證明：?$(1)$?在正五邊形?$ABCDE$?中，?$AB = BC，$??$∠ABM=∠C。$?
?$ $?在?$\triangle ABM$?和?$\triangle BCN$?中，
?$ \begin {cases}AB = BC \\∠ABM=∠C \\BM = CN\end {cases}$?
∴?$\triangle ABM\cong \triangle BCN(\mathrm {SAS})。$?
?$ (2) $?解：∵?$\triangle ABM\cong \triangle BCN，$
?∴?$∠BAM=∠CBN。$?
∵?$∠BAM+∠ABP=∠APN，$?
∴?$∠APN=∠CBN+∠ABP=∠ABC。$?
∵正五邊形內(nèi)角和為?$(5 - 2)×180°=540°，$?
∴?$∠ABC=\frac {(5 - 2)×180°}{5}=108°，$?即?$∠APN = 108°。$?
