Nice problem! Let $H_A$ denote the humpty point, let $G \in (AEF)$ such that $GH_A \perp BC$, It is well known $G,H_A,K$ are collinear. It suffices to show $KGQF$ concyclic, $$\measuredangle FQA = \measuredangle FAH_A = \measuredangle FGH_A$$, as desired.

Why is Iran Orthocenter geometry really good lol. Anyway let $H_M$ the A-humpty point, let $H$ orthocenter of $\triangle ABC$, let $J$ a point in $(AH)$ such that $AHH_MJ$ is an Isosceles Trapezoid. It's well known that $H_M$ is symetic with the interesction of the A-symedian with $(ABC)$ w.r.t. $BC$ (just use ratios to prove this), and as a result $K,J,H_M$ are colinear. Now to finish by 2 Reim's we get $J$ lies in both circles $(JFK), (PEK)$ thus we are done .

Nice problem Let J be the second instrection of circumcircles of PKE and QKF , Let Y be the instrection point of KJ and AM and KJ insrects circumcircle of ABC in point S , let O and H be the circumcenter and orthocenter of circumcirle of ABC let H' be the reflection of H respect to M , clearly H' lies on circumcircle of ABC and in the end suppose that AY insrect circumcircle of ABC in X for the second time Now we want to prove that , why SX and BC are parallel, (because KSX=90) first of all by some easy angle chasing you can see that points A.F.H.Y.J.E are concylic now we prove that MX=MY or we can prove that XYHH' is parallelogram , for this it's enough to show that 2 angle YHM and MH'X are equal first of all clearly H'HBC is parallelogram so 2 angles CHM and MH'B are equal so its enough to show that BH'X=YHC or BAX=BAY=YHC and its enough to show that AFHY is cyclic ( that is clear) now we proof that MX=MY so we can say that M is the circumcenter of XYS , and its well-known that OM is perpendicular to SX also its clear that OM is perpendicular to BC so BC and SX are parallel and the proof is complete.

Let $M$ be the midpoint of $AA'$, and let $X$ be the foot of $H$ onto $AM$. Notice that $(A'BCHX)$ is cyclic with diameter $HA'$. The translation that sends $H$ to $A$ also sends $(BCH)$ to $(ABC)$ and line $HX$ to line $AK$ so it must also send $X$ to $K$. Thus $AHKX$ is a parallelogram. Let $KX$ meet $(AEFHX)$ again at $Y$. Then $$\measuredangle (EY,KY)=\measuredangle (EY,XY)=\measuredangle (EH,HX)=\measuredangle (EP,KP)$$so $Y$ lies on $(EKP)$ and similarly also lies on $(FKQ)$. This means $KX$ is the radical axis of the two circles, finishing the problem.

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