Question

[Phạm Ngân Hà]

Cho \(\Delta ABC\) đường trung tuyến AM. Chứng minh: \(AB^2+AC^2=2AM^2+\frac{BC^2}{2}\)

Answer 1

[bảo nam trần]

\(AB^2+AC^2=AH^2+BH^2+AH^2+CH^2\) (pytago)

\(=2AH^2+\left(BM+MH\right)^2+\left(CM-MH\right)^2\)

\(=2AH^2+BM^2+MH^2+2BM.MH+CM^2-2CM.MH+MH^2\)

\(=2AH^2+2MH^2+BM^2+CM^2\) (do BM=CM)

\(=2\left(AH^2+MH^2\right)+\left(\frac{BC}{2}\right)^2+\left(\frac{BC}{2}\right)^2\)

\(=2AM^2+\frac{BC^2}{2}\) (đpcm)

Answer 2

[Phạm Ngân Hà]

BH = BM - MH; CH = CM - MH chứ nhỉ