Let the line through $D$ perpendicular to $OD$ cut $AB, AC$ at $X, Y$. By Butterfly's theorem, $XD = DY$ hence $$-1 = A(X, Y; D, {\infty}_{XY}) \stackrel{\text{Rotate } 90^{\circ}}{=} H(C, B; {\infty}_{BC}, {\infty}_{\perp XY})$$or $HM\perp XY\implies OD\parallel HM$. Hence we are done.

$B'$ and $C'$ are exactly the same.We choose a point $O'$ in the middle perpendicular to $BC$ where $DHMO'$ is a parallelogram.Suppose that a circle with center $O$ and radius $OB$ intersects $AB$and $AC$ at $X$ and $Y$. To prove that $X$ and $Y$ are $B'$ and $C'$ in the case, It suffices to prove that the lines lie on the line. This can be proved from the Menelaus theremas by a simple cross-sectional calculation.

A short solution without any computation: Let $H', H''$ be the reflections of $H$ over $D$ and $M$, respectively. Clearly, the feet of the perpendicular lines from $H''$ onto $AB$ and $AC$ are just $B,C$. On the other hand (Simson line), since $H'$ is on the circumcircle of $ABC$, the feet from $H'$ are just $B',C'$. Hence $O$ is just the midpoint of $H'H''$ which is easily seen to be equivalent to the claim.