Question

已知，如图，△ABC中，∠ABC的平分线与∠ACE的平分线交与点D，若∠A=80°，求∠D的度数

Answer 1

∠ACE=∠A+∠ABC
∠DCE=1/2∠ACE=∠DBC+∠D
∠D=1/2∠ACE-∠DCB
∠D=1/2∠A+1/2∠ABC-∠DBC
∠D=1/2∠A
∠D=1/2*80°=40°

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追答

∠D=∠DCE-∠DBC

Answer 2

解：
∵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°

Answer 3

角D=1/2角A=40度

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