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AE, EB, EC are therefore equal to one another; wherefore E Book III. is the centre of the circle. From the centre E, at the distance

of any of the three AE, EB, EC, describe a circle, this shall d 9. 3.
pass through 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 centre E falls with-
out the segment ABC, which therefore is less than a semicir-
cle; but if the angle ABD be less than BAD, the centre 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.

PROP, XXVI. THEOR.

IN equal circles, equal angles stand upon equal cir cumferences, whether they be at the centres or circumferences.

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Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, 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 centres are equal: therefore the two sides BG, GC are equal to the two EH, HF; and

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the angle at G is equal to the angle at H; therefore the base

BC is equal to the base EF: and because the angle at A is a 4. 1.

a

equal to the angle at D, the segment BAC is similar to the b 11. def. 5 segment EDF; and they are upon equal straight lines BC, EF;

but similar segments of circles upon equal straight lines are

C

equal to one another; therefore the segment BAC is equal c 24. $.

to the segment EDF: but the whole circle ABC is equal to

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Book III the whole DEF; therefore the remaining segment BKC is equal to the remaining segment ELF, and the circumference BKC to the circumference ELF. Wherefore, in equal circles, &c. Q. E. D.

a 20.3.

PROP. XXVII. THEOR.

IN equal circles, the angles which stand upon equal circumferences 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 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 the angle BAC is also

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b 23. 1.

26.3.

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 centre; 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

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PROP. XXVIII. THEOR.

IN equal circles, equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less.

Let ABC. DEF

Book III.

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 a K, L, the centres of the circles, and join BK, KC, a 1.3. EL, LF: and because the circles are equal, the straight lines

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c 26.3.

from their centres 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. 1. stand upon equal e circumferences, when they are at the centres; 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 tircles, &c. Q. E. D.

PROP. XXIX. THEOR

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

Book III. Take a K, L, the centres of the circles, and join. BK, KC, ~~EL, LF: and because the circumference BGC is equal to the

a 1. 3.

A

D

b 27.3.

€ 4. 1.

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circumference EHF, the angle BKC is equal to the angle. ELF: and because the circles ABC, DEF are equal, the straight lines from their centres 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.

a 10. 1.

b 4. 1. c 28.3.

PROP. XXX. PROB.

TO bisect a given circumference, that is, to divide it into two equal parts.

Let ADB be the given circumference, it is required to bisect it.

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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 CB common to the triangles
ACD, BCD, the two sides AC, CD

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

b to the base BD: but equal straight A

D

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lines cut off equale circumferences, the greater equal to the greater, and the less to the less, and AD, DB are each of them d Cor. 1, 3. less than a semicircle; because DC passes through the centred: wherefore the circumference AD is equal to the circumfer ence DB: therefore the given circumference is bisected in D. Which was to be be done.

PROP. XXXI. THEOR.

Book III.

In 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 centre 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.

a

F

Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal to EBA; also, because AE a 5.1. is equal 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 b 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 where B fore the angle BAC in a semicirale is a right angle.

And because the two angles ABC, BAC, of the triangle ABC are toge

d

D

b 32. 1.

E

c 10. def. 1.

ther less than two right angles, and that BAC is a right d 17. 1. angle, ABC must be less than a right angle: and therefore

the angle in a segment 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 equale to two right angles; there-e 22, S. fore the angles ABC, ADC are equal to two right angles; 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 segment ABC falls without the right angle CAB, but the circumference of the less segment ADC falls within the right angle CAF. And this is all that is meant, when in the Greek text

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