1 2 a 17.3. b18. 3. d 34. I. e 15. def. 1. PROP. VI. Plate II. Fig. 57. In a circle given EABCD to infcribe a square ABCD. (a) Draw the diameters AC, BD cutting each other at right-angles in the center E. Join the extremes of these diameters with the right-lines AB, BC, CD, DA. And the thing is done. Now because the four angles at E are right, the (6) arches and (c) fubtended lines AB, BC, CO, DA, are equal; therefore is the figure ABCD equilateral, and all the angles in femicircles, and so (d) right. (e) Therefore ABCD is a square inscribed in a circle given. Which was to be done. PROP. VII. Fig. 58. About a circle given EABCD, to describe a square FHIG. Draw the Diameters AC, BD, cutting one the other at right-angles; through the extremes of these diameters (a) draw tangents meeting in F, H, I, G. And the thing is done. For because (b) the angles A and Care right, (c) therefore is FG parallel to HI. After the same manner is FH parallel to GI, and therefore FHIG is a Pgr. and also right-angled. It is equilateral because FG (d) =HI (d) =DB (e) =CA (d) =FH (d) =GI. Wherefore FHIG is a (f) square circumscribed to the f 10. def. 1. circle given. Which was to be done. a 7. ax. b byp. 33.1. Schol. Fig. 59. A square ABCD described about a circle is double of the square EFGH inscribed in the fame circle. For the rectangle HB=2 HEF and HD=2 HGF, by the 41. 1. PROP. VIII. Fig. 58. In a square given FGHI, to infcribe a circle EABCD. Bisect the sides of the square in the points B,D,A,C, cutting one the other in E, a circle drawn from the center E, thro' A, shall be inscribed in the square. For because FA and HC are (a) equal and (b) parallel, (c) therefore is FH parallel to AC, parallel to GI. After the fame manner is FG parallel to BD, parallel to HI; therefore EF, EG, EH, EI, are parallelograms. There Therefore FA (d)=FB(e)=AE=BE=CE=ED. The circle therefore described from the center E, through A shall pass through A, B, C, D, and touch the fides of the square, since the angles A, B, C, D, are right. Which was to be doee. PROP. IX. Plate II. Fig. 57. About a square given ABCD, to defcribe a circle EABCD. Draw the diagonals AC, BD, cutting one the other in E. From the center E through A describe a circle; I say this circle is circumscribed to the square. For the angles ABD and BAC are (a) half of rightangles; (b) therefore EA=EB. After the fame manner is EA - ED=EC. The circle therefore described from the center E passes through A, B, C, D the angles of the square given. Which was to be done. PROP. X. Fig. 60. To make an Isocetes triangle ABD, having each angle at the base B, and ADB double to the remaining angle A. d 7. ax. e 34. 1. a 4. cor.. 32. I. b 6. 1. Take any right-line AB, and divide it in C, (a) so that ABXBC may be equal to ACq. From the center A thro' B, describe the circle ABD; and in this circle (6) apply b 1. 4. BD=AC, and join AD; I say ABD is the triangle a II. 2. required. C 5.4. e 32. 3. f 2. ax. For draw DC, and through the points C, D, A, (c) d 37.3. draw a circle. Now because ABxBC=ACq=BDq, (d) it is evident that BD touches the circle ACD which CD cutteth; (e) therefore is the angle BDC=A, and therefore the angle BDC+CDA (f) = A+CDA (g) = BCD. But BDC + CDA = BDA (b) = CBD; (k) therefore the angle BCD = CBD, and therefore DC (1) =DB= (m) AC; (n) wherefore the angle CDA = A =BDC, therefore ADB = 2 A = ABD. Which avas to be done. Coroll. Whereas all the angles A, B, D, (b) make up two right-angles, it's evident that A is one fifth of two right-angles. PROP. XI. Fig. 61, 62. In a circle given ABCDE to infcribe a Pentagon ABCDE equilateral and equiangular. g 32. I. h 5. 1. k 1. ax. 1 6. 1. m constr. n 5. 1. h 32. 1. D 4 (a) Describe a 10. 4. b 2. 4. с 9. 1. d 26. 3. e 29. 3. f 27.3. g 2. ax. (a) Describe an Isosceles triangle FGH, having each angle at the base double to the other; to the circle, (b) infcribe a triangle CAD equiangular to the said triangle FGH. (c) Bisect the angles at the base ACD and ADC with the right-lines DB, CE meeting with the circumference in B and E, join the right-lines CB, BA, AE, ED. Then I say it is done. For it is evident by construction, that the angles CAD, CDB, BDA, DCE, ECA, are equal; wherefore the (d) arches and (e) the lines subtending them DC, CB, BA, AE, DE, are equal. Therefore the pentagon is equilateral, and equiangular, (f) because the angles of it BAE, AED, &c. stand on equal (g) arches BCDE, ABCD, &c. A more easy practice of this problem shall be deliver'd at 10. 