Book III.

Definitions.

I.

Equal circles are thoſe whoſe diameters are equal.

II.

A right line is said to touch a circle when it meets the circle, and being produced does not cut it.

III.

Circles are ſaid to touch one another which meet but do not cut one another.

IV.

Right lines are ſaid to be equally diſtant from the centre of a circle when the perpendiculars drawn to them from the centre are equal.

V.

And the ſtraight line on which the greater perpendicular falls is ſaid to be farther from the centre.

VI.

A ſegment of a circle is the figure contained by a ſtraight line and the part of the circumference it cuts off.

VII.

An angle in a ſegment is the angle contained by two ſtraight lines drawn from any point in the circumference of the ſegment to the extremities of the ſtraight line which is the baſe of the ſegment.

VIII.

An angle is ſaid to ſtand on the part of the circumference, or the arch, intercepted between the right lines that contain the angle.

IX.

A ſector of a circle is the figure contained by two radii and the arch between them.

X.

Similar ſegments of circles are thoſe which contain equal angles.

Circles which have the ſame centre are called concentric circles.

Proposition I. Problem.

To find the centre of a given circle .

Draw within the circle any ſtraight line , make = , draw ; biſect , and the point of biſection is the centre.

For, if it be poſſible, let any other point as the point of concourſe of , and be the centre.

Becauſe in and

= (hyp. and B. 1, def. 15.) = (conſt.) and common, = (B. 1, pr. 8.), and are therefore right angles; but = (conſt.) = (ax. 11.) which is abſurd; therefore the aſſumed point is not the centre of the circle; and in the ſame manner it can be proved that no other point which is not on is the centre, therefore the centre is in , and therefore the point where is biſected is the centre.

Q. E. D.

Proposition II. Theorem.

A ſtraight line () joining two points in the circumference of a circle , lies wholly within the circle.

Find the centre of (B. 3. pr. 1.);
from the centre draw to any point in ,
meeting the circumference from the centre;
draw and .

Then = (B. 1. pr. 5.)
but > or > (B. 1. pr. 16.)
> (B. 1. pr. 19.)
but = ,
> ;
< ;
every point in lies within the circle.

Q. E. D.

Proposition III. Theorem.

If a ſtraight line () drawn through the centre of a circle biſects a chord ( ) which does not paſs through the centre, it is perpendicular to it; or, if perpendicular to it, it biſects it.

Draw and to the centre of the circle.

In and
= , common, and
= = (B. 1. pr. 8.)
and (B. 1. def. 10.)
Again let

Then in and
= (B. 1. pr. 5.)
= (hyp.)
and =
= (B. 1. pr. 26.)
and biſects .

Q. E. D.

Proposition IV. Theorem.

If in a circle two ſtraight lines cut one another, which do not both paſs through the centre, they do not biſect one another.

If one of the lines paſs through the centre, it is evident that it cannot be biſected by the other, which does not paſs through the centre.

But if neither the lines or paſs through the centre, draw from the centre to their interſection.

If be biſected, to it (B. 3. pr. 3.)
= and if be
biſected, (B. 3. pr. 3.)
=
and = ; a part
equal to the whole, which is abſurd:
and
do not biſect one another.

Q. E. D.

Proposition V. Theorem.

If two circles interſect, they have not the ſame centre.

Suppoſe it poſſible that two interſecting circles have a common centre; from ſuch ſuppoſed centre draw to the interſecting point, and meeting the circumferences of the circles.

Then = (B. 1. def. 15.)
and = (B. 1. def. 15.)
= ; a part
equal to the whole, which is abſurd:
circles ſuppoſed to interſect in any point cannot have the ſame centre.

Q. E. D.

Proposition VI. Theorem.

If two circles touch one another internally they have not the ſame centre.

For, if it be poſſible, let both circles have the ſame centre; from ſuch a ſuppoſed centre draw cutting both circles, and to the point of contact.

Then = (B. 1. def. 15.)
and = (B. 1. def. 15.)
= ; a part
equal to the whole, which is abſurd;

therefore the aſſumed point is not the centre of both circles; and in the ſame manner it can be demonſtrated that no other point is.

