# Tetrahedra are Equal iff Bases are Reciprocally Proportional to Heights

## Contents

## Theorem

In the words of Euclid:

*In equal pyramids which have triangular bases the bases reciprocally proportional to the heights; and those pyramids in which the bases are reciprocally proportional to the heights are equal.*

(*The Elements*: Book $\text{XII}$: Proposition $9$)

## Proof

Let $ABCG$ be a tetrahedron whose base is $\triangle ABC$ and whose apex is $G$.

Let $DEFH$ be a tetrahedron whose base is $\triangle DEF$ and whose apex is $H$.

Let $ABCG$ and $DEFH$ be equal in volume.

It is to be demonstrated that the areas of their bases is reciprocally proportional to their heights.

Let the parallelepipeds $BGML$ and $EHQP$ be completed.

We have that $ABCG$ is equal to $DEFH$.

From:

and

it follows that:

- $BGML = 6 \cdot ABCG$

and:

- $EHQP = 6 \cdot DEFH$

Therefore:

- $BGML = EHQP$

- $BM : EQ = h \left({EHQP}\right) : h \left({BGML}\right)$

where:

- $BM$ and $EQ$ are the bases of $BGML$ and $EHQP$ respectively
- $h \left({EHQP}\right)$ denotes the height of $EHQP$
- $h \left({BGML}\right)$ denotes the height of $BGML$.

But from Proposition $34$ of Book $\text{I} $: Opposite Sides and Angles of Parallelogram are Equal:

- $BM : EQ = \triangle ABC : \triangle DEF$

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

- $\triangle ABC : \triangle DEF = h \left({EHQP}\right) : h \left({BGML}\right)$

But:

- $h \left({EHQP}\right) = h \left({DEFH}\right)$
- $h \left({BGML}\right) = h \left({ABCG}\right)$

where:

- $h \left({ABCG}\right)$ denotes the height of $ABCG$
- $h \left({DEFH}\right)$ denotes the height of $DEFH$.

Therefore:

- $\triangle ABC : \triangle DEF = h \left({DEFH}\right) : h \left({ABCG}\right)$

That is the bases of $ABCG$ and $DEFH$ are reciprocally proportional to their heights.

$\Box$

Let $ABCG$ and $DEFH$ be tetrahedra whose bases are reciprocally proportional to their heights:

- $\triangle ABC : \triangle DEF = h \left({DEFH}\right) : h \left({ABCG}\right)$

It is to be shown that $ABCG$ is equal to $DEFH$.

Let the parallelepipeds $BGML$ and $EHQP$ be completed.

We have that:

- $\triangle ABC : \triangle DEF = h \left({DEFH}\right) : h \left({ABCG}\right)$

while:

- $\triangle ABC : \triangle DEF = BM : EQ$

where $BM$ and $EQ$ are the bases of $BGML$ and $EHQP$ respectively.

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

- $BM : EQ = h \left({EHQP}\right) : h \left({BGML}\right)$

- $BGML = EHQP$

But from:

and

it follows that:

- $BGML = 6 \cdot ABCG$

and:

- $EHQP = 6 \cdot DEFH$

Therefore:

- $ABCG = DEFH$

$\blacksquare$

## Historical Note

This theorem is Proposition $9$ 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