To the exterior of side $AB$ of square $ABCD$, we have drawn the regular triangle $ABE$. Point $A$ reflected on line $BE$ is $F$, and point $E$ reflected on line $BF$ is $G$. Let the perpendicular bisector of segment $FG$ meet segment $AD$ at $X$. Show that the circle centered at $X$ with radius $XA$ touches line$ FB$.