E,F是AB上两点,AC⊥CE,BD⊥DF,AE=BF,AC=BD.求证:∠ACF=∠BDE
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![E,F是AB上两点,AC⊥CE,BD⊥DF,AE=BF,AC=BD.求证:∠ACF=∠BDE](/uploads/image/z/7802591-23-1.jpg?t=E%2CF%E6%98%AFAB%E4%B8%8A%E4%B8%A4%E7%82%B9%2CAC%E2%8A%A5CE%2CBD%E2%8A%A5DF%2CAE%3DBF%2CAC%3DBD.%E6%B1%82%E8%AF%81%3A%E2%88%A0ACF%3D%E2%88%A0BDE)
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E,F是AB上两点,AC⊥CE,BD⊥DF,AE=BF,AC=BD.求证:∠ACF=∠BDE
E,F是AB上两点,AC⊥CE,BD⊥DF,AE=BF,AC=BD.求证:∠ACF=∠BDE
E,F是AB上两点,AC⊥CE,BD⊥DF,AE=BF,AC=BD.求证:∠ACF=∠BDE
∵AC⊥CE,BD⊥DF
∴∠ACE=∠BDF=90°
则有:
AE=BF
AC=BD
∠ACE=∠BDF
∴△ACE ≌ △BDF
∴∠A=∠B
又∵AE=BF
∴AE-FE=BF-FE
即AF=BE
故有:
∠A=∠B
AF=BE
AC=BD
∴△ACF≌△BDE
∴∠ACF=∠BDE
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