∠ 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 .