23. Let $A B C$ be a non-isosceles acute-angled triangle with circumcentre $O$, orthocentre $H$ and $\angle C=41^{\circ}$. Suppose the bisector of $\angle A$ passes through the midpoint $M$ of $O H$. Find $\angle H A O$ in degrees.
23. Answer. 38
Let $P$ be the midpoint of the arc $B C$ not containing $A$ on the circumcircle of the triangle $A B C$. Then $O P$ is the perpendicular bisector of $B C$. Since $A M$ bisects $\angle A$, the points $A, M, P$ are collinear. As both $A H$ and $O P$ are perpendicular to $B C$, they are parallel. Thus $\angle H A M=\angle O P M=\angle O A M$. Also $\angle H M A=\angle O M P$. Since $H M=O M$, we have the triangles $A H M$ and $P O M$ are congruent. Therefore $A H=P O=A O$.

Let $L$ be the midpoint of $B C$. It is a known fact that $A H=2 O L$. To see this, extend $C O$ meeting the circumcircle of the triangle $A B C$ at the point $N$. Then $A N B H$ is a parallelogram. Thus $A H=N B=2 O L$. Therefore in the right-angled triangle $O L C$, $O C=O A=A H=2 O L$. This implies $\angle O C L=30^{\circ}$. Since the triangle $A B C$ is acute, the circumcentre $O$ lies inside the triangle. In fact $\angle A=60^{\circ}$ and $\angle B=79^{\circ}$. Then $\angle O A C=$ $\angle O C A=41^{\circ}-30^{\circ}=11^{\circ}$, Consequently, $\angle H A O=2 \angle O A M=2 \times\left(30^{\circ}-11^{\circ}\right)=38^{\circ}$.