kẻ AH\(\perp BC\left(H\in BC\right)\)
ta có: AB2+AC2=AH2+BH2+AH2+HC2
= 2AH2+(MB-MH)2+(MC+MH)2
=2AH2+MB2+MH2-2MB.MH+MC2+MH2+2MC.MH
=2(AH2+MH2)+2MB2(vì MB=MC)
=2AM2+2.\(\frac{BC^2}{4}\)=\(2AM^2+\frac{BC^2}{2}\)(đfcm)
vậy \(AB^2+AC^2=2AM^2+\frac{BC^2}{2}\)