Given: ΔABC such that AD bisects ∠BAC

As AD bisects ∠BAC hence, ∠BAD = ∠CAD

In ΔACD, ∠ADB is exterior angle.

Hence, ∠ADB > ∠DAC (∵ exterior angle is greater than interior angle)

⇒ ∠ADB > ∠BAD (∵ ∠BAD = ∠CAD)

⇒ BA > BD (side opposite to greater angle is greater)

Hence, option B is correct.