3. In the tetrahedron $A-B C D$, $A B \perp$ plane $B C D$. Then the relationship between $\angle C A D$ and $\angle C B D$ is ( ).
(A) $\angle C A D>\angle C B D$
(B) $\angle C A D=\angle C B D$
(C) $\angle C A D<\angle C B D$
(D) Uncertain
3. (D).

A popular misconception is to choose (C), in fact, the relationship between these two angles is uncertain. Consider Example 1, where the right triangle $\triangle ACD$ with side lengths $1, \sqrt{3}, 2$ is folded along the altitude $AB$ from the hypotenuse, forming the relevant edges of the three-story structure as shown in Figure 2, then
$$
\begin{array}{l}
A B=\frac{\sqrt{3}}{2}, A C=1, A D=\sqrt{3}, \\
B C=\frac{1}{2}, B D=\frac{3}{2} .
\end{array}
$$

Applying the cosine rule in $\triangle ACD$ and $\triangle BCD$, we have
$$
\begin{aligned}
C D^{2} & =A C^{2}+A D^{2}-2 A C \cdot A D \cdot \cos \angle C A D \\
& =4-2 \sqrt{3} \cos \angle C A D, \\
C D^{2} & =B C^{2}+B D^{2}-2 B C \cdot B D \cdot \cos \angle C B D \\
& =\frac{5}{2}-\frac{3}{2} \cos \angle C B D .
\end{aligned}
$$

Eliminating $C D$, we get
$$
\cos \angle C A D=\frac{\sqrt{3}}{4}(1+\cos \angle C B D)\angle C A D$,
(2) When $\angle C B D$ is an acute angle less than $30^{\circ}$, $\angle C B D<\angle C A D$;
(3) When $\angle C B D=\arccos \frac{\sqrt{3}}{4-\sqrt{3}}$,
$$
\begin{array}{l}
\cos \angle C A D=\frac{\sqrt{3}}{4}(1+\cos \angle C B D) \\
=\frac{\sqrt{3}}{4}\left(1+\frac{\sqrt{3}}{4-\sqrt{3}}\right)=\frac{\sqrt{3}}{4-\sqrt{3}} \\
=\cos \angle C B D .
\end{array}
$$

Since $\angle C A D$ and $\angle C B D$ are both acute angles, we have
$$
\angle C B D=\angle C A D \text {. }
$$
Therefore, they are equal.