Book III. angles to AE, therefore AD touches the circle; and because AB drawn from the point H Н 'Cor. 16. 3. of contact A cuts the circle, C A F B • 32. 3. AHB5: But the angle DAB is equal to the angle C, there- E PROP. XXXIV. PROB. a 17. 3. 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 D 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. Draw a the straight line EF touching the circle ABC in the point B, and at the point B, in the straight 23. 1. line BF, make b the angle FBC equal to the С angle D: Therefore, be D E B F tact B, the angle FBC * 32. 3. is equal to the angle in the alternate segment 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. Ir two straight lines within a circle cut one an- See N. other, 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, cut one another in the point E; the rectangle contained by AE, EC is equal to the rectangle contained by BE, ED. A ID 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 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 ACwhich does not pass through the centre, at right angles in E, AE, EC are equal a to one another: 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 A E equalb to the square of FB ; that is, to the square of FA: but the squares of AE, EF are equal to the square B of FA; therefore the rectangle BE, ED, together with the square of EF, is equal to the squares of AE, EF: Take away the common square of EF, and the remaining rectangle BE, ED is equal to the remaining square of AE; 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: Then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F 3. 3. 15. 2. c 47. 1. d12. 1. e 3. 3. f Boox III. drawd 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 5. 2. of these equals add the square of GF; therefore the rect angle AE, EC, together with the squares of EG, GF, is D F E G the square of AF; that is, to the B square of FB: But the square of square of ÊF: Take away the common 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 pass Н. F E G PROP. XXXVI. THEOR. If from any point without a circle two straight Let D be any point without the circle ABC, and DCA, ** 18. 3. DB two straight lines drawn from it, of which DCA cuts Book HE. the circle, and DB touclies 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: D And because the straight line AC is bisected in E, and produced to the point D, the rectangle AD, DC, together with the square of EC, is equalb C b 6. 2. to the square of ED, and CE is equal to EB: Therefore the rectangle AD, BA DC, together with the square of EB, is equal to the square of ED: But the square of ED is equal< to the squares € 17. I. of EB, BD, 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 the tangent DB. But if DCA does not pass through the centre of the circle • 1. 3. ABC, taked the centre E, and draw EF perpendiculare to AC, 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 D centre, at right angles, it shall likewise bisect' it; therefore AF is equal 13. 3. 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 equalb to the B. square of FD: To each of these equals add the square of FE; therefore the F rectangle AD, DC, together with the E squares of CF, FE, is equal to the squares of DF, FE But the square of ED is equal to the squares 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 e 12. 1. Book III. squares of EB, BD are equal to the square of ED, be cause EBD is a right angle; therefore the rectangle AD, • 47. 1. 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. Wherefore, if from any point, &c. Q. E. D. Cor. If from any point without a A F B В PROP. XXXVII. THEOR. See N. IF from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; 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 of the line which meets it, the line which meets it shall touch the circle. Let any point D be taken without the circle ABC, and from it let iwo straight lines DCA and DB be drawn, of which DCA cuts the circle, and DB meets it; if the rectangle AD, DC be equal to the square of DB, DB touches the circle. • 17. 3. Drawa the straight line DE, touching the circle ABC, find its centre F, and join FE, FB, FD; then FED is a 18. 3. right b angle : And because DE touches the circle ABC, < 36. 3. and DCA cuts it, the rectangle AD, DC is equal to the square of DE: But the rectangle AD, DC is, hy hypothesis, equal to the square of DB : Therefore the square of DE is equal to the square of DB; and the straight line DE b |