the angles DBF, DBE, being likewise equalf to two right angles, Book III. are equal to the angles BAD, BCD; and DBF has been proved equal to BAD: therefore the remaining angle DBE. is equal to f 13. 1. the gle BCD in the alternate segment of the circle. Wherefore, if a straight line, &c. Q. E, D. PROP. XXXIII. PROB. UPON a given straight line to describe a segment See N. of a circle, containing an angle equal to a given rec. tilineal angle. a 10.1. a 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, containing an angle equal to the angle C. First, Let the angle at C be a H right angle, and bisect a AB in F, с A Bb 31. 3, to the right angle at C. But, if the angle C be not a right angle, at the point A, in the straight line AB, make c the angle BAD equal to the an.c 23. 1. gle C, and from the point A H draw d AE at right angles to AD; bisect a AB in F, and E from F draw d FG at right d 11. 1. angles to AB, and join GB: and because AF is equal to FB, and FG common to the triangles AFG, BFG, the А 13 two sides AF, FG are equal F с to the two BF, FG; and the angle AFG is equal to the angle BFG; therefore the P base AG is equal e to the base GB; and the circle described c 4. 1. from the centre G, at the distance GA, shall pass 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 G Book III. right angles to AE, therefore AD f touches the circle; and be. cause AB drawn from the point C H f Cor. of contact A cuts the circle, the 16. 3. angle DAB is equal to the F angle in the alternate segment B g 32. 3. AHB 8: but the angle DAB is equal to the angle C, therefore G also the angle C is equal to the E angle in the segment AHB : D 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. TO cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and the given rectilineal angle; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the given angle D. a 17. 3. Draw a the straight line EF touching the circle ABC in the point B, and at the point B, A to the angle D; therefore, с point of contact B, the angle ( 32. 3. FBC is equal e to the angle in the alternate segment E B F BAC of the circle : but the angle FBC is equal to the angle D; therefore the angle in the segment BAC is equal to the angle D: wherefore the segment BAC is cut off from the given circle ABC containing an angle equal to the given angle D. Which was to be done. Book III. PROP. XXXV. THEOR. If two straight lines within a circle cut one ano. See N, ther, 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, within the circle ABCD, A E D If AC, BD pass each of them through the centre, so that E is the centre; it is evident, that AE, EC, BE, ED, being all B equal, the rectangle AE, EC is likewise C 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: then, if BD be bisected in F, 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, EC are equal a to one ano D a 3. 3. ther: and because the straight 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 square of EF, is equal to the square b 5.2. of FB, that is, to the square of FA; but A the squares of AE, EF are equal to the Cc 47. 1, E B 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 drawd FG d 12.1, N Book III. perpendicular to AC; therefore AG is equal a to GC; where fore the rectangle AE, EC, together with the square of EG, is a 3. 3. equal b to the square of AG: to each of these equals add the b 5. 2. square of GF; therefore the rectangle AE, EC, together with the squares of EG, GF, is equal to D the squares of AG, GF: but the c 47. 1. squares of EG,GF are equal c to the square of EF; and the squares of F B Lastly, Let neither of the straight lines AC, BD pass through the centre: take the centre F, and H D .F E А G is equal to the rectangle BE, ED. B PROP. XXXVI. THEOR. 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 DCA, DB Book III. two straight lines drawn from it, of which DCA cuts the circle, and DB touches the same : the rectangle AD, DC is equal to the square of DB. Either DCA passes through the centre, or it does not; first, let it pass through the centre E, and join EB ; therefore the angle EBD is a right a angle : and be a 18. 3. cause the straight line AC is bisected D in E, and produced to the point D, the rectangle AD, DC, together with the square of EC, is equal b to the square C b 6.2. of ED, and CE is equal to EB: therefore the rectangle AD, DC, together B 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, be E c 47. 1. cause 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 remain А ing rectangle AD, DC is equal to the square of the tangent DB. But if DCA does not pass through the centre of the circle ABC, take d the centre E, and draw EF perpendiculare to AC, d 1. 3. and join EB, EC, ED: and because the straight line EF, which e 12. 1, passes through the centre, cuts the straight line AC, which does not pass through the centre, at right D angles, it shall likewise bisect fit; f 3. 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: to B each of these equals add the square of FE; therefore the rectangle AD, DC, I E together with the squares of CF, FE, is equal to the squares of DF, FE ; but the square of ED is equal é to the squares A А of DF, FE, because EFD is a right angle ; and 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: and 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 o of ED, be |