Let $G$ be an edge transitive graph. Let $t(G)$ be the number f spanning trees on $G$. Show that each edge lies in exactly $\tfrac{(n-1)t(G)}{m}$ spanning trees. Where $|V(G)|=n$ and $|E(G)=m$.
$\textbf{So Far}$
So far I have used the edge transitivity to show that every edge is contained in the same number of spanning trees. I want to partition $T(G)$ in some fashion and add them all up. The only good partition that I can think of is either a tree contains an edge $e$ or it doesn't. That results in the partition $t(G)=t(G/e)+t(G\backslash e)$ which doesn't get me anywhere.
Or possibly a double count. Count the number of trees containing $e$ for all $e$ and subtract the overlap. Which seems like a lot so I don't know if the double count will work.
A hint would be much appreciated thanks.
It is a known fact that $t(G/e)$ is the number of spanning trees of $G$ containing the edge $e$. In this case $t(G/e)$ is the same for all edges $e$. So we need to prove this equality
$$(n-1)t(G)=\sum_{e\in E(G)}t(G/e).$$
We will count the same set twice. Count $(n-1)t(G)$. RHS is clear. Any spanning tree of $G$ will have $n-1$ edges. Then $T$ is a spanning tree for $G/e$ if and only if $e$ is an edge of $T$. Since $T$ has $n-1$ it will be counted $n-1$ times. And any spanning tree of $G$ contains $n-1$ edges so it will show up. Thus they are the same.