<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] Since 
$$4 = 2\sin^2 A + 2\sin^2 B + 2\sin^2 C$$
we have
$$(1-2\sin^2 A ) + (1-2\sin^2B) + 2 - 2\sin^2C=0.$$
or
$$2\cos^2 C + (\cos(2A) + \cos(2B)) = 0$$
Since $$\cos(2A)+\cos(2B) = 2\cos(A+B) \cos(A-B) = -2\cos C\cos(A-B),$$
we have
$$2\cos C(\cos C - \cos(A-B)) = 0$$
Replace $\cos C = -\cos(A+B)$, we get
$$ \cos C(\cos(A+B) + \cos(A-B)) = 0$$
equivalently
$$\cos A \cos B \cos C = 0.$$
The conclusion follows.
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