ftraight line AB make e the angle BAE equal to the angle Book III. ABĎ, and produce BD, if neceffary, to E, and join EC:

e 23. I.

f 6. I.

d 9. 3.

and because the angle ABE is equal to the angle BAE, the ftraight line BE is equal to EA: and becaufe 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 g to the bafe EC: but AE was fhewn to be equal to 8 4.1. 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 centre of the circle. From the centre E, at the distance of any of the three AE, EB, EC, defcribe a circle, this fhall pass through the other points; and the circle of which ABC is a fegment is defcribed: also 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 defcribed of which it is a fegment. Which was to be done.

PROP. XXVI. THEOR.

IN equal circles, equal angles ftand upon equal arches, whether they be at the centres or circumferences.

Let

Book III.

Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences: the arch BKC is equal to the arch ELF.

Join BC, EF; and because the circles ABC, DEF are equal, the ftraight lines drawn from their centres 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; therea 4. 1. fore the base BC is equal a to the bafe EF: and because the angle at A is equal to the angle at D, the fegment BAC is b 9. def. 3. fimilar b to the fegment EDF; and they are upon equal

ftraight lines BC, EF; but fimilar fegments of circles upon c. 24. 3. equal ftraight lines are equal to one another, therefore the fegment BAC is equal to the fegment EDF: but the whole' circle ABC is equal to the whole DEF; therefore the remaining fegment BKC is equal to the remaining fegment ELF, and the arch BKC to the arch ELF. Wherefore in equal circles, &c. Q. E. D.

N equal circles, the angles which stand upon equal arches 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 ftand upon the equal arches BC, EF: the angle BGC is

equal

equal to the angle EHF, and the angle BAC to the angle Book III. EDF.

If the angle BGC be equal to the angle EHF, it is manifefta that the angle BAC is alfo equal to EDF. But, if not, a 20. 3. one of them is the greater: let BGC be the greater, and at the point G, in the ftraight line BG, make the angle b BGK b 23. I. equal to the angle EHF. And because equal angles stand upon equal arches, when they are at the centre, the arch BK is c 26. 3. equal to the arch EF: but EF is equal to BC; therefore also 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.

N equal circles, equal ftraight lines cut off equal arches, the greater equal to the greater, and the lefs to the lefs.

Let ABC, DEF be equal circles, and BC, EF equal ftraight lines in them, which cut off the two greater arches BAC, EDF, and the two lefs BGC, EHF: the greater BAC is equal to the greater EDF, and the lefs BGC to the lefs. EHF.

Take

Book III.

a 1. 3.

Take a K, L, the centres of the circles, and join BK, KC, EL, LF: and because the circles are equal, the straight lines

from their centres are equal; therefore BK, KC are equal to EL, LF; but the base BC is alfo equal to the base EF; thereb. S. 1. fore the angle BKC is equalb to the angle ELF: and equal c 26. 3. angles ftand upon equal arches, when they are at the centres; therefore the arch BGC is equal to the arch EHF. But the whole circle ABC is equal to the whole EDF; the remaining part, therefore, of the circumference, viz. BAC, is equal to the remaining part EDF. Therefore, in equal circles, &c. Q. E. D.

PROP. XXIX. THEOR.

a 1.3.

IN equal circles equal arches are fubtended by equal ftraight lines.

Let ABC, DEF be equal circles, and let the arches BGC, EHF also be equal; and join BC, EF: the straight line BC is equal to the straight line EF.

Take a K, L the centres of the circles, and join BK, KC, EL, LF and because the arch BGC is equal to the arch EHF, 27.3. the angle BKC is equal b to the angle ELF: alfo because the circles ABC, DEF are equal, their radii are equal: therefore

BK,

EK, KC are equal to EL, LF, and they contain equal angles: Book III.

therefore the bafe BC is equal to the bafe EF. Therefore, c4 1. in equal circles, &c. Q. E. D.

T

it.

O bifect a given arch, that is, to divide it into
two equal parts.

Let ADB be the given arch; it is required to bisect

Join AB, and bifecta it in C; from the point C draw CD a 10. I. at right angles to AB, and join AD, DB: the arch ADB

is bifected in the point D.

Because AC is equal to CB, and CD common to the tri

angles ACD, BCD, the two fides AČ, CD are equal to the two BC,

D

CD; and the angle ACD is equal to the angle BCD, because each of them is a right angle; therefore the bafe AD is equal b to the base

C

b 4. T.

C

BD. But equal ftraight lines cut off equal c arches, c 28.3. the greater equal to the greater, and the lefs to the lefs; and AD, DB are each of them lefs than a femicircle, becaufe DC paffes through the centred: Wherefore the arch d Cor. 1. 3.