Key idea: Basic Proportionality Theorem (BPT / Thales)
**Theorem**: If a line is drawn parallel to one side of a triangle, intersecting the other two sides at distinct points, then it divides them in the same ratio.
**Setup**: ΔABC. Line DE ∥ BC, with D on AB and E on AC.
**Conclusion**: AD/DB = AE/EC.
**Quick proof sketch**:
Draw DE ∥ BC. Triangles ADE and ABC have ∠A common and ∠ADE = ∠ABC (corresponding angles, DE ∥ BC). So ΔADE ~ ΔABC by AA.
Thus AD/AB = AE/AC.
From AD/AB = AE/AC, you can derive AD/DB = AE/EC by simple algebra.
**Converse**: if AD/DB = AE/EC, then DE ∥ BC.
Useful for proving lines parallel and finding lengths in triangles split by parallels.