Q. E. D.

Proposition VII. Theorem.

If from any point within a circle which is not the centre, lines { are drawn to the circumference; the greateſt of thoſe lines is that ( ) which paſſes through the centre, and the leaſt is the remaining part () of the diameter.

Of the others, that () which is nearer to the line paſſing through the centre, is greater than that () which is more remote.

Fig. 2. The two lines ( and ) which make equal angles with that paſſing through the centre, on oppoſite ſides of it, are equal to each other; and there cannot be drawn a third line equal to them, from the ſame point to the circumference.

Figure I.

To the centre of the circle draw and ; then = (B. 1. def. 15.) = + > (B. 1. pr. 20.) in like manner may be ſhewn to be greater than , or any other line drawn from the ſame point to the circumference. Again, by (B. 1. pr. 20.) + > = + , take from both; > (ax.), and in like manner it may be ſhewn that is leſs than any other line drawn from the ſame point to the circumference. Again, in and , common, > , and =

> (B. 1. pr. 24.) and may in like manner be proved greater than any other line drawn from the ſame point to the circumference more remote from .

Figure II.

If = then = , if not take = draw , then in and , common, = and = = (B. 1. pr. 4.) = = a part equal to the whole, which is abſurd:

= ; and no other line is equal to drawn from the ſame point to the circumference; for if it were nearer to the one paſſing through the centre it would be greater, and if it were more remote it would be leſs.

Q. E. D.

Proposition VIII. Theorem.

The original text of this propoſition is here divided into three parts.

I.

If from a point without a circle, ſtraight lines { &c. } are drawn to the circumference; of thoſe falling upon the concave circumference the greateſt is that ( ) which paſſes through the centre, and the line () which is nearer the greateſt is greater than that () which is more remote.

Draw and to the centre.

Then, which paſſes through the centre, is greateſt; for ſince = , if be added to both, = + ; but > (B. 1. pr. 20.) is greater than any other line drawn from the ſame point to the concave circumference.

Again in and , = ,
and common, but > ,
> (B. 1. pr. 24.);
and in like manner may be ſhewn > than any other line more remote from .

II.

Of thoſe lines falling on the convex circumference the leaſt is that () which being produced would paſs through the centre, and the line which is nearer to the leaſt is leſs than that which is more remote.

For, ſince + > (B. 1. pr. 20.)
and = ,
> (ax. 5.)
And again, ſince + >
+ (B. 1. pr. 21.),
and = ,
< . And ſo of others.

III.

Alſo the lines making equal angles with that which paſſes through the centre are equal, whether falling on the concave or convex circumference; and no third line can be drawn equal to them from the ſame point to the circumference.

For if > , but making = ;
make = , and draw .
Then in and we have = ,
and common, and alſo = ,
= (B. 1. pr. 4.);
but = ;
= , which is abſurd.
is not = , nor to any part
of , is not > .
Neither is > , they are
= to each other.

And any other line drawn from the ſame point to the circumference muſt lie at the ſame ſide with one of theſe lines, and be more or leſs remote than it from the line paſſing through the centre, and cannot therefore be equal to it.

Q. E. D.

Proposition IX. Theorem.

If a point be taken within a circle , from which more than two equal ſtraight lines (, , ) can be drawn to the circumference, that point muſt be the centre of the circle.

For if it be ſuppoſed that the point in which more than two equal ſtraight lines meet is not the centre, ſome other point muſt be; join theſe two points by , and produce it both ways to the circumference.

Then ſince more than two equal ſtraight lines are drawn from a point which is not the centre, to the circumference, two of them at leaſt muſt lie at the ſame ſide of the diameter ; and ſince from a point , which is not the centre, ſtraight lines are drawn to the circumference; the greateſt is , which paſſes through the centre: and which is nearer to , > which is more remote (B. 3. pr. 8.); but = (hyp.) which is abſurd.

The ſame may be demonſtrated of any other point, different from , which muſt be the centre of the circle.

Q. E. D.

Proposition X. Theorem.

One circle cannot interſect another in more points than two.

