1.已知正方形ABCD中,E为CD上一点,F为BC延长线的一点,CE=CF2.∠BAC=120°,AB=AC,∠2=∠E=60°,求证∠B=∠1,DE-EC-BD1 .求证△BCE≌DCF,2、若∠BEC=60°,求=∠CDF的度数.
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![1.已知正方形ABCD中,E为CD上一点,F为BC延长线的一点,CE=CF2.∠BAC=120°,AB=AC,∠2=∠E=60°,求证∠B=∠1,DE-EC-BD1 .求证△BCE≌DCF,2、若∠BEC=60°,求=∠CDF的度数.](/uploads/image/z/4536582-6-2.jpg?t=1.%E5%B7%B2%E7%9F%A5%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2CE%E4%B8%BACD%E4%B8%8A%E4%B8%80%E7%82%B9%2CF%E4%B8%BABC%E5%BB%B6%E9%95%BF%E7%BA%BF%E7%9A%84%E4%B8%80%E7%82%B9%2CCE%3DCF2.%E2%88%A0BAC%3D120%C2%B0%2CAB%3DAC%2C%E2%88%A02%3D%E2%88%A0E%3D60%C2%B0%2C%E6%B1%82%E8%AF%81%E2%88%A0B%3D%E2%88%A01%2CDE-EC-BD1+.%E6%B1%82%E8%AF%81%E2%96%B3BCE%E2%89%8CDCF%EF%BC%8C2%E3%80%81%E8%8B%A5%E2%88%A0BEC%3D60%C2%B0%EF%BC%8C%E6%B1%82%3D%E2%88%A0CDF%E7%9A%84%E5%BA%A6%E6%95%B0.)
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