Problem (2016G6): Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC < 90^{\circ}$. The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $AC$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $AC$ and let $\omega$ be the circumcircle of triangle $BPD$. Segments $BM$ and $DM$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $XE$ and $YF$. Prove that $PQ \perp AC$.
AlphaGeometry formalization: a b c = triangle; d = eqangle3 d a c b c a; e = on_line a c, angle_bisector a b c; f = on_line a c, angle_bisector a d c; p = on_line b e, on_line d f; m = midpoint a c; o = circumcenter b p d; x = on_line b m, on_circle o b; y = on_line d m, on_circle o b; q = on_line x e, on_line y f ? perp p q a c
HAGeo formalization: A B C = triangle; ω1 = angle_equal2 A C C B A; D = on_circle ω1; AC = line A C; l1 = angle_bisector A B C; E = intersection AC l1; l2 = angle_bisector A D C; F = intersection AC l2; BE = line B E; DF = line D F; P = intersection BE DF; M = midpoint A C; O = circumcenter B P D; BM = line B M; (BPD) = circle_center_point O B; X = intersection BM (BPD); DM = line D M; Y = intersection DM (BPD); XE = line X E; YF = line Y F; Q = intersection XE YF; PQ = line P Q; Prove: perpendicular PQ AC
Difficulty score: 4.2