圆内接四边形ABCD的两边AB,CD的延长线交于点E,DF经过圆O的圆心,交AB于F,AB=BE,OD=3,FA=FB=

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  • 1、连结BD、AO,

    ∵AF=BF,

    ∴OF⊥AB,

    RT△DAF≌RT△DFB,

    〈DAF=〈FBD,

    〈ABD=〈ACD(同弧圆 周角相等),

    〈ACD=〈DAE,

    〈ADC=〈CDA(公用),

    ∴△DAC∽DEA.

    2、OA=OD=3,AF=√5,

    根据勾股定理,OF=2

    DF=,

    AD=√(DF^2+AF^2)=√30.

    △DAC∽DEA,

    AD/CD=DE/AD,

    AD^2=DE*CD,

    DE*CD=30,.(1)

    AB=BE=2√5,AE=4√5,

    而根据割线定理可知,

    CE*DE=BE*AE=2√5*4√5=40,.(2),

    (1)+(2),

    DE*(CD+CE)=70,

    DE^2=70,

    DE=√70,

    CE=4√70/7,

    CD=3√70/7,

    AC/AE=AD/DE,

    AC=4√105/7.