Given $\triangle ABC$ with incenter $I$ and circumcenter $O$. Let $D$ be the tangency point of $BC$ and ($I$). $AI$ cuts ($O$) at $X$. $DX$ cuts the perpendicular bisector of $AI$ at $S$. $AO$ cuts ($O$) at $K$. Prove that $OS\parallel DK$.
My thought was to show that $AD$ intersects $SO$ at $SO$'s middle point. Isn't so easy. I tried harmonic but failed.
How to deal with such middle points and perpendicular lines? I'm a bit of confused. Any idea welcomed.