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a 23. 3.

Book lit. cause AB is equal to CD: Therefore the straight line AB co.

inciding with CD, the segment AEB multa coincide with the segment CFD, and therefore is equal to it. Wherefore' fimilar segments, &c. Q E. D.

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Segment of a circle being given, to describe the circle of which it is the segment.

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a 10, I. b II. 1.

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Let ABC be the given segment of a circle ; it is required to describe the circle of which it is the segment.

Bisect * AC in D, and from the point D draw o 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 circled : From the center D, at the distance of any of the three DA, DB, DC, describe a circle ; this shall pass thro’ the other points; and the circle of which ABC is a segment is described : And because the center D is in AC, the segment ABC is a seB

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micircle : But if the angles ABD, BAD are not equal to one another, at the point A, in the straight line AB, make the angle BAE cqual to the angle ABD, and produce BD, if necefsary, 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 sides 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, wher». fore also BE is equal to EC : And the three straight lines AE,

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EB,

EB, EC are therefore equal to one another ; wherefore d E is Book III. the center of the circle. From the center E, at the distance of any of the three AE, EB, EC, describe a circle, this shall pass d 9. 3. thro' the other points; and the circle of which ABC is a segment is described : And it is evident that if the angle ABD be greater than the angle BAD, the center E falls without the segment ABC, which therefore is less than a semicircle: Bus if the angle ABD be less than BAD, the center E falls within che segment ABC, which is therefore greater than a semicircle : Wherefore a segment of a circle being given, the circle is described of which it is a segment. Which was to be done.

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IN equal circles, equal angles stand upon equal circum

ferences, whether they be at the centers or circumferences.

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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;

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and the angle at G is equal to the angle at H; therefore the bafe BC is equal to the base EF: And because the angle at A 4 4. 1. is equal to the angle at D, the fegment BAC is similar to the b ir. def. 3 segment EDF; and they are upon equal straight lines BC, EF; but limilar segments of circles upon equal straight lines are equal to one another; therefore the segment BAC is equal C 24.3. to the segment EDF: But the whole circle ABC is equal to the

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Book II. whole DEF; therefore the remaining segment BKC is equal to m the remaining segment ELF, and the circumference BKC to

the circumference ELF. Wherefore, in equal circles, &c.

Q. E. D.

PRO P. XXVII. THE O R.

IN

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 A

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20. 3.

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of them is the greater : Let BGC be the greater, and at the point G, in the straight line BG, make b the angle BGK equal to the angle EHF; but equal angles stand 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 also 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.

b 23. 8.

C 26. 3.

PRO P.

Book III.

PRO P. XXVIII

TH E O R.

IN equal circles, equal straight lines cut off equal cir

cumferences, the greater equal to the greater, and the less to the less.

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 less BGC, EHF: The greater BAC is equal to the greater EDF, and the less BGC to the less EHF.

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

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TT from their centers are equal ; therefore BK, KC, are equal to EL, LF; and the base BC is equal to the base EF; therefore the angle BKC is equal to the angle ELF: But equal angles b 8. . ftand upon equal circumferences, when they are at the cen• c 26. 3. ters; therefore the circumference 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.

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IN equal circles, equal circumferences are subtended 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 straight line EF.

Take

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Book III. Take * K, L the centers of the circles, and join BK, KC

m EL, LF: And because the circumference BGC is equal to the 1 1. 3•

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6 27. 3.

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H circumference IHF, the angle BKC is equal to the angle ELF : And because the circles ABC,DEF are equal, the straight lines from their centers are equal : Therefore BK, KC are equal to EL, LF, and they contain equal angles: Therefore the base BC is equal to the base EF. Therefore, in equal circles, &c. Q. E. D.

PRO P. XXX. PRO B.

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70 bisect a given circumference, that is, to divide it

into two equal parts.

Let ADB be the given circumference; it is required to bi

fect it. a 10. 1. Join AB, and bisect a it in C; from the point C drav CD

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

Because AC is equal to CB, and CD common to the tri-
angles ACD, BCD, the two fides AC,
CD are equal to the two BC, CD;

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and the angle ACD is equal to the
angle BCD, because each of them is a

right angle; therefore the base AD b f. 3. is equal to the base BD: But equal

A С. В flraight lines cut off equal circumferences, the greater equal to the greater, and the less to the less, and AD, DB are each

of them less than a semicircle ; because DC palīts through the d Cor. I. 3, center d: Wherefore the circumference AD is equal to the cirá

cumference DB: Therefore the giren circumference is bilected in D. Which was to be done,

PROP.

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( 28. 3.

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