To solve the problem, it is sufficient to find three end points on the circumference,
not two by two next, of a regular hexagon enscribed.
It is enough to find the end points E, F, B.
To do that,
drow the diameter CB, using Line.
Let Circle2 be a circle centered in C, with radius CA, built using Circle with center point. This circle intersects the given circumference in E and F, that, toghether with B, are the end points of the
desired equilateral triangle.

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