In the given figure, AB > AC and D is any point on BC.

Prove that : AB > AD.

Given: AB > AC and D is point on BC.

To prove: AB > AD.

Proof: AB > AC

⇒ ∠ACB > ∠ABC

⇒ ∠ACD > ∠ABC [∵ From fig, ∠ACB and ∠ACD is the same angle]

In ΔADC,

∠ADB = ∠ACD + ∠DAC [∵ Ext ∠= Sum of two opp. interior ∠s]

⇒ ∠ADB > ∠ACD

⇒ ∠ADB > ∠ACD > ∠ABC [∵ ∠ACD > ∠ABC]

⇒ ∠ADB > ∠ABC

⇒ AB > AD [∵ Side opposite to greater angle is always greater]

Hence, proved that AB > AD.