Question

已知如图,在RT△ABC中,∠ACB=90,AC=BC,点D是BC的中点,CE垂直于AD,垂足为点EBF平行于AC交CE

Answer 1

CF⊥AD, AC⊥BC,∠1=∠2AC=BC, FB⊥BC,

△ACD≌△CBF, FB=CD=DB,

∠3=45=90-45=∠4,

△BDG≌△BFG, FG=DG,FD⊥AB

AB垂直平分DF

附图