解 :作CG⊥AC,交AE的延长线于点F,∵∠BAC=90°,∴CF‖AB,∴∠F=∠则吵FAB=∠ADB.∵∠BAC=90°=∠ACF,AC=AB.∴△ACF≌弊凯△ADB∴CD=CF∵∠ACB=∠孙卜侍ECF=45度,∴△DCE≌△ECF∴∠CDE =∠F.∴∠ADB=∠CDE.