Given : In ABC, DE // AB AD = 8x + 9, CD = x + 3,
BE = 3x + 4, CE = x
By Basic proportionality theorem,
If DE // AB then we should have
\(\frac{CD}{DA}=\frac{CE}{EB}\)
\(\frac{x+3}{8x+9}\) = \(\frac{x}{3x+4}\)
⇒ (x + 3) (3x + 4) = x (8x + 9)
⇒ x (3x + 4) + 3 (3x + 4) – 8x2 + 9x
⇒ 3x2 + 4x + 9x + 12 = 8x2 + 9x
⇒ 8x2 + 9x – 3x2 – 4x – 9x -12 = 0
⇒ 5x2 – 4x – 12 = 0
⇒ 5x2 – 10x + 6x – 12 = 0
⇒ 5x (x – 2) + 6 (x – 2) = 0
⇒ (5x + 6) (x – 2) = 0
⇒ 5x + 6 = 0 or x – 2 = 0
⇒ x =\(\frac{-6}{5}\) or x = 2;
x cannot be negative.
∴ The value x = 2 will make DE // AB.