Let $ABCD$ be a square inscribed in circle $K$. Let $P$ be a point on the small arc $CD$ of circle $K$. The line $PB$ intersects $AC$ in $E$. The line $PA$ intersects $DB$ in $F$. The circle circumscribed to triangle $PEF$ intersects for second time $K$ in $Q$. Prove that $PQ$ is parallel to $CD$.
Denote $AQ \cap BD$ as $G$, and the circumcircle of $PEF$ as $O$. Then we have $G \in O$. This is because $\angle AGB = 135 - \angle GAB = 45 + \angle DAQ = \angle APQ$.
Now we must have $\angle BGE = \angle ABG = \angle BAE = 45$ so $BAGE$ is an isoceles trapezoid. Therefore, $BE = AG$ and $BG = AE$. Furthermore, $GE \parallel CD$ and $\angle ECD = \angle GDC = 45$ so $GECD$ is an isosceles trapezoid. Therefore, we also obtain $GD = EC$.
By POP, we obtain $AG \times GQ = DG \times GB$ and $AE \times EC = PE \times EB$. So $GQ = \frac{DG \times GB}{AG} = \frac{CE \times EA}{BE} = PE$.
Therefore, since $GQPE \in O$, we must have $PQ \parallel GE \parallel CD$.