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

If $8R^2=a^2+b^2+c^2$ then putting $$\begin{cases}R=\dfrac{r^2+s^2}{2}\\a=r^2-s^2\\b=2rs\\c=r^2+s^2\end{cases}$$ it is verified the identity
$$8(\dfrac{r^2+s^2}{2})^2=(r^2-s)^2+(2rs)^2+(r^2+s^2)^2\iff2(r^2+s^2)^2=2(r^2+s^2)^2$$
This show that $a,b,c$ satisfy the well known parametrics of the Pythagorean triples (when $c$ is the diameter of the circumcircle i.e. $2R=r^2+s^2$).
