ATRACTOR

Build the point \(J\), the intersection point of the angle bisectors of the angles \( \measuredangle HFG\) and \(\measuredangle FGI\). This is possible since the angles \(\measuredangle HFG\) and \(\measuredangle FGI\) are not straight angles and their bisectors are both in the half plane delimited by the straight line \(FG\) that contains the point \(A\). In fact, we have: \[\begin{array}{lll} H\hat{F}E & = & 180^{\circ}-H\hat{F}B\\ & = & 180^{\circ}-D\hat{F}B\\ & = & D\hat{F}E\\ & = & 60^{\circ}+E\hat{F}G \end{array}\] \[\begin{array}{lll} G\hat{E}F & = & C\hat{E}B\\ & = & 180^{\circ}-E\hat{B}C-B\hat{C}E\\ & = & 180^{\circ}-\frac{2}{3}A\hat{B}C-\frac{2}{3}B\hat{C}A\\ & = & 180^{\circ}-\frac{2}{3}\left(180^{\circ}-C\hat{A}B\right)\\ & = & 60^{\circ}+\frac{2}{3}C\hat{A}B \end{array}\] \[\begin{array}{lll} H\hat{F}G & = & H\hat{F}E+E\hat{F}G\\ & = & 60^{\circ}+2\cdot E\hat{F}G\\ & = & 60^{\circ}+\left(180^{\circ}-G\hat{E}F\right)\\ & = & 60^{\circ}+\left(120^{\circ}-\frac{2}{3}C\hat{A}B\right)\\ & = & 180^{\circ}-\frac{2}{3}C\hat{A}B<180^{\circ} \end{array}\]

Analogously, \(F\hat{G}I=180^{\circ}-\frac{2}{3}C\hat{A}B<180^{\circ}\)

Since \(\overline{HF}=\overline{FG}\) and \(H\hat{F}J=G\hat{F}J\), the triangles \([HFJ]\) and \([GFJ]\) are congruent, whereby \(\overline{HJ}=\overline{GJ}\). Analogously, the triangles \([GFJ]\) and \([GIJ]\) are also congruent, and \(\overline{FJ}=\overline{IJ}\). Since \(J\hat{F}G=\frac{1}{2}H\hat{F}G=\frac{1}{2}F\hat{G}I=J\hat{G}F\), the triangle \([FJG]\) is isosceles, with \(\overline{FJ}=\overline{GJ}\). Therefore, \(\overline{HJ}=\overline{FJ}=\overline{GJ}=\overline{IJ}\), that is, the point \(J\) is equidistant to the points \(H\), \(G\), \(F\) and \(I\).

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