a. 20. 3.

Book III. rence, and that they have the fame part of the circumference, viz. BCD for their base, therefore the angle BFD is double of the angle BAD. for the fame reason, the angle BFD is double of the angle BED. therefore the angle BAD is equal to the angle BED. But if the segment BAED be not greater than a femicircle, let

T

E. D.

2, 32. I.

gle BED: Wherefore the angles in the same segment, &c.

T

PROP. XXII. THEOR.

HE opposite angles of any quadrilateral figure described in a circle, are together equal to two

right angles.

Let ABCD be a quadrilateral figure in the circle ABCD; any two of its oppofite angles are together 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,

D

C

the angle ABC, therefore the angles ABC, CAB, BCA are equal 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, DCB

may be shewn to be equal to two right angles. Therefore the Book III. opposite angles, &c. Q. E. D.

PROP. XXIII. THEOR.

UP

PON the same straight line, and upon the same See N. fide of it, there cannot be two fimilar segments

of circles, not coinciding with one another.

D

If it be possible, let the two similar segments of circles, viz. ACB, ADB be upon the same side of the same 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. one of the segments 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 similar to the segment A ADB, and that fimilar segments of circles

C

B

a. 10. 3.

contain b equal angles; the angle ACB is equal to the angle b.11.Def.3. ADB, the exterior to the interior, which is impossibles. There- c. 16. 1. fore there cannot be two fimilar segments of a circle upon the

same side of the fame line, which do not coincide. Q. E. D.

S

PROP. XXIV. THEOR.

IMILAR segments of circles upon equal straight See N. lines, are equal to one another.

Let AEB, CFD be fimilar segments of circles upon the equal straight lines AB, CD; the fegment AEB is equal to the segment CFD.

For if the fegment AEB be applied to the seg

AA

ment CFD, so as
the point A be on
C, and the straight A

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

Book III. because AB is equal to CD. therefore the straight line AB coinciding with CD, the segment AEB must coincide with the fegment CFD, and therefore is equal to it. Wherefore fimilar segments, &c. Q. E. D.

a. 23. 3.

See N.

a. IO. I. b. II. 1.

с. б. 1.

d. 9. 3.

A

PROP. XXV. PROB.

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 fegment.

Bifecta AC in D, and from the point D draw & DB at right angles to AC, and join AB. First, let the angles ABD, BAD be equal to one another; then the straight line BD is equal to DA, and therefore to DC. and because the three straight lines DA, DB, DC are all equal, D is the center of the circle d. from the center D, at the distance of any of the three DA, DB, DC defcribe a circle; this shall pass thro' the other points; and the circle of which ABC is a fegment is defcribed. and because the center D is in AC,

the fegment 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 e, 23.1. make the angle BAE equal to the angle ABD, and produce BD 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 base AE is equal f to the base EC. but AE was shewn to be equal to EB, wherefore alfo BE is equal to EC; and the three straight lines AE, EB, EC are therefore equal

to one another; wherefore d E is the center of the circle. from Book III. the center E, at the distance of any of the three AE, EB, EC d. 9. 3. describe a circle, this shall pass thro' the other points; and the circle of which ABC is a fegment is described. and it is evident that if the angle ABD be greater than the angle BAD, the center E falls without the fegment 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 defcribed of which it is a segment. Which was to be done.

PROP. XXVI. THEOR.

N equal circles, equal angles stand upon equal circumferences, whether they beat the centers or circumferences.

Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centers, and BAC, EDF at their circümferences. the circumference BKC is equal to the circumference ELF.

Join BC, EF; and because the circles ABC, DEF are equal, the straight lines drawn from their centers are equal; therefore 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 base BC is equal a a. 4. . to the base EF. and because the angle at A is equal to the angle

at D, the segment BAC is fimilar to the segment EDF; and they b.11.Def.3.

are upon equal straight lines BC, EF; but similar segments of

circles upon equal straight lines are equal to one another; c. 24. 3.

therefore the segment BAC is equal to the segment EDF. but the whole circle ABC is equal to the whole DEF, therefore the

F

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

2. 20. 3.

I

PROP. XXVII. THEOR.

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

Let the angles BGC, EHF at the centers, 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 also equal to EDF. but if not, one of them

is the greater. let BGC be the greater, and at the point G, in the straight line BG, make the angle BGK equal to the angle EHF; €. 26. 3. but equal angles stand upon equal circumferences, when they are at the center; therefore the circumference BK is equal to the circumference EF. but EF is equal to BC, therefore alfo BK is equal to BC, the less to the greater, which is impossible. 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.