It seems to be proposed by Nairi Sedrakyan, very similar to the problems proposed for Mathematical Reflections

$MN$ is parallel to $BC$.

My solution in contest: Let $X$ and $Y$ be intersection point of $BD, CA$ and $DF, AE$ respectively. Let $CF\cap BE=Z$, $BD\cap AF=U$, $CD\cap AB=S$, $AF\cap DE=T$. We know that $S,Z$ and $T$ are collinear in homotety in triangles $\bigtriangleup SCD$ and $\bigtriangleup TFE$. From Pappus theorem In lines $(S,D,V)$ and $(U,A,T)$ we obtain $UD\cap SA=B$, $DT\cap VA=E$, $ST\cap UV=Z^{'}$ are collinear, but $ST$ cuts $BE$ at $Z$. So $Z=Z^{'}$ and $Z\in UV$. From Pappus theorem in lines $(V,D,C)$ and $(U,A,F)$ we have $X,Y$ and $Z$ are collinear. So $\bigtriangleup CBX$ and $\bigtriangleup EFY$ are perspective. From Desargue's theorem: $CX\cap YF=N$, $BX\cap AY=M$ and $CB\cap AY=\infty$ are collinear or $MN||EF$. Then conclusion follows from in triangle altitudes are concurrent.

Official solution: LEMMA: Let $T$ be the common point of extended legs $PS$ and $QR$ of trapezium $PQRS$. Then the radical axis of two circles with diameters $PR$ and $QS$ is the altitude of the triangle $TPQ$ drawn from $T$. Proof: We consider three circles:$\omega _1$ with diameter $PR$, $\omega _2$ with diameter $QS$, and $\omega$ with diameter $PQ$. The common chord of $\omega$ and $\omega _1$ is the altitude drawn from $P$ to $QR$, and the common chord of $\omega$ and $\omega _2$ is the altitude drawn from $Q$ to $PR$. The common point of these altitudes, that is, the orthocenter of $TPQ$, has equal degrees with respect to $\omega _1$ and $\omega _2$, and therefore belongs to their radical axis. Similarly, their radical axis contains the orthocentre of $TRS$. Since the perpendicular drawn from $T$ to $PQ$ contains both the orthocentres, it is the radical axis of $\omega _1$ and $\omega _2$. Applying the lemma to the trapezium $ABDE$, we find that the perpendicular drawn from $M$ to $AB$ is the radical axis of circles with diameters $AD$ and $BE$. Consider the trapezia $CDFA$ and $EFBC$ in the same way, we prove that the perpendiculars drawn from $N$ to $CD$ is the radical axis of the circles with diameters $AD$ and $CF$, and the perpendicular drawn from $K$ to $EF$ is the radical axis of the circles with diameters $CF$ and $BE$.

Juraev wrote: My solution in contest: Let $X$ and $Y$ be intersection point of $BD, CA$ and $DF, AE$ respectively. Let $CF\bigcap BE=Z$, $BD\bigcap AF=U$, $CD\bigcap AB=S$, $AF\bigcap DE=T$. We know that $S,Z$ and $T$ are collinear in homotety in triangles $\bigtriangleup SCD$ and $\bigtriangleup TFE$. From Pappus theorem In lines $(S,D,V)$ and $(U,A,T)$ we obtain $UD\bigcap SA=B$, $DT\bigcap VA=E$, $ST\bigcap UV=Z^{'}$ are collinear, but $ST$ cuts $BE$ at Z. So $Z=Z^{'}$ and $Z\in UV$. From Pappus theorem in lines $(V,D,C)$ and $(U,A,F)$ we have $X,Y$ and $Z$ are collinear. So $\bigtriangleup CBX$ and $\bigtriangleup EFY$ are perspective. From Desargue's theorem: $CX\bigcap YF=N$, $BX\bigcap AY=M$ and $CB\bigcap AY=\infty$ are collinear or $MN||EF$. Then conclusion follows from in triangle altitudes are concurrent. Very nice solution Juraev!

