Q.If \( DE || BC \) and \( BD = (x - 3) \) cm, \( AB = 2x \) cm, \( CE = (x - 2) \) cm, and \( AC = (2x + 3) \) cm, determine the value of \( x \).


\(\because DE \parallel BC\) \(\therefore \frac{AB}{BD} = \frac{AC}{CE}\) \[ \cfrac{2x}{x-3} = \cfrac{2x+3}{x-2} \] \[ 2x(x-2) = (2x+3)(x-3) \] \[ 2x^2 - 4x = 2x^2 - 6x + 3x - 9 \] \[ 2x^2 - 4x - 2x^2 + 3x = -9 \] \[ -x = -9 \] \[ x = 9 \]
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