, Art. 301.
73. Accepting our Axioms as at least exacter than any
experiment we can makc^ we may now easily settle the ques-
56 GEOMETRY. [Th. XXXI.
tion as to the sum of the angles in a A. Let ABC be any
A ; through the vertex C draw the one parallel to the base
AB. Then a = «', ^ = /S' (why?) ; also a' + y + )8' = 6" j
hence a + y-|-/3 = ^/ i.e. (Fig. 45)
Theorem XXXI. — The sui7i of the angles in a A is a
straight angle.
Fig. 4S.
Corollary i. The outer angle E equals the sum of the
inner non-adjacent angles a and y (why?).
Corollary 2. If two angles of a A be known, the third is
also known.
Corollary 3. If two A have two angles, or the sum of two
angles of the one equal to two angles, or the sum of two
angles of the other, then the third angles are equal.
^Corollary 4. To know the three angles of a A is not to
know the A completely, for many A may have the same
three angles. Such A are similar, as we shall see, but are
not congruent; they are alike in shape, but not in size.
73. Next to normality, parallelism is the most important
relation in which rays can stand to each other, and we must
now use the new relation in the generation of new concepts.
Til. XXXIII ]
PARALLELOGRAMS.
57
Theorem XXXII. — Parallel Intercepts between parallels
arc e<]ual.
Data : L and Z', it/and M , two pairs of parallels (Fig. 46).
Fig. 46.
Proof. Draw BD. Then the A ABD and CDB are
congruent (why?), and AB = CD, BC=DA. q. e. d.
Def. The figure ABCD formed by two pairs of parallel
sides is called a parallelogram, and may be denoted by the
symbol O.
A join of opposite vertices, as BD, is called a diagonal.
74. Theorem XXXIII. — Properties of the parallelogram.
A. The opposite sides of a parallelogram are equal.
This has just been proved.
B. The opposite angles of a parallelogram are equal.
Proof. « = /3 (why?); )8 = «' (why?); hence « = «'.
Q. E. D.
Corollary, Adjacent angles of a parallelogram are sup-
plementary.
C. Each diagonal of a parallelogram cuts it into two con-
gruent A. Prove it.
58
GEOMETRY.
[Th. XXXIV.
D. The diagonals of a parallelogram bisect each other
(Fig. 47)
,-''M
FIG. 47.
Proof. The A AMB and CMD are congruent (why?) ;
hence AM^ CM, BM= DM. q. e. d.
75. We may now convert all the foregoing propositions
and obtain as many criteria of the parallelogram.
Theorem XXXIV. — A'. A 4- side with its opposite sides
equal is a parallelogram.
Data : AB ^ CD, AD= CB (Fig 48).
W^-^,
Fig. 48.
Proof. Draw BD. Then ABD and CDB are congruent
(why?); hence ^ = 8; and AD and CB are parallel;
similarly, AB and CD are parallel ; hence ABCD is a par-
allelogram. Q. E. D.
Th. XXXIV.] PARALLELOGRAMS. 59
B'. A 4-side with opposite ant^ks equal is a parailelogram.
Data: « = «', ^ + yS' = y 4- y' (Fig. 49).
'i-A
Fig. 49.
Proof. Since « = «', y3 + y = )3' + y' (why?). Hence
/? = y', ft' = y; i.e. opposite sides are parallel, the 4-side
is a parallelogram, q. e. d.
C, A 4-side that is cut by each diagonal into two congru-
ent A is a parallelogram.
For the opposite angles must be equal (why?) ; hence,
etc. Q. E. D.
D'. A 4-side whose diagonals bisect each other is a
parallelogram.
For the opposite sides are equal, being opposite equal
angles in congruent A ; hence, etc. q. e. d.
E'. A 4-side with one pair 0/ sides equal and parallel is
a parallelogram.
For the other two sides are equal and parallel (why?) ;
hence, etc. q. e. d.
76. The foregoing properties and criteria of the parallel-
ogram illustrate excellently the nature of a definition. This
60 GEOMETRY. [Th. XXXV.
latter defines or bounds off by stating something that is true
of the thing defined, but of nothing else. Accordingly, the
characteristic of every definition or definitive property is
that the proposition that states it may be converted simply.
Thus:
Every parallelogram is a 4-side with opposite angles
equal ; and conversely, every 4-side with opposite angles
equal is a parallelogram.
Not every property is definitive, and hence not every
property may be used as test or criterion.
77. Special Parallelograms.
Def. An equilateral parallelogram is called a rhombus.
Theorem XXXV. — The diagonals of a 7'hombus a7'e nor-
mal to each other.
Let the student conduct the proof suggested by the figure
(Fig- 50)-
Fig. 50.
Conversely, A parallelogram ivhose diagonals are normal
to each other is equilateral, or a rhombus. Let the student
supply the proof.
78. Def. An equiangular parallelogram is called a rect-
angle (for all the angles are right angles) .
Th. XXXVII.] PARALLELOGRAMS.
61
Theorem XXXVI. — The diagonals of a rectangle are equal
(fig- 50-
D
C
A
B
Fig. 51.
For the A ABC and BAD are congruent (why?) ; hence
AC=BD. Q.E.D.
Conversely, A parallelogram with equal diagonals is equi-
angular^ or a rectangle.
For the A ABC and BAD are again congruent, though
for another reason. What reason ?
79. Def. A parallelogram both equilateral and equi-
angular is called a square.
Theorem XXXVII. — The diagonals of a square are equal
and normal to each other.
Fio. 52.
62 GEOMETRY. [Th. XXXVIII.
For the square, being both rhombus and rectangle, has
all the definitive properties of both. Or the student may
prove the proposition directly from the figure (Fig. 52), as
well as its converse :
A parallelogram with diagonals equal and normal to each
other is a square.
80. Can we convert Theorem XXXII. and prove that
equal intercepts between parallels are parallel ? Manifestly
no (Fig. 53), for from the point C we may draw two equal
Fig. 53.
tracts to the other parallel, the one CB parallel to AD, the
other CB^ sloped at the same angle to the parallels but in
opposite ways. We may call CB^ a?iti-parallel to AD, and
the figure AB'CD an anti-parallelogram. Since from any
point C only two equal tracts, or tracts of given length, may
be drawn to the other parallel through A, we have the
Theorem XXXVIII. — Equal intercepts between parallels
are either parallel or anti-parallel.
Corollary i. Adjacent angles of an anti-parallelogram
are alternately equal or supplemental.
Corollary 2. Anti-parallels prolonged meet at the vertex
of an isosceles A.
Th. XL.]
GENERAL QUADRILATERAL.
63
THE GENERAL QUADRILATERAL OR 4-SIDE.
81. A Quadrilateral is determined by four intersecting
rays. These determine six points, the four inner vertices,
C, Z>, Ey Fj and the two outer ones, Aj B. The cross-rays,
CE^ DFy AB, are the diagonals, CE and DF itiner, AB
outer. Commonly the outer diagonal is little used, and the
inner ones are called the diagonals. When none of the
angles C, Z>, E^ F, of the 4-side is greater than a straight
angle, the 4-side is called the normal, as CDEF. It is the
only form ordinarily considered. The other two forms are
(2) the crossed^ ACBE^ and (3) the inverse^ ADBF
(Fig. 54). For all forms let the student prove
Fio. 54.
Theorem XXXIX. — The sum of the inner angles of a
^-side is a round angle.
Corollary. When two angles of a 4-side are supple-
mental, so are the other two.
82 . Theorem XL. — The angles between two rays equal
the angles between two normals to the rays.
64
GEOMETRY.
[Th. XLI.
Data : OL and OM any two rays, PA and PB any two
normals to them (Fig. 55).
/
\
/3'
P
r/^
B
A
M
Fig. 55.
Proof. The angles at A and B are right angles and
therefore supplemental (why?) ; hence a — «', and p = yS'.
Q. E. D.
N.B. The 4-side with its opposite angles supplemental is
very important and has received the name encyclic 4-side,
for reasons to be seen later on (Arts. 126-7).
THREE OR MORE PARALLELS.
83. Theorem XLI. — Three parallels that make equal
intercepts on one transversal, make equal intercepts ofi any
transversal.
Data : Z, M, N, three parallels, and AB = BCy and DEF
any transversal (Fig. 56).
Proof. Draw D'EP parallel to AB C. Then AB = BC
(why?), AB = D^E (why?), and BC=^EF (why?);
hence D^E = EF (why?), hence the A BED' and FEF
are congruent (why?) ; hence DE = EF (why?). Q. e. d.
Til. XLII.] THREE OR MORE PARALLELS. 65
84. Def. A 4-side formed by two parallels and two
transversals is called a trapezoid. Thus ACFD is a trape-
zoid. The parallel sides are called the bases (major and
min3r) ; the parallel through the mid-points of the trans-
verse sides is the mid-parallel.
Fig. 56.
Theorem XLII. — The mid-parallel of a trapezoid equals
the half -sum of its bases.
Let the student elicit the proof from the foregoing figure.
Corollary i . A parallel to a base of a A bisecting one
side bisects also the other. {Hint. Let D fall on A.)
Corollary 2. A ray bisecting two sides of a A is parallel
to the third.
For only one ray can bisect two sides (why?), and we
have just seen (Cor. i) that a ray parallel to the base does
this ; hence, q. e. d.
Corollary 3. The mid-parallel to the base of a A equals
half the base.
85. Def Three or more rays that pass through a point
are said to concur or l)e concurrent.
66 GEOMETRY. [Th. XLIII.
Theorem XLIII. — The me dials of a IS concur.
Data : ABC a A, ^^aiid BQ two medials (Fig. 57).
Proof. Draw a ray from C through (9, the intersection
of the two medials, and lay off 011= CO. Draw AH and
BH; they are parallel io BQ and AF (why?) ; hence
A OBH is a parallelogram (why ?) ; hence AR = BR
(why?). Hence COR is the third medial; />. the three
medials pass through O. q. e. d.
Corollary. Each medial cuts off a third from each of
the other two. For C0 = 20R (why?).
Def. The point of concurrence of the medials is called
the centroid of the A. It is two-thirds the length of each
medial from the corresponding vertex.
86. Theorem XLIV. — The mid-normals of the sides of
a A concur.
Data : ABC a A, Z and M mid-normals to the sides EC
and CA, meeting at S.
Th. XLV.]
CONCURRENTS,
67
Proof. ^ is equidistant from B and C (why?), and from
Cand A (why?) ; hence S is equidistant from A and B
(why?), or is on the mid-normal of AB (why?) ; hence
the mid-normals concur (Fig. 58). g. e. d.
Fig, 58.
Corollary. S is ecjuidistant from A, B, and C, and no
other point in the plane is (why?).
D^/. The point of concurrence of the mid-normals is
called the circumcentre of the A.
87. Z>e/. A tract from a vertex of a A normal to the
opposite side is called an altitude of the A. Sometimes,
when length is not considered, the whole ray is called the
altitude.
Theorem XLV. — T/ie altitudes of a ^ concur.
Proof. Using the preceding figure, draw the A A'B'C.
Its sides are parallel to the sides of ABC (why?) ; hence
its altitudes are the mid-normals L, Af, JV; and these have
just been found to concur. Also, since ABC may be any
A, A'B'C may be any A ; hence the altitudes of any A
concur, q. e. d.
68 ■ GEOMETRY. [Th. XLVL
Def. The point of concurrence of altitudes is called the
orthocentre (or alticentre) of the A.
Def. In a right A the side opposite the right angle is
called the hypotenuse (= subtense = under-stretch).
Queries : Where do circumcentre and orthocentre lie :
(i) in an acute-angled A? (2) in an obtuse-angled A ?
(3) in a right A ?
88. Theorem XL VI. — The inner mid-rays of the angles
of a /S. concur.