First lets turn this hexagon into pentagon and prove for it: Lets move the $DEF$ fragment across $\overrightarrow{DC}$ we will get $ABCEF$ pentagon : (it is trivial to guess that the problem should work for this case as well) $M = BC \cap AE$ $N = C$ $K = CE \cap BF$ In this case it is easy to notice that perpendiculars from $M$ and $K$ coincide at $H(MCK)$ therefore $MK$ should be parallel to $AF$ : from the similarities written below $ \frac{MP}{PE} = \frac{BP}{PF} $ and $\frac{AP}{PE} = \frac{BP}{PK}$ so $\frac{MP}{AP} = \frac{KP}{FP}$ so $AF \parallel MK$ therefore problem is solved for pentagon and those three lines concur at $H(MCK)$

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Lets return to the initial problem, from pentagon case we "guessed" that the concurence point should be $H(MNK)$ therefore we should prove that $AF \parallel MK$ ; $MN \parallel BC$ ; $NK \parallel CE$ : Lets construct the pentagon again: $ABCE_1 F_1$ where $E_1 F_1$ is the translation of $EF$ across $\overrightarrow{DC}$ Let $X = BC \cap AE_1$ and $Y = CE_1 \cap BF_1$ From simple similarities : $\frac{MB}{MD} = \frac {AB}{DE} \: $ and $\frac{XB}{XC} = \frac{AB}{CE_1} \:$ as $CE_1 = DE$ we see that : $\frac{MB}{BD} = \frac{XB}{BC} $ so $MX \parallel CD \parallel XY(from$ $pentagon$ $case) $ therefore $M \in XY $ ; analogously, $N \in XY$ so line $MN$ and line $XY$ are the same, thus ${MN \parallel AF}$ ! similarly $MN \parallel BC$ and $NK \parallel CE$ so those 3 lines will concur at $H(MNK)$ Q.E.D. P.S. I couldn't swap the attachments, it should be vice-versa.

Attachments:

what a wonderful solution, zaurimeshveliani. I already know 3 solutions of this problem, and in my opinion yours is the best

Proof: Let $AB$ and $CD$, $CD$ and $EF$, $EF$ and $AB$ intersect at $X, Y, Z$, respectively. Then by $\dfrac{YE} {ZE} \cdot \dfrac{YF} {ZF} \cdot \dfrac{ZA} {XA} \cdot \dfrac{ZB} {XB} \cdot \dfrac{XC} {YC} \cdot \dfrac{XD} {YD} = 1$ it follows that $A, B, C, D, E, F$ lie on a conic (Carnot's Theorem). Now, applying Pascal's Theorem on $FDBCAE$ we get $MN || BC || YZ$. Hence, the perpendiculars drawn from $M, N, K$ to the sides $AB, CD, EF$ (respectively) are concurrent at orthocenter of the triangle $MNK$.

Today I learned of the following synthetic solution of this problem: For any two points $X,Y$, let $\omega_{XY}$ denote the circle which has $XY$ as its diameter. It's enough to proof that $\ell_M$, the perpendicular from $M$ to $AB$, is the radical axis of $\omega_{AD}$ and $\omega_{BE}$. (For we can, then, consider radical axises of three circles $\omega_{AD},\omega_{BE}$ and $\omega_{CF}$.) Let $H_1$ and $H_2$ be orthocenter of triangle $MED$ and $MAB$ respectively. Let $EH_1\cap MD=E_1,DH_1\cap ME=D_1$ and $AH_2\cap MB=A_1,BH_2\cap MA=B_2$. It's clear that $E_1,B_1\in \omega_{BE}$ and $D_1,A_1\in \omega_{AD}$. Since $H_1E\times H_1E_1=H_1D\times H_1D_1$, we get that $H_1$ lie on the radical axis of $\omega_{AD}$ and $\omega_{BE}$. Similarly, $H_2$ also lie on the radical axis of $\omega_{AD}$ and $\omega_{BE}$. Hence, the radical must be $H_1H_2\equiv \ell_M$, done.

Simplest solution. By converse of Pascal's theorem on $ABCDEF$, we get $A,B,C,D,E,F$ lie on some conic $\mathcal{C}$. By Pascal's theorem on $EACBDF$ we get $MN\parallel EF$. Analogusly, we get $NK\parallel AB, MK\parallel CD$ so the concurrency point is orthocenter if $\Delta MNK$.

