teorema de Jacobi

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In plane geometry, a Jacobi point is a point in the Euclidean plane determined by a triangle △ABC and a triple of angles α, β, γ. This information is sufficient to determine three points X, Y, Z such that ∠ Z A B = ∠ Y A C = α , ∠ X B C = ∠ Z B A = β , ∠ Y C A = ∠ X C B = γ . {\displaystyle {\begin{aligned}\angle ZAB&=\angle YAC&=\alpha ,\\\angle XBC&=\angle ZBA&=\beta ,\\\angle YCA&=\angle XCB&=\gamma .\end{aligned}}} Then, by a theorem of Karl Friedrich Andreas Jacobi, the lines AX, BY, CZ are concurrent, at a point N called the Jacobi point. The Jacobi point is a generalization of the Fermat point, which is obtained by letting α = β = γ = 60° and △ABC having no angle being greater or equal to 120°. If the three angles above are equal, then N lies on the rectangular hyperbola given in areal coordinates by y z ( cot ⁡ B − cot ⁡ C ) + z x ( cot ⁡ C − cot ⁡ A ) + x y ( cot ⁡ A − cot ⁡ B ) = 0 , {\displaystyle yz(\cot B-\cot C)+zx(\cot C-\cot A)+xy(\cot A-\cot B)=0,} which is Kiepert's hyperbola. Each choice of three equal angles determines a triangle center. The Jacobi point can be further generalized as follows: If points K, L, M, N, O and P are constructed on the sides of triangle ABC so that BK/KC = CL/LB = CM/MA = AN/NC = AO/OB = BP/PA, triangles OPD, KLE and MNF are constructed so that ∠DOP = ∠FNM, ∠DPO = ∠EKL, ∠ELK = ∠FMN and triangles LMY, NOZ and PKX are respectively similar to triangles OPD, KLE and MNF, then DY, EZ and FX are...