In \(\triangle ABC\), D and E are points on the sides AB and AC respectively such that \(DE\parallel BC\). If AD \(=x\), DB \(=x-3\), AE \(=x+4\) and EC \(=x-2\) (all in cm), what is the value of \(x\)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
A line parallel to one side of a triangle divides the other two sides in the same ratio (basic proportionality theorem).
\(\displaystyle\frac{AD}{DB}=\frac{AE}{EC}\), so \(\displaystyle\frac{x}{x-3}=\frac{x+4}{x-2}\)
\(x(x-2)=(x+4)(x-3)\)
\(x^2-2x=x^2+x-12\), so \(3x=12\) and \(x=4\)
Check: DB \(=1\) cm and EC \(=2\) cm are positive, and \(\displaystyle\frac{4}{1}=\frac{8}{2}\)
