The answers to this question imply that a line which bisects two sides of a triangle must be parallel to the third side. Why is this true? There must be a simple proof.
More generally: Let D and E be points on $\overline{AB}$ and $\overline{CB}$, respectively, such that $AD:DB=CE:EB$. Then, $\overleftrightarrow{DE} \parallel \overline{AC}$. (Thanks @Isaac)
We could also look at the contrapositive. Suppose that $DF$ is not parallel to $AC$. Then the triangles $DBF$ and $ABC$ won't be similar.