PROPOSITION XIV. THEOREM. Cones and Cylinders, being upon equal Bafes, are LET the Cylinders E B, FD, ftand upon equal For, produce the Axis KL to the Point N; and put L N equal to the Axis GH; and let a Cylinder CM be conceived about the Axis LN: Then, becaufe the Cylinders EB, CM, have the fame Altitude; they are to one another as their Bafes. But * 11 of this. their Bafes are equal; therefore the Cylinders E B, CM, will be alfo equal. And because the Cylinder. FM is cut by a Plane C D, parallel to the oppofite Planes, it fhall be as the Cylinder C M is to the Cylinder FD, fo is the Axis LN to the Axis K L. But the Cylinder C M is equal to the Cylinder EB; and the Axis L N to the Axis GH: therefore the Cylinder E B is to the Cylinder FD, as the Axis GH is to the Axis K L And as the Cylinder E B is to the Cylinder FD, fo is the Cone ABG to the Cone C D K ;‡ 15. 5. for the Cylinders are triple of the Cones. There-10 of this. fore as the Axis GH is to the Axis KL, fo is the Cone A B G to the Cone CDK; and fo the Cylinder EB to the Cylinder FD. Wherefore, Canes and Cyfinders, being upon equal Bafes, are to one another as their Altitudes s; which was to be demonftrated. * PROPOSITION XV. The Bafes and Altitudes of equal Cones and Cylin- ET the Bafes of the equal Cones and Cylinders be the Circles ABCD, EF GH, and their Diameters A C, EG; and Axis KL, MN; which are alfo alfo the Altitudes of the Cones and Cylinders: And let the Cylinders AX, EO be compleated. I fay, the Bafes and Altitudes of the Cylinders A X, E Ó, are reciprocally proportional; that is, the Base ABCD is to the Bafe EF GH, as the Altitude MN is to the Altitude KL. For, the Altitude K L is either equal to the Altitude M N, or not equal. First, let it be equal; and the Cylinder A X is equal to the Cylinder EO. But Cy1x of this. linders and Cones, that have the fame Altitude, are * to one another as their Bafes; therefore the Base ABCD is equal to the Base E F GH: And confequently, as the Bafe ABCD is to the Base EFGH, fo is the Altitude M N to the Altitude KL, But if the Altitude K L be not equal to the Altitude M N, let MN be the greater; and take PM, equal to L K, from MN; and let the Cylinder EO be cut thro' P by the Plane TY S, parallel to the oppofite Planes of the Circles EFGH, RO; and conceive ES to be a Cylinder, whofe Bale is the Circle E F G H, and Altitude PM: Then, because the Cylinder A X is equal to the Cylinder EO, and ES is fome other Cylinder; the Cylinder AX to the Cylinder ES, fhall be as the Cylinder EO is to the Cylinder ES. But as the Cylinder A X is to the Cylinder ES, fo is the Bafe ABCD to the Bafe E F G H; for the Cylinders A X, ES, have the fame Altitude: And as the Cy+11 of this, linder EO is to the Cylinder E S, fo is + the Altitude MN to the Altitude MP; for the Cylinder EO is cut by the Plane TYS, parallel to the oppofite Planes, Therefore as the Base A B C D is to the Bafe EFGH, fo is the Altitude M N to the Altitude MP. Altitude M P is equal to the Altitude K L; wherefore, as the Bafe ABCD is to the Bafe E F G H, fo is the Altitude M N to the Altitude KL; and therefore, the Bafes and Altitudes of the equal Cylinders A X, E O, are reciprocally proportional. 33 of this. But the And if the Bafes and Altitudes of the Cylinders A X, E O, are reciprocally proportional; that is, if the Base A B C D be to the Base E F G H, as the Altitude M N is to the Altitude KL; I fay, the Cy. linder AX is equal to the Cylinder EO. For, the fame Conftruction remaining, because the Bafe ABCD is to the Bafe E F G H, as the Altitude M N is to the Altitude KL; and the Altitude K L is equal to the Altitude M P ; it shall be, as the Base A B C D is to the Base EFGH, fo is the Altitude MN to the Altitude M P. But as the Bafe A BCD is to the Bafe EFGH, fo is the Cylinder A X to the Cylinder ES, for they have the fame Altitude; alfo as the Altitude MN is to the Altitude M P, fo is the Cylinder E O 13 of this. to the Cylinder E S. Therefore as the Cylinder AX is to the Cylinder E S, fo is the Cylinder E O to the Cylinder E'S: Wherefore, the Cylinder A X is equal to the Cylinder EO; which was to be demonftrated. In like manner we prove this in Cones. * PROPOSITION XVI.. Two Circles being about the fame Centre, to in- LET ABCD, EFG H, be two given Circles about the Centre K; it is required to infcribe a Polygon of equal Sides, even in Number in the Circle ABCD, not touching the leffer Circle E F G H. Draw the Right Line B D through the Centre K, as alfo A G, from the Point G, at Right Angles to BD, which produce to C; this Line will touch the Cir- 16. 