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Geometry EXAM II Solutions 1. (a) Assume that 2ABCD is a Saccheri quadrilateral, that is ∢A, and ∢B are right angles and that AD ∼ = BC. Prove that ∢C ∼ = ∢D. Proof: Consider the triangles △DAB and △CBA. By SAS, these are congruent since AB ∼ = CB (by hypothesis) and = BA, DA ∼ ∼ ∢DAB = ∢CBA (both are right angles). Therefore, DB ∼ = CA. Now consider the triangles: △DAC and △CBD. By SSS, these are congruent. Therefore, ∢CDA ∼ = ∢DCB. (b) Let M be the midpoint of AB and M ′ the midpoint of DC. Prove that MM ′ is perpendicular to AB and to DC. Proof: By SAS, △DAM ∼ = △CBM ∼ Therefore DM = CM. By SSS, △DM ′ M ∼ = △CM ′ M, so ∢DM ′ M ∼ = ∢CM ′ M. So MM ′ ⊥DC. By part (a) ∢ADC ∼ = ∢BCD, so we can apply the same argument ′ to show that MM ⊥AB. 1 2. In Theorem 4.1 it was proved in neutral geometry that if alternate interior angles are congruent, then the lines are parallel. Prove that in hyperbolic geometry, such two parallel lines have a common perpendicular. In other words, let m and l be two parallel lines. Let P be a point on m and Q a point on l, such that the alternate interior angles at P and Q are congruent (see the picture). Prove that l and m have a common perpendicular. Hint: Let M be the midpoint of P Q and drop perpendiculars M N and M L to the lines m and l, respectively. Prove that L, M and N are collinear. Proof: By AAS, △NP M ∼ = △LQM. Therefore, ∢P MN ∼ = ∢QML. Then the supplementary angle of ∢P MN: ∢QMN ∼ = ∢LMP , which is supplementary to ∢QML. Therefore (∢LMP )o + (∢P MN)o = 180o . Therefore L, M, N have to be collinear. 3. Prove that every segment has a unique midpoint. Hint: Recall the construction of the midpoint, or exercise 12 in Chapter 4. Proof: In the proof of exercise 12 in chapter 4, we chose a point C on a ←→ side of AB, and then found the point C ′ on the other side of the line, 2 so that: ∢BAC ∼ = BC ′ . Then M was defined as = ∢ABC ′ and AC ∼ the intersection of CC ′ with AB (such point exists because C and C ′ are on opposite sides of AB). Then we showed that AM ∼ = MB. By RAA, assume that there is another point M ′ such that AM ′ ∼ = M ′ B. By SAS: △ACM ′ ∼ = △BC ′ M ′ . Then ∢CM ′ A ∼ = ∢C ′ M ′ B and their supplementary angles are also congruent: ∢BM ′ C ∼ = ∢C ′ M ′ A. ′ ′ Therefore (∢C M B)o + (∢BM ′ C)o = 180o , i.e. C, M ′ , C ′ are collinear. Therefore M and M ′ have to be the same point. 3