In triangle ABC, BC=8. The length of median AD is 5. Let M be the largest possible value of AB^2+AC^2, and let m be the smallest possible value. Find M-m.
\(AB^2=25+16-40cos(180-\theta)\\ AB^2=41+40cos \theta\\ AC^2=41-40cos\theta\\ AB^2+AC^2=82\\ M=m=82\\ M-m=0 \)
+47 
