Question 137
Let $D$ and $E$ be points on the sides $BC$ and $CA$ of the triangle $ABC$ respectively. The circumcircle of the triangle $CDE$ and the line through $C$, which is parallel to $AB$, intersect again in a point $L$. The line $DL$ intersects the side $AB$ in a point $M$. Denote by $N$ the point on the line $AB$ such that $\angle NDA = \angle MEB$ and the point $A$ lies between the points $N$ and $B$. Let $T$ be the intersection of the lines $AD$ and $BE$. Assuming $|DM| = |EN|$, prove that $CT$ is the angle bisector of $\angle ACB$.