解：由折叠性质得$AG = AB = AD = 3，$$EG = BE = 1，$$FG = DF，$$\angle AGE = \angle AGF = 90^\circ，$故$E、G、F$共线。设$DF = x，$则$FG = x，$$FC = 3 - x，$$EC = 3 - 1 = 2，$$EF = 1 + x。$在$Rt\triangle EFC$中，$2^2 + (3 - x)^2 = (1 + x)^2，$解得$x = \frac{3}{2}，$$EF = 1 + \frac{3}{2} = \frac{5}{2}。$答：$EF$的长为$\frac{5}{2}。$
