Tetrahedra are Equal iff Bases are Reciprocally Proportional to Heights

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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

Euclid-XII-9.png

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:

Proposition $28$ of Book $\text{XI} $: Parallelepiped cut by Plane through Diagonals of Opposite Planes is Bisected

and

Proposition $7$ of Book $\text{XII} $: Prism on Triangular Base divided into Three Equal Tetrahedra

it follows that:

$BGML = 6 \cdot ABCG$

and:

$EHQP = 6 \cdot DEFH$

Therefore:

$BGML = EHQP$

But from Proposition $34$ of Book $\text{XI} $: Parallelepipeds are of Equal Volume iff Bases are in Reciprocal Proportion to Heights:

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

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:

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

From Proposition $34$ of Book $\text{XI} $: Parallelepipeds are of Equal Volume iff Bases are in Reciprocal Proportion to Heights:

$BGML = EHQP$

But from:

Proposition $28$ of Book $\text{XI} $: Parallelepiped cut by Plane through Diagonals of Opposite Planes is Bisected

and

Proposition $7$ of Book $\text{XII} $: Prism on Triangular Base divided into Three Equal Tetrahedra

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