Question

已知△ABC,AD为中线,求证AD^2=1/2(AB^2+AC^2)-(BC/2)^2

Answer 1

cos角ADB=(AD^2+BD^2-AB^2)/(2ADBD),cos角ADC=(AD^2+CD^2-AC^2)/(2ADCD),ADC+ADB=180,AD为中线BD=CD.两式相加2AD^2+CD^2+BD^2-AC^2-AB^2=0,2AD^2=AC^2+AB^2-2(BC/2)^2,整理得所证.