Each median in a triangle is shorter than the average of the two adjacent sides.
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
Posted by Jer
on 2011-04-04 15:10:47