作AE⊥BC,设D在B、E间因AB=AC,所以BE=BC/2AB^2=AE^2+BE^2AD^2=AE^2+DE^2AB^2-AD^2=(AE^2+BE^2)-(AE^2+DE^2)=BE^2-DE^2=BE^2-(BE-BD)^2=BE^2-(BE^2-2BE*BD+BD^2)=2BE*BD-BD^2=BD*(2BE-BD)=BD*(BC-DB)=BD*DC 回答人：潇湘诗社 ☆國士無雙卍 有疑问欢迎追问,多谢了~