Median of Trapezoid is Parallel to Bases/Necessary Condition

Theorem

Let $\Box ABCD$ be a trapezoid such that $AB$ and $DC$ are the parallel sides.

Let $E$ be the midpoint of $AD$.

Let $F$ be the midpoint of $BC$.

Then $EF$ is parallel to both $AB$ and $DC$.

Proof

Aiming for a contradiction, suppose $EF$ is not parallel to $DC$.

By Playfair's axiom, there exists a unique straight line through $E$ which is parallel to $DC$.

Let $EF'$ be this line.

From Median of Trapezoid is Parallel to Bases: Sufficient Condition, $F'$ is the midpoint of $BC$.

But by hypothesis $F$ is also the midpoint of $BC$.

That is:

$F = F'$

So:

$EF = EF'$

and so $EF$ is parallel to $DC$.

This contradicts our assertion that $EF$ is not parallel to $DC$.

Hence, by Proof by Contradiction, $EF$ is parallel to $DC$.

By definition of base of trapezoid, $AB$ is parallel to $DC$.

By Parallelism is Transitive Relation, $EF$ is parallel to $AB$.

That is, $EF$ is parallel to both $AB$ and $DC$.

$\blacksquare$