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

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