2.(2024·无锡滨湖二模)已知E,F分别为▱ABCD的边BC,AD上的动点，将▱ABCD沿直线EF折叠，使点C落在边AB上的点C'处,点D的对应点为D'.
(1)如图,当点D'落在BA的延长线上时,求证:四边形EC'D'F为平行四边形;
(2)若AB = BC,∠B = 60°,EC'⊥AB,求$\frac{DF}{AF}$的值;
(3)若AB = 5,BC = 6,▱ABCD的面积为24,求CE长的取值范围.
　　　　　　

（1）∵四边形$ABCD$是平行四边形，∴$AD // BC$，$AB // CD$，$AB = CD$。由折叠的性质，得$CD = C'D'$，$D'F // C'E$。
∴$\angle BC'E = \angle D'$。∵$AD // BC$，∴$\angle B = \angle D'AF$。∵$AB = CD = C'D'$，∴$AB - AC' = C'D' - AC'$，即$BC' = AD'$。
∴$\triangle BC'E \cong \triangle AD'F$。∴$C'E = D'F$。∴四边形$EC'D'F$为平行四边形。（2）如图①，延长$BA$交$D'F$于点$G$。∵四边形$ABCD$为平行四边形，$AB = BC$，∴四边形$ABCD$是菱形。
∴$BC = CD = AB$。设$BC' = 1$。∵$\angle B = 60^{\circ}$，$EC' \perp AB$，∴易得$C'E = \sqrt{3}$，$BE = 2$。由折叠的性质可知，$CE = C'E = \sqrt{3}$，∴$BC = BE + CE = 2 + \sqrt{3}$。由折叠的性质可知，$\angle D' = \angle D = \angle B = 60^{\circ}$，$\angle D'C'E = \angle C = 180^{\circ} - \angle B = 120^{\circ}$，$C'D' = CD = BC = AB = 2 + \sqrt{3}$，$DF = D'F$，∴$\angle D'C'G = \angle D'C'E - \angle EC'G = 120^{\circ} - 90^{\circ} = 30^{\circ}$。∵$C'E // D'G$，∴$D'G \perp AB$。∴$\angle D'GC' = 90^{\circ}$。∴易得$D'G = \frac{2 + \sqrt{3}}{2}$，$C'G = \frac{3}{2} + \sqrt{3}$。∴$AG = BC' + C'G - AB = 1 + \frac{3}{2} + \sqrt{3} - (2 + \sqrt{3}) = \frac{1}{2}$。∵$\angle GAF = \angle B = 60^{\circ}$，∴$\angle AFG = 30^{\circ}$。∴易得$AF = 1$，$GF = \frac{\sqrt{3}}{2}$。∴$DF = D'F = D'G + GF = \frac{2 + \sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3} + 1$。∴$\frac{DF}{AF} = \sqrt{3} + 1$。（3）求$CE$长的取值范围，即求$C'E$长的取值范围。过点$A$作$AH \perp BC$于点$H$。如图②，当$C'E \perp AB$时，$C'E$的长最小。∵$BC = 6$，$\square ABCD$的面积为$24$，∴$AH = 4$。∴$BH = \sqrt{AB^{2} - AH^{2}} = 3$。设$CE = C'E = x$，则$BE = 6 - x$。∵$\angle B = \angle B$，$\angle AHB = \angle EC'B = 90^{\circ}$，∴$\triangle ABH \sim \triangle EBC'$。∴$\frac{AH}{EC'} = \frac{AB}{EB}$，即$\frac{4}{x} = \frac{5}{6 - x}$，解得$x = \frac{8}{3}$。经检验，$x = \frac{8}{3}$是原方程的解，且符合题意。∴$CE$长的最小值为$\frac{8}{3}$。如图③，当点$C'$与点$A$重合时，$C'E$的长最大。∵$CH = BC - BH = 3$，
∴$EH = EC - CH = x - 3$。在$Rt\triangle AEH$中，$AH^{2} + EH^{2} = AE^{2}$，即$4^{2} + (x - 3)^{2} = x^{2}$，解得$x = \frac{25}{6}$。∴$CE$长的最大值为$\frac{25}{6}$。∴$\frac{8}{3} \leq CE \leq \frac{25}{6}$。