Prisms of equal Height with Parallelogram and Triangle as Base

Theorem

In the words of Euclid:

If there be two prisms of equal height, and one have a parallelogram as base and the other a triangle, and if the parallelogram be double the triangle, the prisms will be equal.

(The Elements: Book $\text{XI}$: Proposition $39$)

Proof

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

the parallelogram $HK$ is twice $\triangle GHK$.

Therefore $AF = HK$.

From Proposition $31$ of Book $\text{XI} $: Parallelepipeds on Equal Bases and Same Height are Equal in Volume:

the parallelepipeds $AO$ is equal to the parallelepipeds $GP$.

But:

$ABCDEF$ is half of $AO$

and from Proposition $28$ of Book $\text{XI} $: Parallelepiped cut by Plane through Diagonals of Opposite Planes is Bisected:

$GHKLMN$ is half of $GP$.

Hence $ABCDEF$ is equal to $GHKLMN$.

$\blacksquare$

Historical Note

This proof is Proposition $39$ of Book $\text{XI}$ of Euclid's The Elements.
In the words of Thomas L. Heath:

The phraseology is interesting because we find one of the parallelogrammic faces of one of the triangular prisms called its base, and the perpendicular on this plane from that vertex of either triangular face which is not in this plane its height.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions