Prove that the midpoint of $AH$ lies on the radical axis of $(REC)$ and $(QFB)$. 
The incircle of $\triangle ABC$, centered at $I$, touches $AC$ and $AB$ at $E$ and $F$, respectively. Let $H$ be the foot of the altitude from $A$ and let $R = CI \cap AH$ and $Q = BI \cap AH.$ Prove that the midpoint of $AH$ lies on the radical axis of $(REC)$ and $(QFB)$.

The source is Peru TST.
Progress:

Let $X:=$ midpoint of $AH.$ $J,G$ as point of intersection of $(REC)$ and $(QFB).$ Let the perpendicular to $BC$ from $B$ and $C$ intersect $(QFB)$ and $(REC)$ at $K$ and $L.$
Claim: $G$ lies on $BC$
Proof: For this we will define $G':=(QFB)\cap BC.$ And we will show that $G'\in (RFC).$ For this note that since $BQ$ is angle bisector, we get $FQ=QG'.$ ( using incentre-excentre lemma). But $FQ=QD$ by angle bisector property. So $FQ=QG'=QD.$ So $G'$ and $D$ are reflection of each other wrt altitude. Hence $RG'=RD.$ But $RD=RE.$ So $RG'=RE.$ So by converse of incentre -excentre lemma, we get $G=G'$.
Now note that $K-J-L$ are collinear. This is because by cyclicity, $\angle GBK=90=\angle GJK$ and $GCL=90=\angle GJL.$ So $KJL\perp GJ.$
So enough to show $XJ\perp KJL.$ I also thought of introducing $A-$ excentre, because $X-D-I_A$ are collinear( well known).
Any elementary solutions ( not using trig, vectors, complex or any calculative stuff)?

 A: 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.

