Median to Hypotenuse of Right Triangle equals Half Hypotenuse

Theorem

Let $\triangle ABC$ be a right triangle such that $BC$ is the hypotenuse.

Let $AD$ be the median to $BC$.

Then $AD$ is half of $BC$.

Proof

Construct $BE$ and $CE$ parallel to $AC$ and $AB$ respectively.

Then by definition $ABEC$ is a parallelogram.

By construction, $BC$ is a diagonal of $ABEC$ such that $AD$ is a bisector of it.

Thus by Quadrilateral is Parallelogram iff Diagonals Bisect each other, $AE$ is also a bisector of $ABEC$.

As $\angle BAC$ is a right angle it follows that $\angle BEC$ is also a right angle.

Thus by definition $ABEC$ is a rectangle.

From Diagonals of Rectangle are Equal:

$AE = BC$

From above, $D$ is the midpoint of both $AE$ and $BC$.

Thus $AD = BD$ and hence the result.

$\blacksquare$

Also see

• Thales' Theorem, the converse of this.

Sources

• 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.27$