Kitap 12
15
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.
15
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]
15
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]
15
Therefore also, as the base ABCD is to the base EFGH, so is the height MN to the height QN. [V. 11]
15
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.
15
Therefore in the cylinders AO, EP the bases are reciprocally proportional to the heights.
15
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.
15
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
15
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]
15
Therefore the cylinder AO is equal to the cylinder EP. [V. 9]
