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Answers from famous mathematicians
Proof: connect DF,

∠∠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.