Let $AH$ intersects $BC$ at $T$ and $H'$ be the reflection of $H$ over $M$. First we show that $\angle{HME}=90^{\circ}$. Let $\angle{HME}=\alpha$ and $\angle{HMB}=k$. By law of sines, we have $\frac{MH}{\sin (180^{\circ}-\alpha -C)}=\frac{MC}{\sin (\alpha +k-C)}$. On the other hand, we have $$\frac{MH}{MC}=\frac{MH}{MB}=\frac{\sin (90^{\circ}-C)}{\sin (90^{\circ} +k-C)}\implies \frac{\sin (90^{\circ}-C)}{\sin (90^{\circ} +k-C)} =\frac{\sin (180^{\circ}-\alpha -C)}{\sin (\alpha +k-C)}.$$Not hard to show that the last equation yields $\alpha = 90^{\circ}$. So, $\angle{HME}=90^{\circ}$. Similarly, we get $\angle{HMF}=90^{\circ}$. So $E,M,F$ collinear. Moreover, easy angle chasing gives us $HBTC\sim HEMF$. Also, we have $\frac{HA}{HD}=\frac{HT}{HM}=\frac{HB}{HE}\implies HBTCA\sim HEMFD$. Note that this means $H$ is the orthocenter of triangle $DEF$ with $M$ being the feet of altitude $DM$. So, we get $D,E,H',F$ cyclic. Also, we have $\angle{EH'F}=\angle{EHF}=\angle{BHC}=180^{\circ}-\angle{FAE}$. So, $E,H',F,A$ cyclic. This gives $D,A,E,H',F$ cyclic. From PoP, we get $HD\cdot HH'=2\cdot HD\cdot HM=2\cdot HN\cdot HK=HA\cdot HK$. The above gives $A,D,K,H'$ cyclic. Hence, all of $E,F,K$ lie on $(ADH')$. The conclusion follows immediately.

ThE-dArK-lOrD wrote: First we show that $\angle{HME}=90^{\circ}$.

Nice problem . Here's my solution : WLOG let $\angle B > \angle C$. So, $E \in$ side $AC$ and $F \in$ ray $AB$ beyond $B$. It is well known that $EM , FM$ are perpendicular to $HM$ (for a proof, see post #3) Now, let parallel at $A$ to $BC$ intersect $(AEF)$ at $D'$ Claim 1 : $D = D'$ PROOF: $\angle D'EF = \angle D'AF = \angle B = 90 - \angle HFE$. So, $FH$ is perpendicular to $DE$ and similarly we get $EH$ is perpendicular to $DF$. Therefore, $H$ is the orthocenter of $D'EF$. So, $D' , H , M$ are colinear. Hence, $D' = D$ Claim 2: $HR = RK$ where $R = EF \cap AK$ PROOF: Since $DAMR$ and $DNMK$ are cyclic, from power of point $H$ with respect to both circles we get, $AH.HR = DH.HM = NH.KH \implies KH = 2HR$ $\implies$ $HR = RK$. Now, let $H'$ be the reflection of $H$ about $M$. Notice that $\angle KH'D = \angle RMH = 90 = \angle KAD$. So, $DAKH'$ is cyclic. Also, since $H$ is the orthocenter of $DEF, H'$ lies on $(DFAE)$. So,$ D, F, K, H', E, A $are conclyclic.

