# Proportion of Sizes of Tetrahedra divided into Two Similar Tetrahedra and Two Equal Prisms/Lemma

## Lemma to Proportion of Sizes of Tetrahedra divided into Two Similar Tetrahedra and Two Equal Prisms

In the words of Euclid:

But that, as the triangle $LOC$ is to the triangle $RVF$, so is the prism in which the triangle $LOC$ is the base and $PMN$ its opposite to the base in which the triangle $RVF$ is the base and $STU$ its opposite, we must prove as follows.

## Proof

Let perpendiculars be drawn from $G$ and $H$ to the bases $\triangle ABC$ and $\triangle DEF$.

These are equal because, by definition, the heights of $ABCG$ and $DEFH$ are equal.

We have that the straight lines $GC$ and the perpendicular through $G$ are cut by the parallel planes $ABC$ and $PMN$.

they will be cut in the same ratios.

We have that $GC$ is bisected by the plane $PMN$ at $N$.

Therefore the perpendicular through $G$ will also be bisected by the plane $PMN$.

For the same reason, the perpendicular through $H$ will be bisected by the plane $STU$.

We have that the perpendicular through $G$ and $H$ are equal.

Thefore the perpendiculars from $\triangle PMN$ and $\triangle STU$ to $\triangle ABC$ and $\triangle DEF$ are also equal.

Therefore the prisms in which the $\triangle LOC$ and $\triangle RVF$ are bases, and $\triangle PMN$ and $\triangle STU$ are opposites are of equal height.

the parallelepipeds formed from these prisms are of equal height.

Therefore their halves, that is, the prisms described, are in the same ratio as $\triangle LOC$ is to $\triangle RVF$.

$\blacksquare$

## Historical Note

This theorem is Proposition $4$ of Book $\text{XII}$ of Euclid's The Elements.