II. (20 points) As shown, $AB$ intersects $CD$ at $E$, $DA // EF // BC$, and $AE: EB=1: 2$, $S_{\triangle ADE}=1$. Find $S_{\triangle AEF}$.
$$
\text { II. } \because A D / / B C, \therefore \triangle A E D \backsim \triangle B E C \text {, }
$$

Thus, $S_{\triangle B E C}: S_{\triangle A E D}=4: 1$.
Let $S_{\triangle A E F}=m$, since $\triangle A E F \backsim \triangle A B C, A E$
: $A B=1: 3$, we get
$$
m: S_{\triangle A B C}=1: 9, m: S_{\text { quadrilateral } E F C B}=1: 8 \text {. }
$$
$$
\text { and } S_{\text { quadrilateral } E F C B}=4+S_{\triangle C E F} \text {, }
$$
$$
=4+\frac{4}{5}(1+m) .
$$

Thus, $40 m=20+4+4 m, 36 m=24$.
$\therefore m=\frac{2}{3}$, i.e., $S_{\triangle A E F}=\frac{2}{3}$.