# 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 $AF = HK$.

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

But:

- $ABCDEF$ is half of $AO$

- $GHKLMN$ 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