?$ (1)$?證明：∵?$\triangle ABC$?是等邊三角形
∴?$∠ABC=∠ACB = 60°，$??$AB = AC$?
∴?$∠DBP = 180°-∠ABC=120°，$?
?$∠P CE = 180°-∠ACB = 120°$?
∴?$∠DBP=∠P CE$?
∵?$P A = P D = PE$?
∴?$∠P AD=∠P DA，$??$∠P AE=∠PEA$?
∴?$∠PDB+∠PEC=∠PAD+∠PAE=60°$?
∵?$∠CPE+∠PEC=180°-∠PCE=60°$?
∴?$∠BDP=∠CPE$?
?$ $?在?$\triangle BDP $?和?$\triangle CPE$?中
?$\begin {cases}∠DBP=∠P CE\\∠BDP=∠CPE\\P D = PE\end {cases}$?
∴?$\triangle BDP≌\triangle CPE(\mathrm {AAS})$?
?$ (2)$?解：存在
∵?$\triangle BDP≌\triangle CPE$?
∴?$BD = CP$?
?$ \triangle BDP $?的周長?$=BD + BP + DP=CP + BP + DP=BC + DP$?
∵?$BC = 10，$?∴當(dāng)?$DP $?最小時，?$\triangle BDP $?的周長最小
∵?$P A = P D$?
∴當(dāng)?$AP\perp BC$?時，?$AP {最小}，$?即?$DP $?最小
∵?$\triangle ABC$?是等邊三角形，?$AP\perp BC$?
∴?$BP=\frac 12BC = 5$?
?$ $?即當(dāng)?$BP = 5$?時，?$\triangle BDP $?的周長最小
