证明：过点P作EF⊥AD交AD于点E，BC于点F；过点P作GH⊥AB交AB于点G，CD于点H．则EA=BF，CH=PF，HP=DE．
∴PA²+PC²=EA²+EP²+CH²+HP²
=BF²+EP²+PF²+DE²
=PB²+PD²故：PA²+PC²=PB²+PD²．