Parallel lines and proportional parts

The converse: if D is on AB and E is on AC and AD/DB = AE/EC, then DE ∥ BC. Proof: add 1 to both sides, so AB/DB = AC/EC and then AD/AB = AE/AC. The angle A is common, so △ADE ~ △ABC by SAS. Therefore ∠ADE = ∠ABC, and these are corresponding angles, so DE ∥ BC. To test it, compute both ratios and compare them; both must compare the pieces on the same side of the vertex, for example AD/DB and AE/EC (the upper piece to the lower piece on each side). Mixing AD/DB with AE/AC proves nothing. A special case: when D and E are midpoints, both ratios are 1, so DE ∥ BC and DE = BC/2 (the midsegment theorem). Parallel lines also pass the ratio on from one triangle to the next, so we can follow a chain of parallel lines through a quadrilateral.