Question

In the figure ,AD is bisector of angle A ;prove AB>BD

Answer 1

Given AD is the bisector of ∠A of triangle ABC

Hence ∠DAB=∠DAC

∠BDA is the exterior angle of the ∆DAC

Hence ∠BDA> ∠DAC

or ∠BDA > ∠DAB Since ∠DAC=∠DAB

→ AB > BD In a triangle sides opposite to greater angle is greater.