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

∵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/