Parallelepipeds on Same Base and Same Height whose Extremities are on Same Lines are Equal in Volume

Proof

 * Euclid-XI-29.png

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

Let the endpoints of their vertical sides:
 * $AG, AF, LM, LN, CD, CE, BH, BK$

be on the same straight lines $FN$ and $DK$.

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

We have that each of the figures $CH$ and $CK$ is a parallelogram.

Thus from :
 * $CB$ equals each of the straight lines $DH$ and $EK$.

Thus $DH = EK$.

Let $EH$ be subtracted from both $DH$ and $EK$.

Therefore their remainders $DE$ and $HK$ are equal.

So from:

and:

it follows that:
 * $\triangle DCE = \triangle HBK$

and from :
 * the parallelogram $DG$ equals the parallelogram $HN$.

For the same reason:
 * $\triangle AFG = \triangle MLN$

But the parallelogram $CF$ equals the parallelogram $BM$, as they are opposite.

Therefore:
 * the prism contained by $\triangle AFG$ and $\triangle DCE$ and the three parallelograms $AD, DG, CG$

equals:
 * the prism contained by $\triangle MLN$ and $\triangle HBK$ and the three parallelograms $BM, HN, BN$.

Let there be added to each the solid of which the parallelogram $AB$ is its base and $GEHM$ its opposite.

Therefore the whole parallelepiped $CM$ is equal to the whole parallelepiped $CN$.