In a right-angled triangle, the median from the right angle vertex is divided into 3 parts by the inscribed circle, such that the chord of the circle is equal to the part that ends at the right angle vertex. - Calculate the acute angles of the triangle.
Let the intersection points of the median $C F$ and the inscribed circle with radius $r$ and center $O$ of the right-angled triangle $A B C$ at $C$ be $M_{1}$ and $M_{2}$, starting from $C$, the tangency point of the circle $k$ on the leg $C B$ be $E$, the projection of $O$ onto $C F$ be $G$, and the angle $F C O$ be $\varphi$. We choose the labeling of $A$ and $B$ such that $C A \leq C B$, i.e., with the usual notation for the angles, $\beta \leq \alpha$. Then, since $F C = F B$, we have $\varphi = 45^{\circ} - \beta$.

![](https://cdn.mathpix.com/cropped/2024_05_02_480269d48623f035b7c1g-1.jpg?height=358&width=1130&top_left_y=347&top_left_x=492)

According to the requirement, $C M_{1} = M_{1} M_{2} = 2 M_{1} G$, so from the right-angled triangle $O C G$,

$$
\cos \varphi = \frac{C G}{C O} = \frac{C M_{1} + M_{1} G}{C O} = \frac{3 C M_{1}}{2 C O}
$$

The triangles $C M_{1} E$ and $C E M_{2}$ are similar because the angles $C E M_{1}$ and $C M_{2} E$ are equal as inscribed angles subtending the arc $M_{1} E$ in the circle $k$. From the ratio of the corresponding sides,

$$
C M_{1} : C E = C E : C M_{2} = C E : 2 \cdot C M_{1}
$$

thus $C M_{1} = C E / \sqrt{2}$.

On the other hand, $C E O$ is an isosceles right-angled triangle, since $C O$ bisects the right angle, so $C O = C E \cdot \sqrt{2}$. Substituting these, we get

$$
\cos \varphi = 3 / 4, \quad \varphi = 41^{\circ} 24^{\prime}
$$

thus

$$
\beta = 45^{\circ} - \varphi = 3^{\circ} 36^{\prime} \quad \text{and} \quad \alpha = 45^{\circ} + \varphi = 86^{\circ} 24^{\prime} \text{.}
$$