证明；∵AD⊥AB,BE⊥AB ∴∠DAC＝∠CBE＝90°∵AD=BC,AC=BE∴△DAC≌△CBE﹙SAS﹚∴∠ADC＝∠BCE ,又∵∠ADC+∠ACD＝90°∴∠BCE+∠ACD＝90°∴∠DCE＝180°－90°＝90°,即DC⊥EC