∠ BDA +∠ CDE =180-∠1=135 in the Boxer BDC.

∴∠BAD=∠CDE，

And < b = < c = 45,

∴δabd∽δdce,ab=bc÷√2=√2，

∴BD/CE=AB/CD，

CE=BD(2-BD)/√2

=-√2/2(BD^2-2BD)

=-√2/2(BD- 1)^2+√2/2，

∴ When BD= 1, the maximum CE =√2/2,

(2) ① AD = AE, then ∠ AED = ∠ 1 = 45,

∴∠ DAE = 90, D and B coincide and are discarded.

②AD=DE, then Δ δABD?δDCE,

∴CD=AB=√2，

∴BD=2-√2，

③AE=DE, then ∠ DAE = ∠ 4 = 45,

∴∠AED=90，

∴∠ADB=∠DEC=90，

BD= 1/2BC= 1 .