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
Now ∠ACD be an exterior angle of ∆ ABC
∴ ∠ACD>interior opposite angle ∠ABC
Again, ∠ACB is the external angle of ∆ ACD
Since ∠ACD>∠ACB and ∠ACB>∠ADC
∴ ∠ACD> ∠ADC
Now, in ∆ ACD
∴ AD>AB (proved)