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,
$$8R^2=a^2+b^2+c^2$$
$$8R^2=4R^2 \sin^2 (A)+ 4R^2 \sin^2 (B) + 4R^2 \sin^2 C$$
$$8R^2=4R^2(\sin^2 (A)+\sin^2 (B) +\sin^2 (C)$$
$$2=\sin^2 (A)+\sin^2 (B)+\sin^2 (C)$$

$$F=\sin^2A+\sin^2B+\sin^2C=1-\cos^2A+1-(\cos^2B-\sin^2C)$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
and $\cos(B+C)=\cos(\pi-A)=?$
$$2-F=\cos^2A+\cos(B+C)\cos(B-C)$$
$$=\cos^2A-\cos A\cos(B-C)$$
$$=\cos A\{\cos A-\cos(B-C)\}$$
$$=-\cos A\{\cos(B+C)+\cos(B-C)\}=?$$
