Proof: By Euclid

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 $CD$.

fig09e

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.
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