Book III.

Join BC, EF; then because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal;

a 4. 1.

therefore the two sides BG, GC are equal to the two EH, HF; and the angle at G is equal to the angle at H; therefore the base BC is equala to the base EF. Because the angle at A is equal to the angle at D, the segment BAC is b9. Def. 3. similar to the segment EDF; and they are upon equal straight lines BC, EF; therefore the segment BAC is equal to the segment EDF. But the whole circle ABC is equal to the whole DEF; therefore the remaining segment BKC is equal to the remaining segment ELF, and the arch BKC to the arch ELF. Wherefore, in equal circles, &c. Q. E. D.

c 24. 3.

PROP. XXVII. THEOR.

IN equal circles the angles which stand upon equal arches 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 arches 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, the angle Book III. BAC is also equal to EDF. But, if not, one of them is the

greater. Let BGC be the greater, and at the point G, in the straight line BG, make the angleb BGK equal to the angle b 23. 1. EHF; then the arch BK is equal to the arch EF. But EF c 26. 3. 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. But 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. XXVIII. THEOR.

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

Let ABC, DEF be equal circles, and BC, EF equal straight lines in them, which cut off the two greater arches BAC, EDF, and the two less BGC, EHF; the greater BAC is equal to the greater EDF, and the less BGC is equal to the less EHF.

&

Book III.

Take K, L, the centres of the circles, and join BK, KC, EL, LF. Because the circles are equal, the straight lines

a 8. 1.

b 26. 3.

from their centres are equal; therefore BK, KC are equal to EL, LF; but the base BC is also equal to the base EF; therefore the angle BKC is equal to the angle ELF; therefore the arch BGC is equal to the arch EHF. But the whole circle ABC is equal to the whole circle EDF; therefore the remaining part of the circumference BAC is equal to the re-, maining part EDF. Therefore, in equal circles, &c. Q. E. D..

PROP. XXIX. THEOR.

a 27. 3.

IN equal circles equal arches are subtended by equal straight lines.

Let ABC, DEF be equal circles, and BGC, EHF equal arches; and join BC, EF; the straight line BC is equal to the straight line EF.

Take K, L the centres of the circles, and join BK, KC, EL, LF. Because the arch BGC is equal to the arch EHF, the angle BKC is equal to the angle ELF; and because the

circles ABC, DEF are equal, their radii are equal; therefore Book III.

BK, KC are equal to EL, LF; and they contain equal angles; therefore the base BC is equal to the base EF. Therefore, b4. 1. in equal circles, &c. Q. E. D.

PROP. XXX. PROB.

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

a 10. 1.

Let ADB be the given arch; it is required to bisect it. Join AB, and bisecta it in C; from the point C draw CD at right angles to AB, and join AD, DB; the arch ADB is b 11. 1. 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 angle ACD is equal to the angle BCD, because each of them is a right angle; therefore the base AD is equal to the base BD.

A

C

C 4.1.

Wherefore the arch AD is equald to the arch DB; and AD, d 28. 3. DB are each of them less than a semicircle, because DC

passes through the centree. Therefore the given arch ADB e Cor. 1. 3. is bisected in D. Which was to be done.

N

Book III.

a 5. 1.

b 32. 1.

PROP. XXXI. THEOR.

IN a circle the angle in a semicircle is a right an. gle; 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; 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; 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. Because BE is equal to EA, the angle EAB is equal to EBA; and because AE 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 also equal to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC; therefore each of them is c7. Def. 1. a right angle. Wherefore the angle BAC in a semicircle is a right angle.

d 17. 1.

e 22. 3.

B

C

Because the two angles ABC, BAC of the triangle ABC are together less than two right angles, and BAC is a right angle, ABC must be less than a right angle; therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle.

Because ABCD is a quadrilateral figure in a circle, the angles ABC, ADC are equale to two right angles; and ABC has been shown to be less than a right angle; wherefore