Data : ABC a A, AL, BM, CiVthe inner mid-rays of its
angles (Fig. 59).
Fig. 59.
Proof. Let AL and BM intersect at /. Then / is equi-
distant from AB and AC, and from AB and BC (why?) ;
hence / is equidistant from AC and BC ; hence / is on
the inner mid-ray of the angle C ; i.e. the three inner mid-
rays concur in /. q. e. d.
Def The point of concurrence of the inner mid-rays is
called the in-centre of the A.
Th. XLVII.] EXERCISES I. 69
89. Theorem XLVII. — The outer mid-rays of hvo angles
and the inner mid- ray 0/ the other angle 0/ a A concur.
Let the student conduct the proof (Fig. 60).
I
Fig. 60.
De/. The points of concurrence are called ex-centres of
the A : there are three.
EXERCISES I.
Little by little the student has been left to rely more and
more upon his own resources of knowledge and ratiocination
in the conduct of the foregoing investigations. He has now
possessed himself of a large fund of concepts, and he must
test his ability to wield, combine, and manipulate them in
forging original proofs of theorems. Let him bear always
in mind the fundamental logical principle that ez'ery example
70 GEOMETRY.
of a general concept has all the marks of thai general concept.
Let him begin his proof by stating precisely the data, the
given or known facts, let him draw a corresponding diagram
in order to have a clearer view of the spatial relations in-
volved, let him note carefully what concepts are present in
the proposition, let him draw auxiliary lines and introduce
auxiliary concepts at pleasure. But let him exhaust simple
means before trying more complicated, let him distinguish,
by manner of drawing, the principal from the auxiliary rays,
and especially let him be systematic and consistent in the
literation of his figures.
1. How many degrees in a straight angle? In a right
angle ?
Historical Note. — For purposes of computation the round angle
is divided into 360 equal parts called degrees, each degree into 60
equal minutes (partes minutce primse), each minute into 60 equal
seconds (partes minutae secundcz), denoted by *^, ', " respectively.
This sexagesimal division is cumbrous and unscientific, but is apparently
permanently established. It seems to have originated with the Baby-
lonians, who fixed approximately the length of the year at 360 days, in
which time the sun completed his circuit of the heavens. A degree,
then, as is indicated by the name, which means step in Latin, Greek,
Hebrew {gradus, padfios (or rfXTjixa), ma'a/ak), was primarily the daily
step of the sun eastward among the stars. The Chinese, on the other
hand, determined the year much more exactly at 365} days, and
accordingly, in defiance of all arithmetic sense, divided the circle into
365 1 degrees.
2. The angles of a A are equal ; how many degrees in
each?
Remark. — Such a A is called equiangular, more commonly equi-
lateral, but better still regular.
3. Show that this regular A is equilateral.
4. One angle of a A is a right-angle ; the others are
equal ; how many degrees in each ?
EXERCISES I. 71
5. One angle of a A is twice and the other thrice the
third ; what are the angles ?
6. Two angles of a A are measured and found to be
46° 37' 24" and 52° 48' 39"; what is the third?
7. One angle of a A is measured to be 6i°22'4o"; the
others are computed to be 49° 34' 28" and 69° 2*43" ; what
do you infer ?
8. A half- ray turns through two round angles counter-
clockwise, then through half a right-angle clockwise, then
through a straight angle counter-clockwise, then through \
of a round angle counter-clockwise, then through |^ of a
straight angle clockwise ; what angle does it make in its
final position with its original position ?
9. <9 is a fixed point (called origin) on a ray, A and B
are any pair of points, M their mid-point. Show and state
in words that 2 0M= OA -f OB.
10. Af B^ C are three points on a ray, A\ B\ C are
mid-points of the tracts BC^ CA, AB, and O is any point
on the ray ; show that OA-\-OB-\- 0C= 0A'-{- 0B'-\- OC.
11. Af By C, £>, O are points on a ray; A', B\ C are
mid-points of AB, BC, CD \ A", B", are mid-points of
A'B', B'C; Mis the mid-point of A"B" ; prove SOM=
OA + sOB + sOC+OD.
12. What are the conditions of congruence in isosceles
A ? In right A ?
13. In what A does one angle equal the sum of the
other two?
Z>e/. A number of tracts joining consecutively any number
of points (first with second, second with third, etc.) is called
a broken line, or train of tracts, or polygon. Where the last
72 GEOMETRY.
point falls on the first the polygon is closed ; otherwise it is
open. Unless otherwise stated, the polygon is supposed to
be closed. The points are the vertices, the tracts are the
sides of the polygon. The closed polygon has the same
number of vertices and sides, and we may call it an n-angle
or n-side. The angles between the pairs of consecutive
sides are the angles of the polygon, either inner or outer ;
unless otherwise stated, inner angles are referred to. Inner
and outer angles at any vertex are supplemental. When
each inner angle is less than a straight angle, the polygon
is called convex ; otherwise, re-entrant. Unless otherwise
stated, convex polygons are meant. Sides and angles of a
polygon may be reckoned either clockwise or counter-clock-
wise.
14. Prove that the sum of the inner angles of an «-side
is {n — 2) straight angles. What is the sum of the outer
angles ?
15. Find the angle in a regular {i.e. equiangular and
equilateral) 3-side, 4-side, 5-side, 8-side, 12-side. (For
proof that there is a regular ?z-side, see Art. 137.)
16. Show that a (convex) polygon cannot have more
than three obtuse outer angles, nor more than three acute
inner angles.
17. Two angles of a A are a and ^\ find the angles at
the intersection of their mid-rays.
18. If two A have their sides parallel or perpendicular in
pairs, then the A are mutually equiangular.
19. The medial to the hypotenuse of a right A cuts the
A into two isosceles A.
20. An angle in a A is obtuse, right, or acute, according
as the medial to the opposite side is less than, equal to, or
greater than, half the opposite side.
EXEKCISES I. n
21. A medial will be greater than, equal to, or less than,
half the side it bisects, according as the opposite angle is
acute, right, or obtuse.
22. \{ P and Q be on the mid-normal of ABy then
AAPQ = ABPQ (= indicates congruence).
23. AB is the base, C the opposite vertex of an isosce-
les A; show that ABN=BAM (i) when AM and BN
are altitudes, (2) when they are medials, (3) when they are
mid-rays of angles A and Z?, (4) when MN is normal to the
mid-normal of AB.
24. P is any point within the A ABC; show that
AP+BP<AC^ CB, AP+PB-\- CP> ^ {AB+BC+ CA).
25. ABC"'L and AB'C-'-L are two convex polygons,
not crossing each other, between the same pairs of points,
A and Z ; which is the longer? Give proof.
26. P is a point within AABC ; show that angle APB
>ACB and sum of angles at P= 2{A + B-\- C).
27. Pis equidistant from A, B, and C \ show that angle
APB=2{zxi^^ACB).
28. Conversely, if angle ^/^^ = 2 (angle ACB), angle
BPC= 2 (angle BAC), and angle CPA = 2 (angle CBA),
then P is equidistant from A, B, C.
29. The mid-rays of the angles at the ends of the trans-
verse axis of a kite cut the sides in the vertices of an
anti-parallelogram (Art. 99).
30. The four joins of the consecutive mid-points of the
sides of a 4-side form a parallelogram.
31. The joins of the mid-points of the pairs of opposite
sides and of the pairs of diagonals of a 4-side concur, bisect-
ing each other.
74 GEOMETRY.
52. The mid-parallels to the sides of a A cut it into 4
congruent A.
33. What figures are formed by the mid-parallels when
the A is right? isosceles? regular?
34. A parallelogram is a rhombus if a diagonal bisects
one of its angles.
35. A parallelogram is a square if its diagonals are equal
and one bisects an angle of the parallelogram.
36. From any point in the base of an isosceles A parallels
are drawn to the sides ; the parallelogram so formed has a
constant perimeter ( = measure round = sum of sides) .
37. The sum of the distances of any point on the base of
an isosceles A from the sides is constant.
38. The sum of the distances of any point within a regular
A from the sides is constant. — What if the point be without
the A?
39. /* is on a mid-ray of the angle A in the A ABC)
compare the difference of FB and PC : when B is within
the A, and when B is without.
40. The inner mid-ray of one angle of a A and the outer
mid-ray of another form an angle that is half the third angle
of the A.
41. O is the orthocentre of the A ABC; express the
angles A OB, BOC, CO A, through the angles A, B, C.
42. Do the like for the circum-centre 6" and the in-centre /.
43. The medial to the hypotenuse of a right A equals
one-half of that hypotenuse.
44. The mid-rays of two adjacent angles of a parallelogram
are normal to each other.
EXERCISES I. 75
45. In a 5 -pointed star the sum of the angles at the
points is a straight angle. What is the sum in a 7-pointed
star?
46. Parallels are drawn to the sides of a regular A, tri-
secting the sides ; what figures result ?
47. A side of a A is cut into 8 equal parts, through each
section point parallels are drawn to the other sides ; how are
the other sides cut and what figures result ?
48. Two A are congruent when they have two mid-tracts
of two corresponding angles equal, and besides have equal
( 1 ) these angles and a pair of the including sides ; or
(2) two pairs of corresponding angles; or
(3) one pair of corresponding angles and the correspond-
ing angles of the mid-tract with the opposite side ; or
(4) one pair of including sides and the adjacent segment
of the opposite side.
49. Two A are congruent when they have two corre-
sponding sides and their medials equal, and besides have
equal
( 1 ) another pair of sides ; or
(2) the angles of the medial with its side (in pairs) ; or
(3) a pair of angles of the bisected side with another
side, the angles of the medial with this side being both
acute or both obtuse ; or
(4) a pair of angles of the medial with an including side,
the corresponding angles of the medial with its side being
both acute or both obtuse.
50. Two A are congruent when they have a pair of cor-
responding altitudes equal, and besides have equal
76 GEOMETRY.
( 1 ) the pair of bases and a pair of adjacent angles ; or
(2) the pair of bases and another pair of sides ; or
(3) the pairs of angles of the altitude with the sides ; or
(4) two pairs of corresponding angles ; or
(5) the two pairs of sides, when the altitudes lie both
between or both not between the sides of the A.
SYMMETRY.
90. We have seen that congruent figures are ahke in size
and shape, different only in place, and may be made to fit
point for point, line for line, angle for angle. The parts
that fit one on the other are said to correspond or be corre-
spondent. Plainly only Hke can correspond to like, as point
to point, etc.
Def. The ray through two points we may call the join of
those points, and the point on two rays the join of the rays.
91. It is now plain that if A corresponds to A and B to
B\ then the join of A and B must correspond to the join
of A' and B^ ; for in fitting A on A' and B on B' the ray
AB must fit on the ray A'B' (why?). Also if the ray Z
corresponds to L', and Mto M\ then the join of L and M
must correspond to the join of V and J/' (why?). These
facts are very simple but very important.
We shall think of the plane as a thin double film, the one
figure drawn in the upper layer, the other in the lower.
92. Two congruent figures may be placed anywhere and
any way in the plane, but there are two positions especially
important: (i) the one in which the one figure may be
superimposed on the other by turning the one half of the
plane through a straight angle about a ray called an axis ;
(2) the one in which the one figure may be fitted on the
SYMMETRY.
77
Other by turning the one half of the plane through a straight
angle about a j^oint called a centre.
Congnient figures in either of these two positions are
called symmetric : in the first case axally, as to the axis of
symmetry ; in the second case centrally, as to the centre
of symmetry.
93. In two symmetries, corresponding angles, like all
other correspondents, are of course congruent ; but they
are reckoned oppositely if the symmetry be axal, similarly if
Fig. 61.
78 GEOMETRY.
it be central. To parallels correspond parallels ; to normals,
normals ; to mid-points, mid-points ; to mid-rays, mid-rays ;
to the axis corresponds the axis, each point to itself; to the
centre corresponds the centre itself (Fig. 6i).
