As I think this is not a planar graph. First I deleted the $cd$ edge and kept vertex $b$ onto vertex $e$ then I got a subgraph $K_5$ with vertex $gcade$.Is it correct? Can anyone verify the answer?
2026-03-29 23:42:51.1774827771
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Determine whether my graph is planar or not
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Mathematica:
gg = Graph[{
a \[UndirectedEdge] b, a \[UndirectedEdge] d,
a \[UndirectedEdge] h, a \[UndirectedEdge] c,
b \[UndirectedEdge] c, b \[UndirectedEdge] e,
c \[UndirectedEdge] d, c \[UndirectedEdge] g,
d \[UndirectedEdge] e, d \[UndirectedEdge] f,
d \[UndirectedEdge] g, e \[UndirectedEdge] f,
e \[UndirectedEdge] g, f \[UndirectedEdge] g,
g \[UndirectedEdge] h},
VertexLabels -> "Name"]
PlanarGraphQ[gg]
(* False *)
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Let us start by recalling Kuratowski's theorem https://en.wikipedia.org/wiki/Kuratowski%27s_theorem
So if we can merge distinct blocks of vertices (preserving adjacency in the obvious way) and form $K_5$ or $K_{3,3}$ then a graph is non planar.
The partition $\{ \{a\},\{c\},\{d\},\{e,b\},\{f,g,h\}\}$ is a subgraph that is isomorphic to $K_5$. So your graph is $\color{red}{\text{not}}$ planar.


I'm not sure where you see $K_5$. Here's an alternative argument.
Delete vertex $f$ and all edges connected to it. Delete vertex $dg$. We show that the obtained graph is not planar.
Start by drawing the cylce $adegh$. Next, we wish to place $c$. Without loss, place it inside the cycle. Draw the edges connecting the cycle with $c$.
As $bc$ needs to be drawn, also $b$ is inside the cycle. Either
None of the cases work, proving non-planarity or your graph.