2. In an acute-angled triangle $A B C$, the altitude $B H$ and the bisector $A P$ are drawn. Find the angles of this triangle, given that a circle can be circumscribed around quadrilateral $A B P H$, and $\angle C P H=20^{\circ}$.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_0bfd185293c1e6faf57cg-1.jpg?height=820&width=1053&top_left_y=1549&top_left_x=570)

The angle $\angle A H B$ is an inscribed and right angle, so $A B$ is the diameter of the circle circumscribed around quadrilateral $A B P H$. Therefore, the angle $\angle A P B$ is also a right angle. Since $\angle C P H=20^{\circ}$, then $\angle A P H=70^{\circ}$. From this, we find that $\angle A B H=70^{\circ}$, which subtends the same arc as $\angle A P H$. Now, in the right triangle $A B H$, we know two angles, so we can find the angle $\angle B A H$. We get $\angle B A H=20^{\circ}$. Since $A P$ is the bisector of this angle, it follows that $\angle B A P=\angle H A P=10^{\circ}$. The angles $\angle H A P$ and $\angle H B P$ subtend the same arc, so $\angle H B P=10^{\circ}$. We can find the remaining angles of the triangle. $\angle A B C=\angle A B H+\angle H B P=70^{\circ}+10^{\circ}=80^{\circ}$. In triangle $A B C$, the bisector $A P$ coincides with the altitude, so it is isosceles, and $\angle A B C=\angle A C B=80^{\circ}$. Answer. $\angle A B C=\angle A C B=80^{\circ}, \angle B A C=20^{\circ}$.