Question

In the given fig. DE $\parallel$ BC. If AD = x, DB = x - 2, AE = x + 2 and EC = x - 1, find the value of x.

Answer 1

In $\Delta$ ABC, DE $\parallel$ BC
$\therefore \frac{AD}{DB}=\frac{AE}{EC}$
(By basic proportionality theorem)
$\Rightarrow \frac{x}{x-2}=\frac{x+2}{x-1}$
$\Rightarrow x(x-1)=(x+2)(x-2)$
$\Rightarrow x^2-x=x^2-4$

$\Rightarrow x=4.$