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4.(2024·淮安全湖一模)如图,将一张三角形纸片ABC的边CB与直线l重合放置,$\angle ACB = 90^{\circ}$，$\angle ABC = 30^{\circ}$，AC = 1. 三角形纸片ABC的直角顶点C沿CB方向向终点B运动,运动过程中始终保持点A的对应点A'落在边AB上,记点C,B的对应点分别为C',B',连接BB'.
(1)当A'B'//BC时,$\angle A'C'C =$______°,AA' = ______.
(2)设点A',B'到直线BC的距离分别为a,b,求$a^{2}$与$b^{2}$之间满足的数量关系.
(3)运动过程中$\angle A'BB'$的度数是否为一个定值？如果是,请求出这个定值；如果不是,请说明理由.
(4)当点C'从点C运动到点B时,A'B'的中点P运动的路径长为______.
　　　

(1) 如图①，过点A'作A'E⊥BC于点E. ∵ ∠ACB = 90°，∠ABC = 30°，AC = 1，∴ ∠BAC = 60°，AB = 2AC = 2. 根据题意，得∠B'A'C' = ∠BAC = 60°，∠A'B'C' = ∠ABC = 30°，∠A'C'B' = ∠ACB = 90°，A'C' = AC = 1. ∵ A'B'//BC，∴ ∠B'C'B = ∠A'B'C' = 30°. ∴ ∠A'C'C = 180° - ∠A'C'B' - ∠B'C'B = 180° - 90° - 30° = 60°. ∴ 在Rt△A'EC'中，A'E = A'C'sin∠A'C'E = 1×sin60° = $\frac{\sqrt{3}}{2}$. ∵ ∠A'BE = 30°，∠A'EB = 90°，∴ 在Rt△A'EB中，A'B = 2A'E = $\sqrt{3}$. ∴ AA' = AB - A'B = 2 - $\sqrt{3}$. (2) 如图②，过点A'作A'E⊥BC于点E，过点B'作B'D⊥l于点D，则∠A'EB = ∠B'DC' = 90°. 根据题意，得A'E = a，B'D = b. ∵ ∠ABC = 30°，∴ 在Rt△A'EB中，BE = $\frac{A'E}{\tan30^{\circ}}$=$\sqrt{3}$a. ∵ ∠A'B'C' = 30°，∠A'C'B' = 90°，∴ 在Rt△A'C'B'中，tan30° = $\frac{A'C'}{B'C'}$=$\frac{\sqrt{3}}{3}$. ∵ ∠A'C'E + ∠B'C'D = ∠A'C'E + ∠EA'C' = 90°，∴ ∠EA'C' = ∠DC'B'. ∵ ∠A'EC' = ∠C'DB' = 90°，∴ △A'EC'∽△C'DB'. ∴ $\frac{A'E}{C'D}$=$\frac{EC'}{DB'}$=$\frac{A'C'}{C'B'}$=$\frac{\sqrt{3}}{3}$. ∴ C'D = $\sqrt{3}$A'E = $\sqrt{3}$a，B'D = $\sqrt{3}$EC'. ∴ BE = C'D = $\sqrt{3}$a. ∴ BE - C'B = C'D - C'B，即EC' = BD. ∴ B'D = $\sqrt{3}$EC' = $\sqrt{3}$BD. ∴ EC' = BD = $\frac{\sqrt{3}}{3}$B'D = $\frac{\sqrt{3}}{3}$b. 在Rt△A'EC'中，∵ A'E² + EC'² = A'C'²，∴ a² + ($\frac{\sqrt{3}}{3}$b)² = 1²，即a² + $\frac{1}{3}$b² = 1. (3) ∠A'BB'的度数是定值. 由(2)可知，B'D = $\sqrt{3}$BD. ∵ ∠B'DB = 90°，∴ 在Rt△B'DB中，tan∠B'BD = $\frac{B'D}{BD}$=$\sqrt{3}$. ∴ ∠B'BD = 60°. ∵ ∠ABC = 30°，∴ ∠A'BB' = 180° - ∠ABC - ∠B'BD = 180° - 30° - 60° = 90°，为定值. (4) 如图③，连接C'P，BP. ∵ ∠A'C'B' = ∠A'BB' = 90°，A'B' = AB = 2，P为AB的中点，∴ BP = PC' = $\frac{1}{2}$A'B' = 1. ∴ 当点C'从点C运动到点B时，A'B'的中点P在以点B为圆心，1为半径的圆上. 当点C'在点C处时，△A'B'C'与△ABC重合，此时点P在AB的中点上. 当点C'在点B处时，△A'B'C'的位置如图④所示. ∵ A'C' = 1 = $\frac{1}{2}$AB，∴ A'B'的中点P运动的路径为$\overset{\frown}{A'P}$. 易得△A'BP为等边三角形，∴ ∠A'BP = 60°. ∴ A'B'的中点P运动的路径长为$\frac{60\pi\times1}{180}$=$\frac{\pi}{3}$.