* equal to the two Right Angles. Therefore likewife, * 32. I. the Angles BAC, BDC fhall be equal to two Right Angles. And after the fame Way we prove, that the Angles ABD, ACD, are alfo equal to two Right Angles. Therefore the oppofite Angles of any Quadrilateral Figure defcribed in a Circle, are equal to two Right Angles; which was to be demonstrated.

PROPOSITION XXIII.

THEOREM.

Two fimilar and unequal Segments of two Circles, cannot be fet upon the fame Right Line, and on the fame Side thereof.

FOR

'OR if this be poffible, let the two fimilar and unequal Segments ACB, ADB, of two Circles ftand upon the Right Line AB on the fame Side thereof. Draw ACD, and let CB, BD, be joined. Now because the Segment ACB is fimilar to the Segment AD B, and fimilar Segments of Circles are fuch* Def. 11. which receive equal Angles; the Angle ACB will of this. be equal to the Angle AD B, the outward one to the inward one; which is † abfurd. Therefore fimilar † 16. 1s and unequal Segments of two Circles, cannot be fet upon the fame Right Line, and on the fame Side thereof; which was to be demonftrated.

PROPOSITION XXIV.

THEOREM.

Similar Segments of Circles being upon equal Right
Lines, are equal to one another.

ET AEB, CFD be equal Segments of Circles ftanding upon the equal Right Lines AB, CD. I fay, the Segment AEB is equal to the Segment CFD. For the Segment A E B being applied to the Segment CFD, fo that the Point A co-incides with C, and the Line AB with CD; then the Point B will co-incide with the Point D, fince AB and CD are equal. And fince the Right Line AB co-incides with CD,

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CD, the Segment AEB will co-incide with the Segment CFD. For if at the fame Time that AB co-incides with CD, the Segment AEB fhould not co-incide with the Segment CFD, but be otherwise, as CGD; then a Circle would cut a Circle in more Points than two, viz. in the Points C, G, D; which * 10 of this is impoffible Wherefore if the Right Line AB co-incides with CD, the Segment AEB will co-incide with and be equal to the Segment CF D. Therefore fimilar Segments of Circles being upon equal Right Lines, are equal to one another; which was to be demonftrated.

*

PROPOSITION XXV.

PROBLEM.

A Segment of a Circle being given, to describe the
Circle whereof it is the Segment.

TET ABC be a Segment of a Circle given. It
is required to defcribe a Circle, whereof ABC
is a Segment.

Bifect AC in D, and let DB be drawn + from the Point D at Right Angles to AC, and join AB. Now the Angle ABD is either greater, equal, or less than the Angle BAD. And firft let it be greater, and make the Angle BAE at the given Point A, with the Right Line BA, equal to the Angle ABD; produce DB to E, and join E C.

*

Then because the Angle ABE is equal to the Angle BAE, the Right Line BE will be equal to EA. And becaufe AD is equal to DC, and DE common, the two Sides AD, DE, are each equal to the two Sides CD, DE; and the Angle ADE is equal to the Angle CDE; for each is a Right one. Therefore the Bafe A E is equal to the Bafe EC. But AE has been proved to be equal to EB. Wherefore BE is alfo equal to EC. And accordingly the three Right Lines AE, EB, EC, are equal to each other. Therefore a Circle described about the Center E, with either of the Distances A E, EB, EC, fhall pafs thro' the other Points, and be that required to be described. But it is manifest that the Segment ABC is less than a Semi

a Semicircle, because the Center thereof is without the fame.

But if the Angle ABD be equal to the Angle BAD; then if AD be made equal to BD, or DC, the three Right Lines AD, DB, DC, are equal between themselves, and D will be the Center of the Circle to be described; and the Segment ABC is a Semicircle.

But if the Angle ABD is lefs than the Angle BAD, let the Angle BAE be made, at the given Point A with the Right Line BA, within the Segment AB C, equal to the Angle ABD.

Then the Point E, in the Right Line DB, will be the Center, 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.

THEOREM.

In equal Circles, equal Angles ftand upon equal Circumferences, whether they be at their Centers, or at their Circumferences.

ET ABC, DEF, be equal Circles, and let BGC, EHF, be equal Angles at their Centers, and BAC, EDF, equal Angles at their Circumferences. I fay the Circumference BKC is equal to the Circumference ELF.

For let BC, EF, be joined. Because ABC, DEF, are equal Circles, the Lines drawn from their Centers will be equal. Therefore the two Sides BG, GC, are equal to the two Sides EH, HF; and the Angle G is equal to the Angle at H. Wherefore the Base BC is equal to the Bafe EF. Again, because the * 4. 1. Angle at A is equal to that at D, the Segment BAC will be + fimilar to the Segment EDF; and they are † Def, 11. upon equal Right Lines BC, EF. But thofe fimilar Segments of Circles, that are upon equal Right Lines,

are equal to each other. Therefore the Segment 1 34 of this. BAC, will be † equal to the Segment EDF. But + Def. 1 the whole Circle ABC, is equal to the whole Circle DEF. Therefore the remaining Circumference BKC,

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fhall

23. I.

fhall be equal to the remaining Circumference ELF. Therefore in equal Circles, equal Angles ftand upon equal Circumferences, whether they be at their Centers, 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 Centers or Circumferences.

ET the Angles BGC, EHF, at the Centers

gles BAC, EDF, at their Circumferences, ftand upon the equal Circumferences BC, EF. I fay the Angle BGC is equal to the Angle EHF, and the Angle BAC to the Angle EDF,

:

*

For if the Angle BGC be equal to the Angle EHF, it is manifeft that the Angle BAC is alfo equal to the Angle EDF But if not, let one of them be the greater, as BGC, and make the Angle BGK, at the Point G, with the Line BG, equal to the Angle † 26 of this. EHF. But equal Angles ftand † upon equal Circumferences, when they are at the Centers. Wherefore the Circumference BK, is equal to the Circumference EF. But the Circumference EF is equal to the Circumference BC. Therefore BK 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 one half of the Angle EHF. 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 Centers or Circumferences; which was to be demonftrated,

PRO

PROPOSITION XXVIII.

THEORE M.

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

L

ET ABC, DEF, be equal Circles, in which are the equal Right Lines BC, EF, which cut off the greater Circumferences BAC, EDF, and the leffer Circumferences BGC, EHF. I fay the greater Circumference BAC, is equal to the greater Circumference EDF, and the leffer Circumference BGC, to the leffer Circumference EHF.

For affume the Centers K and L of the Circles, and join BK, KC, EL, LF.

Because the Circles are equal, the Lines drawn from their Centers are alfo equal. Therefore the Def. 1. two Sides BK, KC, are equal to the two Sides EL, LF; and the Bafe BC, is equal to the Bafe EF. Therefore the Angle BKC, is equal to the Angle † 8. 1. ELF. But equal Angles ftand upon equal Circum- ‡ 26 of this. ferences, when they are at the Centers. Wherefore

the Circumference BGC, is equal to the Circumference EHF, and the whole Circle A B C, equal to the whole Circle DEF; and fo the remaining Circumference BA C, fhall be equal to the remaining Circumference EDF. Therefore in equal Circles, equal Right Lines cut off equal Parts of the Circumfe→ rences; which was to be demonstrated.

PROPOSITION XXIX.

THE ORE M.

In equal Circles, equal Right Lines fubtend equal
Circumferences.

LET there be two equal Circles ABC, DEF;
and let the equal Circumferences BGC, EHF,
be affumed in them, and BC, EF, joined. I fay,
the Right Line BC is equal to the Right Line EF.

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For