【模型建立】
\((1)\)如图\(1\),等腰直角三角形\(ABC\)中,\(∠ACB=90^{\circ}\),\(CB=CA\),直线\(ED\)经过点\(C\),过\(A\)作\(AD⊥ED\)于点\(D\),过\(B\)作\(BE⊥ED\)于点\(E\).
求证:\(\triangle BEC\)≌\(\triangle CDA\);
【模型应用】
\((2)①\)已知直线\(l_{1}\):\(y= \dfrac {4}{3}x+4\)与坐标轴交于点\(A\)、\(B\),将直线\(l_{1}\)绕点\(A\)逆时针旋转\(45^{o}\)至直线\(l_{2}\),如图\(2\),求直线\(l_{2}\)的函数表达式;
\(②\)如图\(3\),长方形\(ABCO\),\(O\)为坐标原点,点\(B\)的坐标为\((8,-6)\),点\(A\)、\(C\)分别在坐标轴上,点\(P\)是线段\(BC\)上的动点,点\(D\)是直线\(y=-2x+6\)上的动点且在第四象限\(.\)若\(\triangle APD\)是以点\(D\)为直角顶点的等腰直角三角形,请直接写出点\(D\)的坐标.