证明：因为EF⊥AC于G,DF⊥BC于D所以,∠BCA+∠DEF=90°∠DFE+∠DEF=90°所以,∠BCA=∠DFE(同角的余角相等)因为AB⊥BC,DF⊥BC所以,∠ABC=∠EDF=90°在Rt△ABC和Rt△EDF中,∠BCA=∠DFE(已证),BC=DF(已知),∠ABC=∠EDF(...