Elements, whether points or lines, that correspond to
themselves may be called self-correspondent or double.
It is also manifest that centre and axis are the only self-
correspondents ; hence if a point be self-correspondent, it
must lie on the axis in axal symmetry, or be the centre in
central symmetry ; and if two counter half-rays be corre-
spondent, they (or the ray) must be normal to the axis in
axal symmetry, or go through the centre in central sym-
metry.
94. These facts are all perfectly obvious, but a more
vivid exemplification of the nature of these two kinds of
symmetry may perhaps be found in the following :
Suppose the axis of symmetry to be a perfect plane
mirror ; then either half of the plane may be treated as the
reflection or exact image of the other, and will be the sym-
metric of the other as to the mirror-axis. For the image of
any point A is the point A^ such that the axis is the mid-
normal of AA\ as we know froni Physics ; also, on folding
over the one half of the plane about the axis upon the other
half, the point A falls on A^ (why?) ; hence A^ is the sym-
metric of A as to the axis.
Suppose the centre of symmetry S to be also a reflector ;
then the reflection or image of any point A will be a point
A^ such that 6" is the mid-point of the tract AA\ and on
rotation through a straight angle about 6" the point A falls
on A\ and the half-ray SA fits on the half- ray SA\ Hence
either of two centrally symmetric figures is the exact image
of the other reflected from the centre of symmetry 6".
SYMMETRY.
79
Note carefully that these two species of symmetry depend
upon the two fundamental definitive properties of the plane :
central symmetry upon the homa'oidality of the plane, axal
symmetry upon the reversibility of the plane. Moreover,
axally symmetric figures can 7iot be fitted on each other
without reversion, folding over ; by movement in the plane
their corresponding parts can at best be ^/posed, but never
j«/^rposed ; while on the other hand central symmetries
may be j/y/^rposed, but cannot be <7/posed, along any ray,
by motion in the plane. In central symmetries the corre-
sponding parts follow one another in the same order, but in
axal symmetries they follow in opposite orders.
95. We must now discuss these two symmetries more
minutely, and to exhibit a certain remarkable relation hold-
ing between them we arrange their properties in parallel
columns.
In Axal Symmetry.
1. The axis corresponds to it-
self.
2. Every point of the axis cor-
responds to itself.
3. Every self-correspondent
point lies on the axis.
4. The join of two correspond-
ent rays is on the axis.
(For it is self-correspondent.)
5. Correspondent points are
equidistant from every point on
the axis.
In Central Symmetry.
1. The centre corresponds to
itself.
2. Every ray through the cen-
tre corresponds to itself (each half
to the other).
3. Every self-correspondent ray
goes through the centre.
4. The join of two correspond-
ent points goes through the cen-
tre.
(For it is self-correspondent.)
5. Correspondent rays are
equally inclined (isoclinal) to
every ray through the centre;
hence they are parallel, as is
otherwise manifest.
80
GEOMETRY.
6. The centre is a mid-point of
every tract between correspond-
ent points, and in fact the inner
mid-point.
N.B. The outer mid-point is a
point at infinity.
7. The join of two correspond-
ent rays is at infinity.
(For they are parallel.)
8. Correspondent angles are
contra-posed {i.e. have their arms
extended oppositely).
9. Correspondent rays are
equidistant from the centre.
10. The join of two points and
the join of their correspondents
themselves correspond.
6. The axis is a mid-ray of
every angle between correspond-
ent rays, and in fact the inner
mid-ray.
N.B. The outer mid-ray is a
normal to the axis.
7. The join of two correspond-
ent /o2«/^ is a normal to the axis.
8. Correspondent tracts are
anti-parallel.
9. Correspondent points are
equidistant from the axis.
10. The join of two rays and
the join of their correspondents
themselves correspond.
96. On regarding closely these correlated propositions, it
becomes clear that the one set differs from the other only
in the interchange of certain notions, as poi7it and ray, tract
and angle, etc. Every property of axal symmetry has its
obverse in central symmetry, and vice versa. This most
profound, important, and interesting fact has received the
name of the Principle of Reciprocity. We make this notion
more precise by the following
Def. Two figures such that to every point of each corre-
sponds a ray of the other, and to every ray of each a point
of the other, are called reciprocal. For example :
Suppose rays drawn through a point O to any number of
points. A, B, C, D, E, . . . on 2i ray L. Then the point
O with its ray through it, and the ray L with its point on
it, are two reciprocal figures (Fig. 62). The first is called
a (flat) pencil of rays, O being the centre ; the second is
called a row (or range) of points, L being the axis. Sup-
SYMMETRY.
81
pose we have now a second pencil through O' and a second
row on Z'. These two figures are again reciprocal, and the
two pairs of reciprocals together make up another more
complex pair of reciprocals. In this latter pair we find our
definition fully exemplified. To O and 6>' correspond L and
Z' ; to the rays through O and (9' correspond the points on Z
and Z' ; also, to the join (ray) of O and O' corresponds the
Fig. 62.
join (point) of Z and Z' ; to any point as Pj the join of two
rays {OA^ O'A'), corresponds a ray A A', the join of two
points {A J A'). So Q, R^ S, Tare points corresponding to
the rays BB\ CC\ DD\ EE\ We may notice further
that angle and tract correspond in the reciprocal figures ;
thus the angle AOB corresponds to the tract AB, and the
angle BOC to the tract BC\ while the angle OPO' corre-
sponds to the tract AA^ and the tract RS to the angle
between the rays corresponding to R and S ; namely, between
CO and DD\ Let the student trace out as many corre-
spondences as possible.
82 GEOMETRY,
97. To three points fixing a triangle in either of two
reciprocals must correspond also three rays fixing a triangle
in the other reciprocal; hence, in general, triangle corre-
sponds to triangle in reciprocals. But notice : the sides of
one correspond to the vertices of the other ; hence if the
sides of one all go through the same point, the vertices of
the other all lie on the same ray ; that is, three concurrent
rays in either reciprocal correspond to three collinear points
in the other.
It now appears that axal and central symmetry are recip-
rocal to each other ; the reciprocal of an axal symmetric is
a central symmetric, and the reciprocal of a central sym-
metric is an axal symmetric ; the reciprocal properties of
axal symmetry are the properties of central symmetry, and
the reciprocal properties of central symmetry are the proper-
ties of axal symmetry.
Very often the two symmetric figures may be regarded
as the two halves of one figure ; this one figure is then said
to be symmetric as to the axis of symmetry or as to the
centre of symmetry, as the case may be.
98. If our figure be two points, A and A\ then the mid-
normal X of the tract A A is the axis of symmetry, mani-
festly. If, now, any double point D on the axis be joined
with A and A\ there results the isosceles A ADA\ whence
it appears that (Fig. ^2>)
The isosceles A is a symj?tetric A.
It is plain that any two points on the ray AA^ equidistant
from N are symmetric as to X, that all points on the ray,
and indeed in the whole plane, may be arranged in sym-
metric pairs, the members of each pair equidistant from the
axis X.
SYMMETRY.
83
99. Now take two points on the axis, as D and D\ or
D and Z>", and consider the 4-side DAD'A^ It is com-
posed of two A, ADiy and ADD\ symmetric with each
other as to the axis X, and opposed along that axis. Hence
the 4-side is itself symmetrical as to X.
Def. Such a 4-side, with an axis of symmetry, is called a
kite.
If we hold D fast, and let /)' glide along Jf, the 4-side
ADA'iy remains a kite. We see that there are two kinds
of kites, the convex kite, as ADA'Z>\ and the re-entrant, as
ADA'D". As the gliding point passes through TV the kite
changes from one kind to the other, passing through the
intermediate form of the symmetrical A.
When the gliding point reaches a position £>' such that
JVD = JVD', then the four sides of the kite are all equal
(why?), and the kite becomes a rhombus (why?). In this
case £> and D' are symmetric as to A A' as an axis of syni-
84
GEOMETRY.
metry. Hence the rhombus has two axes of syf?itnetry ;
namely, its two diagonals.
In all cases the diagonals, A A' and DD\ of the kite are
normal to each other (why?).
100. Now consider a pair of points, B and B\ symmetric
as to the axis X (Fig. 64). Then X is mid-normal oi BB'.
C
Fig. 64,
If C and C be any other pair of symmetric points, then X
is also mid-normal of CC ; hence BB^ and CC are parallel
(why?). Also the tracts BC and B' C are symmetric as
to X (why?), and the 4-side BB'CC is itself symmetric as
to the axis X. Hence the angles at C and C are equal,
SYMMETRY.
85
also the angles at B and B' are equal (why ?) ; hence the
angles at B and C and at B' and C are supplemental
(why?), and the 4-side BB'CC is an anti-parallelogram
(why?). Hence we see that another sy nunc trie ^-side is an
anti-parallelogram.
It is plain that every anti- parallelogram is symmetric, for
we know that the oblique sides prolonged yield an isosceles A.
Let the student complete the proof.
loi . There is only one kind of symmetric A, the isosceles.
For, let ABA' (Fig. 65) be symmetric and A' correspondent
Fig. 65.
to A. Then B must correspond to itself (why?) ; hence
B must lie on the axis (why?) ; hence BA = BA' (why?).
Now let the student prove that
(i) /n a symtnetric A the axis of symmetfy is a medial ;
(2) // is also a mid-ray ; (3) /'/ is also a mid-normal.
86 GEOMETRY.
Conversely, let him show that
A medial that is a mid-ray, or a mid-normal, is an axis
of symmetry.
102. There are only two axally symmetric 4-sides ; namely,
the kite and the anti- parallelogram. For, in a symmetric
4-side a vertex must correspond to a vertex (why?). Also,
not all vertices can be on the axis (why?). Also, a vertex
on the axis is a double point (why?). Also, the vertices
not on the axis must appear in pairs (why ?) ; hence there
must be either two or four of them. If there be two only,
then the other two are on the axis and the 4-side is a kite ;
if there be four of them, we have just seen that the 4-side is
an anti-parallelogram.
103. Now let us turn to the reciprocals. The reciprocals
of the two points A and A^ symmetric as to the axis X will
be two rays L, L\ symmetric as to the centre S. But rays
symmetric as to a centre are parallel (why ?) ; hence we have
two parallels symmetric as to S, which is midway between
them. The rays are symmetric as to any other point S' mid-
way between them (why?). The piece of plane between
these parallels is called a parallel strip, or band (Fig. 66).
R,'
y^
L
•s'
L
Fig. 66.
SYMMETRY. 87
But what corresponds to the point D on the axis X? The
answer is : a ray R through 6" (why?). Hence to the sym-
metric A of the three points A, A\ D, there corresponds
the figure formed by two parallels Z, Z', and a transverse R
through Sf — a so-called half-strip. This is truly a three-
side, but not apparently a A (3-angle), for the parallels do
not meet in finity, in regions accessible to our experience.
Hence, instead of saying that the reciprocal of a A in axal
symmetry is a A (3-angle or 3-point) in central sym-
metry, we should have said, accurately, that the recipro-
cal of a A in axal symmetry is a 3-side (or trilateral) in
central symmetry, which will always be a A except when
sides are parallel or all concur. In higher Geometry it is
very convenient to remove this apparent exception by using
this form of expression : the parallels meet not in finity y but
/// infinity.
104. It is indeed plain that
A A can have no centre of symmetry.
For, since vertex corresponds to vertex, and since corre-
spondents appear in pairs, one vertex must be a double point ;
hence it would have to be the centre .S" (why?). But the
other two vertices would have to lie on a ray through 6", being
correspondents ; hence the three vertices would be collinear,
and the A would be flattened out to a triply-laid ray.
