證明：?$(1)∵$?正五邊形?$ABCDE，$?
?$∴AB=BC，$??$∠ABM=∠C，$?
∴在?$△ABM$?和?$△BCN$?中
?$\begin{cases}{AB=BC}\\{∠ABM=∠C}\\{BM=CN}\end{cases}$?
?$∴△ABM≌△BCN(\mathrm {SAS}).$?
(2)解：∵△ABM≌△BCN，
?$∴∠BAM=∠CBN，$?
?$∵∠BAM+∠ABP=∠APN，$?
?$∴∠CBN+∠ABP=∠APN=∠ABC=\frac {(5-2)×180°}{5}=108°.$?
即?$∠APN$?的度數(shù)為?$108°.$?
