證明: (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$l_2$上取點$D，$使$AD = AB，$過點$D$作$DE\perp OA，$垂足為$E。$$\because$直線$y=\frac{4}{3}x + 4$與坐標(biāo)軸交于點$A，$$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)。$設(shè)直線$l_2$的函數(shù)表達式為$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$直線$l_2$的函數(shù)表達式為$y = -\frac{3}{4}x-\frac{9}{4}。$
(3) 分三種情況：
①當(dāng)$\angle CPD = 90^{\circ}$時，過$P$作$MH// x$軸，過$D$作$DH// y$軸，$MH$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。$設(shè)$PH = a，$則$D(6 + a,a - 8 - 6)。$$\because$點$D$是直線$y=-2x + 2$上的動點且在第四象限內(nèi)，$\therefore a - 8 - 6=-2(6 + a)+2，$解得$a=\frac{4}{3}，$$\therefore D\left(\frac{22}{3},-\frac{38}{3}\right)。$
②當(dāng)$\angle PCD = 90^{\circ}$時，此時點$P$與點$A$重合，過$D$作$DE\perp y$軸于$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當(dāng)$\angle CDP = 90^{\circ}$時，過點$D$作$MQ// x$軸，延長$AB$交$MQ$于$Q，$則$\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，$$DQ = CM。$設(shè)$CM = b，$則$DM = 6 - b，$$AQ = 8 + b，$$\therefore D(6 - b,-8 - b)。$$\because$點$D$是直線$y=-2x + 2$上的動點且在第四象限內(nèi)，$\therefore -8 - b=-2(6 - b)+2，$解得$b=\frac{2}{3}，$$\therefore D\left(\frac{16}{3},-\frac{26}{3}\right)。$
綜上，點$D$的坐標(biāo)為$\left(\frac{22}{3},-\frac{38}{3}\right)$或$(8,-14)$或$\left(\frac{16}{3},-\frac{26}{3}\right)。$
