设A1A2A3.A7是圆内接正七边形,求证:1/(A1A2)等于1/(A1A3)+1/(A1A4)

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等同于证AC*AD=AB*(AC+AD)设圆的半径为1所以AB=2sin(π/7)AC=2sin(2π/7)AD=2sin(3π/7)AC*AD=4sin(2π/7)sin(3π/7)=(-2)(cos(5π/7)-cos(π/7))AB*(AC+AD)=4sin(π/7)*(sin(2π/7)+sin(3π/7))=(-2)(cos(3π/7)-cos(π/7)+cos(4π/7)-cos(2π/7))=(-2)(cos(π-(4π/7))-cos(π/7)+cos(4π/7)-cos(π-(5π/7)))=(-2)(cos(5π/7)-cos(π/7))=AC*AD故命题得证

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