Problem on distances in a polygon

Let me prove a bit more general statement.

Let $P=[v_1\dots v_n]$ and $P'=[v_1'\dots v_n']$ be two solid polygons such that if $[vw]$ is a side of $P$ or a diagonal which lies in $P$ completely then $|v-w|\ge |v'-w'|$. Then there is a short map $P\to P'$ which sends any vertex $v$ of $P$ to the corresponding vertex $v'$ of $P'$.

Note that if there is one diagonal $[vw]$ such that $|v-w|=|v'-w'|$ then we can cut $P$ in two polygons and reduce the question to a simpler case.

First note that we can assume that $|v-w|=|v'-w'|$ for any side of $[vw]$ of $P$. If this is note the case choose a vertex $x'$ so that the triangle $[x'v'w']$ lies in $P'$. Put joints in these vertexes and increase $|v'-w'|$ up to $|v-w|$ so that the rest of diagonals stay the same or increase. The new poygon $P''$ admits a short map to $P'$ so we can make each side to be the same as in $P$ one by one and reduce the question to this case.

Note that the convex angles get smaller in $P'$. Therefore one concave angle, say $a$, should go up, in particular it stays concave. Choose 4 vertexes: $a$, its neighbors $x$ and $y$ and a vertex $b$ visible from $a$. Put joints in these vertexes. We can move the polygon straighting the angle at $a$, keeping the distances $|a-x|$, $|a-y|$, $|b-x|$ and $|b-y|$ and nondecreasing the rest of diagonals. Repeating if necessary, we will get one more diagonal fixed.

P.S. Doing these operations, you can end in a generalized solid polygon --- a flat disc with polygonal boundary, when it is mapped to the plane you can get overlaps.