4. In triangle $ABC$, the bisector of angle $C$ intersects side $AB$ at point $M$, and the bisector of angle $A$ intersects segment $CM$ at point $T$. It turns out that segments $CM$ and $AT$ divide triangle $ABC$ into three isosceles triangles. Find the angles of triangle $ABC$.
Solution. Since the sum of angles $A$ and $C$ of triangle $ABC$ is less than $180^{\circ}$, then $\angle TAC + \angle TCA < 90^{\circ}$, so angle $\angle ATC$ is obtuse (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_5378193a6b9f83c6a864g-2.jpg?height=405&width=554&top_left_y=240&top_left_x=814)

This means that in the isosceles triangle $ATC$, side $AC$ is the base. Therefore, $\angle TAC = \angle TCA = \alpha$, so $\angle BAC = \angle BCA = 2\alpha$. Angle $\angle ATM$ is the exterior angle of triangle $ATC$, hence $\angle ATM = 2\alpha$.

Triangle $ATM$ is also isosceles. If $T$ is its vertex, then $\angle TMA = \angle TAM = \alpha$, so the sum of the angles of this triangle is $4\alpha$. But $4\alpha$ is the sum of angles $A$ and $C$ of the original triangle, which is less than $180^{\circ}$. Therefore, $TM$ is the base of triangle $ATM$, and the sum of its angles is $5\alpha$. From this, $\alpha = 36^{\circ}$.

Triangle $CMB$ also turns out to be isosceles, as $\angle MCB = \angle MBC = 36^{\circ}$. Answer: $36^{\circ}, 72^{\circ}, 72^{\circ}$.