Question
In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides AB and AC respectively such that DE || BC.If AD = 8x - 7, DB = 5x - 3, AE = 4x - 3 and EC = (3x - 1), find the value of x.
✓
Answer
We have,DE || BC
Therefore, by basic proportionality theorem, we have,
$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
$\Rightarrow\frac{8\text{x}-7}{5\text{x}-3}=\frac{4\text{x}-3}{3\text{x}-1}$
$\Rightarrow(8\text{x}-7)(3\text{x}-1)=(4\text{x}-3)(5\text{x}-3)$
$\Rightarrow24\text{x}^2-8\text{x}-21\text{x}+7=20\text{x}^2-12\text{x}-15\text{x}+9$
$\Rightarrow24\text{x}^2-20\text{x}^2-29\text{x}+27\text{x}+7-9=0$
$\Rightarrow4\text{x}^2-2\text{x}-2=0$
$\Rightarrow2[2\text{x}^2-\text{x}-1]=0$
$\Rightarrow2\text{x}^2-\text{x}-1=0$
$\Rightarrow2\text{x}^2-2\text{x}+1\text{x}-1=0$
$\Rightarrow2\text{x}(\text{x}-1)+1(\text{x}-1)=0$
$\Rightarrow(2\text{x}+1)(\text{x}-1)=0$
$\Rightarrow2\text{x}+1=0\ \text{or x}-1=0$
$\Rightarrow\text{x}=-\frac{1}{2}\ \text{or x}=1$
$\text{x}=-\frac{1}{2}$ is not possible
$\therefore\text{x}=1$
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