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如图,点

求证:
参考答案:证明:正五边形\(ABCDE\)中,\(AB = BC\),\(\angle ABM = \angle C\),在\(\Delta ABM\)和\(\Delta BCN\)中\(\left\{ {\begin{array}{*{20}{l}}
{AB = BC} \\
{\angle ABM = \angle C} \\
{BM = CN}
\end{array}} \right.\),\(\therefore \Delta ABM \cong \Delta BCN(SAS)\);