Let $A'$ and $B'$ be the feet of the altitudes drawn from $A$ and $B$ respectively in the triangle $AOB$, and $C'$, $D'$ the feet of the altitudes drawn from $C$ and $D$ in the triangle $COD$.
Obviously, $A'$ and $D'$ belong to the circle $C_1$ of diameter $AD$, while $B'$ and $C'$ belong to the circle $C_2$ of diameter $BC$.
It is easy to see that triangles $H_1AB$ and $H_1A'B'$ are similar. It follows that $H_1A \cdot H_1A' = H_1B \cdot H_1B'$. (Alternatively, one could notice that the quadrilateral $ABA'B'$ is cyclic and obtain the previous relation by writing the power of $H_1$ with respect to its circumcircle.)
So, $H_1$ has the same power with respect to circles $C_1$ and $C_2$. Hence, $H_1$ (and similarly, $H_2$) 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 $H_1H_2$ is perpendicular to $PQ$, where $P$ and $Q$ are the midpoints of the sides $AD$ and $BC$, respectively. ($P$ and $Q$ are the centers of circles $C_1$ and $C_2$.)
The condition $H_1H_2 \parallel MN$ is equivalent to $MN \perp PQ$. As $MPNQ$ is a parallelogram, we conclude that $H_1H_2 \parallel MN \perp PQ \parallel MPNQ$ is a rhombus $\Leftrightarrow MP = MQ \Leftrightarrow AC = BD$.