Triangle $ABC$ has side-lengths $AB = 15, BC = 25,$ and $AC = 20.$ The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N.$ What is the perimeter of $\triangle AMN?$

Since $MN\parallel BC$, $\angle MIB=\angle IBC=\angle MBI$ ($BI$ bisects $\angle B$), so $MB=MI$; similarly $NC=NI$.
So $AM+MN+AN=AM+MI+IN+AN=(AM+MB)+(AN+NC)=AB+AC=15+20=\boxed{35}$.