* I of this.

+ Def. 1.

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

Then because the Circumference BGC is equal to the Circumference EHF; the Angle BKC fhall be 27 of this. † equal to the Angle ELF. And because the Circles ABC, DEF, are equal, the Lines drawn from their Centers fhall be equal. Therefore the two Sides BK, KC, are equal to the two Sides EL, LF; and they contain equal Angles: Wherefore the Base BC is equal to the Bafe EF. And fo in equal Circles equal Circumferences fubtend equal Right Lines; which was to be demonstrated.

+ 4. I.

* IO. I.

4. I.

PROPOSITION XXX.

PROBLEM.

To cut a given Circumference into two equal Parts.

ET the given Circumference be AD B. It is required to cut the fame into two equal Parts. Join AB, which bifect * in C; and let the Right Line CD be drawn from the Point C at Right Angles to AB; and join AD, DB.

Now because AC is equal to CB, and CD is common, the two Sides AC, CD, are equal to the two Sides BC, CD; but the Angle ACD is equal to the Angle BCD; for each of them is a Right Angle: Therefore the Bafe AD is equal to the 28 of this. Bafe BD. But equal Right Lines cut off equal Circumferences. Wherefore the Circumference AD fhall be equal to the Circumference BD. Therefore a given Circumference is cut into two equal Parts; which was to be done.

PRÓ

PROPOSITION XXXI.

THE QRE M.

In a Circle, the Angle that is in a Semicircle, is a Right Angle; but the Angle in a greater Segment, 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 Line.

ET there be a Circle ABCD, whofe Diameter is BC, and Center E; and join BA, AC, AD, DC. I fay, the Angle which is in the Semicircle BAC is a Right Angle, that which is in the Segment ABC being greater than a Semicircle, viz. the Angle ABC, is less than a Right Angle; and that which is in the Segment ADC, being lefs than a Semicircle, that is, the Angle ADC is greater than a Right Angle.

For join AE, and produce BA to F.

*

5. I.

Then because BE is equal to EA, the Angle EAB fhall be equal to the Angle EBA. And because * AE is equal to EC, the Angle ACE will be* equal to the Angle CAE. Therefore the whole Angle BAC is equal to the two Angles ABC, ACB; but the Angle FAC being without the Triangle ABC, is equal to the two Angles ABC, ACB: † 32. I. Therefore the Angle BAC is equal to the Angle FAC; and fo each of them is a Right Angle. ‡ Def. 10. 1. Wherefore the Angle BAC in a Semicircle is a Right Angle. And because the two Angles ABC, BAC, of the Triangle ABC*, are less than two Right * 17. 1. Angles, and BAC is a Right Angle; then ABC is less than a Right Angle, and is in the Segment ABC greater than a Semicircle.

And fince ABCD is a quadrilateral Figure in a Circle, and the oppofite Angles of any quadrilateral Figure described in a Circle, are + equal to two Right † 22 of this. Angles; the Angle ABC, ADC are equal to two Right Angles, and the Angle ABC is lefs than a Right Angle: Therefore the remaining Angle ADC

will

will be greater than a Right Angle, and is in the Segment ADC, which is lefs than a Semicircle.

I fay, moreover, the Angle of the greater Segment contained under the Circumference ABC, and the Right Line AC, is greater than a Right Angle; and the Angle of the leffer Segment, contained under the Circumference ADC, and the Right Line AC is lefs than a Right Angle. This manifeftly appears; for because the Angle contained under the Right Lines BA, A C, is a Right Angle, the Angle contained under the Circumference ABC, and the Right Line A C, will be greater than a Right Angle. Again, because the Angle contained under the Right Line CA, AF, is a Right Angle, therefore the Angle which is contained under the Right Line A C, and the Circumference ADC, is less than a Right Angle. Therefore, in a Circle, the Angle that is in a Semicircle, is a Right Angle; but the Angle in a greater Segment, is lefs 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; which was to be demonstrated.

PROPOSITION XXXII.

THEOREM.

If any Right Line touches a Circle, and a Right Line be drawn from the Point of Contact cutting the Circle; the Angles it makes with the Tangent Line, will be equal to those which are made in the alternate Segments of the Circle.

