①略证∵AD//BC∴△GED∽△GBC(AA)∴GE/GB=ED/BC∵E是AD的中点∴AE=ED∴GE/GB=AE/BC②∵AD//BC∴△AEF∽△CBF(AA)∴AE/BC=EF/BF∴GE/GB=EF/BF∵GE=2,BF=3,GB=GE+EF+BF=5+EF2/(5+EF)=EF/3EF²+5EF-6=0(EF-1)(EF+6)=0EF=1或EF=-6(舍去)