Each median in a triangle is shorter than the average of the two adjacent sides.
Prove it.

Call the triangle ABC and M the midpoint of BC so that AM is the median from A to BC.  We are to prove AM<(AB+AC)/2

Construct point D such that ABDC is a parallelogram. 
So we have BD=AC, CD=AB, and AM=MD.

The diagonals of a parallelogram bisect each other so A, M, and D are on a line and AD=2AM

Looking at triangle ABD we see AB+BD>AD by the triangle inequality.  Substituting we get
AB+AC>2AM.
or
AM<(AB+AC)/2

Posted by Jer on 2011-04-04 15:10:47