Proof: By Euclid
(related to Proposition: Prop. 11.32: Parallelepipeds of Same Height have Volume Proportional to Bases)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"