General enunciation: If D is the middle point of the side BC of ∆ABC, prove that AB+AC>2AD.
Particular enunciation: Suppose, in triangle ∆ABC, D is the mid-point of BC. Let us join (A, D). Let us prove that AB+AC>2AD.
Construction: Let us expand AD up to DE such that, AD = DE. (E, C) are joined.
Proof: In triangle ∆ABD and ∆CDE we get,
BD = CD [∵ D is the mid-point of BC]
AD = DE [∵ D is the mid-point of BC]
and ∠ADB =∠CDE [vertically opposite angle]
So the two triangles are congruent
∴ AB = CE
Now in ∆ACE, AC+CE>AE
⟹AC+AB>AD+DE [∵AB =CE]
⟹AB+AC>AD+AD [∵ DE = AD]