First note that $DE\perp AM$ is equivalent to $N$ being on the $A$-symmedian, so $N$ is the midpoint of the symmedian chord. So, we need to show that the perpendicular to $AN$ at $N$, $DE$, and $AL$ concur. We $\sqrt{bc}$ invert, i.e. invert with radius $\sqrt{bc}$ and reflect over the $A$-angle bisector. Now, the problem becomes: Let $ABC$ be a triangle so that $ABNC$ is a parallelogram. The line $\ell$ through $N$ perpendicular to $AN$ meets $AB, AC$ at $D, E$, and $BC$ at $L$. Show that $AL$, $(ADE)$, and $(AN)$ concur. Let $X$ be the foot from $N$ to $AL$, let $D$ be the foot of the $A$-median, $M$ the midpoint of $BC$ (discard the $M$ we received in the original problem), $D'$ be the reflection of $D$ over $M$, the foot from $N$ to $BC$. We see that since $NX$ is an altitude of the right triangle $ANL$, then \[ LX\cdot LA = LN^2 \]Now, we claim that $LN^2 = LD\cdot LE$. In particular, we claim that there is an involution on the line $DE$ swapping $(N,N)$, $(D,E)$, $(L,\infty)$, which will have to be inversion at $L$. Now, projecting through $A$ onto $BC$, we want to show that here is an involution on $BC$ swapping $(M,M)$, $(B,C)$, and $(L,P)$, where $AP\perp AM$. But this is simply reflection across $M$. Thus $LN^2 = LD\cdot LE$, so \[ LX\cdot LA = LD\cdot LE, \]i.e. $(AXDE)$ are concyclic and so $(ADE)$, $(AN)$, $AL$ concur at $X$ as desired. Inverting back $K$ lies on the perpendicular bisector of $AF$ as desired.

Similar solution as above, though phrased without involutions. Do the same stuff to get Quote: We $\sqrt{bc}$ invert, i.e. invert with radius $\sqrt{bc}$ and reflect over the $A$-angle bisector. Now, the problem becomes: Let $ABC$ be a triangle so that $ABNC$ is a parallelogram. The line $\ell$ through $N$ perpendicular to $AN$ meets $AB, AC$ at $D, E$, and $BC$ at $L$. Show that $AL$, $(ADE)$, and $(AN)$ concur. Let $LA$ meet $(ADE)$ again at $K$, which we need to show lies on $(AN)$. As $(AN)$ tangent to $DE$ and $LK \cdot LA = LD \cdot LE$, it suffices to show that $LN^2 = LD \cdot LE$. Let $X, Y$ be the feet of the perpendiculars from $D, E$ to $AE, AD$, respectively, and let $Z = DE \cap XY$. Clearly $(Z, N; D, E) = -1$. Then \[-1 = (Z, N; D, E) \stackrel{A}{=} (AZ \cap BC, M; B, C),\]but as $M$ is the midpoint of $BC$ we must have $AZ \cap BC= \infty$ or equivalently $AZ \parallel BC$. But since $M$ is the midpoint of $AN$, by homothety $L$ is the midpoint of $ZN$. So from $(Z, N; D, E) = -1$ we conclude $LN^2 = LD \cdot LE$, as desired. $\Box$

Here's a non-inversive solution: Denote $\odot (ADE)$ by $\omega$, and let $P$ be its center. As $AN$ and $AM$ are isogonal in $\triangle ADE$, so we get that $AN$ is the $A$-symmedian of $\triangle ABC$. Redefine $K$ as the point where the tangent to $\omega$ at $N$ (i.e. line $ON,$ taking $O$ as the center of $\Gamma$) meets $DE$, and let $AL \cap ON=K'$. Also suppose the tangent to $\Gamma$ at $A$ meets $\omega$ again at $X$. We wish to show that $K=K'$. Then we have, $$-1=(A,F;B,C) \overset{A}{=} (X,N;D,E) \Rightarrow KX \text{ is tangent to } \omega$$This gives that $KP \perp NX,$ which means that $$\angle KPN=\angle XAN=\angle ACF=\angle KOF \Rightarrow KPOF \text{ is cyclic}$$ Now, $K'$ lies on the radical axis of $\omega$ and $\Gamma$, so we get that $$K'N^2=OK^2-OA^2 \Rightarrow OA^2=(OK'-K'N)(OK'+K'N)=ON(OK'+K'N) $$$$\Rightarrow ON \cdot OK'=OA^2-ON \cdot OK'=(ON^2+AN^2)-ON \cdot OK'=ON(ON-OK')+AN^2=AN^2-ON \cdot OK'$$$$\Rightarrow ON \cdot OK'=\frac{AN^2}{2}=NF \cdot NP \Rightarrow K'POF \text{ is cyclic} \Rightarrow K' \equiv K \text{ } \blacksquare$$

