4. In triangle $ABC$, the angle bisectors $\mathrm{AK}$ and $\mathrm{CL}$ and the median $\mathrm{BM}$ are drawn. It turns out that $\mathrm{ML}$ is the bisector of angle AMB, and $\mathrm{MK}$ is the bisector of angle CMB. Find the angles of triangle ABC.
Answer: $30^{\circ}, 30^{\circ}, 120^{\circ}$.

Solution. By the property of the angle bisector, AM:MB=AL:LB=AC:CB. But AC=2AM, so $C B=2 M B$. Similarly, $A B=2 M B$, and thus, triangle $A B C$ is isosceles. But then $B M$ is the altitude, triangle $B M C$ is right-angled, and its leg BM is half of its hypotenuse BC. Therefore, angle C is $30^{\circ}$ (and equal to angle A), and then angle B is $120^{\circ}$.

Grading. Full solution - 7 points.