How can I prove that Equiangular triangles are equilateral in plane geometry using Law of cosine? How is it that equiangular triangles are equilateral is a direct consequence of law of cosine?
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ápplying the law of cosines we obtain $$a^2=b^2+c^2-bc$$ $$b^2=a^2+c^2-ac$$ $$c^2=a^2+b^2-ab$$ adding these equations we get $$a^2+b^2+c^2-ab-bc-ac=0$$ and this is $$\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)=0$$ and this is what you want.
Deepak
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Dr. Sonnhard Graubner
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How did the cosine parts vanished? – Jyotishraj Thoudam May 14 '16 at 04:58
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That last bit should be $\frac{1}{2}\left((a-b)^2+(b-c) ^2+(c-a)^2\right)=0$ – Deepak May 14 '16 at 04:59
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@jyotishrajthoudam Because $\cos 60^{\circ} = 0.5$ – Deepak May 14 '16 at 05:01
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What if we are to assume that we are not given that all the angles are 60? – Jyotishraj Thoudam May 14 '16 at 05:02
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sorry it was a typo – Dr. Sonnhard Graubner May 14 '16 at 05:02
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There's still an error in the last term on the LHS. I'll edit to correct. – Deepak May 14 '16 at 05:04
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if we have given that not all angles are $60^{\circ}$ then the triangle isn't equilateral. – Dr. Sonnhard Graubner May 14 '16 at 05:05
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where is the error? – Dr. Sonnhard Graubner May 14 '16 at 05:05
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yes i have found it thank you it is early in morníng ing Germany – Dr. Sonnhard Graubner May 14 '16 at 05:06
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I've already edited it. Your sum of squared differences wasn't cyclic like it should've been. – Deepak May 14 '16 at 05:06
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But for the argument's sake if we are asked to prove it without the use of 60 degree? – Jyotishraj Thoudam May 14 '16 at 05:10
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but the problems states that we have a equiangular triangle – Dr. Sonnhard Graubner May 14 '16 at 05:12
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Yes if we put a constant "k" instead of cos60? – Jyotishraj Thoudam May 14 '16 at 05:29