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Proposition 24.-Theorem.

Similar segments of circles upon equal straight lines are equal to one another.

Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD.

The segment AEB shall be equal to the segment CFD.

PROOF. For if the seg-A ment AEB be applied to the

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segment CFD, so that the point A may be on the point C, and the straight line AB on the straight line CD,

Then the point B shall coincide with the point D, because AB is equal to CD.

And the straight line AB coinciding with CD, the segment AEB must coincide with the segment CFD (III. 23); and is therefore equal to it.

Therefore, similar segments, &c.

Q.E.D.

Proposition 25.-Problem.

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

Let ABC be the given segment of a circle.

It is required to describe the circle of which ABC is a segment.

CONSTRUCTION.-Bisect AC in D (I. 10).

From the point D draw DB at right angles to AC (I. 11), and join AB.

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CASE I. First, let the angles ABD, BAD be equal to one another.

Then D shall be the centre of the circle required.

They are equal, because they must coincide by Prop. 23.

When LABD = < BAD,

DA = DB

= DC.

and ... D the centre.

Make

< BAE= 4 ABD.

.. EA = EB

and EA = EC.

... EA =

PROOF.-Because the angle ABD is equal to the anglo
BAD (Hyp.);

Therefore DB is equal to DA (I. 6).
But DA is equal to DC (Const.);

Therefore DB is equal to DC (Ax. 1).

Therefore the three straight lines DA, DB, DC are all equal;

And therefore D is the centre of the circle (III. 9).

Hence, if from the centre D, at the distance of any of the three lines, DA, DB, DC a circle be described, it will pass through the other two points, and be the circle required.

CASE II.-Next, let the angles ABD, BAD be not equal to one another.

CONSTRUCTION.-At the point A, in the straight line AB, make the angle BAE equal to the angle ABD (I. 23); Produce BD, if necessary, to E, and join EC.

Then E shall be the centre of the circle required.

PROOF.-Because the angle BAE is equal to the angle ABE (Const.), EA is equal to EB (I. 6).

And because AD is equal to CD (Const.), and DE is common to the two triangles ADE, CDĖ,

The two sides AD, DE are equal to the two sides CD, DE, each to each;

And the angle ADE is equal to the angle CDE, for each of them is a right angle (Const.);

Therefore the base EA is equal to the base EC (I. 4).
But EA was shown to be equal to EB;

Therefore EB is equal to EC (Ax. 1).

Therefore the three straight lines EA, EB, EC are all

EBEC, equal;

and therefore E is

the centre.

And therefore E is the centre of the circle (III. 9).

Hence, if from the centre E, at the distance of any of the three lines EA, EB, EC, a circle be described, it will pass through the other two points, and be the circle required.

In the first case, it is evident that, because the centre D is in AC, the segment ABC is a semicircle.

In the second case, if the angle ABD be greater than BAD, the centre E falls without the segment ABC, which is therefore less than a semicircle;

But if the angle ABD be less than the angle BAD, the

centre E falls within the segment ABC, which is therefore greater than a semicircle.

Therefore, a segment of a circle being given, the circle has been described of which it is a segment.

Q.E.F.

Proposition 26.—Theorem.

In equal circles, equal angles stand upon equal arcs, whether they be at the centres or at the circumferences.

Let ABC, DEF be equal circles, having the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences.

The arc BKC shall be equal to the arc ELF.
CONSTRUCTION.-Join BC, EF.

PROOF.-Because the circles ABC, DEF are equal (Hyp.), the straight lines from their centres are equal (III. def. 1); Therefore the two sides BG, GC are equal to the two sides Triangles

EH, HF, each to each;

And the angle at G is equal to the angle at H (Hyp.);
Therefore the base BC is equal to the base EF (I. 4).
And because the angle at A is equal to the angle at D
(Hyp.),

BGC and EHF are equal in

every re

spect.

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The segment BAC is similar to the segment EDF (III. def. 11),

... seg ments BAC and EDF are similar and on

And they are on equal straight lines BC, EF. But similar segments of circles on equal straight lines are equal equal to one another (III. 24);

straight lines.

Therefore the segment BAC is equal to the segment EDF... are But the whole circle ABC is equal to the whole circle equal. DEF (Hyp.);

Therefore the remaining segment BKC is equal to the remaining segment ELF (Ax. 3).

... arc BKC =

arc ELF.

Therefore the arc BKC is equal to the arc ELF.
Therefore, in equal circles, &c.

Q.E.D.

Proposition 27.-Theorem.

In equal circles, the angles which stand upon equal arcs are equal to one another, whether they be at the centres or at the circumferences.

Let ABC, DEF be equal circles, and let the angles BGC, EHF, at their centres, and the angles BAC, EDF, at their circumferences, stand on equal arcs BC, EF.

The angle BGC shall be equal to the angle EHF, and the angle BAC equal to the angle EDF.

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CONSTRUCTION.-If the angle BGC be equal to the angle EHF, it is manifest that the angle BAC is also equal to the angle EDF (III. 20, ax. 7).

But, if not, one of them must be the greater.

Let BGC If one is be the greater, and at the point G, in the straight line BG, make the angle BGK equal to the angle EHF (I. 23).

greater

than the

other, the

corre

sponding arc is greater.

PROOF. Because the angle BGK is equal to the angle EHF, and that in equal circles equal angles stand on equal arcs, when they are at the centres (III. 26);

Therefore the arc BK is equal to the arc EF.

But the arc EF is equal to the arc BC (Hyp.);

Therefore the arc BK is equal to the arc BC (Ax. 1); the less to the greater, which is impossible.

Therefore the angle BGC is not unequal to the angle EHF; that is, it is equal to it.

And the angle at A is half of the angle BGC, and the angle at D is half of the angle EHF (III. 20);

Therefore the angle at A is equal to the angle at D

(Ax. 7).

Therefore, in equal circles, &c. Q.E.D.

Proposition 28.-Theorem.

In equal circles, equal chords cut off equal arcs, the greater equal to the greater, and the less equal to the less.

Let ABC, DEF be equal circles, and BC, EF equal chords in them, which cut off the two greater arcs BAC, EDF, and the two less arcs BGC, EHF.

The greater arc BAC shall be equal to the greater arc EDF, and the less arc BGC equal to the less arc EHF.

CONSTRUCTION.-Take K, L, the contres of the circles Take K (III. 1), and join BK, KC, EL, LF.

and L the centres.

G

PROOF. Because the circles ABC, DEF are equal, their radii are equal (III. def. 1).

Therefore the two sides BK, KC are equal to the two sides EL, LF, each to each;

And the base BC is equal to the base EF (Hyp.);

Therefore the angle BKC is equal to the angle Triangles ELF (I. 8).

BKC and
ELF are

But in equal circles equal angles stand on equal arcs, equal in when they are at the centres (III. 26);

Therefore the arc BGC is equal to the arc EHF.

But the whole circle ABC is equal to the whole circle DEF (Hyp.);

Therefore the remaining arc BAC is equal to the remaining arc EDF (Ax. 3).

Therefore, in equal circles, &c. Q.E.D.

every respect.

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