Galperin G.A.

Point $P$ lies inside isosceles triangle $ABC (AB = BC)$, and $\angle ABC = 80^{\circ}, \angle PAC = 40^{\circ}$, $\angle ACP = 30^{\circ}$. Find the angle $BPC$.
Let $Q$ be the point of intersection of segment $CP$ with the height $BH$ of the given isosceles triangle, dropped to the base.

![](https://cdn.mathpix.com/cropped/2024_05_06_2f46c6bf14fbb7f478d4g-50.jpg?height=520&width=780&top_left_y=1604&top_left_x=665)

Then $\angle QAC = \angle ACQ = 30^{\circ}$, $\angle BAQ = 20^{\circ}$, $\angle BAP = 10^{\circ} = \frac{1}{2} \angle BAQ$. Therefore, $AP$ is the angle bisector of $\angle BAQ$. By the exterior angle theorem of a triangle, $\angle AQP = \angle ACQ + \angle CAQ = 60^{\circ}$, $\angle AQB = \angle QAC + \angle AHQ = 120^{\circ}$. Therefore, $QP$ is the angle bisector of $\angle AQB$. Thus, $P$ is the point of intersection of the angle bisectors of triangle $ABQ$. Consequently, $\angle BPC = \angle BPQ = 90^{\circ} + \frac{1}{2} \angle BAQ = 90^{\circ} + 10^{\circ} = 100^{\circ}$.

## Answer

$100^{\circ}$.