1. 如图,在$\triangle ABC$中,$D是边BC$上一点,$AD = AB$.
(1)请用尺规作图法作$\triangle ABC绕点A旋转后得到的\triangle ADE$,使旋转后的$AB边与AD$边重合(保留作图痕迹,不写作法);
(2)连接$CE$,若$\angle B = 60^{\circ}$,求证:$CE = AE$.

解: (1) 如图, △ADE即为所求;

(2) ∵ AD = AB, ∠B = 60°,
∴ △ABD是等边三角形,
∴ ∠BAD = 60°,
由旋转可知,
∠CAE = ∠BAD = 60°,
AC = AE,
∴ △ACE为等边三角形,
∴ CE = AE.

2. 如图,已知$\triangle ABC$.
(1)在平面内将$\triangle ABC绕点C逆时针旋转60^{\circ}得到\triangle DEC$(点$D对应点A$)(要求用尺规作图,不写作法,保留作图痕迹);
(2)在(1)的条件下,若$DC\perp BC$,$AC = AB$,求证:直线$BA经过点D$.

解: (1) 如图所示, △CDE即为所求;

(2) 连接AD.
∵ CD ⊥ BC,
∴ ∠DCB = 90°,
∵ ∠DCA = 60°, CD = CA,
∴ △ACD是等边三角形,
∠ACB = 90° - 60° = 30°,
∴ ∠CAD = 60°,
∵ AC = AB,
∴ ∠ACB = ∠B = 30°,
∴ ∠CAB = 180° - 30° - 30° = 120°,
∴ ∠DAB = ∠DAC + ∠CAB = 60° + 120° = 180°,
∴ 直线BA经过点D.

3. 如图,将$\triangle ABC绕点C顺时针旋转得到\triangle DEC$,点$A的对应点D落在边BC$上.
(1)用无刻度的直尺和圆规作出$\triangle DEC$;
(2)连接$BE$,设旋转角为$\alpha(0^{\circ}＜\alpha＜180^{\circ})$,用含$\alpha的式子表示\angle BEC$的度数.

解: (1) 如图, △DEC为所作;

(2) ∵ 将△ABC绕点C顺时针旋转α得到△DEC,
∴ ∠BCE = α, CE = CB,
∴ ∠BEC = ∠EBC,
∴ ∠BEC = $\frac{1}{2}$(180° - α),
即 ∠BEC = 90° - $\frac{1}{2}$α.

4. 如图,在$Rt\triangle ABC$中,$\angle C = 90^{\circ}$,$AC = BC = \sqrt{2}$,将$\triangle ABC绕点A顺时针方向旋转60^{\circ}得到\triangle AB'C'$,点$B$,$C的对应点分别为B'$,$C'$.
(1)依题意,用尺规作图补全图形(保留作图痕迹,不写作法与证明);

(2)连接$C'B$,求$C'B$的长.

解: (1) 补全图形如图所示;

(2) 连接BB',
延长BC'交AB'于点D,
∵ AB = AB', ∠BAB' = 60°,
∴ △BAB'为等边三角形,
∴ BB' = BA,
∵ C'A = C'B',
∴ △BAC' ≌ △BB'C',
∴ ∠ABC' = ∠B'BC' = 30°,
∴ BD垂直平分AB',
又 ∵ AB = $\sqrt{2}$AC = 2,
∴ AD = $\frac{1}{2}$AB = 1,
∵ ∠B'AC' = 45°, ∴ C'D = AD = 1,
∵ BD = $\sqrt{AB^2 - AD^2} = \sqrt{2^2 - 1^2} = \sqrt{3}$,
∴ C'B = BD - C'D = $\sqrt{3}$ - 1.