In the given figure, $\dfrac{EA}{EC} = \dfrac{EB}{ED}$, prove that $\triangle EAB \sim \triangle ECD$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:23 · grounding rag
Model Answer
Proof:
Given: $\dfrac{EA}{EC} = \dfrac{EB}{ED}$
This can be rewritten as:
$$\frac{EA}{EB} = \frac{EC}{ED}$$
Also, $\angle AEB = \angle CED$ (Vertically opposite angles)
Therefore, by SAS similarity criterion, $\triangle EAB \sim \triangle ECD$.
$\blacksquare$
Source: Chapter 6, Section 6.4 (Theorem 6.5 – SAS Similarity Criterion)
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Explanation
- Key step: Rearrange the given ratio so the sides are about the included angle — i.e., sides on either side of $\angle E$ are proportional.
- Vertically opposite angles at E are the included equal angles for both triangles.
- This satisfies SAS similarity (one angle equal + sides including it proportional).
- Examiners look for: correct rearrangement of ratio, naming vertically opposite angles, and citing SAS criterion explicitly. Missing any one step costs a mark.