Let $ABCD$ be a parallelogram. A line $\ell$ intersects lines $AB,~ BC,~ CD, ~DA$ at four different points $E,~ F,~ G,~ H,$ respectively. The circumcircles of triangles $AEF$ and $AGH$ intersect again at $P$. The circumcircles of triangles $CEF$ and $CGH$ intersect again at $Q$. Prove that the line $P Q$ bisects the diagonal $BD$.
Problem
Source: Saudi Arabia BMO TST Day II Problem 3
Tags: geometry, circumcircle, radical axis, geometry unsolved