7. As shown in Figure $2, \triangle A B C$ has an area of 1, points $D, G, E,$ and $F$ are on sides $A B, A C, B C$ respectively, $B D < D A$, $D G \parallel B C, D E \parallel A C, G F \parallel A B$. Then the maximum possible area of trapezoid $D E F G$ is $\qquad$
7. $\frac{1}{3}$.

Let $\frac{A D}{A B}=x$, then $\frac{B D}{B A}=1-x=\frac{C G}{C A}$.
Thus, by the similarity of triangles, we have
$$
\frac{S_{\triangle D C}}{S_{\triangle A B C}}=x^{2}, \frac{S_{\triangle B D E}}{S_{\triangle B C}}=(1-x)^{2}=\frac{S_{\triangle C F C}}{S_{\triangle A B C}} \text {. }
$$

Since $S_{\triangle B C}=1$, we have
$$
\begin{array}{l}
S_{\text {quadrilateral DEFG }}=1-x^{2}-2(1-x)^{2} \\
=-3 x^{2}+4 x-1=-3\left(x-\frac{2}{3}\right)^{2}+\frac{1}{3} .
\end{array}
$$

When $x=\frac{2}{3}$, $S_{\text {quadrilateral DEFG }}$ reaches its maximum value $\frac{1}{3}$.