In the adjoining figure, AC > AB and AD is the bisector of ∠A. Show that : ∠ADC > ∠ADB

We know that,

In a triangle an exterior angle is equal to the sum of two opposite interior angles.

In △ADC,

⇒ ∠ADB = ∠CAD + ∠C …..(1)

In △ADB,

⇒ ∠ADC = ∠BAD + ∠B …..(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.

∴ ∠BAD + ∠B > ∠CAD + ∠C ……(3)

From eq.(1), (2) and (3), we get:

⇒ ∠ADC > ∠ADB

Hence, proved that ∠ADC > ∠ADB.