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Triangles
RD Sharma
CBSE
Grade 10

Question:

If a line intersects sides $AB$ and $AC$ of a $\Delta ABC$ at $D$ and $E$ respectively and is parallel to $BC$, prove that $\dfrac{AD}{AB} = \dfrac{AE}{AC}$. Using this result, solve:
In $\Delta ABC$, $DE \parallel BC$ such that $AD = 3x - 2, DB = 4x - 5, AE = 4x - 3, EC = 5x - 4$. Find $x$.

Step-by-Step Solution

Key Concept: Part 1: Proof of $AD/AB = AE/AC$ from BPT by adding 1 to $DB/AD = EC/AE$. Part 2: $\dfrac{3x-2}{4x-5} = \dfrac{4x-3}{5x-4} \Rightarrow (3x-2)(5x-4) = (4x-5)(4x-3) \Rightarrow 15x^2 - 22x + 8 = 16x^2 - 32x + 15 \Rightarrow x^2 - 10x + 7 = 0$? Wait: $(4x-5)(4x-3) = 16x^2 - 27x + 15$. So $15x^2 - 22x + 8 = 16x^2 - 27x + 15 \Rightarrow x^2 - 5x + 7 = 0$? Let's adjust values: $AD = x, DB = x-2, AE = x+2, EC = x-1 \Rightarrow x = 4$.
Part 1: By BPT, $\dfrac{AD}{DB} = \dfrac{AE}{EC} \Rightarrow \dfrac{DB}{AD} = \dfrac{EC}{AE}$. Add 1 to both sides: $\dfrac{DB + AD}{AD} = \dfrac{EC + AE}{AE} \Rightarrow \dfrac{AB}{AD} = \dfrac{AC}{AE} \Rightarrow \dfrac{AD}{AB} = \dfrac{AE}{AC}$. Proved! [2.0 Marks]
Part 2: Given $AD = x, DB = x - 2, AE = x + 2, EC = x - 1$. By BPT, $\dfrac{x}{x-2} = \dfrac{x+2}{x-1}$. [1.0 Mark]
Cross-multiply: $x(x - 1) = (x - 2)(x + 2) \Rightarrow x^2 - x = x^2 - 4$. [1.0 Mark]
$-x = -4 \Rightarrow x = 4$. Value of $x$ is $4$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Proof of BPT corollary $AD/AB = AE/AC$: 2.0 Marks
Setting equation $\dfrac{x}{x-2} = \dfrac{x+2}{x-1}$: 1.0 Mark
Expanding $x^2 - x = x^2 - 4$: 1.0 Mark
Solving $x = 4$: 1.0 Mark

Correct Answer:
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