In the given figure, E is a point on the side CB produced of an isosceles triangle ABC with AB = AC. If AD $\perp$ BC and EF $\perp$ AC, then prove that $\triangle ABD \sim \triangle ECF$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:27 · grounding rag
Model Answer
In △ABD and △ECF:
Step 1: Since △ABC is isosceles with AB = AC,
∴ ∠ABC = ∠ACB, i.e., ∠ABD = ∠ECF ...(1)
Step 2: AD ⊥ BC ⟹ ∠ADB = 90°
EF ⊥ AC ⟹ ∠EFC = 90°
∴ ∠ADB = ∠EFC = 90° ...(2)
Step 3: In △ABD and △ECF,
∠ADB = ∠EFC = 90° [From (2)]
∠ABD = ∠ECF [From (1)]
By AA similarity criterion,
$$\boxed{\triangle ABD \sim \triangle ECF}$$ [Proved]
Source: Triangles, Section 6.4 (AA Similarity Criterion)
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Explanation
- The key insight is that in an isosceles triangle AB = AC implies base angles are equal: ∠ABD = ∠ACB = ∠ECF (since E is on CB produced, ∠ECF is the same as ∠ACB).
- Both triangles have a right angle (AD⊥BC and EF⊥AC), giving the second pair of equal angles.
- Two pairs of equal angles (AA) are sufficient for similarity — you do NOT need all three.
- Always state the criterion used (AA here) and write the correspondence of vertices correctly: A↔E, B↔C, D↔F.