Book IV. Let ABC be the given circle, and DEF the given triangle, w it is required to infcribe in the circle ABC a triangle equiangular to the triangle DEF. 2 17.3. b 23. I.. Draw the straight line GAH touching the circle in the point A, and at the point A, in the ftraight line AH, make the angle HAC equal to the angle DEF; and at the point A, in the ftraight line G AG, make the angle GAB equal to the D angle DFE, and join BC: Therefore, be the point of contact, A H C € 32.3. d 32. I. a 23. I. b 17. 3: c 18. 3, the angle HAC is A PROP. III. PROB. BOUT 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 defcribe 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 ftraight line KB, make 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 ftraight lines LAM, MBN, NCL touching the circle ABC: Therefore, becaufe 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 points A, B, C, are right angles: And because the four angles of the quadrilateral figure AMBK с AMBK are equal to four right angles, for it can be divided in- Book IV. to 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 totworight A angies; therefore the angles AKB, AMB are equal to the angles DEG, M B 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 demonftrated to be equal to DFE; and therefore the remaining angle MLN is equal to the remaining© 32. X. 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. To infcribe a circle in a given triangle. Let the given triangle be ABC; it is required to infcribe a circle in ABC. Sce N. Bifect the angles ABC, BCA by the ftraight lines BD, CD a 9. 1. meeting one another in the point D, from which, drawb DE, DF, 12. 1. DG perpendiculars to AB, BC, E A D F c 26. I. fore Book IV. fore DE is equal to DF: For the fame reafon, DG is equal to w DF; therefore the three ftraight lines DE, DF, DG are equal to one another, and the circle described from the centre D, at the diftance of any of them, fhall pass through the extremities of the other two, and touch the ftraight lines AB, BC, CA, because the angles at the points E, F, G are right angles, and the ftraight line which is drawn from the extremity of a diameter at right angles to it, touches the circle: Therefore the ftraight 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. d 16. 3. PROP. V. PROB. See N: a IO. I. b II. I. To O defcribe a circle about a given triangle. Let the given triangle be ABC; it is required to defcribe a circle about ABC. 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 pa rallel; 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 bafe AF is equal to the base FB: In like manner, it may be fhown that CF is equal to FA; and therefore BF is equal to FC; and FA, FB, FC are equal to one another; wherefore wherefore the circle defcribed from the centre F, at the di- Book IV. flance of one of them, fhall pass through the extremities of the other two; and be defcribed about the triangle ABC, which was to be done. COR. And it is manifeft, that, when the centre of the circle falls within the triangle, each of its angles is lefs than a right angle, each of them being in a fegment greater than a femicircle; but, when the centre is in one of the fides of the triangle, the angle oppofite to this fide, being in a femicircle, is a right angle; and, if the centre 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 centre of the circle falls within it; if it be a right angled triangle, the centre is in the fide opposite to the right angle; and, if it be. an obtufe angled triangle, the centre falls without the triangle, beyond the fide oppofite to the obtufe angle. PROP. VI. PRO B To inferibe a square in a given circle. Let ABCD be the given circle; it is required to infcribe a fquare in ABCD. A a 4. si D 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 centre, and that EA is common, and at right angles to BD; the bafe BA is equal to the bafe AD; and, for the fame reason, BC, CD are each of them equal to BA or AD; therefore the quadri- B lateral figure ABCD is equilateral. It is also rectangular; for the ftraight line BD, being the diameter of the circle ABCD, BAD is a femicircle; wherefore the angle BAD is a right E C angle; for the fame reafon each of the angles ABC, BCD, b 31. 3CDA is a right angle; therefore the quadrilateral figure ABCD is rectangular, and it has been shown to be equilateral; therefore it is a fquare; and it is infcribed in the circle ABCD. Which was to be done. PROP. Book IV. b 18. 3. c 28. I. PROP. VII. PRO B. To do cribe a square about a given circle. Let ABCD be the given circle; it is required to describe fquare about it. G B E A F D Draw two diameters AC, BD of the circle ABCD, at righ angles to one another, and through the points A, B, C, D draw a 17. 3. FG, GH, HK, KF touching the circle; and because FG touche the circle ABCD, and EA is drawn from the centre E to the point of contact A, the angles at A are right angles; 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 likewife 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 demonftrated to be parallel to BED; therefore the figures GK, GC, AK, FB, BK are parallelograms; and GF is therefore equal to HK, and GH to FK; 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 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 shown that the angles at H, K, F are right angles; therefore the quadrilateral figure FGHK is rectangular, and it was demonftrated to be equilateral; there fore it is a fquare; and it is described about the circle ABCD. Which was to be done. 34. I. a 10. I. H PROP. VIII. PROB. To infcribe a circle in a given square. C K 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 b 31. 1. through E drawb EH parallel to AB or DC, and through F draw FK |