Problem

Source: 21st Philippine Mathematical Olympiad 2019 p4

Tags: geometry, Concyclic, equal angles, circumcircle



In acute triangle $ABC $with $\angle BAC > \angle BCA$, let $P$ be the point on side $BC$ such that $\angle PAB = \angle BCA$. The circumcircle of triangle $AP B$ meets side $AC$ again at $Q$. Point $D$ lies on segment $AP$ such that $\angle QDC = \angle CAP$. Point $E$ lies on line $BD$ such that $CE = CD$. The circumcircle of triangle $CQE$ meets segment $CD$ again at $F$, and line $QF$ meets side $BC$ at $G$. Show that $B, D, F$, and $G$ are concyclic