Answer: A

Small Hint:

The parallels give $\triangle XEY\sim\triangle XAD$ and $\triangle XEF\sim\triangle XAC$

Big Hint:

Combine those ratios with Power of a Point to get $XF\cdot XG=XB\cdot XY;$ find $BD^2$ via Law of Cosines

Solution:

Because $YE\parallel AD$ and $EF\parallel AC,$ we get $\triangle XEY\sim\triangle XAD$ and $\triangle XEF\sim\triangle XAC,$ giving $\dfrac{XY}{XE}=\dfrac{XD}{XA}$ and $\dfrac{XF}{XE}=\dfrac{XC}{XA}.$ Hence $\dfrac{XC}{XD}=\dfrac{XF}{XY},$ so $XF\cdot XD=XC\cdot XY.$

Power of a Point at $X$ gives $XC\cdot XG=XD\cdot XB,$ and combining yields $XF\cdot XG=XB\cdot XY.$ With $d=BD,$ $DX=\dfrac14 d$ and $BY=\dfrac{11}{36}d,$ so $$\begin{aligned} XF\cdot XG &=\left(d-\tfrac14 d\right) \\ &\quad {}\cdot\left(d-\tfrac14 d-\tfrac{11}{36}d\right) \\ &=\dfrac34 d\cdot\dfrac49 d=\dfrac{d^2}{3}. \end{aligned}$$

Since $ABCD$ is cyclic, $\angle BAD$ and $\angle BCD$ are supplementary. The Law of Cosines on $\triangle ABD$ and $\triangle CBD$ gives $\dfrac{73-d^2}{48}=\dfrac{d^2-40}{24},$ so $d^2=51.$ Therefore $XF\cdot XG=\dfrac{51}{3}=17.$

Thus, the correct answer is A.