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.
Let $(WXYZ)$ denote the circumcircle of any cyclic quadrilateral $WXYZ$.
$EFCB$ is cyclic as $\angle OEA=\frac{\pi}{2}-\angle BAO=\angle ACB.$ Therefore, $PQ\|BC $ and $EF\|MN$. Also, it follows that $MNPQ$ is cyclic.
Now, $\angle DBM=\angle DCN$ as $EFCB$ is cyclic. Therefore, $DM=DN$. Similarly, $\angle DEP=\angle DFQ$ as $EFCB$ is cyclic. Consequently, $DP=DQ$. Hence, $D$ is the centre of $(MNPQ)$.
Let $MN$ intersect $PQ$ at $T$. By construction, $S$ is the radical center of $(PQEF), (MNCB)$ and $(EFCB)$. So, $SD$ is the radical axis of $(PQEF)$ and $(MNCB)$. By construction, $T$ is the radical center of $(PQEF), (MNCB)$ and $(MNPQ)$. So, $TD$ is the radical axis of $(PQEF)$ and $(MNCB)$, whence, $S,\ T, D$ are collinear. Thus, $AK\perp SD \iff AK \perp TD.$
Applying Brokard's theorem on $MNPQ$, we have, $AK \perp TD$, as $D$ is the centre of $(MNPQ)$ and $A, K$ and $T$ are the three diagonal points of the complete quadrangle $MNPQ$.
$\blacksquare$
Note that for any $E,\ F$ on $AB$ and $AC$ respectively such that $AEF\sim ACB$, the problem statement is true. The circumcentre of $ABC$ is irrelevant.