In the interior of triangle $ABC$, point $P$ is positioned such that $\angle PAC = 10^{\circ}, \angle PCA = 20^{\circ}, \angle PAB = 30^{\circ}$, and $\angle ABC = 40^{\circ}$. What is the measure of $\angle BPC$ in degrees?
The triangle ABC is clearly isosceles. Let's draw its axis of symmetry $t$, and reflect point $P$ over $t$, with its reflection being $P^{\prime}$. Due to the reflection, $\angle ACP = \angle BCP^{\prime} = 20^{\circ}$. From this, using the fact that $\angle ACB = 100^{\circ}$, we get $\angle PCP^{\prime} = 60^{\circ}$. Since $CP = CP^{\prime}$, it follows that triangle $CPP^{\prime}$ is equilateral. Let the intersection of line $BP^{\prime}$ with $t$ be $E$. $\angle CP^{\prime}E = 30^{\circ}$, since it is the exterior angle of triangle $CP^{\prime}B$. Therefore, $BP^{\prime}$ is the angle bisector in the equilateral triangle $CPP^{\prime}$, and thus also a line of symmetry, so $\angle P^{\prime}CB = \angle P^{\prime}PB = 20^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e5b60acfbdf78e0882eag-1.jpg?height=401&width=832&top_left_y=382&top_left_x=641)

For the desired $\angle CPB$, we thus have

$$
\angle CPB = \angle CPP^{\prime} + \angle P^{\prime}PB = 60^{\circ} + 20^{\circ} = 80^{\circ}
$$