如图△ABC中,AC=BC,以BC为直径做圆O交AB于点D吗,交AC于点G做直线DF⊥AC交AC于点F,交CB延长线于点

连接BG,DO.cosC=(AC^2+BC^2-AB^2)/2AC*BC=(36+36-48)/2*6*6=1/3 sinC=2√2/3CG=BC*cosC=6*1/3=2 AG=6-2=4 BG=BC*sinC=4√2BO=DO ∠OBD=∠ODB AC=BC ∠A=∠OBD ∠ODB=∠A OD//AC BD/AB=BO/BC=1/2 BD=AD BG//EF AD/AB=AF/AG=1/2 AF=FG=AG/2=4/2=2 AD/AB=DF/BG=1/2 DF=BG/2=2√2FC=FG+CG=2+2=4 OD=BC/2=3ED/EF=OD/FC=3/43EF=4DE3(DE+DF)=4DEDE=3DF=6√2