$23 \cdot 8$ In the regular tetrahedron $ABCD$, $E$ is on edge $AB$, $F$ is on edge $CD$, and it is given that
$$
\frac{AE}{EB}=\frac{CF}{FD}=\lambda \quad(0<\lambda<+\infty).
$$

Let $f(\lambda)=\alpha_{\lambda}+\beta_{\lambda}$, where $\alpha_{\lambda}$ is the angle between $EF$ and $AC$, and $\beta_{\lambda}$ is the angle between $EF$ and $BD$. Then
(A) $f(\lambda)$ is monotonically increasing on $(0,+\infty)$.
(B) $f(\lambda)$ is monotonically decreasing on $(0,+\infty)$.
(C) $f(\lambda)$ is monotonically increasing on $(0,1)$ and monotonically decreasing on $(1,+\infty)$.
(D) $f(\lambda)$ is a constant on $(0,+\infty)$.
(China High School Mathematics League, 1997)
[Solution] Since $ABCD$ is a regular tetrahedron, $AC \perp BD$. Draw $EG \parallel AC$ intersecting $BC$ at $G$, and connect $GF$.
Then $\alpha_{\lambda}=\angle GEF$, and $\frac{CG}{GB}=\frac{AE}{EB}=\frac{CF}{FD}$.
Thus, $GF \parallel BD$.
Therefore, $GF \perp EG$, and $\beta_{\lambda}=\angle EFG$.
$$
\begin{aligned}
\therefore \quad f(\lambda) & =\alpha_{\lambda}+\beta_{\lambda}=\angle GEF+\angle EFG \\
& =180^{\circ}-\angle EGF=90^{\circ} .
\end{aligned}
$$

Hence, the answer is $(D)$.