[1] The conic passes through the vertices of the triangle.

[2] Its perspector is the harmonic conjugate S of D with respect to {A,K}, K being the intersection of AD with BC.

[3] The tangents to the conic at {B,C} pass through D. Line BC is the polar of D with respect to the conic.

[4] The tangents to the conic at A and Q, Q being the intersection of AD with the conic, meet at a point on BC.

[5] The tangent to the conic at P and line XY meet at a point on the tangent to the conic at A.

The results can be proved directly. They are also consequences of the discussion of the specific example, consisting of the equilateral and its centroid D. For the general triangle and arbitrary location of D one can define the projectivity mapping the vertices of the equilateral to the vertices of the triangle and the centroid of the equilateral to D. All the properties stated above correspond to properties shown in this particular example (see ConicsMaclaurin.html ) and transfer to ABC through the projectivity.

[1] The conic passes through the vertices of the triangle.

[2] Its perspector is the middle S of AK, K being the intersection of the parallel to [D] from A with BC.

[3] The tangents to the conic at {B,C} are parallel to [D] (pass through [D]). Hence the conic is either degenerate or central (no parabolas).

[4] Line BC is the polar of [D] with respect to the conic. This means that BC is a diameter of the conic, the middle M of BC being the center of the conic. Hence the conic passes also through the symmetric of A with respect to M.

[5] The tangents to the conic at A and Q, Q being the intersection of AD with the conic, meet at a point on BC. In fact BC is the conjugate diameter to the diameter of the conic which is parallel to [D]. Thus, AK meets the conic at Q which is symmetric to A with respect to A.

[6] The tangent to the conic at P and line XY meet at a point on the tangent to the conic at A.

[7] All the conics constructed in this way coincide with five-point-conics (conics passing through five points) the four points being {A,B,C,A'} and the fifth being an arbitrary point Q on the parallel to BC from A'.

Nothing to prove here, since the results are special cases of the previous theorem. A special case of such a conic is the one for which the direction [D] coincides with the

In fact, fix the side BC of the quadrangle and consider the lines BA, CD in the directions of the given angles at B and C correspondingly. The various possible quadrangles having the given angles result by intersecting lines {BA,BD} with lines parallel to a given direction (e). The intersections of the diagonals of these quadrangles move along a conic (c), constructed as in the previous paragraph. The middles of diagonals AC move along a line (f) passing through the middle M of BC, which is also the center of the conic (c). The desired quadrangles will be such that their intersection of diagonals coinicides with one of the two intersection points {H

ConicsMaclaurin.html

ConicsMaclaurin3.html

TriangleConics.html

Trilinears.html

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