11 Ax. +3 Ax. +Const. angle: but ABF is likewise at right angle; therefore the angle ABF is equal to the angles BAD, ABD: take from these equals the common angle ABD: therefore the remaining angle DBF is equal to the angle BAD, which is in the alternate segment of the circle. And because ABCD is a quadrilateral figure in a circle, the opposite angles *22 3. BAD, BCD are equal* to two right angles: but the angles DBF, DBE are likewise equal to two right angles; therefore the angles DBF, DBE 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 segment of the circle. Wherefore, if a straight line, &c. Q. E. D. 13. 1. +1 Ax. +3 Ax. PROP. XXXIII. PROB. Upon a given straight line to describe a segment of a circle, capable of containing an angle equal to a given rectilineal angle. Let AB be the given straight line, and the angle at C the given rectilineal angle; it is required to describe upon the given straight line AB a segment of a circle, which shall contain an angle equal to the angle C. First, let the angle at C be a right angle: #10. 1. * 31. 3. #23. 1. 11. 1. #10. 1. #11. 1. +12 Ax. bisect AB in F, and from the C A H But, if the angle C be not a right angle; at the point A, in a straight line AB, make* the angle BAD equal to the angle C, and from the point A draw AE at right angles to AD; bisect* AB in F, and from F draw* FG at right angles to AB, FG will meet AE in some point G, because GAF is less than a right angle;t and if from G, as a centre with the distance GA, a circle AHB be H described, it will pass through the point B, and AHB will be the segment required, and join GB. Then because AF is equal to FB, and FG common to the triangles AFG, BFG, the two sides AF, FG are equal to the two BF, FG, each to each: and the angle AFG is equal to the angle BFG; therefore the base AG is equal to the base GB; and therefore the circle AHE described from the centre G, at the distance GA, shall pass +8 Def. 1. *4. 1. through the point B: moreover the segment AHB shall contain an angle equal to the given rectilineal angle C. #32. 3. H B G D segment AHB: but the angle DAB is equal to +Const. the angle C; therefore the angle C is equal to the +1 Ax. angle in the segment AHB. Wherefore, upon the given straight line AB the segment AHB of a circle is described, which contains an angle equal to the given angle at C. Which was to be done. PROP. XXXIV. PROB. From a given circle to cut off a segment, which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off from the circle ABC a segment that shall contain an angle equal to the given angle D. #17. 3. Draw the straight line EF touching the circle ABC in the point B, and at the point B, in the *23. 1. straight line BF, make* the angle FBC equal to the angle D: the segment BAC shall contain an angle equal to the given angle D. #32. 3. E B Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal to the angle in the alternate segment BAC of the circle: but the *Const. angle FBC is equal to the angle D; therefore the +1. Ax. angle in the segment BAC is equal to the angle D. Wherefore from the given circle ABC, the segment BAC is cut off, containing an angle equal to the given angle D. Which was to be done. PROP. XXXV. THEOR. If two straight lines cut one another within a circle, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other. Let the two straight lines AC, BD, cut one another in the point E within the circle ABCD: the rectangle contained by AE, EC shall be equal to the rectangle contained by BE, ED. If AC, BD pass each of them through the centre, so that E is the centre; it is evident +12 Def. 1. BE-ED. A B that AE, EC, BE, ED, being allt equal, the rectangle AE⚫EC is likewise equal to the rectangle But let one of them BD pass through the centre, and cut the other AC, which does not pass through the centre, at right angles, in the point E: then, if BD be bisected in F, L F is the centre of the circle ABCD: join AF; and because BD which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E, AE is equal to EC: and #3.3. because the straight line BD is cut into two equal parts in the point F, and into two unequal parts in the point E, the rectangle BE ED, together with the *5. 2. *47. 1. +1 Ax. +3 Ax. *12. 1. *3. 3. E B square of EF, is equal* tothesquare Next, let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in E, but not at right angles: then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw FG perpendicular to AC; therefore AG is equal to GC; wherefore the rectangle AE⚫EC, together with the square of EG, is equal to the square of AG: to each of these equals add the square of GF; therefore the rectangle AE·EC, together with the squares of EG, GF, is equal† to the squares of AG, GF: but the *5.2. +2 Ax. #47. 1. squares of EG, GF are equal* to D the square of EF; and the squares of AG, GF are equal to the square of AF: therefore the rectangle AE·EC, together with the square of EF, is equal to the square of AF; that is, to the square of FB: *5.2. tl Ax. but the square of FB is equal to the rectangle BE ED, together with the square of EF; therefore the rectangle AE EC, together with the square of EF, is equal to the rectangle BE ED, together with the square of EF; take away the common +3 Ax. square of EF, and the remaining rectangle AE⚫EC, is therefore equal to the remaining rectangle BE-ED. Lastly, let neither of the straight lines AC, BD +1. 3. pass through the centre: taket H F the centre F, and through E, the intersection of the straight 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 same reason the rectangle BE ED is equal to the same rectangle GE EH; therefore the rectangle AE-EC is equal to the rectangle BE-ED. Wherefore if two straight lines, &c. +1 Ax. Q. F. D. B NOTE. This proposition may be concisely demonstrated as follows: Let ABD be a circle, and P any point within it. It is required to prove that whatever chord AB be drawn through, P, the rectangle AP PB will be always the same. Find the centre C, and from it draw CM at right angles to AB; draw also CP, CB. Then *3.3. *5. 2. #47.1. because AB is divided equally A M B D squares of PM, MC, that is, together with the square of PC, are equal to the squares of MB, MC, that is, to the square of BC. Therefore, taking away the square of PC, the rectangle AP.PB alone, is equal to the difference of squares of BC, PC. But this difference remains the same whatever chord be drawn through P, because BC is always the semidiameter, and PC is the distance of the fixed point P from the centre. Hence if two chords, &c. |