4.
提出问题:如图\(①\),在四边形\(ABCD\)中,\(P\)是\(AD\)边上任意一点,\(\triangle PBC\)与\(\triangle ABC\)和\(\triangle DBC\)的面积之间有什么关系?
探究发现:为了解决这个问题,我们可以先从一些简单的、特殊的情形入手:
\((1)\)当\(AP= \dfrac {1}{2}AD\)时\((\)如图\(②)\):
\(∵AP= \dfrac {1}{2}AD\),\(\triangle ABP\)和\(\triangle ABD\)的高相等,
\(∴S_{\triangle ABP}= \dfrac {1}{2}S_{\triangle ABD}\).
\(∵PD=AD-AP= \dfrac {1}{2}AD\),\(\triangle CDP\)和\(\triangle CDA\)的高相等,
\(∴S_{\triangle CDP}= \dfrac {1}{2}S_{\triangle CDA}\).
\(∴S_{\triangle PBC}=S_{四边形ABCD}-S_{\triangle ABP}-S_{\triangle CDP}\)
\(=S_{四边形ABCD}- \dfrac {1}{2}S_{\triangle ABD}- \dfrac {1}{2}S_{\triangle CDA}\)
\(=S_{四边形ABCD}- \dfrac {1}{2}(S_{四边形ABCD}-S_{\triangle DBC})- \dfrac {1}{2}(S_{四边形ABCD}-S_{\triangle ABC})\)
\(= \dfrac {1}{2}S_{\triangle DBC}+ \dfrac {1}{2}S_{\triangle ABC}\).
\((2)\)当\(AP= \dfrac {1}{3}AD\)时,探求\(S_{\triangle PBC}\)与\(S_{\triangle ABC}\)和\(S_{\triangle DBC}\)之间的关系,写出求解过程;
\((3)\)当\(AP= \dfrac {1}{6}AD\)时,\(S_{\triangle PBC}\)与\(S_{\triangle ABC}\)和\(S_{\triangle DBC}\)之间的关系式为: ______ ;
\((4)\)一般地,当\(AP= \dfrac {1}{n}AD(n\)表示正整数\()\)时,探求\(S_{\triangle PBC}\)与\(S_{\triangle ABC}\)和\(S_{\triangle DBC}\)之间的关系,写出求解过程;
问题解决:当\(AP= \dfrac {m}{n}AD(0\leqslant \dfrac {m}{n}\leqslant 1)\)时,\(S_{\triangle PBC}\)与\(S_{\triangle ABC}\)和\(S_{\triangle DBC}\)之间的关系式为: ______ .