to the angle B DC; for they are
both in the fame fegment B A D C,
and the angle ACB is equal to the
angle A D B, fince they are both in
the fame segment A DCB: There-
fore the whole angle A D C is equal
to the angles BA ̊C, A CB.
the common angle A B C to both;

Add

Then the angles A B C, B AC, ACB, will be equal to the angles ABC, ADC. But A B C, BA C, AC B, are equal to two right angles: Therefore the angles A B C, A D C also are equal to two right angles. We prove, after the fame manner, that the angles B A D, D C B are also equal to two right angles.

Therefore the oppofite angles of any quadrilateral figure described in a circle, are equal to two right angles. Which was to be demonstrated.

PROP. XXIII.

THEOR.

Two fimilar and unequal fegments of circles cannot be fet upon the fame right line, on the fame fide of it.

For if this be poffible, let two unequal fegments, ACB, A D B, of two circles ftand upon the fame right line AB, both on the fame fide of it. Draw A CD, and join

CB, D B.

Then because the segment ACB is fimilar to the fegment ADB, and those segments of circles are fimilar [by 11. def. 3.] which receive equal angles; the angle A C B will be equal to the angle A D B, the outward angle equal to the in-A

ward one. Which [by 16. 1.] is impoffible.

D

B

Therefore two fimilar and unequal fegments of two circles cannot be fet upon the fame right line, on the fame fide of it. Which was to be demonstrated.

PROP. XXIV. THEOR.

Similar fegments of circles, which are upon equal right lines, are equal to one another.

For let A EB, C F D be fimilar fegments of circles ftanding upon the equal right lines A B, CD: I fay, the fegments AE B, C F D are equal to one another.

For the fegment A E B being applied to the fegment CFD fo as the point A agrees

E

A

F

with C, and the line AB with

G

H.

BC

CD; the point B will agree with the point D, because DAB is equal to CD: But fince the right line AB agrees with C D, the fegment A E B will alfo agree with the fegment CF D. For if AB agrees with CD, and at the fame time the fegment A E B fhould not agree with the fegment CF D, but falls into a different fituation, as CH GD; then as one circle does not cut another [by 10. 3.] in more points than two. But the circle CHGD cuts the circle C F D in more points than two, viz, in the points C, G, D; which is impoffible: Therefore when the right line A B agrees with the right line & D, the fegment A E B cannot but agree with the fegment C F D, and fo will be equal to it.

Therefore fimilar fegments of circles, which are upon equal right lines, are equal to one another. Which was to be demonftrated.

PROP. XXV. PROBL. 1

A Jegment of a circle being given, to defcribe a circle whofe fegment it is.

Let A B C be a given segment of a circle it is required to describe the circle whofe fegment A B C is.

Divide [by 10. 1.] A C into two equal parts in D, and [by 11. 1. draw DB from the point D, at right angles to AC, and join Aa B. Then the angle ABD is either greater than the angle B A D, or equal to it, or lefs than it.

First let it be greater. And [by 23. 1.] make the angle BAB at the given point A with the line B A equal to the

angle

because the angle ABE is equal to

the angle B A E,

E

D

the right line B E [by 6. 1.] will be equal to E A. And becaufe A D is equal to DC, and D E is common, the two fides A D, DE are equal to the two fides c D, D E, each to each, and the angle A D E is equal to the angle C DE; for each of them is a right angle: Therefore [by 4. 1.] the base A E is equal to the bafe E C. But A E has been proved to be equal to EB: Wherefore B E is alfo equal to EC, and fo the three right lines A E, E B, E C, are equal to one another. Therefore a circle described about the centre E with either of the distances A E, E B, E C, will pafs through the other points, and [by 9. 3.] will be the circle required to be described. And it is alfo manifeft that the segment ABC is less than a femicircle; because the centre of the circle is without the segment.

