**General enunciation:** ABC is an isosceles triangle and AB=AC. The side BC is extend up to D. Prove that AD>AB.

**Particular enunciation:** Given that, ABC is an isosceles triangle and AB=AC. The side BC is extend up to D. It is required to prove that AD>AB.

Proof: In ∆ ABC

AB=AC

∴ ∠ABC=∠ACB

Now ∠ACD be an exterior angle of ∆ ABC

∴ ∠ACD>interior opposite angle ∠ABC

⟹∠ACD>∠ACB [∵∠ABC=∠ACB]

Again, ∠ACB is the external angle of ∆ ACD

∴ ∠ACB>∠ADC

Since ∠ACD>∠ACB and ∠ACB>∠ADC

∴ ∠ACD> ∠ADC

Now, in ∆ ACD

∠ACD> ∠ADC

⟹AD>AC

⟹AD>AB [AB=AC]

∴ AD>AB ** (proved)**