Proof: ∫∠ACD =∠BCE = 90

∴∠ACD+∠DCE=∠BCE+∠DCE, that is ∠ACE=∠DCB.

∫△ACD and △BCE are isosceles right triangles,

∴AC=CD,CE=CB，

∴△ACE≌△DCB(SAS)，

∴ae=bd,∠cae=∠cdb；

∠∠AFC =∠dfh is a right angle, ∴∠AFC=∠DFH.

∴∠DHF=∠ACD=90

∴AE⊥BD.