For if it be poſſible, let it interſect in three points;
from the centre of draw ,
and to the points of interſection;
= =
(B. 1. def. 15.),
but as the circles interſect, they have not the ſame centre (B. 3. pr. 5.):
the aſſumed point is not the centre of , and

as , and are drawn from a point and not the centre, they are not equal (B. 3. prs. 7, 8); but it was ſhewn before that they were equal, in which is abſurd; the circles therefore do not interſect in three points.

Q. E. D.

Proposition XI. Theorem.

If two circles and touch one another internally, the right line joining their centres, being produced, ſhall paſs through a point of contact.

For if it be poſſible, let join their centres, and produce it both ways; from a point of contact draw to the centre of , and from the ſame point of contact draw to the centre of .

Becauſe in ; + >
(B. 1. pr. 20.),
and = as they are radii of
,
but + > ; take
away which is common,
and > ;
but = ,
becauſe they are radii of ,
and > a part greater than the whole, which is abſurd.

The centres are not therefore ſo placed, that a line joining them can paſs through any point but a point of contact.

Q. E. D.

Proposition XII. Theorem.

If two circles and touch one another externally, the ſtraight line joining their centres paſſes through the point of contact.

If it be poſſible, let join the centres, and not paſs through a point of contact; then from a point of contact draw and to the centres.

Becauſe + > (B. 1. pr. 20.),
and = (B. 1. def. 15.),
and = (B. 1. def. 15.),

+ > , a part greater than the whole, which is abſurd.

The centres are not therefore ſo placed, that the line joining them can paſs through any point but the point of contact.

Q. E. D.

Proposition XIII. Theorem.

One circle cannot touch another, either externally or internally in more points than one

Fig. 1. For if it be poſſible, let and touch one another internally in two points; draw joining their centres, and produce it until it paſs through one of the points of contact (B. 3. pr. 11.);

draw and ,
But = (B. 1. def. 15.),
= + ;
but = (B. 1. def. 15.),
and + = ; but
+ > (B. 1. pr. 20.),
which is abſurd.

Fig. 2. But if the points of contact be the extremities of the right line joining the centres, this ſtraight line muſt be biſected in two different points for the two centres; becauſe it is the diameter of both circle, which is abſurd.

Fig. 3. Next, if it be poſſible, let and touch externally in two points; draw joining the centres of the circles, and paſſing through one of the points of contact, and draw and .

= (B. 1. def. 15.);
and = (B. 1. def. 15.):
+ = ; but
+ > (B. 1. pr. 20.),
which is abſurd.

There is therefore no caſe in which two circles can touch one another in two points.

Q. E. D.

Proposition XIV. Theorem.

Equal ſtraight lines ( ) inſcribed in a circle are equally diſtant from the centre; and alſo, ſtraight lines equally diſtant from the centre are equal.

From the centre of draw
to and
, join and .

Then = half (B. 3. pr. 3.)
and = 1 / 2 (B. 3. pr. 3.)
ſince = (hyp.)
= ,
and = (B. 1. def. 15.)
2 = 2;
but ſince is a right angle
2 = 2 + 2 (B. 1. pr. 47.)
and 2 = 2 + 2 for the ſame reaſon,
2 + 2 = 2 + 2
2 = 2,
= .

Alſo, if the lines and be equally diſtant from the centre; that is to ſay, if the perpendiculars and be given equal then = .

For, as in the preceding caſe,
2 + 2 = 2 + 2;
but 2 = 2:

2 = 2, and the doubles of theſe
and are alſo equal.

Q. E. D.

Proposition XV. Theorem.

The diameter is the greateſt ſtraight line in a circle: and, of all others, that which is neareſt to the centre is greater than the more remote.

Figure I.

The diameter is > any line .
For draw and .
Then =
and = ,
+ =
but + > (B. 1. pr. 20.)
> .

Again, the line which is nearer the centre is greater than the one more remote.

Firſt, let the given lines be and , which are at the ſame ſide of the centre and do not interſect; draw { , , , . } In and ,
and = and ;
> ,
> (B. 1. pr. 24.)

Figure II.

Let the given lines be and which either are at different ſides of the centre, or interſect; from the centre draw and and , make = , and draw .

