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 C D d 8. 1. E e 16. 3. F 1 THE ELEMENTS OF EUCLID. BOOK IV. DEFINITIONS. I. See Note. 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 III. A rectilineal figure is said to be inscribed scribed figure are upon the circumference IV. A rectilineal figure is said to be described about a circle, when V. In like manner, a circle is said to be inscrib- 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. 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. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle. Draw BC the diameter of the circle ABC; then, if BC be equal to D, the thing required is done; for in the circle ABC a straight line BC is placed e qual to D; but, if it be not, BC A 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. 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 an G A gle GAB equal to the с 32.3. angle DFE, and join D E F B A H C d 32. 1. 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 is equal d 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. a 23. 1. b 17.3. с 18.3. 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 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 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, КВ, КС, the angles at the points A, B, C are right angles: and because the four angles of the quadrilateral figure AMBK are 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, L of which AKB is M B N equal to DEG; wherefore the remaining angle AMB is equal to the remaining angle DFE: in like manner, the angle LNM may be demonstrated to be equal to DFE; and therefore the remaining angle MLN is equal to the remaining angle EDF: wherefore the triangle LMN is equiangular to the triangle DEF: and it is described about the circle ABC. Which was to e 32. 1. be done. PROP. IV. PROB. TO inscribe a circle in a given triangle. Let the given triangle be ABC; it is required to inscribe a circle in ABC. See Note. Bisecta the angles ABC, BCA by the straight 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 side BD, which is opposite to one of the equal angles in each, is common to both; therefore their other sides shall be e A |