Mathematics — Standard (041) — CBSE Class 10 board question

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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• 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.