Question

Cho tam giác ABC, AM là đường trung tuyến.

CMR: \(AM^2=\dfrac{AB^2+AC^2}{2}-\dfrac{BC^2}{4}\)

Answer 1

Kẻ đường cao AH

\(AM^2=AH^2+HM^2\)

\(=\dfrac{AB^2+AC^2-BH^2-HC^2}{2}+HM^2\)

\(=\dfrac{AB^2+AC^2}{2}+\dfrac{2HM^2-BH^2-HC^2}{2}\)

\(=\dfrac{AB^2+AC^2}{2}+\dfrac{\left(HM-BH\right)\left(HM+BH\right)-\left(HC-HM\right)\left(HM+HC\right)}{2}\)

\(=\dfrac{AB^2+AC^2}{2}+\dfrac{\left(HM-BH\right).MB-MC\left(HM+HC\right)}{2}\)

\(=\dfrac{AB^2+AC^2}{2}+\dfrac{BC}{2}.\dfrac{\left(HM-BH\right)-\left(HM+HC\right)}{2}\)

\(=\dfrac{AB^2+AC^2}{2}-\dfrac{BC^2}{4}\)