Problem

Source: Serbia 2018 TST day 2 p5

Tags: geometry, Inversion



Let $H $ be the orthocenter of $ABC $ ,$AB\neq AC $ ,and let $F $ be a point on circumcircle of $ABC $ such that $\angle AFH=90^{\circ} $.$K $ is the symmetric point of $H $ wrt $B $.Let $P $ be a point such that $\angle PHB=\angle PBC=90^{\circ} $,and $Q $ is the foot of $B $ to $CP $.Prove that $HQ $ is tangent to tge circumcircle of $FHK $.