Cho ∆ABC, trung tuyến AM Chứng minh rằng:$A{B^2} + A{C^2} = 2A{M^2} + \frac{{B{C^2}}}{2}$.

Kẻ AH ⊥ BC (H ∈ BC)

Ta có: $A{B^2} + A{C^2} = A{H^2} + B{H^2} + A{H^2} + H{C^2}$

$= 2A{H^2} + {\left( {MB - MH} \right)^2} + {\left( {MC + MH} \right)^2}$

$= 2A{H^2} + M{B^2} + M{H^2} - 2MB.MH + M{C^2} + M{H^2} + 2MC.MH$

$= 2\left( {A{H^2} + M{H^2}} \right) + 2M{B^2}$(vì MB = MC)

$= 2A{M^2} + 2\frac{{B{C^2}}}{4} = 2A{M^2} + \frac{{B{C^2}}}{2}$ (ĐPCM)

Vậy $A{B^2} + A{C^2} = 2A{M^2} + \frac{{B{C^2}}}{2}$.