Does adding more edge to the Triakis Tetrahedral Graph make it non-planar because it contains the a subgaph homeomorphic to $K_{3,3}$?

Question

I know that a graph is planar if and only if it contains no subgraph homeomorphic to $K_5$ or $K_{3,3}$, and I know that the Triakis Tetrahedral Graph is planar (http://mathworld.wolfram.com/TriakisTetrahedralGraph.html).

An additional edge added to the Triakis Tetrahedral will result in a non-planar graph (visually this is apparent). But what is the foundation of this? Is there now a $K_{3,3}$ subgraph? Is there an easy way to show which graphs can be superimposed to form this new graph (with the additional edge)?

Answer 1

The easiest way to prove this is not to worry about $K_5$ or $K_{3,3}$, but instead use the fact that a planar graph on $n$ vertices has at most $3n-6$ edges (as proven, e.g., here). Since the triakis tetrahedral graph has $8$ vertices and $18=3(8)-6$ edges, adding any edge to it will give you a graph with too many edges to be planar.

Kuratowski's theorem will then tell you that the resulting graph has a subgraph homeomorphic to either $K_5$ or $K_{3,3}$, but I don't know of an easy way to find it. If you really want this subgraph and not just a proof of non planarity, you might have to consider each possible way of adding an edge separately.