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 = α = 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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This note was uploaded on 09/12/2011 for the course MAP 4305 taught by Professor Deleenheer during the Summer '06 term at University of Florida.
 Summer '06
 DeLeenheer

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