在三角形ABC中,AD平分角BAC,EF垂直AD交AB、AC于E、F,交BC延长线于M,求证：角M=1/2(角ACB-角

设EF交AD于O
∠BAD=∠DAC,∠AOE＝∠AOF,AO=AO ==》⊿AOE≌⊿AOF
==》∠AEO＝∠AFO＝∠CFM
∠AEO＝∠B＋∠M,∠ACB＝∠CFM＋∠M ==》∠ACB＝∠B＋∠M＋∠M
==》∠M＝1/2（∠ACB－∠B）