In a triangle ABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.

∵ ∠C > ∠B                          [Given]

⇒ ∠C + x > ∠B + x              [Adding x on both sides]

⇒ 180° - ∠ ADC > 180° -  ∠ADB

⇒ - ∠ ADC > - ∠ADB

⇒ ∠ADB > ∠ADC

Hence proved.