but the angles DBF, DBE are likewise equal to two right angles; (1. 13.) therefore the angles DBF, DBE are equal to the angles BAD, BCD, (ax. 1.) 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. (ax. 2.) Wherefore, if a straight line, &c. Q. E.D. PROPOSITION XXXIII. PROBLEM. Upon a given straight line to describe a segment of a circle, which shall contain 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. and from the centre F, at the distance FB, describe the semicircle AHB. Therefore the angle AHB in a semicircle is equal to the right angle at C. (111. 31.) 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, (1. 23.) and from the point A draw AE at right angles to AD; (1. 11.) bisect AB in F, (1. 10.) and from F draw FG at right angles to AB, (1. 11.) and join GB. And 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; (1. def. 10.) therefore the base AG is equal to the base GB; (1.4.) and therefore the circle described from the centre G, at the distance GA, shall pass through the point B: let this be the circle AHB. The segment AHB shall contain an angle equal to the given rectilineal angle C. Because from the point A, the extremity of the diameter AE, AD is drawn at right angles to AE, therefore AD touches the circle: (111. 16. Cor.) and because AB, drawn from the point of contact A, cuts the circle, the angle DAB is equal to the angle in the alternate segment AHB: (III. 32.) but the angle DAB is equal to the angle C; (constr.) therefore the angle C is equal to the angle in the segment AHB. (ax. 1.) 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. Q. E. F. PROPOSITION XXXIV. 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 given angle D. Draw the straight line EF touching the circle ABC in the point B, (III. 17.) and at the point B, in the straight line BF, make the angle FBC equal to the angle D. (1.23.) Then the segment BAC shall contain an angle equal to the given angle D. Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, therefore the angle FBC is equal to the angle in the alternate segment BAC of the circle: (111. 32.) but the angle FBC is equal to the angle D; (constr.) therefore the angle in the segment BAC is equal to the angle D. (ax. 1.) Wherefore from the given circle ABC, the segment BAC is cut off, containing an angle equal to the given angle D. Q.E.F. 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. Then 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 that since AE, EC, BE, ED, being all equal, (1. def. 15.) therefore the rectangle AE, EC, is equal to the rectangle BE, ED. 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. 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, therefore AE is equal to EC: (III. 3.) and because the straight line BD is cut into two equal parts in the point F, and into two unequal parts in the point E, therefore the rectangle BE, ED, together with the square of EF, is equal to the square of FB; (11. 5.) that is, to the square of FA: but the squares of AE, EF, are equal to the square of FA; (1. 47.) therefore the rectangle BE, ED, together with the square of EF, is equal to the squares of AE, EF: (ax. 1.) take away the common square of EF, and the remaining rectangle BE, ED is equal to the remaining square of AE; (ax. 3.) that is, to the rectangle AE, EC. 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. DA F E G B Then, as before, if BD be bisected in F, Join AF, and from F draw FG perpendicular to AC; (1. 12.) wherefore the rectangle AE, EC, together with the square of EG, is equal to the square of AG: (11.5.) 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; (ax. 2.) but the squares of EG, GF, are equal to the square of EF; (1.47.) 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: but the square of FB is equal to the rectangle BE, ED, together with the square of EF; (11. 5.) therefore the rectangle AE, EC, together with the square of EF, is equal to the rectangle BE, ED, together with the square of EF; (ax. 1.) take away the common square of EF, and the remaining rectangle AE, EC, is therefore equal to the remaining rectangle BE, ED. (ax. 3.) Lastly, let neither of the straight lines AC, BD pass through the centre. H D G B Take the centre F, (111. 1.) and through E the intersection of the straight lines AC, DB, 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. (ax. 1.) Wherefore if two straight lines, &c. Q. E.D. PROPOSITION XXXVI. THEOREM. If 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, shall be equal to the square of the line which touches it. Let D be any point without the circle ABC, and let DCA, DB be two straight lines drawn from it, of which DCA cuts the circle, and DB touches the same. Then the rectangle AD, DC shall be equal to the square of DB. Either DCA passes through the centre, or it does not: first, let it pass through the centre E. D B Join EB, therefore the angle EBD is a right angle. (III. 18.) And because the straight line AC is bisected in E, and produced to the point D, therefore the rectangle AD, DC, together with the square of EC, is equal to the square of ED: (11. 6.) but CE is equal to EB; therefore the rectangle AD, DC, together with the square of EB, is equal to the square of ED: but the square of ED is equal to the squares of EB, BD, (1.47.) because EBD is a right angle: therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: (ax. 1.) take away the common square of EB; therefore the remaining rectangle AD, DC is equal to the square of the tangent, DB. (ax. 3.) But if DCA does not pass through the centre of the circle ABC. Take E the centre of the circle, (III. 1.) Ꭰ draw EF perpendicular to AC, (1. 12.) and join EB, EC, ED. And 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. Therefore EF also bisects AC; (III. 3.) 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 square of FC is equal to the square of FD: (11.6.) to each of these equals add the square of FE; therefore the rectangle AD, DC, together with the squares of CF, FE is equal to the squares of DF, FE: (ax. 2.) but the square of ED is equal to the squares of DF, FE, (1.47.) because EFD is a right angle; and for the same reason, the square of EC is equal to the squares of CF, FE; therefore the rectangle AD, DC, together with the square of EC, is equal to the square of ED: (ax. 1.) but CE is equal to EB; therefore the rectangle AD, DC, together with the square of EB, is equal to the square of ED: but the squares of EB, BD are equal to the square of ED, (1.47.) because EBD is a right angle: therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD; take away the common square of EB ; therefore the remaining rectangle AD, DC is equal to the square of DB. (ax. 3.) Wherefore, if from any point, &c. Q. E.D. COR. If from any point without a circle, there be drawn two |