The following remarks are due to Nikolaos Dergiades. They identify the center of the inscribed hexagon with X(37) and give an easy construction of this center. Besides they depict another triangle (A*B*C*), naturally related to ABC and perspective to it, with respect to X(37).

Extend the sides AB, AC to BC', CB', so that BC'=BC=CB'= a. From C' draw C'B" parallel and equal to CB'= a and from B' draw B'C" parallel and equal to BC'= a. The segments BC", CB", B'C' have the same midpoint A*. Similarly define the points B*, C*. The distance of B from AC is a*sin(C). The distance of C" from AC is a*sin(A). Hence the distance of A* from AC is a*(sin(A)+sin(C))/2. Similarly the distance of A* from AB is a*(sin(A)+sin(B))/2. This means that the line AA* passes through the point with trilinears (b+c : c+a : a+b) and this is X(37). Hence the triangles ABC, A*B*C* are perspective and the perspector is X(37). Look at HexadivisionSymmetric.html , for the origin of the subject.

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