Book IV. Let ABC be the given circle, and DEF the given triangle; it ~ is required to infcribe in the circle ABC a triangle equiangular to the triangle DEF. a. 17. 3. b. 23. I. e. 32. 3. d. 32. 1. 2. 23. I. b. 17. 3. c. 18. 3. b Draw the ftraight 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- drawn from the point E nate fegment of the circle. but IFAC is equal to the angle DEF, therefore alfo the angle ABC is equal to DEF. for the fame rea fon, the angle ACB is equal to the angle DFE; therefore the remaining angle BAC is equal to the remaining 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. PROP. III. PROB. ABOUT 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 defcribe a triangle about the circle ABC equiangular to the triangle DEF. a Produce EF both ways to the points G, H, and find the center K of the circle ABC, and from it draw any ftraight 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, 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 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. can be divided into two triangles; and 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 likewife equal d to two right angles; therefore the angles AKB, AMB are e qual to the angles DEG, DEF, of. which AKB is equal to DEG; where-M B N fore 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 remaininge. 32. angle EDF. wherefore the triangle LMN is equiangular to the triangle DEF. and it is defcribed about the circle ABC. Which was to be done. PROP. IV. PRO B. To infcribe a circle in a given triangle. Let the given triangle be ABC; it is required to infcribe a cir cle in ABC. See N. Bifect 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, DF, b. 12. 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 fight 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 équal angles in each, is common to both; therefore their other fides fhall be equal; wherefore e 26. 1. Ꮐ Book IV. DE is equal to DF. for the fame reason, DG is equal to DF; there fore the three straight lines DE, DF, DG are equal to one another, and the circle defcribed from the the extremity of a diameter at right E A d. 16. 3. angles to it, touches the circle. B F therefore the straight lines AB, BC, See N. 3. 10. I. b. II. I. c 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. To defcribe a circle about a given triangle. Let the given triangle be ABC; it is required to describe a circle about ABC. a Bifect AB, AC in the points D, E, and from these points draw DF, EF at right angles to AB, AC; DF, EF produced C. 4. 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 abfurd. let them meet in F, and join FA; alfo, 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 to the base FB. in like manner it may be fhewn that CF is equal to FA; and therefore BF is equal to FC; and FA, FB, 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, fhall pass thro' the extremities of the other two; and be described about the triangle ABC. Which was to be done. COR. And it is manifeft that when the center of the circle falls within the triangle, each of its angles is lefs than a right angle, each of them being in a segment greater than a femicircle. but when the center is in one of the fides of the triangle, the angle opposite to this fide, being in a femicircle, is a right angle. and if the center falls without the triangle, the angle oppofite to the fide beyond which it is, being in a fegment lefs than a femicircle, 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 fide oppofite to the right angle; and if it be an obtuse angled triangle, the center falls without the triangle, beyond the fide oppofite to the obtuse angle. PROP. VI. PROB. To infcribe a square in a given circle. То Let ABCD be the given circle; it is required to infcribe 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 at right angles to BD; the base BA is A a. 4. 3. E the angle BAD is a right bangle. for the fame reafon 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 fhewn to be equilateral, therefore it is a fquare; and it is infcribed in the circle ABCD. Which was to be done. Book IV. b. 18. 3. PROP. VII. PRO B. To defcribe a square about a given circle. T° Let ABCD be the given circle; it is required to describe a fquare about it. a 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 circle ABCD, and EA is drawn from the center E to the point of contact A, the angles at A are right bangles. for the fame reason, the angles at the points B, C, D are right angles. and because the angle AEB is a right angle, as likewise G is EBG, GH is parallel to AC. for the fame reason, AC is parallel to FK. and in like manner GF, HK may each of them be demonstrated to be parallel to BED. B therefore the figures GK, GC, AK, FB, BK are parallelograms, and GF is therefore equal to HK, and GH to FK. H and because AC is equal to BD, and c. 28. I. d. 34. I. a. IC. I. A E K 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 alfo rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB is likewife a right angle. in the fame manner it may be fhewn that the angles at 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 defcribed about the circle ABCD. Which was to be done. PROP. VIII. PRO B. To inscribe a circle in a given square. Let ABCD be the given fquare; it is required to inscribe a circle in ABCD. Bifect each of the fides AB, AD, in the points F, E, and thro'E b. 31. 1. drawb EH parallel to AB or DC, and thro' F draw FK parallel to I |