证明：在CB的延长线上取点E，使BE＝AB，连接AE∵BE＝AB∴∠E＝∠BAE∴∠ABC＝∠E+∠BAE＝2∠E∵∠ABC＝2∠C∴∠E＝∠C∴AE＝AC∵AD⊥BC∴ED＝DC （三线合一）∵ED＝BE+BD＝AB+BD∴AB+BD＝DC
