如图,四棱锥\(P-ABCD\)中,底面\(ABCD\)为梯形,\(PD⊥\)底面\(ABCD\),\(AB/\!/CD\),\(AD⊥CD\),\(AD=AB=1\),\(BC= \sqrt {2}\).
\((\)Ⅰ\()\)求证:平面\(PBD⊥\)平面\(PBC\);
\((\)Ⅱ\()\)设\(H\)为\(CD\)上一点,满足\( \overrightarrow{CH}=2 \overrightarrow{HD}\),若直线\(PC\)与平面\(PBD\)所成的角的正切值为\( \dfrac { \sqrt {6}}{3}\),求二面角\(H-PB-C\)的余弦值.