Very nice problem. Let $PN\cap EF=T $,$MK\cap EF=X $,$HX\cap (HEF)=L'$. By Pappus Theorem we have that $A,F,E $ are collinear.By Thales's $\frac {EA}{EF}=\frac {EN}{EM}=\frac {EK}{EB} $ so $\angle EKA=\angle EBF $ similarly $\angle FKA=\angle FCE $ since $AK $ bisects $\angle BKC $ $\angle EKA=\angle FKA $ which means $BCEF $ is cyclic.By some easy angle chasing we get that $PN $ is tangent to $(KEF) $.So $TK^2=TE\cdot TF $.Since $AK $ bisects $\angle FKE $ $A $ lies on K-Apollonius circle of $FKE $.So $TA=TK$ since $T $ lies on perp.bisector of $AH $ we have $TA=TH=TK$ so $TH^2=TE\cdot TF $.So $TH $ is tangent to $(HEF) $.Project the below pencil from $K $ to $EF $ and then project the resulting pencil from $H $ to $(HEF) $ $-1=(B,C;M,\infty)=(E,F;X,T) =(E,F;L',H) $ $\implies $ $TL'$ is tangent to $(HEF) $.So $AHKL'$ is cyclic $\implies$ $L'\equiv L $ and we are done.

By Pappus it is easy to get $E,F,A$ are collinear.And with Thales it is easy to get $\frac{FA}{AE}= \frac{FP}{PM}= \frac{FK}{KC}$ and $\frac{EA}{AF}= \frac{EN}{NM}= \frac{EK}{KB}$.So, $FB,AK,EC$ are paralel.$\angle FBE= \angle FKA= \angle EKA= \angle ECF$and it means that $EFBC$ is cyclic.Let's say that $BC$ and circumcircle of $AHK$ intersect at $Z$ and its circumcenter is $O$. By easy angle chasing $OA=OK=OH$ , so $E,F,A,O$ are collinear.Since $\angle AHB=90$ we get $E,F,A,Z$ are also collinear.$MK$intersect with $EF,BC$ and  circumcircle of $AHKZ$ at $G$,$T$ and $R$,respectively.If we show $E,K,F,R$ is cyclic,then we are done by radical center of $EKFR,ELFH,ALKHR$.And it is equivalent to show $GF \cdot GE=GR\cdot GK$ or $GF \cdot GE=GA \cdot GZ$. If we say $\angle BKT=\angle CKT=\alpha$,$\angle KCB=\beta$ and $\angle TKM=x$,then by above cyclics and collinearities. We can get all angles of triangles $ZKG,FKG,AKG,EKG,BKT$ and $CKT$. So If we use Sine Law first four trinagles for common side $KG$ and $GZ,GF,GA,GE$ and after combine these, we should show that, $$\frac{sin2x}{sin(2\alpha+2\beta)}= \frac{sin(\alpha-x)sin(\alpha+x)}{sin\beta sin(2\alpha+\beta)}= \frac{cos2x-cos2\alpha}{cos2\alpha-cos(2\alpha+2\beta)}$$Or $$cos2\alpha(sin2x+sin(2\alpha+2\beta))=cos2x sin(2\alpha+2\beta)+sin2x cos(2\alpha+2\beta)=sin(2\alpha+2\beta+2x)=2sin(\alpha+\beta+x)cos(\alpha+\beta+x)$$Or $$ 2sin(\alpha+\beta+x)cos(\alpha+\beta+x)=2cos2\alpha sin(\alpha+\beta+x)cos(\alpha+\beta-x)$$Or $$2cos(\alpha+\beta+x)=cos(3\alpha+\beta-x)+cos(\alpha+x-\beta)$$. But if we look sine law in triangles BKT and CKT ,then we get $\frac{sin\beta}{sin(\alpha-x)}=\frac{sin(2\alpha+\beta)}{sin(\alpha+x)}$. Or $sin\beta sin(\alpha+x)=sin(2\alpha+\beta)sin(\alpha-x)$.Or $2cos(\alpha+\beta+x)=cos(3\alpha+\beta-x)+cos(\alpha+x-\beta)$. But it is above equation which we want to show.

