# Sizes of Tetrahedra of Same Height are as Bases

## Theorem

In the words of Euclid:

*Pyramids which are of the same height and have triangular bases are to one another as the bases.*

(*The Elements*: Book $\text{XII}$: Proposition $5$)

## Proof

Let there be two tetrahedra of the same height whose bases are $ABC, DEF$ and whose apices are $G$ and $H$.

It is to be demonstrated that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of tetrahedron $ABCG$ to tetrahedron $DEFH$.

Suppose the ratio of $ABCG$ to $DEFH$ is not equal to the ratio of $\triangle ABC$ to $\triangle DEF$.

Then the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure $W$ which is either smaller than $DEFH$ or larger than $DEFH$.

First suppose that $W$ is smaller than $DEFH$.

- let $DEFH$ be divided into two equal tetrahedra which are similar to $DEFH$ and two equal prisms.

Then the two prisms are greater than half of $DEFH$.

Again, let those tetrahedra be similarly divided.

Proposition $1$ of Book $\text{X} $: Existence of Fraction of Number Smaller than Given:

- Let this be done continually until there are left over from $DEFH$ some tetrahedra which are less than the excess by which $DEFH$ is greater than $W$.

Let these, for the sake of argument, be $DQRS$ and $STUH$.

Therefore their remainders, the prisms in $DEFH$, are greater than $W$.

Let $ABCG$ be divided similarly and the same number of times as $DEFH$.

- the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of the prisms in $ABCG$ to the prisms in $DEFH$.

But:

- $\triangle ABC : \triangle DEF = ABCG : W$

So from Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:

So from Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:

But $ABCG$ is greater than the prisms in it.

Therefore $W > DEFG$.

But by hypothesis $W < DEFG$.

Therefore it is not the case that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure smaller than $DEFH$.

Now suppose that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure $W$ greater than $DEFH$.

Thus:

- $\triangle DEF : \triangle ABC = W : ABCG$

But using the same technique as Lemma to Proposition $2$ of Book $\text{XII} $: Areas of Circles are as Squares on Diameters:

- $W : ABCG = DEFH : X$

where $X$ is some solid figure less than $ABCG$.

Therefore from Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:

- $\triangle DEF : \triangle ABC = DEFG : X$

But this has been demonstrated to be absurd.

Therefore it is not the case that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure greater than $DEFH$.

Hence the result.

$\blacksquare$

## Historical Note

This proof is Proposition $5$ of Book $\text{XII}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{XII}$. Propositions