AC has to EG: therefore as the cone of which the base is the circle ABCD, and vertex L, is, to the solid X, so is the pyramid the base of which is the polygon DQATBYCV, and vertex L, to the pyramid the base of which is the polygon HSEOFPGR, and vertex N: but the said cone is greater than the pyramid contained in it, therefore the solid X is greater (14. 5.) than the pyramid, the base of which is the polygon HSEOFPGR, and vertex N; but it is also less; which is impossible; therefore the cone, of which the base is the circle

ABCD, and vertex L, has not to any solid which is less than the cone of which the base is the circle EFGH and vertex N, the triplicate ratio of that which AC has to EG. In the same manner it may be demonstrated that neither has the cone EFGHN to any solid which is less than the cone ABCDL, the triplicate ratio of that which EG has to AC. Nor can the cone ABCDL have to any solid which is greater than the cone EFGHN, the triplicate ratio of that which AC has to EG: for, if it be possible, let it have it to a greater, viz. to the solid Z: therefore, inversely, the solid Z has to the cone ABCDL, the

triplicate ratio of that which EG has to AC: but as the solid Z is to the cone ABCDL, so is the cone EFGHN to some solid, which must be less (14. 5.) than the 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 cone ABCDL has to the cone EFGHN the triplicate ratio of that which AC has to EG: but as the cone is to the cone, so (15. 5.) the cylinder to the 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.

If a cylinder be cut by a plane, parallel to its opposite 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 plane GH, parallel to the opposite planes AB, O CD, meeting the axis EF in the point K, and let the line GH be the common section of the plane GH and the surface of the cy- R linder AD let AEFC be the parallelogram, in any position of it, by the revolution of which about the straight line EF the cylinder AD is described; and let GK be the A common section of the plane GH, and the plane AEFC and because the parallel planes AB, GH are cut by the plane AEKG, AE, KG, their common sections G with it, are parallel (16. 11.): wherefore AK is a parallelogram, and GK equal to C EA the straight line from the centre of the circle AB: for the same reason each of the T straight lines drawn from the point K to the line GH may be proved to be equal to those V

• See Note.

P

N

S

E

B

K

H

F

D

X

Y

M

Q

which are drawn from the centre of the circle AB to its circumference, and are therefore all equal to one another. Therefore the line GH is the circumference of a circle, (15. def. 1.) 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.

L

N

S

E

B

Produce the axis EF both ways; and take any number of straight lines EN, NL, each equal to EK; and any number FX, XM, each equal to FK; and let planes parallel to AB, CD pass through 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, X, M, as was proved of the plane R GH: and these planes cut off the cylinders, PR, RB, DT, TQ: and because the axes LN, NE, EK are all equal, therefore the A cylinders PR, RB, BG are (11. 12.) to one another as their bases: but their bases are equal, and therefore the cylinders PR, RB, BG are equal and because the axes LN, G NE, EK are equal to one another, as also the cylinders PR, RB, BG, and that there are as many axes as cylinders; therefore, whatever multiple the axis KL is of 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 KL 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 PG; and of the axis KF and cylinder GD, any equimultiples whatever, viz. the axis KM and cylinder GQ and it has been demonstrated, if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder GQ; and if equal, equal;

C

K

H

T

D

X

Y

V

M

Q

T

and if less, less: therefore (5. def. 5.) the axis EK is to the axis KF, as the cylinder BG to the cylinder GD. Wherefore, if a cylinder, &c. Q. E. D.

PROP. XIV. THEOR.

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

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

E

K

F

D

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 base is CD, and axis LN: and because the cylinders EB, CM, have the same altitude, they are to one another as their bases: (11. 12.) but their bases are equal: therefore also the cylinders EB, CM are equal. And because the cylinder FM is cut by the plane CD parallel to its opposite planes, as the cylinder CM to the cylinder FD, so is (13. 12.) 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 cylinder FD, so is the axis GH to the axis KL: and as the cylinder EB to the cylinder FD, so is (15. 5.) the cone ABG to the cone CDK, because the cylinders are triple (10. 12.) of the cones: therefore also the axis GH 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.

A H

B

M

N

PROP. XV. THEOR.

THE bases and altitudes of equal cones and cylinders 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.

R N

0

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 cylinders of the same altitude being to one another as their bases, (11. 12.) therefore the base ABCD is equal (A. 5.) to the base EFGH; 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 be unequal, and MN the greater of the two, and from MN take MP equal to KL, and, through the point P, cut the cylinder EO by the plane TYS, parallel to the opposite planes of the circles EFGH,

L

X

Y

S

H

RO; therefore the common A

C

K

section of the plane TYS and

M

the cylinder EO is a circle,

B

F

and consequently ES is a cylinder, the base of which is the circle EFHG, and altitude MP: and because the cylinder AX is equal to the cylinder EO, as AX is to the cylinder ES, so (7. 5.) is the cylinder EO to the same ES. But as the cylinder AX to the cylinder ES, so (11. 12.) is the base ABCD to the base EFGH; for the cylinders AX, ES are of the same altitude; and as the cylinder EO to the cylinder ES, so (13. 12.) is the altitude MN to the altitude MP, because the cylinder EO is cut by the plane

* See Note.