13. Coroll. Hence, each angle of an equilateral and equiangular pentagon is equal to three-fifths of two right-angles, or fix-fifths of one right-angle. Schol. Fig. 63. Generally all figures whose number of fides is odd, are Pet. Herig. inscribed in circles by the help of Isosceles triangles, whose angles at the base are multiples of those at the top: and figures whose number of fides is odd, are infcribed in a circle by the help of Isosceles triangles, whose angles at the base are multiples sesquialter of those at the top. a 11.4. As in the Isosceles triangle CAB if the angle A=3C =B, then will AB be the fide of a Heptagon. If A=4C, then is AB the fide of an Enneagon. But if AC, then is AB the fide of a square. And if A=2 CAB will fubtend the fixth part of the circumference, and likewife if A=3C then will AB be the fide of an Octagon. PROP. XII. Plate III. Fig. 1. About a circle given FABCDE, to describe a pentagon HIKLG, equilateral and equiangular. (a) Inscribe a pentagon ABCDE in the circle given; and from the center draw the right-lines FA, FB, FC, FD, FE; and to those lines draw fo many perpendiculars GAH, HBI, ICK, KDL, LEG, meeting in the points H, I, K, L, G, and the thing is done. For because cause GA, GE from the fame point G (6) touch the circle, (c) therefore is GA=GE, and (d) therefore the angle GFA =GFE, therefore the angle AFE=2 GFA. After the fame manner is the angle AFH =HFB, and confequently the angle AFB = 2 AFH. (e) But the angle AFE AFB, (f) therefore the angle GFA AFH. But also the angle FAH (g) =FAG, and the fide FA is common, (b) therefore HA=AG=GE=EL, &c. (k) Therefore HG, GL, LK, KI, IH, the fides of the pentagon are equal, and so also are the angles, because double of the equal angles AGF, AHF, therefore, &c. Coroll. After the same manner, if any equilateral and equiangled figure be described in a circle, and at the extreme points of the semi-diameters, drawn from the center to the angles, be drawn perpendicular lines to the said diameters; I say, that these perpendiculars shall make another figure of as many equal fides and equal angles, circumfcribed to the circle. PROP. XIII. Plate III. Fig. 2. In an equilateral and equiangular pentagon given ABCDE to infcribe a circle FGHK. (a) Bisect two angles of the pentagon A and B, with the right-lines AI, BK, meeting in the point F. From F draw the perpendiculars FG, FH, FI, FK, FL. Then a circle described from the center F through G will touch all the fides of the pentagon. Draw FC, FD, FE. Because BA (6) =BC, and the fide BF common, and the angle FBA(c)=FBC, (d) therefore is AF=FC, and the angle FAB = FCB, but the angle FAB (e) = BAE= BCD. Therefore the angle FCB=BCD. After the same manner are all the whole angles C, D, E bisected. Now whereas the angle FGB (f) =FHB, and the angle FBH=FBG, and the fide FB is common, (g) therefore is FG=FH. In like manner are all the right-lines FH, FI, FK, FL, FG equal. Therefore a circle described from the center F, through G, paffes through the points H, I, K, L, and (b) touches the side of the pentagon, because the angles at those points are right. Which was to be done. Coroll. Hence, if any two nearest angles of an equilateral and equianglar figure are bisected, and from that point in which bear.16.3. c2.cor. 36. d 8. 1. e 27. 3. f 7. ax. g 12. ax. h 26.1. k 2. ax. a 9. I. b byp. f 12. ax. g 26. 1. h cor.16.3. Γ which the lines meet that bisect the angles, be drawn. right-lines to the remaining angles of the figure, all the angles of the figure shall be bisected. Schol. By the fame method may a circle be inscribed in any equilateral and equiangular figure. PROP. XIV. Plate III. Fig. 1. About a pentagon given ABCDE, equilateral and equiakgular, to describe a circle FABCDE. Bisect any two angles of the pentagon with the rightlines AF, BF, meeting in the point F; the circle described from the center F through A shall be described about the pentagon. For let FC, FD, FE be drawn. (a) Then the angles C, D, E are bisected; (b) and therefore FA, FB, FC, FD, FE are equal; therefore the circle described from the center F passes through A, B, C, D, E, all the angles of the pentagon. Which was to be done. Schol. By the same method is a circle described about any figure which is equilateral and equiangular. PROP. XV. Fig. 3. In a circle given GABCDEF to infcribe an Hexagon (or fix-fided figure) ABCDEF equilateral and equiangular. Draw the diameter AD; from the center D through the center G describe a cirrle cutting the circle given in the points C and E. Draw the diameters CF, EB; and join AB, BC, CD, DE, EF, FA. Then I say it's done. a cor. 32. 1. For the angle CGD (a)= of 2 right-angles (a)=DGE (6)=AGF (8)=AGB. (c) Therefore BGC=+ of 2 right d 26. 3. e 29.3. Corol. 1. Hence, the side of an hexagon inscribed in a circle is equal to the femidiameter, 2. Hereby |