3. As shown in Figure 1, in the convex quadrilateral $A B C D$, $A D=D C$ $=C B, A C \neq B D$, and $\angle D A B+\angle C B A=$ $120^{\circ}$. A circle $\omega$ is drawn through points $A, B, D$. Then the point $C'$, which is the reflection of $C$ over $A B$, is ( ).
(A) on the circle $\omega$
(B) outside the circle $\omega$
(C) inside the circle $\omega$
(D) cannot be determined
3. A.

As shown in Figure 4, let
$$
\begin{array}{c}
\angle D A C \\
=\angle D C A=x, \\
\angle C D B \\
=\angle C B D=y . \\
\text { By } \angle D A B+ \\
\angle C B A=120^{\circ} \text {, we get } \\
\angle A D C+\angle D C B=240^{\circ} . \\
\text { Therefore }, 2(x+y)=120^{\circ}, x+y=60^{\circ} .
\end{array}
$$

Then $\angle A D B+\angle A C B=180^{\circ}$, which means
$$
\angle A D B+\angle A C^{\prime} B=180^{\circ} \text {. }
$$

Thus, points $A, D, B, C^{\prime}$ are concyclic.