## Circumconics tangent to a line

Consider a triangle ABC and a line L. The set of circumconics of ABC which are tangent to line L has the following structure.
[1] Construct as usual the trilinear pole (tripole) of L with respect to the triangle:
- Let {A',B',C'} be the intersection points of L with the side-lines of the triangle.
- Join {A',B',C'} with vertices {A,B,C} respectively to define the side-lines of the precevian triangle A0B0C0.
- Lines {AA0, BB0, CC0} intersect at a point PL. The trilinear polar of PL with respect to either of triangles ABC and A0B0C0 is line L.
[2] Consider an arbitrary point Q on L and the lines {QA0,QB0,QC0} joining Q to the vertices of A0B0C0. The intersection points {A'',B'',C''} of these lines with the opposites sides of the triangle and points {A,B,C} are on a conic c(Q).
[3] c(Q) passes through Q and is also tangent to L at Q. Thus, it is the unique conic passing through the vertices of ABC and tangent to L at its point Q.
[4] The tripole of L with respect to triangle A''B''C'' is a point Q' on line QPL.
[5] The tangential triangle A1B1C1 of ABC relative to the conic c(Q) is line perspective to A''B''C'', the perspectrix being line L.
[6] The locus of perspectors of conics c(Q) is the inscribed conic of the triangle with perspector PL.

The clue fact is the relation of the triangle ABC to line L, expressed through the tripole PL of L with respect to the triangle. This basic configuration depends on the four points {A,B,C,PL}. We can restart the whole construction by selecting another location for PL. Since there is a unique projectivity mapping four points in general position to any other four points in general position we can use such a device to transfer the proof to a particular case.
The particular case of choice is the one for which PL is the centroid G of triangle ABC. In this case line L, which is the tripolar of PL coincides with the line at infinity and point Q on that line corresponds to a direction of lines. Besides the corresponding precevian triangle becomes the antiparallel triangle A0B0C0 of ABC.
This particular case is studied in AnticomplementaryAndCircumparabola.html . There it is shown that for every point Q on L (i.e. every direction of lines) there is a parabola passing through the vertices of triangles ABC and A''B''C''. Later triangle has as vertices the intersections of lines {QA0,QB0,QC0} with the opposite sides correspondingly {B0C0,C0A0,A0B0} of triangle A0B0C0.
All the stated properties are trivially verified in this particular case and transfer to the general case through the projecivity fixing the vertices of ABC and mapping the centroid G onto PL.

Remark If follows an easy construction of a conic passing through three non-collinear points {A,B,C} and tangent to a line L at a point Q of it.
One has only to draw the precevian triangle A0B0C0 (depending on L), whose vertices result through line intersections and then find points {A'',B'',C''} on its sides, which result also through simple line intersections. Then one can use the tool which constructs the unique conic through five out of the six points {A,B,C,A'',B'',C''}.