which are together less than two right angles; let them meet in F, and join FA: then a circle described with centre F, and #1. 1. +1 Ax. distance FA will circumscribe the triangle. For 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 shown that CF is equal to FA; and therefore BF is equal to FC; and 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. 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. *1. 3. and 11. 1. Draw any 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 like reason, BC, CD are respectively equal to BA, 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; *4. 1. wherefore the angle BAD is a #31. 3. right angle: for a like reason, each of the angles ABC, BCD, CDA is a right angle: therefore the quadrilateral figure ABCD is rectangular: B and it has been shown to be equilateral; therefore it is at square; and it +30 Def. 1. is inscribed in the circle ABCD. Which was to be done. 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 #17. 3. circle the figure GHKF shall be the square required. 18. 3. #28. 1. G P 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 a similar 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 B a similar reason AC is parallel to 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 parallelograms; and therefore GF is 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; FK #34. 1. H C K *Cor. 46. 1. 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, the angle at G is a right angle:* therefore the quadrilateral figure FGHK, being a parallelogram with a right angle at G, is rectangular and it was demonstrated to be equilateral; +30 Def. 1. therefore it is at square; and it is described 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. 131. 1. Bisect each of the sides AB, AD in the points #10. 1. F, E, and through E draw† EH parallel to AB or DC, and through F draw FK parallel to AD or BC; then if, with their point of intersection G as centre, and GF or GE as distance, a circle be described, it will be that required. 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 +30 Def. 1. that AE is the half of AD, and AF the half of AB, AE is equal to AF; #34. 1. 17 Ax. F K wherefore the sides opposite to these are equal, viz. FG to GE: in a similar manner it may be demonstrated that GH, GK are equal, as also GH, GF: 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 angles, and that the straight line which is drawn #29. 1. B H C from the extremity of a diameter, at right angles *Cor. 16. 3. to it, touches the circle: therefore each of the straight lines AB, BC, CD, DA touches the circle, which therefore is inscribed in the square ABCD. Which was to be done. PROP. IX. PROB. To describe a circle about a given square. Let ABCD be the given square; it is required to describe a circle about ABCD. Join AC, BD, cutting one another in E: then if with E as centre, and EA as distance, a circle be described, it will be that required. Because DA is equal to AB, A D +30 Def. 1. and AC common to the triangles DAC, BAC, the two sides DA, AC are equal to the two BA, AC, each to each; and the base DC is equal to the base BC; wherefore the angle DAC is equal* to the angle BAC, and the angle DAB is bisected by the straight line AC: in a similar manner it may be demonstrated that #8. 1. +30 Def. 1. +7 Ax. #6. 1. B the angles ABC, BCD, CDA are severally bisected by the straight lines BD, AC: therefore, because the angle DAB is equal to the angle ABC, and that the angle EAB is the half of DAB, and EBA the half of ABC; the angle EAB is equalf to the angle EBA; wherefore the side EA is equal* to the side EB : in a similar manner it may be demonstrated that the straight lines EC, ED are respectively equal to EA, EB; therefore the four straight lines EA, EB, EC, ED are equal to one another; and the circle described from the centre E, at the distance of one of them, will pass through the extremities of the other three, and be described about the square ABCD. Which was to be done. PROP. X. PROB. To describe an isosceles triangle, having each of the angles at the base double of the third angle. #11. 2. Take any straight line AB, and divide* it in the point C, so that the rectangle AB BC may be equal to the square of CA; and from the centre A, at the distance AB, describe the circle BDE, in which place the straight line BD equal to AC, which is not greater than the diameter of the circle BDE; and join DA: the triangle ABD shall be such as is required, that is, each of the angles ABD, ADB shall be double of the angle BAD. #1. 4. #5. 4. Join D, C ; and about the triangle ADC describe the circle ACD: and because the rectangle AB⚫BC is equal to the square of AC,† and that AC is equal +1 Ax. to BD, the rectangle AB BC is equal to the square of BD: and because +Const. B E from the point B, without the circle ACD, two straight lines BCA, BD are drawn to the circumference, one of which cuts, and the other meets the circle, and that the rectangle AB BC, contained by the whole of the cutting line, and the part of it without the circle, is equal to the square of BD which meets it; the straight line BD touches the circle ACD: and because BD touches the circle, and DC is drawn from the point of contact D, the angle BDC is equal to the angle DAC in the alternate segment of the circle to each of these add the angle CDA; therefore the whole angle BDA is equal to the two angles CDA, DAC: but the exterior angle BCD is equal to the angles CDA, DAC; therelore also BDA is equalf to BCD: but BDA is *37. 3. *32. 3. 12 Ax. #32. 1. +1 Ax |