105. But there is a centrally symmetrical 4-side ; namely,
the parallelogram. For, consider once more the kite
AXA'X' and let us reciprocate it into a centrally symmetric
figure (Fig. 67). To the axis XX' will correspond the cen-
tre S; to the symmetric pair of rays AX and A'X w'\\\ corre-
spond a symmetric pair of points Pand P \ to the join of
those on the axis A' will correspond the join of these through
88
GEOMETRY.
the centre {PP^). Similarly, to the symmetric rays AX' diud
A'X' will correspond the symmetric points Q and Q', and
to the join X' will correspond the join QQ'. Also, AX
and AX' have a join A while A'X and A'X' have a join A',
and these joins are symmetric as to the axis XX' ; recipro-
FlG. 67.
cally, Pand Q have a join PQ, and /" and Q' have a join
/"^'j and these joins are symmetric as to .S"; that is, they are
parallel (why?). Similarly, PQ' and P'Q correspond to B
{AX, A'X') and B' {A'X, AX') ; but B and B' are sym-
metric as to JTX' (why?) ; hence PQ' and P'Q a.re symmetric
as to S, i.e. are parallel. Hence PQ PQ is symmetric as
to S, and is 2, parallelogram. Q. e. d.
106. We may indeed see at once that since any two par-
allels are centrally symmetrical as to any mid-point, a pair
SYMMETRY,
89
of parallels or a parallelogram is symmetric as to the common
mid-way point, the intersection of the diagonals. But the
foregoing reciprocation is instructive, as illustrating in detail
the method to be pursued, and as showing the intimate rela-
tion of the different symmetric quadrilaterals ; namely, the
parallelogram is the common reciprocal of both kite <///// anti-
parallelogram, which are thus seen to be really one,
107. Central symmetry does not in general imply any-
thing at all with respect to axal symmetry in a figure. We
may draw through any point S any number of rays and lay
off on each from 5 a pair of counter tracts SP and SP\
SQ and SQ^ etc. No matter how PQ^ etc., be chosen, the
figure so obtained will be centrally symmetric as to 6"; but
it may have no axal symmetry whatever. Neither does axal
symmetry in general imply any central symmetry, but we
may establish the following important
Theorem. — Any figure with two rectangular axes of
symmetry has also a centre of symmetry; namely^ the inter-
section of those axes.
p
^
f
p
'^^
S/-^
X
^^
1
w-
p
Y
Fiu. 68.
90 GEOMETRY.
Data : XX^ and KF' two rectangular axes, P any point
of a figure symmetric as to these axes (Fig. 68).
Proof. The point /" symmetric with P as to XX is a
point of the figure (why?) ; also /*" symmetric with /" as
to yy is a point of the figure (why?) ; so too is /*'"
(why?) ; the figure /'/"/'"P"' is a rectangle (why?), its
diagonals halve each other, and SP= SP" = SP' = SP'".
Hence 6* is a centre of symmetry, q. e. d.
THE CIRCLE.
io8. We have already discovered the existence of a
homceoidal plane curve not reversible and have named it
circle.
Defs. A ray cutting a curve is called a secant, as L ; the
part of the secant intercepted by the curve, or the tract
between two points of the curve, is called a chord, as AB.
A finite part of a curve is called an arc. A chord and an
arc with the same two ends are said to subtend each other.
Also, the intercept of any line between the ends of an angle
is said to subtend \h.& angle. Thus .^Cand DE subtend the
angle O (Fig. 69).
Fig. 69.
Tn. XLIX.J
THE CIRCLE.
91
109. Theorem XLVIII. — Congruent arcs subtend con-
gruent chords.
Proof. Let the arcs AB and CD be congruent ; then we
may fii A on C and at the same time B on £>; then the
chords AB and CB fit throughout (why?), q. e. d.
N.B. We can no/ convert this proposition at once (why?)
(Fig. 70).
Fig. 70.
1 10. Theorem XLIX. — A closed curve is cut by a ray in
an eiien number of points (Fig. 71).
Proof. Let Z be a ray, C any closed curve. Suppose a
point P to trace out the ray Z. At first P is without the
curve, at last it is also without the curve ; hence P has
crossed the curve going out as often as it has crossed the
curve going in, for every entrance there is an exit ; hence
the points of intersection appear in pairs, their number is
even, as o, 2, 4, 6, ... 2 //. q. e. d.
92
GEOMETRY.
[Til L.
These preliminary or auxiliary theorems, which prepare
the way for a theorem to follow, are sometimes called
lemmas {Xrjfxfxa = assumption, premise, support, prop).
*iii. Theorem L. — A chrle has aft axis of symmetry
through every one of its points (Fig. 72).
Q
D
R
Q
Fig. 72.
Proof. Let D be any point of a circle. Take any arc
DP, and slip it round till P falls on D and D on /" ; this
is possible (why ?) . Then PDP' is a symmetrical A (why ?) ;
and its axis of symmetry DR halves normally the chord
PP', and also halves the angle PDP' (why?). Now take
any other arc DQ and slip it round till Q falls on D and D
on Q\ so that DQ and QD are congruent. Then the
chords DQ and DQ are congruent (why?). Also, on
taking away the congruents DP and DP' we have left PQ
and P'Q as congruent remainders. Hence the chords PQ
Th. LI.] THE CIRCLE. 93
and PQ are congruent (why?). Hence the A PDQ and
PDQ' are congruent (why?) ; hence the angles PDQ and
P'DQ' are equal (why?) ; hence DR halves also the angle
QDQ (why?). But the A QDQ is symmetric (why?) ;
hence DR is also its axis of symmetry, and Q and Q are
symmetric points of the circle ; hence any point of the
circle has its symmetric point as to DR ; i.e. DR is an
axis of symmetry of the circle. Moreover, D was any point
of the circle ; hence through any point of the circle passes
an axis of symmetry, q. e. d.
Def. A ray halving a system of parallel chords is called a
diameter ; the chords and diameter are called conjugate to
each other.
Corollary i. In a circle a diameter is normal to its con-
jugate chords.
Corollary 2. Every mid-normal to a chord in a circle is
a diameter and halves the subtended arcs.
*ii2. Theorem LI. — A circle has a centre of symmetry
(Fig. 12).
For the ray through D must cut the circle in some second
point, as R (why?), and as the ray turns round from the
position DR to the reversed position RD, through a straight
angle, it must pass through some position, QQy normal to
its original position (why ?) . Hence for any axis of symmetry
there is another normal thereto and their intersection is a
centre of symmetry (why ?) . q. e. i>.
N.B. There is only one centre of symmetry (why?).
Def. This centre of symmetry is named centre of the
circle. It is often convenient to call the whole ray through
the centre a centre ray or line, and to restrict the term
diameter to the centre chord.
94
GEOMETRY.
[Th. LII.
Corollary i. All diameters go through the centre, and
halve each other there ; conversely, chords halving each
other are diameters.
Def. Two diameters each halving all the chords parallel
to the other are called conjugate.
Corollary 2. In the circle two diameters normal to each
other are conjugate ; and conversely, two conjugate diameters
are normal to each other.
N.B. Other curves, as ElHpse and Hyperbola, have
conjugate diameters not in general normal to each other
(Fig- 73)-
*ii3. Theorem LII. — All diameters of a circle are equal
(Fig. 74).
Fig. 73. Fig. 74.
Proof. Let DR and D^R^ be two diameters. The figure
DD^RR^ is a parallelogram (why?), and DD^ is parallel to
RR^ ; hence the mid-normal of these parallels is a diameter
Th. LI I.] THE CIRCLE. 95
through the centre ^ ; hence SD and SD^ are symmetric
and equal ; hence DR = D'R'. q. e. d.
De/. A half- diameter, from centre to circle, is called a
radius.
Corollary i. All radii of a circle are equal ; or, all points
of a circle are equidistant from the centre.
Corollary 2. Every parallelogram inscribed in a circle is
a rectangle.
N.B. By help of this important property the circle is
commonly defined as a plane curiae all points of which arc
equidistant from a point within called the centre. The com-
mon distance of all points of the circle from the centre is
often called the radius. We have deduced this property
from the homoeoidality ; conversely ^ we may deduce the
homoeoidality from this property taken as definition. But
if there were no such surface as the plane, at least for our
intuition, the circle might still exist on the sphere-surface,
without centre, but with the body of its properties unimpaired.
Hence it seems better to define the circle by its intrinsic
homoeoidality than by its extrinsic centrality.
Corollary i. All points within the circle are less, and all
points without are more, than the radius distant from the
centre.
Defs. The two symmetric halves into which a diameter
cuts a circle are called semicircles. The part of the plane
bounded by an arc and its chord is called a segment ; the
part bounded by an arc and the two radii to its ends is
called a sector. If the sum of two arcs be a circle, we may
call them explemental, the one minor y the other major;
every chord belongs equally to each of two explemental arcs,
but in general, unless otherwise stated, it is the minor that
96 GEOMETRY. [Th. LIII.
is referred to. Two arcs whose sum is a half-circle are
called supplemental ; two whose sum is a quarter-circle or
quadrant are called complemental.
Corollary 2. All circles of the same radius are congruent ;
also, all semicircles of the same radius are congruent, and
all quadrants of the same radius are congruent.
Corollary 3. Any circle may be shpped round at will
upon itself about its centre as a pivot, like a wheel about its
axle, without changing in the least the position of the whole
circle.
114. From the foregoing it is clear that if we hold one
point of a ray fixed, and turn the ray in the plane about the
fixed point, every other point of it will trace out a circle
about the fixed point as a centre. An instrument, one point
of which may be fixed while the other is movable about in a
plane, is called a compass or pair of compasses, and is both
the simplest and the most important of all instruments for
drawing.
115. Theorem LIII. — Through any three points not col-
linear one, and only one, circle f?iay be drawn.
Proof. Let A, B, C be the three points not collinear
(Fig. 75). We have already seen that the mid-normals to
the tracts AB, EC, CA concur in a point S equidistant from
A, B, and C ; hence a circle about S with radius d passes
through A, B, C. Also there is only one point thus equi-
distant from A, B, C (why?) ; hence there is only one
circle through A, B, C. Q. e. d.
Def. The circle through the vertices A, B, C, of a A is
called the circum-circle of the A.
Corollary i. A A, or a triplet of points, or a triplet of
rays, determines one, and only one, circle.
Th. Mil.]
THE CIRCLE.
97
Corollary 2. Through two points, A and B, any number
of circles may be drawn. Their centres all lie on the mid-
normal of AB.
Corollary 3. ks BC turns clockwise about i9 as a pivot,
the intersection S, the centre of the circle through A^ B, C,
retires upward ever faster and faster along the mid-normal N
of AB ; when C becomes collinear with A and B, the inter-
FIG. 75.
section of the raid-normals of AB and BC vanishes from
finity, or rf/ires to infinity, as the phrase is. As BC keeps
on turning, S reappears in finity below and moves slower and
slower upward along the mid-normal. Moreover, a circle
passes through A, By and C, no matter how close C may lie
to the ray AB, nor on which side of it : only as C falls upon
the ray does the centre of the circle vanish into infinity ; that
is, we may draw a circle that shall fit as close to the ray AB
as we please f though not upon it, by retiring the centre far
98 GEOMETRY. [Th. LTV.
enough. Hence a ray may be conceived as a circle with
centre retired to infinity ; it is strictly the limit of a circle
whose centre has retired, along a normal to it, without limit.
1 1 6. Theorem LIV. — A circle can cut a ray in only two
points.
For there are only two points on a ray at a given distance
from a fixed point (why ?) . Q. e. d.
117. Theorem LV. — Secants that make equal angles with
the centre ray (or axis) through their intersection intercept
equal arcs on the circle.
Proof. For both the two semicircles and the two secants
are symmetric as to the axis IS (why ?) ; hence, on folding
over the one half-plane upon the other, A falls on A\ B on
B\ arc a fits on arc a\ and chord c on chord c^ (Fig. 76).
