Question
In $\triangle ABC , D$ and $E$ are the points on the sides AB and AC respectively such that DE||BC. If AD = 6x - 7, DB = 4x - 3, AE = 3x - 3 and EC = 2x - 1, find the value of x.

Answer


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Given: In $\triangle ABC , DE \| BC$. Also $AD =6 x -7, DB =4 x -3, AE =3 x -3$ and $EC =2 x -1$
By basic proportionality theorem,
$\frac{A D}{D B}=\frac{A E}{E C}$
$\Rightarrow \frac{6 x-7}{4 x-3}=\frac{3 x-3}{2 x-1}$
$\Rightarrow(6 x-7)(2 x-1)=(3 x-3)(4 x-3)$
$\Rightarrow 12 x^2-6 x-14 x+7=12 x^2-9 x-12 x+9$
$\Rightarrow-20 x+7=-21 x+9$
$\Rightarrow-20 x+21 x=9-7$
$\Rightarrow x=2$

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