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 triangle ABC, DE is parallel to BC with AD = $x$ , DB = $x - 2$ , AE = $x + 2$ , and EC = $x - 1$ . Using the Basic Proportionality Theorem, $\frac{A D}{D B} = \frac{A E}{EC}$ , we solve to get $x = 4$ .

Step-by-step working

1
Since DE $∥$ BC, by BPT: $\frac{A D}{D B} = \frac{A E}{EC}$ .
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2
Substitute: $\frac{x}{x - 2} = \frac{x + 2}{x - 1}$ .
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3
Cross-multiply: $x (x - 1) = (x - 2) (x + 2) \Rightarrow x^{2} - x = x^{2} - 4$ .
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4
Solve: $- x = - 4 \Rightarrow x = 4$ .
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Concepts used

similar trianglesbasic proportionality theoremparallel linesfind value of x

Approach

Step-by-step working

Concepts used

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