Book III, PROP. VII. PROB. To defcribe a square about a given circle. Let ABCD be the given circle; it is required to describe a square about it. 2 17.3. b18.3. Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D draw 2 FG, GH, HK, KF touching the circle; and because FG touches the circle ABCD, and EA is drawn from the center 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 G 6 28. 1. E D 34. Ι. & Io. 1. . 31. I. angle, as likewife is EBG, GH is pa- 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 d is likewise a right angle : In the fame 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 a square; and it is described about the circle ABCD; Which was to be done. PROP. VIII. PROB. To infcribe 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 fides AB, AD, in the points F, E, and through E draw EH parallel to AB or DC, and through F draw FK FK parallel to AD or BC; therefore each of the figures AK, Book IV. KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite fides are equal; and because AD is equal to AB, and c 34 1. that AE is the half of AD, and AF the half of AB, AE is equal A D to AF; wherefore the fides oppofite to these are equal, viz. FG to GE; in the fame manner, it may be demonftrated that GH, GK are each of them equal to FG or GE; therefore the four straight lines GE, GF, GH, GK, F - are equal to one another; and the circle described from the center G, at the distance of one of them, shall pass thro' the extremities of the other three, and touch the straight lines AB, BC, CD, E G K B HC DA; because the angles at the points E, F, H, K are right d d 29. 1. angles, and that the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle; e 1б. 3. 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. PRO В. 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 fides DA, AC are equal to the two BA, AC; and the base DC is equal A D 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 the same manner, it may be demonstrated that the angles ABC, B 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 equal to the angle EBA; wherefore the fide EA is equal to the fide EB: In the fame manner, it may be b 6. 1. demon. Book IV. 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 center 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. a II, 2. b 1. 4. € 5.4. PROP. X. PROB. To defcribe an isofceles triangle, having each of the angles at the base double of the third angle. Take any ftraight line AB, and divide it in the point C, fo that the rectangle AB, BC be equal to the square of CA; and from the center 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; 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 E 37.3 € 32.3. f 32. 1. from the point B without the A C B D but but BDA is equal to the angle CBD, because the fide AD Book IV. is equal to the fide AB; therefore CBD, or DBA is equal to BCD; and confequently the three angles BDA, DBA, BCD, are g5.. equal to one another; and because the angle DBC is equal to the angle BCD, the fide BD is equal to the fide DC; but BD 6. 3, was made equal to CA; therefore also CA is equal to CD, and the angle CDA equal & 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. T O inscribe an equilateral and equiangular pentagon Let ABCDE be the given circle; it is required to infcribe an equilateral and equiangular pentagon in the circle ABCDE. Describe 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 the triangle ACD equiangular to the tri-b2.4 angle FGH, so that the angle CAD be equal to the angle A at F, and each of the angles ACD, CDA equal to the F angle at G or H; wherefore each of the angles ACD, CDA is double of the angle CAD. Bisect the angles c 9. I. ACD, CDA by the straight C lines CE, DB; and join AB, D BC, DE, EA. ABCDE GH is the pentagon required. 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 d circumferences; therefore the d 26. 3. five circumferences AB, BC, CD, DE, EA are equal to one another: Book IV. € 29.3.1 f 27.3. e another: And equal circumferences are subtended by equal 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 angle BAE is equal f to the angle AED: For the fame reason, each of the angles ABC, BCD, CDE is equal to the angle BAE, or AED: Therefore the pentagon ABCDE is equiangular; and it has been shewn that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Which was to be done. 2 11.4. b 17. 3. с 18. 3. 47. I. PROP. XII. PROB. O 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 the circumferences AB, BC, CD, DE, EA are equal; and thro' the points A, B, C, D, E draw GH, HK, KL, LM, MG touching the circle; take the center 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 is drawn from the center 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 equal d to the squares of FC, CK: For the fame 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 |