BALKAN-SL 2018 G6

In a triangle $A B C$ with $A B=A C, \omega$ is the circumcircle and $O$ its center. Let $D$ be a point on the extension of $B A$ beyond $A$. The circumcircle $\omega_{1}$ of triangle $O A D$ intersects the line $A C$ and the circle $\omega$ again at points $E$ and $G$, respectively. Point $H$ is such that $D A E H$ is a parallelogram. Line $E H$ meets circle $\omega_{1}$ again at point $J$. The line through $G$ perpendicular to $G B$ meets $\omega_{1}$ again at point $N$ and the line through $G$ perpendicular to $G J$ meets $\omega$ again at point $L$. Prove that the points $L, N, H, G$ lie on a circle. (Cyprus)