Question 128
An arbitrary point $P$ lies on side $BC$ of triangle $ABC$. Angle bisectors of $\widehat{APB}$ and $\widehat{APC}$ intersect the external angle bisector of $\widehat{A}$ at $X$ and $Y$, respectively. Circumcircle of triangle $PXY$ meets $BC$ for the second time at $Q$. Prove that $\widehat{BAP} = \widehat{CAQ}$.