Problem

Source: BMO 2018 Shortlist G6

Tags: geometry, circumcircle, parallelogram, cyclic quadrilateral, moving points



In a triangle $ABC$ with $AB=AC$, $\omega$ is the circumcircle and $O$ its center. Let $D$ be a point on the extension of $BA$ beyond $A$. The circumcircle $\omega_{1}$ of triangle $OAD$ intersects the line $AC$ and the circle $\omega$ again at points $E$ and $G$, respectively. Point $H$ is such that $DAEH$ is a parallelogram. Line $EH$ meets circle $\omega_{1}$ again at point $J$. The line through $G$ perpendicular to $GB$ meets $\omega_{1}$ again at point $N$ and the line through $G$ perpendicular to $GJ$ meets $\omega$ again at point $L$. Prove that the points $L, N, H, G$ lie on a circle.