1. 如图,将$\triangle ABC绕点C顺时针旋转得到\triangle DEC$,点$D落在线段AB$上.
(1)求证:$DC平分\angle ADE$;
(2)连接$BE$,求证:$\angle A= \angle CBE$.

证明: (1) 由已知得$\triangle ABC \cong \triangle DEC$,
$\therefore \angle A = \angle CDE$,$AC = DC$,
$\therefore \angle A = \angle ADC$.
$\therefore \angle CDE = \angle ADC$,
$\therefore DC$平分$\angle ADE$;
(2)$\because CA = CD$,$CB = CE$,
$\therefore \angle A = \angle CDA$,$\angle CBE = \angle CEB$,
又$\because \angle ACB = \angle DCE$,
$\therefore \angle ACD = \angle BCE$,
$\therefore \angle A = \angle CBE$.

2. 如图,将$\triangle ABC绕点A逆时针旋转得到\triangle ADE$,点$D在BC$上,$\angle B= 70^{\circ}$.
(1)求$\angle CDE$的大小;
(2)若$\angle C= 30^{\circ}$,$AC与DE交于点O$,求证:$OE= DC$.

解: (1) 由旋转得$\triangle ABC \cong \triangle ADE$,
$\therefore AD = AB$,$\angle B = \angle ADE = 70^{\circ}$,
$\therefore \angle ABD = \angle ADB = 70^{\circ}$,
$\therefore \angle CDE = 40^{\circ}$;
(2)$\because \angle C = 30^{\circ}$,
$\angle ADB = \angle B = 70^{\circ}$,
$\therefore \angle BAD = 40^{\circ}$,$\angle BAC = 80^{\circ}$,
$\therefore \angle DAC = 40^{\circ}$,
$\because \angle BAC = \angle DAE$,
$\therefore \angle CAE = \angle BAD = 40^{\circ} = \angle DAC$,
$\because AC = AE$,$\angle E = \angle C$,
$\therefore \triangle AOE \cong \triangle ADC$,$\therefore OE = DC$.

3. (2025 黄冈)如图,在$\triangle ABC$中,$AB= AC$,$\angle BAC= \alpha$,$D为\triangle ABC$内一点,将$AD绕点A逆时针旋转\alpha得到AE$,连接$DE$,$BD$,$CE$.
(1)求证:$BD= CE$;
(2)若$\alpha =42^{\circ}$,$DE\perp AC$,求$\angle BAD$的度数.

解: (1) 由旋转得$AD = AE$,
$\angle DAE = \angle BAC$,
$\therefore \angle CAE = \angle BAD$,
$\because AB = AC$,$\therefore \triangle ABD \cong \triangle ACE$,
$\therefore BD = CE$;
(2)$\because AD = AE$,
$\angle DAE = 42^{\circ}$,$DE \perp AC$,
$\therefore \angle CAE = \frac{1}{2} \angle DAE = 21^{\circ}$.
由(1)得$\triangle ABD \cong \triangle ACE$,
$\therefore \angle BAD = \angle CAE = 21^{\circ}$.

4. (2025 沈阳)如图,将$\triangle ABC绕着点A顺时针旋转得到\triangle ADE$,延长$CB$,$DE交于点F$.
(1)求证:$\angle CAE+\angle CFD= 180^{\circ}$;
(2)若$\angle BAC= 50^{\circ}$,$AD// CE$,求$\angle DFC$的度数.

解: (1) 由旋转得$\triangle ABC \cong \triangle ADE$,
$\therefore \angle AED = \angle ACB$.
$\because \angle AED + \angle AEF = 180^{\circ}$,
$\therefore \angle AEF + \angle ACF = 180^{\circ}$,
$\because \angle ACB + \angle AEF + \angle DFC + \angle CAE = 360^{\circ}$,
$\therefore \angle DFC + \angle EAC = 180^{\circ}$;
(2) 由旋转得$AE = AC$,
$\angle BAC = \angle DAE$.
$\because AD // CE$,
$\therefore \angle AEC = \angle DAE = 50^{\circ}$,
$\therefore \angle AEC = \angle ACE = 50^{\circ}$,
$\therefore \angle EAC = 80^{\circ}$.
由(1)得$\angle DFC + \angle EAC = 180^{\circ}$,
$\therefore \angle DFC = 100^{\circ}$.