Cool problem. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=red; pen sec=orange; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair K,B,C,funny,A,M,NN,P,EE,F,T,H,D,SS; K=dir(125); B=dir(170); C=dir(10); funny=extension(B,C,K,incenter(K,B,C)); A=2K-funny; M=(B+C)/2; NN=(C+A)/2; P=(A+B)/2; EE=extension(B,K,M,NN); F=extension(C,K,M,P); T=extension(NN,P,EE,F); H=foot(A,B,C); D=reflect(T,circumcenter(K,EE,F))*K; SS=extension(EE,F,M,K); filldraw(circumcircle(A,K,H),tfil,tri); draw(T--D--M,tri); filldraw(circumcircle(K,EE,F),sfil,sec); draw(EE--T,sec); draw(A--K,sec+dashed); draw(B--F,sec+dashed); draw(C--EE,sec+dashed); filldraw(circumcircle(B,C,F),fil,pri); filldraw(A--B--C--cycle,fil,pri); draw(F--M--NN--EE,pri); draw(NN--T,pri); dot("$A$",A,dir(80)); dot("$B$",B,SW); dot("$C$",C,SE); dot("$K$",K,NE); dot("$M$",M,S); dot("$N$",NN,dir(15)); dot("$P$",P,SW); dot("$E$",EE,NE); dot("$F$",F,W); dot("$T$",T,W); dot("$H$",H,SE); dot("$D$",D,NW); dot("$S$",SS,dir(75)); [/asy][/asy] First by Pappus' theorem on $BKCNMP$, we have $A\in\overline{EF}$, and by Pappus' theorem on $ABCENK$ and $ACBFPK$, we have $\overline{AP}\parallel\overline{BF}\parallel\overline{CE}$. Since $\overline{AK}$ bisects $\angle BKC$ and $\angle EKF$, we have \[\measuredangle FBE=\measuredangle FBK=\measuredangle AKB=\measuredangle CKA=\measuredangle KCE=\measuredangle FCE,\]whence $BCEF$ is cyclic. Let $T=\overline{EF}\cap\overline{NP}$ and $S=\overline{MK}\cap\overline{EF}$. Note that \[-1=(BC;M\infty_{BC})\stackrel K=(EF;ST)\stackrel M=(NP;KT).\]Since $\overline{BC}$ and $\overline{EF}$ are antiparallel wrt.\ $\angle K$, $\overline{KS}$ is the $K$-symmedian of $\triangle KEF$, so $\overline{KT}$ is tangent to $(KEF)$. But $\measuredangle TAK=\measuredangle FEC=\measuredangle ECB=\measuredangle AKT$, whence $TA=TK$. Since $A$ and $H$ are reflections across $\overline{NP}$, $T$ is the circumcenter of $\triangle AKH$. Let $(AKH)$ and $(KEF)$ intersect again at $D$. Since $(AKH)$ and $(KEF)$ are orthogonal, $\overline{TD}$ is tangent to $(KEF)$, so $T$ is the pole of $\overline{KD}$ and $D\in\overline{MKS}$. The concurrence follows from Radical Axis theorem on $(HEF)$, $(AKH)$, $(KEF)$. @below looks like I don't know how to type P. Substitute P for F and M in the first and third, respectively. I believe the second is correct, however. @2below Sure, you could say ``Pappus on $\overline{BCM}$ and $\overline{KNP}$,'' but note that Pappus' theorem is a special case of Pascal's theorem, and it's more convenient to write it as such. The affine variant of Pappus' theorem is the special case where the intersections lie on the line at infinity.

I thought Pappus theorem only applies if you have $A,B,C$ on one line, and $D,E,F$ on another line. Only then can you show $X=AE \cap BD, Y=BF \cap CE, Z=AF \cap BD$ are collinear. Can someone explain how the Ultimate123 did Pappus' theorem on $BKCNMF$, $ABCENK$, and $ACBFMK$?

@above I don’t know why he expressed it in such a complicated way, but you could also get the same conclusion by FM parallel to AC, AB parallel to EM, which is also named as Pappus theorem (they’re two slightly different ones, one is the more general Pappus hexagon theorem which involves two set of points in a straight line just like what you mentioned)

In my opinion, this problem has two parts: 1) Characterize points $E$ and $F$; and 2) Add in the circles and finish. Part I. We claim that $BFEC$ is an isosceles trapezoid. By Pappus, $E,A,F$ are collinear. Since $\frac{BK}{KE} = \frac {MN}{NE} = \frac{FA}{AE}$, $\frac{CK}{KF} = \frac {MP}{PF} = \frac{EA}{AF}$, lines $KA$, $BF$ and $CE$ are parallel. Hence $\angle BFK = \angle FKA = \angle EKA = \angle KEC$, implying that $BFEC$ is cyclic (and is an isosceles trapezoid). Part II. Let $X=MK\cap EF$, we will show that $X$ lies on the radical axis of $(HKA)$ and $(HEF)$. We first claim that the center of $(HKA)$ lies on $EF$. Suppose $G=BC\cap EF$. Since $BFEC$ is an isosceles trapezoid, we have $\angle GHA = \angle GKA=90^{\circ}$, so $GA$ is a diameter of $(HKA)$, which proves the claim. Observe that we only need to prove $XA\cdot XG = XE\cdot XF \Longleftrightarrow \frac{XA}{XE} = \frac{XF}{XG}.$ Let $J=MX\cap CE$, then this is equivalent to $\frac{XK}{XJ} = \frac{XF}{XG}\Longleftrightarrow FC\parallel GJ$. Finally, since $M$ is the midpoint of $BC$, we have $\frac{JC}{JE} = \frac{BK}{KE} = \frac{BF}{CE} = \frac{GF}{GE}$, which finishes the problem.

