EB, EC are therefore equal to one another; wherefore E is the centre of the circle. From the centre E, at the distance of any of the three AE, EB, EC, defcribe a circle, this shall pass through the other points; and the circle of which ABC is a fegment is described: And it is evident that if the angle ABD be greater than the angle BAD, the centre E falls without the fegment ABC, which therefore is lefs than a femicircle: But if the angle ABD be lefs than BAD, the centre E falls within the fegment ABC, which is therefore greater than a femicircle: Wherefore a segment of a circle being given, the circle is defcribed of which it is a fegment. Which was to be done.

d 9.3.

N equal circles, equal angles ftand upon equal circumferences, whether they be at the centres or circumfe

rences.

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 ftraight lines drawn from their centres are equal: There fore the two fides BG, GC, are equal to the two EH, HF;

and the angle at G is equal to the angle at H; therefore the bafe BC is equal to the bafe EF: And because the angle at A 4. X. is equal to the angle at D, the fegment BAC is fimilar to the b II. def.3. fegment EDF; and they are upon equal ftraight lines BC, EF; but fimilar fegments of circles upon equal ftraight lines are equal to one another; therefore the fegment BAC is equal € 24. 3. to the fegment EDF: But the whole circle ABC is equal to the whole

с

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

a 20. 3.

IN

N equal circles, the angles which ftand 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 ftand 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 manifeft that the angle BAC is alfo equal to EDF: But, if not, one

b 23. I.

c 26. 3.

of them is the greater: Let BGC be the greater, and at the point G, in the ftraight line BG, make the angle BGK equal to the angle EHF; but equal angles ftand upon equal circumferences, when they are at the centre; therefore the circum ference BK is equal to the circumference EF: But EF is equal to BC; therefore alfo BK is equal to BC, the lefs to the greater, which is impoffible: 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 EIIF: Therefore the angle at A is equal to the angle at D. Wherefore, in equal circles, &e. Q. E D.

PROP.

Book III.

PROP. XXVIII.

THEOR.

N equal circles, equal ftraight lines cut off equal circumferences, the greater equal to the greater, and the efs to the lefs.

Let ABC, DEF be equal circles, and BC, EF equal ftraight lines in them, which cut off the two greater circumferences BAC, EDF, and the two lefs BGC, EHF: The greater BAC is equal to the greater EDF, and the lefs BGC to the lefs EHF.

Take K, L the centres of the circles, and join BK, KC, a 1. 3. EL, LF And because the circles are equal, the ftraight lines

from their centres are equal; therefore BK, KC are equal to EL, LF; and the bafe BC is equal to the bafe EF; therefore the angle BKC is equal b to the angle ELF: But equal angles ь 8. I. ftand upon equal circumferences, when they are at the cen- c 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. There fore, in equal circles, &c. Q. E. D.

IN equal circles equal circumferences are fubtended by equal straight lines.

Let ABC, DEF be equal circles, and let the circumferences BGC, EHF alfo be equal; and join BC, EF: The straight line BC is equal to the ftraight line EF.

Take

Book III.

a I. 3.

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

b 27.3.

C 4. I.

circumference EHF, the angle BKC is equal to the angle ELF: And becaufe the circles ABC, DEF are equal, the ftraight lines from their centres are equal: Therefore BK, KC are equal to EL, LF, and they contain equal angles: Therefore the bafe BC is equal to the bafe EF. Therefore, in equal circles, &c. Q. E. D.

10. I.

T

PROP. XXX. PROB.

O bifect a given circumference, that is, to divide it into two equal parts.

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

Join AB, and bifecta it in C; from the point C draw CD at right angles to AB, and join AD, DB: The circumference ADB is bifected in the point D.

A

D

C

B

Because AC is equal to CB, and CD common to the triangles ACD, BCD, the two fides 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 bafe BD: But equal ftraight lines cut off equal circumferences, the greater equal to the greater, and the lefs to the lefs, and AD. DB are each of them less than a femicircle; becaufe DC paffes through the d Cor. 1. 3. centre 4: Wherefore the circumference AD is equal to the circumference DB:: Therefore the given circumference is bifected in D. Which was to be done.

4. I. с 28. 3

PROP.

Book III.

IN

a circle, the angle in a femicircle is a right angle; but the angle in a fegment greater than a femicircle is lefs than a right angle; and the angle in a fegment less than a femicircle 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 fegments ABC, ADC, and join BA, AD, DC; the angle in the semicircle BAC is a right angle; and the angle in the fegment ABC, which is greater than a femicircle, is less than a right angle; and the angle in the fegment ADC, which is less than a femicircle, 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; alfo, becaufe 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 to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, B and each of them is therefore a

A

D

b 32. I.

C

E

c 10. def. I.

And because the two angles ABC, BAC of the triangle ABC are together lefs d than two right angles, and that BAC d 17. I. is a right angle, ABC must be lefs than a right angle; and therefore the angle in a fegment ABC greater than a femicircle, is lels than a right angle,

And because ABCD is a quadrilateral figure in a circle, any two of its oppofite 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 lefs than a right angle; wherefore the other ADC is greater than a right angle.

Befides, it is manifeft, that the circumference of the greater fegment ABC falls without the right Angle CAB, but the circumference of the lefs fegment ADC falls within the right angle CAF. And this is all that is meant, when in the

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