Bisector of angle formed at the orthocentre passes through the circumcentre BdMO 2012
In an acute angled triangle $ABC$, $\angle A= 60$. We have to prove that the bisector of one of the angles formed by the 
altitudes drawn from $B$ and $C$ passes through the center of the circumcircle of the triangle $ABC$ 
Here,I tried to draw the perpendicular bisectors of sides AC and AB and then note that there is a parallelogram is formed at the centre.If we can now prove that the parallelogram is a rhombus,we will be done.I am thinking about using the properties of $30-60-90$ triangle somehow.I also found four concyclic points in the figure.A hint will be appreciated.
All credits to Mick for the picture above.
 A: Let $P$ and $Q$ be, respectively, the orthocenter and circumcenter of $\triangle ABC$. Let $R$ be the intersection of the perpendicular bisector of $\overline{AB}$ with the altitude from $B$; and let $S$ be the intersection of the perpendicular bisector of ${AC}$ with the altitude from $C$.

Introduce $B^\prime = \overleftrightarrow{AB}\cap\overleftrightarrow{QS}$ and $C^\prime = \overleftrightarrow{AC}\cap\overleftrightarrow{QR}$. Then $\triangle ABC^\prime$ and $\triangle AB^\prime C$ are equilateral.

Observe that $R$ is orthocenter, circumcenter, and incenter for $\triangle ABC^\prime$; likewise, $S$ for $\triangle AB^\prime C$. Therefore, $R$ and $S$ lie on the bisector of $\angle A$. A little angle-chasing shows that $\triangle PRS$ and $\triangle QRS$ are equiangular, hence equilateral, so that $\square PSQR$ is a rhombus, and the result follows.
A: I don't have the solution.
I am just trying to help by uploading the rough sketch according to the description in your comment.
A: 
Here, $\angle BAC = 60^\circ$. So, $\angle BHC = \angle BA'C = 180^\circ - \angle A = 120^\circ = 2\times\angle A = \angle BOC$. So, $B,C,H,O$ are concyclic.
Thus, $\angle OHC=\angle OBC = 30^\circ$ . 
Also, $\angle CHH_b = \angle CBH + \angle BCH = (90^\circ - \angle C) + (90^\circ - \angle B) = \angle A =60^\circ$
So, $\angle OHH_b = \angle OHC = 30^\circ$, proving $OH$ bisects $\angle CHH_b$.
