Cones or Cylinders are Equal iff Bases are Reciprocally Proportional to Heights

Proof

 * Euclid-XII-15.png

Let there be cones and cylinders which are similar.

Let the circles $c \left({ABCD}\right)$ and $c \left({EFGH}\right)$ be their bases.

Let $KL$ and $MN$ be the axes of the cones and cylinders.

Let $L$ and $N$ be the apices of the cones.

Thus:
 * let $KL$ be the height of the cone $c \left({ABCDL}\right)$ and the cylinder $AO$
 * let $MN$ be the height of the cone $c \left({EFGHM}\right)$ and the cylinder $EP$.

It is to be shown that:
 * $c \left({ABCD}\right) : c \left({EFGH}\right) = MN : KL$

That is, the bases of the cones and cylinders are reciprocally proportional to their heights.

Either $LK = MN$ or $LK \ne MN$.

First suppose $LK = MN$.

We have that $AO = EP$.

From :
 * $c \left({ABCD}\right) = c \left({EFGH}\right)$

Thus:
 * $c \left({ABCD}\right) : c \left({EFGH}\right) = MN : KL$

Suppose WLOG that $LK \ne MN$.

Suppose $MN > LK$.

Let $QN$ be cut off from $MN$ equal to $KL$.

Let the cylinder $EP$ be cut by the plane $TUS$ through $Q$ parallel to the planes holding the circles $c \left({EFGH}\right)$ and $c \left({RP}\right)$.

Let the cylinder $ES$ be described with the circle $c \left({EFGH}\right)$ as its base and with height $NQ$.

We have that the cylinder $AO$ equals the cylinder $EP$.

Therefore from :
 * $AO : ES = EP : ES$

But from :
 * $AO : ES = c \left({ABCD}\right) : c \left({EFGH}\right)$

and from :
 * $EP : ES = MN : QN$

Therefore from :
 * $c \left({ABCD}\right) : c \left({EFGH}\right) = MN : QN$

But:
 * $QN = KL$

Therefore:
 * $c \left({ABCD}\right) : c \left({EFGH}\right) = MN : KL$

Therefore in the cylinders $AO$ and $EP$, the bases are reciprocally proportional to their heights.

Let the bases of the cylinders $AO$ and $EP$ be reciprocally proportional to their heights:
 * $c \left({ABCD}\right) : c \left({EFGH}\right) = MN : KL$

It is to be proved that cylinders $AO$ and $EP$ are equal.

Let $QN$ be cut off from $MN$ equal to $KL$.

Let the cylinder $EP$ be cut by the plane $TUS$ through $Q$ parallel to the planes holding the circles $c \left({EFGH}\right)$ and $c \left({RP}\right)$.

Let the cylinder $ES$ be described with the circle $c \left({EFGH}\right)$ as its base and with height $NQ$.

We have that $KL = QN$.

Thus:
 * $c \left({ABCD}\right) : c \left({EFGH}\right) = MN : QN$

But from :
 * $c \left({ABCD}\right) : c \left({EFGH}\right) = AO : ES$

and from :
 * $MN : QN = EP : ES$

Therefore from :
 * $AO : ES = EP : ES$

Therefore from :
 * $AO = EP$