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thus by Theorem 39 sinceAO > EQwe have thatOP < QR.5. SinceOP < QRandAO > EQthen again by Theorem 39,]D <]HsinceDfurtherfromPOthanHfromRQ.1
page 59, problem 7:•Prove that two Saccheri quadrilaterals are congruent if the summit and summit angles ofone are equal, respectively, to the summit and summit angles of the other.Proof:1. Suppose we have 2 Saccheri quadrilaterals -ABCDandEFGHwhereDC=FGand]C=]D=]G=]H.2.Let’s first prove that the the arms are equal.We prove it by contradiction.SupposeEH < AD.Then take pointIonADsuch thatAI=EHand pointJonBCsuch thatBJ=FG. ThenABJI∼=EFGHby SASAS and then we have that]I=]D=]J=]Cthus follows that the angle sum ofIJCDis360◦which contradicts the Saccheri quadrilateralsince]D+]C <180◦. Similary, we get contradiction if we assume thatEH > AD. Thus,EH=AD3. Now thatEH=AD, summits are equal, summit angles are equal and the base anglesare90◦, we get thatABCD∼=EFGHby ASASAS.2
page 59, problem 9:•Prove that an exterior angle of a triangle is always greater than the sum of the two oppositeinterior angles.Proof:1. Angles]1and]2are supplementary, thus]1 +]2 = 180◦and thus]2 = 180-]1.2.]2 +]3 +]4<180◦by Theorem 42.3. From 1 and 2:180◦-]1 +]3 +]4<180◦follows that]3 +]4<]13