题目是BD平分角ABC CD平分角ACE 求证角D等于角A的2分之1

∠DBC=∠ABC/2 ∠DCB=180°-∠DCE =180-∠ACE/2 =180-(∠A+∠ABC)/2 ∠D=180°-∠DBC-∠DCB =180°-∠ABC/2-180°+(∠A+∠ABC)/2 =∠A/2