Angle bisector contains the Nine Point Centre

Question

I got a question recently, and have been unable to solve it.

$\Delta ABC$ is a non-isosceles triangle with $\angle C=41°$. $M$ is the midpoint of the segment joining the orthocentre $H$ and circumcentre $O$. If $AM$ bisects $\angle A$, find $\angle HAO$.

I proved that $AM\perp HO$, and $M$ is the nine-point-centre, but I can't find the required angle.

Please help.

Answer 1

By your work we obtain $AO=AH$ or in the standard notation $$R=a|\cot\alpha|$$ or $$\frac{a|\cos\alpha|}{\sin\alpha}=\frac{a}{2\sin\alpha}$$ or $$|\cos\alpha|=\frac{1}{2}.$$ Can you end it now?

Answer 2

Hint

• $O, H$ are the isogonal conjugate points with respect to $\triangle ABC$.
• $ON=\dfrac{1}{2}AH$,where $N$ is the midpoint of $BC$.