过P点作PE⊥AC，PF⊥BD，∵四边形ABCD是矩形，∴AD⊥CD，∴△PEA∽△CDA，∴PECD＝PACA，∵AC=BD=32+42=5，∴PE3＝PA5…①，同理：△PFD∽△BAD，∴PFAB＝PDBD，∴PF3＝PD5…②，∴①+②得：PE+PF3＝PA+PD5＝AD5＝45...