Proof: By Euclid
(related to Proposition: Prop. 11.24: Opposite Planes of Solid contained by Parallel Planes are Equal Parallelograms)
 For let the solid (figure) $CDHG$ have been contained by the parallel planes $AC$, $GF$, and $AH$, $DF$, and $BF$, $AE$.

I say that its opposite planes are both equal and parallelogrammic.

For since the two parallel planes $BG$ and $CE$ are cut by the plane $AC$, their common sections are parallel [Prop. 11.16].
 Thus, $AB$ is parallel to $DC$.
 Again, since the two parallel planes $BF$ and $AE$ are cut by the plane $AC$, their common sections are parallel [Prop. 11.16].
 Thus, $BC$ is parallel to $AD$.
 And $AB$ was also shown (to be) parallel to $DC$.
 Thus, $AC$ is a parallelogram.
 So, similarly, we can also show that $DF$, $FG$, $GB$, $BF$, and $AE$ are each parallelograms.
 Let $AH$ and $DF$ have been joined.
 And since $AB$ is parallel to $DC$, and $BH$ to $CF$, so the two (straight lines) joining one another, $AB$ and $BH$, are parallel to the two straight lines joining one another, $DC$ and $CF$ (respectively), not (being) in the same plane.
 Thus, they will contain equal angles [Prop. 11.10].
 Thus, angle $ABH$ (is) equal to (angle) $DCF$.
 And since the two (straight lines) $AB$ and $BH$ are equal to the two (straight lines) $DC$ and $CF$ (respectively) [Prop. 1.34], and angle $ABH$ is equal to angle $DCF$, the base $AH$ is thus equal to the base $DF$, and triangle $ABH$ is equal to triangle $DCF$ [Prop. 1.4].
 And parallelogram $BG$ is double (triangle) $ABH$, and parallelogram $CE$ double (triangle) $DCF$ [Prop. 1.34].
 Thus, parallelogram $BG$ (is) equal to parallelogram $CE$.
 So, similarly, we can show that $AC$ is also equal to $GF$, and $AE$ to $BF$.
 Thus, if a solid (figure) is contained by (six) parallel planes then its opposite planes are both equal and parallelogrammic.
 (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"