We have,
$$
\begin{array}{cc}
& 8 R^2=a^2+b^2+c^2 \\
\Rightarrow \quad & 8 R^2=4 R^2\left(\sin ^2 A+\sin ^2 B+\sin ^2 C\right) \\
& {\left[\begin{array}{l}
\because R=\frac{a}{2 \sin A}=\frac{b}{2 \sin B}=\frac{c}{2 \sin C}, \\
\text { where } R \text { is circumradius }
\end{array}\right]} \\
\Rightarrow & \sin ^2 A+\sin ^2 B+\sin ^2 C=2 \\
\Rightarrow & 1-\cos ^2 A+1-\cos ^2 B+\sin ^2 C=2 \\
\Rightarrow & \left(\cos ^2 A-\sin ^2 C\right)+\cos ^2 B=0
\end{array}
$$

$$
\begin{array}{rrr}
\Rightarrow & \cos (A+C) \cos (A-C)+\cos ^2 B=0 \\
\Rightarrow & -\cos B\{\cos (A-C)-\cos B\}=0 \\
\Rightarrow & -\cos B\{\cos (A-C)+\cos (A+C)\}=0 \\
\Rightarrow & -2 \cos A \cos B \cos C=0 \\
\Rightarrow & \cos A=0 \text { or } \cos B=0 \text { or } \cos C=0 \\
\Rightarrow & A=\frac{\pi}{2} \text { or } B=\frac{\pi}{2} \text { or } C=\frac{\pi}{2}
\end{array}
$$
Hence, $\triangle A B C$ is a right angled triangle.