Question

in figure IF AD is a bisector of angle A show that AB is greater than BD and AC is greater than CD

Answer 1

If AD is the bisector of ∠A of triangle ABC, show that AB>DB. 
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.

Answer 2

Step-by-step explanation:

In ΔABC

AD is the bisector of ∠BAC

∠1=∠2 …(i)

Also in ΔADC

∠3=∠2+∠C (ext. angle is equal to sum of opposite interior angles)

∠3>∠2 (ext. angle is greater than one of the interior angles)

But ∠1=∠2

∠3>∠1

AB>BD

Hence proved.