problems-ex2 - . Problem 1.6.4 on p.18 Referring to Fig....

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Solutions to a couple of problems: MTG 3214 October 29, 2008 Problem 1.8.5 on p.22 (assume problem 1.6.4 on p.18 has been solved; I’ll do this later) First a remark: The statement could have been more precise. For instance: “the 9 point circle cuts the sides at angles | B C | , | A B | and | A C | , seen from the points A , C and B respectively”. An equivalent statement is: “the 9 point circle cuts the sides at angles 2 | B C | , 2 | A B | and 2 | A C | , seen from its center N ”. I will prove the last statement. Refer to Fig. 1.8B. The chord DA is seen from N under angle: DNA = 2 DKA ( since D,K,A’ are on circle with center N) = 2 HKN ( same angles ) = 2 HAO ( since HKN ∼ △ HAO, reason below) = 2 | B C | ( by problem 1.8.5) , There holds that HKN ∼ △ HAO by SAS, since AH = 2 KH , the angles at H are the same, and HO = 2 HN
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Unformatted text preview: . Problem 1.6.4 on p.18 Referring to Fig. 1.6A, we need to show that HAO = | B C | . We have A = BAC = BAD + HAO + OAC. (1) We know that BAD = 90 o B ( consider the right-angled BAD ) (2) and that OAC = OCA ( AOC is isosceles because O is circumcenter) = 90 o B (3) where the last equality holds for a similar reason that in Fig. 1.6.A, OBC = = 90 o A holds. Plugging (2) , (3) in (1), and solving for HAO , yields: HAO = A 180 o + 2 B = B C, which is | B C | since here B > C (if B were < C instead, you would nd that HAO = C B , which is of course equal to | B C | ). * Instructor: Patrick De Leenheer. 1...
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