Proof: By Euclid

Let $AB$ and $CD$ be parallelepiped solids (having) the same height.
I say that the parallelepiped solids $AB$ and $CD$ are to one another as their bases.

fig32e

That is to say, as base $AE$ is to base $CF$, so solid $AB$ (is) to solid $CD$.
For let $FH$, equal to $AE$, have been applied to $FG$ (in the angle $FGH$ equal to angle $LCG$) [Prop. 1.45].
And let the parallelepipedal solid $GK$, (having) the same height as $CD$, have been completed on the base $FH$.
So solid $AB$ is equal to solid $GK$.
For they are on the equal bases $AE$ and $FH$, and (have) the same height [Prop. 11.31].
And since the parallelepipedal solid $CK$ has been cut by the plane $DG$, which is parallel to the opposite planes (of $CK$), thus as the base $CF$ is to the base $FH$, so the solid $CD$ (is) to the solid $DH$ [Prop. 11.25].
And base $FH$ (is) equal to base $AE$, and solid $GK$ to solid $AB$.
And thus as base $AE$ is to base $CF$, so solid $AB$ (is) to solid $CD$.
Thus, parallelepiped solids which (have) the same height are to one another as their bases.
(Which is) the very thing it was required to show.
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