Question 45
Two circles $\omega_1(O)$ and $\omega_2$ intersect each other at $A, B$. And $O$ lies on $\omega_2$. Let $S$ be the foot of perpendicular line to $AB$ from $O$. Line $OS$ intersects $\omega_2$ for the second time at $P$. The bisector of $\widehat{ASP}$ intersects $\omega_1$ at $L$ ($A$ and $L$ are on the same side of the line $OP$). Let $K$ be a point on $\omega_2$ such that $PS = PK$ ($A$ and $K$ are on the same side of the line $OP$). Prove that $SL = KL$.
