证明：连接CD．∵在Rt△ABC中，AD=BD．∴CD=12AB=AD．∵AC=BC．∴∠A=45°．∵PE⊥AC，PF⊥BC．∴四边形PECF为矩形．∴CF=PE=AE．又∵CD=AD．∴∠ACD=∠BCD=45°．在△AED和△CFD中AD＝CD∠A＝∠DCFAE＝CF，∴△AED...