Nice projective practice. Let $K'=AK\cap BC$, let $T'$ be the intersection of the perpendicular bisector of $AK'$ with $BC$, and let $T$ be the midpoint of $AT'$. Note that we have $T'A=T'K'$, so by homothety, we have $TA=TK$. Since $H$ is the reflection of $A$ in $TK$, we have that $TA=TK=TH$, so $T$ is the circumcenter of $(AKH)$. By Pappus on $CMB$ and $PKN$, we have $A,E,F$ collinear. Call this line $\ell$. We claim that $\ell$ passes through $T'$ (and thus $T$ as well). Indeed, since $KK'$ and $KT'$ are the angle bisectors of $\angle BKC$, we have $(BC;K'T')=-1$. Projecting through $K$ onto $\ell$, we have \[(E,F;A,KT'\cap\ell)=-1.\]Let $U$ be the midpoint of $NP$. Projecting $(NP,U\infty)=-1$ through $M$ onto $\ell$, we get \[(EF;A,BC\cap\ell)=-1.\]Thus, $KT'\cap\ell=BC\cap\ell$, so $T'\in\ell$, as desired. Note that $AT'$ is a diameter of $(AKH)$, and that $(AT';EF)=-1$, so $(AKH)$ and $(HEF)$ are orthogonal. Thus, $LH\cap\ell$ is the polar of $T$ with respect to $(HEF)$ intersected with $\ell$, so \[(T,LH\cap\ell;EF)=-1.\]Projecting $(BC;M\infty)=-1$ from $K$ onto $\ell$ gives \[(EF;MK\cap\ell,T)=-1,\]so $MK\cap\ell=LH\cap\ell$, as desired. This completes the proof. Remark: [Constructing the Diagram] The key is the construction of $T'$ and $T$. To construct the diagram, construct an arbitrary isosceles triangle $AT'K'$, and let $B$ and $C$ be harmonic conjugates in $T'K'$. The rest of the construction is straightforward.

CLAIM 1. $FBCE$ is an isoceles trapezoid. Proof. Let $FE$ and $PN$ meet at $X$, $AK$ meet $BC$ at $D$. then $$\frac{FK}{KC}=\frac{FP}{PM}=\frac{FP}{AN}=\frac{ZP}{ZN}=\frac{AP}{EN}=\frac{NM}{EM}=\frac{BK}{KE}$$Hence $FB\|EC$. Now by Pappus's theorem, $F,A,E$ are collinear. Let the line through $K$ parallel to $FB$ meet $FE$ and $BC$ at $A'$ and $D'$. We show that $A=A'$. Suppose on the contrary that $A\neq A'$, then since $AK=DK$ and $A'K=D'K$($K$ lies on the median of $\triangle ZEC$). We have $\triangle AKA'\cong\triangle DKD'$. Hence $\angle AA'D=\angle A'D'D$ which implies $FE\|BC$. Hence $FP=PM$. Therefore $$\frac{BD}{BC}=\frac{PK}{MC}=\frac{FP}{FM}=\frac{1}{2}$$Since $AK$ bisects $\angle BKC$ we have $BK=BC$. Hence $FBCE$ is a rectangle. Since $AD\perp BC$ we have $AD\|FB$. Contradiction. Now we $AD\| FB$. Hence $\angle FBK=\angle BKD=\frac{1}{2}(\angle BFK+\angle FBK)$ Hence $\angle KFB=\angle FBK$ so $FBCE$ is an isoceles trapezoid. Now Let $MK$ meet $EF$ at $X$. $KX$ meet $(HEF)$ again at $Y$. CLAIM 2. $H,E,F,L$ are concyclic. Proof. Since $\triangle BKC$ and $\triangle FKE$ are oppositely similar, $KX$ is the $K$-symmedian in $\triangle KFE$. Hence $KF\cdot YE=KE\cdot FY$. Hence $YA$ is also the angle bisector of $\angle FYE$. Since $KA=KD$ and $\angle AHD=90^{\circ}$ we have $AK=HK=KD$. Hence $$\angle KHA=90^{\circ}-\angle KHD=90^{\circ}-\angle FAD=90^{\circ}-\frac{1}{2}\angle FKE-\angle KEF=\angle FYA-\angle FYK=\angle KYA$$as desired. Now $XK\times XY= XF\times XE$. Therefore $X$ lies on the radical axis of $(HEF)$ and $(HAK)$. Hence $H,X,L$ are collinear as desired.

