Let ABC be the given circle, and DEF the given triangle; it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF. * D * 23. 1. Draw the straight line GAH touching the circle in 17. 3. 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 angle GAB equal to the angle DFE; and join BC: ABC shall be the triangle required. E F H B Because HAG touches the circle ABC, and AC is drawn from the point of contact, the angle HAC is * equal to the angle ABC in the alternate segment of 32. 3. the circle: but HAC is equal to the angle DEF; + Constr. therefore also the angle ABC is equal† to DEF: for †1 Ax. the same reason, the angle ACB is equal to the angle DFE: therefore the remaining angle BAC is equal to 32. 1. & the remaining angle EDF: wherefore the triangle ABC 1Ax. 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; find † † 1. 3. the centre 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, 23. 1. and the angle BKC equal to the angle DFH; and through the points A, B, C, draw the straight lines LAM, MBN, NCL, touching the circle ABC: LMN 17. 3. shall be the triangle required. 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 points A, B, C, are right angles: and because the four angles of the drilateral figure AMBK are equal to four right angles, for it can be divided into two triangles; and that two * qua *18. 3. + 3 Ax. 13. 1. † 1 Ax. +Constr. † 3 Ax. 3 Ax. * L of them KAM, KBM are right angles, therefore the M-B N * to the remaining angle DEF. In like manner, the angle LNM may be demonstrated to be equal to DFE; 32. 1. & and therefore the remaining angle MLN is equal to the remaining angle EDF: therefore the triangle LMN is equiangular to the triangle DEF: and it is described about the circle ABC. Which was to be done. See N. *9. 1. 12. 1. † 11 Ax. 26. 1. † 1 Ax. * 16. 3. 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. B Bisect* the angles ABC, BCA by the straight lines BD, CD meeting one another in the point D, from which draw* 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 E angle BFD; therefore the two triangles EBD, FBD have two angles of the one equal to two angles of the other, each to each; and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore their other sides are equal *; wherefore DE is equal to DF: for the same reason, DG is equal to DF: therefore DE is equal + to DG: 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, will pass through the extremities of the other two, and touch the straight lines AB, BC, CA, 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 the circle: therefore the straight lines AB, BC, CA do each of them touch the circle, and therefore the circle EFG is inscribed in the triangle ABC. Which was to be done. PROP. V. PROB. To describe a circle about a given triangle. Let the given triangle be ABC; it is required to describe a circle about ABC. See N. Bisect* AB, AC, in the points D, E, and from these * 10. 1. points draw DF, EF at right angles* to AB, AC; DF, *11. 1. EF, produced meet one another: for, if they do not meet, they are pa rallel, wherefore AB, AC, which are at right angles to them, are parallel; which is ab B C B F surd 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 to the base FB.4.1. In like manner, it may be shewn that CF is equal to FA; and therefore BF is equal to FC; and FA, FB, †1 Ax. FC, are equal to one another: wherefore the circle described from the centre F, at the distance of one of them, will pass through 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 centre of the circle falls within the triangle, each of its angles is less than a right anglet, each of them being in a segment † 31. 5. greater than a semicircle; but, when the centre is in one of the sides of the triangle, the angle opposite to this side, being in a semicirclet, is a right angle; and, if the centre falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicirclet, is greater than a right angle: therefore, conversely, 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 side opposite to the right angle; and if it be an obtuse angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle. † 1. 3. and 11. 1. *4:1. * 31. 3. + 30 Def. 1. * 17. 3. #18. 3. 28. 1. PROP. VI. PROB. To inscribe a square in a given circle. Let ABCD be the given circle; it is required to inscribe a square in ABCD. Draw the diameters, AC, BD+, at right angles to one another, and join AB, BC, CD, DA: the figure ABCD shall be the square required. Because BE is equal to ED, for E is the centre, and that EA is common, and at right angles to BD; the base BA is equal* to the base AD: and for the same reason, BC, CD are each of them equal to BA, or AD; therefore the quadrilateral 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* angle for the same reason, each of the : PROP. VII. PROB. To describe a square about a given circle. Let ABCD be the given circle; it is required to describe a square about it. Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D, draw* FG, GH, HK, KF touching the circle: the figure GHKF shall be the square required. G D Because FG touches 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 same reason, the angles at the points B, C, D are right angles: and because the angle AEB is a right angle, as likewise is EBG, GH is parallel* to AC: for the same reason AC is parallel to FK and in like manner GF, HK may each of them be demonstrated to be parallel to BED: therefore the figures GK, GC, AK, FB, BK are pa B H rallelograms; and therefore GF is equal to HK, and S4. 1. 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 also rectangular : for GBEA being a parallelogram, and AEB a right angle, AGB✶ is likewise a right angle: and in the same 34. 1. manner it may be shewn 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 at square: and it is de- +30 Def. 1. scribed about the circle ABCD. Which was to be done. PROP. 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. F A E D G K 10. 1. +31. 1. 34. 1. † 7 Ax. Bisect each of the sides AB, AD in the points F, E, and through E draw+ EH parallel to AB or DC, and through F draw FK parallel to AD or BC: therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD, is a parallelogram; and their opposite sides are equal: and because AD is equal to AB, and that AE is the half of AD, and AF the half of AB, AE is +30 Def. 1. equal to AF; wherefore the sides opposite to these are equal, viz. FG to GE: in the same manner it may be demonstrated that GH, GK are each of them equal to FG or GE: therefore the four straight lines GE, GF, GH, GK are equal to one another; and the circle described from the centre G at the distance of one of them, will pass through the extremities of the other three, and touch the straight lines AB, BC, CD, DA; because the angles at the points E, F, H, K, are right* right* 99. 1. angles, and that the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle: therefore each of the straight lines Cor. 16, AB, BC, CD, DA touches the circle, which therefore 3. is inscribed in the square ABCD. Which was to be done, B H C |