Question 43
In triangle $ABC$ the points $N$ and $M$ are on the side $BC$ such that $\angle BAM = \angle MAN = \angle NAC$. The points $P$ and $Q$ are on the angle bisector of $\angle BAC$ such that $A$, $P$, and $Q$ are all on the same side with respect to $BC$ and $\frac{1}{3}\angle BAC = \frac{1}{2}\angle BPC = \angle BQC$. Let $AM$ and $CQ$ meet at $E$ and $AN$ and $BQ$ meet at $F$. Prove that the common tangents of the circumcircles of $FPE$ and $FQE$ intersect on the circumcircle of $ABC$.