Q. E. D.
Tn. LVI.] THE CIRCLE. 99
Conversely, Secants thai intercept equal arcs make equal
angles with the axis through their intersection.
Proof. Let L and V intersect equal arcs AB and A^B\
Draw the mid-normal of AA^ ; it is an axis of symmetry
(why?). On folding over the left half- plane upon the right
half-plane, A falls on A and B onB' (why?) ; hence AB
and A'B' are symmetric ; hence they meet on the axis and
make equal angles with it (why?), q. e. d.
Corollary i. Equal chords are equidistant from the cen-
tre ; and conversely ^ Chords equidistant from the centre are
equal.
Corollary 2. The greater of two unequal chords is less
distant from the centre.
Corollary 3. A diameter is the greatest chord.
Corollary 4. Arcs intercepted by two parallel chords are
equal.
Corollary 5. Equal chords or arcs subtend equal central
angles (angles at the centre), and conversely.
Corollary 6. Of two unequal chords or arcs, the greater
subtends the greater central angle.
What figure is determined by two parallel chords and the
chords of the intercepted arcs ? By two secants that inter-
cept equal arcs and the central normals thereto ?
118. Theorem LVI. — A central angle subtended by a cer-
tain arc (or chord) is double the peripheral angle subtended
by the same (or an equal) arc (or chord) (Fig. 77).
Proof. Let ASB be a central angle, and APB be a
peripheral angle (periphery = circumference, the circle
itself), subtended by the same arc or chord AB. Draw the
100 GEOMETRY. [Th. LVI.
diameter PD. Then the A ASP and BSP are isosceles
(why?) ; hence the angle ASD = 2 angle APD, and angle
BSD = 2 angle BPD (why ?) ; hence angle ASB = 2 angle
APB. Q.E.D.
Fig. 'jj.
Corollary i . All peripheral angles subtended by (or stand-
ing on) the same or equal chords or arcs are equal. Hence,
as P moves round from A to B^ the angle APB remains
unchanged in size.
Def. An angle with its vertex on a certain arc, and its
arms passing through the ends of that arc, is said to be
inscribed in that arc. Hence for an angle to be inscribed
in a certain arc, and for it to stand on the explernental arc,
are equivalent.
Corollary 2. All angles inscribed in the same or equal
arcs of the same or equal circles are equal.
Corollary 3. As the vertex /* of a peripheral angle sub-
tended by an arc (or chord) AB^ in passing round a circle
goes through either end of the arc (or chord), the angle
itself leaps in value, changes to its supplement.
'P^ (2^^f(>y^
Tn. LVIII.] THE CIRCLE. 101
1 19. Theorem LVII. — The locus of the vertex of a given
angle standing on a given tract ii two symmetric circular
arcs through the ends of the tract (Fig. 78).
\0
Fig. 78.
Proof. Let P be the vertex of the given angle, in any
position, standing on the tract AB. Through A, Py and B
draw a circular arc subtended by AB. We have just seen
that as long as P stays on this arc, the angle P remains the
same in size. Moreover, the point P cannot be without the
arc, as at O, because the angle A OB is less than APB
(why?) ; neither can it come within the arc, as to /, because
the angle A IB is greater than APB (why ?) ; hence so long
as the angle is constant in size the vertex must remain on
the arc APB or on its symmetric arc AP'By of which plainly
the same may be said. q. e. d.
120. Theorem LVIII. — The angle inscribed in a semi-
circle (or standing on a semicircle or diameter) is a right-
angle (Fig. 79).
102
GEOMETRY.
[Th. LVIII.
Proof. Let ABC\>^ any angle in a semicircle. Then it
is half of the central angle ASC (why?), which is a straight
angle (why?). Q. e. d.
U
Fig. 79.
Now let the vertex B, the intersection of the rays L and
N, move round the circle toward C; the angle ABC re-
mains a right angle, 710 matter how close B approaches to C ;
moreover, when B passes C, into the lower semicircle, the
angle remains a right angle (why?). That is, the angle at^
remains a right angle, no matter from which side nor how
close B approaches to C. Hence it is a right angle even
when B falls on C. But then the ray L falls on the diame-
ter A C, hence the ray N takes the position T normal to the
diameter (or radius) at its end. Such a normal to a radius
at its end is called a tangent to the circle at the point of
tangence (or touch or contact^ C.
Def. A ray normal to a tangent to a curve at the point
of touch is called normal to the curve itself. Hence
Corollary. All radii of a circle are normal to the circle ;
and conversely, all normals to a circle are radii of the circle.
Th. LX.]
THE CIRCLE.
103
121. Theorem LIX. — All points on a tangent y except the
point of contact, lie outside of the circle.
Proof. For the point of touch is distant radius from the
centre (why?), and all other points, as D, of the tangent
are further from the centre (why ?) ; hence all other points
of the tangent are without the circle (why?), q. e. d.
122. Theorem LX. — The point of tangence is a double
point.
Proof. For it is on a diameter, or axis of symmetry, of
the circle, and every such point is a double point with
respect to that axis.
Independently of this consideration, it is seen that the
chord CB becomes the tangent CT when, and only when,
the points B and C fall together in C.
Fig. 8o.
Still otherwise, let AB be any chord of a circle about
(Fig. 8o) O. Draw the mid-normal OD. Now let the
circle shrink about the centre O : the points A and B move
104
GEOMETRY.
[Th. LXI.
towards each other, and as D is ahvays mid-way between
them they finally fall together in D, and their join is tan-
gent at D to the circle of radius OD.
Def. Two points thus falling together in a double point
are called consecutive points. Accordingly we may define
a tangent to a circle (or to any curve) as a ray through two
consecutive points of the circle (or curve). Adopting this
definition, let the student prove
123. Theorem LXI. Every tangent to a circle is normal
to a radius at its e?id ; conversely, Every normal to a radius
at its end is tangent to the circle.
124. Theorem LXII. The angle between a tangent and
a chord equals the peripheral angle on the same chord, or
equals half the angle of the chord (Fig. 81).
Proof. For if Z>r be a diameter, then the angles BDT
and BTA are equal, being complements of the same angle
BTD (why?). Or thus : TB '\?> sl chord, and TA is also a
Th. LXIV.l
THE CIRCLE,
105
chord, through the double point T\ hence the angle BTA
is a peripheral angle standing on the arc TB. q. e. d.
125. Theorem LXIII. — The angle behveen two secants
is half the sum or half the difference of the angles of the
intercepted arcSy according as the secants intersect within or
without the circle.
Proof. For on drawing AB^ the angle / is seen (Fig.
82) to be the sum, and the angle O the difference, of the
Fig. 82.
angles at A and B^ standing on the arcs AA' and BB\
Q. E. D.
126. Theorem LXIV. — An encyclic quadrangle has its
opposite angles supplemental.
Proof. For the angles B and D are halves of the two
central angles ASC and CSAj whose sum is a round angle.
Hence the sum of B and Z> is a straight angle, g. e. d.
106 GEOMETRY. [Th. LXV.
127. Theorem LXV. — Conversely, A quadrangle with
its opposite angles supplemental is encyclic (Fig. Zt^),
Fig. 83.
Proof. Let ABCD be the quadrangle with the angles
A and C, B and D, supplemental. About the A ABC draw
a circle. If P be any point on the arc of this circle exple-
mental to ABC, then the angle AFC is the supplement of
ABC ; but if B be not on* this arc, then the angle ABC is
either greater or less than that supplement (why?). Now
the angle Z> is that supplement; hence B> is on the arc.
Q. E. D.
128. Relations of circles to each other.
" Suppose two circles X and K' of radii r and r' to be
concentric, i.e. to have the same centre O. Then, plainly,
the distance between them measured on any half-axis OR
is r—r\ the difference of the radii. Draw tangents AT,
A^T\ where 00^ cuts the circles. They are parallel (why ?).
Now let the centre of K^ move out on (9(9' a distance r — r\-
then A falls on A^ and A'T' on AT; the circles have a
common tangent at A and are said to touch each other
inner ly at A (Fig. 84).
Tn. LXVII.]
THE CIRCLE
107
Now let (9' move still further along 00* ; then the circles
will lie partly within, partly without, each other ; they will
intersect at two i)oints, and only two (why?), symmetric as
to 00* (why?), namely /'and P \ hence
Fig. 84.
Theorem LXVI. — The common axis of two circles is the
mid- normal of their common chord.
When 6>' is distant ;■ -f / ' from Oy the circles lie without
each other, but still have a common tangent (why?) and are
said to touch outerly.
As O* moves still further away from O, the circles cease to
touch and henceforth lie entirely without each other.
Thus we find that there are three critical positions depend-
ing on the distance d between the centres O and (9' :
d=Oy when the circles are concentric.
d—r—r'j when the circles touch innerly.
d= r-\- r', when the circles touch outerly.
There are alio three intermediate positions :
For o < d < r — / the one circle is i,<iihin the other.
F'or r—r'< d< r -\- r the circles intersect.
For r 4- r < // < 00 the circles lie ivithout each other.
129. Theorem LXVII . — From a ny point without a circle
tivOf and only two^ tangents may be drawn to the circle (Fig.
85)-
108
GEOMETRY.
[Th. LXVII.
Proof. Let O be the centre of the circle K^ and F be
the point without. On OP as a diameter draw a circle K^ ;
only one such circle is possible (why?), and it cuts K in two,
and only two, points, T and T. Draw PT and PV : they
are tangent to ^ at T and V (why ?) . Moreover, no other
ray through P, as PU, is tangent to K^ because OUP is not
a right angle (why?). Q. e. d.
Fig. 85.
Def. The chord TT^ through the points of contact of the
tangents is called the chord of contact for the point P or
the polar of the pole P (see Art. ) .
The angle between the tangents to two curves at the
intersection of the curves is called the angle between the
curves themselves. When this is a right angle, the curves
are said to intersect orthogonally.
The distance PT or PT^ is called the tangent-length
from P to the circle.
Corollary i . Two circles, one having as radius the tangent-
length from its centre to the other, intersect orthogonally.
Corollary 2. Two tangents are symmetric as to the axis
through their intersection ; hence, also, the tangent-lengths
are equal.
Th. LXTX.]
THE CIRCLE.
109
130. Theorem LXVIII. — All tangent-lengths to a circle
from points on a concentric circle are equal, and intercept
equal arcs 0/ the circle (Fig. 86).
Fig. 86.
Proof. For if/* be any point without the circle K\ we
may turn F round about the centre (9 on a concentric circle
AT' without affecting any of the relations obtaining (why ?) .
Or thus : the right A TOP and T'OP^ are plainly con-
gruent (why?); hence PT= P^T (why?), q. e. d.
*i3i. Theorem LXIX. — The intercept between tioo fixed
tangents on a third tangent subtends a constant central angle
(Fig. 87).
Proof. Let PT and PT be the fixed tangents, VV^ the
intercept on the variable ray tangent at U. Then TPT' is
a constant angle, and VOV^ is half of TOT (why?), and
hence is constant. Q. e. d.
110
GEOMETRY.
[Tii. LXX.
Frc. 87.
132. Theorem LXX. — If the central (or peripheral) angles
of the common chord of two intersecting circles be equal, the
circles are equal.
Let the student conduct the proof suggested by the figure
(Fig. 88), and let him prove the converse.
*I33. Theorem LXXI. — The circumcircle of a A equals
the circumcircle of the orthocentre and any two vertices of
the l^ (Fig. 89).
Proof. Let K be the circumcircle of the A ABC, K' the
circumcircle of A, B, and O the orthocentre. The angles
Th. LXXII.] the circle. Ill
Cand B'OA^ are supplemental (why?) ; also the angles D
and BOA are supplemental (why?) ; and the angles BOA
and B'OA' are equal (wliy?) ; hence the angles Z> and C
are equal ; hence A' = A'' (why ?) q. e. d.
