摘要:如图.DG⊥BC AC⊥BC.EF⊥AB.∠1=∠2.求证:CD⊥AB A 证明:∵DG⊥BC,AC⊥BC E ∴∠DGB=∠ACB=90º 1 F ∴DG//AC( ) D ∴∠2= ( ) 2 ∵∠1=∠2( )∴∠1=∠DCA B C ∴EF//CD( ) G ∴∠AEF=∠ADC( ) ∵EF⊥AB ∴∠AEF=90º ∴∠ADC=90º 即CD⊥AB
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