(References-Prop. I. 16; III. 10, def. 11.)

If it be possible, let the two similar segments of circles, viz. ACB, ADB, be upon the same side of the same straight line AB, not coinciding with one another.

D

B

CONSTRUCTION

Then, because the circle ACB cuts the circle ADB, in the two points A, B,

they cannot cut one another in any other point; (III. 10)

therefore one of the segments ACB, ADB, must fall within the other. Let ACB fall within ADB, and draw the straight line BCD, and join CA, DA.

DEMONSTRATION

Because the segment ACB is assumed to be similar to the segment ADB, (hyp.)

and that similar segments of circles contain equal angles; (III. def. 11)

the angle ACB must be equal to the angle ADB,

the exterior to the interior, which is impossible; (I. 16)

Therefore there cannot be two similar segments of circles upon the same side of the same line, which do not coincide.

Q. E. D.

PROP. XXIV.- THEOREM.

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

(References-Prop. III. 23.)

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

Then the segment AEB shall be equal to the segment CFD.

DEMONSTRATION

For if the segment AEB be applied to the segment CFD, so that the point A may be on C, and the straight line AB on CD;

then, because AB is equal to CD,

the point B shall coincide with the point D ;

and the straight line AB coinciding with CD,

the segment AEB coincides with the segment CFD, (III. 23)

and therefore the segment AEB is equal to the segment

CFD. (ax. 8.)

Wherefore, similar segments, &c.

Q. E. D.

PROP. XXV.- PROBLEM.

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

(References-Prop. I. 4, 6, 10, 11, 23; III. 9.)

Let ABC be the given segment of a circle.

It is required to describe the circle of which it is the segment.

Bisect AC in D, (1. 10) and from the point D draw DB at right angles to AC; (1. 11) and join AB.

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

Then D shall be the centre of the circle required.

DEMONSTRATION

Because the angle ABD is equal to the angle BAD, (hyp.)

therefore DA is equal to DB; (1. 6)

but DA is equal to DC, (constr.)

therefore DB is equal to DC,

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

and because from the point D more than two equal straight lines DA, DB, DC, can be drawn to the circumference ABC, therefore D is the centre of the circle. (III. 9.)

Hence, if with the centre D, at the distance of any of the three, DA, DB, DC, a circle be described,

it will pass through the other two points, and be the circle required.

Secondly, let the angles ABD, BAD, be not equal to one another.

At the point A, in the straight line BA, make the angle BAE equal to

the angle ABD, (1. 23)

produce BD, if necessary, to E, and join EC.

Then E shall be the centre of the circle required.

DEMONSTRATION

Because the angle ABE is equal to the angle BAE, (constr.)

the straight line BE is equal to EA; (1. 6)

and in the triangles ADE, CDE,

because AD is equal to DC, and DE common to both,

the two sides AD, DE, are equal to the two CD, DE, each to each; and the angle ADE is equal to the angle CDE, for each of them is a right angle; (constr.)

therefore the base AE is equal to the base EC; (1. 4)

but AE was shown to be equal to EB,

wherefore also EB is equal to EC; (ax. 1)

and the three straight lines EA, EB, EC, are therefore equal to one another;

and because from the point E, more than two equal straight lines, EA, EB, EC, can be drawn to the circumference ABC,

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

Hence, if with the centre E, at the distance of any of the three AE, EB, EC, a circle be described,

it will pass through the other point, and be the circle required.

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

In the second case, if the angle ABD be greater than the angle BAD, the centre E falls without the segment ABC,

which is therefore less than a semicircle;

but if the angle ABD be less than BAD, the centre E falls within the segment ABC,

which is therefore greater than a semicircle.

Wherefore a segment of a circle being given, the circle is described of which it is a segment.

Q. E. F.

PROP. XXVI.- THEOREM.

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

(References-Prop. 1. 4; III. 24, def. 1, 11.)

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

Then the circumference BKC shall be equal to the circumference ELF.

Because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal; (III. def. 1)

therefore the two sides BG, GC, are equal to the two 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. (1. 4.)

And because the angle at A is equal to the angle at D, (hyp.) 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 the segment BAC is equal to the segment EDF;

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

therefore the remaining segment BKC is equal to the remaining segment ELF, (ax. 3)

and the circumference BKC to the circumference ELF. Wherefore, in equal circles, &c.

Q. E. D.

PROP. XXVII.-THEOREM.

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

(References-Prop. 1. 23; III. 20, 26.)

Let the angles BGC, EHF, at the centres, and BAC, EDF, at the circumferences of the equal circles ABC, DEF, stand upon equal circumferences 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 ax. 7.)

But, if not, one of them is greater than the other; let BGC be the greater.

CONSTRUCTION

At the point G, in the straight line BG, make the angle BGK equal to the angle EHF. (1. 23.)