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

Proof

 * Euclid-XI-30.png

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 $FR$.

So from :
 * 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 $NR$.

So from :
 * 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 $CN$.