已知在△ABC中,AD为∠BAC外角∠EAC平分线,D为平分线与BC延长线交点,求证：AB/AC=DB/DC

先做∠EDA=∠ADC,∵AD平分∠EAC,∴∠DAE=∠DAC,且AD为△DAE和△DAC的公共边,∴△DAE≌△DAB,∴∠AED=∠ACD,DE=CD,再于直线BAE上取一点F,使得DF=DE,则△DFE为等腰三角形,∠DEF=∠DFE,∴∠DEA+∠DFA=180=∠ACD+ACB,∴...