Beautiful Problem Though it might have been tough to do it on the test Quote: Consider triangle $ABC$ with orthocenter $H$. Let points $M$ and $N$ be the midpoints of segments $BC$ and $AH$. Point $D$ lies on line $MH$ so that $AD\parallel BC$ and point $K$ lies on line $AH$ so that $DNMK$ is cyclic. Points $E$ and $F$ lie on lines $AC$ and $AB$ such that $\angle EHM=\angle C$ and $\angle FHM=\angle B$. Prove that points $D,E,F$ and $K$ lie on a circle. Solution:Let $A'$ be the $A-$antipode in $\odot (ABC)$ $\implies$ $A'$ is the reflection of $H$ over $M$. Let the perpendicular bisector of $HA'$ intersect $AC$ at $X$, then, $MXCA'$ is cyclic $$\implies \angle MA'X=\angle ACB=\angle MHX=\angle MHE \implies \boxed{X \equiv E}$$Hence, $ME \perp HM$ and Similar argument shows $MF \perp HM$ $\implies$ $F$ $-$ $M$ $-$ $E$. Now reflect $H$ over $H_A$ to $H'$ where $H_A$ is the foot of perpendicular from $A$ to $BC$, then, $\implies$ $H'$ $\in$ $\odot (ABC)$ and $AD||H'A'$ $\implies$ $\Delta H'HA'$ $\sim $ $\Delta AHD$, But since, $N,H_A$ are the midpoints of $AH,$ $HH'$ $\implies$ $\Delta DNH $ $\sim$ $\Delta A'H_AH$ $\implies$ $DN$ $||$ $A'H_A$ $$\angle NKM=\angle MDN=\angle MA'H_A \implies KH_AMA' \text{ cyclic } \implies \angle MA'K=90^{\circ} $$Hence, $A' \in \odot (DAK)$ $\implies$ $\angle FHE$ $=$ $\angle FA'E$ $=$ $180^{\circ}-A$ $\implies$ $AFA'E$ is cyclic. Let $AK$ $\cap$ $EF$ $=$ $P$, then, By Midpoint Theorem $\implies$ $HP=PK$. Also it's quite evident that, $\odot (FAH)$, $\odot (AHE)$ are tangent to $EF$ $\implies$ $P$ is also the midpoint of $EF$ $\implies$ $FHEK$ is a parallelogram $\implies$ $FKA'E$ is isosceles trapezium $\implies$ $K$ $\in$ $\odot (AFA'E)$, but, $D$ $\in$ $\odot (A'AK)$ $\implies $ $DFKE$ is cyclic

Quite similar to the previous post. Let $A'$ be the $A-$antipode and show that $EF$ is indeed the perpendicular bisector of $HA'$. Then from here simple angle chasing finishes the proof: $\angle MA'E = \angle ACB = \angle CAD $ yields $E \in \odot DAA'$. Similarly, $\angle FA'M = \angle FBM = 180 - \angle FAD$ yields $F \in \odot DAA'$. Finally, $MN \parallel AA'$ yields $K \in \odot DAA'$.

I don't read the previous posts so probably is the same like others. We take $A'$ to be the antipode of $A$ in $ \odot ABC$.Now take $E'$=$\odot DAA’ \cap AC$.Now $DA'E'$=$C$ so $MA'CE'$ is cyclic so $E'MA'=ACA'=90$.But $M$ is the midpoint of $HA'$ so $HE'A'$ is isosceles so $E'HM=C$ so $E'=E$ so $E \in \odot DAA'$.Analog $F \in \odot DAA'$ and $K \in \odot DAA'$ because $DKA=DMN=DA'A$($MN\parallel AA'$).So $K,D,E,F$ lie on a circle.

Some other facts and ideas in this figure: $MN$ is the simson line of $D$ WRT $AEF$. Actually at the first step it was the problem that was proposed and then became this. So with this fact we can find out $D$ lies on circumcircle of $AEF$ and if we consider points $P$ and $Q$ lies on $AB$ and $AC$ so that $P,Q$ and $H$ are collinear and $PQ\perp HM$, it's well known that $HP=HQ$ so $AH$ bisects $EF$. Now if we let $K'$ be the reflection of $H$ about midpoint of $EF$, $K'$ lies on circumcircle of $AEF$ and by PoP we can find out $DNMK'$ is cyclic so $K'\equiv K$ and $A,E,F,D,K$ lie on a circle.

