Three, (15 points) In $\triangle A B C$, $B C=a, A C=$ $b, A B=c$, and it satisfies
$$
a^{4}+b^{4}+\frac{1}{2} c^{4}=a^{2} c^{2}+b^{2} c^{2} \text {. }
$$

Determine the shape of $\triangle A B C$.
Three, from $a^{4}+b^{4}+\frac{1}{2} c^{4}=a^{2} c^{2}+b^{2} c^{2}$, we get $\left(a^{2}-\frac{1}{2} c^{2}\right)^{2}+\left(b^{2}-\frac{1}{2} c^{2}\right)^{2}=0$.

It is easy to see that $a^{2}=\frac{1}{2} c^{2}$ and $b^{2}=\frac{1}{2} c^{2}$.
Therefore, $a=b$ and $a^{2}+b^{2}=c^{2}$.
Thus, $\triangle A B C$ is an isosceles right triangle.