Question

如图，梯形ABCD，怎么做，有图有题目~

Answer 1

延长AG交伏租橘滚DC延长线于H
可知 CH=AB DFH相似BFA => BF/DF=AB/(AB+DC)
又缺伍兆DCE相似BAE => (BF+EF)/(DF-EF)=AB/DC
联立两式可证

Answer 2

证明：过E作EH‖CB交AF于点H

            ∵G是BC的中点

    禅兄   ∴CG=BG

     誉知      ∵DC‖AB，EH‖CB,

     贺虚袭  ∴BE/BD=AE/AC=EH/CG=EH/BG=EF/FB

            ∴BE/BD=EF/FB

            ∵BE=EF+FB,BD=DF+FB

             ∴(EF+FB)/(DF+FB)=EF/FB

             ∴(EF+FB)*FB=(DF+FB)*EF

             ∴EF*FB+FB*FB=DF*EF+EF*FB

             ∴FB*FB=DF*EF