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Theorem: OBCD is a rhombus, CHED and GEDF are similar crossed-parallelogram. A is a point on the extension of DF, such that DA=OP, then ∠OBC= ∠BOA.
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Pf: Let ∠BOC=θ=∠COD ∴∠CDE=2θ=∠EDA ∴∠EDA=∠DOA+∠DAO=2∠DOA ∴∠DOA=θ ∴∠BOC= 1/3∠BOA
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Use for linkage: If we put the point O to the vertex of any angle BOA, the point B and A on each side of this angle, then ∠OBC= 1/3∠BOA. In other words, we trisect the angle BOA. |
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Reference: Robert C. Yates Line Motion and Trisection NonationalMathematicsMagazine,Volumn13,Issue2(Feb.,1938),63-66 Robert C. Yates A trisector NationalMathematicsMagazine,Volumn12,Issue7(Apr.,1938),323-324 Robert C. Yates Geometrical Tools 1949 P.184-185 |
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