35
In $\triangle ABC$, $DE \parallel BC$. If $AD = x$, $DB = x - 2$, $AE = x + 2$ and $EC = x - 1$, then find the value of $x$.
Show SolutionHide Solution↓
Since $DE \parallel BC \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}$ (I) (1 Mark)
$\frac{x}{x-2} = \frac{x+2}{x-1}$ (II) (1/2 Mark)
Solving, we get $x = 4$ (III) (1/2 Mark)
$\frac{x}{x-2} = \frac{x+2}{x-1}$ (II) (1/2 Mark)
Solving, we get $x = 4$ (III) (1/2 Mark)