23. Given equal circles $\odot A$ and $\odot B$ intersect at points $E$ and $F$, and each circle passes through the center of the other, line $A B$ intersects the two circles at points $C$ and $D$. Then $\angle C E D$ is ( ).
(A) $90^{\circ}$
(B) $105^{\circ}$
(C) $120^{\circ}$
(D) $135^{\circ}$
(E) $150^{\circ}$
23. C.

As shown in Figure 3, connect $A E$ and $B E$.
$$
\begin{array}{l}
\text { Given } A B=A E=B E \Rightarrow \angle E A B=60^{\circ} \\
\Rightarrow \angle E C D=30^{\circ} .
\end{array}
$$

Similarly, $\angle E D C=30^{\circ}$.
Thus, $\angle C E D=180^{\circ}-30^{\circ}-30^{\circ}=120^{\circ}$.