延长AG交圆O于点H,连接BH.
BD是直径,AF⊥BD,那么根据垂径定理,有弧AB=弧BC,AB=BH
于是∠C=∠H,∠H=∠BAH,有∠C=∠BAH.∠ABC是公共角,证得△ABC∽△GBA,有AB/BG=BC/AB,即AB^2=BG.BC.