In the given figure, DE || BC such that AD = x cm, DB = (3x + 4)cm, AE = (x + 3)cm and EC = (3x + 19)cm. Find the value of x.
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In $\triangle\text{ABC,}$
DE || BC
By Basic proportionality theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{x}}{\text{3x}+4}=\frac{\text{x}+3}{\text{3x}+19}$
$\Rightarrow\text{x}(3\text{x}+19)=(\text{x}+3)(3\text{x}+4)$
$\Rightarrow3\text{x}^2+19\text{x}=3\text{x}^2+4\text{x}+9\text{x}+12$
$\Rightarrow6\text{x}=12$
$\Rightarrow\text{x}=2$
DE || BC
By Basic proportionality theorem,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{\text{x}}{\text{3x}+4}=\frac{\text{x}+3}{\text{3x}+19}$
$\Rightarrow\text{x}(3\text{x}+19)=(\text{x}+3)(3\text{x}+4)$
$\Rightarrow3\text{x}^2+19\text{x}=3\text{x}^2+4\text{x}+9\text{x}+12$
$\Rightarrow6\text{x}=12$
$\Rightarrow\text{x}=2$
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