In parallelogram ABCD, side AD is produced to a point E and BE intersects CD at F. Prove that $\triangle ABE \sim \triangle CFB$
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
In parallelogram ABCD, AB ∥ CD (opposite sides).
∠ABE = ∠CFB (alternate interior angles, since AB ∥ CF and BE is transversal)
∠AEB = ∠CBF (alternate interior angles, since AD ∥ BC, i.e., AE ∥ BC and BE is transversal)
Therefore, by AA similarity criterion:
$$\triangle ABE \sim \triangle CFB \quad \text{(Proved)}$$
Source: Chapter 6, Section 6.4 – AA Similarity Criterion
Explanation
- The key is identifying the two pairs of alternate angles using the two pairs of parallel sides of the parallelogram (AB ∥ DC and AD ∥ BC, with AD produced to E).
- ∠ABE = ∠FBC follows from AB ∥ DC with transversal BE.
- ∠AEB = ∠CBF follows from AE ∥ BC with transversal BE.
- Two angle pairs are enough → AA criterion. Write the similarity statement in correct vertex correspondence: A↔C, B↔F, E↔B.