Continuing generalization of the Simson line

Question

In 2014, I found a nice result in plane geometry, the result is a generalization of the Simson line theorem, and there are nine proofs for this result were published in [1]-[7]. Continuing, I find a new generalization of the old result as follows:

Problem: Let $ABC$ be a triangle, let $P$ be a point in the circumcircle, the circumcenter is $O$. Let $Q$ be the point in the plane. The circles $(APQ), (BPQ), (CPQ)$ meet $OQ$ again at $A', B', C'$ respectively. Let $A_1, B_1, C_1$ be the the projections of $A', B', C'$ onto $BC, CA, AB$ respectively. Then $A_1, B_1, C_1$ are collinear, and the new line through a fixed point on the Nine point circle when $Q$ be moved on the given line (or $P$ be moved in the circumcircle). When $Q$ in infinity we get the old result (the result I found in 2014).

My question, could you give a proof for this problem?

Note: Check the result is true with applet by click here

Note: You can see some problem around this configuration in

** References:**

[1]-Nguyen Van Linh, Another synthetic proof of Dao's generalization of the Simson line theorem, Forum Geometricorum, 16 (2016) 57--61.

[2]-Nguyen Le Phuoc and Nguyen Chuong Chi (2016). 100.24 A synthetic proof of Dao's generalisation of the Simson line theorem. The Mathematical Gazette, 100, pp 341-345. doi:10.1017/mag.2016.77.

[3]-Leo Giugiuc, A proof of Dao’s generalization of the Simson line theorem, tạp chí Global Journal of Advanced Research on Classical and Modern Geometries, ISSN: 2284-5569, Vol.5, (2016), Issue 1, page 30-32

[4]-Tran Thanh Lam, Another synthetic proof of Dao's generalization of the Simson line theorem and its converse, Global Journal of Advanced Research on Classical and Modern Geometries, ISSN: 2284-5569, Vol.5, (2016), Issue 2, page 89-92

[5]-Ngo Quang Duong, A generalization of the Simson line theorem, to appear in Forum Geometricorum.

[6]-Three other proofs by Telv Cohl, Luis Gonzalez, Tran Quang Huy A Generalization of Simson Line

[7]-Another proof https://www.artofproblemsolving.com/community/c6h1075523p5181203

Answer 1

Let $CC'$ meet a circle $\omega=(ABC)$ in a point $S\ne C'$. Then $\angle (CP,CS)=\angle (CP,CC')=\angle (QP,QC')=(QP,QO)$. Thus $AA'$, $BB'$ pass through the same point $S$. The following argument is not synthetic, but it explains what is this fixed point on an Euler circle and what is another point in which $A_1B_1C_1$ meets Euler circle. Thus it hopefully may help with a synthetic argument too.

Consider the complex coordinates for which $\omega=\{z:|z|=1\}$, $A,B,C$ correspond to complex numbers $a,b,c$, $OQ$ to a real line, $S$ to $s$. Then $C'$ corresponds to $c'=(c+s)/(1+cs)$ (this is a formula for central projection from $\omega$ to a real line from the point $s$, as may be checked for three points $1,-1,-s$). Next, a projection of $z$ to a line between $a,b$ is $(z-\bar{z}ab+a+b)/2$, as may be checked for points $a,b,0$. So, $C_1$ corresponds to $c_1=(c'(1-ab)+a+b)/2$. Denote $c_2=2c_1-(a+b+c)$. Note that $z\rightarrow 2z-(a+b+c)$ is a homothety which sends Euler circle of $ABC$ to $\omega$. Thus for points $a_2,b_2,c_2$ we should prove that they are collinear and the line passes through a point on $\omega$ not depending on $s$. We get $c_2=c'(1-ab)-c$ and I claim that $c_2$ lies on a line between $s$ and $-abc$. Indeed, the direction between $s$ and $-abc$ is a direction of $s+abc$. The direction between $s$ and $c_2$ is a direction of $c_2-s=-c'(ab+cs)$, that is, direction of $ab+cs$, but the ratio of $s+abc$ and $ab+cs$ is indeed real.

Answer 2

This is an old post, already answered, but this comment (posted as an answer, since I am not entitled) might be of interest. There is an alternative, computational proof which gives the result in a quantitative manner and thus more details (e.g., about possible converses). One can assume that the circumcircle is the unit circle (and then $A=(\cos\theta_1,\sin \theta_1)$, etc.) and that $P=(x_P,y_P)$, $Q=(x_Q,0)$. One can then compute explicitly the further points in the construction and the area of the resulting triangle $A_1B_1C_1$ (disclosure: I used Mathematica). The resulting formula is tractable and reveals the required result immediately, plus more details.