Prisms of equal Height with Parallelogram and Triangle as Base

Proof

 * Euclid-XI-39.png

Let $ABCDEF$ and $GHKLMN$ be triangular prisms.

Let $ABCDEF$ be positioned so that the parallelogram $AF$ can be identified as its base, so that its height is defined as the height of one of its triangular faces.

Let $GHK$ be the triangular (conventionally defined) lower base of $GHKLMN$.

Let the height of $ABCDEF$ equal the height of $GHKLMN$.

Let the area of parallelogram $AF$ be twice the area of $GHK$.

It is to be demonstrated that $ABCDEF$ and $GHKLMN$ are equal in volume.

Let the parallelepipeds $AO$ and $GP$ be completed.

We have that $AF$ is twice $\triangle GHK$.

From :
 * the parallelogram $HK$ is twice $\triangle GHK$.

Therefore $AF = HK$.

From :
 * the parallelepipeds $AO$ is equal to the parallelepipeds $GP$.

But:
 * $ABCDEF$ is half of $AO$

and from :
 * $GHKLMN$ is half of $GP$.

Hence $ABCDEF$ is equal to $GHKLMN$.