II. (16 points) As shown in the figure, extend the sides $AB, BC, CD, DA$ of quadrilateral $ABCD$ to $E, F, G, H$ respectively, such that $\frac{BE}{AB} = \frac{CF}{BC} = \frac{DG}{CD} = \frac{AH}{DA} = m$. If $S_{EFGH} = 2 S_{ABCD}$ ($S_{EFGH}$ represents the area of quadrilateral $EFGH$), find the value of $m$.

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Please note that the figure mentioned in the problem is not provided here.
II. Connect $G A, H B, E C, F D$ and the diagonals $A C, B D$. Then, by the given conditions, we have
$S_{\triangle H A B}=(m+1) S_{\triangle H A B}=(m+1) m S_{\triangle A B D}$.
Similarly, $S_{\triangle F C C}=(m+1) m S_{\triangle B C D}$.
Therefore, $S_{\triangle H A E}+S_{\triangle F C C}=m(m+1) S_{\triangle B C D}$.
Similarly, $S_{\triangle E B F}+S_{\triangle G D H}=m(m+1) S_{\triangle B C D D}$.
Thus, $S_{E F G H}=[2 m(m+1)+1] S_{A B C D}$.

From the given conditions, we know $2 m(m+1)+1=2$.
Solving for the positive value, we get $m=\frac{\sqrt{3}-1}{2}$.