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.)