PROPOSITION 26. THEOREM.

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

Let ABC, DEF be equal circles; and let BGC, EHF be equal angles in them at their centres, and BAC, EDF equal angles at their circumferences: the arc BKC shall be equal to the arc ELF.

Then, 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 EH, HF, each to each;

and the angle at G is equal to the angle at H; [Hypothesis. therefore the base BC is equal to the base EF.

[I. 4.

And because the angle at 4 is equal to the angle at D,[Hyp. the segment BACis similar to the segment EDF;[III.Def.11. and they are on equal straight lines BC, EF.

But similar segments of circles on equal straight lines are equal to one another; [III. 24. therefore the segment BAC is equal to the segment EDF. But the whole circle ABC is equal to the whole circle DEF; [Hypothesis. therefore the remaining segment BKC is equal to the remaining segment ELF;

therefore the arc BKC is equal to the arc ELF.

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

[Axiom 3.

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

B

H

E

D

If the angle BGC be equal to the angle EHF, it is manifest that the angle BẮC is also equal to the angle EDF. [III. 20, Axiom 7.

But, if not, one of them must be the greater. Let BGC be the greater, and at the point G, in the straight line BG, make the angle BGK equal to the angle EHF. [I. 23.

Then, 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.

[Hypothesis. [Axiom 1.

therefore the arc BK is equal to the arc EF. But the arc EF is equal to the arc BC; therefore the arc BK is equal to the arc BC, 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. [4x. 7. Wherefore, in equal circles &c. Q.E.D.

PROPOSITION 28. THEOREM.

In equal circles, equal straight lines 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 straight lines 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.

the straight lines from their centres 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;

[Hypothesis.

therefore the angle BKC is equal to the angle ELF. [I. 8. But in equal circles equal angles stand on equal arcs, when they are at the centres,

therefore the arc BGC is equal to the arc EHF.

[III. 26.

But the circumference ABGC is equal to the circumference DEHF;

[Hypothesis.

therefore the remaining arc BAC is equal to the remaining arc EDF

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

[Axiom 3.

In equal circles, equal arcs are subtended by equal straight lines.

Let ABC, DEF be equal circles, and let BGC, EHF be equal arcs in them, and join BC, EF: the straight line BC shall be equal to the straight line EF.

Take K, L, the centres of the circles,

and join BK, KC, EL, LF.

(III. 1.

Then, because the arc BGC is equal to the arc EHF,

the angle BKC is equal to the angle ELF.

[Hypothesis. [III. 27.

And because the circles ABC, DEF are equal, [Hypothesis.

the straight lines from their centres are equal; [III. Def. 1. therefore the two sides BK, KC are equal to the two sides EL, LF, each to each;

and they contain equal angles;

therefore the base BC is equal to the base EF.

[I. 4.

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

PROPOSITION 30. PROBLEM.

To bisect a given arc, that is, to divide it into two equal parts.

Let ADB be the given arc: it is required to bisect it.

and CD is common to the two triangles ACD, BCD;

the two sides AC, CD are equal to the two sides BC, CD, each to each;

and the angle ACD is equal to the angle BCD, because each of them is a right angle; [Construction. therefore the base AD is equal to the base BD. [I. 4. But equal straight lines cut off equal arcs, the greater equal to the greater, and the less equal to the less; [III. 28. and each of the arcs AD, DB is less than a semi-circumference, because DC, if produced, is a diameter; [III. 1. Cor. therefore the arc AD is equal to the arc DB.

Wherefore the given arc is bisected at D. Q.E.F.

PROPOSITION 31. THEOREM.

In a circle the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

Let ABCD be a circle, of which BC is a diameter and E the centre; and draw CA, dividing the circle into the segments ABC, ADC, and join BA, AD, DC: the angle in the semicircle BAC shall be a right angle; but the angle in the segment ABC, which is greater than a