**Question:**

In following figure if $A D$ is the bisector of $\angle B A C$, then prove that $A B>B D$.

**Solution:**

Given ABC is a triangle such that AD is the bisector of ∠BAC. To prove AB > BD.

Proof Since, AD is the bisector of ∠BAC.

But ∠BAD = CAD …(i)

∴ ∠ADB > ∠CAD

[exterior angle of a triangle is greater than each of the opposite interior angle]

∴ ∠ADB > ∠BAD [from Eq. (i)]

AB > BD [side opposite to greater angle is longer]

Hence proved.