△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,（接下面的补充 ）
1个回答

1)

∠A1

=∠A1CD-∠A1BC(外角定理)

=∠ACD/2-∠ABC/2 (角平分线)

=(∠ACD-∠ABC)/2

=∠A/2 (外角定理)

=θ/2

2)

同理

∠An

=∠AnCD-∠AnBC(外角定理)

=∠An-1CD/2-∠An-1BC/2 (角平分线)

=(∠An-1CD-∠An-1BC)/2

=∠An-1/2 (外角定理)

=∠An-2/(2^2)

=...

=∠A1/(2^(n-1))

=∠A/(2^n)

=θ/(2^n)

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