Book III. Greek text, and the tranflations from it, the angle of the greater fegment is faid to be greater, and the angle of the leís fegment is faid to be lefs, than a right angle.' COR. From this it is manifeft, that if one angle of a triangle be equal to the other two, it is a right angle, because the angle adjacent to it is equal to the fame two; and when the adjacent angles are equal, they are right angles. 1 a II. I. b 19. 3. c 31.3. d 32. I. e 22. 3. IF F a ftraight line touches a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle, fhall be equal to the angles which are in the alternate fegments of the circle. Let the ftraight line EF touch the circle ABCD in B, and from the point B let the flraight line BD be drawn cutting the circle: The angles which BD makes with the touching line EF fhall be equal to the angles in the alternate fegments of the circle; that is, the angle FBD is equal to the angle which is in the fegment DAB, and the angle DBE to the angle in the fegment BCD. a From the point B draw BA at right angles to EF, and take any point C in the circumference BD, and join AD, DC, CB; and because the straight line EF touches the circle ABCD in the point B, and BA is A D C drawn at right angles to the touch- E B F two two right angles; therefore the angles DBF, DBE, being like wife equal to two right angles, are equal to the angles BAD, BCD; and DBF has been proved equal to BAD: Therefore the. remaining angle DBE is equal to the angle BCD in the alternate fegment of the circle. Wherefore, if a straight line, &c. Q. E. D. f Book III. w 13. 1. PROP. XXXIII. PROB. PON a given ftraight line to defcribe a segment of see N. a circle, containing an angle equal to a given rec UPON tilineal angle. Let AB be the given ftraight line, and the angle at C the given rectilineal angle; it is required to defcribe upon the given ftraight line AB a fegment of a circle, containing an angle eequal to the angle C. Firft, Let the angle at C be a right angle, and bifect AB in F, and from the centre F, at the diftance FB, defcribe the femicircle AHB; therefore the angle AHB in a femicircle is bequal to the right angle at C. C H a ro. I. A F B b 31. 30 But, if the angle C be not a right angle, at the point A, in the ftraight line AB, make the angle BAD equal to the angle c 23. 1. C, and from the point A draw d AE at right angles to AD; bifect AB in F, and from F draw FG at right angles to AB, and join GB: And becaufe AF is equal to FB, and FG common to the triangles AFG, BFG, the two fides AF, FG are equal to the two BF, FG; and the angle AFG is equal to the angle BFG; therefore the H d II. I. E bafe AG is equal to the bafe GB; and the circle defcribede 4. I. from the centre G, at the diftance GA, fhall pafs through the point B; let this be the circle AHB: And because from the point A the extremity of the diameter AE, AD is drawn at right J Book III. right angles to AE, therefore ADf touches the circle; and becaufe AB drawn from the point f Cor. 16. 3. of contact A cuts the circle, g 32. 3. a 17.3. 23. I' € 32.3. the angle DAB is equal to the C to the angle in the fegment D given ftraight line AB the feg H ment AHB of a circle is described which contains an angle equal to the given angle at C. Which was to be done. 10 cut off a fegment from a given circle which fhall contain an angle equal to a given rectilineal angle. T Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a fegment from the circle ABC that fhall contain an angle equal to the angle D. a A Draw the ftraight line EF touching the circle ABC in the point B, and at the point B, in the ftraight line BF, make b the angle FBC equal to the angle D: Therefore, because the ftraight line EF touches the circle ABC, and BC is D drawn from the point of contact B, the angle FBC is equal to the angle in the alternate segment BAC E BF of the circle: But the angle FBC is equal to the angle D; therefore the angle in the fegment BAC is equal to the angle D: Wherefore the fegment BAC is cut off from the given cirele ABC containing an angle equal to the given angle D: Which was to be done. Book III IF PRO P. XXXV. THE OR. two ftraight lines within a circle cut one another, the See N. rectangle contained by the fegments of one of them is equal to the rectangle contained by the fegments of the other. Let the two ftraight lines AC, BD, within the circle ABCD, cut one another in the point E; the rectangle contained by AE, EC is equal to the rectangle contained by BE, ED. A If AC, BD pafs each of them through the centre, fo that E is the centre; it is evident, that AE, EC, BE, ED, being all equal, the rectangle AE, EC is likewife B equal to the rectangle BE, ED. a D E But let one of them BD pafs through the centre, and cut the other AC, which does not pafs through the centre, at right angles, in the point E: Then, if BD be bifected in F, F is the centre of the circle ABCD; and join AF: And because BD, which paffes through the centre, cuts the ftraight line AC, which does not pafs through the centre, at right angles in E, AE, EC are equal to one another: And because the ftraight line BD is cut into two equal parts in the point F, and into two unequal in the point E, the rectangle BE, ED together with the fquare of EF, is equal to the fquare of FB; that is, A to the fquare of FA; but the squares of AE, EF are equal to the fquare of FA; therefore the rectangle BE, ED, together with the fquare of EF, F a 3.3. b 5.2. E B is equal to the fquares of AE, EF: Take away the common fquare of EF, and the remaining rectangle BE, ED is equal to the remaining fquare of AE; that is, to the rectangle AE, EC. © 47. I. Next, Let BD which paffes through the centre, cut the other AC, which does not pafs through the centre, in E, but not at right angles: Then, as before, if BD be bifected in F, F is the centre of the circle. Join AF, and from F draw FG perpen- d 12. 1. dicular G a 3. 3. Book III. dicular to AC; therefore AG is equal to GC; wherefore the rectangle AE, EC, together with the fquare of EG, is equal to the fquare of AG: To each of these equals add the fquare of GF; therefore the rectangle AE, EC, together with the fquares of EG, GF, is equal to b 5. a. the fquares of AG, GF: But the D of FB is equal to the rectangle BE, ED, together with the fquare of EF; therefore the rectangle AE, EC, together with the fquare of EF, is equal to the rectangle BE, ED, together with the fquare of EF: Take away the common fquare of EF, and the remaining rectangle AE, ÉC is therefore equal to the remaining rectangle BE, ED. H Laftly, Let neither of the ftraight lines AC, BD pafs through the centre: Take the centre F, and through E, the interfection of the ftraight lines AC, DB, draw the diameter GEFH: And because the rectangle AE, EC is equal, as has been shown, to the rectangle GE, EH; and, for the fame reafon, the A rectangle BE, ED is equal to the fame rectangle GE, EH; therefore the rectangle AE, EC is equal to the rectangle BE, ED. Q. E. D. IF D C G B Wherefore, if two straight lines, &c. F from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, fhall be equal to the fquare of the line which touches it. Let D be any point without the circle ABC, and DCA, DB two ftraight lines drawn from it, of which DCA cuts the circle, |