This geometrical problem was proposed in a Mathematics Contest for high school students of my country. It is truly hard to find its solution.
Let $ABC$ be an acute triangle inscribed in the circle with its center $O$. The line which is perpendicular to $AO$ at $O$ intersects $AB$ and $AC$ at $E$ and $F$ respectively.
Let $D$ be the intersection point of $BF$ and $CE$. The circumscribed circle of triangle $BDC$ intersects $AB$ and $AC$ at $M$ and $N$ respectively and the circumscribed circle of triangle $DEF$ intersects $AB$ and $AC$ at $P$ and $Q$ respectively.
Let $S$ be the intersection point of $BC$ and $EF$, and $K$ be the intersection point of $PN$ and $MQ$.
Prove that $AK\perp SD$.
I am happy if someone could propose some fresh ideas to attack this problem.