another; wherefore E is the center of the circle. from the center Book III. E at the distance of any of the three AE, EB, EC describe a circle, this fhall pass thro' the other points; and the circle of which ABC d. 9. 3.

is a fegment is defcribed. 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 less that BAD; the center E falls within the fegment ABC which is therefore greater than a femicircle: wherefore a feg ment of a circle being given, the circle is defcribed of which it is a fegment. Which was to be done.

IN equal circles, equal angles stand upon equal circum ferences, whether they be at the centers, or circum ferences.

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

a

is equal to the angle at H; therefore the base BC is equal to the a. 4. 1, bafe 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 are upon b. 11. Def. 3. equal ftraight lines BC, EF; but fimilar fegments of circles upon equal ftraight lines are equal to one another; therefore the feg- c. 24. 3. ment BAC is equal to the fegment EDF, but the whole circle ABC is equal to the whole DEF, therefore the remaining fegment BKC

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

2.20.3.

PROP. XXVII. THEOR.

IN 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 * that the angle BAC is alfo equal to EDF. but if not, one of them

is the greater. let BGC be the greater, and at the point G, in the t. 23. 1. ftraight line BG, make the angle BGK equal to the angle EHF; 6. 6. 3. but equal angles ftand 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 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

PROP. XXVIII. THEOR.

In circles, equal fer equal to

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

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

Book III.

Take K, L the centers of the circles, and join BK, KC, EL, a. 1. 3 LF. and because the circles are equal, the straight lines from their

benters are equal, therefore BK, KC, are equal to EL, LF; and the bafe BC is equal to the bafe EF; therefore the angle BKC is equal to the angle ELF. but equal angles ftand upon equal circum- b.. ferences, when they are at the centers; therefore the circumference c. 16. § BGC is equal to the circumference 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.

N equal circles equal circumferences are fubtended by equal straight lines.

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

F 2

Take

Book III.

a

Take K, L the centers of the circles, and join BK, KC, EL, in LF. and because the circumference BGC is equal to the circumfe

b

b. 27.3. fence EHF, the angle BKC is equal to the angle ELF. and because the circles ABC, DEF are equal, the ftraight lines from their centers are equal; therefore BK, KC are equal to EL, LF, and C. 4. . they contain equal angles. therefore the bafe BC is equal to the bafe EF. Therefore in equal circles, &c. Q. E. D.

4. 10. 1.

PRO P. XXX. PRO B.

To bifect a given circumference, that is to divide it into two equal parts.

Let ADB be the given circumference; it is required to bifect it. Join AB, and bifect it in C; from the point C draw CD at right angles to AB, and join AD, DB. the circumference ADB is bifected in the point D.

Because AC is equal to CB, and CD common to the triangles ACD, BCD, the two fides AC, CD are

equal to the two BC, CD; and the

D

c. 28. 3.

c

equal circumferences, the greater equal to the greater, and the lefs to the lefs, and AD, DB are each of them lefs than a femicircle;

wherefore the circumference therefore the given circum

4. Cor. 1.3. because DC paffes thro' the center 4.
AD is equal to the circumference DB.
ference is bifected in D. Which was to be done.

PROP.

Naci Na circle, the angle in a femicircle is a right angle; but the angle in a fegment greater than a femicircle is lefs than a right angle; and the angle in a fegment less than a femicircle is greater than a right angle.

Book III.

Let ABCD be a circle, of which the diameter is BC, and center E; and draw CA dividing the circle into the fegments ABC, ADC, and join BA, AD, DC. the angle in the femicircle BAC is a right angle; and the angle in the fegment ABC, which is greater than a femicircle, is lefs than a right angle; and the angle in the fegment ADC which is lefs than a femicircle is greater than a right angle. Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal to EBA; alfo, because AE is equal a. §. x. to EC, the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to the two angles ABC, ACB. but FAC the exterior angle of the triangle ABC, is equal to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, and each of them is therefore a right angle. wherefore the angle BAC in a femicircle is a right angle.

C

B

F

A

D

b. 32. 1

C

c. 1o. Def.1.

And because the two angles ABC, BAC of the triangle ABC are together lefs than two right angles, and that BAC is a right angle, d. 17. ABC must be lefs than a right angle; and therefore the angle in a fegment ABC greater than a femicircle is lefs than a right angle.

And because ABCD is a quadrilateral figure in a circle, any two

of its oppofite angles are equal to two right angles; therefore the e. 22. 3• angles ABC, ADC are equal to two right angles; and ABC is lefs than a right angle, wherefore the other ADC is greater than a right angle.

Befides, it is manifeft, that the circumference of the greater fegment ABC falls without the right angle CAB, but the circumfe rence of the lefs fegment ADC falls within the right angle CAF. And this is all that is meant, when in the Greek text, and the tranflations

F 3