Question 50
Given the triangle $ABC$ with orthocenter $H$. The point $S$ is on the circumcircle of $AHC$ such that $\angle ASB = 90^\circ$. The point $P$ on the ray $AC$ is such that $\angle APS = \angle BAS$. Prove that the circumcircle of $BPC$, the line $CS$, and the circle with the diameter $AC$ have another common point different from $C$.