Problem 3. In triangle $ABC, \measuredangle ACB=40^{\circ}$. The bisectors of the interior and exterior angles at vertex $C$ intersect the line $AB$ at points $D$ and $E$, such that $\triangle CDE$ is isosceles. Determine the angles of triangle $ABC$.

---

The problem states that in triangle $ABC$, the angle $\measuredangle ACB$ is $40^{\circ}$. The angle bisectors of the interior and exterior angles at vertex $C$ intersect the line $AB$ at points $D$ and $E$, respectively, forming an isosceles triangle $\triangle CDE$. We need to determine the angles of triangle $ABC$.
Solution. Let $CD$ be the bisector of the exterior angle, and $CE$ the bisector of the interior angle at vertex $C$. The bisectors $CD$ and $CE$ are perpendicular, meaning they are the bisectors of two consecutive angles, $\measuredangle DCE = 90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_9d1696c7fd1a2cbf5bb5g-08.jpg?height=421&width=1029&top_left_y=640&top_left_x=377)

From the condition, $\triangle CDE$ is isosceles, so it follows that it is an isosceles right triangle with hypotenuse $DE$. Therefore, $\measuredangle DEC = 45^{\circ}$. This is an exterior angle for $\triangle CEB$, i.e., $45^{\circ} = 20^{\circ} + \beta, \beta = 25^{\circ}$, and the angle $\alpha = 115^{\circ}$.