<s>[INST] If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. Where $a,b,c$ are the sides of triangle and $R$ is the circum radius 
My Attempt 
From sine law, 
$$\dfrac {a}{\sin A}=\dfrac {b}{\sin B}=\dfrac {c}{\sin C}=2R$$
So,
$$a=2R \sin A$$
$$b=2R \sin B$$
$$c=2R \sin C$$
Then,
$$8R^2=a^2+b^2+c^2$$
$$8R^2=4R^2 \sin^2 (A)+ 4R^2 \sin^2 (B) + 4R^2 \sin^2 C$$
$$8R^2=4R^2(\sin^2 (A)+\sin^2 (B) +\sin^2 (C)$$
$$2=\sin^2 (A)+\sin^2 (B)+\sin^2 (C)$$
 [/INST] HINT:
Use the equality valid for any $A$, $B$, $C$ with sum $\pi$
$$1-(\cos^2 A + \cos^2 B + \cos^2 C)- 2 \cos A \cos B \cos C = 0$$
ADDED: The equality follows from the following formula valid for any angles $\alpha$, $\beta$, $\gamma$ with sum $2s$
$$1-(\cos^2\alpha + \cos^2 \beta + \cos^2 \gamma)- 2 \cos \alpha \cos \beta \cos \gamma = - 4 \cos s\cos (s-\alpha) \cos (s-\beta) \cos (s - \gamma)$$
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