Is a line associated with antipodal points (the fact, it is the generalization of Simson line) known?

First time, I found a line associated with antipodal points, detail:

Let $$ABC$$ be a triangle, $$(C)$$ is circumconic of $$ABC$$. $$P$$ and $$P'$$ are two antipodal points. Construct three lines through $$P'$$ and parallel to $$PA$$, $$PB$$, $$PC$$ meets $$BC$$, $$CA$$, $$AB$$ respectively at three collinear points, the new line through the center of circumconic.

Question: Is a line associated with antipodal points above known?

Update: But the fact, the result is generalization of the Simson line, I reformulate as follows:

Let $$ABC$$ be a triangle, $$P$$ be a point in the plane, let $$(C)$$ is the Nine point conic of $$A$$, $$B$$, $$C$$, $$P$$. Let $$O$$ be arbitrary point on $$C$$, $$P'$$ is the reflection of $$P$$ in $$O$$. Then three lines through $$P'$$ and parallel to $$PA$$, $$PB$$, $$PC$$ meet three lines $$BC$$, $$CA$$, $$AB$$ respectively at three collinear point.

When $$P$$ is the orthorcenter, the line is the Simson line of $$P'$$

Question again: The generalization of the Simson line above is known?

• This is one those problems that can be solved using the so-called $p,q$ method, after two reductions: 1) by affine invariance, we cann suppose that the circumconic is, in fact, the circumcentre. 2) we can assume that the vertices are $(0,0)$, $(1,0)$ and $(p,q)$ . We can then compute the coordinates of the auxiliary points and so apply the standard collinearity condition that the area of the corresponding triangle be zero. Oct 10, 2021 at 11:59
• Yes, the solution maybe not hard. But my question that is it known? Oct 10, 2021 at 12:02
• Sorry— misunderstood. No idea if it is known. Oct 10, 2021 at 12:03
• Thank you for your comment. Oct 10, 2021 at 12:04
• The simson line is very well-known, and have some nice properties, I hope that the line will know and have some nice properties Oct 10, 2021 at 12:05

Consider the diagram below. Start with a conic $$\gamma$$(green), a triangle $$ABC$$ inscribed in $$\gamma$$, and a line $$\omega$$ (black dot-dashed). Let $$X$$ be the polar of $$\omega$$ wrt the conic, and draw a line (dotted) through $$X$$ that meets $$\omega$$ at $$P,P'$$. Let the dashed lines through $$P'$$ meet the respective lines from $$P$$ to $$A,B,C$$ at $$\omega$$. Then the dashed lines meet the triangle sides at collinear points (red), and $$X$$ lies on this line.
The OP is the special case when $$\omega$$ is the projective line at infinity.