In fig., LM $∥$ AB. 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 $△$ ABC, we have

LM $∥$ AB

$∴$ $\frac{A L}{L C}$ = $\frac{M B}{M C}$ [By Thale'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)

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

$\Rightarrow$ $x$ = 9

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