PROP. XXXVII. THEOR. 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. 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 he equal to the square of DB; DB shall touch the circle. Draw* the straight line DE, touching the *17.3. circle ABC, find t its centre F, and join FE, †1. 3. FB, FD: then FED is a right * angle: _and * 18. 3 because DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal * to the *35. 3. square of DE : but the rectangle AD, DC is by hypothesis equal to the square of DB; therefore the square of DE is equalt to the square +1 Ax. of DB; and the straight line DE equal to the straight line DB: and FE is equal † to FB; +15Dei. wherefore DE, EF are equal to DB, BF, each to each: and the base FD is common to the two triangles D DEF, DBF; therefore the angle DEF is equal* to the angle DBF: hut DEF was shown to B/ E be a right angle; therefore also DBF ist a right angle: and BF, I + 1 Ax. if produced, is a diameter; and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches * the circle : there- * 16. 3. fore DB touches the circle ABC. Wherefore, if from a point, &c. Q.E.D. 1. * 8. I. * I. another rectilineal figure, when all the angles II. described about another figure, when all the JII. scribed in a circle, when all the IV. a circle, when each side of the circumscribed V. inscribed in a rectilineal figure, when the VI. a rectilineal figure, when the cir- VII. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle. PROP. I. PROB. In a given circle to place a straight line, equal to a given straight line which is not greater than the diameter of the circle. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle: it is required to place in the circle ABC a straight line equal to D. done; *3. 1. Draw BC | the diameter of the circle ABC; then, if BC is equal to D, the thing required is for in the circle ABC a straight line BC is placed equal to D: but, + Hyp. if it is not, BC is greatert than D: make CE equal* B F Because C is the centre of the circle AEF, #15 Def. CA is equal | to CE: but D is equal t to CE; therefore D is equalt to CA. Wherefore in the + Const. +1 Ax. circle ABC, a straight line CA is placed equal to the given straight line D, which is not greater than the diameter of the circle. Which was to be done. 1. PROP. II. PROB. * 17.3. In a given circle to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle; it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF. Draw* the straight line GAH touching the circle in the point A, and at the point A, in the *23. 1. straight line AH, make* the angle HAC equal to the angle DEF; and G. at the point A, in the D straight line AG, make H B +1.3. | Find + the centre, and through it draw any straight line #17 Def, BC terminated both ways by the circumference; this linet is 1. a diameter, Because HAG touches the circle ABC, and AC is drawn from the point of contact, the angle HAC is equal * to the angle ABC in the alter- *32. 3. nate segment of the circle: but HAC is equal + + Conet. to the angle DEF; therefore also the angle ABC is equal † to DEF: for the same reason, the +1 Ax. angle ACB is equal to the angle DFE: therefore the remaining angle BAC is equal * to the *32. 1. remaining angle EDF: wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Which was to be done. and PROP. III. PROB. angular to a given triangle. Produce EF both ways to the points G, H; find t the centre K of the circle ABC, and from +1.8 it draw any straight line KB; at the point K in the straight line KB, make* the angle BKA *23. 1. equal to the angle DEG, and the angle BKC equal to the angle DFH; and through the points A, B, C, draw the straight lines LAM, MBN, NCL, touching * the circle ABC: LMN * 17. 3. shall be the triangle required. Because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, the angles at the points A, B, C are right * angles: and because *18. 3. the four angles of the quadrilateral figure AMBK are equal to four right angles, for it can be I Suppose the line MK drawn: then, because the three angles of every triangle are together equal fto two right angles; +32. L. therefore all the angles of the two triangles AMK, BMK are together equal to four right angles; but all the angles of these triangles are together equal to the angles of the figure AMBK therefore all the angles of the figure AMBK are together equal to four right angles. * L |