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 $F=(0,{2bc\over a})$.
What about this solution (instead of searching for a credible and/or official solution that you may never find to give it the bonus reputation)?
Like the capitals you've associated to points, let $I$ be the center of inscribed circle inside $\bigtriangleup ABC$. Take $S$ the intersection of $AD$ and $BC$, then we have:
$$\dfrac {AI} {IS} = \dfrac {\sin \angle DAB}{\sin \angle CSA}= \dfrac {\sin \angle SCD}{\sin \angle CSD}= \dfrac{DC}{DS} =\dfrac {DI} {DS}$$
Let $P$ be the intersection of line $\overline {IE}$ and $DL$ and $Q$ be the intersection of $IH$ and $DF$. By Seva's theorem in $\bigtriangleup IDH$ and $IP \| AH $ and above equation we have:
$$\dfrac {IQ}{QH}= \dfrac {PD}{HP} . \dfrac {IS}{DS}= \dfrac {DI}{AI} . \dfrac {AI}{DI}=1$$
So $Q$ is the midpoint of $IH$. let $R$ the intersection of $AH$ and $DF$, then $IEHR$ is a rectangle. From the fact that the quadrilateral $ASEF$ is cyclic (why?) and $RI \| BC$, we conclude $AIRF$ is cyclic, so as you mentioned $IF \bot AK$, therefore $IEKF$ and then $AIHK$ is cyclic and we have $KI\bot AD$ and $\angle HID = \angle AKJ$.
Ultimately, $\bigtriangleup HID \sim \bigtriangleup JKF$ and the tangent of $(ABC)$ at $F$ bisects $KJ$ like the way $DF$ bisects $IH$ at $Q$.
I think other solutions may use the cyclic characteristics of quadrilaterals $ASEF$ and $FEHL$.
Best Answer
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.