Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=4$, $BC=3$, $CD=5$, and $DA=6$. Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD}=\frac{1}{3}$ and $\frac{BY}{BD}=\frac{1}{6}$. Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$. Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$. Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$. What is $XF\cdot XG$?