This problem was originally proposed by me for the Iberoamerican MO 2017. One of the official solutions is presented below. Solution. It suffices to prove that $K,\ N$ and $O$ are collinear. First, observe that $DE\perp AM$ implies that $AF$ is the $A$-symmedian of $\bigtriangleup ABC$, so $N$ is the center of spiral similarity mapping $\overline{BA}$ to $\overline{AC}$ and thus $\bigtriangleup BNA\sim \bigtriangleup ANC$. In fact, it's easy to prove that we actually have $\bigtriangleup BNA\sim \bigtriangleup DNE\sim \bigtriangleup ANC$. Let $P,\ Q = \overline{DE}\cap \Gamma$, $PD<QD$; $\omega$ the circumcircle of $\bigtriangleup PNQ$ and $R=\overline{ND}\cap \omega,\ S=\overline{NE}\cap \omega, \ N\neq R,\ N\neq S$. Note that $SE\cdot EN=PE\cdot EQ=AE\cdot EC$, so $ANCS$ is cyclic. Similarly, $ANBR$ is cyclic. Therefore, $$\angle FAS=\angle NAS=180^\circ-\angle ACN-\angle ACS=180^\circ-\angle BAN-\angle ANE$$and analogously, $\angle FAR=180^\circ-\angle NAC-\angle AND$, from which it's easy to conclude that $R,\ A$ and $S$ are collinear; hence, $\angle RSN=\angle ASN=\angle ACN=\angle DAN=\angle DEN$, then $RS\parallel DE$. We conclude that $\bigtriangleup RNS$ and $\bigtriangleup DNE$ are homothetic with center $N$, so ther circumcircles are tangent at $N$ and because $ON$ touches $(DEN)$ at $N$, we have that $ON$ is their radical axis. Finally, by radical axis theorem applied to $\Gamma,\ \omega$ and $(ADE)$ we conclude that $AL,\ DE$ and $ON$ are concurrent, so $K$ lies on $ON$, as desired. $\blacksquare$

Other solution. Again, it's enough to show that $AL,\ DE$ and $ON$ concur at $K$. Let $X=\overline{NE}\cap \overline{AB},\ Y= \overline{ND}\cap \overline{AC},\ S=\overline{ON}\cap \overline{AC}$. We'll prove that $STAL$ and $STED$ are cyclic, which together to the radical axis theorem applied to their circumcircles and $(ADE)$ yields the required concurrency. Observe that $N$ is the orthocenter of $\bigtriangleup XAY$, so $AN\perp XY$ and thus $ST\parallel XY$. Since $XDEY$ is cyclic, this implies that $STED$ is cyclic as well. Note also that $L$ takes $\overline{DB}$ to $\overline{EC}$; showing that $L$ lies on $(SAT)$ requires to prove that $\frac{DS}{SB}=\frac{ET}{TC}$ (1). According to the ratio lemma, we have, $$\frac{DS}{ST}=\frac{DN}{BN}\cdot \frac{\sin \angle DNS}{\sin \angle BNS};\qquad \frac{ET}{TC}=\frac{EN}{CN}\cdot \frac{\sin \angle ENT}{\sin \angle CNT}$$Moreover, $$\angle BNS=90^\circ-\angle DBN-\angle DNS=90^\circ-\angle NAC-\angle DAN=90^\circ-\angle ENT-\angle ENC=\angle CNT$$Therefore, (1) is true if and only if, $$\frac{DN}{BN}\cdot \sin \angle DNS=\frac{EN}{CN}\cdot \sin \angle ENT\iff \frac{DN}{BN}\cdot \frac{CN}{EN}=\frac{\sin \angle ENT}{\sin \angle DNS}$$But $\frac{DN}{EN}=\frac{AB}{AC}$ due to $AN$ being the $A$-symmedian of $\bigtriangleup ABC$. Hence, we just need to show that, $$\frac{AB}{AC}\cdot\frac{CN}{BN}=\frac{\sin\angle ENT}{\sin \angle DNS}=\frac{\sin \angle NAC}{\angle NAB}$$which is evident when applying the sine rule to $\bigtriangleup BAN$ and $\bigtriangleup CAN$, taking into account that $\angle ANB=\angle ANC$ because these triangles are similar. We are done. $\blacksquare$

Note also that $L$ takes $\overline{DB}$ to $\overline{EC}$; showing that $L$ lies on $(SAT)$ requires to prove that $\frac{DS}{SB}=\frac{ET}{TC}$ Sorry.but I don't know what do you mean here.