3. cle EFGH: Then, bifecting the Circumference BAD, and again bifecting the Half thereof, and doing this continually, we fhall have a Circumference left, at laft, less than A D. Let this Circumference ‡ Lemma. be LD, and draw L M from the Point L, perpendicular to BD, which produce to N; and join LD, DN: And then LD is + equal to D N. And fincet 29. 3. LN is parallel to A C, and A C touches the Circle EFGH, LN will not touch the Circle E FGH; and much lefs do the Right Lines L D, D N, touch the Circle. And if Right Lines, each equal to LD, be applied round the Circle ABCD, we shall have a Polygon infcribed therein of equal Sides, even in Number, that does not touch the leffer Circle EFGH; which was to be done." PRO * Def. 14. † 15.3. PRO və LON XVII. PROBLEM. To defcribe a folid Polyhedron, in the greater of two Spheres, having the fame Centre, which shall not touch the Superficies of the lesser Sphere. LET two Spheres be fuppofed about the fame Centre A; it is required to defcribe a folid Polyhedron in the greater Sphere, not touching the Superficies of the leffer Sphere. Let the Spheres be cut by fome Plane paffing thro' the Centre; then the Sections will be Circles: For, because a Sphere is made by the turning of a Semicircle about the Diameter, which is at Reft; in whatfoever Pofition the Semicircle is conceived to be, the Plane in which it is fhall make a Circle in the Superficies of the Sphere. It is alfo manifeft, that this Circle is a greater Circle, fince the Diameter of the Sphere, which is likewife the Diameter of the Semicircle, is + greater than all Right Lines that are drawn in the Circle or Sphere. Now let BCDE be that Circle of the greater Sphere, and F G H of the leffer Sphere; and let BD, CE be two of the Diameters drawn at Right Angles to one another; let BD meet the Jeffer Circle in the Point G, and let G L be drawn at Right Angles to A G, and A L be joined: Then, bifecting the Circumference E B, as alfo the Half thereof, and doing thus continually, we shall have left, at laft, a certain Circumference less than that Part of the Circumference of the Circle BL, which is fubtended by a Right Line equal to G L. Let this be the Circumference B K; then the Right Line B K is lefs than * 16 of this GL; and BK fhall be the Side of a Polygon of equal Sides, even in Number, not touching the leffer Circle Now, let the Sides of the Polygon, in the Quadrant of the Circle B E, be the Right Lines B K, KL, DM, ME; and produce the Line joining the Points K, A, to N; and raife ‡ AX from the Point A, perpendicular to the Plane of the Circle B C D E, meeting the Superficies of the Sphere in the Point X ; and let Planes be drawn thro' A X and BD, and thro' 11. 11. * A X and KN; which, from what has been faid, will make great Circles in the Superficies of the Sphere; and let B XD, K X N, be Semicircles on the Diameters B D, KN: Then, because X A is perpendicular to the Plane of the Circle B C D E, all Planes that pass through X A shall also be perpendicular to * 18. 11. that fame Plane. Therefore the Semicircles B X D, KXN, are perpendicular to that fame Plane. And because the Semicircles BED, B X D, K X N, are equal; for they ftand upon equal Diameters BD, KN; their Quadrants B E, BX, K X, fhall be alfo equal. And therefore, as many Sides as the Polygon in the Quadrant BE has, fo many Sides may there be in the Quadrants BX, K X, equal to the Sides BK, KL, LM, ME. Let thofe Sides be BO, OP, PR, RX, K S, ST, TY, Y X ; and join SO, TP, Y R; and let Perpendiculars be drawn from O, S, to the Plane of the Circle B C DE: Thefe will fall on B D, K N, the common Sections + 38. 11. of the Planes; because the Planes of the Semicircles BXD, K X N, are perpendicular to the Plane of the Circle BCDE: Let the faid Perpendicular be OV, SQ, and join VQ; then, fince the equal Circumferences BO, SK, are taken in the equal Semicircles BX D, K X N, and O V, SQ are Perpendiculars; O V fhall be equal to S Q, and BV to KQ. But the Whole B A is equal to the Whole K A ; therefore the Part remaining V A is equal to the Part remaining QA Therefore, as BV is to VA, fo is KQ to QA; and fo V Qis ‡ parallel to B K. An 12.6. fince O V and S Qare both perpendicular to the Plane of the Circle BCDE, OV fhall be parallel to SQ. But it has alfo been proved equal to it; wherefore QV, S O, are † equal and parallel. And because † 33.1. QV is parallel to SO, and alfo parallel to K B; SO fhall be alfo parallel to KB: Join B O, K S, and then K BOS 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 faid Points is in the fame Plane as the Parallels are. And for the fame Reafon each of the quadrilaterál Figures SOP T, TPR Y, are in one Plane. And the Triangle YR X is † in one Plane; therefore if Right † 2. 12. Lines be fuppofed to be drawn from the Points O, S, P, * 9. 11. 7. 11. |