cone ABCDL, the triplicate ratio of that which EG has to Book XII. AC: But as the solid Z is to the cone ABCDI, so is the cone EFGHN to some solid, which must be lessa than the 14. 5. cone ABCDL, because the solid Z is greater than the cone EFGHN: Therefore the cone EFGHN has to a solid which is less than the cone ABCDL the triplicate ratio of that which EG has to AC, which was demonstrated to be impossible; therefore the cone ABCDL has not to any solid greater than the cone EFGHN, the triplicate ratio of that which AC has to EG; and it was demonstrated, that it could not have that ratio to any solid less than the cone EFGHN: Therefore the coné ABCDL has to the cone EFGHN, the triplicate ratio of that which AC has to EG: But as the cone is to the cone, sob the cylinder to the u 15.5. cylinder; for every cone is the third part of the cylinder upon the same base, and of the same altitude: Therefore also the cylinder has to the cylinder, the triplicate ratio of that which AC has to EG: Wherefore similar cones, &c. Q. E. D.

PROP. XIII. THEOR. Ir a cylinder be cut by a plane parallel to its oppo-See N. site planes, or bases, it divides the cylinder into two cylinders, one of which is to the other as the axis of the first to the axis of the other. Let the cylinder AD be cut by the

0

P plane GH parallel to the opposite planes AB, CD, meeting the axis EF in the point K, and let the line GH be thre common section of the plane GH RK

N and the surface of the cylinder AD: Let AEFC be the parallelogram in any position of it, by the revolution of which about the straight line EF the A E B cylinder AD is described: and let GK be the common section of the plane GH, and the plane AEFC: And be

GEK HH cause the parallel planes AB, GH, are cut by the plane AEKG; AE, KG, their common sections with it, are pa-.

D rallela ; wherefore AK is a parallelo

T gram, and GK equal to EA the straight

区 line from the centre of the circle AB: For the same reason, each of the V M

116. 11.

İN

Boox XII. straight lines drawn from the point K to the line GH may

be proved to be equal to those which are drawn from the centre of the circle AB to its circumference, and are there

fore all equal to one another. Therefore the line GH is * 15 Def. 1. the circumference of a circlea; of which the centre is the

point K: Therefore the plane GH divides the cylinder AD into the cylinders AH, GD; for they are the same which would be described by the revolution of the parallelograms AK, GF, about the straight lines EK, KF: And it is to be shown, that the cylinder AH is to the cylinder HC, as the axis EK to the axis KF.

Produce the axis EF both ways: and take any number of straight lines EN, NL, each equal to EK; and any numberFX, XM, cach equal to FK; and let planes parallel to AB, CD, pass through

P the points L, N, X, M: Therefore the common sections of these planes with the cylinder produced are circles the centres of which are the points, L, N,

R

S
X, M, as was proved of the plane GH;
and these planes cut off the cylinders
PR, RB, DT, TQ: And because the

ΑΙ

HE

E B axes LN, NE, EK, are all equal ;

therefore the cylinders PR, RB, BG, 11. 1%. are to one another as their bases; but

their bases are equal, and therefore the G K ЯН
cylinders PR, RB, BG, are equal: And
because the axes LN, NE, EK, are C E

D
equal to one another, as also the cylin-
ders PR, RB, BG, and that there are T X Y
as many axes as cylinders; therefore,
whatever multiple the axis KL is of v

Q

M the axis KE, the same multiple is the cylinder, PG of the cylinder GB: For the same reason, whatever multiple the axis MK is of the axis KF, the same multiple is the cylinder QG of the cylinder GD: And if the axis KC be equal to the axis KM, the cylinder PG is equal to the cylinder GQ; and if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder QG; and if less, less: since therefore there are four magnitudes, viz. the axes EK, KF, and the cylinders BG, GD, and that of the axis EK and cylinder BG there has been taken any equimultiples whatever, viz. the axis KL and cylinder VG; and of the axis KF and cylinder CD, any equimultiples whatever, viz, the axis KM and cylinder GQ; and it has been demonstrated, if the axis KL Boox XII. be greater than the axis KM, the cylinder PG is greater than the cylinder GQ; and if equal, equal; and if less, less : Therefored the axis EK is to the axis KF, as the cylinder a 5 Def. 5. BG to the cylinder GD. Wherefore, if a cylinder, &c. Q.E.D.

