ATRACTOR

Build the circle with center at \(J\) passing through the points \(H\), \(F\), \(G\) and \(I\). Since the triangles \([HFJ]\), \([GFJ]\) and \([GIJ]\) are congruent, we have that \(H\hat{J}F=F\hat{J}G=G\hat{J}I\). Then: \[\begin{array}{lll} F\hat{J}G & = & 180^{\circ}-2\cdot J\hat{F}G\\ & = & 180^{\circ}-H\hat{F}G\\ & = & 180^{\circ}-\left(180^{\circ}-\frac{2}{3}C\hat{A}B\right)\\ & = & \frac{2}{3}C\hat{A}B \end{array}\] \[\begin{array}{lll} H\hat{J}I & = & H\hat{J}F+F\hat{J}G+G\hat{J}I\\ & = & 3\cdot F\hat{J}G\\ & = & 2\cdot C\hat{A}B\\ & = & 2\cdot H\hat{A}I\Rightarrow H\hat{A}I=\frac{1}{2}H\hat{J}I \end{array}\]

Then, we can conclude that the circle also passes through the point \(A\) and, as the chords \([HF]\), \([FG]\) and \([GI]\) are congruent, we have that the correspondent arcs are also congruent. Hence \(H\hat{A}F=F\hat{A}G=G\hat{A}I\), that is, the half-lines \(AF\) and \(AG\) are in fact the angle trissectors of the angle \(\measuredangle CAB\).