1. As shown in the figure, in quadrilateral $ABCD$, points $E, F$ are on $BC, CD$ respectively, $\frac{DF}{FC}=1, \frac{CE}{EB}=2$. If the area of $\triangle ADF$ is $m$, and the area of quadrilateral $AECF$ is $n(n>m)$, then the area of quadrilateral $ABCD$ is
1. $\frac{3}{2} n+\frac{1}{2} m$

Solution: Connect $A C$. According to the area formula of a triangle, we have
$$
S_{\triangle A F C}=S_{\triangle A D F}=m,
$$
$$
\begin{array}{l}
\text { then } S_{\triangle A C E}=n-m, \\
S_{\triangle A E B}=\frac{1}{2} S_{\triangle A C E}=\frac{n-m}{2},
\end{array}
$$

Therefore, $S_{A B C D}=m+n+\frac{n-m}{2}=\frac{3}{2} n+\frac{1}{2} m$.