* 31. 3o mícircle. Wherefore the Angle BAD is * a Right Angle. And for the same Reason every one of the Angles ABC, BCD, CDA, is a Right Angle. Therefore ABCD is a rectangular quadrilateral Figure: But it has also been proved to be equilateral. Wherefore it shall necessarily be a Square, and is described in the Circle ABCD; which was to be done. PROPOSITION VII, PROBLEM to describe a Square about the fame. Then because F G touches the Circle ABCD, and EA is drawn from the Center É to the point of Contact A, the Angles at A will be + Right Angles. For the fame Reason, the Angles at the Points B, C, D, are Right Angles. And since the Angle AEB is a Right Angle, as also E BG, GH fhall be $ parallel to A C; and for the fame Reason, A C to KF. In this Manner we prove likewise, that GF and HK are parallel to BED; and so' GF is parallel to HK. Therefore GK, GC, AK, FB, BK, are Parallelograms; and so GF is * equal to HK, and GH to FK. And since AC is equal to BD, and AC* equal to either GH, or FK, and BD equal to either GF, or HK; GH, or FK, is equal to GF, or HK. Therefore FGHK is an equilateral quadrilateral Figurę: I say it is also equiangular. For because GBE A is a Parallelogram, and Ą EB is a Right Angle, then AGB shall be also a Right Angle. In like manner we demonstrate, that the Angles at the Points H, K, F, are Right Angles. Therefore the quadrilateral Figure FGHK is rectangular; but it has been proved to be equilateral likewise. Wherefore it must neceffarily be 2 Square, and is described about the Circle A B CD; which was to be done. PRO 34. . PROPOSITION VIII. а PROBLEM. . Bisect * the Sides A B, AD, in the Points F, E;* 10. 1, and draw + EH thro' E, parallel to AB, or DC; and t 31. 19 FK thro' F, parallel + to BC, or AD. Then AK, KB, AH, HD, AG, GC, BG, GD, are all Parallelograms, and their opposite Sides are I equa!. And be- † 34 34 cause D A is equal to A B, and AE is half of AD, and AF half of AB, AE shall be equal to AF; but · the oppofitę Sides are also equal. Therefore F G is equal to GE. In like manner we demonstrate, that GH, or GK, is equal to either F G, or GE. Therefore GE, GF, GH, GK, are equal to each other : And so a. Circle being described about the Center G, with either of the Distances GE, GF, GH, GK, will also pass thro' the other Points, and shall touch the Sides A B, BC, CD, DA, because the Angles at i E, F, H, K, are Right Angles. For if the Circle e should cut the sides of the Square, a Right Line, drawn from the End of the Diameter of a Circle at Right Angles, will fall within the Circle; which is * abfurd. Wherefore a Circle described about the * 16. Center G, with çither of the Distances GĘ, GF, GH, GK, will not cut A B, BC, CD, DA, the Sides of the Square. Wherefore it shall necessarily touch them, and will be described in the Square ABCD; which was to be done. PROPOSITION IX. PROBLEM. L to circumscribe a Circle about the fame. And 8. 1. And since DA is equal to AB, and AC is common, the two Sides DA, A C, are equal to the two Sides BA, AC; but the Bafe DC is equal to the Base BC. Therefore the Angle DAC will * be equal to the Angle BAC: And consequently the Angle D AB is bifected by the Right Line AC. In the fame Manner we prove, that each of the Angles ABC, BCD, CDA, are bisected by the Right Lines AC, DB. Then because the Angle DAB is equal to the Angle ABC, and the Angle EAB is half of the Angle DAB, and the Angle EBA half of the Angle ABC; the Angle E AB shall be equal to the Angle EBA: And so the Side EA is equal to the Side EB. In like manner we demonstrate, that each of the Right Lines, EC, ED, is equal to each of the Right Lines EA, E B. Therefore the four-Right Lines E A, EB, EC, ED, are equal between themselves. Wherefore a Circle being described about the Center E, with either of the Distances EA, EB, EC, ED, will also pass thro' the other Points, and will be described about the Square ABCD; which was to be done. 76. 1. PROPOSITION X. * 11. 