In the following I use $A,B,C,D$ for the vertices and $\alpha,\beta,\gamma,\delta$ for the respective adjacent angles and $a,b,c,d$ for the 'adjacent' (in the sense that $a$ originates from $A$, and so on) edges in a general quadrilateral.
I am searching for a simple algebraic characterisation in form of a set of non-redundant equations ($\mathbf{E}$) for the following specific types of quadrilaterals, namely the square ($S$), rectangle ($Re$), parallelogram ($P$), rhombus ($Rh$), trapez *$^1$ ($T$) and kite ($K$) in such a way that the defintion of all of these can be obtained by simply excluding some of the full set of equations which are given for the $S$.
What I am seeking differs from 'the common way'. For example in using $\alpha=\beta=\gamma=\delta=\pi/2$ for the square there is no way to 'lower the symmetry' to say a parallelogram by just dropping these but at the same time without introducing new equations.
I found a way to do this for all $S,Re,P,Rh,T$ except $K$. This works like the following:
$\mathbf{E}=\{$I, II, III, IV, V$\}$ with
$$\begin{align} \alpha+\beta=\pi \tag{I} \\ \alpha+\gamma=\pi \tag{II} \\ \alpha+\delta=\pi \tag{III}\\ \beta+\gamma=\pi \tag{IV} \\a=b \tag{V} \end{align}$$
With that $S$ is characterized by $\mathbf{E}$ (the whole set of equations I-V exactly define the square).
$R$ is defined by $\mathbf{E}\setminus \{V\}$, $Rh$ is $\mathbf{E}\setminus \{II\}$, $P$ is $\mathbf{E}\setminus \{II,V\}$ and $T$ is $\mathbf{E}\setminus \{II,III,IV,V\}$. That looks all very nice, the problem here as mentioned above is $K$, for $K$ we need exactly the condition that one pair of opposite angles is equal ($\alpha=\gamma$) which cannot be expressed by simply removing some equations from $\mathbf{E}$. However I have found that the constraint for a kite could be encoded for example like
$$\begin{align} \frac{\beta + \delta}{2} + \alpha = \pi \tag{Ia} \\ \frac{\beta + \delta}{2} + \gamma = \pi \tag{IIa} \end{align}$$
which looks somewhat similar to those in $\mathbf{E}$. So now it would be necessary to kind of 'project out' Ia and IIa from $\mathbf{E}$ and then include those into the new equations, but here I get somehow stuck. Can anyone help here?
($\mathbf{E}\setminus \{I,III,IV,V\}$ gives by the way a (general) cyclic quadrilateral).
$^1$) to avoid language confusion I call the quadrilateral with two sides parallel a trapez *
Additional comment
I have found meanwhile another way to do the job similarly, but without restricting to conditions on purely angles and edge lengths, as pointed out in the title. For example one can use the intersection of the diagonals very elegantly to characterise the quadrilaterals by an array of three ratios and an angle, but that would be a different story.