Question
In a $\triangle ABC , D$ and E are points on the sides AB and $A C$ respectively such that $D E \| B C$.If $A D=x, D B=x-2, A E=x$ +2 and $E C=x-1$, find the value of $x$.
✓
Answer
In the figure, $AD = x, DB = x - 2, AE = x + 2$ and $EC = x - 1$
$\because$ DE || BC
$\therefore\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{x}}{\text{x}-2}=\frac{\text{x}+2}{\text{x}-1}$
$\Rightarrow x(x - 1) = (x + 2)(x - 2)$
$\Rightarrow x^2 - x = x^2 - 4$
$\Rightarrow x^2 - x - x^2 = -4$
$\Rightarrow -x = -4$
$\Rightarrow x = 4$
$\therefore$ $x = 4$
$\because$ DE || BC
$\therefore\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{x}}{\text{x}-2}=\frac{\text{x}+2}{\text{x}-1}$
$\Rightarrow x(x - 1) = (x + 2)(x - 2)$
$\Rightarrow x^2 - x = x^2 - 4$
$\Rightarrow x^2 - x - x^2 = -4$
$\Rightarrow -x = -4$
$\Rightarrow x = 4$
$\therefore$ $x = 4$
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