证明：作CF⊥AD,交AD延长线于F,连接AC∵CE⊥AB∴∠AEC=∠AFC=90°∵CD=CB∴∠DAC=∠BAE（等弦对等角）又∵AC=AC∴△AFC≌△AEC（AAS）∴AF=AE,CF=CE又∵∠CFD=∠CEB=90°,CD=CB∴Rt△CFD≌Rt△CEB（HL）∴DF=BE∵AE=...