Since and are equally diſtant from
the centre, = (B. 3. pr. 14.);
but > (Pt. 1. B. 3. pr. 15.),
> .

Q. E. D.

Proposition XVI. Theorem.

The ſtraight line drawn from the extremity of the diameter of a circle perpendicular to it falls without the circle.

And if any ſtraight line be drawn from a point within that perpendicular to the point of contact, it cuts the circle.

Part I

If it be poſſible, let , which meets the circle again, be , and draw .

Then, becauſe = ,
= (B. 1. pr. 5.),
and each of these angles is acute (B. 1. pr. 17.)
but = (hyp.), which is abſurd, therefore
drawn does not meet
the circle again.

Part II.

Let be and let be drawn from a point between and the circle, which if it be poſſible, does not cut the circle.

Becauſe = ,
is an acute angle; ſuppoſe
, drawn from the centre of the circle, it muſt fall at the ſide of the acute angle.
which is ſuppoſed to be a right angle, is > ,
> ;
but = ,

and > , a part greater than the whole, which is abſurd. Therefore the point does not fall outside the circle, and therefore the ſtraight line cuts the circle.

Q. E. D.

Proposition XVII. Theorem.

To draw a tangent to a given circle from a given point, either in or outſide of its circumference.

If the given point be in the circumference, as at , it is plain that the ſtraight line the radius, will be the required tangent (B. 3. pr. 16.)

But if the given point be outſide of the circumference, draw
from it to the centre, cutting ; and
draw , deſcribe
then will be the tangent required.

For in and
= , common,
and = ,
(B. 1. pr. 4.) = = a right angle,
is a tangent to .

Q. E. D.

Proposition XVIII. Theorem.

If a right line be a tangent to a circle, the ſtraight line drawn from the centre to the point of contact, is perpendicular to it.

For if it be poſſible, let be ,
then becauſe = , is acute (B. 1. pr. 17.)
> (B. 1. pr. 19.);
but = ,
and > ,
a part greater than the whole, which is abſurd.

is not ; and in the ſame manner it can be demonſtrated, that no other line except is perpendicular to .

Q. E. D.

Proposition XIX. Theorem.

If a ſtraight line be a tangent to a circle, the ſtraight line , drawn perpendicular to it from a point of the contact, paſſes through the centre of the circle.

For, if it be poſſible, let the centre be without , and draw from the ſuppoſed centre to the point of contact.

Becauſe (B. 3. pr. 18.)

= , a right angle;
but = (hyp.), and = ,
a part equal to the whole, which is abſurd.

Therefore the aſſumed point is not the centre; and in the ſame manner it can be demonſtrated, that no other point without is the centre.

Q. E. D.

Proposition XX. Theorem.

The angle at the centre of a circle, is double the angle at the circumference, when they have the ſame part of the circumference for their baſe.

Figure I.

Let the centre of the circle be on
a ſide of .

Becauſe = ,
= (B. 1. pr. 5.).

But = + ,
= twice (B. 1. pr. 32).

Figure II.

Let the centre be within , the angle at the circumference; draw from the angular point through the centre of the circle;
then = , and = ,
becauſe of the equality of the ſides (B. 1. pr. 5).

Hence + + + = twice .
But = + , and
= + ,
= twice .

Figure III.

Let the centre be without and
draw , the diameter.

Becauſe = twice ; and
= twice (caſe 1.);
= twice .

Q. E. D.

Proposition XXI. Theorem.

The angles ( , ) in the ſame ſegment of a circle are equal.

Figure I.

Let the ſegment be greater than a ſemicircle, and draw and to the centre.

= twice or twice = (B. 3. pr. 20.);
= .

Figure II.

Let the ſegment be a ſemicircle, or leſs than a ſemicircle, draw the diameter, alſo draw .

= and = (caſe 1.)
= .

Q. E. D.

Proposition XXII. Theorem.

The oppoſite angles and , and of any quadrilateral figure inſcribed in a circle, are together equal to two right angles.

Draw and the diagonals; and becauſe angles in the ſame ſegment are equal = ,
and = ;
+ = + + =
two right angles (B. 1. pr. 32.). In like manner it may be ſhown that,
+ = .

