In the given figure, DE || BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

We know that the basic proportionality theorem states that

“If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.”

So if DE || BC,

Then AD/DB = AE/EC

By substituting the given values,

⇒ x cm/(3x + 4)cm = (x + 3)cm/(3x + 19)cm

Cross multiplying, we get

⇒ 3x² + 19x = 3x²+ 9x + 4x + 12

⇒ 3x² + 19x - 3x²- 9x - 4x = 12

⇒ 6x = 12

⇒ x = 2

x = 2