Demonstration. Then the segment BADC is greater than a semicircle, therefore

1. The angles BAC, BEC in the segment BADC are equal by the first case. For the same reason, because CBED is greater than a semicircle,

therefore

2. The angles CAD, CED are equal;

3. The whole angle BAD is equal to the whole angle BED (Ax. 2).

Wherefore, the angles in the same segment, &c. QE.D.

PROPOSITION 22.-Theorem.

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.

B

Construction. Join AC, BD.

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

but

1. The angles CAB, ABC, BCA are equal to two right angles;

2. The angle CAB is equal to the angle CDB (III. 21), because they are in the same segment CDAB; and

3. The angle ACB is equal to the angle ADB; because they are in the same segment ADCB; therefore

4. 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

5. The three angles ABC, CAB, BCA are equal to the two angles ABC, ADC (Ax. 2) ;

but ABC, BAC, BCA are equal to two right angles; therefore also 6. The angles ABC, ADC are equal to two right angles.

In the same manner it may be shown, that

7. The angles BAD, DCB are equal to two right angles. Therefore, the opposite angles, &c. Q.E.D.

PROPOSITION 23.-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, ACB, ADB, not coinciding with one another.

B

Construction. 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

1. One of the segments ACB, ADB, must fall within the other;

let ACB fall within ADB; draw the straight line BCD, and join CA, DA.

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

2. The angle ACB is equal to the angle ADB;

the exterior angle to the interior, which is impossible (I. 16). Therefore, there cannot be two similar segments of circles upon same side of the same line, which do not coincide.

Q.E.D.

the

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.

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 upon CD, then

1. The point B shall coincide with the point D,

because AB is equal to CD; therefore, the straight line AB coinciding with CD,

2. The segment AEB must coincide with the segment CFD (III. 23);

and therefore

3. The segment AEB is equal to the segment CFD (I. Ax. 8). Wherefore, 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 it is the segment.

B

D

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

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

Proof. Then because the angles ABD, BAD are equal to one another,

1. The straight line DA is equal to DB (I. 6), and therefore to DC;

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

2. D is the centre of the circle (III. 9).

From the centre 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 centre D is in AC,

3. The segment ABC is a semicircle.

Secondly. 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), and produce BD, if necessary, to meet AE in E, and join EC.

Proof. Because the angle ABE is equal to the angle BAE, therefore

1. The straight line EA is equal to EB (I. 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

2. The base EA is equal to the base EC (I. 4); but EA was shown to be equal to EB; wherefore also

3. EB is equal to EC (Ax. 1),

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

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

From the centre 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 centre E falls without the segment ABC, which therefore is 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.

PROPOSITION 26.-Theorem,

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

Let ABC, DEF be equal circles, and let the angles BGC, EHF at their centres, and BAC, EDF at their circumferences be equal to each other.

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

Demonstration. And because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal (III. Def. 1) ; therefore