It should be line $MK$ instead of $HK$.

I have read several geometric solutions. However, I think the complex number approach is possible for candidates taking the first day TST of Vietnam. Let $AD$ be the altitude. Take circle $(DEF)$ to be the unit circle with $D(d^2),E(e^2), F(f^2)$. We get $$h=-ef-fd-de,a=-ef+fd+de,b=ef-fd+de,c=ef+fd-de.$$Let $U(u)$ and $V(v)$ be two points on the $(DEF)$ such that $UV\perp DH$, we get $$uv=d^2ef.$$Dilate $U$ and $V$ center $H$ ratio $2$ to $P$ and $Q$ (thus $P$ and $Q$ lies on $(ABC)$ and $PQ\parallel BC$), respectively, we get $$p=2u-h=de + df + ef + 2u$$and $$q=2v-h=de + df + ef + \frac{2d^2ef}{u}.$$Circumcenter $S$ of $BPF$ is given by $$s=\frac{\begin{vmatrix} b & b\bar{b} & 1\\ p & p\bar{p} & 1\\ f^2 & 1 & 1\\ \end{vmatrix}}{\begin{vmatrix} b & \bar{b} & 1\\ p & \bar{p} & 1\\ f^2 & \frac{1}{f^2} & 1\\ \end{vmatrix}}=\frac{e^2f^2 + def^2 + de^2f + u(f^2 + 2ef)}{ef + u}.$$Circumcenter $T$ of $CQE$ is given by $$t=\frac{\begin{vmatrix} c & c\bar{c} & 1\\ q & q\bar{q} & 1\\ e^2 & 1 & 1\\ \end{vmatrix}}{\begin{vmatrix} c & \bar{c} & 1\\ q & \bar{q} & 1\\ e^2 & \frac{1}{e^2} & 1\\ \end{vmatrix}}=\frac{d^2e^2 + 2d^2ef + u(de + df + ef)}{d^2 + u}.$$Since tangent at $E$ and $F$ of $(BPF)$ and $(CQE)$ meet at $X$, we have $$XE\perp SE,\,XF\perp SF$$which is equivalent to $$\frac{x-e^2}{s-e^2}+\frac{\bar x-\frac{1}{e^2}}{\bar s-\frac{1}{e^2}}=0,\,\frac{x-f^2}{t-f^2}+\frac{\bar x-\frac{1}{f^2}}{\bar t-\frac{1}{f^2}}=0.$$We get $$x=\frac{de^3f^2 + 2d^2ef^3 + d^2e^2f^2 + d^2e^3f - 2d^3ef^2 - d^3e^2f + u(df^3 + ef^3 - d^2f^2 + def^2 + 2de^2f - 2d^2ef)}{-d^2f^2 + d^3f + def^2 + d^2ef + u(d^2 - de + df + ef)}.$$Midpoint $M$ of $AH$ is also the midpoint of arc $EF$, thus $m=-ef$. Since $K$ is the projection of $H$ on $EF$, $\frac{h-k}{e^2-f^2}=\frac{\bar h-\bar k}{\frac{1}{e^2}-\frac{1}{f^2}}$ or $k=\frac{de^2 + df^2 + ef^2 - d^2e - d^2f + e^2f}{2d}$. From coordinates of $X,M,K$ we can see $$\frac{x-m}{k-m}=\frac{2d^2ef + 2dfu}{-d^2f^2 + d^3f + def^2 + d^2ef + u(d^2 - de + df + ef)}=-\frac{\bar x-\bar m}{\bar k-\bar m}.$$This means $X,M,K$ are collinear. We are done.

Interesting fact about this problem: This problem stands out like a revolution. It is the first time after decades when Viet Nam has only 1 independent geometry problem in the test (Before there were 2 parts in one problem and very complicated). Tổng quát - Generalization: Let $ABCD$ be a cyclic quadrilateral. Suppose $AB$ cut $CD$ at $E$. $P,Q$ lie on $(EBC)$ such that $PQ//BC$. Tangents from $A,D$ to $(ABP),(DCQ)$ intersects at $W$. Prove that: $W$ lie on a fixed line.

