In the figure (ii) given below, D is any point on the side BC of △ABC. If AB > AC, show that AB > AD.

Given,

AB > AC

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

From figure,

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

⇒ ∠ADB > ∠ACB

⇒ ∠ADB > ∠ABC [∵ ∠ACB > ∠ABC]

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

Hence, proved that AB > AD.