Solution using only inversion and angle chase: After inversion around the circle with radius $\sqrt{bc}$ and reflection over the angle bisector of $\angle BAC$: Inverted problem wrote: Let $ABC$ be a triangle, $M$ the midpoitn of $BC$, $N$ the reflection of $A$ over $M$; $D,E$ on $AB,AC$ resp. such that $N\in DE$ and $DE \perp AN$. $L:=BC\cap DE$ and $AL \cap (ADE)$ again at $K\ne A$. Prove that $AM = MK$. Since $AM = MN$, $AM = MK \iff AK\perp KN \iff Q:=KN \cap (ADE)\setminus K$ is concyclic with $A,N,G:=AH\cap DE$ where $H$ is the point on $(ADE)$ such that $AK, AH$ are isogonal and $\angle NQA = \angle KHA = \angle NGA \implies ANGQ$ cyclic. Now $GQ$ is tangent to $(ADE)$ from $\angle KQG = \angle GAN = \angle QAK$. Since $AQ \perp QG$ and $QG$ is tangent it follows that $Q$ is the antipode of $ADE$. Therefore $AK \perp KQ \equiv KN$, as desired.

hyx wrote: Note also that $L$ takes $\overline{DB}$ to $\overline{EC}$; showing that $L$ lies on $(SAT)$ requires to prove that $\frac{DS}{SB}=\frac{ET}{TC}$ Sorry.but I don't know what do you mean here. $L$ is the center of spiral similarity mapping $\overline{BD}$ to $\overline{CE}$. Thus, if $DS/SB=ET/TC$ is true, we can say that $S$ and $T$ are corresponding points of such a transformation, i.e. we would have that $L$ also carries $\overline{DS}$ to $\overline{ET}$, which implies that $L$ is on $(SAT)$.

hyx wrote: Note also that $L$ takes $\overline{DB}$ to $\overline{EC}$; showing that $L$ lies on $(SAT)$ requires to prove that $\frac{DS}{SB}=\frac{ET}{TC}$ Sorry.but I don't know what do you mean here. More information and useful results about spiral similarity can be found at https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2019-01/mr_1_2019_spiral_similarity.pdf, recently published.

Jafet98 wrote: hyx wrote: Note also that $L$ takes $\overline{DB}$ to $\overline{EC}$; showing that $L$ lies on $(SAT)$ requires to prove that $\frac{DS}{SB}=\frac{ET}{TC}$ Sorry.but I don't know what do you mean here. More information and useful results about spiral similarity can be found at https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2019-01/mr_1_2019_spiral_similarity.pdf, recently published. Thanks a lot.

By assumption N is midpoint of AF with F is A-symmedian cuts (O). Let A-tangent cuts BC at I, AI cuts (AN) at J, H is midpoint of IN, T is midpoint of AN then TH//AI⟂NJ so HJ tangent (T). From A(JNDE)=A(IFBC)=-1 so D,E,H collinear. HA cuts (T) at F' then HO^2-HN^2=ON.OI=OA^2 so P(H/O)=P(H/T) so F' lies on (O) so F=F'. At all, H=K so KO⟂AF so KA=KF q.e.d

