难题一起想!在△ABC中,∠A=96°,延长BC到D∠ABC与∠ACD平分线叫于A1,∠A1BC与∠A1CD的平分线交于A2,依次类推,∠A4BC与∠A4CD的平分线交于A5．求：∠A5的大小．

Question

难题一起想!
在△ABC中,∠A=96°,延长 BC到D∠ABC与∠ACD平分线叫于A1,∠A1BC与∠A1CD的平分线交于A2,依次类推,∠A4BC与∠A4CD的平分线交于A5．求：∠A5的大小．

Answer 1

∠A1=180-(∠A1BC+∠A1CA+∠BCA)
=180-(1/2∠ABC+1/2∠ACD+∠BCA)
=180-[1/2∠ABC+1/2(∠ABC+∠CAB)+∠BCA]
=180-( ∠ABC+∠BCA+1/2∠CAB)
=180-∠ABC-∠BCA-1/2∠CAB
= ∠BAC-1/2∠BAC
=1/2 ∠BAC
由此推论出∠A5=1/32∠BAC=1/32*96=3度