第1个回答 2011-09-09
证明:过A作AG⊥BC于G,交CD于H
∵三角形ABC是等腰直角三角形
∴∠CAG=∠B=45 ···············①
CA=AB ··························②
∵∠ACD+∠ADC=∠ADC+∠BAF=90°
∴∠ACD=∠BAF ················③
根据①、②、③,有△ACH≌△CBE
∴AH=BF
在△ADH和△BDF中
BD=AD,∠BAG=∠B=45°,AH=BF
∴△ADH≌△BDF
∴∠ADC=∠BDF