Let $A'$ be reflection of $A$ through circumcenter of $\triangle ABC$ We define again points $E$, $F$ is intersections of perpendicular bisector of $HA'$ with $AC$, $AB$ It's easy to see that: $C$, $E$, $M$, $A'$ lie on a circle; $B$, $F$, $M$, $A'$ lie on a circle So: $\angle{EHM} = \angle{EA'M} = \angle{ACB}$ and $\angle{FHM} = \angle{FA'M} = 180^o - \angle{FBM} = \angle{ABC}$ Then: $\angle{CAD} = 180^o - \angle{ACB} = 180^o - \angle{EA'M}$ or $E$, $A$, $D$, $A'$ lie on a circle Similarly: $F$, $D$, $A$, $A'$ lie on a circle So: $E$, $A$, $D$, $F$, $A'$ lie on a circle But: $\angle{KDA'} = \angle{KNM} = \angle{KAA'}$ then: $K$, $D$, $A$, $A'$ lie on a circle Hence: $E$, $A$, $D$, $F$, $K$, $A'$ lie on a circle or $K$, $F$, $D$, $E$ lie on a circle

Let the antipode of $A$ be $P$. Since $MN\parallel AP$, we have $DAPK$ concyclic. Denote the circle $(DAPK)$ be $\omega$. We will show that $E$ also lies on $\omega$, hence completing the proof (the other half follows by symmetry). Let $BH$ intersect $AC$ at $G$, $AH$ intersect $BC$ at $F$ and intersects $(ABC)$ at $I$. Since $\angle EHM=\angle ACB=\angle EGM$, points $E,G,M,H$ are concyclic. Thus $EM\perp HM$. This means $\triangle HBF \sim \triangle HEM \Longrightarrow \triangle HBI \sim \triangle HEP$, so $\angle EPD=\angle HIB=\angle ACB=180^{\circ}\angle DAC$, which means that $D,A,E,P$ are concyclic. Thus $E$ lies on $\omega$.

let $A'$ be the reflection of $H$ in $M$ and define $E'=(DAK) \cap AC$ we'll prove that $\angle E'HC=\angle ACB$ since $2.HN.HK=2.HM.HD \implies HA.HK=HD.HA'$ so $DAA'K$ is cyclic $\angle HA'E' = 180 -\angle DAC=\angle ACB \implies E'CA'M$ is cyclic so $\angle E'MA'=\angle E'CA'=90 \implies E'H=E'A' \implies \angle E'HM=\angle EA'M=\angle ACB$ a similar argument for $F$ and we win

Solved with Rg230403 Let $X,Y$ be the feet of altitudes from $B,C$ onto $AC, AB$. Since $\angle FYM = \angle BYM = \angle ABC = \angle FHM$, we see that $LHMF$ is cyclic and so $\angle HMF = 90^\circ$, similarly, $\angle HME = 90^\circ$ and so $F,M,E$ are collinear. Also see that $\triangle HFE \sim \triangle HCB$ Redefine $D$ as the orthocenter of $\triangle HFE$. Since $\angle FDE = 180 - \angle FHE = 180 - (\angle ABC + \angle ACB) = \angle BAC$, $DAEF$ is cyclic. Since $\angle ADM = \angle ADE + \angle MDE = \angle AFE + \angle MFH = 180 - (\angle YHM - \angle HCB) = 180 - \angle HMC = \angle HMB$ and so $AD || BC$ and so $D$ is indeed the point in the problem statement. Now, let $J = AH \cap EF$.Then, $\angle JHF = \angle FHM - \angle JHM = \angle CHG - \angle JHM = \angle CHM$ and so due to the similarity of the triangles, we see that $J$ must be the midpoint of $EF$. Redefine $K$ as the point such that $FHEK$ is a parallelogram, since we showed that midpoint of $EF$ lies on $AH$ ,we see that $K$ lies on $AH$. And since $\angle FKE = \angle FHE = 180 - \angle FDE$, $K \in (FDAE)$ and so we are done. $\blacksquare$

