through the other Points, and will be a Circle described about the Triangle A B C. Secondly, Let DF, E F, meet each other in the Point F, in the Side BC, as in the second Figure ; and join A F Then we prove, as before, that the Point F is the Centre of a Circle described about ibe Triangle ABC. Lastly, Let ihe Right Lines D F, EF, meet one another again in the Point F, without the Triangle, as, in the third Figure ;, and join AF, F.B, F C. And because A D is equal to D B, and D F is common, and at Right Angles, the Bafe A F fhall be equal to the Bale B.F. So likewise we prove, that C'F is alio equal to A F. Wherefore B F is equal to CF. And fo again, if a Circle be described on the Centre F, with either of the Distances FÁ, FB, FC, it will pass through the other Peints, and will be described about the Triangle A B C which was to be done. Corall, If a Triangle be Right-angled, the Centre of the Circle falls in the Side opposite to the Right PROBLEM inscribe a Square within the same. Then, because B £ is equal to E D (for Ę is the Centre), and E A is common, and at Right Angles to B D, the Base B A shall be * equal to the Base A D; and for the fame Reason BC, CD, BA, and A D, are all equal to cach other. Therefore the quadrilateral Figure A B C D is equilateral, I say, it is also Rectangular. For, because the Right Line DB is a Diameter of the Circle ABCD, therefore B A D †11.1. 4. 1. BAD will be a Semicircle. Wherefore the Angle PROBLEM. to describe a Square about the same. * Draw A C, BD, two. Diameters of the Circle, cutting each other at Right Angles † ; and thro’ the + 13.1. Points A, B, C, D, draw * FG, G H, HK, KF, *17.3. Tangents to the Circle ABCD. Then, becaufe F G touches the Circle ABCD, and E A is drawn from the Centre E to the point of Contact A, the Angles at A will be + Right Angles +18.3. For the fame Reason, the Angles at the Points B, C, D, are Right Angles. And fince the Angle 1 28. I. A E B is a Right Angle, as also EBG, G H Ihall I be parallel to A C, and for the same Reason, A C to KF. In this manner we prove likewise, that GF and H K are parallel to BED; and so G F is parallel to H K. Therefore GK, GC, AK, F B, BK, + 34. 1. are Parallelograms; and fo GF + is equal to HK, and GH to FK. And since AC is equal to BD, and AC equal to either G H or FK; and BD equal to either G F, or HK;GH, or F K is equal to GF, or H K. Therefore F GHK is an equilateral quadrilateral Figure : I say, it is also equiangu. lar. For, because G B E'A is a Parallelogram, and A E B is a Right Angle ; then A G B 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 F G H K is rectan. gular; but it has been proved to be equilateral likewise. Wherefore, it must necessarily be a Square, and is described about the Circle ABCD; which was to be done. PRO PR PROPOSITION VIII. PROBLEM. to describe a Circle within the fame. # 16. 3. PROPOSITION IX. PROBLEM. ET ABCD be a Square given. It is required to circumscribe a Circle about the fame. Join AC, BD, mutually cutting one ancther in the Point E. And And since D'A is equal to A B, and AC is common, the two Sides DA, A C, are equal to the two Sides BA, AC; but the Base D C is equal to the Base B C. Therefore the Angle DAC will * be * 8.'1. equal to the Angle BAC : And consequently the Angle D A B is bilected by the Right Line A C. In the lame manner we prove, that each of the other Angles ABC, BCD, CDA, are bisected by the Right Lines A C, DB. Then, because the Angle DAB is equal to the Angle A B C, and the Angle E AB is half of the Angle D A B, and the Angle A E B half of the Angle ABC; the Angle E AB Ihall be equal to the Angle E BA: and so the Side E A is f equal co the † 6. . Side E B. In like manner we demonstrate, that each of the Right Lines E C, ED, is equal to each of the Right Lines EA, EB. Therefore the four Right Lines EA, EB, EC, ED, are equal between themselves. Wherefore, a Circle being described about the Centre E, with either of the Distances É A, EB, EC, ED, will also pass through the other Points, and will be described about ihe Square ABCD; which was to be done. PROPOSITION X. PROBLEM. Angles at ibe Base double to the other Angle. UT* any given Right Line A B in the Point C, * 11.2. . so that the Rectangle contained under A B and B C be equal to the Square of A C; then about the Centre A, with the Distance A B, let the Circle BDE be described; and t in the Circle BD E apply the Right + softbis. Line B D equal to AC; which is not greater than the Diameter. This being done, join DA, DC, and describe I a Circle ACD about the Triangle ADC. I 5 of bio. Then, becaus: the Triangle under A B and B C is equal to the Square of A C, and A C is equal to BD, the Rectangle under A B and B C shall be equal to the Square of BD. And because some Point B, is taken without the Circle ACD, and from that Point there fall two Right Lines, BCA, BD, to the Cir. cle, 5. 1. cle, one of which cuts the Circle, and the other falls on it, and since the Redtangle under A B and B.C is equal to the Square of BD; the Right Line B D fall 37. 3. * touch the Circle A C D. And since D.B touches it, and C D is drawn from the Point of Corradt D, the Angle BDC is equal to the Angle in the alter+ 32. 3. nate Segment of the Circle, viz. equal t to the Angle DAC. And since the Angle BDC is equal to the the whole Angle B D A is equal to the two Angles * 32. 1. CDA, DAC. But the outward Angie BCĐ is I equal to CDA and D AC. Therefore BD A is equal to BCD. But the Angle BDA* is equal to the Angle CBD, because the Side A D is equal to the Side AB.Wherefore D BA fhall be equal to BCD: And so the three Angles B D A, O B A, BCD, are equal to each other. And since the An. gle D B C is equal to the Angle BCD), the Side BD +6.1. is t equal to the Side D C. But B D is put equal to CA." Therefore C A is equal to CD. And fo the I s. 1, Angle CDA is I equal to the Angle D A C. There fore the Angles CDA, DAC, taken together, are double to the Angle DAC. But the Angle BCD is equal to the Angles CDA, and DAC. Therefore the Angle BCD is double to the Angle D A C. But BCD is equal to B D A, or DBA. Wherefore B DA, or D B A, is double to DAB. Therefore, the Ifoceles Triangle ABD is made, having each of the Angles at the Baje double to the other Angle; which was to be done. PROPOSITION XI. PROBLEM. gon in a given Circle. LETABCDE be a Circle given. It is required to describe an equilateral and equiangular Penta gon in the fame. * 10 of Ibis. Make an Isoceles Triangle F GH, having * each of the Angles at the Bale GH double to the other Angle F; and describe the Triarigle ADC in the * 2 of tbis. Circle ABCDE, cquiangular † to the Triangle FGH; |