如图,已知AE⊥AD,AF⊥AB,AB∥CD,AF=CD,AE=AD,求证:AC垂直EF.
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如图,已知AE⊥AD,AF⊥AB,AB∥CD,AF=CD,AE=AD,求证:AC垂直EF.
如图,已知AE⊥AD,AF⊥AB,AB∥CD,AF=CD,AE=AD,求证:AC垂直EF.
如图,已知AE⊥AD,AF⊥AB,AB∥CD,AF=CD,AE=AD,求证:AC垂直EF.
证明:
∵AE⊥AD,AF⊥AB
∴∠DAE+∠BAF=90º+90º=180º
∴∠EAF+∠DAB=180º
∵AB//CD
∴∠ADC+∠DAB=180º
∴∠EAF=∠ADC
又∵AE=AD,AF=CD
∴⊿EAF≌⊿ADC(SAS)
∴∠E=∠DAC
延长CA交EF于M
∵∠DAE=90º
∴∠DAC+∠EAM=90º
∴∠E+∠EAM=90º
∴∠CME=90º
即AC⊥EF
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