Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=4$, $BC=6$, $CD=7$, and $DA=12$. Let $X$ and $Y$ be points on $\overline{BD}$ such that $DX/BD=1/5$ and $BY/BD=3/8$. 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 $XF\cdot XG$?

1. $XF\cdot XG=\left(1-\frac15-\frac38\right)\cdot\frac45\cdot DB^2=\frac{17}{50}DB^2$.
2. By the law of cosines (two ways, via $\angle A$): $DB^2=4^2+12^2-2(4)(12)\cos A=6^2+7^2+2(6)(7)\cos A$, giving $DB^2=120$.
3. $XF\cdot XG=\dfrac{17}{50}(120)=\boxed{\dfrac{204}5}$.