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In the figure given below $DE || BC$. If $AD = x,\ DB = x-2,\ AE = x + 2$ and $EC = x-1$, find the value of $x$.
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Given: In the given figure $DE||BC$, $AD = x,\ DB = x-2,\ AE = x + 2$ and $EC =x -1$
To do: To find the value of $x$.
Solution:
In $\vartriangle ABC,\ DE||BC$
$\therefore \frac{AD}{DB}=\frac{AE}{EC}$ [By basic proportionality theorem]
$\Rightarrow \frac{x}{x-2}=\frac{x+2}{x-1}$
$\Rightarrow x(x−1)=(x+2)(x−2)$
$\Rightarrow x^2−x=x^2−4$
$\Rightarrow x=4$.
Thus, the value of $x=4$.
Updated on: 10-Oct-2022
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