is less than a Semicircle, because the Centre thereof is without the same.

But if the Angle ABD be equal to the Angle BAD; then if AD be made equal to B D, or DC, the three Right Lines A D, BD, DC, are equal between themselves, and D will be the Centre of the Circle to be defcribed, and the Segment A B C is a Semicircle.

But if the Angle ABD is less than the Angle BAD; let the Angle BAE be made, at the given Point A, with the Right Line B A, within the Segment A B C, equal to the Angle A B D.

Then the Point E, in the Right Line D B, will, by arguing as before, appear to be the Centre, and ABC a Segment greater than a Semicircle. Therefore, a Circle is defcribed, whereof a Segment is given ; which was to be done.

PROPOSITION XXVI.
THEORE M.

In equal Circles, equal Angles ftand upon equal
Circumferences, whether they be at their Cen-
tres, or at their Circumferences.

L

ET ABC, DEF, be equal Circles; and let BGC, EHF, be equal Angles at their Cen-` tres; and BA C, EDF, equal Angles at their Circumferences. I fay, the Circumference BKC is equal to the Circumference ELF.

**

Because A B C,

For, let BC, EF be joined. DEF, are equal Circles, the Lines drawn from their Def. 1. Centres will be equal. Therefore the two Sides BG, GC, are equal to the two Sides E H, HF;' and the Angle G is equal to the Angle H. Wherefore the Bafe BC is equal to the Bafe E F. Again, because the Angle at A is equal to that at D, the + Def. 11. Segment BAC will be + fimilar to the Segment EDF; and they are upon equal Right Lines B C, EF. But thofe fimilar Segments of Circles, that are 124 of this. upon equal Right Lines, are equal to each other. + Def. 11. Therefore the Segment B A C will be + equal to the Segment EDF. But the whole Circumference ABKCA is equal to the whole Circumference DELFD,

DELFD. Therefore the remaining Circumference BKC fhall be equal to the remaining Circumference EL F. Therefore, in equal Circles, equal Angles fland upon equal Circumferences, whether they be at their Centres, or at their Circumferences; which was to be demonftrated.

PROPOSITION XXVII.

THEOREM.

Angles that fand upon equal Circumferences in equal Circles, are equal to each other, whether they be at their Centres, or Circumferences.

LET the Angles BGC, EHF, at the Centres of the equal Circles A B C, DEF, and the Angles BAC, EDF, at their Circumferences, ftand upon the equal Circumferences B C, E F. I fay, the Angle BGC is equal to the Angle EHF; and the Angle B A C to the Angie E D F.

23. 1.

For if the Angle BG C be equal to the Angle EHF, it is manifeft that the Angle B A C is alfo equal to the Angle EDF: But if the Angle G B C be not equal to the Angle E H F, let one of them be the greater, as BG C, and make the Angle B GK, at the Point G, with the Line B G, equal to the Angle EHF. But equal Angles ftand + upon equal † 26 of this. Circumferences, when they are at the Centres. Wherefore the Circumference BK is equal to the Circumference EF. But the Circumference EF is equal to the Circumference B C. Therefore RK is equal to BC, a lefs to a greater, which is abfurd. Wherefore the Angle BGC is not unequal to the Angle EHF, and fo it must be equal to it. But the Angle at A is one half of the Angle BGC; and the Angle at D is one half of the Angle E HF. Therefore the Angle at A is equal to the Angle at D. Wherefore, Angles, that stand upon equal Circumferences in equal Circles, are equal to each other, whether they be at their Centres, or Circumferences; which was to be demonftrated.

PRO.

Def. 1.

PROPOSITION XXVIII.

THEOREM.

In equal Circles, equal Right Lines cut off equal Parts of the Circumferences; the greater Part of the one Circumference equal to the greater Part of the other, and the leffer equal to the leffer.

ET ABC, DEF, be equal Circles, in which

are the equal Right Lines BC, EF, which cut off the greater Circumferences B A C, E D F, and the leffer Circumferences B G C, EHF. I fay, the greater Circumference B A C is equal to the greater Circumference EDF, and the leffer Circumference BGC to the leffer Circumference EHF.

For affume the Centres K and L of the Circles; and join B K, K C, E L, L F.

