如图\(1\),在矩形\(ABCD\)中,点\(E\)为\(AD\)边中点,点\(F\)为\(BC\)边中点;点\(G\),\(H\)为\(AB\)边三等分点,\(I\),\(J\)为\(CD\)边三等分点\(.\)小瑞分别用不同的方式连接矩形对边上的点,如图\(2\),图\(3\)所示\(.\)那么,图\(2\)中四边形\(GKLH\)的面积与图\(3\)中四边形\(KPOL\)的面积相等吗?
\((1)\)小瑞的探究过程如下:在图\(2\)中,小瑞发现,\(S\)\({\,\!}_{四边形GKLH}\)\(=\)________\(S\)\({\,\!}_{四边形ABCD}\);在图\(3\)中,小瑞对四边形\(KPOL\)面积的探究如下\(.\)请你将小瑞的思路填写完整:设\(S\)\({\,\!}_{\triangle DEP}\)\(=a\),\(S\)\({\,\!}_{\triangle AKG}\)\(=b\),\(∵EC/\!/AF\),\(∴\triangle DEP\)∽\(\triangle DAK\),且相似比为\(1︰2\),得到\(S\)\({\,\!}_{\triangle DAK}\)\(=4a\).\(∵GD/\!/BI\),\(∴\triangle AGK\)∽\(\triangle ABM\),且相似比为\(1︰3\),得到\(S\)\({\,\!}_{\triangle ABM}\)\(=9b\).又\(∵\)\(S∆DAG=4a+b= \dfrac{1}{6}S四变形ABCD \),\({S}_{∆ABF}=9b+a= \dfrac{1}{4}{S}_{\;四边形ABCD} \),\(∴S\)\({\,\!}_{四边形ABCD}\)\(=24a+6b=36b+4a\).
\(∴a=\)________\(b\),\(S_{四边形ABCD}=\)________\(b\),\(S_{四边形KPOL}=\)________\(b\).\(∴S\)\({\,\!}_{四边形KPOL}\)\(=\)________\(S\)\({\,\!}_{四边形ABCD}\),则\(S\)\({\,\!}_{四边形KPOL}\)________\(S\)\({\,\!}_{四边形GKLH}\)\((\)填写“\( > \)”“\( < \)”或“\(=\)”\()\).
\((2)\)小瑞又按照图\(4\)的方式连接矩形\(ABCD\)对边上的点,则\(S_{四边形ANML}=\)________\(S_{四边形ABCD}\).