A. 3• b 21. 3 Let ABC be a triangle, and from the angles at B, C draw Book. III. BD, CE perpendiculars to AC, AB, and let them meet in F, and join AF meeting BC in G: AC is perpendicular to BC. Join DE: and because ADF, AEF are right angles, the cir. cumference of the circle, of which AF is the diameter, shall pass through the points E, D *; there А a 2. Cor. fore the angle DEF is equal to the angle DAF, because they are in the fame segment DAEF. Again, because BDC, BEC are right angles, the circumference of which BC is the diameter, shall pass through D, E a; therefore the angle DEC is equal to D the angle DBC, because they are in the same segment DEBC: but it has B been proved, that the angle DEC is equal to the angle DAF; therefore the angle CBF is equal to DAF: and the angle BFG is cqual c to DFA; C15. I. therefore the remaining angle BGF is equal d to the remaining angle ADF; that is, to a right angle; d 32. I. and AFG is therefore perpendicular to BC. Wherefore, &c. R. E. D. DEFINITIONS. I. Book tv. RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angles of the inscribed II. about another figure, when all the sides of III. in a circle, when all the angles of the in- IV. about a circle, when each side of the cir- V. In like manner, a circle is said to be inscri. bed in a rectilineal figure, when the cir- VI. O VI. Book IV. A circle is said to be described about a rectilineal figure, when the circumference of the circle paffes through all the angular VII. mities of it are in the circumference of the circle. PROP. I. PROB. IN given straight line, not greater than the diameter of the circle. a 3. I. 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 A D: But, if it is not, BC is greater than D; make CE equal to D, and from the centre G, at the distance CE, describe the B C circle AEF, and join CA: Therefore, because C 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 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. to a given triangle. Draw a the itraight line GAH touching the circle in the a 17. 1. 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, b 23. . 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 Book IV. the angle HAC is equal o to the angle ABC in the alternate fegmment of the circle : But HAC is equal to the angle DEF; C 32. 3. therefore also the angle ABC is equal to DEF: For the same H D d 32. 1. equal d to the remaining angle EDF: Wherefore the triangle B E F in the circle ABC. Which was to be done. A PROP. III. PROB. See N. 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 a 23. 1. 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 b 17. 3. lines LAM, MBN, NCL, touching the circle ABC: There fore, because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, C 18. 3. the angles at the points A, B, C, are right © angles : And be cause the angles AKB, BKC are equal to DEG, DFH, they d 16. 1. are greater than DEG, DEF, that is, greater than two right angles °; therefore, if AC be joined, it shall fall between K and f 2. Cor. L*: and the angles ACL, CAL shall therefore be less than two 15. 1. right angles; wherefore AL, CL shall meet 8, if produced : For g 12.Ax... the same reason, they shall, each of them, meet MN: let them meet in the points L, K. F n B N e 13. I. D А two two AKB, AMB are equal to two right angles : But the angles Book. IV. DEG, DEF are likewise equal to two right angles; therefore w the angles AKB, AMB are equal to the angles DEG, DEF, € 13. 1. 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 demonftrated to be equal to DFE; and therefore the remaining angle MLN is equal to the re- h 32. 1. maining angle EDF: 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. T. O inscribe a circle in a given triangle. Let the given triangle be ABC; it is required to inscribe a circle in ABC. Bife&t a 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. r. DF, DG perpendiculars to AB, BC, CA: And because the angle EBD is equal to the angle FBD, for the angle ABC is bifected 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 B both; therefore their other fides shall be equal €; wherefore c 26. 1. DE is equal to DF: For the farne reason, DG is equal to DF; therefore the three straight lines DE, DF, DG are equal to one another, and the circle described from the centre D, at the distance of any of them, shall pass through the extremities of the other two; and because the angles at the points E, F, G, are right angles, and the straight line which is drawn from the extremity of a diameter at right angles to it, touches d the circle; d 16.3. therefore the Ntraight lines AB, BC, CA do each of them touch the circle, and the circle EFG is infcribed in the triangle ABC. Which was to be done. |