解: (2)$PA = PE。$理由如下：如圖①，延長$EB$至點$H，$使$BH = BP，$連接$PH。$
$\because\triangle ABC$是等邊三角形，$\therefore AB = BC，$$\angle ACB=\angle ABC = 60^{\circ}。$
$\because BM// AC，$$\therefore\angle ACB=\angle CBH = 60^{\circ}。$又$\because BP = BH，$$\therefore\triangle BPH$是等邊三角形，$\therefore PH = BP = BH，$$\angle H = 60^{\circ}=\angle ABC=\angle APE=\angle BPH，$$\therefore\angle APB=\angle EPH，$$\therefore\triangle APB\cong\triangle EPH(ASA)，$$\therefore PA = PE。$
(3)當點$P$在$BC$上時，$BC = BP + BE；$當點$P$在線段$CB$的延長線上時，$BE = BP + BC。$理由如下：當點$P$在$BC$上時，由
(2)可知，$\triangle APB\cong\triangle EPH，$$\therefore AB = EH，$$\therefore BC = EH = EB + BH = BE + BP。$當點$P$在線段$CB$的延長線上時，如圖②，在$BE$上截取$BH = BP，$連接$PH。$
$\because\triangle ABC$是等邊三角形，$\therefore AB = BC，$$\angle ACB=\angle ABC = 60^{\circ}。$
$\because BM// AC，$$\therefore\angle ACB=\angle PBH = 60^{\circ}。$又$\because BP = BH，$$\therefore\triangle BPH$是等邊三角形，$\therefore PH = BP = BH，$$\angle BHP = 60^{\circ}=\angle ABC=\angle APE=\angle BPH，$$\therefore\angle APB=\angle EPH，$$\angle EHP=\angle ABP = 120^{\circ}，$$\therefore\triangle APB\cong\triangle EPH(ASA)，$$\therefore EH = AB，$$\therefore BE = BH + EH = BP + BC。$
