In △ABC, DE ∥ AB, where D and E are the points on sides AC and BC, respectively. If AD = x - 3, AC = 2x, BE = x - 2 and BC = 2x + 3, then what is the value of x ?

9

AC = DC + AD

= CD = 2x - x + 3 = (x + 3)

then,

BC = CE + BE

= CE = 2x + 3 - x + 2 = x + 5

\(\frac{CD}{AE}\) = \(\frac{CE}{BE}\)

= \(\frac{x\;+\;3}{x\;-\;3}\) = \(\frac{x\;+\;5}{x\;-\;2}\)

= \( {x }^{ 2} \) + x - 6 = \( {x }^{ 2} \) + 2x - 15

= x = 9,

Therefore, x is 9.