Well, when there is no idea then coordinate system comes at handy. And actualy it is an easy problem with c.s.
Let $B=(2b,0)$, $C= (2c,0)$, $A= (0,2a)$ and $A'= (2t,0)$, for some fixed $a,b,c$ and variable $t$. The midpoint of $A'B$ is $N = (b+t,0)$. Since $B'$ is on a line $$AC:\;\;\;{x\over 2b}+{y\over 2c}=1$$ we have $$B' = (b+t,{a(b-t)\over b}) $$
and analougly we get $$C' = (c+t,{a(c-t)\over c}) $$
Now the slope of segment $B'C'$ is $$k= {at\over bc}$$ so the slope of $d$ is $$k' = -{1\over k} = -{bc\over at}$$
So the line $d$ has equation $$ y= {bc\over at}x +{2bc\over a}$$
which means that this line goes always through the point $F=(0,{2bc\over a})$.
Ignoring vertex $A$, this becomes a problem on the inscriptable $\square BCNM$. Let $L$ be the midpoint of $\overline{BC}$. Also, let $M'$ (instead of $P$), $N'$ (instead of $Q$), $H'$, $K'$ be the projections of $M$, $N$, $H$, $K$ onto $\overline{BC}$. Let the tangent segments from $B$ and $C$ to the incircle have length $d$; let the tangent segments from $M$ and $N$ have length $m$ and $n$. Finally, define $m' := |MM'|$, $n':=|NN'|$, $m'':=|M'L|$, $n'':=|N'L|$. Without loss of generality, $m\leq n$ so that $m'\leq n'$.
Certainly, if $m=n$, then $\overleftrightarrow{HK}$ meets $\overline{BC}$ at $L$. For $m \neq n$, we'll prove that $L$ is on $\overleftrightarrow{HK}$ by showing $\triangle HH'L\sim \triangle KK'L$ via
$$|HH'||K'L|=|KK'||H'L| \tag{$\star$}$$
Parallelism and proportionality rules tell us that
$$\frac{|M'H'|}{|M'N'|}=\frac{|MH|}{|MN|}=\frac{m}{m+n} \qquad |HH'|=m'+\frac{m}{m+n}(n'-m')=\frac{m'n+mn'}{m+n} \tag{1}$$
The Crossed Ladders Theorem tells us that
$$\frac{1}{|KK'|}=\frac{1}{m'}+\frac{1}{n'} \quad\to\quad |KK'| = \frac{m'n'}{m'+n'} \tag{2}$$
(and, in fact, $K$ is the midpoint of the extension of $\overline{KK'}$ that meets $\overline{MN}$), whereupon some proportional thinking then yields $|M'K'|:|K'N'|=m':n'$, so that we have
$$\frac{|M'K'|}{|M'N'|}=\frac{m'}{m'+n'} \tag{3}$$
Therefore,
$$\begin{align}
|H'L|&=|M'L|-|M'H'| = m'' - \,\frac{m}{m+n}(m''+n'') \;= \frac{m''n-mn''}{m+n} \\[6pt]
|K'L|&=|M'L|-|M'K'| = m'' - \frac{m'}{m'+n'}(m''+n'')=\frac{m''n'-m'n''}{m'+n'}
\end{align} \tag{4}$$
Substituting in $(\star)$, and clearing denominators, we need only verify that
$$(m'n+mn')(m''n'-m'n'') = m'n'(m''n-mn'') \tag{5} $$
That is,
$$\frac{m}{n}\cdot\frac{m''}{n''} = \left(\frac{m'}{n'}\right)^2 \tag{6}$$
It seems like there's a geometric mean argument to be made, but I'm not seeing it. So, writing $\theta$ for the common angle at $B$ and $C$, we have
$$\frac{m}{n}\cdot\frac{d-(m+d)\cos\theta}{d-(n+d)\cos\theta} = \left(\frac{m+d}{n+d}\right)^2 \quad\to\quad (d+m)(d+n)\cos\theta = d^2 - m n \tag{7}$$
This same relation results (for $\theta \neq 0$) from the observation that there's a right triangle with hypotenuse $|MN|$ and legs $|m'-n'|$ and $m''+n''$.
$$(m+n)^2 = (m'-n')^2 + (m''+n'')^2 \qquad\to\qquad (7) \tag{8}$$
This equality establishes $(\star)$ and completes the proof. $\square$
I believe there's a cleaner way to link $(6)$ and $(8)$ (or to demonstrate $(6)$ some other way) without having to show equality through $(7)$. Again, I'm not seeing it. Perhaps I'll return to this question.
Best Answer
WLOG, suppose $XY$ is on the $x$-axis. Let us define points $X,Y$ as $(-a,0)$ and $(a,0)$, respectively. Then, $Z$ should be defined as point $(b,z)$. Additionally, point $Q=\left(q,-\frac{z}{a+b}\left(q+a\right)\right)$. For $a,b,z,q \in \Bbb R$.
We have the following equations for lines $XZ, QY$ and $ZY$
$$XZ: y=\frac{z}{a+b}\left(x+a\right)\qquad m_1=\frac{z}{a+b}\\ ZY:y=-\frac{z}{a-b}\left(x-a\right)\qquad m_2=-\frac{z}{a-b}\\ QY:y=-\frac{z(a+q)}{(a+b)(a-q)}\left(x-a\right))\qquad m_3=-\frac{z(a+q)}{(a+b)(a-q)}$$
Since we should have $\angle ZYQ=\angle ZXY$, we want the angle between lines $QY$ and $ZY$ be equal to $\angle ZXY$ Thus, we have: $$\frac{m_2-m_3}{1+m_2m_3}=m_1$$ Substituting the slopes from the equations above, we get that $m_2$ should be: $$m_2\to-\frac{2 a z (a+b)}{(a+b)^2 (a-q)-z^2 (a+q)}$$ However, we already know that $m_2=-\frac z{a-b}$, therefore: $$-\frac{2 a z (a+b)}{(a+b)^2 (a-q)-z^2 (a+q)}=-\frac{z}{a-b}\\ q\to -\frac{a \left((a+b) (a-3 b)+z^2\right)}{(a+b)^2+z^2}\tag{1}$$
This gives us $q$ such that $\angle ZXY=\angle ZYQ$.
Now the line perpendicular to $XZ$ through $Q$ is defined by the line: $$y=-\frac{a+b}{z}\left(x-q\right)+\frac{z}{a+b}\left(q+a\right)$$ Which has a zero of (i.e. $R=(x,0)$): $$x\to \frac{z^2 (a+q)}{(a+b)^2}+q\tag{2}$$ Since we already know $q$, substituting $(1)$ in $(2)$, and we get: $$\bbox[20px,border:1px black solid]{x\to-\frac{a (a-3 b)}{a+b}\implies \therefore R \text{ is definitely invariant to } Z}$$
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