Book 12
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Since then the cylinders EB, CM are of the same height, they are to one another as their bases. [XII. 11]
14
But the bases are equal to one another; therefore the cylinders EB, CM are also equal.
14
And, since the cylinder FM has been cut by the plane CD which is parallel to its opposite planes, therefore, as the cylinder CM is to the cylinder FD, so is the axis LN to the axis KL. [XII. 13]
14
But the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; therefore, as the cylinder EB is to the cylinder FD, so is the axis GH to the axis KL.
14
But, as the cylinder EB is to the cylinder FD, so is the cone ABG to the cone CDK. [XII. 10]
14
Therefore also, as the axis GH is to the axis KL, so is the cone ABG to the cone CDK and the cylinder EB to the cylinder FD. Q. E. D.
PROPOSITION 15.
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.
15
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.
15
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.
15
For the height LK is either equal to the height MN or not equal.
15
First, let it be equal.
15
Now the cylinder AO is also equal to the cylinder EP.
15
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.
15
Hence also, reciprocally, as the base ABCD is to the base EFGH, so is the height MN to the height KL.
