Prove that the sum of any two sides of a triangle is greater than twice the length of the median drawn to the third side.

Given: In triangle ABC, AD is the median drawn from A to BC.

To prove: AB + AC > AD

Construction: Produce AD to E so that DE = AD, Join BE.

Proof:

Now in

∆ADC and ∆EDB,

AD = DE (by const)

DC = BD(as D is mid-point)

∠ADC =∠EDB (vertically opp.∠s)

Therefore,

In

∆ABE, ∆ADC∆EDB(by SAS)

This gives, BE = AC.

AB + BE > AE

AB + AC > 2AD

∵ AD = DE and BE = AC)

Hence the sum of any two sides of a triangle is greater than the median drawn to the third side.