Proof: By Euclid
(related to Proposition: Prop. 11.25: Parallelepiped cut by Plane Parallel to Opposite Planes)
 For let $AH$ have been produced in each direction.
 And let any number whatsoever (of lengths), $AK$ and $KL$, be made equal to $AE$, and any number whatsoever (of lengths), $HM$ and $MN$, equal to $EH$.
 And let the parallelograms $LP$, $KV$, $HW$, and $MS$ have been completed, and the solids $LQ$, $KR$, $DM$, and $MT$.
 And since the straight lines $LK$, $KA$, and $AE$ are equal to one another, the parallelograms $LP$, $KV$, and $AF$ are also equal to one another, and $KO$, $KB$, and $AG$ (are equal) to one another, and, further, $LX$, $KQ$, and $AR$ (are equal) to one another.
 For (they are) opposite [Prop. 11.24].
 So, for the same (reasons), the parallelograms $EC$, $HW$, and $MS$ are also equal to one another, and $HG$, $HI$, and $IN$ are equal to one another, and, further, $DH$, $MY$, and $NT$ (are equal to one another).
 Thus, three planes of (one of) the solids $LQ$, $KR$, and $AU$ are equal to the (corresponding) three planes (of the others).
 But, the three planes (in one of the soilds) are equal to the three opposite planes [Prop. 11.24].
 Thus, the three solids $LQ$, $KR$, and $AU$ are equal to one another [Def. 11.10] .
 So, for the same (reasons), the three solids $ED$, $DM$, and $MT$ are also equal to one another.
 Thus, as many multiples as the base $LF$ is of the base $AF$, so many multiples is the solid $LU$ also of the solid $AU$.
 So, for the same (reasons), as many multiples as the base $NF$ is of the base $FH$, so many multiples is the solid $NU$ also of the solid $HU$.
 And if the base $LF$ is equal to the base $NF$ then the solid $LU$ is also equal to the solid $NU$.^{1}
 And if the base $LF$ exceeds the base $NF$ then the solid $LU$ also exceeds the solid $NU$.
 And if ($LF$) is less than ($NF$) then ($LU$) is (also) less than ($NU$).
 So, there are four magnitudes, the two bases $AF$ and $FH$, and the two solids $AU$ and $UH$, and equal multiples have been taken of the base $AF$ and the solid $AU$  (namely), the base $LF$ and the solid $LU$  and of the base $HF$ and the solid $HU$  (namely), the base $NF$ and the solid $NU$.
 And it has been shown that if the base $LF$ exceeds the base $FN$ then the solid $LU$ also exceeds the [solid] $NU$, and if ($LF$ is) equal (to $FN$) then ($LU$ is) equal (to $NU$), and if ($LF$ is) less than ($FN$) then ($LU$ is) less than ($NU$).
 Thus, as the base $AF$ is to the base $FH$, so the solid $AU$ (is) to the solid $UH$ [Def. 5.5] .
 (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"
Footnotes