1. The two sides BG, GC, are equal to the two EH, HF,
to each;

each

and the angle at G is equal to the angle at H (hyp.); therefore 2. The vase BC is equal to the base EF (I. 4).

And because the angle at A is equal to the angle at D (hyp.),

3. The segment BAC is similar to the segment EDF (III. Def. 11);

and they are upon equal straight lines BC, EF; but similar segments of circles upon equal straight lines, are equal to one another (III. 24); therefore

4. The segment BAC is equal to the segment EDF;

but the whole circle ABC is equal to the whole DEF (hyp.); therefore

5. The remaining segment BKC is equal to the remaining segment ELF (I. Ăx. 3), and the arc BKC to the arc ELF.

Wherefore, 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 circumferences.

B

H

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

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

If the angle BGC be equal to the angle EHF, it is manifest that the angle BAC is also equal to EDF (III. 20 and I. Ax. 7). But, if not, one of them must be greater than the other; if possible, let the angle BGC be greater than EHF.

Construction. At the point G, in the straight line BG, make the angle BGK equal to the angle EHF (I. 23).

Demonstration. Then because the angle BGK is equal to the angle EHF, and that equal angles stand upon equal arcs, when they are at the centres (III. 26); therefore

1. The arc BK is equal to the arc EF;

but the arc EF is equal to the arc BC (hyp.); therefore also

2. The arc BK is equal to the arc BC,

the less equal to the greater, which is impossible (I. Ax. 1); therefore

3. The angle BGC is not unequal to the angle EHF; that is, it is equal to it; but the angle at A is half of the angle BGC (III. 20), and the angle at D, half of the angle EHF; therefore 4. The angle at A is equal to the angle at D (I. Ax. 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 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 BGC, EHF

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

Construction. Take K, L, the centres of the circles (III. 1), and join BK, KC, EL, LF.

Demonstration. Because the circles ABC, DEF are equal, the straight lines from their centres are equal (III. Def. 1); therefore 1. BK, KC are equal to EL, LF, each to each;

and the base BC is equal to the base EF, in the triangles BCK, EFL; therefore

2. The angle BKC is equal to the angle ELF (I. 8); but equal angles stand upon equal arcs, when they are at the centres (III. 26); therefore

3. The arc BGC is equal to the arc EHF;

but the whole circumference ABC is equal to the whole EDF (hyp.) ; therefore the remaining part of the circumference—viz.,

4. The arc BAC, is equal to the remaining part EDF (I. Ax. 3).

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

PROPOSITION 29.-Theorem.

In equal circles, equal arcs are subtended by equal straight lines. Let ABC, DEF be equal circles, and let the arcs BGC, EHF also be equal, and joined by the straight lines BC, EF.

Then the straight line BC shall be equal to the straight line EF. Construction. Take K, L (III. 1), the centres of the circles, and join BK, KC, EL, LF.

Demonstration. Because the arc BGC is equal to the arc EHF,

1. The angle BKC is equal to the angle ELF (III. 27); and because the circles ABC, DEF ate equal, the straight lines from their centres are equal (III. Def. 1); therefore

2. BK, KC are equal to EL, LF each to each;

and they contain equal angles in the triangles BCK, EFL; therefore 3. The base BC is equal to the base EF (I. 4).

Therefore, 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.

Construction. Join AB, and bisect it in C (I. 10); from the point

C draw CD at right angles to AB (I. 11).

Then the arc ADB shall be bisected in the point D.

Join AD, DB.

Proof. And because AC is equal to CB, and CD common to the triangles ACD, BCD,

and

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

because each of them is a right angle; therefore

3. The base AD is equal to the base BD (I. 4).

But equal straight lines cut off equal arcs (III. 28), the greater arc equal to the greater, and the less arc to the less; and the arcs

4. AD, DB are each of them less than a semicircle;

because DC, if produced, passes through the centre (III. 1, Cor.); therefore

5. The arc AD is equal to the arc DB.

Therefore the given arc ADB is bisected in 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 the diameter is BC, and centre E; and let CA be drawn, dividing the circle into the segments ABC, ADC. Join BA, AD, DC.

Then the angle in the semicircle BAC shall be a right angle; and the angle in the segment ABC, which is greater than a semicircle, shall be less than a right angle; and the angle in the segment ADC, which is less than a semicircle, shall be greater than a right angle.

F

B