59th Ukrainian National Mathematical Olympiad

Since AL is a bisector, then $\angle CAL=\angle LAB=\angle CBA$ (Fig. 4). Then $△ALB$ is isosceles, so LM is its altitude and a median. Thus $\angle LMA=90^{\circ }$. Consider the triangles AML and ALC. Let C' be the projection of L on AC. Then right triangles AML and AC'L are equal by hypothenuse and the angle. Thus, $L{C}^{\prime }=LM=LC$. Therefore, $C=C^{\prime }$, since there exists only one projection.

Thus $△ABC$ is a right triangle for which $2AC=AB$, hence $\angle ABC=30^{\circ }$.