Let's determine the angles of a triangle where the height, angle bisector, and median from one of its vertices divide the angle into four equal parts.
I. solution. Let $A$ be the vertex from which the drawn notable lines divide the angle into 4 equal parts. The sides $A B$ and $A C$ are clearly of different lengths, so we can assume that $A B < A C$. Let the intersection points of the altitude, the angle bisector, and the median with side $B C$ be $M, F, S$. Due to the assumption $A B < A C$, the order of the points on side $B C$ is: $B, M, F, S, C$. Using the additional notations on the diagram, $\angle M B A = 90^{\circ} - \alpha, \angle A C M = 90^{\circ} - 3 \alpha$.

By the Law of Sines:

$$
\begin{aligned}
& \frac{A S}{B S}=\frac{\sin \left(90^{\circ}-\alpha\right)}{\sin 3 \alpha} \quad \text { and } \\
& \frac{A S}{S C}=\frac{\sin \left(90^{\circ}-3 \alpha\right)}{\sin \alpha}
\end{aligned}
$$

From these, considering that $B S = S C$,

$$
\frac{\sin \left(90^{\circ}-\alpha\right)}{\sin 3 \alpha}=\frac{\sin \left(90^{\circ}-3 \alpha\right)}{\sin \alpha}
$$

and thus $\sin \alpha \cdot \cos \alpha = \sin 3 \alpha \cdot \cos 3 \alpha$. Multiplying this equation by 2: $\sin 2 \alpha = \sin 6 \alpha$. From triangle $A M C$, we can see that $3 \alpha < 90^{\circ}$, so $\alpha < 30^{\circ}$, and from the latter equation, $6 \alpha = 2 \alpha$ or $6 \alpha = 180^{\circ} - 2 \alpha$, and thus $\alpha = 0^{\circ}$ or $\alpha = 22.5^{\circ}$. The solution $\alpha = 0^{\circ}$ is not valid. The angles of the triangle are: $90^{\circ}, 67.5^{\circ}$, and $22.5^{\circ}$.

II. solution. It is known that one of the angle bisectors of a triangle and the perpendicular bisector of the opposite side intersect on the circumcircle. On the diagram, this point is $T$. It is easy to show that $\angle F T S = \angle M A F = \alpha$, so triangle $A T S$ is isosceles. Then the perpendicular bisector of $A T$ passes through point $S$. Since the perpendicular bisector of $B C$ also passes through $S$, point $S$ is the center of the circumcircle. Therefore, by Thales' theorem, $4 \alpha = 90^{\circ}$, and thus the angles of the triangle are $90^{\circ}; 67.5^{\circ}$, and $22.5^{\circ}$.