过P点作EF//AD,交AB于E,交CD于F.则AE= DF,EB = FC.
由勾股定理,有:PA^2 =PE^2 + AE^2,PC^2 =PF^2 + FC^2,
PB^2 = PE^2 +EB^2 ,PD^2 = PF^2 +DF^2.
即得:
PA^2 +PC^2 = PE^2 + AE^2+ PF^2 + FC^2,
PB^2 +PD^2 = PE^2 + EB^2+ PF^2 + DF^2.
注意到:AE= DF,EB = FC.
即知:PA^2 +PC^2 = PB^2 + PD^2