a. 17. 3. Book IV. Let ABC be the given circle, and DEF the given triangle ; it nis required to inscribe in the circle ABC a triangle equiangular to the triangle DEF. Draw * the straight line GAH touching the circle in the point D. 23. 1. A, and at the point A, in the straight line AH, make the angle G H B В of contact, the angle HAC is equal to the angle ABC in the alternate segment of the circle. but IIAC is equal to the angle DEF, therefore also the angle ABC is equal to DEF. for the same rea fon, the angle ACB is equal to the angle DFE; therefore the red. 32. 1. maining 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. e. 32. 3. PRO P. III. PRO B. ABOUT a given circle to describe a triangle equi angular to a given triangle. Let ABC be the given circle, and DEF tlie 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 center K of the circle ABC, and from it draw any straight line KB; at the point K in the straight line KB, make * the angle BKA equal to the angle DEG, and the angle BKC equal to the angle DFH; and thro' the points A, B, C draw the straight lines LAM, b. 17. 3. 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 center are drawn KA, KB, KC, the angles at the points C. 18. 3. A, B, C are right © angles. and because the four angles of the quadrilateral figure AMBK are equal to four right angles, for it Book IV. L D d. 13. 1. A к. GE FH M B N 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 equal to the remaininge. 3a: 67 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 To inscribe a circle in a given triangle. See N. Let the given triangle be ABC ; it is required to infcribe a circle in ABC. Bisect * the angles ABC, BCẢ by the straight lines BD, CDa. 9. 1. meeting one another in the point D, from which draw 5 DE, DF, b. 11., DG perpendiculars to AB, BC, А E B F Book IV. DE is equal to DF. for the same reason, DG is equal to DF; there fore the three straight lines DE, DF, DG are equal to one another, А G the extremity of a diameter at right d. 16. 3. angles to it, touches the circle . BF AB, CA do each of them touch the circle, and the circle EFG is inscribed in the triangle ABC. Which was to be done. PROP. V. PROB. See N. To describe a circle about a given triangle. a. 10. I. Let the given triangle be ABC; it is required to describe a circle about ABC. Bisect ' AB, AC in the points D, E, and from these points draw DF, EF at right angles b to AB, AC; DF, EF produced A A b. II. I. meet one another. for if they do not meet they are parallel, wherefore AB, AC which are at right angles to them are parallel ; which is absurd. let them meet in F, and join FA ; also, if the point F be not in BC, join BF, CF. then because AD is equal to DB, and DF common, and at right angles to AB, the base AF is equal o to the base FB. in like manner it may be shewn that CF is equal to FA ; and therefore BF is equal to FC; and FA, FB, C. 4. I. FC are therefore equal to one another. wherefore the circle de- Book IV. scribed from the center F, at the distance of one of them, shall pass thro' the extremities of the other two; and be described about the triangle ABC. Which was to be done. Cor. And it is manifest that when the center of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a semicircle. but when the center is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, is a right angle, and if the center falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, is greater than a right angle. Wherefore, if the given triangle be acute angled, the center of the circle falls within it; if it be a right angled triangle, the center is in the side opposite to the righe angle; and if it be an obtuse angled triangle, the center falls without the triangle, beyond the side opposite to the obtufe angle. PROP. VI. PROB. To inscribe a square in a given circle. a. 4. a; Let ABCD be the given circle ; it is required to inscribe a fquare in ABCD. Draw the diameters AC, BD at right angles to one another ; and join AB, BC, CD, DA. because BE is equal to ED, for E is the center, and that EA is common, and А. at right angles to BD; the base BA is equal to the base AD. and for the fame reason, BC, CD are cach of them equal to BA or AD; therefore the qua-B drilateral figure ABCD is equilateral. It is also rectangular; for the straight line BD being the diameter of the circle ABCD, BAD is a semicircle; wherefore the angle BAD is a right bangle. for the same reason each of the b. 31. 3. angles ABC, BCD, CDA is a right angle, therefore the quadrilateral figure ABCD is rectangular. and it has been shewn to be equilateral, therefore it is a square; and it is inscribed in the circle ABCD. Which was to be done. Book IV. PROP. VII. PROB. To describe a square about a given circle. b. 18. 3. c. 28. 1. Let ABCD bc the given circle ; it is required to describe a square about it. Draw two diameters AC, BD of the circle ABCD, at right a. 17. 3. angles to one another, and thro' the points A, B, C, D draw • FG, GH, HK, KF touching the circle. and because FG touches the BK are parallelograms, and GF is-there- H Н K and because AC is equal to BD, and that AC is equal to each of the two GH, FK; and BD to each of the two GF, HK; GH, FK are each of them equal to GF or HK. therefore the quadrilateral figure FGHK is equilateral. It is also rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB is d likewise a right angle. in the same manner it may be shewn that the angles ar H, K, F are right angles. therefore the quadrilateral figure FGHK is rectangular. and it was demonstrated to be equilateral ; therefore it is a square ; and it is described about the circle ABCD. Which was to be done. PRO P. VIII. PROB. TO inscribe a circle in a given square. Let ABCD be the given square ; it is required to inscribe a circle in ABCD. Bisect'each of the sides AB, AD, in the points F, E, and thro'E 5.31. 1. draw EH parallel to AB or DC, and thro' F draw FK parallel 80 a. IC. I. |