Solution from Twitch Solves ISL: Claim: Points $A$, $E$, $F$ are collinear. Proof. Pappus theorem on $BMC$ and $NKP$. $\blacksquare$ [asy][asy] size(12cm); pair E = dir(60); pair F = dir(155); pair C = conj(E); pair B = conj(F); pair Z = extension(F, E, B, C); pair K = extension(E, B, F, C); pair A = extension(E, F, K, incenter(K, E, F)); pair T = extension(A, K, B, C); pair M = midpoint(B--C); pair G = extension(E, F, M, K); pair N = midpoint(A--C); pair P = midpoint(A--B); draw(B--E--M--F--C, deepgreen); filldraw(A--B--C--cycle, invisible, red); draw(P--N, red); filldraw(circumcircle(E, F, B), invisible, deepgreen); draw(E--Z, blue); draw(Z--B, red); pair H = foot(A, B, C); filldraw(circumcircle(A, K, H), invisible, blue); draw(A--H, red); draw(A--T, orange); draw(B--F, orange); draw(C--E, orange); draw(M--G, deepcyan); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$C$", C, dir(C)); dot("$B$", B, dir(B)); dot("$Z$", Z, dir(Z)); dot("$K$", K, dir(K)); dot("$A$", A, dir(A)); dot("$T$", T, dir(T)); dot("$M$", M, dir(M)); dot("$G$", G, dir(G)); dot("$N$", N, dir(N)); dot("$P$", P, dir(P)); dot("$H$", H, dir(H)); /* TSQ Source: !size(12cm); E = dir 60 F = dir 155 C = conj(E) B = conj(F) Z = extension F E B C K = extension E B F C A = extension E F K incenter K E F T = extension A K B C M = midpoint B--C G = extension E F M K N = midpoint A--C P = midpoint A--B B--E--M--F--C deepgreen A--B--C--cycle 0.1 lightred / red P--N red circumcircle E F B 0.1 lightgreen / deepgreen E--Z blue Z--B red H = foot A B C circumcircle A K H 0.1 lightcyan / blue A--H red A--T orange B--F orange C--E orange M--G deepcyan */ [/asy][/asy] Let $T = \overline{AK} \cap \overline{BC}$. Claim: Lines $BF$, $CE$, $AK$ are pairwise parallel. Proof. We get $BF$ and $CE$ are parallel by Desargue theorem on $\triangle BPF$ and $\triangle ENC$. However, Desargue theorem on $\triangle BAC$ and $\triangle FTE$ gives $BF$, $AT$, $CE$ concurrent. $\blacksquare$ Claim: The quadrilateral $BFEC$ is an isosceles trapezoid. Proof. Since $BK/KC = BT/TC = FA/AE = FK/KE$, we get $BFEC$ is cyclic. $\blacksquare$ We let $Z = \overline{EF} \cap \overline{BC}$. Claim: $ZAKH$ is cyclic, with diameter $\overline{AZ}$. Proof. Because $\angle ZHA = 90^{\circ} = \angle ZKA$. $\blacksquare$ Finally, let $G = \overline{KM} \cap \overline{EF}$. Claim: The point $G$ satisfies \[ GE \cdot GF = GA \cdot GZ. \]Proof. Since $\overline{KM}$ is $K$-median of $\triangle KBC$, it follows $\overline{KG}$ is $K$-symmedian of $\triangle KEF$. However, the circle with diameter $\overline{AZ}$ is the $K$-Apollonian circle of $\triangle KEF$, so it passes through the point $Q$ (not shown) such that $EQFK$ is a harmonic quadrilateral. Since $G$ lies on $KQ$, the conclusion follows $\blacksquare$ Hence the point $G$ is the concurrency point described in the problem, as $HL$ is the radical axis of $(AKH)$ and $(HEF)$.

Solved with Iliyas Noman and Justin Hua The problem can be thought of as two parts : 1. Understand $E,F$ 2. Spot the Appolonian circle and finish First note that $A\in EF$ by Pappus on $\overline {BMC}$ and $\overline {PKN}$. Claim : $ BF \parallel AK \parallel CE$ Proof : We have : $$ \frac {EA}{FA} = \frac {EN}{MN} = \frac {EK}{KB} \implies BF \parallel AK$$ Similarly $AK \parallel CE$. Claim : $BFEC$ is a isosceles trapezoid Proof : $\angle EBF= \angle EBK = \angle BKM = \angle MKC =\angle KCE $ so $BFEC$ is cyclic. Let $MK$ hit $EF$ at $R$. We claim that $R$ is the desired concurrency point. Let $S=EF \cap BC$. Note that $SK$ externally bisects $\angle BKC$, so $\angle (SK,AK) = 90$ and obviously $\angle AHS=90$, so $S$ is $A$-antipode in $(AKH)$. Claim : $R$ has equal powers wrt $(AKH)$ and $(HEF)$. Proof : We need to prove that $RA \cdot RS = RE \cdot RF$. Note that $KA$ and $KS$ are angle bisectors of $\angle EKF$, so $(KASH)$ is the $K$ appolonius circle of $(KEF)$. Furthermore, $KM$ is a medium in $\angle BKC$, so $KR$ is a symmedian in $\triangle EKF$. So if $G= KR \cap (KEF)$, then $G\in (KASH)$. Hence we have : $$ RE \cdot RF = RK \cdot RG = RA \cdot RS$$ Conclude by radical axes on $(BFEC)$, $(AKH)$ and $(HEF)$. $\square$

