∵AB∥DE,
∴△ABC∽△DEC,
∴△D'E'C∽△ABC,
∴D'C/E'C=AC/BC=1/√2,
∵∠D'CE'=∠ACB=45°,
∴∠D'CA=∠E'CB,
∴△AD'C∽△BE'C,
∴BE'/AD'=BC/AC=1/√2,
∵CE=√2CD=4,
∴CE'=CE=4,
由勾股定理得AE'=2,
∴BE'=2√3-2
∴AD'=BE'/√2=√6-√2
∵AB∥DE,
∴△ABC∽△DEC,
∴△D'E'C∽△ABC,
∴D'C/E'C=AC/BC=1/√2,
∵∠D'CE'=∠ACB=45°,
∴∠D'CA=∠E'CB,
∴△AD'C∽△BE'C,
∴BE'/AD'=BC/AC=1/√2,
∵CE=√2CD=4,
∴CE'=CE=4,
由勾股定理得AE'=2,
∴BE'=2√3-2
∴AD'=BE'/√2=√6-√2