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 the Parallel Transversal Theorem.

$\blacksquare$

Sources

• 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.26$