Fig. 89.
♦134. Theorem LXXII. — T/if mid-points of the sides of a
A, the feet of its a/titi/deSy and the mid-points between its
orthocentre and vertices^ are nine encyclic points.
Proof. I^t a circle through X^ V, Z, the mid-points of the
sides, cut the sides in three other points, 0] F, IV. Then
the angle ZXy= angle A (why?), and also = angle ZFV
(why?) ; therefore the A AZF is symmetrical. Hence the
A Zf^B is also symmetrical, Z is equidistant from A, F, and
Bf and the angle A FB is a. right angle (why?) ; so also the
angles at 6^ and IV; i.e. the circle through the mid-points of
the sides goes through the feet of the altitudes (Fig. 90).
Again, if the circle cuts the altitudes at /*, Q^ R, then the
angle F/'[r= angle FZ/r(why?) = 2 angle FA IV (why?).
Moreover, A, F, O, IV, are encyclic (why?) ; hence AO is
a diameter of the circle through them (why?) ; and FA IF
is a peripheral angle standing on the arc FIV-, hence the
112
GEOMETRY.
[Th. LXXIII.
double angle yPJV must be the central angle of the same
arc; i.e. /'is the mid-point between a vertex and orthocen-
tre : so, also, are Q and /?, similarly, q. e. d.
De/. This remarkable circle is called the 9-point circle,
or a'r^/e of Feuerbach, of the A ABC.
Corollary. The radius of the 9-point circle is half the
radius of the circumcircle.
135- Def. A Polygon all of whose sides touch a circle is
said to be circumscribed about it, and the circle is said to be
inscribed in the polygon.
Theorem LXXIII. — A circle may be inscribed in any A.
Proof. Let ABC be any A (see Fig. 59). Draw the
inner mid-rays of the angles at A, B, C ; they concur in the
in-centre /of the A, equidistant from the three sides (why?).
About this point as centre with this common distance as
radius draw a circle ; it will touch the three sides of the A
(why and where ?) . Q. e. d.
N.B. We have seen that the outer mid-rays of the angles
concur in pairs with the inner mid-rays of the angles in the
three ex-centres Ei, Eo, E^, also equidistant from the sides
Th. LXXIV.]
THE CIRCLE.
113
(Fig. 60). The circles about these touch only two sides
innerly, but the third side outerly, and hence are called
escribed, or ex-circles.
Corollary. Four, and only four, circles touch, each, all the
sides of a A.
135 a. Theorem LXXIV. — /// a 4- side circumscribed
about a circle the sums of the tivo pairs of opposite sides are
equal (Fig. 91).
Fig. 91.
Proof. The sum of the four sides is plainly 2t-{- 2u-\- 2v
-f 20/, and the sum of either pair of opposites is t-\-u-\-v^'w.
Q. E. D.
Conversely, If the sums of two pairs of opposite sides of a
4-side be equals the 4-side is circumscribed about a circle.
Proof. Let two counter sides, AB and DC meet in /,
and inscribe a circle K in the triangle ADI. Through B
114
GEOMETRY.
[Th. LXXV.
draw a tangent (Fig. 92) to A' at U, and let it cut DI 2X C.
Then since ABCD is circumscribed about K, we have
or
Fig. 92.
AB^CD^BC ^DA,
Also AB+CD ==BC + nA (why ?) .
Whence CD-CD = BC-B C,
CC=^BC-BC.
Hence C and C fall together (why? Art. 56). q. e. d.
136. Theorem LXXV. — The tan gent- length from a ver-
tex of a A to the in-circle equals half the perimeter of the A
less the opposite side (Fig. 93).
Th. LXXVI.]
THE CIRCLE.
lis
Proof. For the sum of CE ^ CD -\- BD + BF is plainly
2a (why?) ; subtract this from the whole perimeter, a -\- b -\- c,
and there remains AE -\- AF= a -\- d + c — la^ or AE —
d + c-
=^AF.
FIG. 93.
It is common and convenient to denote the perimeter
( Fig. 93) ( = measure round = sum of sides) by 2s \ then
we see that the tangent-lengths from Ay B, C, are s — a^
s — b, s — c.
Corollary. The tangent-length from any vertex, A, of a
A to the opposite ex-circle and the two adjacent ex-circles
are s, s—by s — c. Hence s — a, s^ s — b, s — c, are the
four tangent-lengths from any vertex, ^, of a A to the in-
circle and the three ex-circles.
These relations are useful and important.
137. Theorem LXXVI. — There is a regular n-side.
Proof. For the angle is a continuous magnitude (why?) ;
hence there are angles of all sizes from zero to a round
116
GEOMETRY.
[Th. LXXVI.
angle ; hence there is an angle, the - part of a round angle,
such that, taken 71 times in addition, the sum will be a round
angle. Suppose such an angle drawn, whether or not we
can actually draw it, and suppose ;/ such angles placed con-
secutively around any point O, so as to make a round angle.
In other words, suppose n half-rays drawn cutting the
round angle about O into n equal angles. Draw a circle
about (9, with (Fig. 94) any radius, and draw the ;/ chords
Fig. 94.
subtending the n equal central angles. These chords are
all equal (why?), and subtend equal arcs, and they form an
?z-side. Moreover, the angle between two consecutive sides
Th. LXXVIIL] the circle. 117
is constant in size, because it stands on the part of
n
the circle. Hence the //-side is both equilateral and equian-
gular ; that is, it is regular, q. e. d.
Corollary. The inner angle of a regular w-side is the
part of a straight angle.
(^■)
Find the value in degrees of the inner angles of the first
ten regular «-sides.
N.B. The foregoing demonstration merely settles the
question of the existence or lo^^ical possibility of the regular
w-side. The problem of actually drawing such a figure is
one of the most intricate in all mathematics, and has been
solved only for certain very special classes of values of n.
Hut in order to discover the properties of the figure, it is by
no means necessary to be able to draw it accurately. It is
only since 1864 that we have known how to draw a straight
line or ray exactly.
137a. Theorem LXXVII. — The vertices of a regular
n-side are encyclic (Fig. 94).
Proof. Through any three vertices, as A, B, C, of a regular
«-side, draw a circle K ; about C with radius CB draw
another circle. The fourth vertex D must lie on this circle
(why?). If it lie on the circle A', then the angle BCZ> =
angle ABC, as is the case in the regular «-side. Neither
can it lie off of X, as at D' or Z>", because then the angle
BCD' or BCD" would not equal angle BCD (why?), and
hence would not equal angle ABC. Hence the next vertex
must lie on the same circle K, and so on all around, q. e. d.
138. Theorem LXXVIII. — T/te sides of a regular n-side
are pericyclic (that is, they all touch a circle).
118
GEOMETRY.
[Th. LXXIX.
Proof. For, on drawing the radii of the circumcircle K
(Fig. 95) to the vertices, we get n congruent symmetric A
A
O Fig. 95.
(why?). The altitudes of all are the same (why?) ; with
this common altitude as radius draw another circle, K\
about the same centre. It will touch each of the sides
(why?). Q. E. D.
Corollary. The points of touch of the sides of the regular
circumscribed ;/-side are mid-points of the sides.
139. Theorem LXXIX. — The points of touch of a regular
circujnscj'ibed 11- side are the vertices of a regular inscribed
n-side.
Proof. Connect the points of touch consecutively. Then
the A so formed are all congruent (why?) ; hence the
joining chords are equal ; hence the arcs are equal ;
hence the Theorem, q. e. d.
CIRCLE AS ENVELOPE, 119
THE CIRCLE AS ENVELOPE.
*i4oa. Thus far we have regarded the circle from various
points of view ; from the most familiar it was seen to be the
locus of a point in a plane at a fixed distance from a fixed
point. An almost equally important conception of the curve
treats it not as the locus of a point, but as the envelope of a
ray. If the point P moves in the plane always equidistant
from O, then its locus is the circle, on which it may always
be found ; also, if the ray R moves about in the plane always
equidistant from O^ then its etwelope is the circle, on which
it may always be found, on which it lies, which it continually
touches. The point traces the circle, the ray envelops the
circle, which is accordingly called the envelope (/>. the
enveloped curve — French enveloppee) of the ray. In higher
mathematics the notion of the ray, instead of the point, as
the determining element in the nature of a curve, attains
more and more significance. In this text we are confined
to the circle — the envelope of a ray in a plane, at a fixed
distance from a fixed point.
*i40b. It is not only rays, however, that may envelop a
curve ; but circles, and in fact any other curves. Thus, let
the student draw a system of equal circles, having their
centres on another circle ; the envelope will at once be seen
to be a pair of concentric circles. Let him also find the
envelope of a system of circles equal and with centres on a
given ray. In general, let him find the envelope of a circle
whose centre moves on any given curve. Lastly, let him
draw a large number of circles all of which pass through a
fixed point, while their centres all lie on a fixed circle, and
let him observe what curve they shadow forth as envelope.
Show that as the pole of a chord (or ray) traces a circle,
120 GEOMETRY.
the chord itself envelops a concentric circle, and con-
versely.
Show that tangents from two points on a centre ray form
a kite, and conversely. Also the chords of contact are
parallel, and conversely.
O is the centre of a circle, P any point without it. Show
how to find the point of touch of the tangents from /*, by
drawing a circle about O through P and a tangent where
OP cuts the given circle.
CONSTRUCTIONS.
140. Hitherto, in our reasoning about concepts, figures
have not been at all necessary, though exceedingly useful in
making sharp and precise our imagination of the relations
under consideration, in furnishing sensible examples of the
highly general notions that we dealt with. The conclusions
reached thus far all he wrapt up in axioms and in our defi-
nitions of point, ray, and circle, and our work has been one
of explication only ; we have merely brought them forth to
light. Our demonstrations have not presumed ability to draw
accurately, and would remain unshaken if we could not draw
at all. Nevertheless, for many practical purposes, it is ex-
tremely important and even indispensable that we actually
make the constructions and draw the figures that thus far we
have merely supposed made and drawn.
141. What is meant by drawing a ray, circle, or any line?
Any mark, whether of ink or chalk, though a solid, may be
treated as a line by abstraction. Only its length, not its
width nor thickness, concerns us. How to make not just
any mark, but some particular mark called for, is our prob-
lem (7rpoft\7)ixa = anything thrown forward as a task), and
CONS TR UC TIONS. 1 2 1
its solution consists accordingly of two parts, the logical and
the mechanical. The first is accomplished by fixing exactly
in thought the position of all the geometric elements (points,
rays, circles) in question ; the second, by making marks that
by abstraction may be treated as these elements. Now, a
point is fixed as the join of two rays, a ray as the join of two
points (by what axiom?) ; a circle is fixed or determined by
its centre and radius (why?), or by three points on it (why?).
Accordingly, when we know two rays through a point, or two
points on a ray, or centre and radius, or three points of a
circle, we know the point, or ray, or circle completely. The
logical part of our work is finished, then, when we determine
every point as the join of two known rays, every ray as the
join of two known points, every circle as drawn through three
known points or about a known centre with a known radius.
