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.
Complete text
Book I