Doubt! Could somebody help me?

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Hexagon ABCDEFABCDEF is inscribed in the circle of radius RR . AB=CD=EF=RAB=CD=EF=R. Points II, JJ, KK are the midpoints of segments BC\overline{BC}, DE\overline{DE}, FA\overline{FA} respectively. Then prove ΔIJK\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 ΔIJK\Delta IJK and since it is symmetric, we can claim that the triangle is equilateral.

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

I need to find OIOI and OJOJ. Then using cosine rule, in ΔOIJ\Delta OIJ I'll find IJIJ and then simplify it until it becomes symmetrical in uu, vv and ww and RR. Then we can say that the triangle is equilateral.

I got OI=Rcos(u)OI=R\cos(u) and OJ=Rcos(v)OJ = R\cos(v). Also, in ΔOIJ\Delta OIJ, IOJ=60+u+v\angle IOJ = 60^{\circ} + u + v. So using cosine rule, IJ2=R2cos2(u)+R2cos2(v)2(Rcos(u))(Rcos(v))(cos(60+u+v))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 uu, vv and ww and RR, of course.

Two helpful points are that u+v+w=90u+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! :)

Note by Omkar Kulkarni
6 years, 6 months ago

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