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another; wherefore d 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 shall pass thro' the other points; and the circle of which ABC d. 9. 3. 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 fegment ABC, which therefore is less than a semicircle. but if the angle ABD be less that BAD, the center E falls within the 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 circumiferences.

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

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is equal to the angle at H; therefore the base BC is equal to the 2. 4. f. base EF. and because the angle at A is equal to the angle at D, the segment BAC is similar to the segment EDF; and they are upon b. 21. Def. 3. equal straight lines BC, EF; but similar fegments of circles upon equal straight lines are equal to one another; therefore the leg. C.24. 3. ment BAC is equal to the segment EDF. but the whole circle AEC is equal to the whole DEF, therefore the remaining fegment BKC

Book 111. is equal to the remaining segment ELF, and the circumference mu BKC to the circumference ELF. Wherefore in equal circles, &c.

Q. E. D.

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IN cqual circles, the angles which stand upon equal cir

cumferences, 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 also equal to EDF. but if not, one of them

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is the greater. let BGC be the greater, and at the point G, in the t. 23. 1. straight line BG, make the angle BGK equal to the angle EHF; 6. 36. 3. but equal angles stand 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 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.

PROP

Book int.

PROP. XXVIII.

THEOR:

IN cqual 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, EL, LF. and because the circles are equal, the straight lines from their

A

D

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G

H tenters 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 stand upon equal circum- b.8.13 ferences, when they are at the centers; therefore the circumference c. 26. şi 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

N equal circles equal circumferences are subiended 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.
F 2

Take

Book Iir. Take * K, L the centers of the circles, and join BK, KC, EL, WLF. and because the circumference BGC is equal to the circumfe

A

D

a. I. 3

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G

H 6. 29. 3. rence EHF, the angle BKC is equal o to the angle ELF. and be

cause the circles ABC, DEF are equal, the straight lines from their

centers are equal ; therefore BK, KC are equal to EL, LF, and 6. 4. 1. they contain equal angles. therefore the base BC is equal to the

base EF. Therefore in equal circles, &c. Q. E. D.

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TO bisect a given circumference, that is to divide it

into two equal parts.

Let ADB be the given circumference; it is required to bifećt it.

Join AB, and bisect * it in C; from the point C draw CD at right angles to AB, and join AD, DB. the circumference ADB is bisected in the point D.

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

D angle ACD is equal to the angle BCD,

because each of them is a right angle; 8. 4. 1. therefore the base AD is equal to the A

С B base BD. but equal straight lines cut off C. 28. 3. 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; d. Cor. 1. 3. because DC passes thro' the center d. wherefore the circumference:

AD is equal to the circumference DB. therefore the given circumference is bisected in D. Which was to be done.

PROP

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Book III,

PROP. XXXI. THEOR. a circle, the angle in a semicircle is a right angle; but

the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

Let ABCD be a circle, of which the diameter is BC, and center E; and draw CA dividing the circle into the segments ABC, ADC, and join BA, AD, DC. the angle in the semicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC which is less than a semicircle 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; also, because AE is equal a. s. r. to EC, the angle E AC is equal to ECA; wherefore the whole angle

F BAC is equal to the two angles

AM ABC, ACB, but FAC the exterior angle of the triangle ABC, is e

D b. 32. II qual o to the two angles ABC, ACB; therefore the angle BAC is equal to

B

С the angle FAC, and each of them is therefore a right angle. wherefore

c. 1o.Def... the angle BAC in a semicircle is a righ: angle.

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

And because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal to two right angles; therefore the e. 22. 8. angles ABC, ADC are equal to two right angies; and ABC is less than a right angle, wherefore the other ADC is greater than a right angle.

Besides, it is manifest, that the circumference of the greater seg. ment ABC falls without the right angle CAB, but the circunfe. rence of the less fegment ADC falls within the right angle CAF. And this is all that is meant, when in the Gretk text, and the F 3

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