But if the Altitude KL be not equal to the Altitude MN, let MN be the greater. And take PM equal to L K from MN; and let the Cylinder E O be cut thro' P by the Plane TYS, parallel to the opposite Planes of the Circles EFGH, RO, and conceive ES to be a Cylinder, whose Base is the Circle EFGH, and Altitude PM. Then, because the Cylinder AX is equal to the Cylinder E O, and ES is some other Cylinder, the Cylinder AX to the Cylinder ES, Ihall be as the Cylinder EO, is to the Cylinder ES. But as the Cylinder AX is to the Cylinder ES, fo is * the Base ABCD to the Base * 11 of this. EFGH; for the Cylinders AX, ES have the same Altitude. And as the Cylinder EO is to the Cylinder ES, fo is † the Altitude MN to the Altitude' MP; † 13 of this. for the Cylinder EO is cut by the plane TYS parallel to the opposite Planes. Therefore, as the Base ABCD is to the Base EFGN, fo is the Altitude MN to the Altitude MP. But the Altitude MP is cqual to the Altitude KL. Wherefore as the Base ABCD is to the Base EFGH, so is the Altitude MN to the Altitude KL. And therefore the Bases and Altitudes of the equal Cylinders AX, EO, are reciprocally proportional. And if the Bases and Altitudes of the Cylinders AX, EO, are reciprocally proportional, that is, if the Base ABCD be to the Base EFGH, as the Altitude MN is to the Altitude KL. I say, the Cylinder AX is equal to the Cylinder EO. For the fame Construction remaining; because the Bafe ABCD is to the Base EFGH, as the Altitude MN is to the Altitude KL; and the Altitude KL is equal to the Altitude MP. It fhall be as the Base ABCD is to the Base EFGH, so is the Altitude MN to the Altitude MP. But as the Base ABCD is to the Base EFGH, fo is the Cylinder AX to the Cylinder ES; for they have the same Altitude. And as the Altitude MN is to the Altitude MP, so is the Cylinder 11 of this EO to the Cylinder ES. Therefore, as the Cylinder AX is to the Cylinder ES, so is the Cylinder EO to the Cylinder Es. Wherefore the Cylinder AX is equal to the Cylinder EO. In like Manner we prove this in Cones; wbich was to be demonstrated. PRO PROPOSITION XVI. PROBLEM. in the greater a Polygon of equal Sides even in # 16. 3. ET ABCD, EFGH, be two given Circles about the Center K. It is required to inscribe a Polygon of equal Sides even in Number in the Circle ABCD, not touching the leffer Circle EFGH. Draw the Right Line BD through the Center K, as also AG, from the Point G at Right Angles to BD, which produce to C; this Line will * touch the Circle EFGH. Then bisecting the Circumference BAD, and again bisecting the half thereof, and doing this continually, we shall have a Circumference left at last less than AD. Let this Circumference be LD, and draw LM from the Point L perpendicular to BD, which produce to N; and join LD, DN. And then L D is + equal to DN. And since LN is parallel to AC, and AC touches the Circle EFGH, LN will not touch the Circle EFGH. And much less do the Right Lines LD, DN, touch the Circle. And if Right Lines, each equal to LD, be applied round the Circle ABCD, we shall have a Polygon inscribed therein of equal Sides, even in Number, that does not touch the leffer Circle EFG. which was to be demonstrated. PROPOSITION XVII. PROBL’IM. two Spheres, having the same Center, which shall not touch the Superficies of the lesser Sphere. L ET two Spheres be supposed about the same Center A. It is required to describe a solid Polyhedron in the greater Sphere, not touching the Superficies of the lefler Sphere, Let Let the Spheres be cut by fome Plane passing thro’ the Center. Then the Sections will be Circles; for because a Sphere is * made by the turning of a Semi- * Def. 14circle about the Diameter which is at rest: In what- 11. soever Position the Semicircle is conceived to be, the Plane in which it is shall make a Circle in the Superficies of the Sphere. It is also manifest that this Circle is a great Circle, since the Diameter of the Sphere, which is likewise the Diameter of the Semicircle, is + greater than all Right Lines that are drawn in the † 15. 