Responding to the above, I actually didn't actually do anything. Aryan-23 and Iliyas did all the work. However, I tried following along and learned a lot from them. I wrote up a solution so that I could better understand the ideas to the problem. This solution is probably exactly the same as Aryan's but I just wanted to share. By Pappus' theorem on $P,K,N$ and $C,M,B$, we get that $F-A-E$ are collinear. Next, notive that $\frac{EA}{AF}=\frac{EN}{NM}=\frac{EK}{KB}$, so $FB \parallel AK$ Similarly,$\frac{FA}{AE}=\frac{FP}{PM}=\frac{FK}{KC}$, so $AK \parallel EC$. Thus, we get that $FB \parallel AK \parallel EC$. Next, we claim that BFEC is a isoceles trapezoid. Let $AK \cap BC=D$. This follows since $\angle FBK=\angle BKD=\angle BKC=\angle DKC=\angle KCE$, as desired. This gives us a way to construct the diagram. We draw a cyclic isoceles trapezoid with diagonals intersecting at $K$.We draw in the midpoint of $BC$, $M$ and then draw $MF$ and $ME$. Next, using the fact that $PN$ is parallel to $BC$ by midlines, we can easily find points $P,N$. Lastly, we find $A$ through $BP \cap CN$. We use the radical axis theorem on $(BFEC),(AKH), (EFH)$. We get that $HL,EF$ and the radical axis of $(BFEC), (AKH)$ are concurrent. Let $KM \cap EF=R$. Since $EF$ and $BC$ are antiparallel, $KR$ is a symmedian in $\triangle FKE$. Let $BC \cap EF=S$. Then $\angle AHS=90^{\circ}$ and since $SK$ bisects $\angle BKC$ externally, $SK \perp AK$, since the internal angle bisector is perpendicular to the external bisector. Thus, $SA$ is the diameter. Now, let $KR \cap (KEF)=G$. WE claim that $G$ lies on $(AKH)$. This follows due to the fact that $G$ is well known to be the second intersection of the circumcircle and A-appolonius circle. Now, notice that the A-appolonius circle of $(KEF)$ is $(AKS)=(AKH)$, as desired. Now, using power of a point, we see that: $$RF \cdot RE=RG \cdot RK=RA \cdot RS $$ So, R lies on the radical axis of $(BFEC),(AKH)$, as desired.

Firstly, obseve that from Pappus' Theorem on $NKP,BMC$, we have that $A,E,F$ are collinear. Now, we prove the following claim: Claim: Let $Q$ be the circumcenter of $AKH \implies Q$ lies on $PN$. Proof: Let $R$ be the intersection of the line through $K$ perpendicular to $AK$ with $BC$. Since $AK$ bisects $\angle BKC \implies (KA,KR;KB,KC)=-1$, and $\angle AKR= 90º= \angle AHR$, so $R \in (AKH)$, and $AR$ is diameter, hence $Q$ is the midpoint of $AR$, and since $R \in BC \implies Q \in PN$, as desired. $\square$ Now, observe that $-1=(B,C;M,\infty_{BC}) \stackrel{K}= (E,F;\{MK \cap EF\}, \{PN \cap EF\}) \stackrel{M}= (N,P;K,\{PN \cap EF\}) (\star)$, but since $(KA,KR;KB,KC)=-1=(\{KA \cap BC\},R;B,C) \implies$ projecting from $A$ onto $PN$, we have that $$(K,Q;P,N)=-1$$$\implies$ from $(\star)$, $Q= PN \cap EF$. $\implies$ let $S=MK \cap EF \implies$ since $-1=(\{KA \cap BC\},R;B,C) \stackrel{K}=(A,R;E,F) \implies$ the circle with diameter $AR$ is the Apollonius' circle WRT $EF,A \implies$ since $(AKH)$ has diameter $AR$, $$\frac{HE}{HF}=\frac{AE}{AF}=\frac{LE}{LF}$$$\implies LEHF$ is a harmonic quadrialateral. Then, since $(A,R;E,F)=-1 \implies$ the midpoint of $AR$, which is $Q$, satisfies $QA^2=QE.QF \implies QH^2=QE.QF=QL^2 \implies Q= \Pi_{(HEF)}(HL) \implies HL \cap EF$ satisfies $(\{HL \cap EF\},Q;E,F)=-1=(S,Q;E,F)$ (from $(\star)$), so $HL \cap EF= S \implies MK,EF,HL$ are concurrrent, as desired. $\blacksquare$

