《2025年新课程问题解决导学方案八年级数学上册华师大版》

阅读材料：如图 1，在$\triangle ABC$中，$AB = AC$，$P$为$BC$上一点，点$P$到两腰的距离分别为$r_{1}$、$r_{2}$，腰上的高为$h$，连结$AP$，则$S_{\triangle ABP} + S_{\triangle ACP} = S_{\triangle ABC}$，即$\dfrac{1}{2}AB\cdot r_{1} + \dfrac{1}{2}AC\cdot r_{2} = \dfrac{1}{2}AB\cdot h$，$\therefore$ $r_{1} + r_{2} = h$(定值)，即$PE + PF$为定值.
(1)深入探究
将“在$\triangle ABC$中，$AB = AC$，$P$为$BC$上一点”改成“$P$为等边三角形$ABC$内一点”，如图 2，作$PE\perp AB$，$PF\perp AC$，$PM\perp BC$，$BG\perp AC$，垂足分别为点$E$、$F$、$M$、$G$，有类似结论吗？请写出结论并证明.
(2)理解与应用
如图 3，当点$P$在$\triangle ABC$外时，(1)中的结论是否仍成立？若成立，请给予证明；若不成立，$PE$、$PF$、$PM$和$BG$之间又有怎样的关系？说明理由.

答案：

(1) $PE + PF + PM = BG$.
证明：如图1，连结$PA$、$PB$、$PC$，

则$S_{\triangle ABP} + S_{\triangle BCP} + S_{\triangle ACP} = S_{\triangle ABC}$.
$\because \triangle ABC$是等边三角形，
$\therefore AB = AC = BC$.
$\because PE \perp AB$，$PF \perp AC$，$PM \perp BC$，$BG \perp AC$，
$\therefore \frac{1}{2}AB \cdot PE + \frac{1}{2}AC \cdot PF + \frac{1}{2}BC \cdot PM = \frac{1}{2}AC \cdot BG$.
$\therefore \frac{1}{2}AB \cdot PE + \frac{1}{2}AB \cdot PF + \frac{1}{2}AB \cdot PM = \frac{1}{2}AB \cdot BG$.
$\therefore PE + PF + PM = BG$.
(2) $PE + PF - PM = BG$.
理由：如图2，连结$PA$、$PB$、$PC$，

则$S_{\triangle ABP} + S_{\triangle ACP} - S_{\triangle BCP} = S_{\triangle ABC}$.
$\because \triangle ABC$是等边三角形，
$\therefore AB = AC = BC$.
$\because PE \perp AB$，$PF \perp AC$，$PM \perp BC$，$BG \perp AC$，
$\therefore \frac{1}{2}AB \cdot PE + \frac{1}{2}AC \cdot PF - \frac{1}{2}BC \cdot PM = \frac{1}{2}AC \cdot BG$.
$\therefore \frac{1}{2}AB \cdot PE + \frac{1}{2}AB \cdot PF - \frac{1}{2}AB \cdot PM = \frac{1}{2}AB \cdot BG$.
$\therefore PE + PF - PM = BG$.