divided into two triangles; and that two of them KAM, KBM are right angles, therefore the +3 Ax. other two AKB, AMB are equal to two right angles: but the angles DEG, DEF are like * *13.1. wise equal to two. right angles; therefore the angles AKB, +1 Ax. AMB are equal L K to the angles DEG, D + Const. is equal to DEG; wherefore the remaining †3 Ax. angle AMB is equal† to the remaining angle DEF. In like manner, the angle LNM may be demonstrated to be equal to DFE; and there*32. 1. fore 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. was to be done. and 3 Ax. 9. 1. PROP. IV. PROB. Which To inscribe a circle in a given triungle. Let the given triangle be ABC; it is required to inscribe a circle in ABC. A E D Bisect the angles ABC, BCA by the straight lines BD, CD meeting one another in the point *12. 1. 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 +11 Ax. BED is equal to the right 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 B F sides are equal; * wherefore DE is equal to DF: * 26. 1. for the same reason, DG is equal to DF: therefore DE is equal† to DG: therefore the three †1 Ax. 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 *16. 3. 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. * Bisect* AB, AC in the points D, E, and from *10. i. these points draw DF, EF at right angles to *11. 1. AB, AC; DF, EF produced meet one another: B АС В B F 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 to the base FB. In *4. 1. like manner, it may be shown that CF is equal to FA; and therefore BF is equal to FC; and +1 Ax. * FA, FB, 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 +31.3. angles is less than a right angle,† each of them being in a segment 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 +31. 3. a semicircle,† 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 +31. 3. less than a semicircle,† 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. 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, DÅ: 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 To describe a square about a given circle. Let ABCD be the given circle; it is required to describe a square about it. *31.3. † 30 Def. 1. 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, *17. 3. HK, KF touching the circle: the figure GHKF shall be the square required. 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: *18. 3. for the same reason, the angles at the points B, C, D are right angles : and because the E *28. 1. C of them be demonstrated to be parallel to BED: therefore the figures GK, GC, AK, FB, BK are parallelograms; and therefore GF is equal* *81 L 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 also rectangular; for GBEA being a parallelogram, and AEB a 34. 1. right angle, AGB* is likewise a right angle: and in the same 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 demonstrated to be equila+50Def. teral; therefore it is at square; and it is described about the circle ABCD. Which was to be done. 1. 10.1. *34. 1. 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. * Bisect each of the sides AB, AD in the points +31.1. 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 +30 Def. because AD is equal to AB, and that AE is the half of AD, and AF the half of AB, AE is equal to AF; wherefore the sides opposite to these are equal, A 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 17 Ax. +29. 1. +30 Def. E D G K the four straight lines GE, GF, B H C other; and the circle described from the centre Because AB is parallel to EH, the two angles BAE, AEH are together equal to two right angles: but BAE is a fright angle; therefore AEH is a right angle: and in the same manner the angles at the points F, H, K, may be proved to be right angles. |