Suppose that a simple graph $G$ on $n$ vertices drawn in the plane. Prove that if every edge crosses at most one another edge, then the number of edges in $G$ is at most $4n − 8$
What I have tried: I tried to take a graph with finite vertices let's say 12 and drew it in a criss-cross manner so that every edge have at most one crossing. Here since it not planar then it won't satisfy the Euler formula or edge requirement conditions. So I thought of removing edges one by one it won't have any crossings. And then I can apply one of the edge conditions. But it did not result in a meaningful way. Then I do not know any other approach. Could you please help me to solve this?
Cheers!!
For each edge $f = uv$ that crosses an edge $e$, if there are not already edges from the endpoints of $e$ to the endpoints of $f$, add them. (These edges can always be added without crossing other edges, since $f$ meets only $e$ on its path from $u$ to $v$.)
Thus we obtain a graph $H$ with (potentially) more edges than $G$. Consider a planar subgraph $H'$ obtained by deleting one of each pair of edges that cross. $H'$ has at most $3n-6$ edges, and at most $2n - 4$ faces. Each removed edge crosses two faces of $H'$, and at most one of the removed edges appears in each face of $H'$, so there are at most $n - 2$ removed edges.