smc

Let $O$ be the incenter of $△ABC$. Because $\overline{MO}\parallel \overline{BC}$ and $\overline{BO}$ is the angle bisector of $\angle ABC$, we have $$\angle MBO=\angle CBO=\angle MOB=\frac{1}{2}\angle MBC$$ It then follows due to alternate interior angles and base angles of isosceles triangles that $MO=MB$. Similarly, $NO=NC$. The perimeter of $△AMN$ then becomes $$\begin{array}{rl}AM+MN+NA& =AM+MO+NO+NA\\ & =AM+MB+NC+NA\\ & =AB+AC\\ & =30\to \boxed{30}\end{array}$$