It is known that am is the middle line of BC in triangle ABC AM^2=1/2(AB^2+AC^2)-BM^2

It is known that am is the middle line of BC in triangle ABC AM^2=1/2(AB^2+AC^2)-BM^2

Let vector AB = a, vector AC = B, vector am = C, vector BM = D, extend am to d so that am = DM, connect BD and CD, then ABCD is a parallelogram, then vector a + B = 2C (a + b) square = 4C square a + 2Ab + b square = 4C square (1) vector B-A = 2D (B-A) square = 4D square a-square-2ab + b square = 4D square (2) (1) +