Let $T$ be the orthocenter of $\triangle HMN$. 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,
$$HO \perp PQ. \qquad ①$$
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 $\triangle HCR$ is regular.
By $\angle HPC = \angle HQB$ and $\angle HCP = \angle HBQ$, we see that $\triangle PHC \sim \triangle QHB$. Then $\triangle PHR \sim \triangle QHB$, so $\triangle QHP \sim \triangle BHR$.
Denote by $\angle (UV, XY)$ the angle between $\vec{UV}$ and $\vec{XY}$ (positive anticlockwise).
Since $\angle PHR = 150^\circ$, we have $\angle (PQ, RB) = \angle (HP, HR) = 150^\circ$, $\triangle BCD$ and $\triangle RCH$ are regular. So $\triangle BRC \cong \triangle DHC$. 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
$$DH \perp PQ. \qquad \textcircled{2}$$
By ① and ②, points $H$, $O$ and $D$ are collinear. □