13. (15 points) As shown in the figure, the area of square $ABCD$ is $1$, $M$ is the midpoint of side $CD$, and $E, F$ are two points on side $BC$ such that $BE=EF=FC$. Connecting $AE$ and $DF$ intersect $BM$ at $H$ and $G$ respectively. Find the area of quadrilateral $EFGH$.

保留源文本的换行和格式，直接输出翻译结果如下：

13. (15 points) As shown in the figure, the area of square $ABCD$ is $1$, $M$ is the midpoint of side $CD$, and $E, F$ are two points on side $BC$ such that $BE=EF=FC$. Connecting $AE$ and $DF$ intersect $BM$ at $H$ and $G$ respectively. Find the area of quadrilateral $EFGH$.
【Analysis】Draw $M Q$ parallel to $B C$ intersecting $D F$ at $Q$, and draw $E P$ parallel to $A B$ intersecting $B M$ at $P$. Using the proportional relationships between line segments, find the ratio of the areas of the triangles, and finally determine the area of the shaded part.

【Solution】According to the analysis, as shown in the figure, draw $M Q$ parallel to $B C$ intersecting $D F$ at $Q$, and draw $E P$ parallel to $A B$ intersecting $B M$ at $P$,
$\because M$ is the midpoint of $C D$, so $Q M: P C=1: 2, \therefore Q M: B F=1: 4$, so $G M: G B=1: 4$, $\therefore B G: B M=4: 5$; also, since $B F: B C=2: 3, S_{\triangle \mathrm{BFG}}=\frac{4}{5} \times \frac{2}{3} \mathrm{~S}_{\triangle \mathrm{ABC}}=\frac{2}{15}$;
$\because E$ is the trisection point on side $B C$, so $E P: C M=1: 3, \therefore E P: A B=1: 6$,
$\therefore B H: H P=6: 1, \therefore B H: H M=6: 15=2: 5, B H: B G=2: 7$,
also $\because G M: \quad G B=1: 4, \therefore B H: B G=5: 14, \therefore \mathrm{S}_{\triangle \mathrm{BEH}}=\frac{5}{14} \times \frac{1}{2} \times \mathrm{S}_{\triangle \mathrm{BFG}}=\frac{1}{42}$,

Therefore, the answer is: $\frac{23}{210}$.