Opposite Planes of Solid contained by Parallel Planes are Equal Parallelograms

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Theorem

In the words of Euclid:

If a solid be contained by parallel planes, the opposite planes in it are equal and parallelogrammic.

(The Elements: Book $\text{XI}$: Proposition $24$)


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 Proposition $16$ of Book $\text{XI} $: Common Sections of Parallel Planes with other Plane are Parallel:

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 Proposition $10$ of Book $\text{XI} $: Two Lines Meeting which are Parallel to Two Other Lines Meeting contain Equal Angles:

$\angle ABH = \angle DCF$

From Proposition $34$ of Book $\text{I} $: Opposite Sides and Angles of Parallelogram are Equal:

$AB$ and $BH$ are equal to $DC$ and $CF$.

and because:

$\angle ABH = \angle DCF$

it follows that:

$AH = DF$

So from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

$\triangle ABH = \triangle DCF$

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

From Proposition $34$ of Book $\text{I} $: Opposite Sides and Angles of Parallelogram are Equal:

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 $BF$.

$\blacksquare$


Historical Note

This proof is Proposition $24$ of Book $\text{XI}$ of Euclid's The Elements.


Sources