试题解析(1)证明:∵四边形ABCD是矩形 ∴∠A=∠D=∠C=90° ∵⊿BCE沿BE折叠为⊿BFE ∴∠BFE=∠C=90° ∴∠AFB+∠DFE=180°-∠BFE=90° 又∠AFB+∠ABF=90° ∴∠ABF=∠DFE ∴⊿ABE∽⊿DFE (2)解:在Rt⊿DEF中,sin∠DFE== ∴设DE=a,EF=3a,DF==2a ∵⊿BCE沿BE折叠为⊿BFE ∴CE="EF=3a,CD=DE+CE=4a,AB=4a," ∠EBC=∠EBF 又由(1)⊿ABE∽⊿DFE,∴=== ∴tan∠EBF== tan ∠EBC=tan∠EBF= |