Let $\overline{AK}$ intersect $\overline{BC}$ at $D$. By radical axes, it suffices to show that $MK$ is the radical axis of $(EFK)$ and $(AKH)$. Now by the angle bisector theorem, we have \[\frac{FK}{KC}=\frac{PK}{KN}=\frac{BD}{DC}=\frac{BK}{KC}\]so $FK=BK$. Similarly, $EK=CK$. Therefore, $\triangle KEF\cong\triangle KCB$, and since $K$ is the midpoint of $AD$ it follows that $A$ lies on $\overline{EF}$. Now let $\overline{EF}$ intersect $\overline{BC}$ at $T$. Then by symmetry, $LK\perp AD$, and so $\measuredangle KTH=90^\circ-\measuredangle TDK=\measuredangle KAH$, and so $T$ lies on $(AKH)$. Let $\overline{ME}$ and $\overline{MF}$ each cut $(EFK)$ at $Q$ and $R$. Claim 1: $M,Q,R,H$ are concyclic. Proof. It is easy to see that $M,H,P,N$ are concyclic. Since $PN\parallel BC$, $\measuredangle FEK=\measuredangle KCB=\measuredangle FKP$, $(EFK)$ is tangent to $\overline{PN}$ at $K$. Therefore, it follows that $\triangle PKR\sim\triangle PFK\sim\triangle NCK$, and similarly, $\triangle NKQ\sim\triangle KPB$. Consequently, \[PR\cdot CN=PK\cdot KN=QN\cdot PB \Longleftrightarrow \frac{PR}{PB}=\frac{NQ}{NC}\Longleftrightarrow \frac{PR}{RH}=\frac{QN}{NH}\]Thus $\triangle RPH\sim\triangle QNH$, and so $\measuredangle HRM=\measuredangle HQN$ as desired. $\blacksquare$ Now by Miquel's theorem in $\triangle MET$, it follows that $T,F,R,H$ are concyclic. Therefore, \[\text{Pow}(M,(EFK))=MR\cdot MF=MH\cdot MT=\text{Pow}(M,(AKH))\]and we're done. [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen betterdash = linetype(new real[] {5,5}); pair K = dir(120); pair B = dir(170); pair C = dir(10); pair D = extension(K,incenter(K,B,C),B,C); pair A = 2*K-D; pair H = foot(A,B,C); pair P = midpoint(A--B); pair N = midpoint(A--C); pair M = midpoint(B--C); pair E = extension(B,K,M,N); pair F = extension(C,K,M,P); pair T = extension(E,F,B,C); pair R = 2*foot(circumcenter(E,F,K),F,M)-F; pair Q = 2*foot(circumcenter(E,F,K),E,M)-E; pair S = 2*foot(circumcenter(E,F,K),K,M)-K; draw(K--B--C--cycle); draw(A--B--C--cycle, black+1); draw(M--F); draw(M--E); draw(C--F); draw(B--E); draw(T--B); draw(T--E); draw(A--H); draw(A--D); draw(P--N); draw(M--S, betterdash); draw(circumcircle(E,F,K)); draw(circumcircle(A,K,H)); draw(circumcircle(M,Q,R), betterdash); dot("$A$", A, dir(75)); dot("$B$", B, dir(135)); dot("$C$", C, dir(0)); dot("$H$", H, dir(270)); dot("$D$", D, dir(270)); dot("$M$", M, dir(270)); dot("$E$", E, dir(0)); dot("$F$", F, dir(180)); dot("$T$", T, dir(180)); dot("$P$", P, dir(190)); dot("$N$", N, dir(20)); dot("$R$", R, dir(220)); dot("$Q$", Q, dir(0)); dot("$K$", K, dir(20)); [/asy][/asy]

[asy][asy] size(9cm); defaultpen(fontsize(10pt)); pair K = dir(110), B = dir(170), C = dir(10), A = extension(2K-B,2K-C,K,incenter(K,B,C)), H = foot(A,B,C), M = (B+C)/2, N = (A+C)/2, P = (A+B)/2, E = extension(B,K,M,N), F = extension(M,P,C,K), Q = extension(E,F,N,P), R = extension(E,F,M,K), L = R+dir(H--R)*abs(E-R)*abs(F-R)/abs(H-R); draw(A--B--C--A, heavyblue); draw(M--E--F--M^^N--Q--F^^C--F^^B--E^^Q--H, heavycyan); draw(circumcircle(E,F,H), purple); draw(arc(circumcenter(A,K,H),circumradius(A,K,H),-30,45), purple); draw(H--L, red); draw(M--R, red+dotted); draw(F--H--P^^E--H--N, heavygreen); dot("$A$", A, dir(75)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$M$", M, dir(270)); dot("$N$", N, dir(0)); dot("$P$", P, dir(210)); dot("$K$", K, dir(285)); dot("$H$", H, dir(300)); dot("$L$", L, dir(75)); dot("$Q$", Q, dir(180)); dot("$E$", E, dir(30)); dot("$F$", F, dir(135)); dot("$R$", R, dir(150)); [/asy][/asy] First, note that $E$, $A$, $F$ are collinear by Pappus. Let $R = \overline{HL} \cap \overline{EF}$ and $Q = \overline{NP} \cap \overline{EF}$, so that we want $R$, $K$, $M$ collinear. Claim: $(HKAL)$ is the $H$-Apollonian circle in $\triangle HEF$. Proof. We are given that $\overline{KA}$ bisects $\angle EKF$, so it suffices to show that $\overline{HA}$ bisects $\angle EHF$. From $\overline{MN} \parallel \overline{AB}$ and $\overline{MP} \parallel \overline{AC}$, we deduce $\triangle EAN \sim \triangle AFP$, whence $$\frac{PH}{PF} = \frac{PA}{PF} = \frac{NE}{NA} = \frac{NE}{NH}.$$Furthermore $$\measuredangle FPH = \measuredangle MPH = \measuredangle MNH = \measuredangle ENH,$$so $\triangle FPH \overset{-}\sim \triangle HNE$ by SAS. Thus $$\measuredangle EHC = \measuredangle EHN + \measuredangle NHC = \measuredangle MFH + \measuredangle BMF = \measuredangle BHF$$as desired. $\square$ It follows that $\overline{HL}$ is a symmedian in $\triangle HEF$. Also $\overline{NP}$ is the perpendicular bisector of $\overline{HA}$, so $\overline{HQ}$ is tangent to $(HEF)$ from $$\measuredangle QHF = \measuredangle QHA - \measuredangle FHA = \measuredangle HAQ - \measuredangle AHE = \measuredangle HEF.$$Therefore $$-1 = (EF;HL) \overset{H}= (EF;QR) \overset{K}= (B,C;\infty_{BC},\overline{KR} \cap \overline{BC})$$which implies the conclusion. Remark: It can be reverse engineered by projecting through $K$ that we just want $(HKA)$ to be the Apollonian circle, after which it's not too hard.

