【摘 要】
：
笔者在文献[1]中获得的命题3如下:命题3设G是△ABC的重心,AG、BG、CG与△ABC的外接圆分别交于点A_1、B_1、C_1,则GA+GB+GC≤GA_1+GB_1+GC_1.1证法探究重新审视该命题,笔者发
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笔者在文献[1]中获得的命题3如下:命题3设G是△ABC的重心,AG、BG、CG与△ABC的外接圆分别交于点A_1、B_1、C_1,则GA+GB+GC≤GA_1+GB_1+GC_1.1证法探究重新审视该命题,笔者发现文献[1]给出的证明技巧性太强,可以对证明方法进行优化.下面给出图1两种证法,其中证法2大部分证明与文献[1]类似,为节省篇幅,前面证明部分省略.
Proposition 3 in the literature [1] is as follows: Proposition 3 Let G be the center of gravity of ABC, and the circumcircles of AG, BG, CG and △ ABC are respectively intersected with points A_1, B_1 and C_1, then GA + GB + GC ≤GA_1 + GB_1 + GC_1.1 Research on the Law of Evidence To revisit this proposition, I found that the evidence given in [1] is too strong and can be used to optimize the methods of proof. Most of the proofs of Law 2 are similar to those of [1]. In order to save space, some of the previous proofs are omitted.
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