PROP. XIV. THEOR.

a

Cones and cylinders upon equal bases are to one another as their altitudes.

Let the cylinders EB, FD, be upon equal bases AB, CD: As the cylinder EB to the cylinder FD, so is the axis GH to the axis KL.

Produce the axis KL to the point N, and make LN equal to the axis GH, and let CM be a cylinder of which the bay is CD, and axis LN; and because the cylinders EB, CM, have the same altitude, they are to one another as their basesa : But their bases are equal, therefore also the cylin- * 11. 12. ders EB, CM, are equal :

K
And because the cylinder
FM is cut by the plane CD
parallel to its opposite planes,
as the cylinder CM to the

G
с

1 13. 12. cylinder FD, so is b the axis LN to the axis KL: But the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH: Therefore as the cylinder EB to the cylin- 4

B

M der FD, so is the axis GH to the axis Kl: And as the cylinder EB to the cylinder FD, so is the cone ABG to the cone CDK, because the cylin-2 15.5. ders are tripled of the cones: Therefore also the axis GH a ro. 12. is to the axis KL, as the cone ABG to the cone CDK, and the cylinder EB to the cylinder FD. Wherefore cones, &c. Q. E. D.

d

Book XII.

PROP. XV. THEOR.

a

a

See n. The bases and altitudes of equal cones and cylin

ders are reciprocally proportional; and if the bases and altitudes be reciprocally proportional, the cones and cylinders are equal to one another.

Let the circles ABCD, EFGH, the diameters of which are AC, EG, be the bases, and KL, MN, the axes, as also the altitudes, of equal cones and cylinders; and let ALC, ENG be the cones, and AX, EO the cylinders ; The bases and altitudes of the cylinders AX, EO are reciprocally proportional: that is, as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL.

Either the altitude MN is equal to the altitude KL, or these altitudes are not equal. First let them be equal; and the cylinders AX, EO being also equal, and cones and cy

linders of the same altitude being to one another as their * 11. 12. bases a, therefore the base ABCD is equalb to the base A. 5. EFGÁ; and as the base ABCD is to the base EFGH, so

is the altitude MN to the altitude KL. But let the altitudes
KL, MN, he une-
qual, and MN the

R
greater of the two,
and from MN take
MP equal to KL,

L
and through the

X

S point P cut the cylinder EO by the plane TYS, parallel to the op- A posite planes of

NI the circles FFGH, RO; therefore the common section of the plane TYS and the cylinder EO is a circle, and consequently ES is a cylinder, the base of which is the circle EFGH, and altitude

MP: And because the cylinder AX is equal to the cylinder 7.5. EO, as AX is to the cylinder ES, so is the cylinder EO to

the same ES: But as the cylinder AX to the cylinder ES, soa is the base ABCD to the base EFGH: for the cylin

ders AX, ES are of the same altitude; and as the cylinder * 13. 12. EO to the cylinder ES, sod is the altitude MN to the alti

tude MP, because the cylinder EO is cut by the plane TYS

с

1

parallel to its opposite planes. Therefore as the base ABCD Book XII. to the base EFGH, so is the altitude MN to the altitude MP: But MP is equal to the altitude KL; wherefore as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL; that is, the bases and altitudes of the equal cylinders AX, EO, are reciprocally proportional.

But let the bases 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 base EFGH; then because as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL; MN is cqualb to KL, "A. 5. and therefore the cylinder AX is equal a to the cylinder EO. « 11. 12.

But let the bases ABCD, EFGH be unequal, and let ABCD be the greater; and because as ABCD is to the base EFGH, so is the altitude MN to the altitude KL, therefore MN is greater than KL. Then, the same construction being made as before, because as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL; and because the altitude KL is equal to the altitude MP; therefore the base ABCD isa to the base EFGH as the cylinder AX to the cylinder ES, and as the altitude MN to the altitude MP or KL, so is the cylinder EO to the cylinder ES: Therefore the cylinder AX is to the cylinder ES, as the cylinder EO is to the same ES: Whence the cylinder AX is equal to the cylinder E0; and the same reasoning bolds in cones. Q. É, D.

PROP. XVI. PROB.

To describe in the greater of two circles that have the same centre, a polygon of an even number of equal sides, that shall not meet the lesser circle.

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

Through the centre K draw the straight line BD, and from the point G, where it meets the circumference of the lesser circle, draw GA at right angles to BD, and produce it to