Tetrahedra are Equal iff Bases are Reciprocally Proportional to Heights
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 proof 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