**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]

⟹AB+AC>2AD ** (proved)**