An amazing configuration! P.S. This is my 200th post on HSO! Iran TST 2019, Day 1, P3 wrote: In triangle $ABC$, $M,N$ and $P$ are midpoints of sides $BC,CA$ and $AB$. Point $K$ lies on segment $NP$ so that $AK$ bisects $\angle BKC$. Lines $MN,BK$ intersects at $E$ and lines $MP,CK$ intersects at $F$. Suppose that $H$ be the foot of perpendicular line from $A$ to $BC$ and $L$ the second intersection of circumcircle of triangles $AKH, HEF$. Prove that $MK,EF$ and $HL$ are concurrent. Solution.(Solved with Psyduck909) First, by Pappus on $BMC$ and $NKP$, we obtain that $F,A,E$ are collinear. Define the points $U=AK \cap BC,V=EF \cap BC$. We now have the following important claim: Claim.$AK \parallel BF \parallel CE.$ In particular, each of $FAUB,AECU,FECB$ is an isoceles trapazoid. Proof. We have $$ \frac{EA}{AF}=\frac{EN}{NM}=\frac{EK}{KB}\implies AK \parallel BF,$$and similarly $CE \parallel AK$. Now simple angle chasing gives $$ \measuredangle FBE=\measuredangle FBK=\measuredangle UBK=\measuredangle CKU=\measuredangle FCE,$$which implies that $BFEC$ is cyclic. The other two follow now due to parallel lines.$\square$ Note that $\overline{PN}$ is the perpendicular bisector of $\overline{AH}$. Since $K$ is the midpoint of $\overline{AU}$ we have $\overline{VK} \perp \overline{AU}.$ In particular, this implies that $\odot (AKVH)$ is the $K$-apollonius circle in $\triangle FKE$. Call this circle $\Gamma$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -36.18, xmax = 11.545, ymin = -2.01, ymax = 20.34; /* image dimensions */ pen ffqqff = rgb(1.,0.,1.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen xfqqff = rgb(0.4980392156862745,0.,1.); pen zzttqq = rgb(0.6,0.2,0.); pen qqffff = rgb(0.,1.,1.); /* draw figures */ draw((-10.,6.)--(-2.,6.), linewidth(0.5) + ffqqff); draw((-12.,2.)--(-7.27,6.), linewidth(0.5) + red); draw((-7.27,6.)--(4.,2.), linewidth(0.5) + red); draw((-8.,10.)--(-7.27,6.), linewidth(0.5) + qqwuqq); draw((1.86080586080586,13.721611721611719)--(-13.108159392789375,8.072106261859583), linewidth(0.5) + ffqqff); draw((-13.108159392789375,8.072106261859583)--(-4.,2.), linewidth(0.5) + blue); draw((1.86080586080586,13.721611721611719)--(-4.,2.), linewidth(0.5) + blue); draw((1.86080586080586,13.721611721611719)--(-7.27,6.), linewidth(0.5) + red); draw((-13.108159392789375,8.072106261859583)--(-7.27,6.), linewidth(0.5) + red); draw((-13.108159392789375,8.072106261859583)--(-12.,2.), linewidth(0.5) + qqwuqq); draw((1.86080586080586,13.721611721611719)--(4.,2.), linewidth(0.5) + qqwuqq); draw((-8.,10.)--(-8.,2.), linewidth(0.5) + uququq); draw((-7.27,6.)--(-6.54,2.), linewidth(0.5) + qqwuqq); draw((-13.108159392789375,8.072106261859583)--(-29.196850393700803,2.), linewidth(0.5) + ffqqff); draw((-29.196850393700803,2.)--(-12.,2.), linewidth(0.5) + ffqqff); draw((-10.,6.)--(-18.598425196850403,6.), linewidth(0.5) + ffqqff); draw((-4.,2.)--(-9.941099989212452,9.267400598425018), linewidth(0.5) + blue); draw(circle((-18.593904109589037,6.), 11.323904109589035), linewidth(0.5) + ffxfqq); draw(circle((-5.132936314718871,9.596590000764575), 8.119620299938545), linewidth(0.5) + ffxfqq); draw((-11.406851531016088,14.750833075541744)--(-8.,2.), linewidth(0.5) + xfqqff); draw((-8.,10.)--(-12.,2.), linewidth(0.5) + zzttqq); draw((-12.,2.)--(4.,2.), linewidth(0.5) + ffqqff); draw((-8.,10.)--(4.,2.), linewidth(0.5) + zzttqq); draw(circle((-7.270291337378625,15.259739332457546), 9.259739337040692), linewidth(0.5) + qqffff); draw((-16.34563344349581,17.09847211534778)--(-9.941099989212452,9.267400598425018), linewidth(0.5)); /* dots and labels */ dot((-8.,10.),linewidth(4.pt) + dotstyle); label("$A$", (-8.2,10.19), NE * labelscalefactor); dot((-12.,2.),linewidth(4.pt) + dotstyle); label("$B$", (-12.43,1), NE * labelscalefactor); dot((4.,2.),linewidth(4.pt) + dotstyle); label("$C$", (4.12,0.9), NE * labelscalefactor); dot((-4.,2.),linewidth(4.pt) + dotstyle); label("$M$", (-4.105,1), NE * labelscalefactor); dot((-2.,6.),linewidth(4.pt) + dotstyle); label("$N$", (-1.555,5.99), NE * labelscalefactor); dot((-10.,6.),linewidth(4.pt) + dotstyle); label("$P$", (-11,5.29), NE * labelscalefactor); dot((-7.27,6.),linewidth(4.pt) + dotstyle); label("$K$", (-7.3,6.365), NE * labelscalefactor); dot((1.86080586080586,13.721611721611719),linewidth(4.pt) + dotstyle); label("$E$", (2.07,13.79), NE * labelscalefactor); dot((-13.108159392789375,8.072106261859583),linewidth(4.pt) + dotstyle); label("$F$", (-13.955,8.015), NE * labelscalefactor); dot((-8.,2.),linewidth(4.pt) + dotstyle); label("$H$", (-8.2,0.99), NE * labelscalefactor); dot((-6.54,2.),linewidth(4.pt) + dotstyle); label("$U$", (-6.655,1), NE * labelscalefactor); dot((-18.598425196850403,6.),linewidth(4.pt) + dotstyle); label("$G$", (-18.9,6.19), NE * labelscalefactor); dot((-9.941099989212452,9.267400598425018),linewidth(4.pt) + dotstyle); label("$X$", (-9.95,9.5), NE * labelscalefactor); dot((-29.196850393700803,2.),linewidth(4.pt) + dotstyle); label("$V$", (-30.1,1.09), NE * labelscalefactor); dot((-11.406851531016088,14.750833075541744),linewidth(4.pt) + dotstyle); label("$L$", (-11.9,15.2), NE * labelscalefactor); dot((-16.34563344349581,17.09847211534778),linewidth(4.pt) + dotstyle); label("$T$", (-17.03,17.3), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We are now ready to finish the problem. Claim.$KM,LH,EF$ concur. Proof. Let $EF \cap KM=X,$ and note that $KX$ is the $K$-symmedian in $\triangle FKE$ due to anti-parallel lines $\overline{EF}$ and $\overline{BC}$. In particular, the point $T=KX \cap \odot (FKE)$ lies on $\Gamma$ since $\frac{KF}{KE}=\frac{TF}{TE}.$ Finally, by radical axis on $\odot (FEKT),\odot (TKHLAV),\odot (HELF),$ we have that $\overline{EF},\overline{LH},\overline{TK} \equiv \overline{KM}$ concur.$\square$$\blacksquare$

