In the adjoining figure, AB > AC and D is any point on BC. Show that AB > AD

Given,

AB > AC

∴ ∠ACB > ∠ABC (As angle opposite to greater side is greater)

From figure,

∠ADB = ∠ACD + ∠DAC (As exterior angle is equal to sum of two opposite interior angles)

⇒ ∠ADB > ∠ACD

⇒ ∠ADB > ∠ACB

⇒ ∠ADB > ∠ABC [∵ ∠ACB > ∠ABC]

⇒ ∠ADB > ∠ABD

∴ AB > AD (As side opposite to greater angle is greater)

Hence, proved that AB > AD.