---
book: 12
number: 15
id: "XII.15"
kind: "theorem"
uses: ["[[book-12/proposition-11]]", "[[book-5/proposition-7]]", "[[book-12/proposition-13]]", "[[book-5/proposition-11]]", "[[book-5/proposition-9]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:12.prop.15"
license: "CC-BY-SA-4.0"
---

# XII.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.*

## Proof

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; [[book-12/proposition-11|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*. [[book-5/proposition-7|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; [[book-12/proposition-11|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. [[book-12/proposition-13|XII. 13]]

Therefore also, as the base *ABCD* is to the base *EFGH*, so is the height *MN* to the height *QN*. [[book-5/proposition-11|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; [[book-12/proposition-11|XII. 11]] and, as the height *MN* is to *QN*, so is the cylinder *EP* to the cylinder *ES*; [[book-12/proposition-13|XII. 13]] therefore, as the cylinder *AO* is to the cylinder *ES*, so is the cylinder *EP* to the cylinder *ES*. [[book-5/proposition-11|V. 11]]

Therefore the cylinder *AO* is equal to the cylinder *EP*. [[book-5/proposition-9|V. 9]]

And the same is true for the cones also. Q. E. D.