See attachment for my solution. Harmonic Division is the vein of this problem.

Nice! Here's the solution I found. Let $K$ be the midpoint of $BC$, and let $A_1$ be the reflection of $A$ over $K$. Note that $F$ is the reflection of $D$ over $OK$, so we find that $DFA_1A$ is an isosceles trapezoid. Then, \[ \angle MED = \angle TFD = \angle MA_1D. \] Therefore, $MDA_1E$ is cyclic. But then by Power of a Point, we see that \[ AD \cdot AE = AM \cdot AA_1 = AN \cdot AK \] so $DKEN$ is cyclic, as desired. [asy][asy] /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(10cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair B = (-4.0, 2.0); pair O = (-2.5, 1.0); path w = CirclebyPoint(O,B); pair D = (-3.192795835973656, 2.664341890855831); pair C = (-0.9999999999999998, 2.0); pair K = midpoint(B--C); pair E = (-3.9853897408833356, -0.021575899127684872); path DE = D--E; path BC = B--C; pair F = 2*foot(D,relpoint(Line(K,O,lisf),0.5-10/lisf),relpoint(Line(K,O,lisf),0.5+10/lisf))-D; pair A = IntersectionPoint(DE,BC); pair T = IntersectionPoint(w,Line(A,F,lisf),1); pair M = IntersectionPoint(Line(B,C,lisf),Line(E,T,lisf)); pair N = (2)*(M)-A; pair A_1 = (2)*(K)-A; /* Draw objects */ draw(w, rgb(0.0,0.0,0.8) + linewidth(1.0)); draw(DE); draw(D--F); draw(T--F); draw(C--N, rgb(0.0,0.6,0.0) + linetype("4 4")); draw(E--M); draw(circumcircle(D,E,N), rgb(1.0,0.0,0.0) + linewidth(0.6)); draw(F--A_1); draw(circumcircle(D,M,E), rgb(0.0,0.8,0.0) + linewidth(1.0) + dashed); draw(D--A_1); draw((abs(dot(unit(D-E),unit(M-E))) < 1/2011) ? rightanglemark(D,E,M) : anglemark(D,E,M), rgb(0.0,0.4,0.0)); draw((abs(dot(unit(D-A_1),unit(A-A_1))) < 1/2011) ? rightanglemark(D,A_1,A) : anglemark(D,A_1,A), rgb(0.0,0.4,0.0)); draw((abs(dot(unit(D-F),unit(A-F))) < 1/2011) ? rightanglemark(D,F,A) : anglemark(D,F,A), rgb(0.0,0.4,0.0)); /* Place dots on each point */ dot(B); dot(O); dot(D); dot(C); dot(K); dot(E); dot(F); dot(A); dot(T); dot(M); dot(N); dot(A_1); /* Label points */ label("$B$", B, lsf * dir(115)); label("$O$", O, lsf * dir(80)); label("$D$", D, lsf * dir(100)); label("$C$", C, lsf * dir(45)); label("$K$", K, lsf * dir(45)); label("$E$", E, lsf * dir(-135)); label("$F$", F, lsf * dir(45)); label("$A$", A, lsf * dir(140)); label("$T$", T, lsf * dir(-30)); label("$M$", M, lsf * dir(135)); label("$N$", N, lsf * dir(135)); label("$A_1$", A_1, lsf * dir(45)); [/asy][/asy]

My solution sketch for the contest: Note that $\angle MTA=\angle MAE$ since $\stackrel{\frown}{MF}=\stackrel{\frown}{DC}$. So $\triangle MAT\sim\triangle MEA$, so $(MA)^2=(MT)(ME)=(MB)(MC)$. But $(MA)^2=(MB)(MC)$ is equivalent with $(NA)(AK)=(BA)(AC)$, and since $(BA)(AC)=(DA)(AE)$ the result follows from POP.

As $\widehat{MAT}=\widehat{CAF}=\widehat{AFD}=\widehat{TED}$, we get that $MA^2=MT\cdot ME=MB\cdot MC$, hence $(N,A,B,C)=-1$. Now let $\{R\}=(DEN)\cap \omega $; as $A$ is on the radical axis of $(DEN)$ and $\omega$, we have $AB\cdot AC=AN\cdot AR$, which together with the fact that $(N,A,B,C)$ is a harmonic division yields the conclusion.