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Here, let me present to you the solution of the ULTIMATE geo god. Haozhe Yang. 杨皓哲. Ha-O-R-Zhe. Extend the tangent to $(ECQ)$ at $E$ so that it hits $(BCEF)$ at $X$. Extend the tangent to $(FBP)$ at $F$ so that it hits $(BCEF)$ at $Y$. Let $T = XY \cap EF$, $S = FX \cap EY$, and $Z = FY \cap EX$. The problem asks us to show that $M$, $K$, $Z$ are collinear. It is well known that $ME$ and $MF$ are tangent to $(BCEF)$. Thus, we apply Pascal's theorem on $FFYEEX$, and we find that $M$, $Z$, and $S$ are collinear. Moreover, by brokards, the polar of $T$ in $(BCEF)$ is the line $MZ$, so it suffices to show that $K$ also lies on the polar. In other words, $-1 = (EF; KT)$. Now, we introduce two magical points: $G = CY \cap (ABC)$ and $L = BX \cap (ABC)$. The main claim is that Claim. $P$, $F$, and $G$ are collinear. As well as $Q$, $E$, and $L$. Proof.Redefine $G' = PF \cap CY$, then $\measuredangle PG'C = \measuredangle FG'C = \measuredangle FG'Y = \measuredangle G'FY + \measuredangle FYC = \measuredangle PBA + \measuredangle ABC = \measuredangle PBC$. Thus, $G' \in (ABC)$, meaning $G'=G$ indeed. Define $H_B$ and $H_C$ as the reflection of $H$ over $AC$ and $BC$. Now, we being the messy Ratio Chase will finish: \[\frac{TE}{TF} = \frac{TE}{TX}\cdot \frac{TX}{TF} = \frac{EY}{XF} \cdot \frac{EX}{FY}\]Continuing: \begin{align*} \frac{EY}{FY} &= \frac{\sin\angle YFE}{\sin\angle YEF} = \frac{\sin \angle GCE}{\sin \angle GCF} = \frac{GA}{GH_C} \\ & = \frac{GA/AF}{GH_C/H_CF} \cdot \frac{AF}{H_CF} = \frac{BP/PF}{CP/PF} \cdot \frac{AF}{HF} \\ & = \frac{BP}{CP}\cdot \frac{AF}{HF} \end{align*}Likewise, \[\frac{EX}{FX} = \frac{BQ}{CQ}\cdot\frac{HE}{AE}\]Thus, \[\frac{TE}{TF} = \frac{EY}{XF}\cdot\frac{EX}{FY} = \frac{BP}{CP}\cdot\frac{AF}{HF}\cdot\frac{BQ}{CQ}\cdot\frac{HE}{AE} = \frac{AF}{AE} \cdot \frac{HE}{HF}\]Very lastly, it suffices to show that \[\frac{AF}{AE} \cdot \frac{HE}{HF} = \frac{KE}{KF} \]Indeed, we have $\triangle FKH \sim \triangle AEH$ and $\triangle EKH \sim \triangle AFH$. Therefore, \[\frac{FK}{KE} = \frac{FK}{KH}\cdot \frac{KH}{KE} = \frac{AE}{EH} \cdot \frac{FH}{AF}\]

Did anyone solve this on the contest?

nobodyknowswhoIam wrote: Did anyone solve this on the contest? I redacted my message due to controversial...

Paramizo_Dicrominique wrote: nobodyknowswhoIam wrote: Did anyone solve this on the contest? I think this problem is not hard so sure there are many people can solve. This problem is difficult for candidates in a room with a combinatorial problem and an algebra problem before that. It won't be a problem for the outside solver. I rate it as a P3 of IMO.

