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