Consider $\phi$ the inversion around $A$ with raduis $r=\sqrt{AD.AE}$ and let $S=(DNE)\cap BC$ clearly, $\phi$ fixes both $(ABCD)$ and $(DNE)$. also, $\phi(DF)=(ATE)$ since $FD\parallel BC$ and $\phi(BC)=BC$ it follows that $BC$ is tangent to $(ATE)$. therefore: $MA^2=MT.ME=MC.MB=MN^2$, i.e $(B,C;N,A)=-1$ now denote by $S'$ the midpoint of $BC$. wellknown that $AC.AB=AN.AS'=AN.AS$which yellds $S\equiv S'$ as desired.  

See also http://artofproblemsolving.com/community/c6h1431350p8074805.

We see that $\angle{ETA}=\angle{FDA}=\angle{DAB}=\angle{MAE} \implies$ MA is tangent to $\odot{ETA}$.This means that $MN^2=MA^2=ME.MT=MC.MB$.Since M is the midpoint of AN this means that $(N,A;C,B)=-1$.Thus if X is the midpoint of BC then NA.AX=AC.AB=AE.AD $\implies$ X lies on $\odot{DEN}$ as desired.

let $L$ be the midpoint of $BC$ it is enough to prove : $AL.AN=AD.AE$ or $AL.AN=AB.AC$ which is true iff $(C,B;A,N)=-1$ now consider the projective transformation that takes the circle to itself and maps $A$ into the center of the circle $CT\parallel BC\implies M$ is the point at infinity as well as $N$ now it is clear that $(C,B;A,N)=-1$ and so we are done

Note that $\angle MED=\angle TFD=\angle TAM$, so triangles $EAM$ and $ATM$ are similar. Let $X$ be the midpoint of $BC$ This means $$AM^2=MT\cdot ME=MA\cdot MB$$ Now, by PoP, we have $$AE\cdot AD=AB\cdot AC=(BM-AM)(AM-CM)=AM(BM+CM)-AM^2-BM\cdot CM=AM(BM+CM-2AM)$$ Then, we have $$AM(BM+CM-2AM)=AM[(BM-AM)-(AM-CM)]=AM(AB-AC)=AM(BC-2AC)=2AM(\frac{BC}{2}-AC)=2AM(XC-AC)=AN\cdot AX$$ Thus, we have $$AE\cdot AD=AN\cdot AX$$, so by PoP, $XEND$ lies on a circle.

Let $K$ be the midpoint of $BC$ and $A'$ be the reflection of $A$ in $K$. By parallel lines and symmetry, it's easy to see $AA'FD$ is a cyclic isosceles trapezoid. Claim: $A'FMT$ is cyclic. Proof. Notice $$\measuredangle FTM = \measuredangle FTE = \measuredangle FDE = \measuredangle FDA = \measuredangle FA'A = \measuredangle FA'M$$as desired. $\square$ Thus, $$AD \cdot AE = Pow_{\omega}(A) = AF \cdot AT = Pow_{(A'FMT)}(A) = AA' \cdot AM$$$$= \frac{AA'}{2} \cdot (2 \cdot AM) = AK \cdot AN$$implying the conclusion by POP. $\blacksquare$ Remark: Overall, I feel like this question motivated contestants to reflect $A$ in $K$, as reflecting $A$ over $M$ in the problem statement seems a bit odd (as it only deals with lengths). This led me to work backwards with POP, eventually settling on $A'FMT$ as my (desired) cyclic quadrilateral of choice.

