# Prove or disprove that If $G$ is a graph with one or more triangles and contains no subdivision of$K_5$ as a subgraph, then $G$ is planar.

I'm working in the following graph theory excercise.

Prove or disprove that If $$G$$ is a graph with one or more triangles and contains no subdivision of$$K_5$$ as a subgraph, then $$G$$ is planar.

I'm thinking about the graph $$K_6$$ and how is non-planar by Kuratowski's theorem, so the answer would be that the statement is false. But I'm not sure about what does "no subdivision of$$K_5$$ as a subgraph" means, any hint or help will be really appreciated.

Subdivisions are defined here; essentially you can subdivide a graph by adding extra vertices along edges (as you choose). This adds a bunch of extra vertices with degree $$2$$.
When the question says the graph "has no subdivision of $$K_5$$", it means that no subgraph of the graph is of this form. As a non-example, $$K_6$$ indeed has a subdivision of $$K_5$$, as if we remove $$3$$ edges coming from a single vertex (so that it now has degree $$2$$), then the resulting graph is a subdivision of $$K_5$$.
To give you a hint, if this were true, then we could take any non-planar graph without a subdivision of $$K_5$$, add in three extra vertices connected in a triangle but disconnected from the rest of the graph, and suddenly it would be planar. That is, every graph with a subdivision of $$K_5$$ would have to be non-planar. Compare this with Wagner's Theorem (often mistakenly attributed to Kuratowski), to find a non-planar graph without a $$K_5$$ subdivision, and use it as above to form a counterexample.