AI' such that AI' is parallel to a and perpendicular to a. Construct another line A'I such that A'I is parallel to a and perpendicular to a'. Let the intersection of AI' and A'I be B. Because ᗉ IBI' > /2 radians, the case now plays out like case 1, allowing construction of the common parallel to BI and BI'. Case 3: a and a' are ultraparallel to each other Using a hyperbolic ruler, construct BI' such that BI' is perpendicular to a and parallel to a' and construct a line B'I such that B'I is perpendicular to a''' and parallel to a in a way that puts BI' and B'I on the same side of the common perpendicular to a and a' , which can be found using the ultraparallel theorem. Let the intersection of BI' and B'I be C. Then ᗉ ICI' ≠ /2 radians, which allows you to finish the construction like the other two cases. Line perpendicular to another line at a point Suppose you have a line a and a point A on that line, and you want to construct a line perpendicular to a and through A. Then let a' be a line through A where a and a are two distinct lines. Then you will have one of two cases.Case 1: a is perpendicular to a In this case, we already have the line perpendicular to a through A.Case 2: a and a' are not perpendicular to each otherUsing a hyperbolic ruler, construct a line BI such that BI is perpendicular to a and parallel to a'. Also, construct a line CI' such that CI' is perpendicular to a and parallel to a' but in the opposite direction of BI. Now draw a line II" so that II" is the common parallel to BI and I'C. The ultraparallel theorem now allows us to create the common perpendicular to II" and a because these two lines are ultraparallel. This common perpendicular is now the line perpendicular to a and through A. Midpoint of a line segment Suppose you are trying to find the midpoint of a line segment AB. Then construct the line AI such that AI is through A and perpendicular to AB. Also, construct a line BI' such that BI' intersects AB at B and is perpendicular to AB. Now construct the line II' such that II' is the common parallel to