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