13. As shown in the right figure, point $M$ is a point on side $C D$ of parallelogram $A B C D$, and $D M: M C=1: 2$. Quadrilateral $E B F C$ is a parallelogram, and $F M$ intersects $B C$ at point $G$. If the difference in area between triangle $F C G$ and triangle $M E D$ is $13 \mathrm{~cm}^{2}$, find the area of parallelogram $A B C D$.
【Answer】 60
【Solution】
Connect $B D$
$$
\begin{array}{l}
\because D E / / B C \\
\therefore \frac{D E}{B C}=\frac{E M}{M B}=\frac{D M}{M C}=\frac{1}{2} \\
\therefore \frac{S_{\triangle D E M}}{S_{\square C E M}}=\frac{S_{\triangle C E M}}{S_{\triangle C B M}}=\frac{S_{\triangle D E M}}{S_{\triangle B D M}}=\frac{1}{2} \\
\text { Let } S_{\triangle D E M}=a
\end{array}
$$
$$
\begin{array}{l}
\text { Then } S_{\triangle C E M}=S_{\triangle B D M}=2 a, S_{\triangle C B M}=4 a \\
\therefore S_{\triangle B C F}=S_{\triangle B C E}=2+4=6 a \\
\because M B / / C F \\
\therefore \frac{C G}{G B}=\frac{C F}{M B}=\frac{E B}{M B}=\frac{3}{2} \\
\therefore \frac{S_{\square G C F}}{S_{\triangle B G F}}=\frac{C G}{G B}=\frac{3}{2} \\
\therefore S_{\square G C F}=\frac{3}{3+2} \times S_{\triangle B C F}=\frac{3}{5} \times 6=\frac{18}{5} a \\
\because S_{G G C F}-S_{\square D E M}=13 \\
\therefore \frac{18}{5} a-a=13 \\
\therefore a=5 \\
\because S_{\triangle B C D}=S_{\triangle B D M}+S_{\triangle B C M}=2 a+4 a=6 a \\
\therefore S_{\triangle A B C D}=2 \times S_{\triangle B C D}=2 \times 6 a=12 a=12 \times 5=60 \mathrm{~cm}^{2}
\end{array}
$$