[alogo] The antiparallel hyperbolas of a triangle

The antiparallels B'C' to the base BC of triangle ABC are created by intersecting sides AB, AC with circles (c') passing through vertices B and C. By its definition, quadrangle BCC'B' is cyclic, the direction (e) of segments B'C' is constant and equal to the direction of the tangent to the circumcircle (c) of ABC at A. Thus, all these segments are orthogonal to line AO, O being the circumcenter of ABC. The discussion here is on properties of the intersection point P of the diagonals of the cyclic quadrangle BCC'B'.

[1] The locus of P is a rectangular hyperbola passing through the vertices of ABC.
[2] The hyperbola has its center at the middle M of BC and its axes parallel to the bisectors of angle A.
[3] Side BC and line (e) are conjugate directions of the hyperbola.
[4] Let A1 be the intersection point of BC with the tangent to the ABC-circumcircle (e) at A. Take the symmetric A2 of A with respect to A1. The hyperbola passes through the vertices of triangle ACA2 and is invariant under the isotomic conjugation with respect to the median CA1 of this triangle.
[5] The tangent to the hyperbola at A coincides with the symmedian line of triangle ABC through vertex A.

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]
[2_0] [2_1] [2_2] [2_3]

[1, 2] From the middle M of BC draw the parallel B''C'' to the outer bisector of angle A. From the cyclicity of quadrangle BCC'B' follows that the two small (yellow) triangles BB1B'' and CC''C3 are equal and A'B''C'', PB''C'' are isosceli. The first isosceles is a translation of the fixed isosceles AB1C3 by B1B''.
Project P on the parallels to the bisectors of angle A passing from M. The projections create rectangle PM'MM'' has the same area as C1C2C''C3. The basis-length of last parallelogram is C''C3=d*cot(phi) and its height PM'=f*tan(phi). Hence its area is equal to (d*f). This proves that P lies on a rectangular hyperbola, centered at M and having for asymptotes the parallels to the bisectors of angle A. By varying the position of B'C' we see easily that the hyperbola passes through the vertices of triangle ABC.
[3, 4] Consider point P'=f(P). The middle of PP' is on line BC and P' is a point of the hyperbola. To see it draw P'C and P'B and their intersections B'', C'' with the sides of the triangle. It suffices to show that B''C'' is parallel to B'C', or equivalently, parallel to PP'. This is a consequence of an easy exercise on Thales theorem (see ThalesApplication.html ).
[5] Is a consequence of the fact that {P,P'} are harmonic conjugate with respect to {P1,P2}, later being the intersections of PP' with the sides of the triangle. Thus, AP is the harmonic conjugate of line AP' with respect to the pair of sides (AB, AC) of triangle. As P approaches A, line AP tends to the tangent at A and PP' to e=AA2. The property follows from the well known fact on the symmedian and the tangent (e) to the circumcircle at A.

Remark-1 Because of the conjugacy of the directions of BC and (e), the tangents at points B, C (which are symmetric with respect to the center of the hyperbola) are parallel to line (e).
Remark-2 As every rectangular hyperbola passing through the vertices does, the hyperbola passes also through the orthocenter H. The symmetric of H with respect to the middle of the side meets AO at point A3 on the circumcircle. Thus A3 is the antipodal of A and the hyperbola is easily constructible by passing a conic through the five points {A,B,C,H,A2}.
Remark-3 The generation of a conic by translating a line, like B'C', parallel to itself and considering the intersection point P of lines BC' and CB' is a special case of the Maclaurin generation of conics (see Maclaurin.html ) through lines pivoting about a point X. Here X is a point on the line at infinity. In fact, the antiparallel direction of BC determines the unique point at infinity X0, for which the corresponding conic is a rectangular hyperbola. Each side of the triangle defines an analogous rectangular hyperbola.

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