If LM $∥$ AB, AL=x-3, AC=2x, BM=x-2, BC=2x+3. What is value of AC?

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Solution

In $△$ ABC,

It's given LM $∥$ AB,
So, $\frac{A L}{C A} = \frac{B M}{C B}$ [By BPT]
$\frac{(x - 3)}{2 x} = \frac{(x - 2)}{(2 x + 3)}$

$(x - 3) \times (2 x + 3)$ = $(x - 2) \times 2 x$

$2 x^{2}$ - $3 x$ - 9 = $2 x^{2}$ - $4 x$

$x = 9$
$S o A C = 2 x = 18$

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