已知:如图12,∠BCD=∠CDE,AB//EF,求证:∠ABC+∠FED=180°

∵∠BCD=∠CDE,∴BC‖DE
做BC延长线与EF交与G点,
∵BC‖DE,∴BG‖DE,∠ABC=∠BGE,∠FED=∠FGB
∴∠ABC+∠FED=∠BGE+∠FGB=180°