In $\triangle ABC$, $R$ is circumradius and $8R^2=a^2+b^2+c^2$. Then $\triangle…

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In $\triangle ABC$, $R$ is circumradius and $8R^2=a^2+b^2+c^2$. Then $\triangle ABC$ is:

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In any triangle: $a^2+b^2+c^2 = 8R^2$ ⇔ right-angled triangle identity.

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