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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/23 12:07:36
![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](/uploads/image/z/2574900-36-0.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%E2%88%A0ABC%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%E4%B8%8E%E2%88%A0ACE%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8ED%E7%82%B9%2C%E8%8B%A5%E2%88%A0A%3D80%C2%B0%2C%E6%B1%82%E2%88%A0D%E7%9A%84%E5%BA%A6%E6%95%B0)
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
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
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
∵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°