But if the angle A B D be equal to the angle B AD; if A D be made equal to BD or DC, the three right lines D A, DB, DC will be equal to one another; and (by 9. 3.] D will be the centre of the circle to be defcribed; the fegment A B C in this cafe being a femicircle,

Laftly, if the angle A B D be lefs than the angle BAD, and the angle BAE be made at the given point A, with the right line BA, equal to the angle ABD, the centre E will be in the right line D B, within the fegment A B C, which fegment will be greater than a femicircle.

Therefore the fegment of a circle being given, the circle is described whofe fegment it is. Which was to be demonftrated.

PROP. XXVI. THEOR.

In equal circles equal angles ftand upon equal parts of their] circumferences, whether they be at their centres or their circumferences.

Let ABC, DEF be equal circles, and let the angles BG C, E HF at their centres be equal, and the angles BA C, EDF at their circumferences be equal: I fay, the part

K 4

BK C

BK C of the circumference of one circle, is equal to the part ELF of the circumference of the other,

For join BC, EF

Then because ABC, DEF are equal circles, the lines drawn from their centres will be equal: Therefore the two fides BG, GC are equal to the two fides E H, HF, and the angle at G is equal to the angle at H: Therefore [by 4. 1.] the bafe BC is equal to the

D

AA

E

K

CE

H

F

angle at A is equal to the angle at D, the fegment BAC [by 11. def. 3.] will be fimilar to the fegment EDF, and they are upon the equal right lines BC, EF: But fimilar fegments of circles which are upon equal right lines [by 24. 3.] are equal to one another: Therefore the fegment B A C is equal to the fegment ED F. But the whole circle ABC is likewife equal to the whole circle DEF: Therefore the remaining fegment B K C is equal to the remaining fegment ELF. And fo the part BKC of the circumference of one circle will be equal to the part ELF of the circumference of the other.

Therefore in equal circles, equal angles ftand upon equal parts of their circumferences, whether they ftand at their centres, or at their circumferences. Which was to be demonftrated.

PROP, XXVII. THEOR.

In equal circles, the angles that stand upon equal [parts of their] circumferences, are equal to one another, whether they be at their centres or at their circumferences.

B

For in the equal circles A B C, DEF, let the angles BGC, EHF at their centres, and the angles BA C, E D F, at their circumferences, ftand upon the equal parts E C, EF of the circumferences: I fay, the angle BGC is equal to the angle EHF, and the angle B AC, equal to the angle

ED F.

For if the angle BG C be equal to the angle EHF, it is manifeft [by 20. 3.] that the angle BAC is alfo equal to the

angle

angle EDF. But if not, one of them is the greater, which

let be BG C, and [by 23. 1.] make the angle BG K, at the point G, with the line B G, equal to the angle EHF. But equal angles [by 26. 3.] ftand upon equal parts of the circum

ferences of the circles when they are at the centres. Therefore the part B K of the one circumference, is equal to the part E F of the other circumference. But the part EF of one circumference is equal to the part B C of the other: Therefore B K is equal to в C, the lefs to the greater; which cannot be. Therefore the angle BGC is not unequal to the angle EHF; and fo they must be equal. But [by 20. 3.] the angle at A is one half the angle BG C, and the angle EDF one half the angle at H: Therefore the angle at A is equal to the angle at D.

Therefore in equal circles, the angles that ftand upon equal [parts of their] circumferences, are equal to one another, whether they ftand at their centres or circumferences. Which was to be demonftrated.

PROP. XXVIII. THEOR. In equal circles, equal right lines cut off equal parts of their circumferences, the greater part equal to the greater, and the lesser part equal to the lesser.

Let A B C, DEF be equal circles, and let the equal right lines BC, E F cut off the greater parts B A C, E DF, and the leffer parts B G C, E HF of the circumferences: I fay, the greater part B A C of the one circumference, will be equal to the greater part EDF of the other circumference, and the leffer part B G C equal to the leffer part E HF.

For [by 1. 3.] find the centres K, L of the circles, and join BK, KC, EL, LF.

Then because the circles are equal, the lines drawn from their centres will be equal: Wherefore the two fides B K, KC are equal to B the two fides E L, LF,

and

D

A

K

F

H

the