Question

如图，在△ABC中，∠ABC的平分线与∠ACE的平分线交于D点，若∠A=80°，求∠D的度数.

Answer 1

解：
∵∠ACE＝∠A+∠ABC，CD平分∠ACE
∴∠DCE＝∠ACE/2＝（∠A+∠ABC）/2
∵BD平分∠ABC
∴∠DBC＝∠ABC/2
∴∠DCE＝∠D+∠DBC＝∠D+∠ABC/2
∴∠D+∠ABC/2＝（∠A+∠ABC）/2
∴∠D＝∠A/2＝80/2＝40°

Answer 2

∵BD平分∠ABC，CD平分∠ACE
∴∠DBC=1/2∠ABC
∠ACD=1/2∠ACE=1/2(180°-∠ACB)
∴∠D=180°-∠DBC-∠ACB-∠ACD
=180°-1/2∠ABC-∠ACB-1/2(180°-∠ACB)
=180°-1/2∠ABC-∠ACB-90°+1/2∠ACB
=90°-1/2∠ABC-1/2∠ACB
=90°-1/2(∠ABC+∠ACB)
=90°-1/2(180°-∠A)
=1/2∠A
=40°