3. Circle, or Sphere. Now, let BCDE be that Circle of the greater Sphere, and FGH of the lesser Sphere; and let BD, CE be two of their Diameters drawn at Right Angles to one another. Let BD meet the lefser Circle in the Point G, from which to AG let GL be drawn at Right Angles, and AL joined. Then bisecting the Circumference EB, as also the half thereof, and doing thus continually, we shall have left at last a certain Circumference less than that Part of the Circumference of the Circle BCD, which is fubtended by a Right Line equal to GL. Let this be the Circumference BK. Then the Right Line BK is less than GL; and BK shall be the side of a Polygon of equal Sides, even in Number, not touching the lefser Circle. Now, let the Sides of the Polygon in the Quadrant of the Circle BE, be the Right Lines BK, KL, LM, ME, and produce the Line joining the Points K, A, to N: And raise I AX 12. II. from the Point A, perpendicular to the Plane of the Circle BCDE, meeting the Superficies of the Sphere in the Point X, and let Planes be drawn thro' AX, and BD, and thro' AX, and KN, which from what has been said will make great Circles in the Superficies of the Sphere. And let BXD, KXN, be Semicircles on the Diameters BD, KN. Then because XA is perpendicular to the Plane of the Circle BCDE, all Planes * 18. 11. that pass thro’ XA shall also * be perpendicular to that fame Plane. Therefore the Semicircles BXD, KXN are perpendicular to that same Plane. And because the Semicircles BED, BXD, KXN, are equal ; for they stand upon equal Diameters BD, KN; their Quadrants BE, BX, KX, shall be also equal. And therefore, as many Sides as the Polygon in the Quadrant BE has, fo 'many Sides may there be + 38.11. 1 2. 6. . 6. II. + 33. 1. be in the Quadrants BX, KX, equal to the sides BK, KL, LM, ME. Let those Sides be BO, OP, PR, RX, KS, ST, TY, YX: And join SO, TP, YR; and let Perpendiculars be drawn from O, S, to the Plane of the Circle BCDE, There will † fall on BD, KN, the common Sections of the Planes; because the Planes of the Semicircles BXD, KXN, are perpendicular to the Plane of the Circle BCDE. Let the said Perpendiculars be OV, SQ, and join VQ. Then fince the equal Circumferences BO, SK, are taken in the equal Semicircles BXD, KXN, and OV, SQ are Perpendiculars, OV shall be equal to SQ, and B V to KQ. But the Whole B A is equal to the Whole KA. Therefore the Part remaining VA, is equal to the Part remaining QA. Therefore as BV is to VA, so is KQ, to QA: And so VQ is I parallel to BK. And fince O V and SQ are both perpendicular to the Plane of the Circle BCDE, O V shall be * parallel to SQ. But it has also been proved equal to it. Wherefore QV, SO are equal and parallel. And because QV is parallel to SO, and also parallel to KB, SO shall be also I parallel to KB: But BO, KS, join them. Therefore KBOS is * a quadrilateral Figure in one Plane: For if two Right Lines be parallel, and Points be taken in both of them, a Right Line joining the said Points is in the same Plane as the Parallels are. And for the fame Reason, each of the quadrilateral Figures SOPT, TPRY, are in one Plane, And the Triangle YRX, is † in one Plane. There's fore, if Right Lines be supposed to be drawn from the Points O, S, P, T, R, Y, to the Point A, there will be constituted a certain folid polyhedrous .Figure within the Circumferences BX, KX, composed of Pyramids, whose Bases are the quadrilateral Figures KBOS, SOPT, TPRY, and the Triangle YŘX; and Vertices the Point A. And if there be made the same Construction on each of the Sides KL, LM, ME, like as we have done on the Side KB, and also in the other three Quadrants, and the other Hemisphere, there will be conftituted a polyhedrous Figure described in the Sphere, composed of Pyramids whose Bases are the aforesaid quadrilateral Figures, and the : Triangle YRX, being of the fame Order, and and Vertices the Point A. I say, the said Polyhe- from the Point A, perpendicular to the Plane of the DI quadrilateral Figure KBSO, meeting it in the Point Ż, and join BZ, ZK. Then fince AZ is perpendi- fore, |