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对费尔马问题有一组有趣的等比关系式.性质1如图1,OX轴上方是圆x~2+y~2=a~2的上半部分,OX轴下方是正方形ABCD,边长等于圆的直径2a.在半圆上任取一点P(端点A、B除外),连结PC、PD分别交AB于点E、F.则|AF|·|BE|=|EF|~2.证明A(a,0),B(-a,0),C(-a,-2a),D(a,-2a).
There is an interesting set of equi-relational relations for the Fermat problem. Property 1 As shown in Figure 1, the upper part of the OX axis is the upper half of the circle x2 + y2 = a2, the square ABCD below the OX axis, Equal to the diameter of the circle 2a. Take a point P on the semicircle (except for the endpoints A and B), and connect the PC and PD to AB respectively at points E and F. Then | AF | · | BE | = EF | ~ 2. (a, 0), B (-a, 0), C (-a, -2a), D (a, -2a).