On the sides AC and BC of \(\triangle\)ABC, two points L and M are positioned respectively such that \(LM \parallel AB\), and \(AL = (x - 2)\) units, \(AC = 2x + 3\) units, \(BM = (x - 3)\) units, and \(BC = 2x\) units. Then, find the value of \(x\).

Q.On the sides AC and BC of \(\triangle\)ABC, two points L and M are positioned respectively such that \(LM \parallel AB\), and \(AL = (x - 2)\) units, \(AC = 2x + 3\) units, \(BM = (x - 3)\) units, and \(BC = 2x\) units. Then, find the value of \(x\).

Since \(LM \parallel AB\)
\(\therefore \cfrac{CL}{AL} = \cfrac{CM}{BM}\)
Or, \(\cfrac{AC - AL}{AL} = \cfrac{BC - BM}{BM}\)
Or, \(\cfrac{(2x + 3) - (x - 2)}{(x - 2)} = \cfrac{2x - (x - 3)}{(x - 3)}\)
Or, \(\cfrac{2x + 3 - x + 2}{(x - 2)} = \cfrac{2x - x + 3}{(x - 3)}\)
Or, \(\cfrac{(x + 5)}{(x - 2)} = \cfrac{(x + 3)}{(x - 3)}\)
Or, \((x + 5)(x - 3) = (x + 3)(x - 2)\)
Or, \(x^2 - 3x + 5x - 15 = x^2 - 2x + 3x - 6\)
Or, \(x^2 + 2x - x^2 - x = -6 + 15\)
Or, \(x = 9\)

∴ The required value is \(x = 9\) (Answer)