#2) Construct a Saccheri Quadrilateral

Saccheri Quadrilateral

I figured out what I was doing wrong. I was measuring the distance with the Euclidean tools not the hyperbolic tools. Given any two points A and B construct the line segment between them. Then construct three perpendiculars to this line. One at A and one at B and a perpendicular bisector. Put a point on the bisector E and construct a perpendicular line segment to E. Label the intersection points of this perpendicular line with the perpendiculars at A and B call it J and I and that concludes the construction of the Saccheri quadrilateral.

b. Uniqueness

N.T.S. The common perpendicular bisector of the base and the summit is unique. Suppose to the contrary that there are two common perpendicular bisectors of the base and the summit. The perpendicular bisector of either the base or the summit is the midset of, in this case, A and B or I and J. W.L.O.G. we will focus our proof on the segment AB. If there are two points one point on one perpendicular bisector and the second point of the other perpendicular bisector then there exists a point on the segment formed by these two points by the betweenness axioms that is equal distance from both A and B. Thus there is no line because the two points are the same point because the perpendicular bisector is unique. We can form an identical argument for the perpendicular bisector of the summit and safely conclude that the common perpendicular bisector is unique.