Without loss of generality assume $AB < AC$. Let $W$ be the intersection of the angle bisector $AL$ with the circumcircle of $\Delta ABC$. Then $MW = OW-OM = R-r = \sqrt{OI^2 + r^2} = MI$ and $\measuredangle BMI = 90^\circ - 2\measuredangle AWO = \angle A + 2\angle C - 90^\circ$. Let $AH_A$ and $BM_B$ be the altitude and the median drawn from the corresponding vertices. Then $\measuredangle MM_BH_A = \measuredangle MM_BA - \measuredangle H_AM_BA = 180^\circ - \angle A - 2\angle C$. Since all three points lie on the nine-point circle (denote its center by $O_9$), by angle chase we get $\measuredangle H_AMO_9 = \angle A + 2\angle C - 90^\circ = \measuredangle H_AMI$, which concludes the proof.