Paramizo_Dicrominique wrote: nobodyknowswhoIam wrote: Did anyone solve this on the contest? I think this problem is not hard so sure there are many people can solve. hmm, questionable... Anyways, it is doable with Method of Moving points right? We can fix the ABC and the 2 tangent lines would have degree of one each, which means the intersection would have degree of 2. All we need to prove is this problem is true in three specific positions. The two can be easily proven: $P = B$ and $C$. We will fix the last one such that $P = Q$ on the smaller arc of $BC$. Then it would be fairly easy to write trigonometry bash on those conditions.

CheshireOrb wrote: Let $ABC$ be an acute, non-isosceles triangle with circumcircle $(O)$. $BE, CF$ are the heights of $\triangle ABC$, and $BE, CF$ intersect at $H$. Let $M$ be the midpoint of $AH$, and $K$ be the point on $EF$ such that $HK \perp EF$. A line not going through $A$ and parallel to $BC$ intersects the minor arc $AB$ and $AC$ of $(O)$ at $P$, $Q$, respectively. Show that the tangent line of $(CQE)$ at $E$, the tangent line of $(BPF)$ at $F$, and $MK$ concur. Quite easy, solved it on paper in my English lesson . Let $EF$ cut $(CQE)$ and $(BPF)$ at $X,Y$, respectively. Let $CF$ and $BE$ cut $(CQE)$ and $(BPF)$ at $U,V$, respectively. The tangent line of $(CQE)$ at $E$ and the tangent line of $(BPF)$ at $F$ cuts $MK$ at $L_1,L_2$, respectively. It suffices to prove $L_1\equiv L_2$. Since $\angle XQC=\angle FEC=180^\circ -\angle ABC=\angle AQC$, $A,X,Q$ are collinear. Note that $\angle KEL=\angle AQE$ and $\angle MEL=\angle MEF+\angle KEL=\angle ACF+\angle XCA=\angle XCU$, we have $$\frac{L_1K}{L_1M}=\frac{\sin KEL}{\sin MEL}\cdot\frac{KE}{ME}=\frac{\sin XQE}{\sin XCU}\cdot\frac{KE}{ME}=\frac{XE}{XU}\cdot\frac{KE}{ME},$$so it remains to prove $$\frac{KF}{KE}=\frac{XE}{XU}\cdot\frac{YV}{YF}.$$By Steiner's formula, RHS equals $$\frac{XE}{YF}=\frac{FB\cdot\frac{EY}{EB}}{EC\cdot\frac{FX}{FC}}=\frac{XE}{XF}\cdot\frac{YE}{YF}\cdot\frac{BF}{CE}\cdot\frac{CF}{BE}=\frac{AE^2}{AF^2}\cdot\frac{AC}{AB}\cdot\frac{BF}{CE}=\frac{DB}{DC}=\frac{KF}{KE}.$$

Paramizo_Dicrominique wrote: nobodyknowswhoIam wrote: Did anyone solve this on the contest? I think this problem is not hard so sure there are many people can solve. I disagree with you. Both problems 1 and 2 are easier than the IMO 1 and 2, and I believe this is to give time for this super difficult problem 3. But I'm not a contestant...

HaO-R-Zhe wrote: Paramizo_Dicrominique wrote: nobodyknowswhoIam wrote: Did anyone solve this on the contest? I think this problem is not hard so sure there are many people can solve. I disagree with you. Both problems 1 and 2 are easier than the IMO 1 and 2, and I believe this is to give time for this super difficult problem 3. But I'm not a contestant... teacher buratino sent a clarification up there, so we can ignore this, I guess,...

