证明: (1) $\because AD\perp ED，$$BE\perp ED，$$\therefore\angle ADC=\angle CEB = 90^{\circ}。$又$\because\angle ACD+\angle ACB+\angle BCE = 180^{\circ}，$$\angle ACB = 90^{\circ}，$$\therefore\angle ACD+\angle BCE = 90^{\circ}。$又$\because\angle ACD+\angle DAC = 90^{\circ}，$$\therefore\angle DAC=\angle ECB。$在$\triangle BEC$和$\triangle CDA$中，$\begin{cases}\angle CEB=\angle ADC\\\angle ECB=\angle DAC\\BC = CA\end{cases}，$$\therefore\triangle BEC\cong\triangle CDA(AAS)。$
(2) 在$l_2$上取点$D，$使$AD = AB，$过点$D$作$DE\perp OA，$垂足为$E。$$\because$直线$y=\frac{4}{3}x + 4$与坐标轴交于点$A，$$B，$$\therefore A(-3,0)，$$B(0,4)，$$\therefore OA = 3，$$OB = 4。$由
(1)同理得$\triangle BOA\cong\triangle AED，$$\therefore DE = OA = 3，$$AE = OB = 4，$$\therefore OE = 7，$$\therefore D(-7,3)。$设直线$l_2$的函数表达式为$y = kx + b，$$\therefore\begin{cases}-7k + b = 3\\-3k + b = 0\end{cases}，$解得$\begin{cases}k = -\frac{3}{4}\\b = -\frac{9}{4}\end{cases}，$$\therefore$直线$l_2$的函数表达式为$y = -\frac{3}{4}x-\frac{9}{4}。$
(3) 分三种情况：
①当$\angle CPD = 90^{\circ}$时，过$P$作$MH// x$轴，过$D$作$DH// y$轴，$MH$和$DH$交于$H。$$\because\triangle CPD$是等腰直角三角形，$\therefore\angle CPD = 90^{\circ}，$$CP = PD。$同
(1)得$\triangle CMP\cong\triangle PHD，$$\therefore DH = PM = 6，$$PH = CM。$设$PH = a，$则$D(6 + a,a - 8 - 6)。$$\because$点$D$是直线$y=-2x + 2$上的动点且在第四象限内，$\therefore a - 8 - 6=-2(6 + a)+2，$解得$a=\frac{4}{3}，$$\therefore D\left(\frac{22}{3},-\frac{38}{3}\right)。$
②当$\angle PCD = 90^{\circ}$时，此时点$P$与点$A$重合，过$D$作$DE\perp y$轴于$E。$$\because\triangle CPD$是等腰直角三角形，同
(1)得$\triangle AOC\cong\triangle CED，$$\therefore OA = CE = 6，$$OC = DE = 8，$$\therefore D(8,-14)。$
③当$\angle CDP = 90^{\circ}$时，过点$D$作$MQ// x$轴，延长$AB$交$MQ$于$Q，$则$\angle Q=\angle DMC = 90^{\circ}。$$\because\triangle CDP$是等腰直角三角形，同
(1)得$\triangle PQD\cong\triangle DMC，$$\therefore PQ = DM，$$DQ = CM。$设$CM = b，$则$DM = 6 - b，$$AQ = 8 + b，$$\therefore D(6 - b,-8 - b)。$$\because$点$D$是直线$y=-2x + 2$上的动点且在第四象限内，$\therefore -8 - b=-2(6 - b)+2，$解得$b=\frac{2}{3}，$$\therefore D\left(\frac{16}{3},-\frac{26}{3}\right)。$
综上，点$D$的坐标为$\left(\frac{22}{3},-\frac{38}{3}\right)$或$(8,-14)$或$\left(\frac{16}{3},-\frac{26}{3}\right)。$
