[ The bisector divides the arc in half ]

It is known that in a certain triangle, the median, bisector, and altitude drawn from vertex $C$ divide the angle into four equal parts. Find the angles of this triangle.

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Let $\alpha=\angle A<\angle B$. According to the previous problem, $\angle C=90^{\circ}$. The median $C M$ divides triangle $A B C$ into two isosceles triangles. $\angle A C M=\angle A=\alpha, \angle M C B=3 \alpha$, hence, $\alpha+3 \alpha=90^{\circ}$, i.e., $\alpha=22.5^{\circ}$. Therefore, $\angle A=22.5^{\circ}, \angle B=67.5^{\circ}, \angle C=90^{\circ}$.