25. (本小题满分10分)
如图,在矩形$ABCD$中,$AB = 6$,$AD = 11$. 三角尺的直角顶点$P在AD$上滑动时(点$P与点A$,$D$不重合),一直角边始终经过点$C$,另一直角边与$AB交于点E$.
(1) $\triangle CDP与\triangle PAE$相似吗？如果相似,请写出证明过程.
(2) 当$\angle PCD = 30^{\circ}$时,求$AE$的长.
(3) 是否存在这样的点$P$,使$\triangle CDP的周长等于\triangle PAE$周长的2倍？若存在,求出$DP$的长；若不存在,请说明理由.

(1)
(2)
(3)

(1)$\triangle CDP\backsim\triangle PAE$.证明:$\because$四边形ABCD是矩形,$\therefore \angle D=\angle A=90^{\circ}$,$CD=AB=6$,$\therefore \angle PCD+\angle DPC=90^{\circ}$.又$\because \angle CPE=90^{\circ}$,$\therefore \angle EPA+\angle DPC=90^{\circ}$,$\therefore \angle PCD=\angle EPA$,$\therefore \triangle CDP\backsim\triangle PAE$.
(2)在$Rt\triangle PCD$中,由$\tan\angle PCD=\frac{PD}{CD}$,$\therefore PD=CD\cdot\tan\angle PCD=6\cdot\tan30^{\circ}=6×\frac{\sqrt{3}}{3}=2\sqrt{3}$,$\therefore AP=AD-PD=11-2\sqrt{3}$.解法1:由$\triangle CDP\backsim\triangle PAE$,知$\frac{PD}{AE}=\frac{CD}{AP}$,$\therefore AE=\frac{PD\cdot AP}{CD}=\frac{2\sqrt{3}×(11-2\sqrt{3})}{6}=\frac{11}{3}\sqrt{3}-2$.解法2:由$\triangle CDP\backsim\triangle PAE$,知$\angle EPA=\angle PCD=30^{\circ}$,$\therefore AE=AP\cdot\tan\angle EPA=(11-2\sqrt{3})\cdot\tan30^{\circ}=\frac{11}{3}\sqrt{3}-2$.
(3)假设存在满足条件的点P,设$DP=x$,则$AP=11-x$.$\because \triangle CDP\backsim\triangle PAE$,$\therefore \frac{CD}{AP}=2$,即$\frac{6}{11-x}=2$,解得$x=8$.此时$AP=3$,$AE=4$.