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> Suppose that $ABCDEF$ is a cyclic hexagon and the radius of a circumcircle is $r$. Suppose further that $|AB|=|CD|=|EF|=r$. Let the midpoints of $BC, DE$ and $FA$ be $M\_1$, $M\_2$ and $M\_3$ respectively. Is the triangle $M\_1M\_2M\_3$ necessary equilateral?
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I tried to use the law of cosines to prove that, but it was hard to prove that those system of equations has a solution only is three variable are equal. I also tried to rotate the triangle but could not found anything useful. I thing it has something to do with the fact that the equal length of hexagon sides is used to perform angle equalities.