证明：$(1)$如图，连接$AP$
∵$PE\bot AB$，$PF\bot AC$，$BG\bot AC$
∴$S_{\triangle ABP}=\dfrac {1}{2}AB·PE$，$S_{\triangle ACP}=\dfrac {1}{2}AC·PF$，
$S_{\triangle ABC}=\dfrac {1}{2}AC·BG$
又∵$S_{\triangle ABP}+S_{\triangle ACP}=S_{\triangle ABC}$
∴$\dfrac {1}{2}AB·PE+\dfrac {1}{2}AC·PF=\dfrac {1}{2}AC·BG$
∵$AB=AC$
∴$PE+PF=BG$
$(2) $有结论$BG=PE+PF+PM$
理由是：如图$2$，连接$PA$、$PB$、$PC$
∵$S_{\triangle ABC}=S_{\triangle APB}+S_{\triangle ACP}+S_{\triangle PBC}$
∴$\dfrac {1}{2}AC×BG=\dfrac {1}{2}AB×PE+\dfrac {1}{2}AC×PF+\dfrac {1}{2}BC×PM$
∵$\triangle ABC$为等边三角形
∴$AC=AB=BC$
∴$BG=PE+PF+PM$
