Problem 3. In triangle $ABC$, the angle bisectors of the angles at vertices $A$ and $B$ intersect at an angle of $140^{\circ}$. The altitude and the angle bisector drawn from the angle at vertex $C$ form an angle of $20^{\circ}$. Determine the angles of the triangle.
Solution. From triangle $ABF$ we have that $\frac{\alpha}{2}+\frac{\beta}{2}+140^{\circ}=180^{\circ}$, i.e., $\frac{\alpha}{2}+\frac{\beta}{2}=40^{\circ}$. From here $\alpha+\beta=80^{\circ}$, so $\gamma=180^{\circ}-80^{\circ}=$ $100^{\circ}$. Since $\angle ECA=\frac{\gamma}{2}=50^{\circ}$ and the angle formed by the angle bisector and the altitude drawn from vertex C follows that

$$
\angle DCA=\angle ECA-\angle ECD=30^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_06e08f885cfce663e921g-06.jpg?height=315&width=483&top_left_y=1164&top_left_x=996)

From the right triangle ADC we get that

$$
\alpha=180^{\circ}-\left(90^{\circ}+30^{\circ}\right)=60^{\circ}
$$

The sum of the angles in a triangle is $180^{\circ}$, from which it follows that

$$
\beta=180^{\circ}-(\alpha+\gamma)=20^{\circ}
$$