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.
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![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/1659473-17-3.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%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.
∵∠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°
∵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°...
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∵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°
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