证明：：ABCD是平行四边形.
过A作AE垂直BC,过D作DF垂直BC于F.
那么,BE=CF,
所以EF=BC AC^2=AE^2+(BC-BE)^2
BD^2=DF^2+(BC+CF)^2
因为 AE=DF,BE=CF,
所以
AC^2+BD^2
=2AE^2+2BC^2+2BE^2
=2(AE^2+BE^2)+2BC^2
=2AB^2+2BC^2
=2(ABˇ2+BCˇ2)