Yes, this problem can be solved by moving points. I haven't found a synthetic solution yet though. Forget about points $P$ and $Q$ (which need not be restricted as in the problem statement), and instead focus on the centers $O_b$ of $(BPF)$ and $O_c$ of $(CQE)$, which makes it easier to hangle the tangency condition anyways. Let $O$ be the center of $(ABC)$, $M_A$ the midpoint of $BC$, and let $l_b$ and $l_c$ be the perpendicular bisectors of $CE$ and $BF$, respectively, so that $O_b\in l_b$ and $O_c\in l_c$. Now, let $X_b$ be the intersection of the tangent to $(CQE)$ at $E$ and $MK$, and $X_c$ be the intersection of the tangent to $(BPF)$ at $F$ and $MK$. We will show that $X_b=X_c$ Sneak into the projective plane, and animate $O_b=(P(t):Q(t):R(t))\in l_b$ linearly, meaning that $\max(\deg(P), \deg(Q), \deg(R))\leq 1$. We will endeavour to show that both $X_b$ and $X_c$ move linearly on $MK$. Point $X_b\in ML$ is defined by $EX_b\perp EO_b$, so, since $E$ and $MK$ are fixed, it will move linearly on $MK$. Now, the condition that $PQ\parallel BC$ translates to $OO_b$ and $OO_c$ being symmetric about line $OM_A$. Hence $O_c$ moves linearly on $l_c$, and, by the same reasoning as for $X_b$, $X_c$ moves linearly as well. To show that $X_b=X_c$, it now suffices to check only 2 cases, since then their defining polynomials in $\mathbb{PR}^2$ will be the same by the fundamental theorem of algebra. The case $P=C$, $Q=B$ gives $X_b=M=X_c$ by well known orthocenter configuration. he case where $P=Q=W$ at the midpoint of arc $BC$ not containing $A$, requires more care, and could in fact be an entire problem by itself.

nobodyknowswhoIam wrote: Paramizo_Dicrominique wrote: nobodyknowswhoIam wrote: Did anyone solve this on the contest? I think this problem is not hard so sure there are many people can solve. hmm, questionable... Anyways, it is doable with Method of Moving points right? We can fix the ABC and the 2 tangent lines would have degree of one each, which means the intersection would have degree of 2. All we need to prove is this problem is true in three specific positions. The two can be easily proven: $P = B$ and $C$. We will fix the last one such that $P = Q$ on the smaller arc of $BC$. Then it would be fairly easy to write trigonometry bash on those conditions. I think the case $P=B$ cannot give any information. We know that $EF$ would be a common tangent to the cirucmference, but I cannot see how to see that the limit is $K$. I might be missing something though.

Ru83n05 wrote: Yes, this problem can be solved by moving points. I haven't found a synthetic solution yet though. Forget about points $P$ and $Q$ (which need not be restricted as in the problem statement), and instead focus on the centers $O_b$ of $(BPF)$ and $O_c$ of $(CQE)$, which makes it easier to hangle the tangency condition anyways. Let $O$ be the center of $(ABC)$, $M_A$ the midpoint of $BC$, and let $l_b$ and $l_c$ be the perpendicular bisectors of $CE$ and $BF$, respectively, so that $O_b\in l_b$ and $O_c\in l_c$. Now, let $X_b$ be the intersection of the tangent to $(CQE)$ at $E$ and $MK$, and $X_c$ be the intersection of the tangent to $(BPF)$ at $F$ and $MK$. We will show that $X_b=X_c$ Sneak into the projective plane, and animate $O_b=(P(t):Q(t):R(t))\in l_b$ linearly, meaning that $\max(\deg(P), \deg(Q), \deg(R))\leq 1$. We will endeavour to show that both $X_b$ and $X_c$ move linearly on $MK$. Point $X_b\in ML$ is defined by $EX_b\perp EO_b$, so, since $E$ and $MK$ are fixed, it will move linearly on $MK$. Now, the condition that $PQ\parallel BC$ translates to $OO_b$ and $OO_c$ being symmetric about line $OM_A$. Hence $O_c$ moves linearly on $l_c$, and, by the same reasoning as for $X_b$, $X_c$ moves linearly as well. To show that $X_b=X_c$, it now suffices to check only 2 cases, since then their defining polynomials in $\mathbb{PR}^2$ will be the same by the fundamental theorem of algebra. The case $P=C$, $Q=B$ gives $X_b=M=X_c$ by well known orthocenter configuration. he case where $P=Q=W$ at the midpoint of arc $BC$ not containing $A$, requires more care, and could in fact be an entire problem by itself. I see. Well, when $P=Q=W$, it is much easier to write trigonometry. When $P=B$ the line will coincide, therefore we could choose every point on the line, which contains $K$ (it might make no sense, I am new to this method so I might be wrong).

