* * Let $E$, $F$, and $G$ be the midpoints of the edges $AB$, $BC$, and $CD$ of the regular tetrahedron $ABCD$ (as shown in Figure a). Then the size of the dihedral angle $C-FG-E$ is ().
(A) $\arcsin \frac{\sqrt{6}}{3}$
(B) $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$
(C) $\frac{\pi}{2}-\arctan \sqrt{2}$
(D) $\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$

Figure a
Figure b
Parse As shown in Figure b, construct the circumscribed cube of the regular tetrahedron, then $E, F, G$ are the centers of the respective square faces. $E G / / B N, F G / / B D$, hence plane $E F G / /$ plane $B N D$. Let the dihedral angle between plane $C B D$ and plane $N B D$ be $\alpha$, the required dihedral angle is $\pi-\alpha$. Take the midpoint $M$ of $B D$, in the equilateral $\triangle C B D$, $C M \perp B D$. Also, the diagonals of a square are perpendicular to each other, i.e., $N M \perp B D$, thus $\alpha=\angle C M N=\operatorname{arccot} \frac{M N}{C N}=\operatorname{arccot} \frac{\sqrt{2}}{2}$, or $\alpha=$ $\arccos \frac{S_{\triangle N B D}}{S_{\triangle C B D}}=\arccos \frac{\sqrt{3}}{3}$. Therefore, the dihedral angle $C-F G-E$ is $\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$. Choose D.

Comment The above solution uses the method of "completing the shape", transforming the regular tetrahedron into a cube, thereby utilizing the line and plane relationships within the cube to achieve the purpose of solving the problem.