The mechanical part of the solution requires us to put and
keep a point in motion along a circle or a ray. Circular
motion is brought about by the compasses already described
(Art. 114), of which the shape is arbitrary, the necessary
parts being merely a fixed point rigidly connected in any way
with a movable point. But in the ruler one edge is supposed
made straight to begin with, so that a pencil point gliding
along it may trace a straight mark. Hence the nse of the
ruler is really illogical, since it assumes the problem of draw-
ing a ray or straight line as already solved in constructing
the straight edge. To say that, in order to draw a straight
liney we must take a straight edge and pass a pencil point
along it, is no better logically than to say that, in order to
draw a circular line, we must take a circular edge and pass
a pencil point along it. The question at once arises. How
make the edge straight or circular in the first place ? It was
not until 1864 that Peaucellier won, though he did not at
once receive, the Montyon prize from the French Academy
122 GEOMETRY.
by solving the thousand-year-old problem of imparting rec-
tilinear motion to a point without guiding edge of any kind
(Page ooo). But, though the ruler is logically valueless, it is
practically invaluable, even after the great discovery of Peau-
cellier. Its edge being assumed as straight and of any
desired length, and a pair of compasses of adjustable size
being given, we now make the following Postulates :
I. About any point may be drawn a circle of any radius.
II. Through any tivo points may be drawn a ray (more
strictly, a tract of any required length) .
Corollary. On any ray from any point on it we may lay
off a tract of any required length.
These are the only instruments used or postulates assumed
in the constructions of Elementary Geometry.
142. The fundamental relations of rays to each other are
two : Normality and Parallelism. Hence
Problem I. — To draw a ray normal to a given ray. Since
there are many rays normal to a given ray, to make the
problem definite we insert the limiting condition, through a
given point. Two cases then arise :
A. When the given point is on the given ray. All we can
do is to draw a circle about the point P. It cuts the ray at
two points, A and A\ symmetric as to P. Hence the mid-
normal of AA^ is the normal sought. Hence any point on
this normal lies on two circles of equal radius about A and
A\ Hence (Fig. 96)
Solution. From the given point P lay off on the given
ray two equal tracts PA, PA\ About A and A^ draw two
equal circles. Through their points of intersection draw
their common chord. It is the normal sought.
CONS 7'A' re IVOiVS.
123
Proof. For it is the mid -normal of A A', since it has two
of its i)oints equidistant from // and A', and P is the mid-
point of A A'.
Fig. 96.
Query : What radius shall we take for the circles about
A and^'?
B. IV/irn the given point is not on the given ray. All we
can do is to draw a circle about the given point P. Let it
cut the ray at A and A\ Then the mid-normal to AA^ is
the normal required (why?). Hence (Fig. 97)
Solution. Determine the points A, A' on the ray by
a circle about the given point F; then proceed as in the
first case (A).
124
GEOMETRY.
Proof. For the mid-normal of AA goes through P
(why?).
P
\
A'
"Ta
Fig. 97.
Query : What radius shall we take for the circle about P ?
143. Problem II. — To draw a parallel to a given ray.
Since there are many parallels to every ray, to make the
problem definite we must insert the limiting condition,
through a given point ; then it becomes perfectly definite
(why?). Manifestly the point must be not on the ray
(why?). We now reflect that a transversal makes equal
corresponding angles with parallels, and we have just learned
to draw a normal transversal. Hence (Fig. 98)
Solution. Through the point draw a normal to the ray ;
through the same point draw a normal to this normal. It
will be the parallel required.
Proof. For it goes through the point and is parallel to the
ray (why?).
These two problems have been discussed at such length
as being the hinges on which nearly all others turn. At
CONSTRUCTIONS. 125
the end of a problem is sometimes written q. e. f. = quod
erat facicTnium — ivhich ivas to he done, and translates the
Euclidean h-ntp iBei irpa^ai.
A'\ ; ^A
Fir,. 98.
144. Problem III. — To bisect a given tracts or to draw
the mid-normal to a given tract , AB,
Proceed as in Problem I.
145. Problem IV. — To bisect a given angle.
Solution. About the vertex draw any circle cutting the
arms at A and A\ and draw the mid-normal of AA\ It is
the mid-ray sought (why?).
Corollary. Show how to bisect any circular arc AB,
146. Problem V. — To bisect the angle between two rays
whose Join is not given (Fig. 99).
126
GEOMETRY.
We reflect that the join AA^ of two corresponding points
on the rays makes equal angles with the two rays that form
the angle. Hence
Solution. From any point P oi L draw the normal to it,
cutting M at Q. From Q draw the normal to M. Bisect
Fig. 99.
the angle at Q between these two normals by the mid-tract
QR. Draw the mid-normal of QR. It is the mid-ray
sought (why?).
147. Problem VI. — To multisect a given tract AB (Fig.
100).
L
CONSTRUCTIONS,
127
Solution. Through either end of the tract, as A, draw any
ray, and lay off on it from A successively n equal tracts, L
being the end of the last. Draw BL. Through the ends
of the equal tracts draw parallels to BL. They cut AB
into n equal parts (why?).
148. Problem VII. — To draw an angle of given size^ i.e.
equal to a given angle (Fig. loi).
Solution. At any point A of either arm of the given angle
O erect a normal to OA cutting the other arm at B. From
any point O on any other ray lay off O' A* = OA^ and normal
to the ray erect A^B' = AB and draw 0^B\ Then angle O*
= angle O (why?).
When does this construction fail ? How proceed then ?
149. Problem VIII. — To draw a tiiu-t cf given length
subtending a given angle and parallel to a given ay.
Data: O the given angle, L the ray, a the length (Fig.
102).
Solution. Through any point P of either arm of the angle
draw a parallel to the ray, and lay off on it towards the other
128
GEOMETKY.
arm a tract PA of the given length a. Through A draw a
parallel to OP, cutting the other angle arm at Q ; through Q
draw a parallel to PA meeting (9/* at P. QR is the subtense
sought (why?).
Fig. I02.
150. Problem IX. — To construct a A :
A. When altertiate parts (three sides or three angles) are
given.
Solution. About the ends of one side AB, with the other
sides for radii, draw circles meeting in C. Then ABC is
the A sought (why?) (Fig. 103).
Fig. 103.
How many such A may be drawn on the same base AB'>
How are they related? When is the solution impossible?
CONSTH UCT/OiVS.
129
When the angles are given, apply the construction of Problem
VII. How many solutions arc possible ? What kind of A ?
B. When three eonsecutive parts (two sides and included
angle or two angles and included side) are given.
Solution. Ai)[)ly the construction in Problem VII.
C. When opposite parts (two angles and an opposite side
or two sides and an opposite angle) are given.
Solution. Apply the constniction in Problem VII. When
is the construction ambiguous ?
D. When two sides and the altitude to the third side are
given.
Solution. Through one end of the altitude draw a normal
to it for the base ; about the other end C as centre, with the
sides as radii, draw circles cutting the base at A and A\
B and B' ; then ACB or A'CB' is the A required. Why?
E. When two sides and the. medial of the third side are
given.
A
Fig. 104.
130 GEOMETRY.
\i SA be the medial of BC, and SA^ be symmetric with
(Fig. 104) SA as to S, then ABA'C is a parallelogram
(why?) ; hence
Solution. Take a tract the double of the medial. About
its ends as centres with the sides as radii draw circles and
then complete the construction. How many A fulfilling the
conditions are possible ? How are they related ?
F. When the three medials are given (Fig. 105).
Solution. Remember that the medials trisect each other ;
construct the A OBC according to (E), and draw OA
counter to OM and twice as long.
151. Problfem X. — To construct an angle of given size
and subtended by a given tract.
Data : O the given angle, AB the given tract (Fig. 106).
Solution. Construct the angle BAD of given size (how?),
draw the mid- normal of AB, meeting AD at P) also the
normal to AD at A, meeting the mid-normal at S. About
CONS TR UC TIONS.
131
^S" as a centre with radius SA draw a circle ; it touches AD
at A (why?). The vertex V of the required angle may be
anywhere on the arc A VB or on its symmetric A V^B (why ?) .
p
/
— "^
"^
\
V
V
Fio. io6.
152. Problem XI. — To draw a circle tangent to a given
ray.
Solution. About any point S with a radius equal to the
distance of S from the ray, Z, draw a circle ; it will be a
circle required (why?). If the circle must touch the ray L
at a given point Py then S must be taken on the normal to
Z through P (why ?) . If, besides, the circle must go through
a given point Q, then S must also be on the mid-normal of
y^^ (why?). Hence the construction.
132 ■ GEOMETRY.
153. Problem XII. — To draw a circle touching two given
rays.
The centre may be anywhere on either mid-ray (why ?) .
If now the circle is to touch a third given ray, the centre
must be also on another mid-ray ; that is, it must be the
intersection of two mid-rays of the three angles of the three
rays. There are four such intersections — what are they ?
Complete the construction. See Fig. 60.
154. Problem XIII. — To draw a circle through two
points.
The centre 6" may be anywhere on the mid-normal of the
tract AB between the points (why?), the radius is — what?
If now the circle is to pass through a third point C, then *S
must also be on the mid-normal oi BC and CA (why?).
There is one, and only one, such point (why ?) ; complete
the construction. When is the construction impossible ?
155. Problem XIV. — To draw a circle through two given
points and tangent to a given ray ; or, tangent to two given
rays and through a given point.
This double problem is mentioned here because it must
naturally present itself to the mind of the student ; but the
solution involves deeper relations than we have yet explored.
See Art. 000.
Several of the foregoing problems were indefinite, admit-
ting any number of solutions : these latter taken all together
form a system or family. Problems concerning parallelo-
grams and other 4-sides may often be solved on cutting the
4-side into two A.
156. Problem XV. — To inscribe a regular 4-side (square)
in a circle (Fig, 107).
CONSTRUCTIONS, 133
Solution. Join consecutively the ends of two conjugate
diameters. The 4-side formed is inscribed (why?) and is
a square (why?).
Fig. 107.
157. Problem XVI. — To inscribe a regular 6-side in a
circle.
Solution. Apply the radius six times consecutively as a
chord to the circle (Fig. 108). The figure formed will
be the regular 6-side (why?).
Fig. 108.
N.B. This seems to have been one of the first geometric
problems ever solved. The Babylonians discovered that
six radii thus applied would compass the circle, and having
134
GEOMETRY.
already divided the circle into 360 steps, they accordingly
divided this number by 6 and thus obtained 60 as the
basis of the famous sexagesimal notation, which long domi-
nated mathematics and still maintains its authority un-
diminished in astronomy and chronometry.
In more difficult problems it is often advisable, or in fact
necessary, to suppose the problem solved, the construction
made, and investigate the relations thus brought to light.
Then the facts thus discovered may be used regressively
in making the construction required. This method is illus-
trated in the following :
158. Problem XVII. — To draw a square with each of
its sides through a given point.
Let A^ B, C, D, be the four given points, and suppose
(Fig. 109) PQRS to be the square properly drawn. Draw
p
c
\
"g
\ /
?K/
s
7
i R~'
1
1
Fig. 109.
AB, cutting a side of the square, and through B draw BE
parallel to the side cut. Through a third point C draw a
normal to AB, meeting QR in F. Also draw FG parallel
to PQ. Then the A ABE and CFG are congruent (why ?) .
cuA\^/A'UC7V0jVS.
135
Hence we discover that CF= AB, This fact is the key to
the
Solution. Join two points A and B ; from a third, C, lay
off CF equal and normal to AB, The join of Z>, the fourth
point, and /'is one side of the square in position (why?).
Let the student complete the construction and show that
four squares are possible.
159. Problem XVIII. — To trisect a given angle.
Suppose the problem solved and the ray OT to make
^ TOB = 2TOA (Fig. no).
Fig. I 10.
From any point A of the one end of the angle draw a
parallel and a normal to the other end ; also draw to the
trisector a tract AS— OA. Then the following relations are
evident :
^ AOS^ ^ ASO=:^ SAT-\- ^ STA ;
but :fAOS=z2^ TOB = 2 STA ;
hence ^ STA ^"^ SAT, and ST= SA.
136 GEOMETRY.
Again, ^ SAR = ^ SRA, being complements of equal
angles ; hence SA = SR, TR = 2 OA. Hence
Solution. From any point A of either arm of the given
angle draw a parallel and a normal to the other arm ; then,
with one point of a straight-edge fixed at the vertex O, turn
the edge until the intercept between the normal and the
parallel equals 2 OA. But to do this we need a graduated
edge, or a sliding length 2 OA on the edge itself. Accord-
ingly, this construction, while simple, useful, and interesting,
is not elementary geometric in the sense already defined.
