举报
guanggzh88
1) ∵∠A1CD是∠A1,∠A1BC的外角 ∴∠A1=∠A1CD-∠A1BC ∵BA1,CA1分别是∠ACD的与∠ABC的平分线 ∴∠A1CD=∠ACD/2,∠A1BC=∠ABC/2 ∴∠A1=∠ACD/2-∠ABC/2 =(∠ACD-∠ABC)/2 又∵∠ACD是∠A,∠ABC的外角 ∴∠ACD-∠ABC=∠A ∴∠A1=(∠ACD-∠ABC)/2=∠A/2=θ/2 2) 同理: 1) ∵∠AnCD是∠An,∠AnBC的外角 ∴∠An=∠AnCD-∠AnBC ∵BAn,CAn分别是∠An-1CD的与∠An-1BC的平分线 ∴∠AnCD=∠An-1CD/2,∠AnBC=∠An-1BC/2 ∴∠An=∠AnCD/2-∠AnBC/2 =(∠An-1CD-∠An-1BC)/2 又∵∠An-1CD是∠An-1,∠An-1BC的外角 ∴∠An-1CD-∠An-1BC=∠An-1 ∴∠An=(∠An-1CD-∠An-1BC)/2=∠An-1/2 依次类推, 得∠An=∠An-1/2=∠An-2/(2^2)=...=∠A1/(2^n-1)=∠A/(2^n) =θ/2^n