$ABCDEF$ 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 $\Delta IJK$ is:

Hexagon
Right angle
Equilateral
Isosceles
Scalene

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This is how far I got. Note that I've taken $\angle IOC = \angle IOB = u$ , $\angle JOD = \angle JOE = v$ , $\angle KOA = \angle KOF = w$ .

The method given to me is to find $OI$ and $OJ$ . Then using cosine rule, in $\Delta OIJ$ I had to 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 $R$ . A helpful point is that $u+v+w=90^{\circ}$ . I don't know how to use it. Help would be appreciated. Thanks! :)