Question

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

Answer 1

In $△ABC$, we have
$DE||BC$

$\therefore$ $\frac{AD}{DB}=\frac{AE}{EC}$ [By Thale's Theorem]

$\Rightarrow$ $\frac{x}{x-2}=\frac{x+2}{x-1}$

$\Rightarrow$ $x(x-1)=(x-2)(x+2)$

$\Rightarrow$ $x^2-x=x^2-4\Rightarrow x=4$