Book IV. a 17.3. b 18.3. c 28. 1. d 34. 1. 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. F Draw two diameters AC, BD of the circle ABCD, at right angles to each other, and through the points A, B, C, D drawa FG, GH, HK, KF touching the circle. 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 bangles. For the same reason the angles at the points B, C, D are right angles. Because the angles AEB, EBG are right angles, GH is parallel to AC. For the same reason AC is parallel G to FK. In like manner GF, HK may be demonstrated to be parallel to BED. Therefore the figures GK, GC, AK, FB, BK are paral- B lelograms; therefore GF is equald to HK, and GH to FK. Because AC is equal to BD, and to each of the two GH, FK; and because BD H is equal to each of the two GF, HK ; E D 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, AGBd is likewise a right angle. 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 equilateral; therefore it is a 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. E D Book IV. Bisecta each of the sides AB, AD in the points F, E; a 10. 1. through E drawb EH parallel to AB or DC, and through F b 31. 1. 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 equals. Because AD is equal c 34. 1. to AB, and AE is the half of AD, and AF the half of AB, AE is equal to AF; wherefore the sides FG, GE opposite to these are equal. 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; therefore the circle described from the centre G, at the distance of one of them, will pass through the extremities of the other three. Because the angles at the points E, F, H, K are rightd angles, each of the straight lines AB, BC, CD, DA touches the inscribed in the square ABCD. F K H C d 29. 1. B circle, which is therefore e 16. 3. Which was to be done. Book IV. a 8. 1. b 6.1. 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 it. Draw AC, BD, cutting each other in E. Because DA is equal to AB, and AC common to the triangles DAC, BAC, the two sides DA, AC are equal to the two BA, AC, and the base DC is equal to the base BC; wherefore the angle DAC is equal B and the angle EAB is the half of DAB, and EBA the half of ABC, the angle EAB is equal to the angle EBA; therefore the side EA is equal to the side EBb. In the same manner it may be demonstrated that the straight lines EC, ED are each of them equal to EA, or EB. Therefore the four straight lines EA, EB, EC, ED are equal to one another; therefore the circle described from the centre E, at the distance of one of them, must pass through the extremities of the other three, and will therefore be described about the square ABCD. Which was to be done. PROP. X. PROB. Book IV. TO describe an isosceles triangle, having each of the angles at the base double of the third angle. Take any straight line AB, and dividea it in the point C, a 11. 2.. so that the rectangle AB.BC may be equal to the square of CA; from the centre A, at the distance AB, describe the circle BDE, in which placeb the straight line BD equal to b 1. 4. AC, which is not greater than the diameter of the circle BDE; join DA, DC, and about the triangle ADC describe c 5. 4. the circle ACD. ABD is the triangle required, having each of the angles ABD, ADB double of the angle BAD. E d 37. 3. Because the rectangle AB.BC is equal to the square of AC, and AC is equal to BD, the rectangle AB.BC is equal to the square of BD; therefore the straight line BD touches the circle ACD. Because BD touches the circle, and DC is drawn from the point of contact D, the angle BDC is equale 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. CDĂ, DAC. But the ex-. B D £ 32. 3. terior angle BCD is equal to the angles CDA, DAC; f 32. 1. therefore also BDA is equal to BCD. But BDA is equals g 5.1. to CBD, because the side AD is equal to the side AB; therefore CBD, or DBA, is equal to BCD. Consequently the three angles BDA, DBA, BCD are equal to one another. Because the angle DBC is equal to the angle BCD, the side BD is equal to the side DC. But BD was made equal h 6. 1. g 5.1. Book IV. to CA; therefore CA is equal to CD; therefore the angle CDA is equals to the angle DAC; therefore the angles CDA, DAC together are double of the angle DAC. But BCD is equal to the angles CDA, DAC; therefore also BCD is double of DAC. But BCD is equal to each of the angles BDA, DBA; therefore each of the angles BDA, DBĂ is double of the angle DAB. Wherefore an isosceles triangle ABD is described, having each of the angles at the base double of the third angle. Which was to be done. COR. 1. The angle BAD is the fifth part of two right angles. For since each of the angles ABD and ADB is equal to twice the angle BAD, they are together equal to four times BAD, and therefore all the three angles ABD, ADB, BAD, taken together, are equal to five times the angle BAD. But the three angles ABD, ADB, BAD are equal to two right angles, therefore five times the angle BAD are equal to two right angles, or BAD is the fifth part of two right angles." OR. 2. Because BAD is the fifth part of two, or the tenth part of four right angles, all the angles about the centre A are together equal to ten times the angle BAD, and may therefore be divided into ten parts each equal to BAD. And as these ten equal angles at the centre must stand on ten equal arches, the arch BD is one-tenth of the circumference; therefore the straight line BD, that is, AC, is equal to the side of an equilateral decagon inscribed in the circle BDE." |