Suppose $AB=3\times AD$ in triangle $ABC$, where $D$ lies on $AB,E$ lies on $AC$, and $DE$ is parallel to $BC$. Then $\frac{DE}{BC}$ equals ____.

$AB=3\times AD$ means $\frac{AD}{AB}=\frac{1}{3}$. Since $DE$ is parallel to $BC$, triangle $ADE$ is similar to triangle $ABC$, and corresponding sides are in the ratio of the sides they come from: $\frac{DE}{BC}=\frac{AD}{AB}=\frac{1}{3}$. The option $\frac{1}{9}$ is the ratio of the AREAS, not of the lengths; $\frac{1}{2}$ is $\frac{AD}{DB}$ (since $DB=2AD$), the classic BPT confusion; $\frac{2}{3}$ is $\frac{DB}{AB}$.