Prove that in a triangle, the length of every median is smaller than the average of the lengths of the two sides meeting at the median's vertex.

Given: Consider a triangle ABC in which AD is median drawn from vertex A.

To prove: AB + AC > AD

Proof: In Δ A B D

 AB + B D> AD    ⇒[In a triangle sum of lengths of two sides is greater than the third side]  .................(1)

In Δ A CD

AC + DC  > AD  [In a triangle sum of lengths of two sides is greater than the third side]  .................(2)

Adding (1) and (2)

AB + AC+ B D + D C > B D + DC

A B+ A C+ B C > 2 B D .................(3)

Also , Considering Δ AB C

AB + B C  > B C ⇒[In a triangle sum of lengths of two sides is greater than the third side]

⇒ AB + B C  - B C >0    ........................(4)

Adding (3) and (4)

A B+ B C+B C+ AB +A C- B C > 2 A D

⇒2 AB + 2 A C> 2 A D

Dividing both side of inequality by 2, we get

A B+ A C> A D

Hence proved.

2nd one in the first row is correct