A square piece of paper has side length $1$ and vertices $A,B,C,D$ in order. The paper is folded so that vertex $C$ meets edge $\overline{AD}$ at $C'$, edge $\overline{BC}$ meets edge $\overline{AB}$ at $E$, and the fold crease meets $\overline{DC}$ at $F$. Suppose $C'D=\frac15$. What is the perimeter of $\triangle DFC'$?

1. Since $FC'=FC=1-DF$ (the crease preserves length), $C'D+DF+FC'=C'D+DF+(1-DF)=C'D+1$, independent of $DF$.
2. With $C'D=\frac15$: perimeter $=\frac15+1=\boxed{\dfrac65}$.