Proof: By Euclid

For let $EB$ and $FD$ be cylinders on equal bases, (namely) the circles $AB$ and $CD$ (respectively).
I say that as cylinder $EB$ is to cylinder $FD$, so $GH$ (is) to $KL$.

fig14e

For let the $KL$ have been produced to point $N$.
And let $LN$ be made equal to $GH$.
And let the cylinder $CM$ have been conceived about $LN$.
Therefore, since cylinders $EB$ and $CM$ have the same height they are to one another as their bases [Prop. 12.11].
And the bases are equal to one another.
Thus, cylinders $EB$ and $CM$ are also equal to one another.
And since cylinder $FM$ has been cut by the plane $CD$, which is parallel to its opposite planes, thus as cylinder $CM$ is to cylinder $FD$, so $LN$ (is) to $KL$ [Prop. 12.13].
And cylinder $CM$ is equal to cylinder $EB$, and $LN$ to $GH$.
Thus, as cylinder $EB$ is to cylinder $FD$, so $GH$ (is) to $KL$.
And as cylinder $EB$ (is) to cylinder $FD$, so cone $ABG$ (is) to cone $CDK$ [Prop. 12.10].
Thus, also, as $GH$ (is) to $KL$, so cone $ABG$ (is) to cone $CDK$, and cylinder $EB$ to cylinder $FD$.
(Which is) the very thing it was required to show.
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