Given: AD is the bisector of ∠A of ∆ABC
To Prove: AB > DB
Proof: In ∆ADC,
Ext. ∠ADB = ∠DAC + ∠ACD
An exterior angle of a triangle is equal to the sum of its two interior opposite angles
⇒ ∠ADB > ∠DAC
⇒ ∠ADB > ∠DAB
| ∵ AD is the bisector of ∠A ∠DAB = ∠DAC
∴ AB > DB
| Side opposite to greater angle is longer.