Prove that the sum of two medians of a triangle is greater than the third median.

Given a ∆ABC in which AD, BE and CF are the medians on the sides BC , AC and AB.

To prove: AD+BE>CF

BE+CF>AD

AD+CF>BE

Proof:

We will extend AD to H to form ∆BHC

Also, AG=GH………………(1)

F is the midpoint of AB and G is the midpoint of AH.

So, by midpoint theorem,

FG||BH

And FG= 1/2 BH

Similarly, GC||BH and BG||CH

So, we can see from the above that

BGCH is a parallelogram.

So, BH=GC …………(2)

And BG=HC…………..(3)

Now, in ∆BGH,

BG+GH>BH

⇒ BG+AG>GC (from (1),(2))

So, BE+AD>CF

Similarly, BE+CF>AD

AD+CF>BE

Proved.