Let $(\ell_1)$ and $(\ell_2)$ be the tangents from $F$ to $(PFB)$ and from $E$ to $(EQC)$, respectively. Let the parallel from $M$ to $FE$ intersect $(\ell_1),(\ell_2)$ at $S,T$ respectively. It suffices to prove that $\dfrac{SM}{MT}=\dfrac{FK}{KE}$. Note that $\angle SFX=\angle FPB$ and $\angle XSF=180^\circ-\angle SXF-\angle SFX=180^\circ-\angle C-\angle FPB=\angle APF$, and so using the Ratio Lemma and the above equalities, $\dfrac{SX}{XF}=\dfrac{\sin \angle SFX}{\sin \angle XSF}=\dfrac{\sin \angle FPB}{\sin \angle APF}=\dfrac{FB}{FA} \cdot \dfrac{AP}{PB},$ and so $SX=XF \cdot \dfrac{FB}{FA} \cdot \dfrac{AP}{PB}$. Similarly, $TE=EY \cdot \dfrac{CE}{EA} \cdot \dfrac{AQ}{QC}$. Thus, we need to prove that $\dfrac{XF \cdot \dfrac{FB}{FA} \cdot \dfrac{AP}{PB}+XM}{EY \cdot \dfrac{CE}{EA} \cdot \dfrac{AQ}{QC}+MY}=\dfrac{FK}{KE}$. Divide both the numerator and the denominator with $BC$, and so we have to prove that $\dfrac{\dfrac{FX}{FA} \cdot \dfrac{FB}{BC} \cdot \dfrac{AP}{PB}+\dfrac{XM}{BC}}{\dfrac{EY}{EA} \cdot \dfrac{EC}{BC} \cdot \dfrac{AQ}{QC}+\dfrac{MY}{BC}}=\dfrac{FK}{KE}, \,\,\, (1)$. Note that since $PQ \parallel BC$, we may set $\angle PAB=\angle QAC=\theta$. In order to compute the ratios that appear in (1), we split the computation in 4 sections: Section 1: In this section, we compute $\dfrac{FX}{FA}=\dfrac{EY}{EA}$. Let $AH$ and $EF$ intersect at point $R$. Note that $\dfrac{FX}{FA}=\dfrac{RM}{RA}=\dfrac{RM/RH}{RA/RH},$ and by the Ratio Lemma, $\dfrac{RM}{RH}=\dfrac{FM}{FH} \cdot \dfrac{\sin \angle MFR}{\sin \angle RFH}=\dfrac{\sin \angle MHF}{\sin \angle FMH} \cdot \dfrac{\sin \angle MFR}{\sin \angle RFH}=\ldots=\dfrac{\cos \angle A}{2\cos \angle B\cos \angle C}$. Moreover, again by the Ratio Lemma, $\dfrac{RA}{RH}=\dfrac{FA}{FH} \cdot \dfrac{\sin \angle AFR}{\sin \angle RFH}=\dfrac{\sin \angle AHF}{\sin \angle FAH} \cdot \dfrac{\sin \angle AFR}{\sin \angle RFH}=\ldots=\tan \angle B \tan \angle C,$ and so in total we obtain $\dfrac{FX}{FA}=\dfrac{EY}{EA}=\dfrac{\cos \angle A}{2\sin \angle B \sin \angle C}$ Section 2: In this section, we compute $\dfrac{XM}{BC}$ and $\dfrac{MY}{BC}$. Let $O$ be the circumcenter of $ABC$ and $N$ the midpoint of $BC$. Note that by a well known Lemma, $AM=ON$. Then, $\dfrac{MX}{BC}=\dfrac{MX}{MA} \cdot \dfrac{MA}{BC}=\dfrac{\cos \angle B}{\sin \angle C} \cdot \dfrac{ON}{2BN}=\dfrac{\cos \angle B}{2\sin \angle C} \cdot \cot \angle A$. Similarly, $\dfrac{MY}{BC}=\dfrac{\cos \angle C}{2\sin \angle B} \cdot \cot \angle A$. Section 3: In this section, we compute $\dfrac{FK}{KE}$. This is an easy application of the Ratio Lemma: $\dfrac{FK}{KE}=\dfrac{FH}{HE} \cdot \dfrac{\sin \angle FHK}{\sin \angle KHE}=\dfrac{\cos \angle B}{\cos \angle C} \cdot \dfrac{\sin \angle C}{\sin \angle B}$. Section 4: In this section, we compute $\dfrac{FB}{BC}$ and $\dfrac{EC}{BC}$. This is trivial; the first one is equal to $\cos \angle B$ and the second one to $\cos \angle C$. Section 5: In this