Graph Theory proving planarity I have these set's of graphs: 

I used Euler's inequality, and the 4 color theorem which resulted in a inconclusive result. Using Kuratowski's theorem I was unable to create a K3 3 or K5 graph. So does this prove that the graphs are planar? How do we really know that there does not exist a K3 3 or K5? I did use planar embedding and was able to put it in a form where the edges do not overlap, but the method does not seem rigorous enough to prove planarity. 

Graph $G$ is not planar. The subgraph you get by deleting the edges $ab,$ $ad,$ $bd,$ $cg,$ $eg$ is homeomorphic to $K_{3,3};$ the vertices $a,b,d$ are connected to the vertices $e,f,g$ by the internally disjoint paths $ace,$ $af,$ $ag,$ $be,$ $bf,$ $bg,$ $de,$ $df,$ $dg.$
Graph $H$ is planar. Plot $a$ at $(0,0),$ $b$ at $(2,0),$ $c$ at $(1,1),$ $d$ at $(1,3),$ $e$ at $(1,4),$ $f$ at $(2,3),$ $g$ at $(1,2),$ and draw all the edges as straight line segments.
