17・11 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 $\angle E C F=$
(A) $60^{\circ}$.
(B) $45^{\circ}$.
(C) $30^{\circ}$.
(D) Uncertain.
(Anhui Province, China Junior High School Mathematics Competition, 1997)
[Solution] Let $\angle E C F=x, \angle A C F=\alpha, \angle B C E=\beta$. Then we have $\angle E F C=x+\beta, \angle F E C=x+\alpha, x+\alpha+\beta=90^{\circ}$.
$$
\begin{array}{l}
\therefore \angle E F C+\angle F E C+x=180^{\circ}, \\
3 x+\alpha+\beta=180^{\circ}, 2 x=90^{\circ}, x=45^{\circ} .
\end{array}
$$

Therefore, the answer is $(B)$.