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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 rightangled 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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 Summer '06
 DeLeenheer

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