Question

如图，已知△ABC的两个外角的角平分线相交于点D，若∠D=60°，求△ABC中∠A的外角度数

如图，帮帮忙

Answer 1

解：
∵∠BAC+∠ABC+∠ACB＝180
∴∠ABC+∠ACB＝180-∠BAC
∵∠FBC＝180-∠ABC，BD平分∠FBC
∴∠DBC＝∠FBC/2＝（180-∠ABC）/2＝90-∠ABC/2
∵∠GCB＝180-∠ACB，CD平分∠GCB
∴∠DCB＝∠GCB/2＝（180-∠ACB）/2＝90-∠ACB/2
∵∠D+∠DBC+∠DCB＝180
∴∠D+90-∠ABC/2+90-∠ACB/2＝180
∴∠D-（∠ABC+∠ACB）/2＝0
∴∠D-（180-∠BAC）/2＝0
∴180-∠BAC＝2∠D
∵∠FAC＝180-∠BAC
∴∠FAC＝2∠D
∵∠D＝60
∴∠FAC＝120