B or a sin A R But in a similar fashion, A 2 b sin B...

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B or a sin A R But in a similar fashion, A 2 b sin B A' Figure 1.35 2R с sin C Proof. In AABC, let A'O be the perpendicular bisector of BC, as in Fig- ure 1.35. Note that ABOC is isosceles, that ZBOA' ZCOA', and that |BA'| = |A'C| = . Note also that, by the Star Trek lemma, ZBOC = 2A, so ZBOA' = A. Thus, R sin A = 2R= b sin B B' a a sin A 2R. C sin C с - Exercise 1.87. Though we have defined O to be the circumcenter, we have not shown that the circumcircle always exists. Prove that the perpendicular bisectors of two sides of a nondegenerate triangle AABC intersect, and that this point is equidistant from all three vertices. Conclude that AABC has a circumcircle and that the perpendicular bisectors are coincident. B or a sin A R But in a similar fashion, A 2 b sin B A' Figure 1.35 2R с sin C Proof. In AABC, let A'O be the perpendicular bisector of BC, as in Fig- ure 1.35. Note that ABOC is isosceles, that ZBOA' ZCOA', and that |BA'| = |A'C| = . Note also that, by the Star Trek lemma, ZBOC = 2A, so ZBOA' = A. Thus, R sin A = 2R= b sin B B' a a sin A 2R. C sin C с - Exercise 1.87. Though we have defined O to be the circumcenter, we have not shown that the circumcircle always exists. Prove that the perpendicular bisectors of two sides of a nondegenerate triangle AABC intersect, and that this point is equidistant from all three vertices. Conclude that AABC has a circumcircle and that the perpendicular bisectors are coincident.