如图,DE平行BC,EF平分∠AED,EF⊥AB,CD⊥AB,垂足分别为点F、D,求证:CD平分∠ACB

1/2∠ACB=1/2∠AED=∠AEF=∠DEF ∠DEF =∠EDC ∠EDC =∠DCB
平行线同位角相等 EF平分∠AED EF∥CD DE∥BC
平行线内错角相等
1/2∠ACB=∠DCB