證明：?$(1)$?如圖，連接?$AP$?
∵?$PE\bot AB$?，?$PF\bot AC$?，?$BG\bot AC$?
∴?$S_{\triangle ABP}=\dfrac {1}{2}AB·PE$?，?$S_{\triangle ACP}=\dfrac {1}{2}AC·PF$?，
?$S_{\triangle ABC}=\dfrac {1}{2}AC·BG$?
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) $?有結論?$BG=PE+PF+PM$?
理由是：如圖?$2$?，連接?$PA$?、?$PB$?、?$PC$?
∵?$S_{\triangle ABC}=S_{\triangle APB}+S_{\triangle ACP}+S_{\triangle PBC}$?
∴?$\dfrac {1}{2}AC×BG=\dfrac {1}{2}AB×PE+\dfrac {1}{2}AC×PF+\dfrac {1}{2}BC×PM$?
∵?$\triangle ABC$?為等邊三角形
∴?$AC=AB=BC$?
∴?$BG=PE+PF+PM$?