Guptaamitu orz Also $5$th.post and $1$st time using latex

Diagram stolen[asy][asy] import olympiad; import cse5; size(10cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(14pt)); /* Initialize Objects */ pair B = (-4.0, 2.0); pair O = (-2.5, 1.0); path w = CirclebyPoint(O,B); pair D = (-3.192795835973656, 2.664341890855831); pair C = (-0.9999999999999998, 2.0); pair K = midpoint(B--C); pair E = (-3.9853897408833356, -0.021575899127684872); path DE = D--E; path BC = B--C; pair F = 2*foot(D,relpoint(Line(K,O,lisf),0.5-10/lisf),relpoint(Line(K,O,lisf),0.5+10/lisf))-D; pair A = IntersectionPoint(DE,BC); pair T = IntersectionPoint(w,Line(A,F,lisf),1); pair M = IntersectionPoint(Line(B,C,lisf),Line(E,T,lisf)); pair N = (2)*(M)-A; pair A_1 = (2)*(K)-A; /* Draw objects */ draw(w, rgb(0.0,0.0,0.8) + linewidth(3.0)); draw(DE); draw(D--F); draw(T--F); draw(C--N, rgb(0.0,0.6,0.0) + linetype("4 4")); draw(E--M); draw(circumcircle(D,E,N), rgb(1.0,0.0,0.0) + linewidth(1.5)); draw(F--A_1); draw(circumcircle(D,M,E), rgb(0.0,0.8,0.0) + linewidth(3.0) + dashed); draw(D--A_1); draw((abs(dot(unit(D-E),unit(M-E))) < 1/2011) ? rightanglemark(D,E,M) : anglemark(D,E,M), rgb(0.0,0.4,0.0)); draw((abs(dot(unit(D-A_1),unit(A-A_1))) < 1/2011) ? rightanglemark(D,A_1,A) : anglemark(D,A_1,A), rgb(0.0,0.4,0.0)); draw((abs(dot(unit(D-F),unit(A-F))) < 1/2011) ? rightanglemark(D,F,A) : anglemark(D,F,A), rgb(0.0,0.4,0.0)); /* Place dots on each point */ dot(B); dot(O); dot(D); dot(C); dot(K); dot(E); dot(F); dot(A); dot(T); dot(M); dot(N); dot(A_1); /* Label points */ label("$B$", B, lsf * dir(115)); label("$O$", O, lsf * dir(80)); label("$D$", D, lsf * dir(100)); label("$C$", C, lsf * dir(45)); label("$K$", K, lsf * dir(45)); label("$E$", E, lsf * dir(-135)); label("$F$", F, lsf * dir(45)); label("$A$", A, lsf * dir(140)); label("$T$", T, lsf * dir(-30)); label("$M$", M, lsf * dir(135)); label("$N$", N, lsf * dir(135)); label("$A_1$", A_1, lsf * dir(45)); [/asy][/asy] Invert at $A$ with radius $=\sqrt{AB.AC}=\sqrt{AD.AE}$. Then $ B \rightarrow C,D \rightarrow E, T \rightarrow K$ and vice versa.We will show that $N$ goes to the mid point of $BC$. Now $\angle MAT =\angle AFD= \angle AEM \implies MA^2 =MT.ME =MB.MC$.Let , under this inversion , $M$ goes to $M'$ on $BC$. Then, $$MB.MC+AM(MB+MC)+AM^2 =AB.AC=AM.AM'=AM^2+AM.AM'\implies AM^2=MB.MC=AM(BM'-MC) \implies AB=CM'$$Since $AN=2AM, AN'=\frac{AM'}{2}\implies N'=\text{md-pt of BC}$

Let $X$ be the midpoint of $\overline{BC}$ and $A'$ the reflection of $A$ over $X.$ Notice that $AA'FD$ is cyclic since the perpendicular bisector of $\overline{AA'}$ is congruent to that of $\overline{DF}.$ Thus, $$\measuredangle DA'M=\measuredangle DFA=\measuredangle DEM$$and $A'DME$ is cyclic. Hence, $$EA\cdot AD=A'A\cdot AM=2XA\cdot \tfrac{1}{2}AN=AX\cdot AN$$by PoP. $\square$

This solution assumes $A$ lies outside the circle, but it should work for $A$ inside with a few (possibly zero) changes if we use directed lengths/angles I think (somehow $A$ lying outside was the first thought that came to mind) Rename $M$ to $P$ and $N$ to $Q$. Also let $M$ (new) denote the midpoint of $\overline{BC}$ and let $A'$ be the reflection of $A$ over $M$. By power of a point on $A$ wrt $(BCDE)$, we have $$AD\cdot AE=AT\cdot AF=AB\cdot AC,$$so we are done if we can prove $$AT\cdot AF=AD\cdot AE=AQ\cdot AM=2AP\cdot \frac{AB+AC}{2}=AP\cdot AA'.$$This is true since $$\measuredangle PTF=\measuredangle ETF=\measuredangle EDF=\measuredangle DAA'=\measuredangle PAF,$$where the last equality comes from the fact that $AA'FD$ is an isosceles trapezoid, hence $PA'FT$ is cyclic and $AT \cdot AF=AP\cdot AA'$. $\blacksquare$

Let $R$ be the midpoint of $\overline{BC}$. The cyclicity condition is equivalent to $AR \cdot AN = AD \cdot AE$, or $AA' \cdot AM = AD \cdot AE$, where $A'$ is the reflection of $A$ over $R$. Thus, it suffices to show that $DA'EM$ is cyclic. This is true because $$\angle BA'M=\angle BEM \iff \angle MAT = \angle AEM \iff \widehat{BF}+\widehat{CT}+\widehat{DT} = 360^\circ.$$But the final condition is clearly true because $DFCB$ and $DFAA'$ are both isosceles trapezoids.

