In Figure, AD is the bisector of ∠BAC. Prove that AB > BD.
Given: AD is the bisector of ∠BAC.
As AD is the bisector of ∠BAC.
∠BAD = ∠CAD …(1)
∠ADB > ∠CAD (exterior angle of a triangle is greater than each of opposite interior angle)
Then ∠ADB > ∠BAD (from (1))
⇒ AB > BD (opposite sides to greater angle is greater)
Hence proved.