Book III. cause 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 DB. 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 circle, are equal to one another, viz. the rectangle BA, AE to the rectangle CA, AF: for each of them is equal to the square of the straight line AD which touches the circle. A F D C B PROP. XXXVII. THEOR. See N. a 17. 3. b 18. 3. c 36. 3. 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. c Draw a the straight line DE touching the circle ABC, find its centre F, and join FE, FB, FD; then FED is a right b 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 : D d 8. 1. FE is equal to FB, wherefore DE, EF are equal to DB, BF; Book III. and the base FD is common to the two triangles DEF, DBF; therefore 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. B at right angles to a diameter, from the extremity of it, touches the circle: therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D. E e 16. 3. F Α THE ELEMENTS OF EUCLID. BOOK IV. DEFINITIONS. I. Book IV. 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. See N. II. In like manner, a figure is said to be described gular points of the figure about which it is described, each III. A rectilineal figure is said to be inscribed in a circle, when all the angles of the in- 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 inscrib- K 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. Book IV. 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. Lét ABC be the given circle, and D the given straight line, not greater than the diameter of the circle. A Draw BC the diameter of the circle ABC; then, if BC is 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 is not, BC is greater than D; make CE equal a to D, and from the centre C, at the distance CE, describe the circle AEF, and join CA: therefore, because Ċ is the centre of the circle AEF, CA is equal to CE; but D is equal to CE; therefore D is equal to CA: wherefore, in D a 3. 1. E C B F 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. Book IV. a 17.3. b 23. 1. 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. D G Α H Draw a the straight line GAH touching the circle in the point A, and at the point A, in the straight line AH, make the angle HAC equal to the angle DEF; and at the point A, in the straight line AG, make the angle GAB equal to the angle DFE, and join BC therefore because HAG touches the circle ABC, and AC is drawn from the point of contact, the angle : c 32. 3. HAC is equal to the angle ABC in the alter nate segment of the cir A E F B C cle: but HAC is equal to the angle DEF; therefore also the angle ABC is equal to DEF: for the same reason, the angle ACB is equal to the angle DFE; therefore the remaining angle BAC d 32. 1. is equal 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; a 23. 1. at the point K, in the straight line KB, make a 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 17.3. lines LAM, MBN, NCL touching 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, c 18.3. KC, the angles at the points A, B, C are right angles: and because the four angles of the quadrilateral figure AMBK are |
