In Acute triangle $ABC$, $PB, PC$ are tangent lines to its circumscribed circle. $E,F$ are on line $AB, AC$ such that $\angle EPF = 2 \angle BAC$. $P$ is within $\triangle AEF$. $A'$ is the symmetrical point of $A$ with respect to $EF$. Prove that the circumscribed circle of $A'EF$ passes through a fixed point.
The condition $\angle EPF = 2 \angle BAC$ is eye-catching. It would have been interesting if $P$ is the center of a certain circle. Also $\angle BPC = 180^{o} - 2 \angle A$. But I didn't go anywhere too interesting.
