∵ABCD是矩形∴∠BAD=∠ABC=∠ADC=∠BCD=90°又∵EF⊥BE∴∠BEF=90°∴∠BAD+∠BEF=180°∴A、B、E、F四点共圆∴∠DFE=∠ABE在Rt△EFD中∠DFE+∠DEF=90°又∠ABE+∠CBE=90°∴∠DEF=∠CBE
