# Prism on Triangular Base divided into Three Equal Tetrahedra

## Theorem

In the words of Euclid:

*Any prism which has a triangular base is divided into three pyramids equal to one another which have triangular bases.*

(*The Elements*: Book $\text{XII}$: Proposition $7$)

### Porism

In the words of Euclid:

*From this it is manifest that any pyramid is a third part of a prism which has the same base with it and equal height.*

(*The Elements*: Book $\text{XII}$: Proposition $7$ : Porism)

## Proof

Let $ABCDEF$ be a prism whose base is $\triangle ABC$ and whose opposite is $\triangle DEF$.

Then it is to be demonstrated that $ABCDEF$ can be divided into three equal tetrahedra.

Let $BD, EC, CD$ be joined.

We have that $ABED$ is a parallelogram.

Therefore $BD$ is the diameter of $ABED$.

So from Proposition $34$ of Book $\text{I} $: Opposite Sides and Angles of Parallelogram are Equal:

- $\triangle ABD = \triangle EBD$

Therefore from Proposition $5$ of Book $\text{XII} $: Sizes of Tetrahedra of Same Height are as Bases:

- the tetrahedron $ABDC$ whose base is $\triangle ABD$ and whose apex is $C$ equals the tetrahedron $DEBC$ whose base is $\triangle DEB$ and whose apex is $C$.

But the tetrahedron $DEBC$ is the same as the tetrahedron $EBCD$ whose base is $\triangle EBC$ and whose apex is $D$.

Therefore the tetrahedron $ABDC$ equals the tetrahedron $EBCD$.

We have that $FCBE$ is a parallelogram.

Therefore $CE$ is the diameter of $FCBE$.

So from Proposition $34$ of Book $\text{I} $: Opposite Sides and Angles of Parallelogram are Equal:

- $\triangle CEF = \triangle CBE$

Therefore from Proposition $5$ of Book $\text{XII} $: Sizes of Tetrahedra of Same Height are as Bases:

- the tetrahedron $BCED$ whose base is $\triangle BCE$ and whose apex is $D$ equals the tetrahedron $ECFD$ whose base is $\triangle ECF$ and whose apex is $D$.

But the tetrahedron $BCED$ was proved equal to the tetrahedron $ABDC$ whose base is $\triangle ABD$ and whose apex is $C$.

Therefore the tetrahedron $CEFD$ whose base is $\triangle CEF$ and whose apex is $D$ equals the tetrahedron $ABDC$.

Therefore the prism $ABCDEF$ has been divided into three equal tetrahedra.

We have that the tetrahedron $ABDC$ is the same as the tetrahedron $CABD$ whose base is $\triangle CAB$ and whose apex is $D$.

We have that the tetrahedron $ABDC$ is a third of the prism $ABCDEF$ whose base is $\triangle ABC$ and whose opposite is $\triangle DEF$.

Therefore tetrahedron $ABCD$ whose base is $\triangle ABC$ and whose apex is $D$ is a third of the prism $ABCDEF$ whose base is $\triangle ABC$ and whose opposite is $\triangle DEF$.

$\blacksquare$

## Historical Note

This theorem is Proposition $7$ of Book $\text{XII}$ of Euclid's *The Elements*.

The treatise *The Method* of Archimedes, discovered in $1906$, tells us that this result was discovered by Democritus.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{XII}$. Propositions - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.3$: Democritus (ca. $\text {460}$ – $\text {370}$ B.C.)