thrice the cones: Therefore as the axis GH is to the axis KL, fo is the cone A B G to the cone CDK, and the cylinder EB to the cylinder FD. Which was to be demonstrated.

PROP. XV. THEOR.

The bafes of equal cones or cylinders are reciprocally proportional to their altitudes: alfo thofe cones and cylinders whose bases are reciprocally proportional to their altitudes, are equal.

Let there be equal cones or cylinders, whofe bafes are the circles ABCD, EFG H, and let A C, EG be their diameters, and KL, MN their axes, which are the altitudes of the cones or cylinders; and complete the cylinders AL, EO: I fay the bafes are reciprocally propor tional to the altitudes; that is, as the bafe ABCD is to the base EFGH, fo is the altitude M N to the altitude

KL.

RON

AK

AK

B

EM

F

For the altitude K L is either equal to the altitude MN, or not equal to it. First let it be equal, and then the cylinder AL is equal to the cyS linder EO. But cones and cylinders [by 11. 12.] having the fame altitude, are to each other as their bafes: ThereGfore the bafe A B C D is equal to the bafe EF GH, and accordingly they are reciprocally proportional, that is, as the base ABCD is to the bafe EFGH fo is the altitude M N to the altitude KL. Now let the altitude KL not be equal to the altitude MN, but let MN be greater. And from MN take the altitude PM equal to the altitude KL, and cut the cylinder EO thro' P by the plane TV s, parallel to the oppofite planes EFGH, RO of the circles, and conceive the cylinder Es to be made, whose base is the circle EFGH, and altitude PM. Then because [by fuppofition] the cylinder A L is equal to the cylinder EO,

and

and Es is fome other cylinder; it will be [by 7. 5.} as the cylinder AL is to the cylinder Es, fo is the cylinder EO to the cylinder E S. But [by 11. 12.] as the cylinder AL is to the cylinder Es, fo is the bate ABCD to the base EFGH. For the cylinders AX Es have the fame altitude. But as the cylinder E o is to the cylinder Es, fo [by 13. 12.] is the altitude MN to the altitude MP; for the cylinder EO is cut by the plane TV s parallel to the oppofite planes: Therefore as the base ABCD is to the bafe E F G H, fo is the altitude MN to the altitude MP. But the altitude MP is equal to the altitude KL: Wherefore as the base ABCD is to the base EFGH, fo is the altitude MN to the altitude KL. Therefore the bafés of equal cylinders AL, EO are reciprocally proportional to their altitudes.

But now let the bafes of the cylinders A L, E O Be reciprocally proportional to their altitudes, and let the base ABCD be to the base EFG H, as the altitude MN is to the altitude KL: I fay the cylinder AL is equal to the cylinder EO.

For the fame conftruction remaining, becaufe as the bafe ABCD is to the base EFG H, fo is the altitude M N to the altitude KL, and the altitude K L is equal to the altitude MP: The bafe ABCD wil be to the base EFGH, as the altitude M N is to the altitude MP. But [by 11. 12.] as the bafe ABCD is to the bafe EFGH, fo is the cylinder AL to the cylinder Es; for they have the fame altitude. But as the altitude MN is to the altitude MP, fo [by 13. 12.] is the cylinder Eo to the cylinder Es: And therefore as the cylinder Ax is to the cylinder Es, fo is the cylinder EO to the cylinder Es: Therefore [by 9. 5.] the cylinder A L is equal to the cylinder EO, and the like in cones. Which was to be demonftrated.

Two circles about the fame centre being given; to defcribe a polygon in the greater of them, of an equal and even number of fides, which does not touch the leffer circle.

Let A B CD, EFGH be two given circles, having the

fame

fame centre K: It is required to defcribe a polygon of^ an equal and even number of fides in the greater circle ABCD, not touching the leffer circle E F G H.

