Parallelepipeds of Same Height have Volume Proportional to Bases

Proof

 * Euclid-XI-32.png

Let $AB$ and $CD$ be parallelepipeds of the same height.

Let $AE$ be the base of $AB$ and $CF$ be the base of $AB$.

Using :
 * Let the parallelogram $FH$ equal in area to $AE$ be applied to $FG$.

Let the parallelepiped $GK$ be constructed with the same height as $CD$ on the base $FH$.

Then from :
 * the parallelepiped $AB$ equals the parallelepiped $GH$.

We have that the parallelepiped $CK$ is cut by the plane $DG$ which is parallel to opposite planes.

Therefore from :
 * the parallelepiped $CD$ is to the parallelepiped $DH$ as the base $CF$ is to the base $FH$.

But:
 * the base $FH$ equals the base $AE$

and:
 * the parallelepiped $GK$ equals the parallelepiped $AB$.

Therefore, as the base $AE$ is to the base $CF$, the parallelepiped $AB$ is to the parallelepiped $CD$.