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 (18. 3.) angle: and because the straight line AC is bisected in E, and produced to the point D, AD.DC+EC=ED (6. 2.). But EC=EB, therefore AD.DC+ EB-ED. Now ED3=(47.1.) EB2+BD, because EBD is a right angle; therefore AD. DC + EB=EB + BD, and taking EB3 from each, AD DC= BD2. B But, if DCA does not pass through the centre of the circle ABC, take (1.3.) the centre E, and draw EF perpendicular (12. 1.) 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 centre, at right angles, it likewise bisects (3. 3.) it; therefore AF is equal to FC; and because the straight line AC is bisected in F, and produced to D (6. 2.), В AD.DC+FC2=FD2; add FE to both, then AD.DC+FC+FE=FD+FE. But (47. 1.) EC=FC+FE3, and ED2=FD+FE, because DFE is a right angle; therefore AD. DC+EC=ED. Now, because EBD is a right angle, ED3=EB+BD*=EC+BD, A and therefore, AD.DC + EC2 = EC2 + BD, and AD.DC=BD2. Wherefore, if from any point, &c. Q.E.D. Cor. If from any point without a circle, 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 D circle, are equal to one another, viz. BA. AE=CA.AF; for each of these rectangles is equal to the square of the straight line AD, which touches the circle. F A 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 he 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 (17. 3.) the straight line DE touching the circle ABC; find the centre F, and join FE, FB, FD; then FED is a right (18. 3.) angle: and because DE touches the circle ABC, and DCA cuts it, the rectangle AD. DC is equal (36. 3.) 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: but FE is equal to FB, wherefore DE, EF are equal to DB, BF; and the base FD is common to the two triangles DEF,DBF; therefore the angle DEF is equal (8.1.) to the angle DBF; and DEF is a right angle, therefore also DBF is a right angle: but FB, B if produced, is a diameter, and the straight line which is drawn at right angles to a diameter from the extremity of it, touches (16.3.) the circle: therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D. E ELEMENTS OF GEOMETRY. BOOK IV. DEFINITIONS. I. A RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angles of the inscribed figure are upon the sides of the figure in which it is inscribed, ✓ each upon each. II. In like manner, a figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each. III. A rectilineal figure is said to be inscribed in a circle, when all the angles of the inscribed figure are upon the circumference of the circle. IV. A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle. V. In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle touches each side of the figure. VI. A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described. 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, not greater than the diameter of the circle. Let ARC be the given circle, and D the given straight line, not greater than the diameter of the circle. Draw BC the diameter of the cir cle ABC; then, if BC be equal to D, the thing required is done; for in the circle ABC a straight line BC is placed equal to D: But, if it be not, BC is greater than D; make CЕ еqual (3. 1.) to D, and from the centre C, at the distance CE, describe the circle AEF, and join CA: Therefore, because C is the centre of the circle AEF, CA is equal to CF; but Dis equal to CE; therefore D is equal to CA: Wherefore, in the circle ABC, a straight line is placed, equal to the given straight line D, which is not greater than the diameter of the circle. Which was to be done. PROP. II. PROB. 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 (17.3.) the straight line GAH touching the circle in the point A, and at the point A, in the straight line AH, make (23. 1.) the angle HAC equal to the angle DEF; and at the point A, in the 1 qual to DEF; for the same rea son, the angle ACB is equal to the angle DFE; therefore the remaining angle BAC is equal (32.1.) to the 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. PROP. III. PROB. About a given circle to describe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle; it is required to describe a triangle about the circle ABC equiangular to the triangle DEF. Produce EF both ways to the points G, H, and find the centre K of the circle ABC, and from it draw any straight line KB; at the point K in the straight line KB, make (23. 1.) the angle BKA 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 (17.3.) the circle ABC: Therefore, 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 (18.3.) angles. And because the four angles of the quadrilateral figure AMBK are equal to four right angles, for it can be divided into two triangles; and because two of them, KAM, KBM |