A graph $G$ is Eulerian path iff degree of each vertex is even with atmost one trivial component

Question

For this proof ,proving

$\Rightarrow$ If Graph $G$ is eulerian then degree of each vertex is even with atmost one trivial component.

As $G$ is Eulerian ,it means it **must not** repeat Edges but can repeat vertices.Now for the Eulerian (path) traversal ,we pass through that vertex using two incident edges,one for entry and other for exit.

Then what is wrong  in this graph?

Here we have Eulerian path traversal as

$C\rightarrow A \rightarrow B \rightarrow D \rightarrow C \rightarrow F \rightarrow E \rightarrow H \rightarrow G \rightarrow F $

but here degree of $C,F =3$ contradictory....

help me out where i am wrong

Comments

there exists eulerian path, but not eulerian circuit

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http://math.stackexchange.com/questions/2077797/a-graph-g-is-eulerian-path-iff-degree-of-each-vertex-is-even-with-atmost-one-t

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Does Eulerain path do not follow property that every vertex of it must have even degree!!!???

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no not