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.

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

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 $CB$.

$\blacksquare$

Sources

• 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.25$: Corollary