Proportion of Sizes of Tetrahedra divided into Two Similar Tetrahedra and Two Equal Prisms

Proof

 * Euclid-XII-4.png

Let there be two tetrahedra of the same height whose bases are $ABC, DEF$ and whose apices are $G$ and $H$.

From :
 * let each of $ABCG$ and $DEFH$ be divided into two equal tetrahedra which are similar and two equal prisms.

It is to be demonstrated that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of all the prisms in $ABCG$ to all the prisms in $DEFH$.

We have that:
 * $BO = OC$

and:
 * $AL = LC$

Therefore:
 * $LO \parallel AB$

and $\triangle ABC$ is similar to $\triangle LOC$.

For the same reason, $\triangle DEF$ is similar to $\triangle RVF$.

We have that:
 * $BC = 2 \cdot CO$

and:
 * $EF = 2 \cdot FV$

Therefore:
 * $BC : CO = EF : FV$

So from :
 * $\triangle LOC : \triangle ABC = \triangle DEF : \triangle RFV$

Therefore from :
 * $\triangle ABC : \triangle DEF = \triangle LOC : \triangle RFV$

But from :
 * the ratio of $\triangle LOC$ to $\triangle RFV$ equals the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite.

Therefore the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite.

But from :
 * the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite equals the ratio of the prism of which the parallelogram $KBOL$ is its "base" and the straight line $PM$ lies opposite to the prism of which $QEVR$ is its "base" and the straight line $ST$ lies opposite.

Therefore, by :
 * the ratio of the two prisms of which the parallelogram $KBOL$ is its "base" and the straight line $PM$ lies opposite and in which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite are to the two prisms of which the the parallelogram to the prism of which $QEVR$ is its "base" and the straight line $ST$ lies opposite and in which $\triangle RFV$ is its base and $\triangle STU$ is its opposite.

Similarly from :
 * the two tetrahedra $PMNG$ and $STUH$ can be divided into two equal tetrahedra which are similar and two equal prisms.

As the ratio of the base $PMN$ is to the base $STU$, so will the ratio of the two prisms in the tetrahedron $PMNG$ be to the two prisms in the tetrahedron $STUH$.

But:
 * $\triangle PMN : \triangle STU = \triangle LOC : \triangle RFV$

Therefore:
 * $\triangle ABC : \triangle DEF$ equals the ratio of the four prisms in the tetrahedron $ABCG$ to the four prisms in the tetrahedron $DEFH$.

Similarly, if the remaining tetrahedra are divided into two tetrahedra and two prisms, then, as $\triangle ABC$ is to $\triangle DEF$, so will all the prisms in $ABCG$ be to all the prisms in $DEFH$.