Prism on Triangular Base divided into Three Equal Tetrahedra

Proof

 * Euclid-XII-7.png

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 :
 * $\triangle ABD = \triangle EBD$

Therefore from :
 * 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 :
 * $\triangle CEF = \triangle CBE$

Therefore from :
 * 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 $\triangle DEF$.