The configuration of this problem consists only of parallel and perpendicular lines (and no circles etc.), thus it makes sense that it could be reasonably nicely coordinate-bashed. What is more, if we notice from an adequate diagram that it is expected that $NK \parallel AB \parallel DE$ and the analogous $MN \parallel EF \parallel BC$ and $MK \parallel CD \parallel AF$ and that having proven this, the conclusion follows immediately (the concurrency point is the orthocenter of $MNK$), life will be even nicer! It remains to prove $MN \parallel BC$ via coordinates. Let $A(0,0), B(1,0), D(d,n), E(e,n)$ (having in mind $AB \parallel DE$); now we derive $c_2$ in $C(p, c_2)$ and $f_2$ in $F(q,f_2)$. From $BC \parallel EF$ we have $\frac{c_2}{p-1} = \frac{n-f_2}{e-q}$, and from $CD \parallel AF$ we have $\frac{n-c_2}{d-p} = \frac{f_2}{q}$. Hence $c_2 = \frac{n(p-1)(p+q-d)}{d(q-e)+ep-q}$ and $f_2 = \frac{nq(e+1-p-q)}{d(e-q) -ep + q}$. The computation of $M(m_1, m_2)$ is via $m_2 = \frac{n}{e}m_1$ and $m_2 = \frac{n}{d-1}(m_1-1)$, i.e. $M\left(\frac{e}{e-d+1}, \frac{n}{e-d+1}\right)$. The computation of $N(n_1,n_2)$ is via $n_2 = \frac{c_2}{p}n_1$ and $\frac{n_2 - n}{f_2-n} = \frac{n_1-d}{q-d}$, i.e. $N\left(\frac{p(nq-df_2)}{c_2(q-d)+p(n-f_2)}, \frac{c_2(nq-df_2)}{c_2(q-d)+p(n-f_2)}\right)$, which after substituing $c_2$ and $f_2$ becomes $N\left(\frac{pq}{p+q-d},\frac{nq(p-1)}{dq-de+ep-q}\right)$. It remains to calculate the slope of $BC$, namely $\frac{c_2}{p-1} = \frac{n(p+q-d)}{dq-de+ep-q}$, and that of $MN$, namely $$ \frac{\frac{nq(p-1)}{d(q-e)+ep-q} - \frac{m}{e-d+1}}{\frac{pq}{p+q-d} - \frac{e}{e-d+1}} = \frac{n(p+q-d)}{dq-de+ep-q} $$and as they are equal, the problem is solved, wohoo!

Radical axis

Let $\omega_1,\omega_2,\omega_3$ be circles with diameters $AD,BE,CF$, respectively. $P,Q$ be intersection points of $\omega_1$ and $\omega_2. H_1$ and $H_2$ be orthocenters in triangles $\triangle MDE$, $\triangle MBA$,respectively Claim: $H_1,H_2 \in PQ$ Let $DX,EY$ be altitudes in $\triangle MDE.$ Then easily note that $X \in \omega_1 , Y \in \omega_2$ and $DYEX$ is cyclic. Thus $H_1$ is radical center of $\omega_1,\omega_2$ and $(DYEX) \implies H_1 \in PQ$.Similarly $H_2 \in PQ \ \ \ \square$ Since $M,H_1,H_2$ are collinear $\implies$ perpendicular from $M$ to $AB$ is radical axis of $\omega_1$ and $\omega_2$. Analagously perpendiculars from $N,K$ to $CD,EF$ are radical axises of $\omega_1$ and $\omega_3, \omega_2 $ and $\omega_3,$ respectively. So perpendiculars drawn from $M, N, K$ to lines $AB, CD, EF$ respectively intersect at radical center of $\omega_1,\omega_2,\omega_3$ so we are done

Apply converse of Pascal to get that $ABCDEF$ is a coconic hexagon. Now apply Pascal on $\bullet$ $AFBDCE$ $\bullet$ $FEACBD$ $\bullet$ $EDFBAC$ And so we get that the concurrency point is basically the orthocenter of $\triangle MNK$.

Projective geo kills! It is well known (and easy to prove via Pascal's) that points $A$ , $B$ , $C$ , $D$ , $E$ and $F$ are coconic. We simply need to prove the following key claim. Claim : Lines $AF$ , $CD$ and $MK$ (and similarly for the other sides) are pairwise parallel. Proof : Note that applying Pascal's Theorem on hexagon $CEAFBD$, we have that $K=CE \cap FB$ , $M=EA \cap BD$ and $\infty = AF \cap DC$ are collinear. Thus, $\overline{KM}$ is parallel to $\overline{AF}$ and $\overline{CD}$ as desired. Now, note that the perpendicular from $N$ to $\overline{CD}$ is simply the perpendicular from $N$ to $\overline{MK}$. Thus, the desired perpendiculars are simply the altitudes of $\triangle KMN$, which are well known to concur at the orthocenter of $\triangle KMN$.