4. As shown in the figure, in $\triangle A B C$, $\angle A C B=90^{\circ}, A C=$ $A E, B C=B F$. Then $\angle E C F$ $=(\quad)$.
(A) $60^{\circ}$
(B) $45^{\circ}$
(C) $30^{\circ}$
(D) Uncertain
4. (B).

Let $\angle E C F=x, \angle A C F=\alpha, \angle B C E=\beta$. Then we have
$$
\begin{array}{l}
\angle E F C=x+\beta, \angle F E C=x+\alpha, x+\alpha+\beta=90^{\circ} . \\
\quad \therefore \angle E F C+\angle F E C+x=180^{\circ}, 3 x+\alpha+\beta==
\end{array}
$$
$$
180^{\circ}, 2 x=90^{\circ}, x=45^{\circ} .
$$