Very nice problem. Without loss of generality, let the diagram be in the figure. Let $H'$ be the reflection of $H$ across $M$ and let $BP$, $CJ$, $AG$ be the altitudes. It is well known that $H' \in (ABC)$ and $AH'$ is a diameter. We have $\angle MPE = \angle MPC = \angle C = \angle MHE \implies HPEM$ is cyclic. Similarly, $HJFM$ is cyclic. Thus, $90 = \angle HME = \angle HMF \implies E,M,F$ are collinear. Note also that $\angle H'ME = 90 = \angle H'CE \implies H'CEM$ is cyclic and similarly, $H'MBF$ is cyclic. So, $\angle DAE = 180 - \angle C$ and $\angle DH'E = \angle MH'E = \angle C \implies DAEH'$ is cyclic. And similarly, $DAH'F$ is cyclic. Thus it suffices to show $K \in (DAEH'F)$. Now note that since $NM$ is a mid-segment in $\triangle HAH'$, we have $\angle AH'D = \angle NMD = \angle NKD = \angle AKD \implies K \in (DAEH'F)$, and thus, we are done. $\blacksquare$

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Let $H'$ be the reflection of $H$ over $M$, by the very famous lemma, we have that $H' \in (ABC)$, let $G$,$T$ and $P$ be the feet from $A,B$ and $C$, respectively. Since we have that $DH.HM=NH.HK$, if we multiply by $2$, we have that: $$2.DH.HM=2.NH.HK$$$$DH.HH'=AH.HK$$this implies that $DAH'K$ is cyclic. Now notice that $\angle MTE = \angle MHE$, i.e. we must have that $HTEM$ is cyclic, similarly we have that $MHPF$ is cyclic. Both imply that $\angle HME = 90 = \angle HMF$, implying that $F,M$ and $E$ are collinear points. This implies: $$\angle H'FM = \angle H'BM = \angle H'BC = \angle H'AC = 90 - \angle B \implies \angle DH'F = \angle FH'M = \angle B = \angle DAF$$implying that $F \in (DAH')$. Similarly we get that $\angle DAE + \angle DH'E = 180$, thus we have that $DFKE$ is cyclic.

Dadgarnia wrote: Consider triangle $ABC$ with orthocenter $H$. Let points $M$ and $N$ be the midpoints of segments $BC$ and $AH$. Point $D$ lies on line $MH$ so that $AD\parallel BC$ and point $K$ lies on line $AH$ so that $DNMK$ is cyclic. Points $E$ and $F$ lie on lines $AC$ and $AB$ such that $\angle EHM=\angle C$ and $\angle FHM=\angle B$. Prove that points $D,E,F$ and $K$ lie on a circle. Proposed by Alireza Dadgarnia just asking, which IMO level question it will be?

@above maybe a G2 Let $P$ be the $A-$ antipode w.r.t $(ABC)$ and $Q= AH \cap BC$.Then we know that $H-M-P$ and also that $HBPC$ is a parallelogram. Thus $PM=HM$.Now , it is given that $(DNMK)$.Thus $$\angle DKA =\angle DMN= \angle DPA \implies (DKPA)\implies \angle KPM =\angle KAD =\angle KQM \implies (KQMP) $$Now let $E'$ be the intersection of the perpendicular to $PH$ at $M$ and $AC$.Then $\angle E'MP= \frac{\pi}{2} \implies E'H=E'P$ . Also then $ \frac{\pi}{2} = \angle E'MP =\angle E'CP \implies (E'MPC)$. Thus $$\angle E'CP =\angle E'PC =\angle MCE' =\angle C \implies E' \equiv E$$.So we have got the said point $E$.Then $$\angle PEC =\angle PMC =\angle QKP =\angle AKP \implies (AKPE )\implies (ADKPE)$$Similarly $(ADFKP)$.Combining , we have that $(DFKE)$ , as desired

