13. (15 points) As shown in the figure, $A B C D$ is a parallelogram, $A M=M B, D N=C N, B E=E F=F C$, the area of quadrilateral $E F G H$ is 1, find the area of parallelogram $A B C D$.


【Analysis】As shown in the figure, draw $E Q / / C D, F P / / C D$, intersecting $B N$ at points $Q$ and $P$, respectively. Then, determine the areas of $\triangle H E Q$, quadrilateral $E F P Q$, and $\triangle F G P$ as fractions of the area of parallelogram $A B C D$. This will help in finding the area of quadrilateral $E F G H$ as a fraction of the area of parallelogram $A B C D$. Finally, using the concept of division of fractions, divide the area of quadrilateral $E F G H$ by the fraction it represents of the area of parallelogram $A B C D$ to find the area of parallelogram $A B C D$.

【Solution】Solution: As shown in the figure,
draw $E Q / / C D, F P / / C D$, intersecting $B N$ at points $Q$ and $P$,
since $\frac{\mathrm{EQ}}{\mathrm{BM}}=\frac{\mathrm{EQ}}{\mathrm{CN}}=\frac{\mathrm{BE}}{\mathrm{BC}}=\frac{1}{3}$,
so $\frac{\mathrm{EH}}{\mathrm{HM}}=\frac{\mathrm{OE}}{\mathrm{BM}}=\frac{1}{3}$;
since the area of $\triangle B E M$ is:
$$
\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{12},
$$

the area of $\triangle H E Q$ as a fraction of the area of parallelogram $A B C D$ is:
$$
\begin{array}{l}
\frac{1}{12} \times \frac{3}{3+1} \times\left(\frac{1}{3}\right)^{2} \\
=\frac{1}{12} \times \frac{3}{4} \times \frac{1}{9} \\
=\frac{1}{144}
\end{array}
$$

since $\frac{\mathrm{FP}}{\mathrm{ND}}=\frac{\mathrm{FP}}{\mathrm{CN}}=\frac{2}{3}$,
the area of $\triangle B F P$ as a fraction of the area of $\triangle B C N$ is: $\left(\frac{2}{3}\right)^{2}=\frac{4}{9}$,
so the area of quadrilateral $E F P Q$ as a fraction of the area of parallelogram $A B C D$ is:
$$
\begin{array}{l}
\left(\frac{1}{2} \times \frac{1}{2}\right) \times \frac{4}{9} \times\left[1-\left(\frac{1}{2}\right)^{2}\right] \\
=\frac{1}{4} \times \frac{4}{9} \times \frac{3}{4} \\
=\frac{1}{12}
\end{array}
$$

since $\frac{\mathrm{FP}}{\mathrm{ND}}=\frac{\mathrm{FG}}{\mathrm{GD}}=\frac{2}{3}$,
the area of $\triangle F G P$ as a fraction of the area of parallelogram $A B C D$ is:
$$
\begin{aligned}
& \left(\frac{1}{3} \times \frac{1}{2}\right) \times \frac{3}{3+2} \times \frac{1}{2} \times\left(\frac{2}{3}\right)^{2} \\
= & \frac{1}{6} \times \frac{3}{5} \times \frac{1}{2} \times \frac{4}{9} \\
= & \frac{1}{45}
\end{aligned}
$$

so the area of parallelogram $A B C D$ is:
$$
\begin{array}{l}
1 \div\left(\frac{1}{144}+\frac{1}{12}+\frac{1}{45}\right) \\
=1 \div \frac{9}{80} \\
=8 \frac{8}{9}
\end{array}
$$

Answer: The area of parallelogram $A B C D$ is $8 \frac{8}{9}$.