(related to Proposition: Prop. 11.09: Lines Parallel to Same Line not in Same Plane are Parallel to each other)

- For let $AB$ and $CD$ each be parallel to $EF$, not being in the same plane as it.
- I say that $AB$ is parallel to $CD$.

- For let some point $G$ have been taken at random on $EF$.
- And from it let $GH$ have been drawn at right angles to $EF$ in the plane through $EF$ and $AB$.
- And let $GK$ have been drawn, again at right angles to $EF$, in the plane through $FE$ and $CD$.
- And since $EF$ is at right angles to each of $GH$ and $GK$, $EF$ is thus also at right angles to the plane through $GH$ and $GK$ [Prop. 11.4].
- And $EF$ is parallel to $AB$.
- Thus, $AB$ is also at right angles to the plane through $HGK$ [Prop. 11.8].
- So, for the same (reasons), $CD$ is also at right angles to the plane through $HGK$.
- Thus, $AB$ and $CD$ are each at right angles to the plane through $HGK$.
- And if two straight lines are at right--angles to the same plane then the straight lines are parallel [Prop. 11.6].
- Thus, $AB$ is parallel to $CD$.
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"