已知A1A2A3A4A5A6A7是圆内接正七边形,求证;1/A1A2=1/A1A3+1/A1A4

来源：学生作业帮助网编辑：作业帮时间：2024/05/07 10:54:22

已知A1A2A3A4A5A6A7是圆内接正七边形,求证;1/A1A2=1/A1A3+1/A1A4
已知A1A2A3A4A5A6A7是圆内接正七边形,求证;1/A1A2=1/A1A3+1/A1A4

已知A1A2A3A4A5A6A7是圆内接正七边形,求证;1/A1A2=1/A1A3+1/A1A4
很简单吧
正七边形的特点是所有边的边长相同
所以A1=A2=A3=A4
所以得证

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