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Download Homework 28 Answers Conditions for #1
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Homework 28 Answers Conditions for #1-6: Assume the axioms, definitions, and theorems of neutral geometry. Let l and m be two distinct lines, and P and Q two distinct points on m that are on the same side of l. Drop perpendiculars from P and Q to the line l, and call the feet P' and Q', respectively. 1. Suppose that P' = Q' (RAA). Then since P and Q are distinct points and do not lie on l, it follows that PQP' is a triangle. P R Q m Let R be a point in the interior of PQ. Choose points A and B on l such that A lies on the same side of P′R as P and B lies on the same of P′R as Q. Note that ∠AP'P, ∠AP'Q, ∠BP'P, l and ∠BP'Q are all right angles. Furthermore, Q is in the A P'=Q' B interior of ∠PP'B (Converse of Crossbar Thm). So µ(∠PP'B) = µ(∠PP'Q) + µ(∠QP'B) (Protractor Thm). But µ(∠AP'P) + µ(∠PP'B) = 180 ⇒ µ(∠AP'P) + µ(∠PP'Q) + µ(∠QP'B) = 180 ⇒ µ(∠PP'Q) = 0, a contradiction. So P' ≠ Q'. 2. We've already shown that P' ≠ Q'. Suppose that PP′ ∩ QQ! ≠ ∅. Let E be the point of intersection. Then E doesn't lie on l, otherwise P' = Q'. So EP'Q' is a triangle. Note that ∠EP'Q' = ∠PP'Q', a right angle, and ∠EQ'P' = ∠QQ'P', a right angle. We know that σ(EP'Q') ≤ 180 ⇒ ∠P'EQ' = 0, a contradiction. So PP′ ∩ QQ! = ∅. m Q P l Q' E P' 3. We know that P, Q, P', and Q' are distinct points by #1. By #2, PP′ and QQ′ do not intersect. Also, because P and Q are on the Q m P same side of l, we know that PQ and P′Q′ do not intersect (Plane Separation). What about adjacent sides, such as PP′ and PQ? Because these segments already intersect at P, if they intersect at l Q' another point, then P, Q and P' would be on the same line, which P' would make m ⊥ l because P' is the foot of the perpendicular from P to l. Then QP'Q' would be a triangle with two right angles, a contradiction. So these two segments cannot intersect at a point other than P. Similarly, all other adjacent sides can intersect only at endpoints. So PQQ'P' is a quadrilateral. 4. If If d(P, l) = d(Q, l), then PP'Q'Q is a Saccheri quadrilateral, since base angles are right angles and the two sides are of equal length. If d(P, l) ≠ d(Q, l), then it is a trapezoid, because the two sides are parallel by the Corresponding Angles Thm. 5. Since d(P, l) = d(Q, l), PP'Q'Q is a Saccheri quadrilateral, and one of the properties of a Saccheri quadrilateral is that opposite sides are parallel. So l || m. 6. If P and Q are on opposite sides of l, then PQ intersects l, which implies that m intersects l, so l and m are not parallel. 7. They must be parallel by the Alternate Angles Thm. 9. Let P and Q be two points on m such that d(P, l) = d(Q, l). Drop perpendiculars from P and Q to l, and call the feet A and B respectively. By #4, ABQP is a Q N Saccheri quadrilateral. Let M be the midpoint of AB and N be the m P midpoint of PQ. Then by the properties of Saccheri quadrilaterals, MN is perpendicular to both l and m. So l and m admit a common l Q' perpendicular. M P' 10. Let l and m admit a common perpendicular n. Let n intersect l at P and m at Q. Suppose there exists n' such that n' ⊥ l and n' ⊥ m. Let n' intersect l at P' and m at Q'. Suppose that P ≠ Q (RAA). We can think of n as the m Q' Q perpendicular dropped from Q to l and n' as the perpendicular dropped from Q' to l. Then by #3, PP'Q'Q is a quadrilateral. P' l P Furthermore, all four angles of PP'Q'Q are right angles, which n means it is a rectangle. But this is a contradiction in hyperbolic n' geometry. Therefore, P = P'. But then Q = Q' by the same proof used in #1. So n = n'.