$\angle BAD =\angle DAM \implies \Delta BAM$ is isosceles triangle, Hence, $\boxed{BD=DM=\frac{1}{2}MC}$ By angle bisector theorem, $$\frac{AD}{AC}=\frac{DM}{MC}=\frac{\tfrac{1}{2}MC}{MC}=\frac{1}{2} \implies \boxed{AC=2AD}$$Therefore, $$\cos 2\angle DAM = \frac{AD}{AC}=\frac{1}{2} \implies \boxed{\angle DAM=30^{\circ}}$$Hence, the angles of $\Delta ABC$ are $$\angle A =90^{\circ} ; \angle B=60^{\circ} ; \angle C=30^{\circ}$$

This problem was on Indian RMO Region 4, 2016 / P5