Let $G$ be a connected graph that is not a tree. Let $u,v \in V(G)$ , $u \not =v$ such that $G-u$ and $G-v$ are both tree. Show that $deg(u)=deg(v)$
Here is what I got so far.
Let $T_1 =G-u$ and $T_2=G-v$.Note that $u$ and $v$ can't be on distinct cycles, otherwise, after delete one vertex, we still have one cycle, which isn't allow in tree. Let $n$ be the order of $G$ then both $T_1$ and $T_2$ have size $m=(n-1)-1=n-2$. Moreover the sum of degree of both $T_1$nad $T_2$ are $2m=2n-4$.
Now I'm stuck, any hint will be much appreciated.
You are just there.
Both $T_1$ and $T_2$ has $n-1$ points and both are trees, so both has $n-2$ edges. Assume $G$ had $m$ edges, so we must have deleted $m-(n-2)$ edges when deleting either $u$ or $v$. Thus we arrive to $$\deg(u)=m-(n-2)=\deg(v)\,.$$