Q. E. D.

Proposition XXIII. Theorem.

Upon the ſame ſtraight line and upon the ſame ſide of it, two ſimilar ſegments of circles cannot be constructed which do not coincide.

For if it be poſſible, let two ſimilar segments
and be conſtructed;
draw any right line cutting both the ſegments,
draw and .

Becauſe the ſegments are ſimilar,
= (B. 3. def. 10.),
but > (B. 1. pr. 16.)
which is abſurd: therefore no point in either of the ſegments falls without the other, and therefore the ſegments coincide.

Q. E. D.

Proposition XXIV. Theorem.

Similar ſegments and , of circles upon equal ſtraight lines ( and ) are each equal to the other.

For, if be ſo applied to ,
that may fall on , the extremities of
may be on the extremities and
at the ſame ſide as ;

becauſe = ,
muſt wholly coincide with ;
and the ſimilar ſegments being then upon the ſame ſtraight line at the ſame ſide of it, muſt alſo coincide (B. 3. pr. 23.), and are therefore equal.

Q. E. D.

Proposition XXV. Problem.

A ſegment of a circle being given, to deſcribe the circle of which it is the ſegment.

From any point in the ſegment draw and biſect them, and from the points of biſection

draw
and
where they meet is the centre of the circle.

Becauſe terminated in the circle is biſected perpendicularly by , it paſſes through the centre (B. 3. pr. 1.), likewiſe paſſes through the centre, therefore the centre is in the interſection of theſe perpendiculars.

Q. E. D.

Proposition XXVI. Theorem.

In equal circles and , the arcs , on which ſtand equal angles, whether at the centre or circumference, are equal.

Firſt, let = at the centre,
draw and .

Then ſince = ,
and have
= = = ,
and = ,
= (B. 1. pr. 4.).

But = (B. 3. pr. 20.);
and are ſimilar (B. 3. def. 10.);
they are alſo equal (B. 3. pr. 24.)

If therefore the equal ſegments be taken from the equal circles, the remaining ſegments will be equal;

hence = (ax. 3.);
= .

But if the given equal angles be at the circumference, it is evident that the angles at the centre, being double of thoſe at the circumference, are alſo equal, and therefore the arcs on which they ſtand are equal.

Q. E. D.

Proposition XXVII. Theorem.

In equal circles, an[d] the angles and which ſtand upon equal arches are equal, whether they be at the centres or at the circumferences.

For if it be poſſible, let one of them
be greater than the other
and make
=

= (B. 3. pr. 26.)
but = (hyp.)
= a part equal
to the whole, which is abſurd; neither angle is greater than the other, and they are equal.

Q. E. D.

Proposition XXVIII. Theorem.

In equal circles and , equal chords , cut off equal arches.

From the centres of the equal circles,
draw , and , ;
and becauſe =
, = ,
alſo = (hyp.)
=
= (B. 3. pr. 26.)
= (ax. 3.)

Q. E. D.

Proposition XXIX. Theorem.

In equal circles and the chords and which ſubtend equal arcs are equal.

If the equal arcs be ſemicircles the propoſition is evident. But if not,
let , , and ,
be drawn to the centres;
becauſe = (hyp.)
and = (B. 3. pr. 27.);
but and = and
= (B. 1. pr. 4.);
but theſe are the chords ſubtending the equal arcs.

Q. E. D.

Proposition XXX. Problem.

To biſect a given arc .

Draw ;
make = ,
draw , and it biſects the arc.
Draw and .
= (conſt.),
is common,
and = (conſt.)
= (B. 1. pr. 4.)
= (B. 3. pr. 28.),
and therefore the given arc is biſected.

Q. E. D.

Proposition XXXI. Theorem.

In a circle the angle in a ſemicircle is a right angle, the angle in a ſegment greater than a ſemicircle is acute, and the angle in a ſegment leſs than a ſemicircle is obtuſe.

Figure I.

The angle in a ſemicircle is a right angle.

Draw and
= and = (B. 1. pr. 5.)
+ = = the half of two
right angles = a right angle. (B. 1. pr. 32.)

Figure II.

The angle in a ſegment greater than a ſemicircle is acute.