H

E

KGM

B

L

For draw the right line BD thro' the centre K, and from the point a draw AG at right angles to BD, which continue out to c: Then [by 16. 3.] Ac touches `the circle EFGH; and fo bifecting the femi circumference' BAD, and again bifecting its half, and doing this continually; [by 1. 10.] there will at last remain fome part of the femi-circumference lefs than A D. Let this remainder be LD. And from the point L'draw LM perpendicular to BD, which continue out to N; and join LD, DN. Therefore L D is equal to 'D N And because L N is parallel to AC, and AC touches the circle EF GH; LN will not touch the circle EFGH, and much lefs will the' right lines LD, DN touch the circle EFG H. Therefore if [by 1.4.] right lines be continually applyed round the circumference of the circle, each equal to L D, there will be described in the circle a polygon of an equal and even number of fides, not touching the leffer circle EFG H. Which was to be demonftrated.

F

D

Two fpheres, having the fame centre, being given; to defcribe a folid polyhedron in the greater, not touching the fuperficies of the leffer Sphere".

Conceive two spheres having the fame centre A: It is required to defcribe a folid polyhedron in the greater fphere, not touching the fuperficies of the leffer sphere.

Let the sphere be cut by fome plane paffing thro' the centre: Then will the fections be circles: Becaufe [by 14. def. 11.] the fphere is made by the revolution of a circle about its diameter while the diameter remains at reft; fo that in whatsoever fituation the femicircle is conceived to be, the plane of it being continued, will make a circle in the fuperficies of the fphere: And it is manifeft that the circle is a great circle, because the diameter of the fphere (and that of the femicircle) [by 15. 3.] is greater than all right

2

lines

lines drawn in the circle or sphere. Therefore let BCDE be a circle of the greater fphere, and FGH one of the leffer fphere; and draw the diameters BD, CE of them at right angles to one another: And [by 16. 12] defcribe the polygon B C D E of equal and an even number of fides in the greater circle BCDE, not touching the leffer circle FGH, both having the fame centre; whofe fides BK, KL, LM, ME are in the quadrant BE of the circle; and joining K A, produce it to N; and from the point A raise Ax at right angles to the plane of the circle BCDE,

Dd

meeting

and

Book XII. meeting the fuperficies of the sphere in the point x; draw a plane thro' Ax and BD, and another thro' AX and K N, which, from what has been faid, make great circles in the fuperficies of the fphere. Let fuch be made, and upon their diameters BD, KN let BX D, KXN be their femicircles. Then because the right line x A is perpendicular to the plane of the circle BCDE, all the planes which pafs by xA will be [by 18. 11.] perpendicular to the plane of the circle BCDE: And so the femicircles BXD, KX N are perpendicular to the same plane. And because the femicircles B XD, KX N are equal, for they are upon the equal diameters BD, KN; the quadrants BE, BX KX are equal: Therefore as many fides of a polygon as are in the quadrant BE, fo many fides there will be in the quadrants B X, K x equal to B K, KL, LM, ME. Defcribe them, and let them be BO, OP, PR, RX, KS, ST, Tv, vx; and join so, TP, VR; and from the points o, s, draw perpendiculars to the plane of the circle BCDE: thefe [by 38. 11.] will fall in the common fections BD, KN of the planes; because the planes of the femicircles BX D, KXN [by constr.] are perpendicular to the plane of the circle BCDE. Let thefe perpendiculars be oq, sz, and join oz. Then because in the equal femicircles BXD, KXN there are taken the equal parts of the circumference BO, Ks, and there are drawn the perpendiculars o Q, sz; oq will be equal to sz, and BQ equal to Kz. But the whole B A is equal to the whole KA: Wherefore the remainder QA is equal to the remainder ZA: Therefore as BQ is to QA, fo is Kz to za; and fo [by 2. 6.] qz is parallel to KB: And because oQ, sz are each perpendicular to the plane of the cirele BCDE; [by 6. 11.]0Q will be parallel to s z. But it has been alfo proved to be equal to it. Wherefore [by 33. 1.] zq, so are equal and parallel. And because zo is parallel to so, and also parallel to KB; [by 9. 11.] so will be parallel to KB; and Bo, Ks join them: Therefore the quadrilateral figure K BO's is in one plane'; for [by 7. 11.] if two right lines be parallel, and any points be taken in each of them, the right line joining thofe points is in the fame plane as the parallels. And by the fame reafon the quadrilateral figures SOPT, TPRV are each in one plane. But [by 2. 11.] the triangle VRX is in one plane: If therefore from the