12. (15 points) As shown in the figure, point $M$ is a point on side $CD$ of parallelogram $ABCD$, and $DM: MC=1: 2$. Quadrilateral $EBFC$ is a parallelogram, and $FM$ intersects $BC$ at point $G$. If the difference in area between triangle $FCG$ and triangle $MED$ is $13 \mathrm{~cm}^{2}$, find the area of parallelogram $ABCD$.
【Analysis】To find the area of the parallelogram, we need to determine the area difference 13 corresponding to the area ratio. We need to establish a relationship between these two triangles. First, we need to know that in $D E M B C$, it is an hourglass model, and $B C F G M$ is also an hourglass model. The two parallelograms have the same base and height, making them two equal-area parallelograms. $\triangle B C F$ is half the area of parallelogram $A B C D$.

【Solution】Solution: Connect $B D$
According to the proportional sides of the hourglass model:
$$
\begin{array}{l}
\because D E / / B C \\
\therefore \frac{\mathrm{DE}}{\mathrm{BC}}=\frac{\mathrm{EM}}{\mathrm{MB}}=\frac{\mathrm{DM}}{M C}=\frac{1}{2} \\
\therefore \frac{S_{\triangle \mathrm{DEM}}}{S \triangle C E M}=\frac{S_{\triangle D E M}}{S_{\triangle C E M}}=\frac{S_{\mathrm{DEM}}}{S_{\triangle \mathrm{BDM}}}=\frac{1}{2}
\end{array}
$$

Let $S_{\triangle D E M}=a$
Then $S_{\triangle C E M}=S_{\triangle B D M}=2 a, S_{\triangle C B M}=4 a$
$$
\begin{array}{l}
\therefore S_{\triangle B C F}=S_{B C E}=2+4=6 a \\
\because M B / / C F \\
\therefore \frac{C G}{G B}=\frac{C F}{M B}=\frac{E B}{M B}=\frac{3}{2} \\
\therefore \frac{S_{\triangle G C F}}{S_{\triangle B G F}}=\frac{C G}{G B}=\frac{3}{2}
\end{array}
$$
$$
\begin{array}{l}
\therefore S_{G C F}=\frac{3}{3+2} \times S_{\mathrm{BCF}}=\frac{3}{5} \times 6=\frac{18}{5} a \\
\because S_{\triangle G C F}-S_{\triangle D E M}=13 \\
\therefore \frac{18}{5} a-\mathrm{a}=13 \\
\therefore a=5 \\
\because S_{\triangle B C D}=S_{\triangle B D M}+S_{\triangle B C M}=2 a+4 a=6 a \\
\therefore S \square A B C D=2 \times S_{B C D}=2 \times 6 a=12 a=12 \times 5=60 \mathrm{~cm}^{2}
\end{array}
$$

Answer: The area of the parallelogram is $60 \mathrm{~cm}^{2}$.