TYS parallel to its oppofite planes. Therefore as the bafe Book XII. ABCD to the bafe EFGH, fo is the altitude MN to the altitude MP: But MP is equal to the altitude KL; wherefore as the bafe ABCD to the bafe EFGH, fo is the altitude MN to the altitude KL; that is, the bafes and altitudes of the equal cylinders AX, EO are reciprocally proportional.

But let the bafes and altitudes of the cylinders AX, EO, be reciprocally proportional, viz. the base ABCD to the base EFGH, as the altitude MN to the altitude KL: The cylinder AX is equal to the cylinder EO.

First, let the base ABCD be equal to the bafe EFGH; then, because as the base ABCD is to the bafe EFGH, fo is the altitude MN to the altitude KL; MN is equal b to KL, and b A. s. therefore the cylinder AX is equal a to the cylinder EO.

But let the bafes ABCD, EFGH be unequal, and let ABCD be the greater; and because, as ABCD is to the base EFGH, fo is the altitude MN to the altitude KL; therefore MN is greater b than KL. Then, the fame construction being made as before, because as the base ABCD to the bafe EFGH, fo is the altitude MN to the altitude KL; and because the altitude KL is equal to the altitude MP; therefore the base ABCD is a to the base EFGH, as the cylinder AX to the cylinder ES; and as the altitude MN to the altitude MP or KL, fo is the cylinder EO to the cylinder ES: Therefore the cylinder AX is to the cylinder ES, as the cylinder EO is to the fame ES: Whence the cylinder AX is equal to the cylinder EO: And the fame reafoning holds in cones. Q. E. D.

Τ TO

PROP. XVI. PRO B.

O describe in the greater of two circles that have the fame centre, a polygon of an even number of equal fides, that shall not meet the leffer circle.

Let ABCD, EFGH be two given circles having the fame centre K: It is required to infcribe in the greater circle ABCD a polygon of an even number of equal fides, that shall not meet the leffer circle.

Through the centre K draw the ftraight line BD, and from the point G, where it meets the circumference of the leffer

circle

a 11. 12.

a 16. 3.

B

Book XII. circle, draw GA at right angles to BD, and produce it to Cr therefore AC touches a the circle EFGH: Then, if the circumference BAD be bifected, and the half of it be again bisected, b Lemma. and so on, there must at length remain a circumference less than AD: Let this be LD; and from the point L draw LM perpendicular to BD, and produce it to N; and join LD, DN. Therefore LD is equal to DN; and because LN is parallel to AC, and that AC touches the circle EFGH; therefore LN does not meet the circle EFGH; And much less fhall the ftraight lines LD, DN meet the circle EFGH: fo that if ftraight lines equal to LD be applied in the circle ABCD from the point L around to N, there fhall be defcribed in the circle a polygon of an even number of equal fides not meeting the leffer circle. Which was to be done.

LEMMA

H

E

KGMD

F

N

C

II.

F two trapeziums ABCD, EFGH be infcribed in the circles, the centres of which are the points K, L; and if the fides AB, DC be parallel, as alfo EF', HG; and the other four fides AD, BC, EH, FG be all equal to one another; but the fide AB greater than EF, and DC greater than HG. The straight line KA from the centre of the circle in which the greater fides are, is greater than the ftraight line LE drawn from the centre to the circumference of the other circle.

If it be poffible, let KA be not greater than LE; then KA 'must be either equal to it, or lefs. Firft, let KA be equal to LE: Therefore, becaufe in two equal circles, AD, BC in the one are equal to EH, FG in the other, the circumferences a 28. 3. AD, BC are equal a to the circumferences EH, FG; but becaufe the faight lines AB, DC are refpectively greater than EF, GH, the circumferences AB, DC are greater than EF, HG: Therefore the whole circumference ABCD is greater than the whole EFGH; but it is alfo equal to it, which is

impoffible a

impoffible: Therefore the ftraight line KA is not equal to Book XII. CLE.

