Question

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

Answer 1

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）