Question

In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC = x + 2, then find the value of x.

Answer 1

In ∆ABC, seg BD bisects ∠ABC. [Given] 

∴ AB/BC = AD/CD [Property of angle bisector of a triangle] 

∴ x/(x + 5) = (x - 2)/(x + 2) 

∴ x(x + 2) = (x – 2)(x + 5) 

∴ x² + 2x = x² + 5x – 2x – 10 

∴ 2x = 3x – 10 

∴ 10 = 3x – 2x 

∴ x = 10