Question

D is any pointvon the base BC of a triangle ABC. If AB>AC then prove that AB>AD

Answer 1

Now, in ΔABC as AB>AC , ∠ACD>∠ABD.

Now it ΔACD the exterior angle 
∠ADB=∠ACD+∠CAD (sum of interior opposite angles) > ∠ABD+∠CAD.

In ΔADB,∠ADB> ∠ABD
therefore AB>AD

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