Question

In $Δ A BC$ , $D E ∥ BC$ . If $A D = x$ , $D B = x - 2$ , $A E = x + 2$ and $EC = x - 1$ , then find the value of $x$ .

Approach

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

Step-by-step working

1
By BPT: $\frac{A D}{D B} = \frac{A E}{EC} \Rightarrow \frac{x}{x - 2} = \frac{x + 2}{x - 1}$
1 mark
2
Cross-multiply: $x (x - 1) = (x - 2) (x + 2) \Rightarrow x^{2} - x = x^{2} - 4$
0.5 marks
3
Solve: $- x = - 4 \Rightarrow x = 4$
0.5 marks

Concepts used

trianglesbasic proportionality theoremsimilar trianglesfind value

Approach

Step-by-step working

Concepts used

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