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -9.263297836702954, xmax = 4.242119148175437, ymin = -3.1075211582676316, ymax = 5.398175644823021; /* image dimensions */ pen fuqqzz = rgb(0.9568627450980393,0,0.6); /* draw figures */ draw(circle((-0.049315284117739694,1.0823705336061704), 2.972642072949428), linewidth(1)); draw((-1.53,3.66)--(2.77,0.14), linewidth(1)); draw((-1.53,3.66)--(-2.81,-0.02), linewidth(1)); draw(shift((-1.318792986922658,2.289632613959226))*xscale(1.3865479346554435)*yscale(1.3865479346554435)*arc((0,0),1,-81.23826622259303,98.76173377740696), linewidth(1)); draw(shift((-1.318792986922658,2.289632613959226))*xscale(1.386547934655444)*yscale(1.386547934655444)*arc((0,0),1,98.76173377740697,278.76173377740696), linewidth(1)); draw((-1.53,3.66)--(-0.685171947690632,-1.8214695441630964), linewidth(1)); draw((-2.3961533177855756,3.1624498299788364)--(-1.53,3.66), linewidth(1)); draw((-2.3961533177855756,3.1624498299788364)--(-2.698036360031409,2.4317694945809474), linewidth(1)); draw((-1.53,3.66)--(-8.205606512267819,-0.17471273153456396), linewidth(1) + blue); draw((-1.53,3.66)--(-4.656596243056576,0.3722762481919425), linewidth(1)); draw((-8.205606512267819,-0.17471273153456396)--(2.77,0.14), linewidth(1) + blue); draw((-0.685171947690632,-1.8214695441630964)--(-8.205606512267819,-0.17471273153456396), linewidth(1) + blue); draw((-4.656596243056576,0.3722762481919425)--(0.06629398757225977,2.3532663171501507), linewidth(1) + fuqqzz); draw((-4.656596243056576,0.3722762481919425)--(-2.3961533177855756,3.1624498299788364), linewidth(1) + fuqqzz); draw((-8.205606512267819,-0.17471273153456396)--(-4.656596243056576,0.3722762481919425), linewidth(1)); draw((-4.656596243056576,0.3722762481919425)--(-1.107585973845316,0.9192652279184519), linewidth(1) + fuqqzz); /* dots and labels */ dot((-1.53,3.66),dotstyle); label("$A$", (-1.4859553316998588,3.780982203306511), NE * labelscalefactor); dot((-2.81,-0.02),dotstyle); label("$B$", (-2.7574891063273492,0.10217574855903026), NE * labelscalefactor); dot((2.77,0.14),dotstyle); label("$C$", (2.8224455163097923,0.26266059390036334), NE * labelscalefactor); dot((0.1983161512027491,-7.553775773195878),linewidth(4pt) + dotstyle); label("$D_{1}$", (-9.263297836702954,5.398175644823021), NE * labelscalefactor); dot((-0.685171947690632,-1.8214695441630964),linewidth(4pt) + dotstyle); label("$F$", (-0.6341511525804722,-1.7248824907115308), NE * labelscalefactor); dot((-1.107585973845316,0.9192652279184519),linewidth(4pt) + dotstyle); label("$N$", (-1.0538807480885757,1.0157048681943108), NE * labelscalefactor); dot((0.06629398757225977,2.3532663171501507),linewidth(4pt) + dotstyle); label("$D$", (0.1188931217134783,2.447723488163129), NE * labelscalefactor); dot((-2.3348241153273275,1.3461306684339338),linewidth(4pt) + dotstyle); label("$E$", (-2.2883795584065276,1.447779451805592), NE * labelscalefactor); dot((-2.698036360031409,2.4317694945809474),linewidth(4pt) + dotstyle); label("$L$", (-2.5,2.3), NE * labelscalefactor); dot((-4.656596243056576,0.3722762481919425),linewidth(4pt) + dotstyle); label("$K$", (-4.9,0.47252539165441426), NE * labelscalefactor); dot((-2.3961533177855756,3.1624498299788364),linewidth(4pt) + dotstyle); label("$G$", (-2.6,3.3), NE * labelscalefactor); dot((-8.205606512267819,-0.17471273153456396),linewidth(4pt) + dotstyle); label("$P$", (-8.152248907416798,-0.07065408488548226), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] As pointed out, $ANF$ is the $A$-symmedian. Note that $L$ is the center of spiral similarity sending $BFC \rightarrow END$. Let tangents at $A,F$ to $\Gamma$ meet at $P$ (which also lies on $BC$), and $AP$ meet $(ADE)$ at $G$, note that $(G,N;E,D)=A(A,F;B,C)=-1$, hence the spiral sends $ABFC$ to $GEND$. Next let $K$ be the midpoint of $PN$, since $AN$ is diameter of $(ADE)$, thus $\angle PGN=90^{\circ}$ so $KN=KP=KG$, and this implies $KG$ is tangent to $(ADE)$ as well. Then $K$ also lies on $DE$ since $GEND$ harmonic. So we have huge spiral figures $LENDGK \sim LBFCAP$, and we are left to show $K,L,A$ collinear. Here we can just angle chase, first note from the similarity that $\angle LKG=\angle LPA=\angle LPG$ thus $KLPG$ cyclic, now $\angle PLK=\angle PGK=\angle KPG = 180^{\circ}-\angle KLG = 180^{\circ} - \angle PLA$ to give $K,L,A$ collinear, where the last angle equality is also from the similarity.