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.

$\blacksquare$


Sources

  • 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.26$: Corollary $2$
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