Harvard-MIT Mathematics Tournament

Solution:

Let $X$ be the intersection of the line tangent to $\omega$ at $B$ with the line tangent to $\omega$ at $C$. Note that $\triangle OMC \sim \triangle OCX$ since $\angle OMC = \angle OCX = \frac{\pi}{2}$. Hence $\frac{OM}{OC} = \frac{OC}{OX}$, or, equivalently, $\frac{OM}{OA} = \frac{OA}{OX}$. By SAS similarity, it follows that $\triangle OAM \sim \triangle OXA$. Therefore, $\angle OAM = \angle OXA$.

We claim now that $\angle OAD = \angle OAX$. By the similarity $\triangle OAM \sim OXA$, we have that $\angle OAX = \angle OMA$. Since $AOMD$ is a cyclic quadrilateral, we have that $\angle OMA = \angle ODA$. Since $OA = OD$, we have that $\angle ODA = \angle OAD$. Combining these equations tells us that $\angle OAX = \angle OAD$, so $A$, $D$, and $X$ are collinear.

Finally, since both $AH$ and $OX$ are perpendicular to $BC$, it follows that $AH \parallel OX$, so $\angle DAH = \angle DXO = \angle AXO = \angle MAO$, as desired.

Solution:
Let $Y$ be the intersection of $BC$ with the line tangent to $\omega$ at $A$. Then the circumcircle of triangle $AOM$ has diameter $OY$, so $AD$ is perpendicular to $OY$ because the radical axis of two circles is perpendicular to the line between their centers. Since $Y$ is on the polar of $A$, it follows that $A$ is on the polar of $Y$, so $AD \perp OX$ implies that $AD$ is the polar of $Y$, i.e. $AD$ is the symmedian from $A$ in triangle $ABC$. Hence $AD$ and $AM$ are isogonal. Since $AH$ and $AO$ are also isogonal, the desired conclusion follows immediately.