$3.0 、 I$ are the circumcenter and incenter of acute $\triangle A B C$, respectively. $O^{\prime} 、 I^{\prime}$ are the reflections of $O 、 I$ over $B C$, respectively. It is known that $A 、 B 、 O^{\prime} 、 C$ are concyclic. Then, the position of point $I^{\prime}$ is ( ).
(A) outside $\odot O$
(B) inside $\odot O$
(C) on $\odot O$
(D) cannot be determined
3.C.

As shown in Figure 4, since points $A$, $B$, $O'$, and $C$ are concyclic, we have
$$
\begin{array}{l}
\angle A + \angle B O' C = 180^{\circ}. \\
\text{Therefore, } \angle A + \angle B O C \\
= 180^{\circ}. \\
\text{Also, } \angle B O C = 2 \angle A, \text{ so }
\end{array}
$$

$\angle A = 60^{\circ}$.
Thus, $\angle B I C = 90^{\circ} + \frac{\angle A}{2} = 90^{\circ} + \frac{60^{\circ}}{2} = 120^{\circ}$.
Hence, $\angle B I' C = \angle B I C = 120^{\circ}$.
At this point, $\angle A + \angle B I' C = 60^{\circ} + 120^{\circ} = 180^{\circ}$, which means points $A$, $B$, $I'$, and $C$ are also concyclic.