证明：设正方形边长为a,
AE=DE=a/2DF=a/4
AB/DE=a/(a/2)=2
AE/DF=(a/2)/(a/4)=2
∴AB/DE=AE/DF
RT△ABE∼RT△DEF
BE/EF=AB/ED=2
∴BE=2EF
∴∠DEF=∠ABE
∠AEB+∠ABE=RT∠
∴∠AEB+∠DEF=RT∠
∴∠BEF=180°-90°=90°
⇒BE⊥EF