Problem on diagonals in a polygon Consider the polygon $P$ in the following picture which has sides drawn in black and internal diagonals drawn in red and blue. $P$ has 4 convex angles and 4 concave angles in alternating order as it's clear from the picture.

I'm trying to understand if it is possible to shorten one or more sides or blue diagonals of $P$ in a continuous process in such a way that both the following conditions are satisfied:


*

*by shortening said sides or blue diagonals of $P$ no other side or blue diagonal of $P$ gets longer

*by shortening said sides or blue diagonals of $P$ the lengths the two red diagonals remain constant
All my attempts at doing so failed, so I guess the answer is no, but I can't produce a proof.
 A: This a classical engineering problem, that falls under the subject of  Static Analysis of Trusses,
as well as under that of Kinematic Chains.
You are probably not familiar with such discipline, and there is no space here to resume its basis.
However, since everybody has common experience of trusses and kinematic chains, I will cite here some facts that should sound
"plausible", while referencing to specialized documentation for a rigourous treatment.

The figure above translates the problem you proposed in the engineering language.
The black and red segments are stiff rods (elements) and the circles represents pin joints (nodes).
Being stiff, all the elements are supposed not to change in length, and can only rotate 
around the joints. Note that at the points where the diagonal elements cross
there is no bound, meaning that the elements can freely slide over each other.
Now, a rigid body in the plane has three degre of freedom to move: two for translation
and one for rotation.
To null them and "keep the truss on the drawing sheet" it is customary to fix the $(x,y)$ position 
of one node (here $A$) and  give another (suitably chosed) only the fredom to move horizontally (here $C$).
In this way the whole truss can only have "internal" movements and not "as a whole".
One of the major scopes of the Static Analysis is to determine infact if the truss can have or not
such "internal movements". If it has, it will be unstable, and under a minimal load it will collapse, in the sense
that it will change its shape macroscopically. If it has not, then it may be Isostatic or OverConstrained.
And this distinction is very important in Construction Design.    
Your question practically translates to ask whether the truss is underconstrained or not.   
A result of Static Analysis, well known to all engineers, says that a necessary condition for the system
to be underconstrained is that
$$ 
m < 2n - 3 
$$
(if $m=2n-3$ it is isostatic, and overdetermined if $m>2n-3$)
where $m$ is the number of elements and $n$ the numbers of nodes (pin joints).
The results comes from analyzing the degrees of freedom of the various members and the degrees
reduced by the constraints.     
Your "truss" gives $m=10<2n-3=13$, and so it is heavily undetermined, having $3$ degrees of freedom
unrestrained.
Thus it can move without changing the length of any member, and since shortening consumes 1 degree
of movement, sure you can short any segment even without changing the length of the other $9$.
Clearly, the amount of movement or shortening is limited to a certain extent.   
The picture below provides an example   
 
---  Addendum   ----
Finally, after the clearing got from the comments and the re-editing of the post, the truss has become highly Hyperstatic.
To solve the problem in this case we can refer to a truss whose members are stiff vs. elongation
and soft vs. compression,  except only the diagonals $AE$ and $GC$ which are stiff  in both directions.
A sort of a truss made by wires around the two main diagonal rods.
If such a truss turns out to be stable, then it means that you cannot shorten a side without stretching some other.
But now the truss is quite complicated to be solved "by hand", and its non-linear stiffness complicates the analysis
(a director's chair is not a proper truss).

If we move the sliding bound to $E$, we get a vertical stiff rod ($GC$) connected at both ends
by wires stranding from the fix nodes, and these strands are then tightly knotted somewhere in between in all
directions (the other connections are not shown to reduce cluttering).
Common experience suggests that it should be sufficient to secure the vertical rod 
(unless of extreme configurations).
But how to provide mathematical evidence of that is out of my reach, at present.
