Plane Geometry - Bruce E. Shapiro

152 SECTION 31. QUADRILATERALS IN NEUTRAL GEOMETRYSince M is a bisector, AM = MB; by definition of the Saccheri quadrilateralAD = BC and ∠A ∼ = ∠B. Hence △DAM ∼ = △CBM by SAS.Hence DM = CM; by definition of bisector, DN = CN; hence △DMN ∼ =△CMN by SSS (they share a common side).By congruence, ∠DNM ∼ = ∠CNM; since they form a linear pair, theymust each be right angles.To show that the angle with the base is right, observe that since △DAM ∼ =△CBM, then∠AMD ∼ = ∠BMCand since △DMN ∼ = △CMN thenHence∠DMN ∼ = ∠CMNm(∠AMN) = m(∠AMD) + m(∠DMN)= m(∠BMC) + m(∠CMN)= m(∠BMN)by angle addition. Since ∠AMN and ∠BMN are a linear pair then180 = m(∠AMN) + m(∠BMN)= 2m(∠AMN)hence the angles are right angles.Theorem 31.18 If □ABCD is a Saccheri quadrilateral, then it is a parallelogram.Proof. See figure 31.7. MN ⊥ AB and MN ⊥ DC. By the alternateinterior angles theorem (theorem 29.4), CD ‖ AB. Similarly, AD ‖ BC.Theorem 31.19 Saccheri quadrilaterals are convex.Proof. This follows because every Saccheri quadrilateral is a parallelogram(theorem 31.18) and every parallelogram is convex (theorem 31.9).Theorem 31.20 The summit angles of a Saccheri quadrilaterals are eitherright angles or acute.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012