Consider $Q$ such that $BDPQ$ is a rectangle. $(N, M, K, S)$ the midpoints of $(AB, BC, BD, CQ)$. $T$ the circumcenter of $\triangle HPQ$ and $R=TK\cap ON$ Facts: - $K$ belongs to the segment $BM$ and then $R$ belongs to the segment $ON$ - let $\phi$ the translation defined by $AH$, then $\phi (B,R,O,D, \triangle ABD)=(Q,T,S,P, \triangle HQP)$. - $M$ is the midpoint of $OS$ - $ \triangle ABD = \triangle HQP$ - $R$ is the circumcenter $ \triangle ABD$ and $T$ is the circumcenter $ \triangle HPQ$. -$\angle HQB=\angle BAH=\angle HCB$, then $BHCQ$ is cyclic and then $\angle BHQ=\angle BCQ\quad (1)$ Now, - $R_{\triangle HPQ}=R_{\triangle ADB}=BR\leq BO=R_{\triangle ABC}\quad (*)$ - $ROST$ is a prallelogram, then $\angle TSO=\angle TRO=\angle B$ - $\angle CSO=\angle SOC=\angle A$ - by the previous $\angle TSC=\angle A+\angle B$, then $TC\geq SC=OC=R_{\triangle ABC}\qquad (**)$ - From $(*),(**)$ we get $TC\geq R_{\triangle HPQ}$, then $C$ is outside circumcircle of $\triangle HPQ$. Consider $L=QC\cap (HPQ)$ and $L$ belongs to the segment $QC$. Now $\angle PCQ=\angle PLQ-\angle CPL<\angle PLQ=\angle PHQ$. Adding $(1)$ we get $\angle PCB<\angle PHB$