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Geometry Difficulty 4.6 AIME Prove it Romania

Let $ABC$ be a triangle with $\angle A < 90^\circ$ and $AB \neq AC$ . Denote by $H$ the ortocenter of triangle $ABC$ , by $N$ the midpoint of $[AH]$ , by $M$ the midpoint of the side $[BC]$ and by $D$ the intersection point of the angle bisector of $\angle BAC$ with $[MN]$ . Prove that $\angle ADH = 90^\circ$ .

Solution

Let $O$ be the circumcircle of triangle $ABC$ . Rays ( $AO$ and ( $AH$ are isogonal. We deduce that $\angle NAD \equiv \angle OAD$ . (1)

But $OM \parallel AH$ and $OM = \frac{1}{2}AH = AN$ , which means that $MOAN$ is a parallelogram. It follows that $MN \parallel OA$ . (2)

Combining (1) and (2) gives $\angle NAD \equiv \angle OAD \equiv \angle NDA$ , hence triangle $NAD$ is isosceles with $NA = ND$ . In triangle $DAH$ , the segment $[DN]$ is a median, and $DN = AN = \frac{1}{2}AH$ leads to $\angle ADH = 90^\circ$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.