10.
如图\(1\),已知四边形\(ABCD\)是菱形,\(G\)是线段\(CD\)上的任意一点时,连接\(BG\)交\(AC\)于\(F\),过\(F\)作\(FH/\!/CD\)交\(BC\)于\(H\),可以证明结论\( \dfrac {FH}{AB}= \dfrac {FG}{BG}\)成立\(.(\)考生不必证明\()\)
\((1)\)探究:如图\(2\),上述条件中,若\(G\)在\(CD\)的延长线上,其它条件不变时,其结论是否成立?若成立,请给出证明;若不成立,请说明理由;
\((2)\)计算:若菱形\(ABCD\)中\(AB=6\),\(∠ADC=60^{\circ}\),\(G\)在直线\(CD\)上,且\(CG=16\),连接\(BG\)交\(AC\)所在的直线于\(F\),过\(F\)作\(FH/\!/CD\)交\(BC\)所在的直线于\(H\),求\(BG\)与\(FG\)的长.
\((3)\)发现:通过上述过程,你发现\(G\)在直线\(CD\)上时,结论\( \dfrac {FH}{AB}= \dfrac {FG}{BG}\)还成立吗?