(1) 如图 1, $C B$ 平分 $\angle A C D$, 求证: 四边形 $A B D C$ 是菱形;
(2) 如图 2 , 将 (1) 中的 $\triangle C D E$ 绕点 $C$ 逆时针旋转 (旋转角小于 $\angle B A C$ ), $B C, D E$ 的延长 线相交于点 $F$, 用等式表示 $\angle A C E$ 与 $\angle E F C$ 之间的数量关系, 并证明;
(3) 如图 3, 将 (1) 中的 $\triangle C D E$ 绕点 $C$ 顺时针旋转 (旋转角小于 $\angle A B C$ ), 若 $\angle B A D=\angle B C D$, 求 $\angle A D B$ 的度数.

\begin{aligned} & \because A B C \cong \triangle D E C, \\ & \therefore A C=D C . \\ & \because A B=A C, \\ & \therefore \angle A B C=\angle A C B, A B=D C . \\ & \because C B \text { 平分 } \angle A C D, \\ & \therefore \angle A C B=\angle D C B, \\ & \therefore \angle A B C=\angle D C B, \\ & \therefore A B / / C D, \\ & \therefore \text { 四边形 } A B D C \text { 是平行四边形. } \\ & \text { 又 } A B=A C, \\ & \therefore \text { 四边形 } A B D C \text { 是茭形. } \end{aligned}

(2) \begin{aligned} \text { 结论: } \angle A C E+\angle E F C=180^{\circ} . \\ & \because \triangle A B C \cong \triangle D E C, \\ & \therefore \angle A B C=\angle D E C . \\ & \because A B=A C, \\ & \therefore \angle A B C=\angle A C B, \\ & \therefore \angle A C B=\angle D E C . \\ & \because \angle A C B+\angle A C F=\angle D E C+\angle C E F=180^{\circ}, \\ & \therefore \angle A C F=\angle C E F . \\ & \because \angle C E F+\angle E C F+\angle E F C=180^{\circ}, \\ & \therefore \angle A C F+\angle E C F+\angle E F C=180^{\circ}, \\ \therefore \angle A C E &+\angle E F C=180^{\circ} . \end{aligned}

(3) 在 $A D$ 上取一点 $M$, 使得 $A M=C B$, 连接 $B M$.
$\because A B=C D, \angle B A D=\angle B C D$,
$\therefore \triangle A B M \cong \triangle C D B$,
$\therefore B M=B D, \angle M B A=\angle B D C$,
$\therefore \angle A D B=\angle B M D$,
$\because \angle B M D=\angle B A D+\angle M B A$,
$\therefore \angle A D B=\angle B C D+\angle B D C$.

\begin{aligned} &\because C A=C D, \\ &\therefore \angle C A D=\angle C D A=\alpha+2 \beta, \\ &\therefore \angle B A C=\angle C A D-\angle B A D=2 \beta, \\ &\therefore \angle A C B=\frac{1}{2}\left(180^{\circ}-\angle B A C\right)=90^{\circ}-\beta, \\ &\therefore \angle A C D=\left(90^{\circ}-\beta\right)+\alpha . \\ &\because \angle A C D+\angle C A D+\angle C D A=180^{\circ}, \\ &\therefore\left(90^{\circ}-\beta\right)+\alpha+2(\alpha+2 \beta)=180^{\circ}, \\ &\therefore \alpha+\beta=30^{\circ} \text {, 即 } \angle A D B=30^{\circ} . \end{aligned}