Sizes of Tetrahedra of Same Height are as Bases

Proof

 * Euclid-XII-5.png

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 $DEFH$.

From :
 * 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.


 * 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 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 $DEFH$.

From :
 * 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 :
 * the ratio of $ABCG$ to $W$ equals the ratio of the prisms in $ABCG$ to the prisms in $DEFH$.

So from :
 * the ratio of $ABCG$ to the prisms in it is equal to the ratio of $W$ to the prisms in $DEFH$.

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 :
 * $W : ABCG = DEFH : X$

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

Therefore from :
 * $\triangle DEF : \triangle ABC = DEFG : X$

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.