$17 \cdot 37$ As shown in the figure, in $\triangle A B C$, $B C$ is the largest side, $A B=A C, C D=B F, B D=$ $C E$, then the range of $\angle D E F$ is
(A) $0^{\circ}<\angle D E F<45^{\circ}$.
(B) $45^{\circ}<\angle D E F<60^{\circ}$.
(C) $60^{\circ} \leqslant \angle D E F<90^{\circ}$.
(D) None of the above.
[Solution] Since $C D=B F, C E=B D, \angle C=\angle B$,
thus $\triangle C D E \cong \triangle B F D$.
Therefore, $D E=D F, \angle C D E=\angle B F D$.
But $\quad \angle C D F=\angle B F D+\angle B$.
So $\angle E D F=\angle B$.
Hence $\angle D E F=\frac{1}{2}\left(180^{\circ}-\angle B\right)=90^{\circ}-\frac{1}{2} \angle B$.
Since $60^{\circ} \geqslant \angle B>0^{\circ}$.
Therefore $30^{\circ} \geqslant \frac{1}{2} \angle B>0^{\circ}$, thus $60^{\circ} \leqslant \angle D E F<90^{\circ}$. Hence the answer is $(C)$.