Prove that any two sides of a triangle are together greater than twice the median drawn to the third side.

Explanation

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: In `triangle ADC and triangle EDB`
AD=DE(constant)
DC=BD ( D is the midpoint )
`angle ADC=angle EDB ` (vertically opposite angles)
`therefore` In` triangle ABE`, `triangle ADC~= triangle EDB` ( by side angle side)
This gives BE=AC
AB+BE>AE
AB+AC>2AD ( `therefore ` AD=DE and BE=AC)
Hence the sum of any two sides of a triangle is greater than twice the median with respect to the third side.