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Given: AD is bisector of ∠BAC

To Prove: AB > BD

Proof: AD is bisector of ∠BAC

⇒ ∠BAD = ∠DAC …..(1)

Now, In ΔADC, ∠ ADB is the exterior angle

∠ADB > ∠DAC ..(2) [exterior angle of a triangle is greater than each

of the opposite interior angles]

Substituting ∠DAC = ∠BAD in (2)

⇒ ∠ADB > ∠BAD

⇒ AB > BD [side opposite to larger angle is larger]