section, we compute $\dfrac{AP}{PB}$ and $\dfrac{AQ}{QC}$. Note that $\dfrac{AP}{PB}=\dfrac{\sin \angle ABP}{\sin \angle PAB}=\dfrac{\sin (\angle C-\angle \theta)}{\sin \angle \theta}$ and similarly $\dfrac{AQ}{QC}=\dfrac{\sin(\angle B-\angle \theta)}{\sin \angle \theta}$. Now, it's time to put these all together. Combining the results of the last 5 sections, it suffices to prove that $\dfrac{\dfrac{\cos \angle A}{2\sin \angle B\sin \angle C} \cdot \cos \angle B \cdot \dfrac{\sin(\angle C-\angle \theta)}{\sin \angle \theta}+\dfrac{\cos \angle B}{2\sin \angle C} \cot \angle A}{\dfrac{\cos \angle A}{2\sin \angle B\sin \angle C} \cdot \cos \angle C \cdot \dfrac{\sin(\angle B-\angle \theta)}{\sin \angle \theta}+\dfrac{\cos \angle C}{2\sin \angle B} \cot \angle A}=\dfrac{\cos \angle B}{\cos \angle C} \cdot \dfrac{\sin \angle C}{\sin \angle B}$ Multiplying both the numerator and the denominator of the LHS of the above... monster with $2\tan \angle A\sin \angle B\sin \angle C\sin \angle \theta,$ it suffices to prove that $\dfrac{\sin \angle A \cos \angle B \sin(\angle C-\angle \theta)+\sin \angle B \cos \angle B \sin \angle \theta}{\sin \angle A\cos \angle C \sin(\angle B-\angle \theta)+\sin \angle C \cos \angle C \sin \angle \theta}=\dfrac{\cos \angle B}{\cos \angle C} \cdot \dfrac{\sin \angle C}{\sin \angle B}$ Note that $\sin(\angle C-\angle \theta)=\sin \angle C\cos \angle \theta-\cos \angle C\sin \angle \theta$ and $\sin(\angle B-\angle \theta)=\sin \angle B\cos \angle \theta-\cos \angle B\sin \angle \theta$, and so it suffices to prove that $\dfrac{x \cos \angle \theta+z\sin \angle \theta}{y\cos \angle \theta+w\sin \angle \theta}=\dfrac{\cos \angle B}{\cos \angle C} \cdot \dfrac{\sin \angle C}{\sin \angle B},$ where $x=\sin \angle A \cos \angle B \sin \angle C, y=\sin \angle A \sin \angle B \cos \angle C$ and $z=\sin \angle A \cos \angle B \cos \angle C-\sin \angle B \cos \angle B, w=\sin \angle A\cos \angle C \cos \angle B-\sin \angle C \cos \angle C$ Since $\dfrac{x}{y}=\dfrac{\cos \angle B}{\cos \angle C} \cdot \dfrac{\sin \angle C}{\sin \angle B},$ it is enough to prove that $\dfrac{\sin \angle A\cos \angle B \cos \angle C-\sin \angle B \cos \angle B}{\sin \angle A \cos \angle C \cos \angle B-\sin \angle C \cos \angle C}=\dfrac{\cos \angle B}{\cos \angle C} \cdot \dfrac{\sin \angle C}{\sin \angle B},$ or equivalently that $\dfrac{\sin \angle A \cos \angle C-\sin \angle B}{\sin \angle A \cos \angle B-\sin \angle C}=\dfrac{\sin \angle C}{\sin \angle B}$. Cross-multiplying, we are left to prove that $\sin^2 \angle B-\sin^2 \angle C=\sin \angle A (\sin \angle B\cos\angle C-\sin \angle C \cos \angle B),$ which is true, since $\sin \angle A (\sin \angle B\cos\angle C-\sin \angle C \cos \angle B)=\sin \angle A\sin(\angle B-\angle C)=$ $=\dfrac{\cos(\angle A-(\angle B-\angle C))-\cos(\angle A+(\angle B-\angle C))}{2}=\dfrac{\cos 2\angle C-\cos 2 \angle B}{2}=\sin^2 \angle B-\sin^2 \angle C,$ and so we are finally done.

