In triangle ABC, side AC is greater than side AB. If the internal bisector of angle A meets the opposite side at point D, prove that : ∠ADC is greater than ∠ADB.

In △ ADC,

⇒ ∠ADB = ∠1 + ∠C [In a triangle an exterior angle is equal to the sum of two opposite interior angles.] ………(1)

In △ ADB,

⇒ ∠ADC = ∠2 + ∠B [In a triangle an exterior angle is equal to the sum of two opposite interior angles.] ………(2)

In △ ABC,

⇒ AC > AB (Given)

⇒ ∠B > ∠C [If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.]

Since, AD is the bisector of angle A.

∴ ∠2 + ∠B > ∠1 + ∠C ………(3)

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

⇒ ∠ADC > ∠ADB.

Hence, proved that ∠ADC is greater than ∠ADB.