2. PROBLEM. Angles at the Base double to the other Angle. so that the Rectangle contained under AB, BC, be equal to the Square of AC; then about the Cen ter A, with the Distance AB, let the Circle B DE + 3 of this. be described ; and † in the Circle BDE apply the Right Line BD equal to AC; which is not greater than the Diameter. This being done, join DA, DC, # 5 of this and describe I a Circle ACĎ about the Triangle ADC. Then because the Rectangle ABC is equal to the Square of A C, and AC is equal to BD, the Rectangle under AB, BC, shall be equal to the Square of B D.' And because some Point B is taken without the Circle A CD, and from that Point there fall two Right Lines BCA, BD, to the Circle, one of which 2 cuts cuts the Circle, and the other falls on it. And since the Rectangle under AB, BC, is equal to the Square of BD, the Right Line BD shall * touch the Circle * 37. 3. ACD. And fince BD touches it, and D C is drawn from the Point of Contact D, the Angle BDC is equal to the Angle in the alternate Segment of the Circle, vix. equal + to the Angle DAC. And since † 32. 3. the Angle BDC is equal to the Angle DAC; if CDA, which is common, be added, the whole Angle BDA is equal to the two Angles CDA, DAC. But the outward Angle BCD is equal to CDA, # 32. 1. DAC. Therefore B D A is equal to B CD. But the Angle BDA* is equal to the Angle CBD, be- * 5. I. cause the Side AD is equal to the Side AB. Wherefore D B A shall be equal to BCD: And so the three Angles BDA, DBA, BCD, are equal to each other. And fince the Angle DBC is equal to the Angle BCD, the Side BĎ is tequal to the Side DC. But † 6. 1. BD is put equal to CA. Therefore CA is equal to CD. And fo the Angle CDA is equal to the Angle DAC. Therefore the Angles CDA, DAC, taken together, are double to the Angle DAC. But the Angle BCD is equal to the Angles CDA, DAC. Therefore the Angle BCD is double to the Angle DAC. But BCD is equal to BDA, or D B A. Wherefore BDA, or DBA, is double to DAB. Therefore the Isosceles Triangle ABD is made, having one of the Angles at the Base, double to the other Angle; which was to be done. PROPOSITION XI, PROBLEM. gon in a given Circle. to Make an Isosceles Triangle F GH, having * each. 10 of this. of the Angles at the Base GH, double to the other Angle F; and describe the Triangle ADC in the Circle ABCDE, equiangular † to the Triangle FGH; so that + 2 of tbiso the z6. 3 the Angle CAD be equal to that at F, and ACD, CDA, each equal to the Angles Gor H. Wherefore the Angles A CD, CDA, are each double to the An gle CAD. This being done, bisect * ACD, CDA, by the Right Lines CE, DB, and join A B, BC, DE, EA. Then because each of the Angles A CD, CDA, is double to CAD, and they are bisected by the Right Lines CE, DB ; the five Angles DAC, ACE, ECD, CDB, BDA, are equal to each other. But equal Angles ftand * upon equal Circumferences. Therefore the five Circumferences AB, BC, CD, DE, EA, are equal to each other. But equal Circumferences † 29. 3. fubtend t equal Right Lines. Therefore the five Right Lines AB, BC, CD, DE, EA, are cqual tą each other. Wherefore ABCDE is an equilatera! Pentagon. I fay, it is also equiangular; for because the Circumference AB is equal to the Circumference DE, by adding the Circumference BCD, which is common, the whole Circumference ABCD is equal to the whole Circumference EDCB, but the Angle AED stands on the Circumference ABCD, and BAE on the Circumference EDCB: Therefore the Angle B A E is equal to the Angle A ED. For the fame Reason, each of the Angles ABC, BAD, CDE, is equal to BAE, or AED. Wherefore the Pentagon ABCDE is equiangular ; but it has been proved to be also equilateral. And consequently there is an equilateral and equiangular Pentagon inscribed in a given Circle; which was to be done. PROPOSITION XII. PROBLEM. gon about a Circle given. L * By 11 ET ABCDE be the given Circle. It is required to describe an equilateral and equiangular Pentą: gon about the fame. Let A, B, C, D, E, be the angular Points of a Pentagon supposed to be inscribed in the Circle; fo that thę Çircumferences AB, BC, CD, DE, E A, be equal; of ibis. |