题目内容
如图，已知∠1 +∠2 =180°，∠DEF =∠A，求证：∠ACB = ∠DEB．
见解析解析:
证明：∵
，
∴
······ 2分
∴ EF∥AB·································· 3分
∴ ∠DEF =∠BDE······························· 4分
∵ ∠DEF=∠A
∴ ∠BDE =∠A································ 6分
∴ DE∥AC·································· 7分
∴ ∠ACB =∠DEB 8分
利用平角可求得EF∥AB，从而求得DE∥AC，最后得出结论
证明：∵
， ∴······ 2分∴ EF∥AB·································· 3分
∴ ∠DEF =∠BDE······························· 4分
∵ ∠DEF=∠A
∴ ∠BDE =∠A································ 6分
∴ DE∥AC·································· 7分
∴ ∠ACB =∠DEB 8分
利用平角可求得EF∥AB，从而求得DE∥AC，最后得出结论
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