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.
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"