L'

ET any Right Line EF touch the Circle ABCD in the Point B, and let the Right Line BD be any how drawn from the Point B cutting the Circle. I fay, the Angles which BD makes with the Tangent Line EF, are equal to thofe in the alternate Segments of the Circle; that is, the Angle FBD is equal to an Angle made in the Segment DAB, viz. to the Angle DAB; and the Angle DBE equal to the Angle DCB, made in the Segment DCB. For

Draw

Draw* BA from the Point B at Right Angles to EF; and take any Point C in the Circumference BD, and join AD, DC, CB.

* II. I.

Then because the Right Line EF touches the Circle ABCD in the Point B; and the Right Line BA is drawn from the Point of Contact B at Right Angles to the Tangent Line; the Center of the Circle ABCD, will be in the Right Line BA; and fo† 19. 3. BA is a Diameter of the Circle, and the Angle ADB,

in a Semicircle, is a Right Angle. Therefore the ‡ 31 of this. other Angles BAD, ABD, are * equal to one Right * 32. 1. Angle. But the Angle ABF, is alfo à Right Angle: Therefore the Angle ABF, is equal to the Angles BAD, ABD; and if ABD, which is common, be taken away, then the Angle DBF remaining, will be equal to that which is in the alternate Segment of the Circle, viz. equal to the Angle BAD. And becaufe ABCD is a Quadrilateral Figure in a Circle, and the oppofite Angles thereof are † equal to two † 22 of this. Right Angles: The Angles DBF, DBE, will be equal to the Angles BAD, BCD. But BAD has been proved to be equal to DBF; therefore the Angle DBE, is equal to the Angle made in DCB, the alternate Segment of the Circle, viz. equal to the Angle DCB. Therefore, if any Right Line touches a Circle, and a Right Line be drawn from the Point of Contact cutting the Circle; the Angles it makes with the Tangent Line, will be equal to those which are made in the alternate Segments of the Circle; which was to be demonftrated.

PROPOSITION XXXIII.

PROBLEM.

To defcribe, upon a given Right Line, a Segment of a Circle, which shall contain an Angle, equal to a given Right-lined Angle.

L

ET the given Right Line be AB, and C the given Right-lined Angle. It is required to defcribe the Segment of a Circle upon the given Right Line AB containing an Angle, equal to the Angle C.

At

23. I.

II. I.

† 10. I.

14. I.

*Cor. 16. of this.

At the Point A, with the Right Line AB, make the Angle BAD equal to the Angle C, and draw *AE from the Point A, at Right Angles to AD. Likewife bifect + AB in F, and let FG be drawn from the Point F, at Right Angles, to AB, and join G B.

Then because AF is equal to FB, and FG is common, the two Sides AF, FG are equal to the two Sides BF, FG; and the Angle AFG, is equal to the Angle BFG. Therefore the Bafe AG is t equal to the Bafe GB. And fo if a Circle be described about the Center G, with the Distance AG, this shall pass through the Point B. Defcribe the Circle, which let be ABE, and join E B. Now because AD is drawn from the Point A, the Extremity of the Diameter A E, at Right Angles to AE, the faid AD will* touch the Circle. And fince the Right Line AD touches the Circle ABE, and the Right Line AB, is drawn in the Circle from the Point of Contact A, +32 of this. the Angle DAB is † equal to the Angle made in the alternate Segment, viz. equal to the Angle AEB. But the Angle DAB, is equal to the Angle C. Therefore the Angle C will be equal to the Angle A E B. Wherefore the Segment of a Circle AEB is defcribed upon the given Right Line AB, containing an Angle A E B, equal to a given Angle C; which was to be done.

1 17 of this. * 23. 1.

PROPOSITION XXXIV.

THEOREM.

To cut off a Sement from a given Circle, that fall contain an Angle, equal to a given Right-lin'dAngle.

ET the given Circle be ABC, and the Rightlined Angle given D. It is required to cut off a Segment from the Circle ABC, containing an Angle equal to the Angle D.

Draw the Right Line EF, touching the Circle in the Point B, and make the Angle F BC at the Point B equal to the Angle D.

Then because the Right Line EF touches the Circle ABC in the Point B, and BC is drawn from

the