24. 模型建立：如图①,在$Rt\triangle ABC$中,$\angle ACB = 90^{\circ}$,$AC = BC$。直线$ED经过点C$,过点$A作AD \perp ED于点D$,过点$B作BE \perp ED于点E$。求证：$\triangle BEC \cong \triangle CDA$。
模型应用：
(1)如图②,在平面直角坐标系中,直线$l_{1}:y = - \frac{4}{3}x - 4交x轴于点A$,交$y轴于点B$,将直线$l_{1}绕着点B逆时针旋转45^{\circ}至l_{2}$。过点$A作AC \perp l_{1}交l_{2}于点C$,过点$C作CD \perp x轴于点D$。求直线$l_{2}$所对应的函数解析式；
(2)如图③,在矩形$ABCO$中,$O$为坐标原点,点$B的坐标为(8, - 6)$,$A$、$C两点分别在x$轴、$y$轴上。$P是线段AB$上的动点,点$D$在第四象限,且是直线$y = - 2x + 6$上的一点。若$\triangle PCD是不以点C$为直角顶点的等腰直角三角形,直接写出点$D$的横坐标。

模型建立：
$\because BE\perp ED$，$AD\perp ED$，
$\therefore\angle E = \angle D = 90^{\circ}$.
$\therefore\angle EBC + \angle BCE = 90^{\circ}$.
$\because\angle ACB = 90^{\circ}$，
$\therefore\angle BCE + \angle ACD = 90^{\circ}$.
$\therefore\angle EBC = \angle ACD$.
$\because AC = BC$，$\therefore\triangle BEC\cong\triangle CDA$.
模型应用：
(1)$\because$直线$l_{1}:y = -\frac{4}{3}x - 4$交$x$轴于点$A$，交$y$轴于点$B$，
$\therefore A(-3,0)$，$B(0,-4)$.
$\therefore OA = 3$，$OB = 4$.$\because AC\perp l_{1}$，
$\therefore\angle BAC = 90^{\circ}$.$\because\angle ABC = 45^{\circ}$，
$\therefore\angle ACB = \angle ABC = 45^{\circ}$.$\therefore AC = AB$.
$\because CD\perp x$轴，$\angle AOB = 90^{\circ}$，
$\therefore$由模型建立，得$\triangle CDA\cong\triangle AOB$.
$\therefore CD = OA = 3$，$AD = OB = 4$.
$\therefore C(-7,-3)$.
设直线$l_{2}$所对应的函数解析式为$y = kx + b$.
将$B(0,-4)$，$C(-7,-3)$代入上式，
得$\begin{cases}b = - 4,\\-7k + b = - 3.\end{cases}$解得$\begin{cases}k = -\frac{1}{7},\\b = - 4.\end{cases}$
$\therefore$直线$l_{2}$所对应的函数解析式为$y = -\frac{1}{7}x - 4$.
(2)点$D$的横坐标为$4$，$\frac{20}{3}$，$\frac{28}{3}$.