Proving a graph has no Hamiltonian cycle Show that $ G = (V, E)$ has no Hamiltonian cycle, where the vertices are $ V = \{ a, b, c, d, e, f, g \} $ and
the edges are $E = \{ ab, ac, ad, bc, cd, de, dg, df, ef, fg \}$.
From my working out, the vertices $ a, b, c, d, e, f$ are odd degrees of 3 and 1. Moreover $g $ being a even vertices of 2. 
There were three points that were made in my textbook to show that a graph does not contain a Hamilton circuit:


*

*A graph with a vertex of degree one cannot have a Hamilton circuit.

*Moreover, if a vertex in the graph has degree two, then both edges that are incident with this vertex must be part of any Hamilton circuit.

*A Hamilton circuit cannot contain a smaller circuit within it.


According to the definition graph G does not have a Hamiltonian cycle because of the first definition. 
However, I am confused about 2 & 3 definitions and I am not sure if this graph involves them or not.  

Illustration of graph G edges
The pic shows the graph G with the according vertices and edges. 
Sorry I couldn't seem to add the desired red lines. 
I am making the assumption that the graph is a Hamiltonian ( I know it isn't)
However, I manage to work out that vertices b, e and g have degrees of 2. 
However, I don't understand how to use fact 3 to contradict this. 
Also I notice that D is a cut-vertex, can this be added to prove that the graph is not a Hamiltonian cycle. 
