Question

In $△$ ABC, DE $∥$ BC. If AD = $x$ , DB = $x - 2$ , AE = $x + 2$ and EC = $x - 1$ , then find the value of $x$ .

Approach

In $△ A BC$ , $D E ∥ BC$ . By Basic Proportionality Theorem, $\frac{A D}{D B} = \frac{A E}{EC}$ . Substituting values: $\frac{x}{x - 2} = \frac{x + 2}{x - 1}$ . Cross-multiplying and solving gives $x = 4$ .

Step-by-step working

1
Since $D E ∥ BC$ , by BPT: $\frac{A D}{D B} = \frac{A E}{EC}$
1 mark
2
$\frac{x}{x - 2} = \frac{x + 2}{x - 1} \Rightarrow x (x - 1) = (x - 2) (x + 2) \Rightarrow x^{2} - x = x^{2} - 4$
0.5 marks
3
Solving: $- x = - 4 \Rightarrow x = 4$
0.5 marks

Concepts used

similar trianglesbasic proportionality theoremparallel linesfind value of x

Approach

Step-by-step working

Concepts used

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