China National Team Selection Test

Let $T$ be the orthocenter of $△HMN$. Extended lines of $HM$ and $CA$ intersect at point $P$. Extended lines $HN$ and $BA$ intersect at point $Q$. It is easy to see that points $N$, $M$, $P$ and $Q$ are concyclic.



By $\angle THM=\angle OHN$, we see that $\angle PQH-\angle OHN=\angle NMH-\angle THM=90^{\circ }$, that is, $$HO\perp PQ.①$$ Let point $R$ be symmetric to point $C$ over $HP$. Then $HC=HR$. By $\angle HPC+\angle HCP=(\angle BAC-60^{\circ })+(90^{\circ }-\angle BAC)=30^{\circ }$, we see that $\angle CHR=60^{\circ }$. Hence $△HCR$ is regular. By $\angle HPC=\angle HQB$ and $\angle HCP=\angle HBQ$, we see that $△PHC\sim △QHB$. Then $△PHR\sim △QHB$, so $△QHP\sim △BHR$. Denote by $\angle (UV,XY)$ the angle between $\overrightarrow{UV}$ and $\overrightarrow{XY}$ (positive anticlockwise). Since $\angle PHR=150^{\circ }$, we have $\angle (PQ,RB)=\angle (HP,HR)=150^{\circ }$, $△BCD$ and $△RCH$ are regular. So $△BRC\cong △DHC$. Hence $\angle (RB,HD)=\angle (CR,CH)=-60^{\circ }$.

Consequently, $$\angle (PQ,HD)=\angle (PQ,RB)+\angle (RB,HD)=150^{\circ }-60^{\circ }=90^{\circ },$$ that is $$DH\perp PQ.\stackrel{◯}{2}$$ By ① and ②, points $H$, $O$ and $D$ are collinear. □