Example 1 Let points $E, F, G$ be the midpoints of the edges $AB, BC, CD$ of tetrahedron $ABCD$, respectively. Then the size of the dihedral angle $C-FG-E$ is ( ).
(A) $\arcsin \frac{\sqrt{6}}{3}$
(B) $\frac{\pi}{2}-\arctan \sqrt{2}$
(C) $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$
(D) $\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$
(1998, National High School Competition)
Solution: As shown in the figure,
1, draw $E E_{1} \perp$
plane $B C D$, then
$E E_{1} \perp B G$.
Since $A B C D$

is a regular tetrahedron,
we have $A C \perp B D$.
Also, since $E F$
$/ / A C, F G$ //
$B D$,
$$
\therefore E F \perp F G \text {. }
$$

Therefore, $E_{1} F \perp F G$. Thus, $\angle E F E_{1}$ is the plane angle of the dihedral angle $E-F G-B$, denoted as $\alpha$.

Assume the edge length of the regular tetrahedron $A B C D$ is 1, then its height
$$
\begin{aligned}
A H & =\sqrt{A B^{2}-B H^{2}} \\
& =\sqrt{1^{2}-\left(\frac{\sqrt{3}}{3}\right)^{2}}=\frac{\sqrt{6}}{3} . \\
\therefore E E_{1} & =\frac{1}{2} A H=\frac{\sqrt{6}}{6} . \\
\therefore E F & =\frac{1}{2} A C=\frac{1}{2}, \\
\therefore \alpha & =\arcsin \frac{\frac{\sqrt{6}}{6}}{\frac{1}{2}}=\arcsin \frac{\sqrt{6}}{3} .
\end{aligned}
$$

Therefore, the size of the dihedral angle $C-F G-E$ is $\pi-\arcsin \frac{\sqrt{6}}{3}=\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$. Hence, the answer is (D).