13. (15 points) In the figure, $A B C D$ is a parallelogram, $F$ is on $A D$, the area of $\triangle A E F = 8 \mathrm{~cm}^{2}$, the area of $\triangle D E F = 12 \mathrm{~cm}^{2}$, the area of quadrilateral $B C D F = 72 \mathrm{~cm}^{2}$, find the area of $\triangle C D E$?
【Analysis】Connect $B D$ (as shown in the figure below). If $\triangle A E F$ has $A F$ as its base and $\triangle E F D$ has $F D$ as its base, and their heights are equal, then the ratio of the bases equals the ratio of the areas, which allows us to find $A F: D F=2: 3$;

If $\triangle A B F$ and $\triangle B F D$ have $A F$ and $F D$ as their bases respectively, and their heights are the same, then $S_{\triangle A B F}=0.2 \times S \square_{A B C D}$ and $S_{\triangle B D F}=0.6 \times S_{\triangle A B D}=0.3 \times S_{\square A B C D}$;
$S_{\triangle B C D F}=S_{\triangle B F D}+S_{\triangle B C D}$, we can find $S \square_{A B C D}$;
From $S_{\triangle A B F}=0.2 \times S \square_{A B C D}$, we can find $S_{\triangle A B F}$; According to $S_{\triangle A E B}=S_{\triangle A B F}-S_{\triangle A E F}$, we can find $S_{\triangle A E B}$;
The sum of $S_{\triangle A E B}$ and $S_{\triangle E C D}$ is half the area of the parallelogram, which allows us to find $S_{\triangle E C D}$.
【Solution】
Solution: Connect $B D$ (as shown in the figure above). Given the area of $\triangle A E F = 8 \mathrm{~cm}^{2}$ and the area of $\triangle D E F = 12 \mathrm{~cm}^{2}$, we find $A F : D F = 8 : 12 = 2 : 3$;
$S_{\triangle B C D F} = S_{\triangle B F D} + S_{\triangle B C D} = 0.5 S \square_{A B C D} + 0.3 S \square_{A B C D} = 0.8 S \square_{A B C D} = 72$, so: $S \square_{A B C D} = 90$;
$S \triangle A B F = 0.2 S \square_{A B C D} = 18, S_{\triangle A B E} = S_{\triangle A B F} - S_{\triangle A E F} = 10$;
$S_{\triangle A B E} + S_{\triangle E C D} = 0.5 \times S \square_{A B C D} = 45$;
Therefore, $S_{\triangle E C D} = 45$.
Answer: The area of $S_{\triangle E C D}$ is $45 \mathrm{~cm}^{2}$.