Algorithm Design Homework: Breadth First Search

The homework problem that I am a bit confused on is as follows:

Suppose that an n-node undirected graph G = (V, E) contains two nodes s and t such that the distance between s and t is strictly greater than n/2. Show that there must exist some node u, not equal to either s or t, such that deleting v from G destroys all s-t paths. (In other words, the graph obtained from G by deleting v contains no path from s to t.) Give an algorithm with running time O(m + n) to find such a node v

I am having trouble with a problem that I have been working on for practice and found a supposed solution using the below algorithm.

      Set Discovered[s] = true and Discovered[v] = false for all other v
      Let L be a queue and initialize with s.
      i = 0
      distance = 0
      While i < n
           For each node u ∈ L
                Consider each pair of nodes (u,v)
                If Discovered[v] = false then
                     Set Discovered[v] = true
                     Add v to queue at next position (position k+1)
                     Distance = distance + 1
                End if
           End for
      End while
      If n/2 >= distance >= (n+1)/2 
           There must be a node v that deleting (along with all incident edges) would result in all paths between s and t to be disconnected
      End if

I know that this is a variation of Breadth first search which makes sense as the BFS will start at node s and explore in level traversal of the graph until it reaches the level with node v and then s. My confusion is in how the above algorithm can determine that because of the inequality in the last if statement means that there is certainly a node v that when deleted will disconnect all paths between s and t. If someone can clarify this or explain a better explained algorithm for this situation it would be greatly appreciated. On a side note I thought that the inequality in itself should be the other way around considering n/2 will always be less than (n+1)/2.