Why is the space of plane quadrilaterals with area center of mass = edges center of mass = vertices center of mass 2-dimensional? A naiv geometric question. 
We consider shapes of triangles and plane quadrilaterals (i.e. these objects up to translation, rotation and scaling).  
The shape of triangles ($T$) is 2-dimensional (two angles resp. 3 points with 6 coordinates minus 2 translationals degree of freedoms  minus 1 rotational degree of freedom minus 1 scaling degree of freedom). 
The shape of plane quadrilaterals ($Q$) is 4-dimensional (one unit side to fix the scale and two adjacent sides defined by two angles and two side lenghts resp. 4 points with 8 coordinates minus 2 translationals degree of freedoms  minus 1 rotational degree of freedom minus 1 scaling degree of freedom). 
For a polygon we can define three natural centers of mass: a) ACM, area center of mass, b) ECM, edges center of mass (edges carry a homogenous mass density per length unit), c) VCM, vertices center of mass (equal masses at vertices). 
For a triangle $t \in T$ we have always $\mathrm{ACM}(t)=\mathrm{VCM}(t)$. 
Regarding ECM: 
Assuming the two 2-dimensional graphs $\{(t, \mathrm{ACM}(t)), t \in T\}  \subset T \times \mathbb{R}^2$ and $\{(t, \mathrm{ECM}(t)), t \in T\} \subset T \times \mathbb{R}^2$ lie in a general position and thus intersect transversally, we expect the space of triangle shapes with ACM=ECM to be 2+2-4=0-dimensional. And in fact, the only triangle shape with ACM=ECM is the equilateral one. 
For plane quadrilaterals we would assume  the two 4-dimensional graphs $\{(q, \mathrm{ACM}(q)), q \in Q\}  \subset Q \times \mathbb{R}^2$ and $\{(q, \mathrm{ECM}(q)), q \in Q\} \subset Q \times \mathbb{R}^2$ also to lie in a general position and thus intersect transversally. And thus we expect the dimension of plane quadrilateral shapes with ACM=ECM to be 4+4-6=2-dimensional. And the same for the other two pairs (ECM=VCM and ACM=VCM). Is this correct? 
Intersecting the 2-dimensional surface ACM=ECM with the 2-dimensional surface ACM=VCM we would expect a 0-dimensional space. 
However the space of all planar quadrilaterals with ACM=ECM=VCM seems to be 2-dimensional, namely all paralellogramms (you can choose one angle and ration of side lengths; they have a $180^{\circ}$-symmetry and so ACM=ECM=VCM. 
Why does dimension counting not work (which subspaces are not in general position?) or where am I mistaken? 
 A: This is not a direct answer to the issue. Just information about different ways to consider the general framework.
1) You describe a quadrilateral shape as an equivalence class of quadrilaterals up to general similitudes, i.e., translations, homotheties and scaling. The presentation given in the "Hamburg school" directed in the '1930s by a genial mathematician called Blaschke (L. Santalo was one of his students) is to consider this kind of space as an "homogeneous space" https://en.wikipedia.org/wiki/Homogeneous_space . For example the space of quadrilaterals is, said in an approximate manner : $\mathbb{R}^8$ (4 points with 2 coordinates) quotiented by the transitive group of similitudes (in the sense of group actions, but once again, I advise you to have a look at Santalo's book, reviewed here https://projecteuclid.org/download/pdf_1/euclid.bams/1183539854).
2) I have been working in the past on similar issues, and I found a pertinent subdomain called centro-affine geometry also developed in the first part of XXth century.
3) Last but not least. See this very recent article  : "Quadrilaterals in_Shape Theory II Alternative Derivations of_Shape Space. Successes_and_Limitations" by Edward Anderson  https://arxiv.org/pdf/1810.05282.pdf that should bring you something (and even more...). 
The bibliography you will find therein has many self-citations... In particular you will see there is a part I and a forthcoming part III. I just discovered this article today : instead of taking days to analyse it at spare time, I preferred to share the information at once.  
