dividing planar regions into congruent parts I've been looking for counterexamples, with no success, to this for a few days running:

Suppose for a planar region $R$ with finite positive area, there is a
  point $x$ where all lines of rational slope, through $x$, divide $R$
  into two congruent parts. 
Does it follow that any line through $x$ with irrational slope also
  divides $R$ into two congruent parts?

My intuition says (which may be wrong) I can approximate a line with irrational slope by a sequence of lines with rational slope, each which divides our region into two congruent parts. I really think I'm missing big things in this approach. 
 A: Here's our setup: We have the plane $\mathbb R^2$, the topological group $G$ of congruence maps $$(x,y)\mapsto(ax+by+u,cx+dy+v)$$
is isomorphic to the subgroup of $GL_3(\mathbb R)$ consisting of matrices
$$\begin{pmatrix}a&b&u\\c&d&v\\0&0&1\end{pmatrix} $$
and thus is a semidirect product of the compact group $O_2(\mathbb R)$ and $\mathbb R^2$. The obstacle we have here is that the factor $\mathbb R^2$ is not compact.
For nonzero $z\in\mathbb R^2$, especially for $z\in S^1$, the line $z^\perp=\{x\in\mathbb R^2\mid \langle x,z\rangle=0\}$ divides the plane into two half planes
$$H_+(z)=\{x\in\mathbb R^2\mid \langle x,z\rangle>0\} \qquad\text{and}\qquad H_-(z)=\{x\in\mathbb R^2\mid \langle x,z\rangle<0\}.$$
Now the claim of the problem statement follows from the 
Proposition. Let $Q$ be a dense subset of $S^1$.
Let $R\subseteq \mathbb R^2$ be a nonempty open set of finite measure.
For each $z\in Q$, let $\alpha_z\in G$ be a congruence with $\alpha_z(R\cap H_+(z))=R\cap H_-(z)$.
Then for each $z\in S^1$, there exists $\alpha_z\in G$ with $\alpha_z(R\cap H_+(z))=R\cap H_-(z)$.
Proof: (Warning: $Q\to G, z\mapsto\alpha_z$ is one of the few things in this discussion that is not continuous).
We may assume wlog. that $Q=-Q$ and that $\alpha_{-z}=\alpha_z^{-1}$ for $z\in Q$.
Assume that $\{\lVert \alpha_z(0)\rVert\mid z\in Q\}$ is not bounded from above.
Then there exists a sequene $(z_n)_{n\in\mathbb N}$ in $Q$ such that $\lVert \alpha_{z_n}(0)\rVert\to\infty$. As $S^1$ is compact we may assume (by switching to a convergent subsequence if necessary) that $z_n\to\zeta$ for some $\zeta\in S^1$.
Select $x\in R\cap H_+(\zeta)$. Then a ball $B_{\epsilon}(x)$ is in $R\cap H_+(\zeta)$ and the smaller ball $B_{\epsilon/2}(x)$ is in $R\cap H_+(z_n)$ for almost all $n$, hence for all $n$ if we switch to a subsequence.
For $\alpha\in G$, we have $\lVert \alpha(x)\rVert\ge\lVert \alpha(0)\rVert-\rVert x\rVert$. Thus if we switch to a subsequence such that $\lVert \alpha_{z_{n+1}}(0)\rVert>\lVert \alpha_{z_n}(0)\rVert+2\lVert x\rVert+\epsilon$, we ensure that the open balls $\alpha_{z_n}(B_{\epsilon/2}(x))=B_{\epsilon/2}(\alpha_{z_n}(x))$ are pairwise disjoint subsets of $R$. Since each of these balls has the same positive measure $\frac{\pi\epsilon^2}4 $, we get a contradiction to the finiteness of the measure of $R$.
Therefore, $\{\lVert \alpha_z(0)\rVert\mid z\in Q\}$ is bounded and hence all $\alpha_z$, $z\in Q$, are in some compact subset $K\subset G$ (of the form $K=O_2(\mathbb R )\times [-M,M]^2$).
Now let $\zeta\in S^1$ be arbitrary.
As $Q$ is dense, there is a sequence $z_n\to\zeta$ with $z_n\in Q$.
As $K$ is compact, we may assume wlog. (i.e. by switching to a subsequence) that the corresponding sequence $\alpha_n$ converges to some $\alpha\in K\subset G$.
Let $x\in  R\cap H_+(\zeta)$ be arbitrary.
As above, some $B_\epsilon(x)$ is in $R\cap H_+(\zeta)$ and $B_{\epsilon/2}(x)$ is in $H_+(z_n)$ for almost all $n$.
For $n$ big enough, $\alpha(x)\in B_{\epsilon/3}(\alpha_{z_n}(x))$, hence $$B_{\epsilon/6}(\alpha(x))\subset B_{\epsilon/2}(\alpha_{z_n}(x)) \subset \alpha_{z_n}(R\cap H_+(z_n))=R\cap H_-(z_n).$$
We conclude $B_{\epsilon/6}(\alpha(x))\subset R\cap \overline{H_-(\zeta)}$ and hence $\alpha(x)\in R\cap H_-(\zeta)$.
This shows $\alpha(R\cap H_+(\zeta))\subseteq R\cap H_-(\zeta)$.
By considering the sequence $-z_n\to-\zeta$, $\alpha_{-z_n}\to\alpha^{-1}$, we similarly obtain $\alpha^{-1}(R\cap H_-(\zeta))\subseteq R\cap H_+(\zeta)$ and ultimately conclude 
$$ \alpha(R\cap H_+(\zeta))= R\cap H_-(\zeta)$$
as desired.$_\square$
Remark: The finiteness of $\mu(R)$ is essential as the following example shows (though it uses a different dense subset $Q\subset S^1$ instead of "all rational slopes"):
Let $\Lambda=\mathbb Z^2+(0,\sqrt 2)$ be the standard lattice, but irrationally displaced. Let $R=\mathbb R^2\setminus \Lambda$. For every line passing through a lattice point of $\Lambda$, the 180° rotation around that point is a congruence between the parts of $R$ defined by this line.
On the other hand, any congruence between half plane parts of $R$ gives us a congruence of the lattice and each of these (apart from translation and skew translations, which can be excluded for our problem) has a fixed point in $\frac12\mathbb Z^2+(0,\sqrt 2)$. This fixedpoint must be on the common boundary, i.e. the dividing line.
Hence any of th euncountably many lines avoiding $\frac12\mathbb Z^2+(0,\sqrt 2)$ does not allow a congruence among the halves of $R$ it defines.
Openness of $R$ is also essential as the example $R=\{e^{i\pi q}\mid q\in\mathbb Q\}$ readily shows. 
