Thus by theorem 39 since ao eq we have that op qr 5

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thus by Theorem 39 since AO > EQ we have that OP < QR . 5. Since OP < QR and AO > EQ then again by Theorem 39, ] D < ] H since D further from PO than H from RQ . 1
page 59, problem 7: Prove that two Saccheri quadrilaterals are congruent if the summit and summit angles of one are equal, respectively, to the summit and summit angles of the other. Proof: 1. Suppose we have 2 Saccheri quadrilaterals - ABCD and EFGH where DC = FG and ] C = ] D = ] G = ] H . 2. Let’s first prove that the the arms are equal. We prove it by contradiction. Suppose EH < AD . Then take point I on AD such that AI = EH and point J on BC such that BJ = FG . Then ABJI = EFGH by SASAS and then we have that ] I = ] D = ] J = ] C thus follows that the angle sum of IJCD is 360 which contradicts the Saccheri quadrilateral since ] D + ] C < 180 . Similary, we get contradiction if we assume that EH > AD . Thus, EH = AD 3. Now that EH = AD , summits are equal, summit angles are equal and the base angles are 90 , we get that ABCD = EFGH by ASASAS. 2
page 59, problem 9: Prove that an exterior angle of a triangle is always greater than the sum of the two opposite interior angles. Proof: 1. Angles ] 1 and ] 2 are 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 < ] 1 3

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