ATRACTOR

Build the point \(F\) over the line segment \([BE]\) such that \(E \hat{D}F = 30^{^{\circ}}\) and the point \(G\) over the line segment \([CE]\) such that \(E \hat{D}G = 30^{^{\circ}}\). Note that \(D \hat{E}F = D \hat{E}G\) and \(E \hat{D}F = E \hat{D}G\). Then the triangles \([EDF]\) and \([EDG]\) are congruent, and \(\overline{DF} = \overline{DG}\). As \(F \hat{D}G = 60^{^{\circ}}\), \([FDG]\) is an equilateral triangle. We need to prove that the half lines \(AF\) and \(AG\) are in fact the angle trisectors of the angle \(\measuredangle CAB\).

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