Draw the diameter, and
= a right angle
is acute.

Figure III.

The angle in a ſegment leſs than ſemicircle is obtuſe.

Take in the oppoſite circumference any point, to which draw and .

Becauſe + = (B. 3. pr. 22.)
but < (part 2.),
is obtuſe.

Q. E. D.

Proposition XXXII. Theorem.

If a right line be a tangent to a circle and from the point of contact a right line be drawn cutting the circle, the angle made by this line with the tangent is equal to the angle in the alternate ſegment of the circle.

If the chord ſhould paſs through the centre, it is evident the angles are equal, for each of them is a right angle. (B. 3. prs. 16, 31.)

But if not, draw from the point of contact, it muſt paſs through the centre of the circle, (B. 3. pr. 19.)

= (B. 3. pr. 31.)
+ = = (B. 1. pr. 32.)
= (ax.).
Again = = + (B. 3. pr. 22.)

= , (ax.), which is the angle in the alternate ſegment.

Q. E. D.

Proposition XXXIII. Problem.

On a given ſtraight line to deſcribe a ſegment of a circle that ſhall contain an angle equal to a given angle , , .

If the given angle be a right angle, biſect the given line, and deſcribe a ſemicircle on it, this will evidently contain a right angle. (B. 3. pr. 31.)

If the given angle be acute or obtuſe, make with the given line, at its extremity,

= , draw and
make = , deſcribe
with or as radius, for they are equal.

is a tangent to (B. 3. pr. 16.)
divides the circle into two ſegments
capable of containing angles equal to
and which were made reſpectively equal
to and (B. 3. pr. 32.)

Q. E. D.

Proposition XXXIV. Problem.

To cut off from a given circle a ſegment which ſhall contain an angle equal to a given angle .

Draw (B. 3. pr. 17.), a tangent to the circle at any point; at the point of contact make

= the given angle;
and contains an angle = the given angle.

Becauſe is a tangent,
and cuts it, the
angle in = (B. 3. pr. 32.),
but = (conſt.)

Q. E. D.

Proposition XXXV. Theorem.

If two chords { } in a circle interſect each other, the rectangle contained by the ſegments of the one is equal to the rectangle contained by the ſegments of the other.

Figure I.

If the given right lines paſs through the centre, they are biſected in the point of interſection, hence the rectangles under their ſegments are the ſquares of their halves and are therefore equal.

Figure II.

Let paſs through the centre, and
not; draw and .
Then × = 2 2 (B. 2. pr. 6.),
or × = 2 2.
× = × (B. 2. pr. 5.).

Figure III.

Let neither of the given lines paſs through the centre, draw through their interſection a diameter ,

and × = × (Part. 2.),
alſo × = × (Part. 2.);
× = × .

Q. E. D.

Proposition XXXVI. Theorem.

If from a point without a circle two ſtraight lines be drawn to it, one of which is a tangent to the circle, and the other cuts it; the rectangle under the whole cutting line and the external ſegment is equal to the ſquare of the tangent .

Figure I.

Let paſs through the centre;
draw from the centre to the point of contact;
2 = 2 minus 2 (B. 1. pr. 47),
or 2 = 2 minus 2,
2 = × (B. 2. pr. 6).

Figure II.

If do not paſs through the centre,
draw and .

Then × = 2 minus 2
(B. 2. pr. 6), that is,
× = 2 minus 2,
× = 2 (B. 3. pr. 18).

Q. E. D.

Proposition XXXVII. Problem.

If from a point outſide a circle two ſtraight lines be drawn, the one cutting the circle, the other meeting it, and if the rectangle contained by the whole cutting line and its external ſegment be equal to the ſquare of the line meeting the circle, the latter is a tangent to the circle.

Draw from the given point
, a tangent to the circle, and draw from the
centre , , and ,
2 = × (B. 3. pr. 36.)
but 2 = × (hyp.),
2 = 2,
and = ;

Then in and
and = and ,
and is common,
= (B. 1. pr. 8.);
but = a right angle (B. 3. pr. 18.),
= a right angle,
and is a tangent to the circle (B. 3. pr. 16.).

Q. E. D.