Suppose CE=x AB=CD=2.
DE= radical sign (4+x 2)
CH=2x/DE divided by the area of triangle CDE.
From CH/BF=CE/BE
2x/ radical sign (4+x 2)/BF = x/(x+2)
Get BF=y=2(x+2)/ radical sign (4+x 2).
BF⊥DE intersects with DE at point F, and BF intersects with edge CD at point G.
The maximum value is 2 x (f coincides with point d) 0.
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