IN A TRIANGLE ABC, AD BISECTS ANGLE A AND ANGLE C > ANGLE B. PROVE THAT ANGLE ADB> ANGLE ADC

In ΔABC,

∠C > ∠B (Given)

∴ ∠ACB > ∠ABC 

⇒ ∠ACB + ∠2 > ∠ABC + ∠1 .....(1)  (AD bisects ∠A ⇒∠1 = ∠2)

In ΔABD,

∠ABC + ∠1 + ∠ADB = 180°

∴ ∠ABC + ∠1 = 180° – ∠ADB ....(2)

In ΔACD,

∠ACD + ∠2 + ∠ADC = 180°

∴ ∠ACB + ∠2 = 180° – ∠ADC ....(3)

From (1), (2) and (3), we get

180° – ∠ADC > 180° – ∠ADB

∴ ∠ADB – ∠ADC > 180° – 180°

⇒ ∠ADB – ∠ADC > 0

⇒ ∠ADB > ∠ADC

Hence proved.