For a regular polygon with sides of length $l$, prove that all points within $l$ from a vertex lie on an incident edge I am trying to prove that all the isometries of a regular polygon that map the polygon back onto itself must map vertices to vertices. I nearly have the proof, but I need to prove one more statement:

Take a regular $n$ sided polygon, with $n>3$ and sides of length $l$.
  Pick any vertex $A$. Then any point on the polygon $x$ such that
  $|A-x| < l$ (i.e. the distance from $A$ to $x$ is less than $l$) must
  lie on an edge incident to $A$ or is equal to $A$.

This statements seems to be true to me, just through visualising polygons, but I can't think of an elegant proof.
 A: I haven't spent any time on the subsidiary Q, but as to the problem of an isometry $f$ of the polygon $P$ to itself, observe that the isometric image $f(S)$ of a closed  bounded line segment $S$ of positive length $l$  must be a closed bounded line segment of equal length, because 
(1)..... For any $\{x_1,x_2,x_3\}\subset S$ and for some $\{i,j,k\}=\{1,2,3\}$ we have $d(x_i,x_j)+d(x_j,x_k)=d(x_i,x_k)$ and hence $d(f(x_i),f(x_j))+d(f(x_j),f(x_k))=d(f(x_i),f(x_k)).$..... Therefore
($\bullet$)  Any 3 points in $f( S)$ are collinear.
(2)..... Let $e_1,e_2$ be the endpoints of $S$. Let $T$ be the closed line segment from $f(e_1)$ to $f(e_2).$ Let $x$ be any  point of $S$.  Then $l= d(e_1,e_2)=d(f(e_1),f(e_2)),$ while $d(f(x),f(e_1))+ d(f(x),f(e_2))=d(x,e_1)+d(x,e_2)=l$ ...... So by ($\bullet$), we have 
($\bullet \bullet)$ $f(S)   \subset T.$  And  $f(S)$ includes the endpoints of $T,$ and the length of $T$ is $l$.
(3)..... For any $y\in T ,$ there exists $x\in S$ with $d(x,e_i)=d(y,f(e_i))$ for $i\in \{1,2\},$ and  by ($\bullet \bullet$) we have $f(x)\in T.$ So we  must have $f(x)=y.$..... Hence $f(S)\supset T.$ Together with $f(S)\subset T$ from (2), we have $$  f(S)=T.$$
For the polygon $P,$ and an isometry $f:P\to P$, let $S$ be any of its sides. Then $f(S)$ is a closed straight-line  segment  of equal length and $f(S)\subset P,$ so $f(S)$ is a side of $P.$ And from ($\bullet \bullet$), $f$ maps the endpoints of $S,$ which are vertices of $P,$ to the endpoints of $f(S),$ which are also vertices of $P$, as they are the endpoints of the side $f(S). $
A: In a regular $n$-gon (considered in your question as a "one-dimensional complex") the inner angle between subsequent edges is $\bigl(1-{2\over n}\bigr)\pi$. If $n=4$ this angle is ${\pi\over2}$, so that the two edges not adjacent to $A$ are just tangents to the circle with center $A$ and radius $\ell$. If $n\geq5$ this angle is $>{\pi\over2}$, so that the next two edges are "running away" from the mentioned circle. The edges "further down" should cause no problems.
