Book III. the bafe B C is equal to the base E F: Therefore the angle BKC [by 8. 1.] is equal to the angle E L F. But [by 26. 3.] equal angles, when they are at the centres, ftand upon equal parts of the circumference: Wherefore the part BGC of the one circumference is equal to the part E HF of the other circumference. But the whole circumference AB C is equal to the whole circumference D E F : Therefore the remaining part BAC of the one circumference, will be equal to the remaining part EHF of the other circumfer

ence.

Therefore in equal circles, equal right lines cut off equal parts of their circumferences, the greater part equal to the greater, and leffer part equal to the leffer. Which was to be demonftrated.

PROP. XXIX. THEOR.

In equal circles the right lines cutting off equal parts of their circumferences are equal.

Let the circles ABC, DEF be equal, and let BG C' EH F be equal parts of their circumferences; and join B C, EF: I fay, the right line в c is equal to the right line E F.

For, find K, L the centres of the circles. And join B K, K C, EL, LF. Then because the part BGC of one Fcircumference is e

qual to the part EHF of the other; the an

gle BKC fhall be equal to the angle ELF [by 27. 3.]. And because the circles A B C, DEF are equal, the lines drawn from their centres will be equal: Wherefore the two fides BK, K C are equal to the two fides EL, LF, and they con tain equal angles: Therefore the bafe B C [by 4. 1.] is equal to the base E F.

Wherefore in equal circles, the right lines cutting off equal parts of their circumferences are equal. Which was to be demonftrated.

PROP.

PROP. XXX. PROBL.

To divide any given part of the circumference of a circle into two equal parts.

Let A DB be a given part of the circumference of a circle: It is required to divide this given part A DB of the circumference of a circle into two equal parts.

Join A B, and [by 10. 1.] bifect the fame in c; from the point c [by II. I.] draw CD at right angles to A B, and join AD, DB.

C

B

Then because AC is equal to CB, and CD is common, the two fides A C, CD are equal to the two fides A BC, CD, and the angle A C D is equal to the angle B C D, as being each of them a right angle: Therefore the base AD [by 4. 1.] is equal to the base DB. But [by 28. 3.] equal right lines cut off equal parts of circumferences, the greater part equal to the greater, and the leffer part equal to the leffer; and each of the parts AD, DB of the circumference, is lefs than half the circumference. Wherefore the part A D of the circumference will be equal to the part B D of it.

Therefore any given part of the circumference of a circle is divided into two equal parts. Which was to be done.

PROP. XXXI. THEOR.

Any angle in a femicircle is a right angle, any angle in a greater fegment is less than a right angle, and any angle in a leffer fegment is greater than a right angle; and moreover the angle of a greater fegment is greater than a right angle, and the angle of a leffer fegment is less than a right angle.

Let ABCD be a circle, whose diameter is B C, and centre E. Take any points A and D, in the circumference, and join B A, A C, A D, DC: I fay, the angle B A C in the femicircle B A C is a right angle: The angle in the segment A B C greater than a femicircle, is less than a right angle; and the angle ADC in the fegment DA C, which is lefs than a femicircle, is greater than a right angle.

Join

A

Б

F

E

D

Join A E, and produce B A to F. Then because B E is equal to E A, the angle EA B is [by 5. 1.] equal to the angle EBA. Also because E A is equal to EC,' the angle A CE Cwill be equal to the angle cAE: and fo the whole angle B A C, is equal to both the angles A B C, A CB. But the angle F A C without the triangle ABC, is [by 32. 1.] alfo equal to both the angles ABC, ACB: Wherefore the angle B A C is equal to the angle F A C ; and fo each of them [by 10. def. 1.] is a right angle: Wherefore the angle C A B in a femicircle is a right angle.

And because the two angles A BC, BAC of the triangle ABC [by 17. 1.] are lefs than two right angles, and B A C is a right angle; the angle ABC will be lefs than a right angle. And it is in the fegment A B C, which is greater than a femicircle.

And fince A B C D is a quadrilateral figure in a circle, and [by 22. 3.] the oppofite angles of this figure are equal to two right angles; the angles ABC, A DC will be equal to two right angles; and the angle A B C is lefs than a right angle: Therefore the remaining angle A D C will be greater than a right angle, and it is in the segment ADC, which is less than a femicircle.

