四边形ABCD中,∠A=∠BCD=90°,BC=CD,E是AB延长线上一点,若DE=AB=3cm,CE=4√2cm,AD=

AD为5
连接 CE,延长EC交AB延长线为F,∠FBC = ∠ADC （外角等于内对角）又因为 CD = BC,DE = AB ,∴△DEC = △ABC ,∴AC = CE = 4√2,∠ECD = ∠ACB,∴∠ECA为直角,∴EA = 8,AD = EA-ED = 5.