Let $ G $ be a tree with $ n $ vertices. Show that the following statements are equivalent:
a) $ G $ is the path with $ n $ vertices.
b) $ G $ has a maximum degree of two.
c) $ G $ has exactly two vertices of degree one.
The usual thing would be to show $ a \rightarrow b \rightarrow c \rightarrow a $. I think it would be easier to show $ a \rightarrow c \rightarrow b \rightarrow a $
If $ a \rightarrow c $ is easy, because if $ G $ is the path with $ n $ vertices, the initial and final vertex are the only ones with degree 1.
For the rest of the implications, what are your suggestions?
You can complete the loop $a\to c\to b\to a$ by using the handshake lemma.
For $c\to b$ you can argue that $T$ has $n-1$ edges, so the sum of the degrees of its vertices must be $2(n-1)=2n-2$. $T$ has $n-2$ vertices of degree at least $2$; if any of these vertices had degree greater than $2$, the sum of the degrees of the vertices of $T$ would be greater than $2(n-2)+2=2n-2$, which is impossible.
For $b\to a$ let $n_1$ be the number of vertices of degree $1$ and $n_2=n-n_1$ the number of vertices of degree $2$. Since $T$ must have $n-1$ edges, we have
$$2(n-1)=n_1+2n_2=n_1+2(n-n_1)=2n-n_1\,,$$
so $n_1=2$, and $n_2=n-2$: $T$ has exactly $2$ vertics of degree $1$.