$E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $\triangle ABE \sim \triangle CFB$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:23 · grounding rag
Model Answer
Given: ABCD is a parallelogram. E is a point on AD produced. BE intersects CD at F.
To Prove: △ABE ~ △CFB
Proof:
Since ABCD is a parallelogram, AB ∥ CD (opposite sides).
Step 1: AB ∥ CF (since CF is part of CD)
Therefore, ∠ABE = ∠CFB (alternate interior angles, with transversal BE) ... (1)
Step 2: AD ∥ BC (opposite sides of parallelogram).
Since E lies on AD produced, AE ∥ BC.
Therefore, ∠AEB = ∠CBF (alternate interior angles, with transversal BE) ... (2)
Step 3: In △ABE and △CFB:
- ∠ABE = ∠CFB [from (1)]
- ∠AEB = ∠CBF [from (2)]
Therefore, by AA similarity criterion,
$$\triangle ABE \sim \triangle CFB \qquad \textbf{(Proved)}$$
Source: Chapter 6, Section 6.4 – Criteria for Similarity of Triangles
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Explanation
- The key is identifying the two pairs of alternate angles correctly using the two pairs of parallel sides of the parallelogram.
- ∠ABE and ∠CFB are alternate angles (AB ∥ FC, transversal BF/BE).
- ∠AEB and ∠FBC are alternate angles (AE ∥ BC, transversal BE).
- Once two pairs of angles are equal, AA criterion immediately gives similarity — no need for a third angle.
- State the correspondence carefully: A↔C, B↔F, E↔B, so write △ABE ~ △CFB in exactly that order.