Given: AD = (7x – 4) cm, AE = (5x – 2), DB = (3x + 4) cm and EC = 3x cm

By Thale’s theorem,

⇒

⇒ 3x(7x – 4) = (5x – 2)(3x + 4)

⇒ 21x² – 12x = 15x² + 20x – 6x – 8

⇒ 21x² – 12x = 15x² + 14x – 8

⇒ 21x² – 15x² – 12x – 14x + 8 = 0

⇒ 6x² – 26x + 8 = 0

⇒ 2×(3x² – 13x + 4) = 0 [Simplifying the equation]

⇒ 3x² – 13x + 4 = 0

⇒ 3x² – 12x – x + 4 = 0

⇒ 3x(x – 4) – (x – 4) = 0

⇒ (3x – 1)(x – 4) = 0

⇒ (3x – 1) = 0 or (x – 4) = 0

⇒ x = 1/3 or x = 4

Now since we’ve got two values of x, that is, 1/3 and 4. We shall check for its feasibility.

Substitute x = 1/3 in AD = (7x – 4), we get

AD = 7×(1/3) – 4 = -1.67, which is not possible since side of a triangle cannot be negative.

Hence, x = 4 cm.