Question 52
Let $AB$ be a diameter of a circle ($\omega$) with centre $O$. From an arbitrary point $M$ on $AB$ such that $MA \lt MB$ 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 $KC = NL$.