答案：
解: (1)证明: $\because \angle D E C=\angle B+\angle B D E=\angle C E F+\angle D E F, \angle D E F=\angle B$,
$$
\begin{aligned}
& \therefore \angle C E F=\angle B D E . \\
& \because A B=A C, \\
& \therefore \angle C=\angle B . \\
& \text { 又 } \because C E=B D, \\
& \therefore \triangle B D E \cong \triangle C E F .
\end{aligned}
$$
(2) $\because \triangle B D E \cong \triangle C E F$
$$
\begin{aligned}
& \therefore D E=F E . \\
& \therefore \triangle D E F \text { 是等腰三角形. } \therefore \angle E D F=\angle E F D \\
& \text { 又 } \because \text { 在 } \triangle A B C \text { 中, } A B=A C, \angle A=40^{\circ} \\
& \therefore \angle B=70^{\circ}, \because \angle D E F=\angle B \therefore \angle D E F=70^{\circ} \\
& \therefore \angle E D F=\angle E F D=\frac{1}{2} \times\left(180^{\circ}-70^{\circ}\right)=55^{\circ} .
\end{aligned}
$$