To discover such a solution for this famous problem, has up
to this time baffled the utmost efforts of mathematicians.
EXERCISES II.
1. State and prove the reciprocals of Exercises 9, 10,
II, page 71.
2. Find a point on a given ray, the sum of whose dis-
tances from two fixed points is a minimum.
3. The same as the foregoing, difference supplacing
sum.
4. A and A\ B and B\ C and C\ are symmetric as to
MN. Show that AABC=AA'B' C.
5. The inner and outer mid-rays of the basal angles of
a symmetric A form a kite.
6. The inner mid-rays of the angles of a trapezium form
a kite with two right angles.
7. The joins, of the mid-points of the parallel sides of an
anti-parallelogram, with the opposite vertices, form a kite.
8. The mid-rays of the angles at the ends of the trans-
verse axis of a kite cut the sides in the vertices of an anti-
parallelogram.
EXERCISES II. 137
9. How must a billiard ball be stnirk so as to rebound
from the four sides of a table and return through its original
place ?
10. Trace a ray of light from a focus P^ to another given
point Qi reflected from a convex polygonal mirror.
11. A ray of light falls on a mirror J/, is reflected along
5 to a second mirror M\ is thence reflected along T. Re-
membering that the angle of incidence equals the angle of
reflection, show that the angle between the original ray R
and its last reflection T is twice the angle between the
mirrors (angle RT= 2 angle MM'). On this theorem is
grounded the use of the sextant.
12. Two mirrors stand on a plane and form an inner
angle of 60° ; a luminous point P\s on the mid-ray of this
angle (or anywhere within it) ; how many images of /*are
formed? How are they placed? What if the angle of the
mirrors be i/« of a round angle?
This theorem is beautifully illustrated in the kaleidoscope.
13. A regular «-side has n axes of symmetry concurring
in the centre of the «-side, which centre is equidistant from
the sides of the //-side.
14. How do these axes lie when n is even? when n is
odd? Show that if // be even, the centre is a centre of
symmetry.
1 5 . The half- rays from centre to vertices of a regular «-side
form a regular pencil of n half-rays, and those from the cen-
tre normal to the sides, another regular pencil ; also the half-
rays of each p>encil bisect the angles of the other.
16. In a figure with two rectangular axes of symmetry
each point, with three others, determines a rectangle, and
each ray, with three others, a rhombus.
138 GEOMETRY.
1 7. Find the axes of symmetry of two given tracts.
18. A regular A, along with the figure symmetric with it
as to its centre, determines a regular 6-angle (6-pointed
star) .
19. Two congruent squares, the diagonals of one lying on
the mid-parallels of the other, form a regular 8-angle ; also
find the lengths of the intercepts at the corners.
20. The outer angle of a regular «-side is in times the
outer angle of a regular ;//«-side.
EXERCISES III.
1. A circle with its centre on the mid-ray of an angle
makes equal intercepts on its arms.
2. Tangents parallel to a chord bisect the subtended
arcs, and conversely.
3. Tangents at the end of a diameter are parallel.
4. A and B are ends of a diameter, C and D any other
two points of a circle ; iS" is on the diameter, and angle
AED — 2 angle CAD ; find the possible positions of £.
5 . From n points are drawn 2 n equal tangent-lengths to
a circle ; where do the points lie ?
6. In a circumscribed hexagon, or any circumscribed
2 n-side, the sums of the alternate sides are equal.
7. If the vertices of a circumscribed quadrangle, hexagon,
or any 2 «-side, be joined with the centre of the circle, the
sums of the alternate central angles will be equal.
8. The sums of the alternate angles of an encyclic 2n-
side are equal, namely, each sum is (n — i) straight angles.
9. The joins of the ends of two parallel chords are sym-
metric as to the conjugate diameter of the chords.
EXERCISES III. 139
10. A centre ray is cut by two parallel tmgents. Show
that the intercepts between tangent and circle are equal.
11. Normals to a chord from the ends of a diameter
make, with the circle, equal intercepts on the chord.
12. The joins of the ends of two diameters are parallel in
pairs, and form a rectangle, and meet any two parallel tan-
gents in points symmetric in pairs as to the centre.
13. The joins of the ends of two parallel chords meet the
tangents normal to the chords in points whose other joins
are parallel to the chords.
14. A chord AB is prolonged to C, making ^C= radius,
and the centre ray CD is drawn ; show that one intercepted
arc is thrice the other.
15. The intercepts, on a secant, of two concentric circles
are equal.
16. A chord through the point of touch of two tangent
circles subtends equal central angles in the circles.
1 7. Two rays through the point of touch of two tangent
circles intercept arcs in the circles whose chords are parallel.
18. The transverse joins of the ends of parallel diameters
in two tangent circles go through the point of tangence.
19. Four circles touch each other outerly in pairs: ist
and 2d, 2d and 3d, 3d and 4th, 4th and ist; show that the
points of touch are encyclic.
20. Show that three circles drawn on three diameters OA^
OBf OC intersect on the sides of the A ABC.
21. Find the shortest and the longest chord through a
point within a circle.
22. In a convex 4-side the sum of the diagonals is greater
than the sum of two opposite sides, less than the sum of all
the sides, and greater than half the sum of the sides.
140 GEOMETRY.
23. Three half-rays trisect the round angle O ; on each
is taken any point, as A, B, C. Find a point M such that
the sum MA -f MB + MC is the least possible (a minimum).
24. Two tangents to a circle meet at a point distant twice
the radius, from the centre ; what angles do they form?
25. The intercept of two circles on a ray through one of
their common points subtends a constant angle at the other.
26. What is the envelope of equal chords of a circle?
27. Two movable tangents to a circle intersect under con-
stant angles; find the envelope of the mid-rays of these
angles.
28. The vertex Fof a revolving right angle is fixed mid-
way between two parallels, and its arms cut the parallels at
A and B ; find the envelope of AB.
29. From a fixed point P a normal FN is drawn to a
movable tangent 7" of a circle, and through the mid-point M
of /W there is drawn a parallel to T; find its envelope.
30. The vertices of a A are Fj, Vo, V^ ; the mid-points of
its sides are J/j, M^, M-^ ; the feet of its altitudes are
Ai, A2, A3 ; the inner bisectors of its angles meet the oppo-
site sides at B^, B.,, B^; and the outer bisectors at
B'l, B'2, B\ ] its centroid is C, its in-centre is /, its circum-
centre is S] its angles are cti, «2, «3, and their complements
are a\, a\, w'g. Express through these six angles the
angles between : (i) Fi^iandFjFa; (2) AiA^Sind V2V3;
(3) AiAoSind ViAi; (4) A1A2 Sind A^A^; (5) M1A2 B.nd
V,V,; (6) J/i^sand V.V,; (7) M,A2SindM,A,; (8) A,M2
and AiM^; (9) A1M2 and AiA^; (10) 6^1 and SV2',
(11) SV, and V.Vo; (12) SF, and V2A2; (13) /Fi and
IF.; (14) /Fiand F^^., ; (15) F^'i and FoB'..
EXERCISES IV. 141
31. Find the locus of the mid-points of chords through a
fixed point upon, within, or without a fixed circle.
32. Find the locus of the mid-points of the intercepts of
a secant between a fixed point and a fixed circle.
33. As the ends of a ruler slide along two grooves normal
to each other, how does its mid-point move?
34. Two equal hoops move along grooves normal to each
other and touch each other ; how does the point of touch
move?
35. Orthocentre O, centroid C, circum-centre S, and cen-
tre F oi Feuerbach's (9-point) circle, of a A are collinear,
and (9C= 2 CS (Euler), OF=FS,
36. Two parallel tangents meet two diameters of a circle
at the vertices of a parallelogram concentric with the circle.
37. The inner mid-rays of the angles of a 4-side form an
encyclic 4-side.
38. The outer mid-rays of the angles of a 4-side form an
encyclic 4-side. How are the 4-sides of 37 and 2>^ related?
39. The circum-centres of the four A into which a 4-side
is cut by its diagonals are the vertices of a parallelogram.
40. The circum-centres of the two pairs of A, into which
a 4-side is cut by its diagonals in turn, are how related to
each other and to the centres in 39 ?
EXERCISES IV.
1. Construct a square, knowing
{a) Its side; or (h) its diagonal.
2. Construct a rectangle, knowing
(a) Two sides; or {b) a side and a diagonal; or (c)
either a side or a diagonal and the angle of either with the
other; or (//) a diagonal and its angles with the other
diagonal.
142 GEOMETRY.
3. Construct a parallelogram, knowing
{a) Two sides and one angle ; or {b) a side, a diagonal,
and the included angle ; or (<r) two sides and the opposite
diagonal; or {d) two sides and the included diagonal; or
{e) two diagonals and a side; or (/) two diagonals and
their angles with each other.
4. Construct an anti-parallelogram, knowing
{a) Its parallel sides and the distance between them;
{h) two adjacent sides and their included angle ; (r) two
adjacent sides and the angle between the non-parallel sides ;
id) a diagonal and two adjacent sides ; {e) a diagonal, a
side, and the included angle.
5. Construct a kite, knowing
{a) Two sides and an axis; {b) two sides and the in-
cluded angle ; (r) .a side and the axes.
6. Construct the rays equidistant from three given
points.
7. Draw a ray through a given point equally sloped to
two given rays.
8. A square has one vertex at a given point, and two
others on two given parallel rays ; draw it.
9. Hypotenuse and sum of sides of a right A are given ;
draw it.
10. Construct a regular 2**-side, and a regular 3.2''-side.
11. Find the centre of a given circular arc.
12. Trisect a right angle.
13. Two points, A and B, of a ray are given; find any
number of points of the ray without drawing it, and without
opening the compasses more than AB.
14. Find a point on a given ray or given circle that has
a given tangent-length with respect to a given circle.
15. Through a given point draw a secant on which a
given circle shall make a given intercept.
EXERCISES IV, 143
1 6. Draw four common tangents to two given circles.
17. Draw a ray touching a given circle and e(iuidistant
from two given points.
18. Draw a ray on which two given < ir( Ics shall make
two given intercepts.
19. With three given radii draw three circles, each touch-
ing the other two.
20. Draw a circle toucliing the radii and the arc of a
given sector.
21. Draw a circle touching two given equal intersecting
circles and their centre ray.
22. On the central intercept of two equal intersecting
circles as diameter draw a circle ; then draw a circle touch-
ing the three circles.
23. Three equal circles touch each other outerly ; draw
a circle touching the three.
24. Find a point from which two given apposed tracts
appear to be equal.
25. Through two given points of a given circle draw a
circle that shall cut a third circle orthogonally.
26. Construct a A, knowing
(rt) The feet of the altituOv^s ; {b) the foot of one altitude
and the mid-points of the other two sides ; {c) the three
ex-centres ; (^/) two. ex-centres and the in-centre.
27. Draw through a given point a ray that shall form with
the sides of a given angle a A of given perimeter. Hint :
Use the properties of ex-circles.
28. Draw a 5 -side, knowing the mid-points of the sides.
29. On a tract AB there is drawn a regular 3-side ; draw
on it a regular 6-side. Generalize the problem, changing
3 into «, 6 into 2«, and solve it.
30. Given a regular //-side ; draw a regular 2«-side having
liie original /; vertices for alternate vertices. Do not use
the circumcircle.
ELEMENTARY SYNTHETIC GEOMETRY
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POINT, LINE, AND CIRCLE IN THE PLANE.
By Nathan F. Dupuis, M.A., FJtC.S., Professor of Mathematics In
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ELEMENTARY ALGEBRA.
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