By midpoint theorem, we have A(V)B // P(A’)B’. Then, $\angle 1 = \angle 2 = \angle 3= \angle 4$.
Let (1) the in-circle touches AC at U; (2) BI extended cut PA’B’ at M. From A’ B = A’M = A’C, we can say that M is also on the circle TIUC with IC as the diameter and $\angle BMC = 90^0$.
[Explanation added: A’ B = A’M = A’C Implies A' is the center of the circle that passes though B, M, C. Then, $\angle BMC = 90^0$ by angle in semi-circle.
Now, $\angle IMC = \angle IUC = 90^0$ implies UICM is cyclic. Since UITC must be cyclic by tangent properties, M is the 5th concyclic point circle UITC.]
Note also that CM // PTNV because $\angle BNT = \angle CMX = 90^0$. Then, $\angle 5 = \angle 1 = \angle 4$ implies P is another con-cyclic point of the circle TIUC.
[Explaination added: $\angle 5 = \angle 4$ is a sufficient condition to say TPCM is cyclic.]
[Please ignore:- Since $\angle AIM$ and $\angle MCT$ are equal to 2x + y independently, IA is the extended line of the cyclic quad MCPI.]
[Newly added:-]
$\angle (1) = \angle (2)$, (angle at center and angle at circumference)
$ = \angle (3)$, (ext. angle, cyclic quad of the red circle)
$ = \angle (4)$, (angles in the same segment)
$ \alpha = \beta$, (angles in the same segment)
$ = \angle (2)$, (ext. angle cyclic quad.)
That is, $ \alpha = \angle (1)$. And this means IA is the extended part of the cyclic quadrilateral UCPI.
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
Let's consider point $P$ such that $APDH$ is a parallelogram (actually, a rectangle since $AH\perp BC$). It's clear that $GH=HD$ (as you've mentioned in the question, we have $RG=RD=RE$), so $APHG$ is a parallelogram and $X$ is its center (recall that $X$ is a midpoint of $AH$).
Now, note that $G$ is one of the intersection points of the circles $(REC)$ and $(QFB)$, so it's sufficient to prove that $GX$ is the radical axis of these circles, or to prove that $P$ lies on this radical axis.
Firstly, note that $P$ lies on the circle $AFE$. Indeed, the incenter $I$ lies on $PD$ (since $ID$ and $PD$ are perpendicular to $BC$), so $\angle IPD=90^{\circ}$. Besides that, $\angle IEA=\angle IFA=90^{\circ}$, so $E$, $F$ and $P$ lie on the circle with diameter $AI$
Secondly, the points $P$, $G$ and $J$ are collinear. In irder to prove that, just recall that $J\in(AFE)$ (it's well-known: circles $(AFE)$, $(BFG)$ and $(CGE)$ has common point). Hence, $$ \angle (PJ, JF)=\angle (PA, AF)=\angle (CB, BA)=\angle (GB, BF)=\angle (GJ, JF). $$ Therefore, $P$, $G$ and $J$ are collinear and $X$ (the midpoint of $GP$) lies on the radical axis of $(REC)$ and $(QFB)$, as desired.