第3个回答 2012-10-05
延长CD与BA的延长线交于点F∵∠BAC=90,AB=AC∴∠ABE+∠AEB=90,∠CAF=∠BAC=90,∠ABC=∠ACB=45∵CD⊥BE∴∠BDC=∠BDF=90∴∠ACF+∠CED=90∵∠AEB=∠CED∴∠ABE=∠ACF∴△ABE≌△ACF (ASA),∴BE=CF∵BE平分∠BAC∴∠CBD=∠ABD=∠ABC/2=22.5∴∠ACF=22.5∵BD=BD∴△CBD≌△FBD (ASA)∴CD=DF=CF/2∴D是CF的中点∴AD=CD∴∠CAD=∠ACF=22.5∴∠ADC=180-∠CAD+∠ACF=135∴∠ADB=∠ADC-∠BDC=45°