a. 20. 3.

A E

Book III. rence, and that they have the fame part of the circumference, viz. BCD for their bafe, therefore the angle BFD is double a 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 fegment BAED be not greater than a femicircle, let BAD, BED be angles in it; thefe also are equal to one another. draw AF to the center, and produce it to C, and join CE. therefore the fegment BADC is greater than a semicircle; and the angles in it BAC, BEC are equal, by the first case. for the fame reafon, the angles CAD, CED are equal. therefore the whole angle BAD is equal to the whole an

a, 32. I.

b. 21.3.

F

C

gle BED. Wherefore the angles in the fame segment, &c. Q. E. D.

THE

HE oppofite 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,

BCA are equal to two right angles.
but the angle CAB is equal to the
angle CDB, because they are in the
fame fegment BADC; and the angle
ACB is equal to the angle ADB, be-

Ꭰ

cause they are in the fame fegment A

B

ADCB. therefore the whole angle
ADC is equal to the angles CAB,

ACB. to each of thefe equals add

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

UPO

PON the fame ftraight line, and upon the fame See N. fide of it, there cannot be two fimilar fegments of circles, not coinciding with one another.

If it be poffible, let the two fimilar fegments of circles, viz. ACB, ADB be upon the fame fide 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 fegments must there

D

a. 10. 3.

fore fall within the other; let ACB fall within ADB, and draw the straight line

BCD, and join CA, DA. and because the fegment ACB is fimilar to the fegment A

B

ADB, and that fimilar segments of circles

contain equal angles; the angle ACB is equal to the angle b.11.Def.3. ADB, the exterior to the interior, which is impoffible . There- c. 16. 1.

fore there cannot be two fimilar fegments of a circle upon the same side of the fame line, which do not coincide. Q. E. D.

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

SIMI

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

For if the fegment AEB be applied to the fegment CFD, fo as the point A be on

ZAA

C, and the ftraight

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

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

a. 23. 3.

See N.

a. 10. I.

b. II. I.

c. 6. 1.

d. 9.3

A

Segment of a circle being given, to defcribe 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.

C

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 ftraight lines DA, DB, DC are all equal, D is the center of the circle . from the center D, at the distance of any of the three DA, DB, DC describe a circle; this fhall pafs thro' the other points; and the circle of which ABC is a fegment is described. and because the center D is in AC,

the fegment ABC is a femicircle. but if the angles ABD, BAD are not equal to one another, at the point A in the ftraight line AB e, 23. 1. make the angle BAE equal to the angle ABD, and produce BD

f. 4. I.

e

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 to the base EC. but AE was fhewn to be equal to EB, wherefore alfo BE is equal to EC; and the three straight lines AE, EB, EC are therefore equal

f

to one another; wherefore a E is the center of the circle. from Book II. the center E, at the distance of any of the three AE, EB, EC d. 9. 3. describe a circle, this fhall 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 lefs than a femicircle. but if the angle ABD be lefs than BAD, the center E falls within the fegment ABC, which is therefore greater than a femicircle. wherefore a segment of a circle being given, the circle is defcribed of which it is a segment. Which was to be done.

In quaerences, whether they be at the centers or

equal circles, equal angles ftand upon equal cir

circumferences.

Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centers, 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, 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. 4. I, to the base EF. and because the angle at A is equal to the angle at D, the fegment BAC is fimilar to the fegment EDF; and they b.11.Def.3. are upon equal straight lines BC, EF; but fimilar fegments of circles upon equal straight lines are equal to one another; c. 24. 3. therefore the segment BAC is equal to the fegment 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 fegment ELF, and the circumference BKC to the circumference ELF. Wherefore, in equal circles, &c. Q. E. D.

a. 20. 3.

[N equal circles, the angles which ftand 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 manifeft that the angle BAC is alfo equal to EDF. but if not, one of them

b. 23. 1.

c. 26. 3.

is the greater. let BGC be the greater, and at the point G, in the ftraight line BG, make b the angle BGK equal to the angle EHF; but equal angles ftand upon equal circumferences o, 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 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.