The summit is greater than the base is due to the construction of the Saccheri quadrilateral. Given segment A'B' we construct perpendicular segments at A' and B'. These are the segments B'B and A'A. B'B and A'A then by construction parallel. Since they are both perpendicular to A'B' then A'B' is a common perpendicular which is the shortest distance between lines B'B and A'A. This segment B'A', the common perpendicular is unique we proved that in homework 4. Thus any other segment AB must be greater than A'B'.