# Parallelepipeds on Same Base and Same Height whose Extremities are not on Same Lines are Equal in Volume

## Theorem

In the words of Euclid:

*Parallelepidedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are not on the same straight lines, are equal to one another.*

(*The Elements*: Book $\text{XI}$: Proposition $30$)

## Proof

Let $CM$ and $CN$ be parallelepipeds on the same base $AB$ and of the same height.

Let the endpoints of their vertical sides:

- $AF, AG, LM, LN, CD, CE, BH, BK$

be not on the same straight lines.

It is to be demonstrated that the parallelepiped $CM$ is equal to the parallelepiped $CN$.

Let $NK$ and $DH$ be produced, and meet each other at $R$.

Let $FM$ and $GE$ be produced to $P$ and $Q$.

Let $AO, LP, CQ, BR$ be joined.

We have that $CM$ and $CP$ are on the same parallelogram $ABCL$ and of the same height, and the endpoints of their vertical sides:

- $AF, AO, LM, LP, CD, CQ, BH, BR$

are on the same straight lines $FP$ and $FR$.

- the parallelepiped $CM$, of which the parallelogram $ABCL$ is the base and $FDHM$ the opposite

is equal to

- the parallelepiped $CP$, of which the parallelogram $ABCL$ is the base and $OQRP$ the opposite.

But we also have that $CP$ and $CN$ are on the same parallelogram $ABCL$ and of the same height, and the endpoints of their vertical sides:

- $AG, AO, LN, LP, CE, CQ, BK, BR$

are on the same straight lines $GQ$ and $NR$.

- the parallelepiped $CN$, of which the parallelogram $ABCL$ is the base and $ACBL$ the opposite

is equal to

- the parallelepiped $CP$, of which the parallelogram $ABCL$ is the base and $OQRP$ the opposite.

Hence the parallelepiped $CM$ equals the parallelepiped $CN$.

$\blacksquare$

## Historical Note

This proof is Proposition $30$ of Book $\text{XI}$ of Euclid's *The Elements*.

## Sources

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