Anonymous
Anonymous asked in 科學數學 · 8 years ago

幾何證明題

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  • 8 years ago
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    Hint: 因為CE=GE= √(半徑^2-OE^2)

    CD=CE-DE, FG=EG-EF

    若能證明DE=EF即可得證CD=FG

    :pf:先假設CG弦交AB直徑於H,(交角∠AHD=∠BHG=θ)

    DH/AH = EH/OH = FH/BH (= cosθ)

    利用和分比性質

    =>(DH+EH)/(AH+OH) = EH/OH = (FH-EH)/(BH-OH)

    =>(DE)/(AO) = HE/HO = (EF)/(BO)

    因為AO=BO=半徑=> 所以 DE = EF

    CE=GE,DE = EF

    => CD=CE-DE=GE-EF=FG得證

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