如图:作辅助线OM⊥BC,ON⊥CD,设CE=a,CF=b
根据直角三角形勾股定理:
EF^2 = OE^2+OF^2 = OM^2+ME^2 + ON^2+NF^2
EF^2 = CE^2 + CF^2
OM=CN=(b/2)+1,ME=2-(a/2),ON=CM=2+(a/2),NF=(b/2)-1
即:
[(b/2)+1]^2 + [2-(a/2)]^2 + [2+(a/2)]^2 + [(b/2)-1]^2 = a^2 + b^2
化简可得:a^2 + b^2 = 20
即:EF^2 = 20
所以:EF = √20 = 2√5