Gary Tupper says: >if AB is thus bisected to AC and CB, the common end point of AC & CB is C.
With a key word in there being "common". Viewed that way AC & CB have a point in common, they are not disjoint. Neighbor would seem to be asking for two disjoint line segments. I think Gary may be reminding us that Euclid (perhaps) never offered any method for bisection that explicitly specified or claimed "disjoint" segments.