Q1. [2] § 6.4 Criteria for Similarity of Triangles
In $\triangle ABC$, DE $\parallel$ BC. If AD = $x$, DB = $x - 2$, AE = $x + 2$ and EC = $x - 1$, then find the value of $x$.
Previously asked in CBSE board exam
2026 30/1/1 Q22(A)
Generated by claude-sonnet-4-6 · 2026-06-15 10:25 · grounding rag
Model Answer
Since DE ∥ BC, by Basic Proportionality Theorem (Thales' Theorem):
$$\frac{AD}{DB} = \frac{AE}{EC}$$
$$\frac{x}{x-2} = \frac{x+2}{x-1}$$
Cross-multiplying:
$$x(x-1) = (x+2)(x-2)$$
$$x^2 - x = x^2 - 4$$
$$-x = -4$$
$$x = 4$$
Source: Chapter 6, Section 6.3 (Theorem 6.1 — Basic Proportionality Theorem)
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Explanation
- The key theorem here is Theorem 6.1 (BPT/Thales' Theorem): If DE ∥ BC, then AD/DB = AE/EC.
- Substitute the given expressions and cross-multiply to get a linear equation (the $x^2$ terms cancel).
- Examiners expect you to state BPT before applying it — don't skip that step.
- Always verify: $x = 4$ gives AD = 4, DB = 2, AE = 6, EC = 3 → 4/2 = 6/3 = 2 ✓
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