Ꭰ d 8. I. FE is equal to FB, wherefore DE, EF are equal to DB, BF Book III. and the bafe FD is common to the two triangles DEF, DBF; therefore the angle DEF is equal d to the angle DBF; but DEF is a right angle, therefore alfo 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 extremity of it, touches the cir cle: Therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D. e 16. gi F A Book IV. See N. THE ELE MEN T S OF EUCL I D. A воок IV. 'DEFINITIONS. I. RECTILINEAL figure is faid to be infcribed in another rectilineal figure, when all the angles of the infibed figure are upon the fides of the figure in which it is infcribed, each upon each. II. In like manner, a figure is faid to be described about another figure, when all the fides of the circumfcribed figure pass through the an gular points of the figure about which it is described, each through each. III. A rectilineal figure is faid to be infcribed IV. A rectilineal figure is faid to be defcribed about a circle, when each fide of the circumfcribed figure touch es the circumference of the circle. V. In like manner, a circle is faid to be in- VI. VI. A circle is faid to be defcribed about a rectilineal figure, when the circumference of the circle paffes through all the angular points of the figure about which it is defcribed. VII. A ftraight line is faid to be placed in a cir cle, when the extremities of it are in the circumference of the circle. 103 Book IV. PRO P. I. PR Ó B. IN a given circle to place a straight line, equal to a given straight line not greater than the diameter of the circle. Let ABC be the given circle, and D the given ftraight line, not greater than the diameter of the circle. 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 ftraight 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, defcribe the circle AEF, and join CA Therefore, because C is the centre of the circle AEF, CA is equal to CE; D A 23. I E B F but D is equal to CE; therefore D is equal to CA: Wherefore, in the circle ABC, a ftraight 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. IN PRO P. II. PRO B. a given circle to infcribe a triangle equiangular to a given triangle. Book IV. a 17. 3. b 23. I. € 32. 3. Let ABC be the given circle, and DEF the given triangle; it is required to infcribe in the circle ABC a triangle equiangu lar to the triangle DEF. G Draw a the ftraight line GAH touching the circle in the point A, and at the point A, in the straight line AH, make b the angle HAC equal to the angle DEF; and at the point A, in the ftraight 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 HAC is e qual to the angle A E F H ABC in the alternate fegment of the circle: But HAC is equal to the angle DEF; therefore alfo the angle ABC is equal to DEF: For the fame reason, the angle ACB is equal to the angle d32. I. DFE; therefore the remaining angle BAC is equal d to the re á 23. I. b 17. 3. € 18.3. maining angle EDF: Wherefore the triangle ABC is equiangular to the triangle DEF, and it is infcribed in the circle ABC. Which was to be done. A PROP. III. PROB. BOUT a given circle to describe a triangle equi angular 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 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 ftraight lines LAM, MBN, NCL, touching b 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 angles: And because the four angles of the quadrilateral figure AMBK are с equal equal to four right angles, for it can be divided into two tri- Book IV. angles and that two of them KAM, KBM are right angles, the other two AKB, AMBare equal to . two right angles: But the angles L D DEG, DEF are likewife equal dto d 13. 1. two right angles; A K therefore the an gles AKB, AMB areequaltothean equal to DEG; wherefore the remaining angle AMB is equal to the remaining angle DEF: In like manner, the angle LNM may be demonftrated to be equal to DFE; and therefore the remaining angle MLN is equal to the remaining angle EDF : ¢ 32 1, Wherefore the triangle LMN is equiangular to the triangle DEF: And it is described about the circle ABC. to be done. Which was Let the given triangle be ABC; it is required to infcribe a circle in ABC. See N. Bifect the angles ABC, BCA by the ftraight lines BD, CD a 9. 1. meeting one another in the point D, from which draw b DE, b 12. 1. DF, DG perpendiculars to AB, BC, CA: And because the angle EBD is equal to the angle FBD, for the angle ABC is bisected by BD, and that the right angle BED is equal to the right angle BFD, the two triangles EBD, FBD have two angles of the one equal to two angles of the other, and the fide BD, which is oppofite to one of the equal angles in each, is common to both; there- B fore their other fides shall be e ; E A D F qual; |