I think this is a good problem for VNTST. It has good categorization. The "generalization" to a point lying on a fixed line is not really a generalization because the original problem specified that fixed line (passing through $M$ and $K$). However, if we consider fixed line problem extension, I have seen some other extensions for it.

Solved with emotional support from crazyeyemoody907. Let $\ell_F, \ell_E$ be the radical axes of $(ABC)$ with the points $F$ and $E$, respectively. Then $PB \cap \ell_F = R$ is on the tangent to $(BPF)$ at $F$, and $QC \cap \ell_E = S$ is on the tangent to $(CQE)$ at $E$, so it suffices to show that $FR \cap ES \in MK$. Now let's identify some other points on $\ell_F, \ell_E$: If the tangents to $(BFEC)$ at $F$ and $E$ intersect $BC$ at $X$ and $Y$ respectively, then $X \in \ell_F, Y \in \ell_E$. Also, if the perpendicular bisectors of $BF, CE$ intersect the tangents to $(ABC)$ at $B$ and $C$ at $U, V$ respectively, then $U \in \ell_F, V \in \ell_E$. By angle chasing, $\angle{AFE} = \angle{ACB} \implies U, V \in EF \implies U, F, K, E, V$ are collinear. Also, $X, F, M$ are collinear since $XF \perp$ the line through $F$ and the midpoint of $BC$, and $MF$ is also perpendicular to this line ($M$ is the circumcenter of $\triangle{AEF}$). ---cue start of length bash--- If $R' = FR \cap MK$, then by the Ratio Lemma on $\triangle{FMK}$ and $\triangle{FUX}$, $\frac{MR'}{R'K} = \frac{FM}{FK} \cdot \frac{XR}{UR} \div \frac{XF}{UF}$, so it suffices to show that this is equal to the symmetric expression for $\frac{MS'}{S'K}$ where $S' = ES \cap MK$, which is $\frac{EM}{EK} \cdot \frac{YS}{VS} \div \frac{EY}{EV}$. Also note that $\angle{UBR} = \angle{VCS}$ and $\angle{XBR} = \angle{YCS}$, so by the Ratio Lemma on $\triangle{BXU}$ and $\triangle{CYV}$, $\frac{XR}{UR} \div \frac{YS}{VS} = \frac{BX}{BU} \div \frac{CY}{CV}$. Substituting this in, we get $$\frac{FM}{FK} \cdot \frac{BX}{BU} \cdot \frac{UF}{XF} = \frac{EM}{EK} \cdot \frac{CY}{CV} \cdot \frac{VE}{YE}.$$Since $UB = UF$, $VC = VE$, and $FM = EM$, this simplifies to $\frac{EK}{FK} = \frac{CY}{EY} \cdot \frac{FX}{BX}$. By Law of Sines on $\triangle{CYE}, \triangle{BXF}$, the RHS equals $\frac{\tan \angle{C}}{\tan \angle{B}} = \frac{dist(C, AH)}{dist(B, AH)} = \frac{EK}{FK}$, as desired. $\square$

Just move points