Let $A^{\prime}$ and $B^{\prime}$ be the feet of the altitudes drawn from $A$ and $B$ respectively in the triangle $A O B$, and $C^{\prime}$ and $D^{\prime}$ are the feet of the altitudes drawn from $C$ and $D$ in the triangle $C O D$. Obviously, $A^{\prime}$ and $D^{\prime}$ belong to the circle $c_{1}$ of diameter $\overline{A D}$, while $B^{\prime}$ and $C^{\prime}$ belong to the circle $c_{2}$ of diameter $\overline{B C}$.

It is easy to see that triangles $H_{1} A B$ and $H_{1} B^{\prime} A^{\prime}$ are similar. It follows that $\overline{H_{1} A} \cdot \overline{H_{1} A^{\prime}}=\overline{H_{1} B} \cdot \overline{H_{1} B^{\prime}}$. (Alternatively, one could notice that the quadrilateral $A B A^{\prime} B^{\prime}$ is cyclic and obtain the previous relation by writing the power of $H_{1}$ with respect to its circumcircle.) It follows that $H_{1}$ has the same power with respect to circles $c_{1}$ and $c_{2}$. Thus, $H_{1}$ (and similarly, $\mathrm{H}_{2}$) is on the radical axis of the two circles.

The radical axis being perpendicular to the line joining the centers of the two circles, one concludes that $\mathrm{H}_{1} \mathrm{H}_{2}$ is perpendicular to $P Q$, where $P$ and $Q$ are the midpoints of the sides $\overline{A D}$ and $\overline{B C}$, respectively. ( $P$ and $Q$ are the centers of circles $c_{1}$ and $c_{2}$.)

The condition $H_{1} H_{2} \| M N$ is equivalent to $M N \perp P Q$. As $M P N Q$ is a parallelogram, we conclude that $H_{1} H_{2} \| M N \Leftrightarrow M N \perp P Q \Leftrightarrow M P N Q$ a rhombus $\Leftrightarrow \overline{M P}=\overline{M Q} \Leftrightarrow \overline{A C}=\overline{B D}$.