Because the Circles are equal, the Lines drawn from their Centres are alfo equal. Therefore the two Sides B K, K C, are equal to the two Sides E L, LF; and the Bafe BC is equal to the Bafe E F. Therefore the Angle BKC is + is equal to the Angle 126 of this. ELF. But equal Angles ‡ stand upon equal Cir

+ 8. 1.

cumferences, when they are at the Centres. Wherefore the Circumference BGC is equal to the Circumference EHF, and the whole Circumference ABGCA equal to the whole Circumference DEHFD; and fo the remaining Circumference B A C fhall be equal to the remaining Circumference E DF. Therefore, in equal Circles, equal Right Lines cut off equal Parts of the Circumferences; which was to be demonftrated.

PROPOSITION XXIX.

THEOREM.

In equal Circles, the Right Lines, which fubtend equal Circumferences, are equal.

LET there be two equal Circles, ABC, DEF; and let the equal Circumferences BGC, EHF, be affumed in them, and B C, E F, joined. I fay,

the

the Right Line BC is equal to the Right Line EF.

For, find * K and L, the Centres of the Circles; * 1 of this. and join B K, KC, EL, LF.

Then, because the Circumference BGC is equal to

the Circumference EHF, the Angle BKC fhall

bet equal to the Angle ELF. And because the † 27 of this.
Circles ABGCA, DEHF D, are equal, the
Lines drawn from their Centres fhall be equal. ‡ Def. 1.
Therefore the two Sides B K, K C, are equal to the
two Sides EL, LF; and they contain equal Angles:
Wherefore the Bafe BC is + equal to the Bafe E F.
† 4. I.
And fo, in equal Circles, the Right Lines which fubtend
equal Circumferences, are equal; which was to be de-
monstrated.

PROPOSITION XXX.

PROBLEM.

To cut a given Circumference into two equal

L'

Parts.

'ET the given Circumference be AD B. It is required to cut the fame into two equal Parts.

Join A B, which bifect in C; and let the Right * 10. 1. Line CD be drawn from the Point C at Right Angles to A B+; and join A D, D B.

Now, because A C is equal to CB, and C D is common, the two Sides A C, C D are equal to the two Sides B C, CD; but the Angle A C D is equal to the Angle BCD; for each of them is a Right Angle: Therefore the Bafe A D is + equal to the Bafe B D. But equal Right Lines cut off equal Circumferences. Wherefore the Circumference A D fhall be equal to the Circumference B D. Therefore, a given Circumference is cut into two equal Parts; which was to be done.

tan

I 28 of rbise

PRO

5.1.

PROPOSITION XXXI.
THEORE M.

In a Circle, the Angle that is in a Semicircle, is a
Right Angle; but the Angle in a greater Seg-.
ment is less than a Right Angle; and the Angle
in a leffer Segment, greater than a Right Angle :
Moreover, the Angle of a greater Segment is
greater than a Right Angle; and the Angle of
a leffer Segment is less than a Right Angle.

LET there be a Circle ABDC, whofe Diameter

is B C, and Centre E; and join BA, A C, A D, DC. I fay, the Angle which is in the Semicircle BAC is a Right Angle; that which is in the Seganent A B C being greater than a Semicircle, viz. the Angle ABC is lefs than a Right Angle, and that which is in the Segment ADC being less than a Semicircle; that is, the Angle ADC is greater than a Right Angle.

For join A E, and produce B A to F.

Then, becaufe BE is equal to E A, the Angle EAB fhall be equal to the Angle EB A. And because A E is equal to EC, the Angle A CE will be * equal to the Angle C A E. Therefore the whole Angle B AC, is equal to the two Angles A B C, ACB; but the Angle F AC, being without the Tri† 32. 1. angle ABC, is + equal to the two Angles ABC, ACB; therefore the Angle BAC is equal to the Def.16. 1. Angle FAC; and fo each of them is a Right An gle. Wherefore, the Angle BA C, in a Semicircle, is a Right Angle. And because the two Angles A B C, BAC, of the Triangle ABC, are lefs than two Right Angles, and B A C is a Right Angle; then, ABC is less than a Right Angle; and is, in the Segment ̧ ABC, greater than a Semicircle.

* 17. I.

And fince ABCD is a quadrilateral Figure in a Circle, and the oppofite Angles of any quadrilate.al +22 of this. Figure defcribed in a Circle are equal to two Right Angles; the Angles A B C, AD C, are equal to two Right Angles; and the Angle ABC is lefs than a Right Angle. Therefore, the remaining Angle ADC