fect A B in F, and from F draw the circle. And because the right line AD touches the circle A B E, and the right line A B drawn from the point of contact A, cuts the circle; the angle D A B [by 32. 3.] will be equal to the angle which is in the alternate fegment of the circle, viz. equal to A E B : But the angle D A B is equal to the angle c; and therefore the angle e will be equal to the angle A EB. Wherefore a fegment A E B of a circle is described upon the right line A B containing an angle A EB equal to a given right-lined angle c. Secondly, Let the given angle c be a right angle: and again, it is required to defcribe a fegment of a circle upon the given right line A B containing an angle equal to the given angle c. Again, in the fecond figure, make the angle B AD [by 23. 1.] equal to the right angle c; and [by 10. I.] bifect AB in F, and with the centre F, and a diftance equal to A F or F B describe the circle A E B. Then [by 16. 3.] the right line A D touches the circle A B E, because the angle at A is a right angle. And the angle B AD is equal to the angle in the fegment A EB: for it is [by 31. 3.] a right angle, because it is in a femicircle. But the angle B AD is equal to the angle c. Therefore again a segment A E B of a circle is defcribed upon the right line A B containing an angle equal to the angle c. Laftly, Let the angle c be obtufe, and at the right line A B, AB, and the point A make [by 23. 1.] the angle B A D equal to the angle c, as in the third figure; and [by 11. 1.] draw the right line A E at right angles to B D, and again [by 10. 1.] bifect AB in F, draw FG at right angles to A B, and join G B. And because AF is equal to FB, and FG is common, the two fides FG, AF are equal to the two fides B F, FG, and the angle AFG is equal to the angle BFG: Therefore [by 4. 1.] the bafe AG is equal to the bafe GB; therefore a circle defcribed with the centre G and distance AG, will pafs thro' B. Let it be defcribed as A E B. And because A D is drawn from the end of the diameter AE at right angles to it, the right line A D [by 16. 3.] will touch the circle A E B, and A B is drawn from the point of contact A: Wherefore the angle B A D is equal to the angle which is in the alternate fegment A H B of the circle. But the angle BAD is equal to the angle c. Therefore the angle c is equal to the angle in the alternate segment A HB of the circle. Wherefore the fegment A HB of a circle is defcribed upon the given right line A B, containing an angle equal to the angle c. Which was to be done. PROP. XXXIV. PROBL. To cut off a fegment from a given circle, containing an angle equal to a given right lined angle. Let ABC be a given circle, and the given right-lined angle D: It is required to cut off a fegment from the given circle A B C, containing an angle equal to the given rightlined angle D. D Draw [by 17. 3.] the right line E F touching the circle A E B ABC in the point B, and at the right line F E, and point B therein make [by 23. 1.] the angle FBC equal to the angle D. Then because the right line EF touches the circle A B C, and the right line B C is drawn F from the point of contact B ; the angle FBC [by 32. 3.] will be equal to that in the alternate segment B A C of the circle. But the angle FBC is equal to the angle D. Therefore the angle in the fegment B A C will also be equal to the angle D. Therefore Therefore the fegment B A C is cut off from the given circle A B C containing an angle equal to the given right lined angle D: Which was to be done. PROP. XXXV. THEOR. If two right lines in a circle mutually cut one another, the rectangle contained under the fegments of the one, will be equal to the rectangle contained under the fegments of the other 1. For let two right lines, AC, B D in a circle A B C D2 mutually cut one another in the point E: I fay the rectangle contained under A E, E C is equal to the rectangle contained under D E, E B. Now if A C, B D país thro' the centre, fo that E is the centre of the circle A B CD: it is manifeft, that A E, E C, DE, EB being then equal, the rectangle contained under AE, EC will be equal to the rectangle contained under DE, E B. But now let the right lines A c, DB, not pass thro' the centre, and [by 1. 3.] find F the centre of the circle ABCD, and [by 12. 1.] from F draw the right lines F G, F H perpendicular to A C, DB, and join F B, FC, A D right line AC And because not drawn thro' the centre at right angles; it will [by 3. 