10. As shown in the figure, $\triangle A B C$ is an isosceles right triangle, $E$ is a point on side $B C$ such that $B E=3 C E$, and points $D, A, F$ are collinear. Let the height from $D$ to $B E$ in $\triangle D B E$ be $h_{1}$, and the height from $F$ to $E C$ in $\triangle F E C$ be $h_{2}$. We have $3 h_{1}+h_{2}=3$ cm. The sum of the areas of $\triangle D B E$ and $\triangle F E C$ is 6 square cm, then the area of $\triangle A B C$ is $\qquad$ square cm.
【Answer】 64

【Solution】Since $B E=3 C E \Rightarrow B E=\frac{3}{4} B C, C E=\frac{1}{4} B C$. And
$$
\begin{aligned}
S_{\triangle D B E}+S_{\triangle F E C} & =\frac{1}{2} B E \times h_{1}+\frac{1}{2} C E \times h_{2} \\
& =\frac{1}{2} \times \frac{3}{4} B C \times h_{1}+\frac{1}{2} \times \frac{1}{4} B C \times h_{2} \\
& =\frac{1}{8} B C \times\left(3 h_{1}+h_{2}\right)
\end{aligned}
$$

Substitute $\left\{\begin{array}{l}3 h_{1}+h_{2}=3 \\ S_{\triangle D B E}+S_{\triangle F E C}=6\end{array}\right.$, we get $B C=16$. Therefore, $S_{\triangle A B C}=\frac{1}{4} B C^{2}=\frac{1}{4} \times 16^{2}=64$ square centimeters.