3. We remind you that the sum of the angles in a triangle equals 180 degrees. In triangle $A B C$, angle $A$ is a right angle. Let $B M$ be the median of the triangle, and $D$ be the midpoint of $B M$. It turns out that $\angle A B D = \angle A C D$. What are these angles?
# Solution.

Draw $A D$. Since triangle $A B M$ is a right triangle, its median $A D$ equals half the hypotenuse $B M$. Therefore, $\angle A B D = \angle B A D = \alpha$, and $\angle A D M$ equals $2 \alpha$, as the exterior angle of triangle $A B D$. However, since $D M = A D$, $\angle D A M = \angle D M A = 90^{\circ} - \alpha$. This means triangle $A D C$ is a right triangle, because $\angle A C D + \angle D A C = 90^{\circ}$. Since $D M$ is the median of the right triangle $A D C$, it equals half the hypotenuse. We previously established that it also equals one of the legs of this triangle $(A D)$. Therefore, $\angle A C D$, which lies opposite this leg, equals 30 degrees.