In the given figure $L M | | A B$ . If AL $= x - 3$ , AC $= 2 x$ , BM $= x - 2$ and BC $= 2 x + 3$ , find the value of $x$ .

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Solution

In $△ A B C$ , we have

$L M | | A B$

$∴$ $\frac{A L}{L C} = \frac{B M}{M C}$ [By Thaley's Theorem]

$\Rightarrow$ $\frac{A L}{A C - A L} = \frac{B M}{B C - B M}$

$\Rightarrow$ $\frac{x - 3}{2 x - (x - 3)} = \frac{x - 2}{(2 x + 3) - (x - 2)}$

$\Rightarrow$ $\frac{x - 3}{x + 3} = \frac{x - 2}{x + 5}$

$\Rightarrow$ $(x - 3) (x + 5) = (x - 2) (x + 3)$ [on cross multiplication]

$\Rightarrow$ $x^{2} + 2 x - 15 = x^{2} + x - 6$

$\Rightarrow$ $x = 9$ [cancelling $x^{2}$ from both sides and solving for $x$ ]

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