Problem — The median to the hypotenuse

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Let $DEF$ be a right triangle with the right angle at the vertex $F$, so that $DE$ is the hypotenuse. Letting $N$ be the midpoint of the hypotenuse $DE$, prove that the median $FN$ is congruent to half the hypotenuse, that is $FN \cong \dfrac{1}{2}\,DE$.

Solution

On the extension of the median $FN$ take the point $G$ such that $N$ is also the midpoint of $FG$; in this way $FN \cong NG$.

The quadrilateral $DFEG$ has its two diagonals $DE$ and $FG$ meeting at $N$, their common midpoint ($DN \cong NE$ because $N$ is the midpoint of $DE$, and $FN \cong NG$ by construction). A quadrilateral whose diagonals bisect each other is a parallelogram, so $DFEG$ is a parallelogram.

Since the angle $\widehat{DFE} = 90^\circ$, the parallelogram $DFEG$ has a right angle: it is therefore a rectangle. In a rectangle the diagonals are congruent, hence $FG \cong DE.$ But $FN = \dfrac{1}{2}\,FG$ by construction, so $FN = \frac{1}{2}\,FG = \frac{1}{2}\,DE.$ The median to the hypotenuse is therefore equal to half the hypotenuse. $\;\blacksquare$