Why half transitive graphs have even degree? A graph is half transitive if it is both vertex and edge transitive, but not arc transitive.

Question: Why is every half transitive graph of even degree?

(there is no trivial example)
The solution seems to be simple because it is stated as an elementary homework.
 A: Let $G$ be half-transitive and $e\in E(G)$ an edge.
Choose an orientation of $e$.
Because $G$ is edge- but not arc-transitive, this induces a unique orientation on every other edge of $G$.
We obtain a directed graph $G'$ with $\mathrm{Aut}(G)=\mathrm{Aut}(G')$.
Now let us count the number $d_{\mathrm{out}}(i)$ of out-going edges and the number $d_{\mathrm{in}}(i)$ of in-going edges at a vertex $i\in V$.
Since $G$ is vertex-transitive, these numbers are the same for every vertex, so let's call them $d_{\mathrm{out}}$ and $d_{\mathrm{in}}$.
Because every edge has a head and a tail, in every directed graph holds
$$\sum_{i\in V} d_{\mathrm{out}}(i)=\sum_{i\in V} d_{\mathrm{in}}(i).$$
But the left side equals $|V| d_{\mathrm{out}}$, and the right side equals $|V| d_{\mathrm{in}}$.
Cancelling out $|V|$ yields that we have the same number of out-going edges as in-going edges at every vertex. Therefore the degree must be even.
By the way, this argument shows that every vertex-transitive directed graph has even degree.
