Line joining Points on Parallel Lines is in Same Plane

Theorem

In the words of Euclid:

If two straight lines be parallel and points be taken at random on each of them, the straight line joining the points is in the same plane with the parallel straight lines.

(The Elements: Book $\text{XI}$: Proposition $7$)

Proof

Let $AB$ and $CD$ be two straight lines which are parallel.

Consider the plane in which they lie to be the plane of reference.

Let $E$ and $F$ be taken at random on both of them.

Then the straight line $EF$ is to be demonstrated to lie in the same plane as $AB$ and $CD$.

Suppose to the contrary that $EF$ is in a more elevated plane: $EGF$.

Let a plane be drawn through $EGF$.

From Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane, the common section of this plane with the plane of reference is a straight line passing through $E$ and $F$.

Therefore $EF$ and $EGF$ enclose an area, which is impossible.

Therefore the straight line $EF$ joining $E$ and $F$ is in the same plane as $AB$ and $CD$.

$\blacksquare$

Historical Note

This proof is Proposition $7$ of Book $\text{XI}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions