D d 8.1. equal to the straight line DB: And FE is equal to FB, Book III. 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 to the angle DBF; but DEF is a right angle, therefore also DBF is a right angle And FB, if produced, is a B diameter; and the straight line which is drawn 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. F E 16. 3. 94. THE ELEMENTS OF EUCLI D. 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 See N. which it is inscribed, each upon each. 11. In like manner a figure is said to be described angular points of the figure about which it is described, III. A rectilineal figure is said to be inscribed in a circle, when all the angles of the IV. A rectilineal figure is said to be described about a circle, V. In like manner, a circle is said to be in- C 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 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 C is the D E a 3. 1. B F centre of the circle AEF, CA is equal to CE: But D is 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 trian BOOK IV. gle; it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF. a 17. 3. b Draw a the straight line GAH touching the circle in the 23. 1-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 straight line AG, G make the angle GAB equal to the angle H because HAG touches the circle ABC, and AC is drawn from the point of contact, the angle © 32. 3. HAC is equal to the c E B angle ABC in the alternate segment of the circle: 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 432. 1. angle BAC 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 * 23. 1. 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, 17. 3. 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, KC, the angles at the 18. 3. 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 triangles; and Book IV. that two of them KAM, KBM are right angles, the other two AKB, AMB are equal to two right angles : But the angles DEG, DEF are, likewise equald to two right angles; therefore the angles AKB, AMB are equal to the angles DEG, DEF, of which AKB is equal to DEG; wherefore the remaining angle AMB is equal to the remaining angle DEF: In like manner, the angle LNM may be demonstrated to be equal to DFE; and therefore the remaining angle MLN is equale 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. Which was to 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 N. a Bisect a the angles ABC, BCA by the straight lines BD, " 9.1. CD meeting one another in the point D, from which drawb 12. 1. DE, 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 B E A G equal angles in each, is common to both; therefore their H |