Question 66
Let $ABC$ be a triangle with the circumcircle $\omega$. The internal angle bisectors of $\angle ABC$ and $\angle ACB$ intersect $\omega$ at $X$ and $Y$, respectively. Let $K$ be a point on $CX$ such that $\angle KAC = 90^\circ$. Similarly, let $L$ be a point on $BY$ such that $\angle LAB = 90^\circ$. Let $S$ be the midpoint of arc $CAB$ of $\omega$. Prove that $SK = SL$.