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MTH 250 Graded Assignment 5 (part 3) Theorems of absolute geometry (questions) Material from Kay, sections 3.3, 3.4, 3.5 Q1 (long question, short answer!): Saccheri-Legendre, which states that the sum of the angles in a triangle in absolute geometry cannot exceed 180, is one of the key theorems of absolute geometry. Its proof involves building up several results. Just to refresh your memory, the first result comes out of the proof of the Exterior Angle Inequality. In particular, there’s a specific construction that takes place, with the result that in the figure below Next, there’s a key lemma, which states: Take a triangle ABC and construct a segment BE from B through the midpoint M of AC , so that BM ME . Form a triangle BCE . [This is the same construction used in the exterior angle inequality.] Prove that the sum of the angles in the triangle formed by this construction, BCE , is equal to the sum of the angles of ABC . That’s the goal of this question – finish the proof of this lemma. Use the setup described in the lecture, assigning variables: And take as a given that AMB CME [you don’t need to re-prove this part, since it’s already been proven – use it as the jumping off point]. All that remains to be shown is that sum of the angles in BCE is equal to the sum of the angles in ABC . Q2: There’s a collection of lemmas about perpendiculars and bisectors related to the symmetry of an isosceles triangle (they come out of the Isosceles Triangle Theorem, and are used to get to the Perpendicular Bisector Theorem). They’re familiar results that are straightforward to prove, and Lemma B and Lemma C are done in the suggested problems. What’s left is Lemma D (p 144), which states If PA PB and M is the midpoint of segment AB , then ray PM bisects APB . Fill in the proof of Lemma D: Proof: Since PA PB (equivalent PA PB ), APB is by definition isosceles, with base AB . By the Isosceles Triangle Theorem, [Fill in 1] Since M is the midpoint of segment AB , by definition of midpoint, [Fill in 2] Therefore, by the [Fill in 3] , AMP BMP . Make a statement about angles APM and BPM (and justify): [Fill in 4] by [Fill in 5] Next, since M AB , M is an interior point of angle addition holds and [Fill in 6] APB . Since M is an interior point of APB , The above gives us the definition of betweenness for rays: PA PM PB . And finally, PA PM PB along with m APM m BPM is the precise [Fill in 7] Therefore, ray PM bisects APB . *Note: Since the structure of this one is already laid out, I'm looking closely at notation/precision of statements. Q3: In the figure to the right, it is given that WS WT , and RS ST TU with R S T U . Prove that RWS TWU . [Kay, Section 3.3 #18, p. 151] As always, be sure to write justifications within Kay’s axiomatic system, and be particular about the fine details. For example (hey, here’s a hint!), it’s obvious “by looks” that WST and WSR form a linear pair, but to set that up, you first need to assert that ST and SR are opposite rays since R S T . That level of picky. Q4: Use the Exterior Angle Inequality to give upper and lower bounds for x and y in the diagram. Be sure to justify your calculations (you can assume betweenness, opposite rays, and so on as it appears in the diagram for this one without it being explicitly given). Q5: In SKL we already know that m1 m2 180 by a corollary of the Exterior Angle Inequality. Show that if ray KS bisects WKL and ray LS bisects WLK , then we can further conclude that m1 m2 90 . [Kay, Section 3.4 #18, p. 165] [Relationships as apparent in diagram; 1 is LKS , 2 is KLS .] Q6: Given QR QT and angle measures as indicated in the diagram, determine which segment in the figure is the shortest, and justify your answer. [Kay, Section 3.5 #12, p. 172] Be careful to NOT assume that the sum of angles in the triangle is exactly 180, because that’s a theorem of Euclidean geometry…and we aren’t in Euclidean geometry. There are two key theorems of Section 3.5 at work here – be sure to cite as used.