2. On the base $AC$ of isosceles $\triangle ABC$, take a point $E$, and on the two legs $AB, BC$, take points $D, F$ respectively, such that $DE \parallel BC, EF \parallel AB$. If $BF: EF=2: 3$, what fraction of the area of $\triangle ABC$ is the area of $\triangle DEF$?
2. As shown in Figure 1, quadrilateral $B D E F$ is a parallelogram, and
$$
\begin{array}{l}
\triangle A D E \\
\sim \triangle A B C, \\
\triangle E F C \\
\sim \triangle A B C,
\end{array}
$$

the similarity ratios are respectively
$$
\begin{array}{l}
\frac{D E}{B C}=\frac{2 x}{2 x+3 x}=\frac{2}{5}, \frac{F C}{B C}=\frac{3 x}{2 x+3 x}=\frac{3}{5} . \\
\text { Therefore, } S_{\triangle D E F}=\frac{1}{2} S_{\text {quadrilateral BDEF }} \\
=\frac{1}{2}\left(S_{\triangle A B C}-S_{\triangle A D E}-S_{\triangle E F C}\right) \\
=\frac{1}{2}\left(1-\frac{4}{25}-\frac{9}{25}\right) S_{\triangle A B C} \\
=\frac{6}{25} S_{\triangle A B C} .
\end{array}
$$