# Midline and Median of Triangle Bisect Each Other

## Theorem

Let $\triangle ABC$ be a triangle.

Let $DE$ be the midline of $\triangle ABC$ which bisects $AB$ and $AC$.

Let $AF$ be the median of $ABC$ which bisects $BC$.

Then $AF$ and $DE$ bisect each other.

## Proof

Construct the midlines $DF$ and $EF$.

Then by the Midline Theorem $DF \parallel AE$ and $EF \parallel AD$.

Thus by Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel, $\Box ADFE$ is a parallelogram.

By construction, $AF$ and $DE$ are the diagonals of $\Box ADFE$.

The result follows from Diameters of Parallelogram Bisect each other.

$\blacksquare$