Proof That A And B Must Lie On The Same Side Of CD

Let A and B be not on the same side of CD. Let AB cuts CD at F.

As ABCD is a Saccheri quadrilateral so BAD =90? and CDA=90?. Now one of the triangles DAF and BFC contains a right angle and an obtuse angle because one of the linear pair of angles of F must be an obtuse or a right angle. In this case, angles FAD=90? and FDA=90?. As a result of which, the sum of the three angles of a triangle will be more than 180?. Contradiction!!!! Hence A and B must lie on the same side of CD

Let C and D be not on the same side of AB. Let CD cuts AB at E. 

As ABCD is a Saccheri quadrilateral so BAD =90? and CDA=90?. Now one of the triangles DAE and BEC contains a right angle and an obtuse angle because one of the linear pair of angles of E must be an obtuse or a right angle. In this case angles EAD=90? and EDA=90?. So sum of three angles in Triangle EAD is more than 180?. Contradiction!!!!
Hence C and D must lie on the same side of AB

Let A and D be not on the same side of BC. Let AD cuts BC at F. 

As ABCD is a Saccheri quadrilateral so BAD =90? and CDA=90?. Now one of the triangles DCF and BAF contains a right angle and an obtuse angle because one of the linear pair of angles of F must be an obtuse or a right angle. In this case angles FDC=90? and CFD≥90?. So sum of three angles in Triangle CDF is more than 180?. Contradiction!!!! Hence A and D must lie on the same side of BC.