Prove that a graph $G$ has an Eulerian orientation if and only if $G$ is Eulerian
Here is what I got so far.
=> Let $G$ has an Eulerian Orientation, then $G$ is an Eulerian digraph. For any digraph , if it is Eulerian then $od(v)=id(v)$ for each $v\in V(G)$, this mean that every vertex in $G$ has even degree, thus $G$ is Eulerian.
<=
Assume that $G$ is Eulerian, then every vertex of $G$ has even degree, meaning for each $v\in V(G)$, $id(v)$ and $od(v)$ must be both odd or both even. From here I got the feeling that I need to use the first theorem of digraph which say $\sum id(v_i)= \sum od(v_i)$, but I'm still not sure how to link this to the orientation of $G$