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,
$$8R^2=a^2+b^2+c^2$$
$$8R^2=4R^2 \sin^2 (A)+ 4R^2 \sin^2 (B) + 4R^2 \sin^2 C$$
$$8R^2=4R^2(\sin^2 (A)+\sin^2 (B) +\sin^2 (C)$$
$$2=\sin^2 (A)+\sin^2 (B)+\sin^2 (C)$$
 A: 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.
A: 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 $2s$
$$1-(\cos^2\alpha + \cos^2 \beta + \cos^2 \gamma)- 2 \cos \alpha \cos \beta \cos \gamma = - 4 \cos s\cos (s-\alpha) \cos (s-\beta) \cos (s - \gamma)$$
A: 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. $2R=r^2+s^2$).
A: By the law of cosines,
\begin{align} 
a^2+b^2+c^2
&=
2ab\cos\gamma+
2bc\cos\alpha+
2ca\cos\beta
\tag{1}\label{1}
\end{align}
Using expressions for the area $S$ of triangle
\begin{align} 
S&=\tfrac12ab\sin\gamma=
\tfrac12bc\sin\alpha=
\tfrac12ca\sin\beta
,\\
S&=2R^2\sin\alpha\sin\beta\sin\gamma
,
\end{align}
we have 
\begin{align} 
a^2+b^2+c^2
&=
2ab\sin\gamma\,\cot\gamma+
2bc\sin\alpha\,\cot\alpha+
2ca\sin\beta\,\cot\beta
\\
&=4S(\cot\alpha+\cot\beta+\cot\gamma)
\\
&=8R^2
\sin\alpha\sin\beta\sin\gamma
(\cot\alpha+\cot\beta+\cot\gamma)
=8R^2
,
\end{align}
\begin{align} 
\cos\alpha\sin\beta\sin\gamma+
\sin\alpha\cos\beta\sin\gamma+
\sin\alpha\sin\beta\cos\gamma
&=1
,\\
(\cos\alpha\sin\beta+\sin\alpha\cos\beta)\sin\gamma+
\sin\alpha\sin\beta\cos\gamma
&=1
,\\
\sin(\alpha+\beta)\sin(\alpha+\beta)
+
\sin\alpha\sin\beta\cos\gamma
&=1
,\\
\sin^2(\alpha+\beta)
+
\sin\alpha\sin\beta\cos\gamma
&=1
,\\
1-\cos^2(\alpha+\beta)
-
\sin\alpha\sin\beta\cos(\alpha+\beta)
&=1
,\\
-\cos(\alpha+\beta)(\sin\alpha\sin\beta+\cos(\alpha+\beta))
&=0
,\\
-\cos(\alpha+\beta)(\sin\alpha\sin\beta+\cos\alpha\cos\beta-\sin\alpha\sin\beta)
&=0
,\\
-\cos(\alpha+\beta)\cos\alpha\cos\beta&=0
,\\
\cos\alpha\cos\beta\cos\gamma&=0
.
\end{align}  
A: $$F=\sin^2A+\sin^2B+\sin^2C=1-\cos^2A+1-(\cos^2B-\sin^2C)$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
and $\cos(B+C)=\cos(\pi-A)=?$
$$2-F=\cos^2A+\cos(B+C)\cos(B-C)$$
$$=\cos^2A-\cos A\cos(B-C)$$
$$=\cos A\{\cos A-\cos(B-C)\}$$
$$=-\cos A\{\cos(B+C)+\cos(B-C)\}=?$$
