#1. 3.

Take F the centre of the circle,* draw BF, FD, as also the diameter AC, and join E, C. Because the angle BFC at the centre is upon the same arc BC as the angle BAC at the circumference, BFC is the

double of BAC.† Again, because 120. 3. the angle BFC at the centre is upon

the same arc BC as the angle BEC at the circumference, BFC is the double of BEC; but it was shown that BFC is also the double of BAC, therefore the double of BAC is equal to the double of BEC, and consequently BAC to BEC. In like manner it may be demonstrated that the angle CAD is equal to the angle CED. Therefore the whole angle BAD is equal to the whole angle BED. Therefore the angles in the same segment, &c. Q. E. D.

#1 Ax.

PROP. XXII. THEOR.

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; any two of its opposite angles shall together be equal to two right angles.

32. 1.

Join AC, BD: and because the three angles of every triangle are equal to two right angles, the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA, are equal to two right angles: but #21.3. the angle CAB is equal to the

angle CDB, 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

12 Ax.

12 Ax.

together equal to the whole

angle ADC: to each of these equals add the angle ABC; therefore the three angles ABC, CAB, BČA are equal to the two angles ABC, ADC: con

K

+1 Ax.

sequently the two angles ABC, ADC are equal† to two right angles. In the same manner the angles BAD, DCB, may be shown to be equal to two right angles. Therefore, the opposite angles, &c. Q. E. D.

PROP. XXIII. THEOR.

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. Then, because the circle ACB cuts the circle

ADB in the two points A, B, they cannot

#10. 3.

+нур.

B

cut one another in any other point: therefore one of the segments must fall within the other: let ACB fall within ADB: draw the straight line BCD, and join CA, DA. And because the segment ACB is similart to the segment ADB, and that similar segments of circles *10 Def. 3. contain* equal angles; therefore the angle ACB is equal to the angle ADB, the exterior to the interior, which is impossible." Therefore, there cannot be two similar segments of circles upon the same side of the same line, which do not coincide. Q. E. D.

#16. 1.

PROP. XXIV. THEOR.

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

F

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, the point B shall coincide with the point D, because AB is equal to CD: A therefore, the straight line AB

coinciding with CD, the segment AEB must* *23. 3. coincide with the segment CFD, and therefore is +8 Ax. equal to it. Wherefore similar segments, &c.

Q. E. D.

It may be proper to add here, as Bonnycastle does, that "Similar segments upon equal chords have equal arcs," as is evident from the demonstration.

PROP. XXV. PROB.

An arc of a circle being given, to describe the circle of which

it is an arc.

Let ABK be the given arc of a circle; it is required to describe the circle of which it is the arc.

Take any point B in the arc, from which draw any two

chords BC, BD; bisect these chords in the points E and F, through which draw EG, FH at right angles to BC, BD, respectively. These perpendiculars will intersect in a point P, which will be the centre of the circle. For the centre of the circle to which

*Cor. 1. 3.

the arc BC belongs is in the line

bisecting its chord at right angles; the same centre is also in the line bisecting the other chord at right angles*; *Cor. 1. 3. the centre being thus in both perpendiculars, must be at the point P, where they intersect. Therefore, with P as a centre, and the distance PM between it and any point M of the arc, if a circle be described it will be that of which the given arc is a part.

PROP. XXVI. THEOR.

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

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

First, let the angles BGC, EHF at the centres be equal. Take any point A in the arc BAC, and any point D in the arc EDF; draw AB, AC, DE, DF, BC, EF: and because the circles ABC, DEF are +12 Def. 1. equal, the straight lines drawn from their centrest are equal: 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; therefore the base BC is equal to the base EF. And +20. 3. and because the angle BAC is equal to the angle at

+Hyp.

#4. 1.

Ax. 7.

#10 Def. 3. EDF, the segment BAC is similar to the segment EDF: and they are upon equal straight lines BC, EF; but similar segments of circles upon equal *24. 3. straight lines are equal to one another, therefore the segment BAC is equal to the segment EDF: but the

+Hyp.

+3 Ax.

#1. 3.

#20. 3.

+ 6 Ax.

whole circle ABC is equal to the whole DEF; therefore the remaining segment BKC is equal to the remaining segment ELF, and therefore the arc BKC to the arc ELF. Next let the angles BAC, EDF at the circumferences be equal. Find the centres G and H, and draw GB, GC, BC; HE, HF, EF; then, if BAC be less than a right angle, the angle BGC at the centre will stand upon the same arc BKC as BAC, and will therefore be the double of BAC. In like manner the angle EHF will be the double of EDF: thereforet the angles BGC, EHF at the centres are equal, and consequently, as before proved, the arcs BKC, ELF, must be equal. But if the angle BAC be not less than a right angle, bisect it*, as also the equal angle EDF, by the straight lines AK, DL, respectively; then the angles BAK, EDL, being equal,t and each less than a right angle, the arc BK is, by the first case, equal to the arc EL. In like manner the arc KC is equal to the arc LF, therefore the arc BKC is equal to the arc ELF. Wherefore, in equal circles, &c. Q. E. D.

#9. 1.

+7 Ax.