连接BF，过B作BO⊥AC于O，过点F作FM⊥AC于M．Rt△ABC中，AB=3，BC=6，AC=AB2+BC2=35BO=AB×BCAC=655∵EF=BG=2BE=2GF，BC=2AB，∴Rt△BGF和Rt△ABC中BGFG＝BCAB＝2，∴Rt△BGF∽Rt△ABC，∴∠FBG=∠ACB∴AC∥BF∴FM=O...