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

## Theorem

In the words of Euclid:

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 cones and cylinders which are similar.

Let the circles $\map c {ABCD}$ and $\map c {EFGH}$ 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 $\map c {ABCDL}$ and the cylinder $AO$
let $MN$ be the height of the cone $\map c {EFGHM}$ and the cylinder $EP$.

It is to be shown that:

$\map c {ABCD} : \map c {EFGH} = MN : KL$

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 $AO = EP$.

$\map c {ABCD} = \map c {EFGH}$

Thus:

$\map c {ABCD} : \map c {EFGH} = MN : KL$

Without loss of generality, suppose $LK \ne MN$.

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 $\map c {EFGH}$ and $\map c {RP}$.

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

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

$AO : ES = EP : ES$
$AO : ES = \map c {ABCD} : \map c {EFGH}$
$EP : ES = MN : QN$
$\map c {ABCD} : \map c {EFGH} = MN : QN$

But:

$QN = KL$

Therefore:

$\map c {ABCD} : \map c {EFGH} = MN : KL$

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

$\Box$

Let the bases of the cylinders $AO$ and $EP$ be reciprocally proportional to their heights:

$\map c {ABCD} : \map c {EFGH} = MN : KL$

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 $\map c {EFGH}$ and $\map c {RP}$.

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

We have that $KL = QN$.

Thus:

$\map c {ABCD} : \map c {EFGH} = MN : QN$
$\map c {ABCD} : \map c {EFGH} = AO : ES$
$MN : QN = EP : ES$
$AO : ES = EP : ES$
$AO = EP$

$\blacksquare$

## Historical Note

This proof is Proposition $15$ of Book $\text{XII}$ of Euclid's The Elements.