Proof: By Euclid

Let $AB$ and $CD$ be two parallel straight lines, and let the random points $E$ and $F$ have been taken on each of them (respectively).
I say that the straight line joining points $E$ and $F$ is in the same (reference) plane as the parallel (straight lines).

fig07e

For (if) not, and if possible, let it be in a more elevated (plane), such as $EGF$.
And let a plane have been drawn through $EGF$.
So it will make a straight cutting in the reference plane [Prop. 11.3].
Let it make $EF$.
Thus, two straight lines (with the same end-points), $EGF$ and $EF$, will enclose an area.
The very thing is impossible.
Thus, the straight line joining $E$ to $F$ is not in a more elevated plane.
The straight line joining $E$ to $F$ is thus in the plane through the parallel (straight lines) $AB$ and $CD$.
Thus, if there are two parallel straight lines, and random points are taken on each of them, then the straight line joining the two points is in the same plane as the parallel (straight lines).
(Which is) the very thing it was required to show.
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