therefore also the angle BEF is double of the angle EAB: for the same reason, the angle FEC is double of the angle EAC: therefore the whole angle BEC is double of the whole angle BAC. Secondly, let the center of the circle be without the angle BAC.

F

B

E

It may be demonstrated, as in the first case, that the angle FEC is double of the angle FAC,

and that FEB, a part of the first, is double of FAB, a part of the other; therefore the remaining angle BEC is double of the remaining angle BAC.

Therefore the angle at the center, &c. Q.E.D.

PROPOSITION XXI. THEOREM.

The angles in the same segment of a circle are equal to one another.
Let ABCD be a circle,

and BAD, BED angles in the same segment BAED.
Then the angles BAD, BED shall be equal to one another.
First, let the segment BAED be greater than a semicircle.

Take F, the center of the circle ABCD, (III. 1.) and join BF, FD. Because the angle BFD is at the center, and the angle BAD at the circumference, and that they have the same part of the circumference, viz. the arc BCD for their base;

therefore the angle BFD is double of the angle BAD: (III. 20.) for the same reason the angle BFD is double of the angle BED: therefore the angle BAD is equal to the angle BED. (ax. 7.) Next, let the segment BAED be not greater than a semicircle.

Draw AF to the center, and produce it to C, and join CE.
Because AC is a diameter of the circle,

therefore the segment BADC is greater than a semicircle;
and the angles in it BAC, BEC are equal, by the first case:

for the same reason, because CBED is greater than a semicircle, the angles CAD, CED, are equal:

therefore the whole angle BAD is equal to the whole angle BED. (ax. 2.) Wherefore the angles in the same segment, &c.

PROPOSITION XXII. THEOREM.

Q. E.D.

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

Let ABCD be a quadrilateral figure in the circle ABCD. Then any two of its opposite angles shall together be equal to two right angles.

A

D

C

B

Join AC, BD.

And because the three angles of every triangle are equal to two right angles, (I. 32.)

the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA, are equal to two right angles:

but the angle CAB is equal to the angle CDB, (III. 21.)

because they are in the same segment CDAB; and the angle ACB is equal to the angle ADB, because they are in the same segment ADCB: therefore the two angles CAB, ACB are together equal to the whole angle ADC: (ax. 2.)

to each of these equals add the angle ABC;

therefore the three angles ABC, CAB, BCA are equal to the two angles ABC, ADC: (ax. 2.)

but ABC, CAB, BCA, are equal to two right angles; therefore also the angles ABC, ADC are equal to two right angles. In the same manner, the angles BAD, DCB, may be shewn to be equal to two right angles.

Therefore, the opposite angles, &c. Q.E.D.

PROPOSITION XXIII. THEOREM.

Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with one another.

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

Then, because the circumference ACB cuts the circumference ADB in the two points A, B, they cannot cut one another in any other point: (III. 10.)

therefore one of the segments must fall within the other:
let ACB fall within ADB:

draw the straight line BCD, and join CA, DA.

Because the segment ACB is similar to the segment ADB, (hyp.) and that similar segments of circles contain equal angles; (III. def. 11.) therefore the angle ACB is equal to the angle ADB,

the exterior angle to the interior, which is impossible. (1. 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.

PROPOSITION XXIV. 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.

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

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 upon CD, then the point B shall coincide with the point D, because AB is equal to CD:

therefore, the straight line AB coinciding with CD,

the segment AEB must coincide with the segment CFD, (III. 23.) and therefore is equal to it. (I. ax. 8.)

Wherefore similar segments, &c. Q. E.D.

PROPOSITION XXV. 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 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 :

B

A D C

then the straight line DA is equal to DB, (1. 6.) and therefore, to DC; and because the three straight lines DA, DB, DC are all equal, therefore D is the center of the circle. (III. 9.)

From the center D, at the distance of any of the three DA, DB, DC, describe a circle;

this shall pass through the other points;

and the circle of which ABC is a segment has been described:

and because the center D is in AC, the segment ABC is a semicircle. But if the angles ABD, BAD are not equal to one another:

at the point A, in the straight line AB,

make the angle BAE equal to the angle ABD, (1. 23.) and produce BD, if necessary, to meet AE in E, and join EC. Because the angle ABE is equal to the angle BAE, therefore the straight line EA is equal to EB: (1.6.)

and because AD is equal to DC, and DE common to the triangles ADE, CDE,

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 EA is equal to the base EC: (1. 4.) but EA was shewn to be equal to EB :

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

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

wherefore E is the center of the circle. (III. 9.)

From the center E, at the distance of any of the three EA, EB, EC, describe a circle;

this shall pass through the other points;

and the circle of which ABC is a segment, is described.

And it is evident, that if the angle ABD be greater than the angle BAD, the center E falls without the segment ABC, which therefore is less than a semicircle :

but if the angle ABD be less than BAD, the center 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.

PROPOSITION XXVI. THEOREM.

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

Let ABC, DEF be equal circles,

and let the angles BGC, EHF at their centers,

and BAC, EDF at their circumferences be equal to each other. Then the arc BKC shall be equal to the arc ELF.

Join BC, EF.

And because the circles ABC, DEF are equal,

the straight lines drawn from their centers 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, (I. ax. 3.)

and the arc BKC to the arc ELF.

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

PROPOSITION XXVII. THEOREM.

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

Let ABC, DEF be equal circles,

and let the angles BGC, EHF at their centers,
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. (111. 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,

and at the point G, in the straight line BG,

make the angle BGK equal to the angle EHF. (1. 23.) Then because the angle BGK is equal to the angle EHF, and that equal angles stand upon equal arcs, when they are at the centers; (III. 26.)

therefore the arc BK is equal to the arc EF:

but the arc EF is equal to the arc BC; (hyp.) therefore also the arc BK is equal to the arc BC, the less equal to the greater, which is impossible: (I. ax. 1.)