Question

△ABC中,外角∠ACD的平分线与∠ABC的平分线交于A1,∠A1BC与∠A1CD的平分线交于点A2,

Answer 1

∵∠A+∠ABC+∠ACB＝180

∴∠ABC+∠ACB＝180-∠A

∵∠ACD＝180-∠ACB,CA1平分∠ACD

∴∠A1CD＝∠ACD/2＝（180-∠ACB）/2＝90-∠ACB/2

∵BA1平分∠ABC

∴∠A1BC＝∠ABC/2

∵∠A1CD是△A1BC的外角

∴∠A1CD＝∠A1+∠A1BC＝∠A1+∠ABC/2

∴∠A1+∠ABC/2＝90-∠ACB/2

∴∠A1＝90-（∠ABC+∠ACB）/2＝90-（180-∠A）/2＝∠A/2

同理可得：

∠A2＝∠A1/2＝∠A/4

∠A2＝∠A/2²

依次类推：∠An＝∠A/2 ⁿ

则当∠A＝64,n＝4时,∠A4＝∠A/2⁴＝64/2⁴＝4°