由已知,|AB|^2-|BD|^2=|AC|^2-|CD|^2|AB|^2-|BD|^2=(AB+BD)·(AB-BD)=AD·(AB-BD).|AC|^2-|CD|^2=(AC+CD)·(AC-CD)=AD·(AC-CD).所以0=AD·[(AB-BD)-(AC-CD)]=AD·[(AB-AC)+(CD-BD)]=2AD·CB.即AD⊥BC.