Let $H'$ be reflection of $H$ across $M$. Claim1 : $\angle HME = \angle 90 = \angle HMF$. Proof : Let $BH$ meet $AC$ at $P$. we have $\angle MHE = \angle C = \angle MPE$ so $HMEP$ is cyclic so $\angle HME = 90$. we prove the other one with same approach. Claim2 : $MECH'$ and $MBFH'$ are cyclic. Proof : $\angle MH'F = \angle MHF = \angle B \implies MBFH'$ is cyclic. $\angle MH'E = \angle MHE = \angle C \implies MECH'$ is cyclic. Claim3 : $ADFH'$ , $ADH'E$ and $ADKH'$ are cyclic. Proof : $\angle DAF = \angle DAB = \angle B = \angle MH'F = \angle DH'F \implies ADFH'$ is cyclic. $\angle DAE = \angle DAC = \angle 180 - \angle C = \angle 180 - \angle DH'E \implies ADH'E$ is cyclic. $DNMK$ is cyclic $\implies DH.HM = KH.HN \implies DH.HH' = KH.HA \implies ADKH'$ is cyclic. Now we have $ADFKH'E$ is cyclic. we're Done.

Anshu-Droid's solve if you know what I mean lol Iran TST 2019 P4 wrote: Consider triangle $ABC$ with orthocenter $H$. Let points $M$ and $N$ be the midpoints of segments $BC$ and $AH$. Point $D$ lies on line $MH$ so that $AD\parallel BC$ and point $K$ lies on line $AH$ so that $DNMK$ is cyclic. Points $E$ and $F$ lie on lines $AC$ and $AB$ such that $\angle EHM=\angle C$ and $\angle FHM=\angle B$. Prove that points $D,E,F$ and $K$ lie on a circle. Let $H'$ be the reflection of $H$ over $M$, its well known that $H'$ is the $A-$antipode of $\odot ABC$. Let the perpendicular bisector of $HH'$ intersect $AC, AB$ at $E', F'$ respectively. So $\measuredangle EMH' = \measuredangle EMH = 90^{\circ} = \measuredangle ACH' = \measuredangle ECH' \implies EMH'C \text{ is cyclic} \implies \angle ACB = \angle ECM = \angle EH'M = \angle EHM \implies E' = E$. Similarly $F' = F$. Now, $$\measuredangle AEH' = \measuredangle CEH' = \measuredangle CMH' = \measuredangle ADH' \implies AEH'D \text{ is cyclic.}$$Similarly, $AFH'D$ is cyclic $\implies AEFH'D$ is cyclic. Now since $DNMK$ is cyclic, so, $$HA \cdot HK = 2 HN \cdot HK = 2 \cdot Pow_{\odot DNMK}(H) = 2 HM \cdot HD = HH' \cdot HD \implies DAH'K \text{ is cyclic.}$$ So, $AEFH'DK$ is cyclic $\implies DEFK$ is cyclic and we are done .

Nice geo Let $H'$ be the reflection of $H$ about $M$ and let the feet of altitudes from $B$ and $C$ in $\triangle ABC$ be $X$ and $Y$ respectively. It is easy to see that $$\angle MXC = \angle BCA = \angle MHE \implies E \in \odot(HXM) \implies EM \perp MH \text{ and similarly } FM \perp MH \implies E-H-F$$Also by PoP $$HD \cdot HH' = 2 \cdot HD \cdot HM = 2 \cdot HN \cdot HK = HA \cdot HK \implies \odot(DAH'K)$$Now by trivial angle chasing and the fact that $H'FHE$ is a kite, one arrives to the conclusion that $$\angle H'AE = \angle MYH = \angle MFH = \angle EFH = \angle H'FE \implies E \in \odot(DAKH') \text{ and similarly } F \in \odot(DAKH') \implies \odot(DEFK) \blacksquare$$

BVKRB- wrote: $\angle H'AE = \angle MYH = \angle MFH = \angle EFH = \angle H'FE \implies E \in \odot(DAKH')$ @above: This is wrong. So your finishing is wrong. It would be like this- Since $H'\in (ADK)$ and $FHEH'$ is kite therefore \[ \angle DAF=\angle DAB=\angle ABC=\angle FHM=\angle MH'F=\angle DH'F\implies F\in (ADKH'). \]Similarly $E\in (ADKH')$. So all the points $A,D,F,K,H',E$ are cyclic.

Beautiful configuration Solved with help from geogebra My solution is same as @aboves