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

## Contents

## Theorem

In the words of Euclid:

*If there be two pyramids of the same height which have triangular bases, and each of them be divided into two pyramids equal to one another and similar to the whole, and into two equal prisms, then, as the base of one pyramid is to the base of the other pyramid, so will all the prisms in the one pyramid be to all the prisms, being equal in multitude, in the other pyramid.*

(*The Elements*: Book $\text{XII}$: Proposition $4$)

### Lemma

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.*

(*The Elements*: Book $\text{XII}$: Proposition $4$ : Lemma)

## Proof

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

- let each of $ABCG$ and $DEFH$ be divided into two equal tetrahedra which are similar to the whole 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 Proposition $22$ of Book $\text{VI} $: Similar Figures on Proportional Straight Lines:

- $\triangle LOC : \triangle ABC = \triangle DEF : \triangle RFV$

Therefore from Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:

- $\triangle ABC : \triangle DEF = \triangle LOC : \triangle RFV$

- 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.

- 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 Proposition $12$ of Book $\text{V} $: Sum of Components of Equal Ratios:

- 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 Proposition $3$ of Book $\text{XII} $: Tetrahedron divided into Two Similar Tetrahedra and Two Equal Prisms:

- 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 $DEFH$.

$\blacksquare$

## Historical Note

This theorem is Proposition $4$ 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