But let KA be lefs than LE, and make LM equal to KA, and from the centre L, and diftance LM defcribe the circle MNOP, meeting the ftraight lines LE, LF, LG, LH, in M, N, O, P; and join MN, NO, OP, PM which are respectively parallel a to, and lefs than EF, FG, GH, HE: Then, becaufe EH is greater than MP, AD is greater than MP; and

22.6.

the circles ABCD, MNOP are equal; therefore the circumference AD is greater than MP; for the fame reafon, the cir cumference BC is greater than NO; and because the straight line AB is greater than EF which is greater than MN, much more is AB greater than MN: Therefore the circumference AB is greater than MN; and for the fame realon, the circumference DC is greater than PO: Therefore the whole circumference ABCD is greater than the whole MNOP; but it is like wife equal to it, which is impoffible: Therefore KA is not lefs than LE; nor is it equal to it; the ftraight line KA muft therefore be greater than LE. Q. E. D.

COR. And if there be an ifofceles triangle the fides of which are equal to AD, BC, but its bafe lefs than AB the greater of the two fides AB, DC; the straight line KA may, in the fame manner, be demonftrated to be greater than the straight line drawn from the centre to the circumference of the circle de fcribed about the triangle.

PROP.

Book XII.

PROP. XVII. PROB.

See N. TO defcribe in the greater of two fpheres which have the fame centre, a folid polyhedron, the fuperficies of which shall not meet the leffer fphere.

a 15. 3.

b 16. 12.

Let there be two spheres about the fame centre A^; it is required to defcribe in the greater a folid polyhedron, the fuperficies of which shall not meet the leffer sphere.

Let the fpheres be cut by a plane paffing through the centre; the common sections of it with the spheres fhall be circles; because the sphere is defcribed by the revolution of a femicircle about the diameter remaining unmoveable: fo that in whatever pofition the femicircle be conceived, the common fection of the plane in which it is with the fuperficies of the sphere is the circumference of a circle; and this is a great circle of the fphere, because the diameter of the sphere, which is likewise the diameter of the circle, is greater than any straight line in the circle or sphere: Let then the circle made by the section of the plane with the greater sphere be BCDE, and with the leffer sphere be FGH; and draw the two diameters BD, CE at right angles to one another; And in BCDE, the greater of the two circles, describe b a polygon of an even number of equal fides not meeting the leffer circle FGH; and let its fides, in BE, the fourth part of the circle, be BK, KL, LM, ME ; join KA and produce it to N; and from A draw AX at right angles to the plane of the circle BCDE meeting the fuperficies of the fphere in the point X; and let planes pafs through AX and each of the ftraight lines BD, KN, which, from what has been faid, fhall produce great circles on the fuperficies of the sphere, and let BXD, KXN be the femicircles thus made upon the diameters BD, KN: Therefore, because XA is at right angles to the plane of the circle BCDE, every plane which 18. 11. paffes through XA is at right angles to the plane of the circle BCDE; wherefore the femicircles BXD, KXN are at right angles to that plane: And because the femicircles BED, BXD, KXN, upon the equal diameters BD, KN, are equal to one another, their halves BE, BX, KX, are equal to one another: Therefore, as many fides of the polygon as are in BE, fo many there are in BX, KX equal to the fides BK, KL, LM, ME; Let thefe polygons be defcribed, and their fides be BO, OP, PR, RX; KS, ST, TY, YX, and join

C

OS

OS, PT, RY; and from the points O, S draw OV, SQ perpen. Book XIL diculars to AB, AK: And because the plane BOXD is at right angles to the plane BCDE, and in one of them BOXD, ÖV is drawn perpendicular to AB the common fection of the planes, therefore OV is perpendicular a to the plane BCDE: For the 4. def. fame reafon SQ is perpendicular to the fame plane, because the plane KSXN is at right angles to the plane BCDE. Join VQ, and because in the equal femicircles BXD, KXN the

C

R

circumferences BO, KS are equal, and OV, SQ are perpendicular to their diameters, therefore d OV is equal to SQ,d 16. 1. and BV equal to KQ But the whole BA is equal to the whole KA, therefore the remainder VA is equal to the remainder Q4: As therefore BV is to VA, fo is KQ to QA, wherefore VO is parallel to BK: And becaufe OV, SQ are each ofe 2. 6. them at right angles to the plane of the circle BCDE, OV is parallel f to SQ; and it has been proved that it is alfo equalf 6. 11. to it; therefore QV, SO are equal and parallel g: And because 33. I. QV is parallel to SO, and alfo to KB; OS is parallel h to BK ; h 9. 11. and therefore BO, KS which join them are in the fame plane

in