On the legs $AC, BC$ of a right triangle $\vartriangle ABC$ select points $M$ and $N$, respectively, so that $\angle MBC = \angle NAC$. The perpendiculars from points $M$ and $C$ on the line $AN$ intersect $AB$ at points $K$ and $L$, respectively. Prove that $KL=LB$. (O. Clurman)