证明：在BD上截取DF=DE，连接AF，∵DF=DE，AD⊥BF，AD=AD，∴△ADF≌△ADE，（3分）∴AE=AF，∠AFD=∠AEF∵∠ABC=2∠C，BE是∠ABC的平分线，∴∠ABE=∠C=∠EBC，∵∠AFE=∠ABE+∠BAF，∠AEF=∠EBC+∠C∴∠FAB=∠ABF...