[Hint. Produce AD to E such that AD = DE and join C and E.]
OR
Prove that the sum of any two sides of a triangle is greater than twice the length of median drawn to the third side.Â
Construction: Produce AD to E such that AD = DE and join C and E.
Proof.: In ∆ADB and ∆EDC,
AD = DE Â Â Â | By const.
BD = DC
| ∵ AD is a median of ∆ABC ∠ADB = ∠EDC
| Vertically Opposite Angles
∴ ∆ADB ≅ ∆EDC    | SAS Axiom
∴ AB = EC    ...(1) | C.P.C.T.
In ∆AEC,
AC + EC >Â AE
∵ The sum of any two sides of a triangle is greater than the third side
⇒ AC + AB > AE    | From (1)
⇒ AB + AC > AE
⇒ AB + AC > 2AD.    | By const.
Prove that the sum of the three sides of a triangle is greater than the sum of its three medians.
OR
Prove that the perimeter of a triangle is greater than the sum of its three medians.