Proof: By Euclid

For let the two parallel planes $AB$ and $CD$ have been cut by the plane $EFGH$.
And let $EF$ and $GH$ be their common sections.
I say that $EF$ is parallel to $GH$.

fig16e

For, if not, being produced, $EF$ and $GH$ will meet either in the direction of $F$, $H$, or of $E$, $G$.
Let them be produced, as in the direction of $F$, $H$, and let them, first of all, have met at $K$.
And since $EFK$ is in the plane $AB$, all of the points on $EFK$ are thus also in the plane $AB$ [Prop. 11.1].
And $K$ is one of the points on $EFK$.
Thus, $K$ is in the plane $AB$.
So, for the same (reasons), $K$ is also in the plane $CD$.
Thus, the planes $AB$ and $CD$, being produced, will meet.
But they do not meet, on account of being (initially) assumed (to be mutually) parallel.
Thus, the straight lines $EF$ and $GH$, being produced in the direction of $F$, $H$, will not meet.
So, similarly, we can show that the straight lines $EF$ and $GH$, being produced in the direction of $E$, $G$, will not meet either.
And (straight lines in one plane which), being produced, do not meet in either direction are parallel [Def. 1.23] .
$EF$ is thus parallel to $GH$.
Thus, if two parallel planes are cut by some plane then their common sections are parallel.
(Which is) the very thing it was required to show.
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