a 3. 3 b s. 2. Book 111. dicular to AC; therefore AG is equal to GC; wherefore mthe rectangle AE, EC, together with the square of EG, is e qual 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 D €47. I. squares of EG, GF are equal to the square of EF; and the squares of F E c EC, together with the square of EF, G is equal to the square of AF; that is, B. to the square of FB : 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 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 through D с G B PRO P. XXXVI. THEOR. IF 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 two straight lines drawn from it, of which DCA cuts the circle, circie, and DB touches the same: The rectangle AD, DC is Book III. equal to the square of DB. Either DCA passes through the center, or it does not ; first, let it pafs through the center E, and join EB; therefore the angle EBD is a right angle: And because a 18. 3. D the straight line AC is bisected in E; and produced to the point D, the rectangle AD, DC, together with the b 6.2. square of EC, is equal to the square of ED, and CE is equal to EB: There fore the rectangle AD, DC, together B with the square of EB, is equal to the square of ED : But the square of ED, C 47. I. is equal to the squares of EB, BD, becaute 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 A 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 center of the circle ABC, taked the center E, and draw EF perpendiculare to d 1.-3. AC, and join EB, EC, ED: And because the straight line EF, e 12. 1. which passes through the center, cuts the straight line AC, which does not pass through the center, at right angles, it shall likewise bisect fit; there: fore AF is equal to FC: And because the straight line AC is bifected in F, and produced to D, the rectangle AD, DC, together with the square of FC, is equal to the square of FD: To each of these equals add the square of FE;there B fore the rectangle AD, DC, together F with the squares of CF, FE, is equal to E 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, DE, together with the square of EB, is equal f 3. 3: 1 G 2 to Book II. to the square of ED: But the squares of EB, BD are equal to the square of ED, because EBD is a right angle; therefore € 17. I. the rectangle AD, DC, together with the square of EB, is e qual to the squares of EB, BD: Take away the common square Cor. If from any point with- EU D с F 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 shall touch the circle. & 17. 3 18. 3. Let any point D be taken without the circle ABC, and from it let two 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. Draw * the straight line DE touching the circle ABC, find its center F, and join FE, FB, FD; then FED is a right angle: And because DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square of DE: But the rectangle AD, DC is, by hypothesis, equal to the square of DB : Therefore the square of DE is equal to the square of DB, and the straight line DE equal to the straight line DB: And € 36. 3. and FE is equal to FB, wherefore DE, EF are equal to DB, Book III. BF; and the base FD is common to the two triangles DEF, DBF; there D d 8.1. fore the angle DEF is equal to the angle DBF, but DEF is a right angle, therefore also DBF is a right angle: And FB, if produced, is a diameter, and the straight line which is drawn at right angles to a diameter, from the B E extremity of it touches - the circle : CIG. Therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D. A F A 1. Rectilineal figure is said to be inscribed in another recti lineal figure, when all the angles of the inscribed figure II. about another figure, when all the fides of III. in a circle, when all the angles of the in- IV. each side of the circumscribed figure V. bed in a rectilineal figure, when the cir- |