设A1A2A3A4A5A6A7是圆内接正七边形,求证：1/A1A2=1/A1A3+1/A1A4 .

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设A1A2A3A4A5A6A7是圆内接正七边形,求证：1/A1A2=1/A1A3+1/A1A4 .

设A1A2A3A4A5A6A7是圆内接正七边形,求证：1/A1A2=1/A1A3+1/A1A4 .
设A1A2A3A4A5A6A7是圆内接正七边形,求证：1/A1A2=1/A1A3+1/A1A4 .

设A1A2A3A4A5A6A7是圆内接正七边形,求证：1/A1A2=1/A1A3+1/A1A4 .
证明：连A1A5,A3A5,并设A1A2=a,A1A3=b,A1A4=c ．
在圆内接四边形A1A3A4A5中,
有A3A4=A4A5=a,A1A3=A3A5=b,A1A4=A1A5=c.
由托勒密定理可知, A1A4×A3A5=A1A3×A4A5+A1A5×A3A4
于是有 ,cb=ba+ca
同除以abc ,即得1/a=1/b+1/c ,也就是1/A1A2=1/A1A3+1/A1A4 .
证毕.

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