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Book III. caufe AB is equal to CD: Therefore the inciding with CD, the fegment AEB muft fegment CFD, and therefore is equal to it. fegments, &c. Q_ E. D.

a 23.3.

straight line AB cocoincide with the Wherefore fimilar

See N.

a 10. I. b II. I.

c 6. I.

d 9.3.

A s

PROP. XXV. PROB.

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: Firft, let the angles ABD, BAD, be equal to one another; then the ftraight line BD is equal to DA, and therefore to DC: And becaufe the three straight lines DA, DB, DC are all equal, D is the center of the circled From the center D, at the distance of any of the three DA, DB, DC, defcribe a circle; this fhall pass 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 fe

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micircle: But if the angles ABD, BAD are not equal to one another, at the point A, in the ftraight line AB, make the angle BAE equal to the angle ABD, and produce BD, if neceffary, to E, and join EC: And because the angle ABE is equal to the angle BAE, the ftraight 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, where. fore 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 center of the circle. From the center E, at the distance of any of the three AE, EB, EC, defcribe a circle, this fhall pafs d 9. 3. 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 he fegment ABC, which is therefore greater than a femicircle: Wherefore a fegment of a circle being given, the circle is described of which it is a fegment. Which was to be done.

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IN equal circles, equal angles stand upon equal circumferences, whether they be at the centers or 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 ftraight lines drawn from their centers are equal: therefore the two fides BG, GC, are equal to the two EH, HF;

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and the angle at G is equal to the angle at H; therefore the bafe BC is equal to the bafe EF: And becaufe the angle at A a 4.1. is equal to the angle at D, the fegment BAC is fimilar to the b 11. def. 3 fegment EDF; and they are upon equal ftraight lines BC, EF; but fimilar fegments of circles upon 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

C 24.3.

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.

20. 3.

IN

PRO P. XXVII. THE OR.

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 ftand 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 a that the angle BAC is also equal to EDF: But, if not, one

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b. 23. I.

c 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, when they are at the center; therefore the circumference BK is equal to the circumference EE: 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.

Book III:

PROP. XXVIII

THEOR.

N equal circles, equal straight 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 equal to the greater EDF, and the lefs BGC to the lefs EHF.

a

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

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from their centers 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 b 8. 1. ftand upon equal circumferences, when they are at the cen- c 26. 3. ters; therefore the circumference BGC is equal to the cir cumference 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.

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

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

Take

Book III.

a 1. 3.

a

Take K, L the centers of the circles, and join BK, KC, EL, LF: And becaufe the circumference BGC is equal to the

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b 27.3.

€ 4. I.

circumference 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 they contain equal angles: Therefore the bafe BC is equal to the bafe EF. Therefore, in equal circles, &c. Q. E. D.

c

a 10. 1.

PRO P. XXX. PROB.

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

Ti

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 åt right angles to AB, and join AD, DB: The circumference ADB is bifected in the point D.

A

D

B

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 angle ACD is equal to the angle BCD, becaufe each of them is a right angle; therefore the bafe AD is equal to the bafe BD: But equal fraight lines cut off equal circumferences, the greater equal to the greater, and the lefs to the lefs, and AD, DB are each of them less than a femicircle; because DC paffes through the d Cor. 1. 3, center: Wherefore the circumference AD is equal to the circumference DB: Therefore the given circumference is bifected in D. Which was to be done.

b 4. 3.

€ 28. 3.

PROP.

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