a € 28. I. d с d cor. 16. bifecting the half BC thereof, and doing this continually, we fhall have a Circumference left a 1. 10. at last less than HC. Let this Circumference be IC, and draw LI from the Point L perpendicular to AC, which produce to K; and join CI, CK. And then ĈI is equal to CK. And 29. 3. fince IK is parallel to HG, and HG touches the Circle EFD, IK will not touch the Circle EFD. And much lefs do the Right def. 34.1, Lines CI, CK, touch the Circle. And if Right Lines, each equal to CI, be applied round the Circle ABC, we fhall have a Polygon infcribed therein of equal Sides, even in Number, that does not touch the leffer Circle EFG. Q. E. D. 3. PROP. XVII. Probl. To defcribe a folid Polyhedron BOPRXYTSK in BCDE the greater of two Spheres, having the fame Centre A, which shall not touch the Superficies of the leffer Sphere FGH. Cut the Spheres by a Plane paffing thro' the Centre A, and the Sections will be concentrick Circles BCDE, FGH. For fince a Sphere is made by turning a Semicircle about the Dia- def. 14. meter at reft; in whatsoever Pofition the Se- 11. micircle is conceiv'd to be, the Plane in which it is, will make a Circle in the Superficies of the Sphere. It is alfo manifeft that this Circle is one of the greatest Sections of the Sphere, becaufe the Diameter of the Sphere, that is, of the Circle, is the greateft Line that can be 15. 3. drawn in the Circle or Sphere. 8 Draw the Diameters BD, CE, at right Angles to each other, and at the Point G draw GL at right Angles to BD, and join AL, and infcribe h the Equilateral Polygon BKLME, &c. in the 16. 12. Cir 1 B K b Circle BCDE, not touching the leffer Circle. 12. 11. Continue out KA to N, and from A raise AX perpend. to the Plane of the Circ. BCDE, and draw Planes thro' AX, BD; AX, NK, making the Semicircles BXD, KXN in the 18. 11. Superficies of the Sphere, which will be perpendicular to the Plane of the Circle BCDE. Again, the Quadrants BE, BX, KX of the equal Semicircles BED, BXD, KXD, are equal; and fo if the Sides of a Polygon BO, OP, PR, RX, KS, ST, TY, YX, may be drawn in the Quadrants BX, KX, fo that their Number be equal to the Number of Sides in the Quadrant BE, each equal to BK, draw them and join SO, TP, YR. = = Again, from the Points O, S, draw Perpendiculars OV, SQ to the Plane of the Circle BCDE, which will a fall in the Sections BD, 38. 11ë KN. join VQ. Now because the Arch BO = Arch SK, and OV, SQ are perpendicular, and OV SQ, and BV KQ, and BAKA, therefore AV AQ. and fo BV : VA :: KQ: QA. Whence VQ is parallel to BK, and fince b 2.6. OV is equal to SQ, and both are perpend. to the fame Plane BCDE, they fhall be pa- 6. 11. rallel, and QV, SO are alfo equal and pa- 33. I. rallel. Whence SO is parallel to BK, and fo9. 11. KBOS is a Quadrilateral Figure in one Plane; f 7. 11. and for the fame Reafon each of the Quadrilateral Figures SOPT, TPRY, and the Triangle YRX are in one Plane. Therefore if right 32. 11. Lines be drawn from the Points O, S, P, T, c d d e . R, Y, to the Point A, there will be conftituted a folid Polyhedron within the Arches BX, KX, compofed of Pyramids, whose Bases are the Quadrilateral Figures KBOS, SOPT, TPRY, and the Triangle YRX, and Vertices at the Point A ; and if there be made the fame Conftruction on each of the Sides KL, LM, ME, as on the Side KB, as alfo in the other three Quadrants, and the other Hemifphere, there will be conftituted a Polyhedron in the Sphere, compofed of Pyramids, and Triangles, which does not touch the Superficies of the leffer Sphere. A с AZ+ZK2. and taking away the common 霍 Square AZ, and then ZK2 ZB2; and fo ZK From the Centre A drawh AZ perpend. to "11. IL the Plane KBOS, and join BZ, ZK. Now fince AZ is perpend. to BZ, ZK, and AB = AK; def. 3.11. i 2 2 therefore AZ + ZB = AB AK = k 47. I. ZKZB. In like manner we demonstrate k CORO L. Alfo if a folid Polyhedron be defcribed in fome other Sphere, fimilar to that which is defcribed in the Sphere BCDE; the folid Polyhedron defcribed in the Sphere BCDE, to the folid Polyhedron described in that other Sphere, fhall have a triplicate Proportion of that which the Diameter of the Sphere BCDE bath to the Diameter of that other Sphere. For the Solids being divided into Pyramids, equal in Number, and of the fame Order, the faid Pyramids fhall be fimilar. But fimilar Pyramids are to each other in a triplicate Propor- . tion of their homologous Sides. Therefore the Pyramid whofe Base is the quadrilateral Figure KBOS, and Vertex the Point A, to the Pyramid of the fame Order in the other Sphere, has a triplicate Proportion of that which the homologous Side of one, has to the homologous 10 I Side Side of the other, that is, which AB, drawn PROP. XVIII Spheres BAC, EDF are to one another in the A B ooo с F E G · For if you deny it, let the Sphere BAC be to the Sphere G in the triplicate Proportion of the Diameter BC to the Diameter EF. I fay GEDF. For if poffible let GEDF. Q And |