Here's a ratio chasing proof: We can easily obtain $\angle C_1HP=\angle C_1HQ$, $\angle HQA_1=\angle C_1A_1H=\angle HA_1B_1$, $\angle HPB_1=\angle HB_1A_1$. Let the line perpendicular to $A_1B_1$ meet $PQ$ at $J$. Thus $J$ is the midpoint of $PQ$ $\iff$ $[PJC]=[QJC]$ $\iff$ $\frac {sin\angle PCJ}{sin\angle QCJ}=\frac {CQ}{CP}$. Since $HB_1\perp CP, HA_1\perp CQ$. By the law of sines, we have $\frac {CQ}{CP}=\frac {sin\angle CHQ}{sin\angle PHC}*\frac {sin\angle HPC}{sin\angle HQC}=\frac {sin\angle HB_1A_1}{sin\angle HA_1B_1}=\frac {sin\angle PCJ}{sin\angle QCJ}$ and we're done. Q.E.D Synthetic: Like the proof above, we can easy obtain $\angle C_1HP=\angle C_1HQ=\angle PCQ$. Let $CH$ meet $(PHQ)$ at $H,X$. Thus $PX=QX$, $\angle XPQ=\angle XQP=\angle XHP=\angle PCQ$ , which means that $X$ is the intersection of the tangents of $(CPQ)$ through $P,Q$. Hence $CX$ is the $C$-symmedian of $CPQ$. Since $CH$ and the line described in the problem are isogonals, thus that line bisects $PQ$. Q.E.D In my opinion the two proofs are not that different...

Let tangents at $P$ and $Q$ to circumcircle of triangle $CPQ$ intersect at point $S$. We must prove that $H$ lies on symmedian of triangle $CPQ$ or $H\in CS$. By angle chasing points $P$, $Q$, $S$ and $H$ lie on the circle. So \[\angle PHS+\angle PHB_{1}+\angle CHB_{1}=\angle ACB+90^{\circ}-\angle C_{1}AH+\angle BAC=180^{\circ},\] thus $H\in CS$.