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AE, EB, EC are therefore equal to one another; wherefored E Book 11r. 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 without the segment ABC, which therefore is less than a semicircle; 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,

a

PROP, XXVI. THEOR.

IN equal circles, equal ongles stand upon equal circumferences, whether they be at the centres or circumferences.

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 : there. fore 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 a to the base EF: and because the angle at A is a 4. 1. equal to the angle at D, the segment BAC is similar to the b 11. def. 3 segment EDF; and they are upon equal straight lines BC, EF; but similar segments of circle's upon equal straight lines are equal c to one another; therefore the segment BAC is equal c 24.3. $o the segment EDF: but the whole circle ABC is equal to

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Book IIL, 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 circļes, &c. Q: E. D.

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

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that the angle BAC IS ALSO

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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-Bķ 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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Book III,

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.

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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 А

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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 c circumferences, when they are at the cen

26. 3. tres; 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,

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

Book III.

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b 27.3.

circumference EHF, the angle BKC is equalb 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 equalé to the base EF. Therefore, in equa! circles,&c. Q. E. D.

c 4. 1.

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 bi

sect it. a 10.1.

Join AB, and bisecte 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 4. 1. b to the base BD: but equal straight A с c 28. 3. lines cut off equal c 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 circumfercnce DB: therefore the given circumference is bisected in D. Which was to be be done.

37

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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.

a

Let ABCD be a circle, of which the diameter is BC, and
centre E; apd draw CA dividing the circle into the segments
ABC, ADC, and join BA, AD, DC; the angle in the semi-
circle 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
40 EA, the angle EAB is equal a to EBA; also, because AE a 5.1.
is equal to EC, the angle EAC, is

F
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 equalb

D

b 32.1, 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

E

c 10. def. I. fore the angle BAC in a semicirale is a right angle.

And because the two angles ABC, BAC, of the triangle ABC are together lessd 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 equal to two right angles; there-e 22. 3. 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 circunference 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 test:

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