∠∠BCE+∠ACE = 90，∠ACE+∠CAE=90，

∴∠BCE=∠CAE.

∵AC⊥BC,BF∥AC.∴BF⊥BC.

∴∠ACD=∠CBF=90，

AC = CB，

∴△ACD≌△CBF.∴CD=BF.

∵CD=BD=？ 12BC,∴BF=BD.

∴△BFD is an isosceles right triangle.

∫∠ACB = 90，CA=CB，

∴∠ABC=45。

∫∠FBD = 90 degrees,

∴∠ABF=45。

∴∠ABC=∠ABF, that is, BA is the bisector of∠∠ ∠FBD.

∴BA is the high line of FD edge, and Ba is the middle line of FD edge.

That is, AB divides df vertically.