如图,在\(\triangle OAB\)中,\( \overrightarrow{OC}= \dfrac {1}{4} \overrightarrow{OA}\),\( \overrightarrow{OD}= \dfrac {1}{2} \overrightarrow{OB}\),\(AD\)与\(BC\)交于点\(M\),设\( \overrightarrow{OA}= \overrightarrow{a}\),\( \overrightarrow{OB}= \overrightarrow{b}\).
\((1)\)用\( \overrightarrow{a}\),\( \overrightarrow{b}\)表示\( \overrightarrow{OM}\);
\((2)\)在线段\(AC\)上取一点\(E\),在线段\(BD\)上取一点\(F\),使\(EF\)过\(M\)点,设\( \overrightarrow{OE}=p \overrightarrow{OA}\),\( \overrightarrow{OF}=q \overrightarrow{OB}\),求证:\( \dfrac {1}{7p}+ \dfrac {3}{7q}=1\).