Prove AB^2+AC^2=2BM^2+2AM^2 for the diagram at http://i.imgur.com/hOYs3.jpg

In the triangle ABC, M is the mid point of
BC.

Draw a perpendicular from AD to BC, D being the feet of
the perpendicular. 

So in the figure  BD =  BM+DM. and CM =
CD-DM.

Angle ADB = angle ADC = 90
deg.

So AB^2 = = BD^2+AD^2 =
(BM+MD)^2.

AB^2 = (BM+MD)^2+AD^2
.....(1)

AC^2 =
(CM-MD)^2+AD^2......(2)

(1)+(2): AB^2+AC^2 =
2BM^2+2MD^2+2AD^2, since CM = BM by data. And (x+y)^2+(x-y)^2 = 2x^2+2y^2, so
(BM+MD)^2+(CM-MD)^2 = 2BM^2+2MD^2.

AB^2+AD^2 = 2BM^2
+2(BM^2+AD^2)

AB^2+AC^2 = 2BM^2+2AM^2, as the triangle ADM
is right angled at D, MD^2+AD^2 = AM ^2.

So
AB^2+AC^2 = 2BM^2+2AM^2.