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In the given figure, $ABC$ is a triangle in which $DE||BC$. If $AD = x$, $DB = x-2$, $AE = x + 2$ and $EC = x-1$, then find the value of $x$.
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In $\triangle ABC$, $DE || BC$
$\therefore \frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{x}{x-2} = \frac{x+2}{x-1}$
$x(x - 1) = (x + 2)(x - 2)$
$x^2 - x = x^2 - 4 \Rightarrow x = 4$
$\therefore \frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{x}{x-2} = \frac{x+2}{x-1}$
$x(x - 1) = (x + 2)(x - 2)$
$x^2 - x = x^2 - 4 \Rightarrow x = 4$