Consider four disjoint points in the plane, $v_{1}$, $v_{2}$, $v_{3}$ and $v_{4}$. The cycle, $C:=v_1v_2v_{3}v_{4}v_{1}$, is the union of the (Jordan) arcs, $A_{12}$, $A_{23}$, $A_{34}$, and $A_{41}$, in which every $A_{ij}$ is an arc between the points $v_{i}$ and $v_{j}$. By the Jordan Curve Theorem (JCT), the region enclosed by $C$, has an interior and an exterior, respectively denoted by $\text{Int}(C)$ and $\text{Ext}(C)$.
In addition, we assume that there exists an arc, $A_{13}$, entirely contained within $\text{Int}(C)$, apart from its endpoints. Clearly, $A_{13}$ appears to partition $\text{Int}(C)$ into two disjoint regions. Such a statement may be written as follows, \begin{equation}\notag \text{Int}(C) = \text{Int}(C_{1231}) \cup \text{Int}(A_{13}) \cup \text{Int}(C_{1341}), \end{equation} where the cycles $C_{1231}$ and $C_{1341}$ are defined as $C_{1jk1}:=v_{1}v_{j}v_{k}v_{1}$, and $\text{Int}(A_{13})$ denotes the arc $A_{13}$, without its endpoints.
When attempting to prove such a statement, however, one keeps switching from the interiors to the exteriors of each sub-region, without much success. Indeed, using the JCT does not appear to be sufficient, since it solely defines the interior of a simple closed curve, relative to its exterior and vice-versa, thereby making it seemingly impossible to combine the interiors $\text{Int}(C_{1231})$ and $\text{Int}(C_{1341})$, in order to produce $\text{Int}(C)$.
Am I missing something obvious? Or is such a fact too elementary to be proved? Could this be part of the definition of the faces of a plane graph, for instance? But if one can prove the JCT, why wouldn't it be possible to prove such a similar claim?
