Unable to show that the graph is planar

Question

How to show that the following graph is planar?

I tried to redraw the graph but I cant draw it without the edges crossing.

Also I find that the edges $3,4,5$ form a complete graph. Also I am unable to find any subdivision of $K_5$ or $K_{3,3}$. So I am confused whether it is planar or not?

Answer 1

Place vertices $5$ and $1$ inside triangle $234$. Graph with $5$ vertices can't be subdivision of $K_{3,3}$, and it can be subdivision of $K_5$ only if it's $K_5$ itself.

Answer 2

It is planar. See the attached image for a planar embedding.

Answer 3

$K_{3, 3}$ requires at least 6 veritces, and $K_5$ would require that all five vertices you have are completely connected. This is not the case, so the graph is planar.

As for redrawing, here is one: