$H$ is an orthocenter of angle $ABC$. Angle $B$ is $60^{\circ}$. Perpendicular bisectors of $AH$ and $CH$ cross line $AC$ at points $A_{1}$ and $C_{1}$. Show that the centre of $A_{1}HC_{1}$'s circumcircle lies on the angle $B$'s bisector.
I tried doing it using angle bisector's property in a triangle. I'm stuck on this one
Consider the circumcenter $M$ of triangle $\triangle AHC$. This is the intersection point of the perpendicular bisectors of $AH$ and $CH$. We have $\angle AMC = 2(180^\circ - \angle AHC) = 120^\circ$, so $M$ lies on the circumcircle of $\triangle ABC$. Since $M$ lies on the perpendicular bisector of $AC$ as well, we find that $M$ lies on the angle bisector of $\angle B$. Angle chasing gives $\angle A_1HC_1 = 60^\circ$, so $\angle A_1JC_1=120^\circ$ and $\angle A_1MC_1=60^\circ$, where $J$ is the circumcenter of $\triangle A_1HC_1$. We find that $A_1JC_1M$ is a cyclic quadrilateral. Since $|JA_1|=|JC_1|$, we have that $J$ lies on the angle bisector of $\angle MA_1C_1$. Since the lines $MA_1$, $MC_1$, $AB$ and $AC$ form a parallellogram, it follows that $B$, $J$ and $M$ are collinear, yielding the desired conclusion.