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14.5.

* 15.5.

† 10. 12.

polygon HSEOFPGR, and vertex N: but it is also less; which is impossible: therefore the cone, of which

to any

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*

the base is the circle ABCD and vertex L, has not to
solid which is less than the cone of which the base
any
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
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, in-
versely, the solid Z has to the cone ABCDL, the tri-
plicate 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 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 impos-
sible: 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 *
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 See N. 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.

R

A

G

B

H

C

T

Y

Let the cylinder AD be cut by the plane GH parallel to the opposite planes AB, CĎ, 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 cylinder 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 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 with it, are parallel*: wherefore AK *16. 11. is a parallelogram, and GK equal to EA the straight line from the centre of the circle AB: for the same reason, each of the 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 therefore all equal to one another; therefore the line GH is the circumference of a circle* of which the centre is the point K: there- 15 Def. 1. fore 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 shewn, 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 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 GH; and these planes cut off the cylinders PR,

*11. 12.

RB, DT, TQ. And because the axes
LN, NE, EK, are all equal, therefore
the cylinders, PR, RB, BG, are* to
one another as their bases: but their
bases are equal, and therefore the cy-
linders PR, RB, BG, are equal: and
because the axes LN, 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,

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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 GQ; and if less, less: therefore, since there are four magnitudes, viz. the axes EK, KF, and the cylinders BG, GD; and that of the axis EK and cylinder BG there have 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 since it has been demonstrated, that if the axis KL be greater than the axis KM, the cylinder FG is greater than the cylinder GQ; and if equal, equal; and if less, *5 Def. 5. less: therefore* as the axis EK is to the axis KF, so is the cylinder BG to the cylinder GD. Wherefore, if a cylinder, &c. Q. E. D.

See N.

* 11. 12.

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 shall the axis GH be 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 base is CD, and axis LN. Then because the cylinders EB, CM, have the same altitude, they are to one another as their bases*: 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 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

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cylinder FD, so is the axis GH to the axis KL: and as the cylinder EB to the cylinder FD, so is the 15.5. cone ABG to the cone CDK, because the cylinders are triple * of the cones: therefore also the axis GH is * 10. 12. to the axis KL, as the cone ABG to the cone CDK, and as the cylinder EB to the cylinder FD. Wherefore cones, &c. Q. E. D.

PROP. XV. THEOR.

The bases and altitudes of equal cones and cylinders are See N. reciprocally proportional; and if the bases and altitudes be reciprocally proportional, the cones and cy linders 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 shall be reciprocally proportional; that is, as the base ABCD to the base EFGH so shall the altitude MN be 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 cylinders of the same altitude being to one another as their bases*, therefore the base ABCD is *11.12. equal to the base EFGH; and as the base ABCD is * A. 5. 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 cireles EFGH, RO; there

R

E

* 7.5.

* 11. 12.

* 13. 12.

* A. 5.

*11. 12.

* A. 5.

*11. 12:

:

cy

*

fore the common section of the plane TYS and the
linder EO is a circle, and consequently ES is a cylin-
der, the base of which is the circle EFGH, and alti-
tude MP and because the cylinder AX is equal to the
cylinder 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, so is the base ABCD to the base
EFGH; for the cylinders AX, ES are of the same alti-
tude; and as the cylinder EO to the cylinder ES, so
is the altitude MN to the altitude MP, because the
cylinder EO is cut by the plane TYS parallel to its
opposite planes; therefore as the base ABCD 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 propor-
tional.

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 shall be 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 equal to KL; and therefore the cylinder AX is equal to the cylinder EO.

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*

*

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*

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 is 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 EO: and the same reasoning holds in cones.

Q. E. D.

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