解∵平行四边形ABCD∴AB∥CD，AD∥BC∵AE⊥BC,AF⊥CD∴AE⊥AD,AF⊥AB∵∠EAF=45°∴∠BAE=45°，∠DAF=45°∴AB＝AE×√2, AD＝AF×√2∵ABCD周长＝2（AB+AD）＝2（AE×√2+ AF×√2）＝2√2（AE+AF）∴AE+AF＝2√2∴ABCD周长＝2√2×2√2＝8
