BALKAN-SL 2015 G6

Let $AB$ be a diameter of a circle $(\omega)$ with centre $O$ . From an arbitrary point $M$ on $AB$ such that $MA < MB$ we draw the circles $(\omega_1)$ and $(\omega_2)$ with diameters $AM$ and $BM$ respectively. Let $CD$ be an exterior common tangent of $(\omega_1), (\omega_2)$ such that $C$ belongs to $(\omega_1)$ and $D$ belongs to $(\omega_2)$ . The point $E$ is diametrically opposite to $C$ with respect to $(\omega_1)$ and the tangent to $(\omega_1)$ at the point $E$ intersects $(\omega_2)$ at the points $F, G$ . If the line of the common chord of the circumcircles of the triangles $CED$ and $CFG$ intersects the circle $(\omega)$ at the points $K, L$ and the circle $(\omega_2)$ at the point $N$ (with $N$ closer to $L$ ), then prove that $KC = NL$ .