Opposite Planes of Solid contained by Parallel Planes are Equal Parallelograms

Proof

 * Euclid-XI-24.png

Let the solid $CDHG$ be contained by the parallel planes $AC, GF, AH, DF, BF, AE$.

It is to be demonstrated that opposite planes are equal parallelograms.

We have that the two parallel planes $BG$ and $CE$ are cut by the plane $AC$.

From :
 * the common sections of $BG$ and $CE$ are parallel lines.

Thus $AB \parallel DC$.

Again, we have that the two parallel planes $BF$ and $AE$ are cut by the plane $AC$.

Thus $BC \parallel AD$.

But $AB \parallel DC$.

Therefore $AC$ is by definition a parallelogram.

Similarly it can be shown that each of $GF, AH, DF, BF, AE$ are parallelograms.

Let $AH$ and $DF$ be joined.

We have that:
 * $AB \parallel DC$

and:
 * $BH \parallel CF$

Thus the two straight lines $AB$ and $BH$ which meet one another are parallel to the two straight lines $DC$ and $CF$ which also meet one another, but not in the same plane.

Therefore by :
 * $\angle ABH = \angle DCF$

From :
 * $AB$ and $BH$ are equal to $DC$ and $CF$.

and because:
 * $\angle ABH = \angle DCF$

it follows that:
 * $AH = DF$

So from :
 * $\triangle ABH = \triangle DCF$

We have that the parallelogram $BG$ is double $\triangle ABH$.

From :
 * the parallelogram $CE$ is double the $\triangle DCF$.

Therefore the parallelogram $BG$ equals the parallelogram $CE$.

Similarly it is shown that:
 * the parallelogram $AC$ equals the parallelogram $GF$

and:
 * the parallelogram $AE$ equals the parallelogram $BF$.