Question 139
Let $D$ be an arbitrary point inside $\triangle ABC$. Let $\Gamma$ be the circumcircle of $\triangle BCD$. The external angle bisector of $\angle ABC$ meets $\Gamma$ again at $E$. The external angle bisector of $\angle ACB$ meets $\Gamma$ again at $F$. The line $EF$ meets the extension of $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that the circumcircles of $\triangle BFP$ and $\triangle CEQ$ always pass through the same fixed point regardless of the position of $D$. (Assume all the labelled points are distinct.)