PROPOSITION 15.
In equal cones and cylinders the bases are reciprocally proportional to the heights; and those cones and cylinders in which the bases are reciprocally proportional to the heights are equal.
Let there be equal cones and cylinders of which the circles
ABCD,
EFGH are the bases; let
AC,
EG be the diameters of the bases, and
KL,
MN the axes, which are also the heights of the cones or cylinders; let the cylinders
AO,
EP be completed.
I say that in the cylinders
AO,
EP the bases are reciprocally proportional to the heights, that is, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
KL.
For the height
LK is either equal to the height
MN or not equal.
First, let it be equal.
Now the cylinder
AO is also equal to the cylinder
EP.
But cones and cylinders which are of the same height are to one another as their bases; [
XII. 11] therefore the base
ABCD is also equal to the base
EFGH.
Hence also, reciprocally, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
KL.
Next, let the height
LK not be equal to
MN, but let
MN be greater; from the height
MN let
QN be cut off equal to
KL, through the point
Q let the cylinder
EP be cut by the plane
TUS parallel to the planes of the circles
EFGH,
RP, and let the cylinder
ES be conceived erected from the circle
EFGH as base and with height
NQ.
Now, since the cylinder
AO is equal to the cylinder
EP, therefore, as the cylinder
AO is to the cylinder
ES, so is the cylinder
EP to the cylinder
ES. [
V. 7]
But, as the cylinder
AO is to the cylinder
ES, so is the base
ABCD to the base
EFGH, for the cylinders
AO,
ES are of the same height; [
XII. 11] and, as the cylinder
EP is to the cylinder
ES, so is the height
MN to the height
QN, for the cylinder
EP has been cut by a plane which is parallel to its opposite planes. [
XII. 13]
Therefore also, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
QN. [
V. 11]
But the height
QN is equal to the height
KL; therefore, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
KL.
Therefore in the cylinders
AO,
EP the bases are reciprocally proportional to the heights.
Next, in the cylinders
AO,
EP let the bases be reciprocally proportional to the heights, that is, as the base
ABCD is to the base
EFGH, so let the height
MN be to the height
KL; I say that the cylinder
AO is equal to the cylinder
EP.
For, with the same construction, since, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
KL, while the height
KL is equal to the height
QN, therefore, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
QN
But, as the base
ABCD is to the base
EFGH, so is the cylinder
AO to the cylinder
ES, for they are of the same height; [
XII. 11] and, as the height
MN is to
QN, so is the cylinder
EP to the cylinder
ES; [
XII. 13] therefore, as the cylinder
AO is to the cylinder
ES, so is the cylinder
EP to the cylinder
ES. [
V. 11]
Therefore the cylinder
AO is equal to the cylinder
EP. [
V. 9]
And the same is true for the cones also. Q. E. D.