Hexagon \(ABCDEF\) is inscribed in the circle of radius \(R\) . \(AB=CD=EF=R\). Points \(I\), \(J\), \(K\) are the midpoints of segments \(\overline{BC}\), \(\overline{DE}\), \(\overline{FA}\) respectively. Then prove \(\Delta IJK\) is equilateral.

I know this proof is really easy using complex numbers and rotation and stuff, but I'm trying to do this using trigonometry. The method is to find a symmetric expression for one of the sides of \(\Delta IJK\) and since it is symmetric, we can claim that the triangle is equilateral.

Note that \(\angle IOC = \angle IOB = u\), \(\angle JOD = \angle JOE = v\), \(\angle KOA = \angle KOF = w\).

I need to find \(OI\) and \(OJ\). Then using cosine rule, in \(\Delta OIJ\) I'll find \(IJ\) and then simplify it until it becomes symmetrical in \(u\), \(v\) and \(w\) and \(R\). Then we can say that the triangle is equilateral.

I got \(OI=R\cos(u)\) and \(OJ = R\cos(v)\). Also, in \(\Delta OIJ\), \(\angle IOJ = 60^{\circ} + u + v\). So using cosine rule, \[IJ^{2} = R^{2}\cos^{2}(u) + R^{2}\cos^{2}(v) - 2(R\cos(u))(R\cos(v))(\cos(60^{\circ}+u+v))\] ahead of which I don't know what to do. I need to make this expression symmetrical in \(u\), \(v\) and \(w\) and \(R\), of course.

Two helpful points are that \(u+v+w=90^{\circ}\) and without loss of generality, we can take the radius of the circle to be one (to simplify calculations). Could someone please finish this for me? Help would be appreciated. Thanks! :)

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