I fay moreover, the angle of a greater fegment contained by the circumference A B C, and the right line AC, is greater than a right angle; and the angle of the leffer segment contained by the circumference ADC, and the right line AC is lefs than a right angle; which is indeed fufficiently evident. For because the angle contained by the right lines BA, AC is a right angle, and the angle contained by the circumference A B C and the right line A c is greater than a right angle. Also because the angle contained by the right lines CA, AF is a right angle; the angle contained by the right line C A and the circumference A D C, is less than a right angle.

Therefore any angle which is in a femicircle is a right angle, any angle in a great fegment is less than a right angle, and any angle, in a leffer fegment is greater than a right angle, and moreover the angle of a greater fegment

is greater than a right angle, and the angle of a leffer feg ment is less than a right angle. Which was to be demonftrated.

Otherwife :

The angle B A C is proved to be a right angle thus. Because the angle AEC is double to the angle BA E, fince [by 32. 1.] it is equal to both the inward oppofite angles; and the angle A E B is double to the angle EAC. Therefore the angles A E B, A EC will be double to the angle BA C. But [by 13. 1.] the angles AEB, AEC are equal to two right angles: Therefore the angle B A C will be one right angle. Which was to be demonftrated.

Corollary. From hence it is manifeft, if one angle of a triangle be equal to the two others, it is a right angle; because the angle which is adjacent to it, is equal to both those angles; and because [by 10. def. 1.] when adjacent angles are equal, they are each of them a right angle.

PROP. XXXII. THEOR. If a right line touches a circle, and a right line be drawn from the point of contact cutting the circle; the angles that this line makes with the tangent, will be equal to the angles that ftand in the alternate Segments of the circle.

For let the right line EF touch the circle A B CD in B, and draw any right line BD from B cutting the circle: I fay, the angles that BD makes with the tangent E F, are equal to the angles which ftand in the alternate fegments of the circle; that is, the angle FBD is equal to any angle in the segment D A B ; and the angle F B D equal to any angle in the fegment D C B.

For [by 11. 1.] draw BA from the point в at right angles to EF, take any point c in the circumference B D, and join AD, D C, C B.

Then because the right line EF touches the circle ABCD in the point B, and B A is drawn from the point of contact в at right angles to the tangent, the centre of the circle E

D

B

F

ABCD

A B CD [by 19. 3.] will fall in B A: Wherefore [by 31. 3.] the angle ADB in a femicircle is a right angle: therefore the other two angles B AD, A B D are both together one right angle. But [by conftr.] A BF is a right angle: therefore [by ro. ax.] the angle ABF is equal to the angles BAD, ABD. Take A B D, which is common, from both; then will the remaining angle D B F be equal to an angle in the alternate fegment of the circle, viz. to the angle BAD. And because A B C D is a quadrilateral figure in a circle, its oppofite angles [by 22. 3.] are equal to two right angles : Therefore [by 13. 1.] the angles DBF, DBE, are equal to the angles B A D, BC D. But B A D has been proved to be equal to DBF: Therefore the remaining angle D B E will be equal to the angle 'D CB in the alternate fegment DCB of the circle.

If therefore a right line touches a circle, and any right line be drawn from the point of contact cutting the circle, the angles that this line makes with the tangent, will be equal to the angles ftanding in the alternate segments of the circle. Which was to be demonstrated.

PROP. XXXIII. PROBL.

To defcribe a fegment of a circle upon a given right line containing an angle equal to a given right-lined angle.

Let AB be a given right line, and c the given rightlined angle: it is required to defcribe the fegment of a circle upon the given right line AB, containing an angle equal to the angle c. This angle is either an acute, a right angle, or an obtufe angle.

F

D

In the first place, let it be E an acute angle, as in the first figure, and at the right line A B and point A therein, make [by 23. 1.] the angle B A D equal to the angle c: Therefore BAD is an acute angle. And from the point A draw [by 11. 1.] the right line A E at right angles to A D, and [by 10. 1.] bifect

B