(1)如图,过E作EM⊥CB于E交AC与M,而AE⊥EF,∴∠AEF=90°,∴∠AEM+∠MEF=∠CEF+∠MEF,∴∠AEM=∠CEF,又∵AC是正方形的对角线,∴∠ACE=45°,∴CE=ME,∵AE=EF,∴△AEM≌△FEC,∴∠CFE=∠CAE,而∠ANE=∠CNF,∴∠ACF=∠AEF=90°,即CF⊥AC;(2)若点E落在BC的延长线上时(如图3),(1)中结论是否仍然成立.过F作FH⊥BC,交BC的延长线于H,∵四边形ABCD、四边形AEFG是正方形,∴∠AEF=∠B=∠EHF=90°,AE=EF,∴∠AEB+∠BAE=∠AEB+∠FEH=90°,∴∠BAE=∠FEH,∴△FEH≌△EAB,∴EH=AB,FH=BE,即EH=AB=BC,FH=BE=BC+CE,∴FH=EH+CE=CH,即∠FCH=45°,而∠ACB=45°,∴AC⊥CF.