1. Problem statement: In triangle ABC a line intersects sides AB and AC at D and E respectively and is parallel to BC. 2. Goal: Prove that $\frac{AD}{AB} = \frac{AE}{AC}$. 3. Because DE is parallel to BC, corresponding angles are equal, so $\angle ADE = \angle ABC$ and $\angle AED = \angle ACB$. 4. Triangles ADE and ABC have two pairs of equal angles, hence they are similar by the AA similarity criterion. 5. From the similarity of triangles ADE and ABC we get equality of corresponding side ratios, in particular $\frac{AD}{AB} = \frac{AE}{AC}$. 6. Therefore $\frac{AD}{AB} = \frac{AE}{AC}$, as required. 7. Final answer: $\frac{AD}{AB} = \frac{AE}{AC}$.