4. As shown in Figure $2, \odot O_{1} 、 \odot O_{2}$ intersect at points $A 、 B$, $O_{1}$ 、 $O_{2}$ are on opposite sides of $A B$, $A C$ is the diameter of $\odot O_{1}$, extending $B C$ intersects $\odot O_{2}$ at point $D, E$ is a point on arc $\overparen{B C}$, extending $E B$ intersects $\odot O_{2}$ at point $F, M 、 N$ are the midpoints of $C D 、 E F$ respectively, $A C=2 C E$. Then $\angle A M N=$ $\qquad$
4. $60^{\circ}$.

Connect $A E, A F$.
Then $\angle A E C=90^{\circ}, \cos \angle A C E=\frac{C E}{A C}=\frac{1}{2}$
$\Rightarrow \angle A C E=60^{\circ}$.
It is easy to prove, $\triangle A C D \backsim \triangle A E F$.
Also, $A M, A N$ are the corresponding medians of $\triangle A C D, \triangle A E F$, then
$$
\begin{array}{l}
\triangle A C M \backsim \triangle A E N \\
\Rightarrow \angle C A M=\angle E A N \\
\Rightarrow \angle C A E=\angle M A N \\
\Rightarrow \triangle A C E \backsim \triangle A M N \\
\Rightarrow \angle A M N=\angle A C E=60^{\circ} .
\end{array}
$$