∵将矩形沿EF折叠，A，C重合，∴∠AOE=∠D=90°，又∵∠OAE=∠DAC，∴△AOE∽△ADC，∵AD=BC=8，CD=AB=6，∴AC= AD2+CD2 =10，∴AO=5，∴ AO AD = EO CD ，∴ 5 8 = EO 6 ，解得：EO= 15 4 ，∴EF=2EO= 15 2 ．故折痕EF的长为 15 2 ．
