证：作EG//AD//BC,则AE=AD,∠3=∠4并且∠2=∠4∵EG//AD//BC, ∴∠1=∠5在RT△EDG和RT△EFG中：∠G=90, （∵DF⊥BC）DG=FG∴RT△EDG≌RT△EFG∴∠1=∠2∴∠5=∠2,∠2=∠4,∠4=∠3∴∠3=∠5∴∠AED=∠EFB...