3.] bifect the fame: fo that A G is equal to G C. the right line AC is divided into equal parts in the point G, and into unequal ones in the point E, the rectangle contained under A E, E C, together with the fquare of GE, will [by 5. 2.] be equal to the fquare of G c. Add the fquareof G F to both: then the rectangle under AE, EC, together with the íquares of G E, G F is equal to the fquares of CG, GF. But [by 47. 1.] the fquare of F E is equal to the squares of EG, GF, and the square of Fc equal to L the Book III. the fquares of CG, GF: Therefore the rectangle under A E, EC, together with the fquare of F E, is equal to the fquare of Fc. But FC is equal to FB: Therefore the rectangle under AE, E C, together with the square of E F, is equal to the fquare of F B. By the fame reason, the rectangle under DE, E B, together with the fquare of F E, is equal to the fquare of F B. But it has been proved that the rectangle under A E, E C, together with the square of FE is equal to the fquare of F B: Therefore the rectangle under AE, E C, together with the fquare of F E, is equal to the rectangle under DE, E B, together with the fquare of F E. Take away from both the fquare of F E, and then there will remain the rectangle under A E, E C equal to the rectangle under DE, EB. Therefore if two right lines in a circle mutually cut one another, the rectangle contained under the fegments of the one, will be equal to the rectangle contained under the fegments of the other. Which was to be demonftrated. i This theorem may be inverted thus: If two right lines fo cut one another, that the rectangle contained under the fegments of the one, be equal to the rectangle contained under the fegments of the other; a circle may be defcribed thro' the extreme points of thofe two eight lines, that is, a circle defcribed thro' any three of thofe extreme points will also pass thro' the fourth. For let the right lines A B, CD mutually cut one another in E, fo that the rectangle under AE, EB be equal to the rectangle under CE, ED: Ifay, the four points A, D, B, C do all fall in the circumference of the fame circle, and fo that circle which paffes thro' the three points, A, D, B, will neceffarily pass thro' the point c. For defcribe fome circle thro' the three points A, D, B [how this is to be done, fee prop. 5. lib. 4.] which if is does not pafs thro' c, will pafs beyond c, on this fide of it, or thro' F. Then because the rectangle under FE, ED is [by 35. 3.] equal to the rectangle under a E, EB; and the rectangles under c E, E D, is also put equal to the fame rectangle under A E, EB; the rectangles under CE, ED, and under FE, ED will be equal. The part and the whole; which is abfurd. Therefore the circle paffes thro' the point c. was to be demonstrated. C F Α C E Which be more This xxxvth propofition, as well as the next, may elegantly demonftrated by the fixth book, chiefly from the propor. proportionality of the fides of equiangular triangles, and by prop. xvi. of that book, But Euclid wanted thefe propofitions before, for demonftrating the tenth and eleventh of the fourth book; and therefore he was obliged to demonstrate these propofitions in the third book, as well as he could: befides, as the third book treats of the circle, he has put them into their proper place. PROP. XXXVI. THEOR. If any point be taken without a circle, and two right lines be drawn from it to the circle, the one whereof cuts the circle, and the other touches the circle; the rectangle contained under the whole fecant for cutting line] and its outward fegment between the affumed point, and the convexity of the circle, will be equal to the fquare of the tangent. For let any point D be taken without the circle A B C, and let two right lines D C A, D B, be fo drawn, that DCA cuts the circle, and D B touches it: I fay, the rectangle under A D, DC is equal to the fquare of D B. Now DCA does either pass thro' the centre or not. First let it pass thro' the centre F of the circle A B C : and join F B. Then [by 18. 3.] the angle F B D will be a right an D D gle; and fo because the right line A c is bifected in F, and the right fine C D is added to it, the rectan gle under AD, DC together with the fquare of FC, will [by 6, 2.] be equal to the fquare of FD; but F C is equal to FB: Therefore the rectangle under A D, DC, together with the fquare of F B, is equal to the fquare of F D. But the fquare of F D [by 47. 1.] is equal to the fquares of FB, BD; for the angle FBD) is a right angle. Therefore the rectangle under A D, DC, to L 2 gether |