or PROOF.—Because from the point A, the extremity of the And AD liameter AE, AD is drawn at right angles to AE (Const.); the circle, Therefore AD touches the circle (III. 16, cor.) Because AB is drawn from the point of contact A, the angle DAB is equal to the angle in the alternate segment A/B (III. 32). But the angle DAB is equal to the angle C (Const.); and .. 4 Therefore the angle in the segment AHB is equal to the in AHB angle C (Ax. 1). Therefore, on the given straight line AB, the segment AHB of a circle has been described, containing an angle equal to the given angle C. Q.E.F. Proposition 34.-Problem. 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 angle D. CONSTRUCTION.—Draw the straight line EF touching the Draw tancircle ABC in the point B (III. gent EBF, 17); And at the point B, in the straight line BF, make the angle FBC equal to the angle D (I. 23). given 2. Then the segment BAC shall contain an angle equal to the given angle D. PROOF.—Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B (Const.); Therefore the angle FBC is equal to the angle in the alternate segment BAC of the circle (III. 32), But the angle FBC is equal to the angle D (Const.); Therefore the angle in the segment BAC is equal to the :: 2 BAC angle D (Ax, 1). Therefore, from the given circle ABC, the segment BAC has been cut off, containing an angle equal to the given angle D. Q.E.F. and E = _ FBC = LD B Proposition 35.-Theorem. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them shall be 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 and EC shall be equal to the rectangle contained by BE and ED. CASE I.—Let AC, BD pass each of them through the centre. PROOF.—Because E is the centre, EA, EB, EC, ED are all equal (I. def. 15); Therefore the rectangle AE, EC is equal to the rectangle BE, ED. CASE II.-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. CONSTRUCTION. — Bisect BD in F, then F is the centre of the circle; join AF. PROOF.—Because BD, which passes through the centre, cuts AC, which does not pass through the centre, at right angles in E (Hyp.); Therefore AE is equal to EC (III. 3). And because BD is cut into two equal parts in the point F, and into two un equal parts in the point E, The rectangle BE, ED, together with the square on EF, is equal to the square on FB (II. 5); that is, the square on AF. But the square on AF is equal to the squares on AE, EF (I. 47); Therefore the rectangle BE, ED, together with the square on EF, is equal to the squares on AE, EF (Ax. 1). Take away the common square on EF; Then the remaining rectangle BE, ED is equal to the remaining square on AE; that is, to the rectangle AE, EC, since AE is equal to EC. AE= EC. B B В BE ED + .. BE ED = AE2= AE.EC. F A C Bisect BD in F the centre. Draw FG at right angles to AC. B GC. EG2 .. AE.EC CASE III.-Let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in the point E, but not at right angles. CONSTRUCTION.—Bisect BD in F, then F is the centre of the circle. Join AF, and from F draw FG perpendicular to AC (I. 12). ProoF.—Then AG is equal to GC .. AG = (III. 3). Therefore the rectangle AE, EC, together with the square Now, AE:ÉC + on EG, is equal to the square on AG (II. 5). To each of these equals add the square on GF; = AG2. Then the rectangle AE, EC, together with the squares on EG, GF, is equal to the squares on AG, GF (Ax. 2). But the squares on EG, GF are equal to the square on EF; And the squares on AG, GF are equal to the square on AF (I. 47). Therefore the rectangle AE, EC, together with the square on EF, is equal to the square on AF; that is, the square on FB. +EF2 But the square on FB is equal to the rectangle BE, ED, together with the square on EF (II. 5); Therefore the rectangle AE, EC, together with the square on EF, is equal to the rectangle BE, ED, together with the square on EF. Take away the common square on EF; And the remaining rectangle AE, ÉC is equal to the .: AE EC remaining rectangle BE, ED (Ax. 3). CASE IV.-Let neither of the straight lines AC, BD pass through the centre. CONSTRUCTION.- Take the centre F Again, (III. 1), and through E, the intersection of the lines AC, BD, draw the diameter A = BE-ED, GEFH. as just PROOF.-Because the rectangle GE, EH is equal, as has been shown, to the rectangle AE, EC, and also to the rectangle BE, ED; Therefore the rectangle AE, EC is equal to the rectangle .AE EC BE, ED (Ax. 1). Therefore, if two straight lines, &c. Q.E.D. AF2 = FB2 = BE ED + EF2. = BE-ED, GE.EH = G shown, = BE.ED. Proposition 36.-Theorem, Let D be any point without the circle ABC, and let DCA, CASE I.—Let DCA pass through the centre E, and join EB. PROOF.—Then EBD is a right angle (III. 18). And because the straight line AC is bisected in E, and produced to D, the rectangle AD, DC, together with the square on EC, is equal to the square on ED (II. 6). But EC is equal to EB; Therefore the rectangle AD, DC, together with the square on EB, is equal to the square D AD DC + HE А on ED. .. ADDC + EB2 = BD2+EB2. .. AD.DC = BD2. B But the square on ED is equal to the squares on EB, BD, because EBD is a right angle (I. 47); Therefore the rectangle AD, DC, together with the square on EB, is equal to the squares on EB, BD. Take away the common square on EB; Then the remaining rectangle AD, DC is equal to the square on DB (Ax. 3). CASE II.—Let DCA not pass through the centre of the circle ABC. CONSTRUCTION.—Take the centre E (III. 1), and draw EF perpendicular to AC (I. 12), and join EB, EC, ED. PROOF.—Because the straight line EF, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles, it also bisects it (III. 3); Draw EF perpendicular to AC. .. AF = FC. + .: AD DC + EC2 = DE2. Therefore AF is equal to FC. And because the straight line AC is bisected in F and produced to D, the rectangle AD, DC, together with the ; AD DC square on FC, is equal to the square on FD (II. 6). To each of these equals add the square on FE; Therefore the rectangle AD, DC, together with the squares on CF, FE, is equal to the squares on DF, FE (Ax. 2). But the squares on CF, FE are equal to the square on CE, because CFE is a right angle (I. 47); And the squares on DF, FE are equal to the square on DE; Therefore the rectangle AD, DC, together with the square on CE, is equal to the square on DE. But CE is equal to BE; Therefore the rectangle AD, DC, together with the square on BE, is equal to the square on DE. But the square on DE is equal to the squares on DB, BE, because EBD is a right angle (I. 47); Therefore the rectangle AD, DC, together with the square ::: AD:DU on BE, is equal to the squares on DB, BE. Take away the common square on BE; Then the remaining rectangle AD, DC is equal to the .:. AD-DC square on DB (Ax. 3). Therefore, if from any point, &c. Q.E.D. COROLLARY.—If from any point without a circle there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of D them without the circle, are equal to one another; namely, the rectangle BA, AE is equal to the rectangle CA, AF; for each of them is equal to the square on the straight line AD, which touches the circle. + BE2 DB2+ BEX = DB2. Proposition 37.—Theorem. If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it, and if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square on |