Triangle Side-Splitters and Parallels

Suppose we have a triangle $\triangle ABC$ with $D$ on $AB$ and $E$ on $AC$ such that $DE\parallel BC$. The Side-splitter theorem states that $\frac{AD}{AB}=\frac{AE}{AC}$.

Here's the proof: Since parallel lines imply congruent angles, $\angle AED=\angle ACB$. By AA similarity, $\triangle ADE\sim\triangle ABC$, so $\frac{AD}{AB}=\frac{AE}{AC}$.