Rephrase the problem as follows: Let $BCDF$ be an isosceles trapezoid inscribed in a circle $\omega$ with $\overline{BC} \parallel \overline{DF}$. Let $A$ be a point on $\overline{BC}$, and let $\overline{DA}$ and $\overline{FA}$ intersect $\omega$ again at $E$ and $T$ respectively. Define $M$ and $N$ as before, and let $P$ be the midpoint of $\overline{BC}$. Let $A'$ be the reflection of $A$ over $A'$. $DENP$ is cyclic if $AP\cdot AN=AD\cdot AE$. By power of a point at $A$ on $BCDE$, this is equivalent to $AB\cdot AC=AA'\cdot AM$, which is equivalent to $A'DEM$ being cyclic. This is true, since $$\measuredangle MED=\measuredangle TED=\measuredangle AFD=\measuredangle FAA'=\measuredangle AA'D=\measuredangle MA'D.~\blacksquare$$ Edit: wait wtf ive done this problem before nahhhh

cute Claim: $(AN)$ orthogonal to $\omega$. Proof: \[\measuredangle MAT=\measuredangle DFT=\measuredangle AET\]so $MT\cdot ME=MA^2$. Evidently this implies $(BC)$ is orthogonal to $(AN)$. Put $B,P,A,C,N$ on a number line, with directed lengths we may show $AB\cdot AC=AP\cdot AN$. (say, with $-1,0,1/p,1,p$) But $AB\cdot AC=AD\cdot AE$, done.

Let $X$ be the midpoint of $\overline{BC}$, and let $Y$ be the reflection of $A$ over $X$. We wish to prove that $AN \cdot AX = AD \cdot AE$, which is equivalent to $DXEN$ cyclic. Note that $AN \cdot AX = AM \cdot AY$, due to midpoints. $ADFY$ is an isosceles trapezoid, due to the fact that the perpendicular bisector of $DF$ coincides with the perpendicular bisector of $AY$. $\angle AFD = \angle DEM = \angle DYA$, so $DYEM$ is cyclic. It follows that $AM \cdot AY = AE \cdot AD$, so $AN \cdot AX = AE \cdot AD$, so $DNEX$ is cyclic. Hence, we are done.

Let $P$ be the midpoint of $BC$. First note that $\measuredangle BAT = \measuredangle DFT = \measuredangle AET \implies BC$ is tangent to $(AET)$. By power of point, $AM^2=EM \cdot TM=BM \cdot CM$. Now $AM^2=BM \cdot CM \implies (B,C;A,N)=-1 \implies AB \cdot AC=AN \cdot AP$, so by power of point again we have $AD \cdot AE = AB \cdot AC = AN \cdot AP \implies D$, $E$, $N$, $P$ are concyclic.

Let $K$ be the midpoint of $BC$. We wish to show $DKEN$ is cyclic. Let $A'$ be the reflection of $A$ over $K$. We claim that $DA'EM$ is cyclic. First, we show $FA'TM$ to be cyclic. Because of the parallel condition and symmetry, $DFA'A$ is an isosceles trapezoid and \[ \angle FA'A = \angle A'FD = -\angle ADF = -\angle ETF = -\angle MTF = \angle FTM, \]so $FA'TM$ is cyclic. Then, the radical axis theorem tells us that $DA'EM$ is cyclic. Finally, we have $AD \cdot AE = AA' \cdot AM = AK \cdot AN$, so we are done.

My solution is slightly different from others.

My solution uses $(ATE)$ is tangent to $BC$.

$ $

We first make an initial claim: Claim: $\triangle MAT \sim \triangle MEA$. Proof: We have that \[\angle MEA = \angle DET = \angle DFT = \angle MAT, \]and \[\angle MTA = \angle ETF = \angle EDF = \angle ADF = \angle DAM = \angle EAM, \]as claimed. $\square$ Then from this similarity, one obtains \[\frac{MA}{MT} = \frac{ME}{MA} \implies MA^2 = MT \cdot ME = MC \cdot MB = MA \cdot MN. \]Therefore $(N, A; B, C) = -1$ is harmonic. Now let $X$ denote the midpoint of $\overline{BC}.$ Then $AX \cdot AN = AB \cdot AC = AE \cdot AD \implies X, N, E, D$ are concyclic, as desired. $\blacksquare$