解题过程,谢谢
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解题过程,如下:
∵∠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°
∵∠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°
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