8. Let the area of $\triangle A B C$ be $1, D$ be a point on side $B C$, and $\frac{B D}{D C}=\frac{1}{2}$. If a point $E$ is taken on side $A C$ such that the area of quadrilateral $A B D E$ is $\frac{4}{5}$, then the value of $\frac{A E}{E C}$ is $\qquad$.
8. $\frac{7}{3}$.

Connect $A D$. According to the problem, we have
$$
\begin{array}{l}
S_{\triangle C D E}=\frac{1}{5}, S_{\triangle A D B}=\frac{1}{3}, \\
S_{\triangle A D E}=\frac{4}{5}-\frac{1}{3}=\frac{7}{15} . \\
\text { Therefore, } \frac{A E}{E C}=\frac{S_{\triangle A D E}}{S_{\triangle C D E}}=\frac{7}{3} .
\end{array}
$$