证明：在AC上取点E,使AE＝AB,连接DE∵AD平分∠BAC∴∠BAD＝∠CAD∵AB＝AE,AD＝AD∴△ABD≌△AED （SAS）∴DE＝BD,∠AED＝∠B∵∠AED＝∠C+∠CDE,∠B＝2∠C∴∠C+∠CDE＝2∠C∴∠CDE＝∠C∴DE＝CE∴BD＝CE∵AC＝AE+CE...