rectangle BE, ED, together with the square of EF; therefore the rectangle A E, 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 A E, EO is therefore equal to the remaining rectangle BE, ED. Lastly, Let neither of the straight lines AC, BD pass through the centre: Take the centre F, and through E, the intersection of the straight lines AC, DB, draw the diameter GEFH: And because the rectangle A E, 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 A same rectangle GE, EH; therefore the rectangle A E, EC is equal to the rectangle BE, ED. Wherefore, if two straight lines, &c. Q.E.D. G в 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 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 angle (III. 18.): And because the straight line Ac is bisected in E, and produced to the point D, the rectangle A D, DC, together with the square of Ec, is equal (11. 6.) to the square of ED, and ce is equal to EB: Therefore the rectangle AD, DC, together with the square of E B, is equal to the square of ED: But the square of Ed is equal (1.47.) to the squares of EB, BD, because EBD is a right angle: Therefore the rectangle AD, DC, together with the square of EB, is A equal to the squares of EB, BD: Take D B A 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 A BC, take (11I. 1.) the centre E, and draw EF perpendicular (1. 12.) 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 (111. 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 (11. 6.) to the B square of FD: To each of these equals add the square of FE; therefore the rectangle' AD, DC, together with the squares of CE, FE, is equal to the squares of DF, FE: But the square of Ed is equal (1. 47.) 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 E B, is equal to the square of ED: But the squares of EB, BD are equal to the square (1. 47.) of Ed, because EBD is a right angle; therefore the rectangle AD, DC, together with the square of E B, is equal to the squares of E B, 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 eircle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circle are equal to one another, viz. the rectangle BA, A E to the rectangle D CA, AF: for each of them is equal to the square of the straight line AD which touches the circle. A B 1 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 be equal to the square of DB; DB touches the circle. Draw (11. 17.) the straight line de touching the circle ABC, find its centre F, and join FE, FB, FD; and then FED is a right angle (111. 18.): And because DE touches the circle ABC and DCA cuts it, the rectangle AD, DC is equal (111. 36.) 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 D E equal to the straight line dB: And Fe is equal to FB, wherefore DE, EF are equal to DB, BF, each to each ; and the base FD is common to the two triangles DEF, DBF; therefore the angle DEF is equal (1. 8.) to the angle dbF; but DEF is a right angle, therefore also DBF is a right angle: And FB, if produced, is a dia eter, and the straight line B which is drawn at right angles to a diameter, from the extremity of it, touches (111. 16.) the circle: Therefore touches the circle ABC. Wherefore, if from a point, &c. Q. E. D. D DB GEOMETRICAL EXERCISES, EXERCISES ON BOOK I. B PROBLEMS. No. I. 1. To trisect a right angle ; that is, to divide it into three equal parts. Let KBC be a right angle: It is required to trisect it. In Bk take any point A; on BA describe (Euclid 1. 1.) the equilateral triangle ABD; and bisect (E. 1. 9.) the angle ABD by the straight line BE; then the angles CBD, DBE, and EBA are equal to one another. For the triangle ABD being equilateral the angle ABD (E. I. 5. and 32.) is one third of two right angles, or two thirds of the right angle ABC; therefore the angle CBD is one third of the right angle CBA; but since the angle DBA is bisected by BE, the angles DBE and EBA are each of them equal to one third of a right angle; wherefore the angles CBD, DBE, and EBA are equal to one another. 2. To trisect a given straight line. Let AB be the given straight line: It is required to divide it into three equal parts. Upon A B describe (E. 1. 1.) the equilateral triangle ABC; bisect (E. I. 9.) the angles CAB and CBA by the lines Ad and BD meeting in D; and from o draw (E. 1. 31.) DE parallel to CA, and DF parallel to CB: Then AB is trisected in the points E and F. А F B E D PI F For since dE is parallel to CA and ad meets them, therefore (E. I. 29.) the angles DAC and ADE are equal; but (by constr.) the angle EaD is equal to the angle DAC; therefore the angle ADE is equal to the angle EAD; wherefore (E. 1. 6.) AE is equal to ED. For the same reason FB is equal to FD. But dE being parallel to CA and DF to CB, the angle der is equal to the angle CA B, and DFE to CBA; and therefore the remaining angle EDF is equal to the remaining angle ACB; hence the triangle EDF is equiangular, and therefore ED, E F, FD are equal to one another ; but Ed has been shown to be equal to A E, and FD to FB, therefore A E, EF, FB are equal to one another. 3. Through a given point to draw a straight line which shall make equal angles with two straight lines given in position. Let p be the given point, and BE and CF the lines given in position. It is required to draw, through the point P, a straight line which shall make equal angles with BE and CF. Produce BE and cF to meet in A; bisect (E. 1. 9.) the angle CAB by the line AD; and from p let fall (E. 1. 12.) the perpendicular pd, and produce it on both sides to E and F. Then FE is the line required. For, by construction, the angle DAE is equal to the angle DAF, and the angles at D are right angles, and moreover Ad is common to both the triangles ADE and ADF, therefore (E. 1. 26.) the angle AED is equal to the angle AFD. 4. From two given points on the same side of a line given in position, to draw two lines which shall meet in that line, and make equal angles with it. Let c and D be the two given points, and as the line given in position. From c let fall (E. 1. 12.) the perpendicular cG, and produce it to F making GF equal to GC; join H T FD, CE; then CE and ED are the lines required. Since cg is equal to GF, and GE is common to the two triangles CGE с D F |