Line Parallel to Side of Triangle which Bisects One Side also Bisects Other Side

Theorem
Let $ABC$ be a triangle.

Let $DE$ be a straight line parallel to $BC$.

Let $DE$ bisect $AB$.

Then $DE$ also bisects $AC$.

That is, $DE$ is a midline of $\triangle ABC$.



Proof
This is a direct application of Parallel Transversal Theorem.