延长CE、BA相交于F1、∵∠BAC＝90∴∠CAF＝∠BAC＝90, ∠ABD+∠ADB＝90∵CE⊥BD∴∠ACF+∠CDE＝90∵∠ADB＝∠CDE∴∠ABD＝∠ACF∵AB＝AC∴△ABD≌△ACF（ASA）∴BD＝CF又∵BD平分∠ABC,CE⊥BD∴CE＝EF＝CF/2（三...