连结DO,DB,则BD⊥AC,AO=BO=BD===>∠ODB=∠OBD,∠ADO=∠A=∠CRt△BED∽Rt△BDC===>∠BDE=∠C=∠ADO又∠ADO+∠ODB=90º===>∠BDE+∠ODB=∠EDO=90º∴DE是圆O的切线DB=AB/2=4DG=2DF=2*(√3/2)*DB=4√3
