equal to the angle ACE, the less to the greater; which † 1 Ax. is impossible; therefore F is not the centre of the circle ABC. In the same manner it may be shewn, that no other point which is not in CA, is the centre; that is, the centre is in CA. Therefore, if a straight line, &c.

PROPOSITION XX.

Q. E. D.

THEOR.-The angle at the centre of a circle is double of the See N. angle at the circumference upon the same base, that is, upon the same part of the circumference.

Let ABC be a circle, and BEC an angle at the centre, and BAC an angle at the circumference, which have the same circumference BC for their base: the angle BEC shall be double of the angle BAC.

First, let the centre of the
Because EA is equal to EB,

Join AE, and produce it to F. circle be within the angle BAC. the angle EAB is equal to the angle EBA; therefore the angles EAB, EBA are double of the angle EAB: but the angle BEF is equal to the angles EAB, EBA; 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.

Again, let the centre of the circle be without the angle BAC: it may be demonstrated, as in the first case, that the angle FEC is double of the angle FAC, and that FEB, a part F of the first, is double of FAB, a part of the

other; therefore the remaining angle BEC is

B

E

double of the remaining angle BAC. Therefore, the angle at the centre, &c. Q. E. D.

PROPOSITION XXI.

5. 1.

* 32. 1.

THEOR.-The angles in the same segment of a circle are See N. equal to one another.

Let ABCD be a circle, and BAD, BED angles in the same segment BAED: the angles BAD, BED shall be equal to one another.

† 1. 3.

* 20.3.

† 7 Ax.

† 2 Ax.

E

First, let the segment BAED be greater than a semicircle. Take F, the centre of the circle ABCD, and join BF, FD: and because the angle BFD is at the centre, and the angle BAD at the circumference, and that they have the same part of the circumference, viz. BCD for their base, therefore the angle BFD is double* of the angle BAD: for the same reason, the angle BFD is double of the angle BED; therefore the angle BAD is equal to the angle BED.

B

Next, let the segment BAED be not greater than a semicircle. Draw AF to the centre, and produce it to C, and join CE: therefore the segment BADC is greater than a semicircle; and the angles in

B

F

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
the whole angle BED. Wherefore, the angles in the same
segment, &c.

Q. E. D.

to

• 32. 1.

* 21. 3.

† 2 Ax.

+2 Ax.

PROPOSITION XXII.

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

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

*

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 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 together equal to the whole angle

ADC: to each of these equals add the angle ABC; therefore the three angles ABC, CAB, BCA are equal † to the two angles ABC, ADC: but ABC, CAB, BCA are equal to two right

angles; therefore also the angles ABC, ADC are equal to † 1 Ax. two right angles. In the same manner the angles BAD, DCB may be shewn to be equal to two right angles. Therefore,

THEOR.-Upon the same straight line, and upon the same side See N.

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

Α

10. 3.

cannot 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 similar to the segment ADB, and that similar segments of † Hyp. circles contain* equal angles, therefore the angle ACB is 11 Def. 3. equal to the angle ADB‡, the exterior to the interior; which is impossible. Therefore, there cannot be two similar seg-⚫ 16. 1. ments of circles upon the same side of the same line, which do not coincide.

Q. E. D.

PROPOSITION XXIV.

THEOR.-Similar segments of circles upon equal straight lines, See N. 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.

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

That is, ACB, the exterior angle of the triangle ACD, is proved to be equal to the interior and opposite angle ADC * ; which is impossible.

16. 1.

* 23. 3.

+8 Ax.

端

to CD: therefore the straight line AB coinciding with CD, the segment AEB must coincide with the segment CFD, and therefore is equal to it. Wherefore, similar segments, &c. †

Q. E. D.

PROPOSITION xxv.

See N.

10. 1. * 11. 1.

PROB.-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, and from the point D, draw * DB at right angles to AC, and join AB. First, let the angles ABD, BAD + See Fig. 1. be equal to one another; then the straight line BD is equal

6. 1.

**

* 9.3.

+ See Fig. 2. and 3.

23. 1.

to DA, and therefore to DC: and because the three straight lines DA, DB, DC are all equal, D is the centre of the circle *. 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 is described : and because the centre 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 + See Fig. 2. angle BAE equal to the angle ABD, and produce + BD, if necessary, to E, and join EC. And because the angle ABE is equal to the angle BAE, the straight line BE is equal * to EA: 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; therefore the base AE is equal to the base EC: but AE was shewn to be equal to EB; wherefore also BE is equal † to EC; and therefore the three straight lines AE, EB, EC are equal to one another: wherefore E is the centre of the circle. From the

* 6. 1.

+ Constr.

4. I. † 1 Ax.

* 9.3.

+1 Ax.

*

*

For since AC was bisected in D, therefore AD is equal to DC: but BD is proved to be equal to AD, therefore BD is equal to DC.

centre E, at the distance of any of the three AE, 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 + See Fig. 2. 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, See Fig. 3. which is therefore greater than a semicircle. Wherefore, a segment of a circle being given, the circle is describedo f which it is a segment.

Which was to be done.

PROPOSITION XXVI.

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

ences.

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

† Hyp.

* 4. 1.

Hyp.

* 11 Def. 3.

Join BC, EF: and because the circles ABC, DEF are equal, the straight lines drawn from their centres † are equal: there- 1 Def. 3. fore 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 because the angle at A is equal to the angle at D, 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 straight lines, are equal * to one another; therefore the segment BAC is equal to the segment EDF: but the whole circle ABC

D

* 24.3.

H

B

CE

K

is equal to the whole DEF;

+ Hyp.

therefore the remaining segment BKC is equal † to the remain- † 3 Ax. ing segment ELF, and the circumference BKC to the circumference ELF. Wherefore, in equal circles, &c. Q. E. D.

PROPOSITION XXVII.

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