58th Ukrainian National Mathematical Olympiad

We can cut $△ABC$ with $\angle BAC=30^{\circ }$ into triangles as shown in Fig. 25.



Here $BC=CD$ and $△DBC$ is isosceles with right angle. Then $BN=MD$ and $△DBM$ is isosceles with angle $\angle DMB=150^{\circ }$. Thus $\angle DMA=30^{\circ }$, $△ADM$ is isosceles with $\angle ADM=120^{\circ }$.