A “harmonic” problem Pappus theorem gives that $A,E,F$ are collinear. Let $MK\cap EF=J,BC\cap EF=T$ Note that $MA$ bisects $NP$, $MT\parallel NP$, then $(MT,MK;MN,MP)=-1\Longrightarrow K(T,A;E,F)=-1$ Since $KA$ bisects $\angle EKF$, then $KA\perp KT \Longrightarrow T\in\odot(AKH)$ So $\triangle KBC\cong\triangle KEF$ So $KJ$ is the $K$-symmedian of $\triangle KEF$ So $JE\cdot JF=JA\cdot JT\Longrightarrow J\in HL$

Nice problem! Sketch since I'm tired. First spam Pappus to get $A \in EF$, $AK \parallel BF \parallel CE$ giving $BCFE$ cyclic, Let $T = EF \cap NP$, $S = MK \cap EF$, use some harmonic chase to get $T$ is the center of $(ADH)$ then use orthogonality of $(ADH)$ and $(DKE)$ with radax to finish.

Iran 2019 Here is a sketch of the solution. Firstly, by pappus theorem we have that $AEF$ are colinear. Secondly,we have that $AK \parallel BF \parallel CE$ by Thales theorem, and some angle chase. Finally we show that $EF$ intersect $NP$ (say at $O$) is the Center of circle $(AHD)$ hence by radicle axis we are done.