# 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 $GH$.

• 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)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"