Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. The perpendicular bisector of segment $CH$ intersects the sides $AC$ and $BC$ in points $X$ and $Y$ , respectively. The lines $XO$ and $YO$ intersect the side $AB$ in points $P$ and $Q$, respectively. Prove that if $XP + Y Q = AB + XY$ then $\angle OHC = 90^o$.