BOOK IV. 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, a 17. 3. C, D, draw a FG, GH, HK, KF touching the circle; and because FG touches the circle ABCD, and EA is drawn from the centre E to the point of contact A, the angles at b 18. 3. 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 с € 28. 1. parallel to AC; for the same rea- K 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 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, 10. 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. Bisecta each of the sides AB, AD, in the points F, E, and 31. 1. through E draw EH parallel to AB or DC, and through K F draw FK parallel to AD or BC; therefore each of the Book IV. figures AK, KB, AH, HD, AG, GC, BG, GD, is a parallelogram, and their opposite sides are equal; and because € 34. 1. AD is equal to AB, and that AE is the half of AD, and AF the half of AB, AE is equal to AF; A 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 F 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, shall 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 a 29. 1. the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle; there- 16. 3. fore 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. B 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. Join AC, BD, cutting one another in E; and because DA is equal to AB, and AC common to the triangles DAC, BAC, the two sides DA, AC are equal a 8. 1. to the two BA, AC, and the base DC A B BOOK IV. 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; and the circle described from the centre E, at the distance of one of them, shall pass through the extremities of the other three, and be described about the square ABCD. Which was to be done. * 11. 2. PROP. X. PROB. 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, so that the rectangle AB, BC be equal to the square of CA; and from the centre A, at the distance AB, describe the 1. 4. circle BDE, in which placeb the straight line BD equal to E AC, which is not greater than the diameter of the circle 5. 4. BDE; join DA, DC, and about the triangle ADC describe the circle ACD; the triangle ABD is such as is required, that is, each of the angles ABD, ADB is double of the angle BAD. Because the rectangle AB, BC is equal to the square of AC, and that AC is equal to BD, the rectangle AB, BC is equal to the square of BD; and because 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 37. 3. line BD touches d the circle B D ACD; and because BD touches the circle, and DC is drawn from the point of contact D, the angle BDC is 32. 3. 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; 32. 1. but the exterior angle BCD is equalf to the angles CDA, DAC; therefore also BDA is equal to BCD; but BDA is 5. 1. equals to the angle CBD, because the side AD is equal Book IV. to the side AB; therefore CBD, or DBA, is equal to BCD; and consequently the three angles BDA, DBA, BCD, are equal to one another; and because the angle DBC is equal to the angle BCD, the side BD is equal to the side DC; 6. 1. but BD was made equal to CA, therefore also CA is equal to CD, and the angle CDA 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, and BCD is equal to each of the angles BDA, DBA; each therefore of the angles BDA, DBA 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. PROP. XI. PROB. To inscribe an equilateral and equiangular pentagon in a given circle. Let ABCDE be the given circle; it is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE. Describe a an isosceles triangle FGH, having each of the a 10. 4. angles at G, H, double of the angle at F; and in the circle ABCDE inscribe b the triangle ACD equiangular to the 2. 4. triangle FGH, so that the angle CAD be equal to the Because each of the angles ACD, CDA is double of CAD, and are bisected by the straight lines CE, DB, the five angles DAC, ACE, ECD, CDB, BDA are equal to one another; but equal angles stand upon equal circum- a 26. 3. ferences; therefore the five circumferences, AB, BC, CD, € 29. 3. BOOK IV. DE, EA are equal to one another: And equal circumfer ences are subtended by equale straight lines; therefore the five straight lines AB, BC, CD, DE, EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is also equiangular; because the circumference AB is equal to the circumference DE: If to each be added BCD, the whole ABCD is equal to the whole EDCB: And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the 27.3. angle BAE is equal to the angle AED: For the same reason, cach of the angles ABC, BCD, CDE is equal to the angle BAE, or AED: Therefore the pentagon ABCDE is equiangular; and it has been shown that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Which was to be done. b PROP. XII. PROB. To describe an equilateral and equiangular pentagon about a given circle. Let ABCDE be the given circle; it is required to describe an equilateral and equiangular pentagon about the circle ABCDE. Let the angles of a pentagon, inscribed in the circle, by the last proposition, be in the points A, B, C, D, E, so that 11. 4. the circumferences AB, BC, CD, DE, EA are equala; and through the points A, B, C, D, E, draw GH, HK, KL, ↳ 17. 3. LM, MG, touching the circle; take the centre F, and join FB, FK, FC, FL, FD: And because the straight line KL touches the circle ABCDE in the point C, to which FC 18. S. is drawn from the centre F, FC is perpendicular to KL, therefore each of the angles at C is a right angle: For the same reason, the angles at the points B, D are right angles: And because FCK is a right angle, the square of FK is 47. 1. equal to the squares of FC, CK: For the same reason, the square of FK is equal to the squares of FB, BK: Therefore the squares of FC, CK are equal to the squares of FB, BK, of which the square of FC is equal to the square of FB; the remaining square of CK is therefore equal to the remaining square of BK, and the straight line CK equal to BK: And because FB is equal to FC, and FK common to |