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

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

人气：289 ℃　时间：2020-04-18 06:39:07

解答

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...