Median of Trapezoid is Parallel to Bases/Sufficient Condition

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


 * Median-of-Trapezoid.png

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

Let $F$ lie on $BC$.

Let $EF$ be parallel to both $AB$ and $DC$.

Then $F$ is the midpoint of $BC$.

Proof

 * Median-of-Trapezoid-Proof.png

Let $DH$ be constructed parallel to $BC$ to cut $AB$ at $H$.

From the Parallel Transversal Theorem:
 * $DG : GH = DE : EA$

and so $G$ is the midpoint of $AH$.

That is:
 * $(1): \quad DG = GH$

Then we have that:
 * $DC$ is parallel to $GF$

and:
 * $DG$ is parallel to $CF$

so, by definition, $\Box GFCD$ is a parallelogram.

Similarly, we have:


 * $GF$ is parallel to $HB$

and:
 * $GH$ is parallel to $FB$

so, by definition, $\Box GFBH$ is also a parallelogram.

By Opposite Sides and Angles of Parallelogram are Equal we have that:
 * $CF = DG$

and:
 * $GH = FB$

But from $(1)$:
 * $DG = GH$

and so:
 * $CF = FB$

and so $F$ is the midpoint of $CB$.