Prove that the depth of every vertex in a cycle is $\le \lfloor \frac{n}{2} \rfloor$ in a breadth first search?
There may be other ways to prove your claim, the simplest seems to me is by induction on $n$.

Base case : $n=3$, that means input graph is a triangle and it is easy to see that in this case maximum depth is $\le \lfloor \frac{n}{2}\rfloor$

Inductive case:  Suppose that depth of every vertex is $\le  \lfloor \frac{k}{2}\rfloor$ in $k$ cycle graph want to prove that depth of every vertex is $\le  \lfloor \frac{k+1}{2}\rfloor$  for $k+1$ cycle graph.

Hint for Inductive case: Going from $k$ cycle to $k+1$ cycle means we have added one vertex to any one of the edge ( as we know its both endpoints have depth  $\le  \lfloor \frac{k}{2}\rfloor$ (due to induction ) ) of the $k$ cycle graph it means depth of both endpoints in $k+1$ cycle will be    $\le  \lfloor \frac{k+1}{2}\rfloor$.