determining whether $H$ is a nonplanar graph 
Determine, with justification, whether the graph $H$ below is nonplanar.


I think it's nonplanar as I can find a $K_{3,3}$ subdivision for it. I can describe the subdivision as follows. Join vertex $8$ to vertex $9$ (as they are connected by the path $812349$). Then add an edge from vertex $4$ to vertex $9$ (as they are connected by the path $439$). Add an edge from $4$ to $7$ as they're connected by the path $4397$. Now delete the edges $18, 12, 23, 34, 14, 39, 37, 79,29$ (since we added edges $49$ and $89$, there are $9$ edges remaining). Delete the vertices of the resulting graph with degree zero, which are $1, 2,3.$ The result is the $K_{3,3}$ graph with vertices $5,7,9$ on one side of the graph and vertices $8,6,4$ on the other, which shows that $H$ has a $K_{3,3}$ subdivision, and hence is nonplanar.

I think my reasoning may be incorrect as I obtained the subdivision incorrectly. If so, is there a way to obtain a correct $K_{3,3}$ subdivision? I know the $5$ vertices of degree $4$ are all joined by a path to each other, so that might be useful.

 A: I know this isn't exactly what you asked for, but there's another, easier way to prove the graph is non-planar, based on Tutte's conflict graph theorem.  Because the graph has a Hamilton cycle, it's easy to tell whether or not it's planar.  Let's redraw the graph:

To attempt to embed the graph in the plane, we can start by placing the graph nodes at the vertices of a regular polygon, in the order of the cycle, and represent the remaining edges as diagonals.  Consider the diagonals of the polygon.  If two of them intersect when they are both drawn inside the polygon, then they will also intersect if both are drawn (as graph edges) outside the polygon.  Now look at the three diagonals colored red.  Each intersects both the others, so at least two must be drawn inside or at least two must be drawn outside, and there is no planar embedding.
The conflict graph is the graph whose vertices are the diagonals, in which two vertices are adjacent if the two diagonals intersect when drawn in the same region.  A graph with a Hamilton cycle is planar if and only if the conflict graph is bipartite.
(I had to change the labelling of the vertices, because Geogebra won't accept a number as the label of a point.)
A: If you contract edges $81$, $12$, $65$, $43$, you will get $K_5$.
