Question 116
Let $ABC$ be a triangle and $\Omega$ its circumcircle. Let $D$ be the foot of the altitude from vertex $A$. The bisector from vertex $A$ meets the segment $[BC]$ at the point $P$ and meets the circle $\Omega$ again at the point $S$. Let $A'$ be the point diametrically opposite vertex $A$ on the circle $\Omega$. Prove that the lines $(SD)$ and $(A'P)$ intersect on the circle $\Omega$.