Pass through point B as BM⊥BF and pass through CF at point M.

⑵∵AF⊥CF,AB⊥BC，

Easy to get: ∠FAB=∠BCF.

And ∠∠FBM =∠ABC = 90, we can get: ∠ABF=∠BCM, AB=AC.

∴△ABF≌△BCM(ASA), so BF=BM, △BFM is an isosceles right triangle.

∴∠BFC=∠BMF=45。

∫BF∨CD

∴∠DCF=∠BFC=45。

From isosceles right angle △ADE and isosceles right angle △ABC, we can get △ ADC △ ace (SAS),

∴∠ACD=∠ACF=22.5，

From the isosceles right angle △ABC: ∠ BCF = 45,

∴∠BCF=∠ACF=∠ACD=22.5。

∴CE sharing∠ ∠ACB

(1) from (1): ∠ BMF = 45,

∴∠CBM=45 -∠BCF=22.5。

∠CBM=∠BCF，

So BM=CM=BF,

Easy to get: EBM = BEM = 90-22.5 = 67.5,

∴EM=BM

Therefore, CE=2BM=2BF.

If you are an eighth-grade student, give the best answer!