Book III. the circumference, viz. BCD for their bafe; therefore the an gle BFD is double a of the angle BAD: For the fame reason, the angle BFD is double of the angle RED: Therefore the angle BAD is equal to the angle BED.

a 20. 3.

But, if the fegment BAED be not greater than a femicircle,

@32 I.

b 21. 3.

BEC are equal, by the firft cafe: For
the fame reason, because CBED is
greater than a femicircle, the angles
CAD, CED are equal: Therefore
the whole angle BAD is equal to the

C.

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

PROP. XXII. THEOR.

HE oppofite angles of any quadrilateral figure defcribed in a circle, are together equal to two right

TH

angles.

Let ABCD be a quadrilateral figure in the circle ABCD ; any two of its oppofite angles are together equal to two right angles.

D

Join AC, BD; and because the three angles of every triangle are equal a 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 fame segment BADC, and the angle ACB is equal to the angle ADB, because they are in the fame fegment ADCB: Therefore the whole angle ADC is equal to the angles CAB, ACB: To each of these equals add the angle ABC; therefore the angles ABC, CAB, BCA are e

B

qual to the angles ABC, ADC: But ABC, CAB, BCA are equal to two right angles; therefore alfo the angles ABC, ADC are equal to two right angles: In the fame manner, the angles

BAD,

BAD, DCB may be fhewn to be equal to two right angles. Book III. Therefore, the oppofite angles, &c. Q. E. D.

PROP. XXIII. THEOR.

PON the fame ftraight line, and upon the fame See N. fide of it, there cannot be two fimilar fegments of

UPON

circles, not coinciding with one another.

D

a 10. 3.

If it be poffible, let the two fimilar segments of circles, viz. ACB, ADB, be upon the fame fide of the fame straight line AB, not coinciding with one another: Then, because the circle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other point a: One of the fegments must therefore fall within the other; let ACB fall within ADB, and draw the straight line BCD, and join CA. DA: And because the segment ACB is fimilar to the fegment ADB, and that fimilar fegments of circles contain b equal angles; the angle ACB is equal b 11.def. 3. to the angle A DB, the exterior to the interior, which is impoffible c. Therefore, there cannot be two fimilar fegments of a c'16. 1. circle upon the fame fide of the fame line, which do not coincide. Q. E. D.

A

PROP. XXIV. THEOR.

B

IMILAR fegments of circles upon equal ftraight see N lines, are equal to one another.

STM

Let AEB, CFD be fimilar fegments of circles upon the equal ftraight lines AB, CD; the fegment AEB is equal to the fegment CFD.

AB upon CD, the point B shall coincide with the point D, be

Book III. caufe AB is equal to CD: Therefore the straight line AB coinciding with CD, the fegment AEB mufta coincide with the fegment CFD, and therefore is equal to it. Wherefore fimilar fegments, &c. Q. E. D.

a 23. 3.

See N.

a 10. I. b II. I.

c 6. I.

d 9. 3.

A

PROP. XXV. PRO B.

SEGMENT of a circle being given, to describe the circle of which it is the fegment.

Let ABC be the given fegment of a circle; it is required to defcribe the circle of which it is the fegment.

Bifecta AC in D, and from the point D draw b DB at right angles to AC, and join AB: First, let the angles ABD, BAD be equal to one another; then the ftraight line BD is equal c 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 d: From the centre D, at the distance of any of the three DA, DB, DC, defcribe a circle; this fhall pass through the other points; and the circle of which ABC is a fegment is defcribed: And because the centre D is in AC, the fegment ABC is a fe

B

A

€ 23. I.

£ 4. I.

e

micircle: But if the angles ABD, BAD are not equal to one another, at the point A, in the straight line AB make e the 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 fides 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 bafe AE is equal f to the bafe EC: But AE was fhewn to be equal to EB, wherefore alfo BE is equal to EC: And the three ftraight lines AE,

EB

EB, EC are therefore equal to one another; wherefore E is Book III. the centre of the circle. From the centre E, at the distance of d

any of the three AE, EB, EC, describe a circle, this fhall pass d 9. 3.
through the other points; and the circle of which ABC is a feg-
ment is described: And it is evident, that if the angle ABD be
greater than the angle BAD, the centre E falls without the
fegment ABC, which therefore is less than a femicircle: But
if the angle ABD be less than BAD, the centre E falls within
the fegment ABC, which is therefore greater than a femicircle:
Wherefore a fegment of a circle being given, the circle is de-
fcribed of which it is a fegment. Which was to be done.

PROP. XXVI. THEOR.

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

Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences: The circumference BKC is equal to the circumference ELF. Join BC, EF; and because the circles ABC, DEF are equal, eftraight lines drawn from their centres are equal: There fore the two fides BG, GC, are equal to the two EH, HF;

and the angle at G is equal to the angle at H; therefore the

bafe BC is equal a to the base EF: And because the angle at A a 4. 1. is equal to the angle at D, the segment BAC is fimilar b to the b 11.def. 3. fegment EDF; and they are upon equal straight lines BC, EF; but fimilar fegments of circles upon equal, ftraight lines are equal c to one another, therefore the fegment BAC is equal to 24. 3. the segment EDF: But the whole circle ABC is equal to the

whole

4

1

Book III. whole DEF; therefore the remaining fegment BKC is equal to the remaining fegment ELF, and the circumference BKC to the circumference ELF. Wherefore, in equal circles, &c. Q. E. D.

2.3.

Ν

PROP. XXVII. THEOR.

IN equal circles, the angles which fland upon equal

circumferences are equal to one another, whether they be at the centres or circumferences.

Let the angles BGC, EHF at the centres, and BAC, EdF at the circumferences of the equal circles ABC, DEF stand upon the equal circumferences BC, EF: The angle BGC is 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 a that the angle BAC is alfo equal to EDF. But, if not, one

b 23. I.

€ 26.3.

of them is the greater: Let BGC be the greater, and at the point G, in the ftraight line BG, make the angle BGK equal to the angle EHF; but equal angles ftand upon equal circumferences c, when they are at the centre; therefore the circumference BK is equal to the circumference EF: But EF is equal to BC; therefore alfo BK is equal to BC, the lefs to the greater, which is impoffible: Therefore the angle BGC is not unequal to the angle EHF; that is, it is equal to it: And the angle at A is half of the angle BGC, and the angle at D half of the angle EHF: Therefore the angle at A is equal to the angle at D. Wherefore, in equal circles, &c. Q. E. D.

PROP.