Cho tam giác ABC, trung tuyến AM Chứng minh rằng: AB^2 + AC^2 = 2AM^2
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 \(A{B^2} + A{C^2} = 2A{M^2} + \frac{{B{C^2}}}{2}\).