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