Proof: By Euclid
(related to Proposition: Prop. 11.16: Common Sections of Parallel Planes with other Plane are Parallel)
 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 $GH$.
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"