In \(\triangle ABC\), D and E are points on the sides AB and AC respectively such that \(DE\parallel BC\). If AD \(=x\), DB \(=x-2\), AE \(=x+2\) and EC \(=x-1\), what is the value of \(x\)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4
A line drawn parallel to one side of a triangle divides the other two sides in the same ratio (basic proportionality theorem), so \(\frac{AD}{DB}=\frac{AE}{EC}\).
\(\displaystyle\frac{x}{x-2}=\frac{x+2}{x-1}\)
Cross-multiplying: \(x(x-1)=(x+2)(x-2)\)
\(x^2-x=x^2-4\), so \(x=4\)
Then AD \(=4\), DB \(=2\), AE \(=6\), EC \(=3\), and both ratios are \(2:1\)
