Apply $\sqrt{bc}$ inversion calling $X'$ the image of point $X$. Then $D'C'B'E'$ are collinear in that order, $K'=\odot(AB'D') \cap \odot(AC'E')$ with $K' \ne A$ and $N'=\odot(B'K'E') \cap \odot(C'K'D')$. Let $L=\overline{AK'} \cap \overline{B'C'}$. Then $LB' \cdot LD'=LC' \cdot LE'$ since $L$ lies on the radical axis of $\odot(AB'D')$ and $\odot(AC'E')$. Thus, $L$ lies on $\overline{K'N'}$ if and only if $LD' \cdot LC'=LB' \cdot LE' \iff LB'=LC' \iff \overline{AK}$ is a symmedian in $\triangle ABC$.

Let the tangents from $B,C$ to $\odot ABC$ intersect at $Z$. Extend $ZB,ZC$ to meet $\odot BEK, \odot CKD$ respectively again at $F,G$. By angle chasing: $$\angle BFK=\angle BEK=\angle BEC=\angle BAC=\angle ZBC \implies FK \parallel BC$$Similarly $GK \parallel BC$ from which it follows: $$BF=CG \implies ZF=ZG \implies ZB \cdot ZF=ZC \cdot ZG$$So $Z$ is on the radical axis of $\odot BEK, \odot CKD$ which is just the line $KN$. Hence if $AKN$ colinear then $AKZ$ colinear $\implies AK$ is the $A-\text{symedian}$ and if $K$ on $A-\text{symedian}$ then as $AKZ$ colinear and $KNZ$ colinear it follows $AKN$ colinear.

Let $X$ be a point of intersection of tangents at $B$ and $C$ to $k_{ABC}$. Let $\angle BAC=a,\angle KBC=x,\angle KCB=z,\angle EBK=y,\angle DCK=y$. From $\triangle BEC,BDC$ we have that $a+x+y+z=180$. Let $XB,XC$ intersect circles around $BKE,DKC$ at $Y,Z$. Since $\angle YKZ=180-a-x-y+180-x-z+180-a-z-y=180$ and $\angle BYK=\angle KYZ=a$ we have that $YZ\parallel BC$ so $XB\cdot XY=XC\cdot XZ$ so $X,K,N$ are collinear and on radical axis. $A$ is on $KN$ if and only if $A$ is on $KX$ which is equivalent to $K$ being on $AX$ which is a symmedian.

Simple problem. Just Pascal. Let the tangents to $(ABC)$ at $B$ and $C$ intersect at $T$ and let $EB \cap DC = P$. Applying Pascal's Theorem for $EBBDCC$, we get that $K, T, P$ are collinear. Now since $P$ lies on the radical axis of $(BKE)$ and $(CKD)$, so, $K, N, T, P$ are collinear. Now if $K$ belongs to the $A$-symmedian i.e. belongs to $AT$, then $A, K, N$ are collinear. And if $A, K, N$ are collinear, then the fact that $K, N, T$ are collinear implies that $K$ belongs to $AT$ i.e. belongs to the $A$-symmedian.

Let the tangents at $B$ and $C$ meet at $L$. $\angle{BNK}=\angle{BEC}=\angle{BDC}=\angle{KNC}=\angle{BAC}$ Thus $\angle{BNC}=2 \angle{BAC}$ and $NK$ is the angle bisector of $\angle{BNC}$. Hence we get $B,N,O,C,L$ concyclic. $KN$ must meet the perpendicular bisector of $BC$ on this circle and that point is $L$. So $K,N,L$ are collinear, and $A,K,N$ are collinear if and only if $A,K,L$ are collinear or equivalently if $K$ belongs to the $A-symmedian$ of $\triangle{ABC} \blacksquare$

Define $X=(BKE)\cap AB$ and $Y=(CKD)\cap AC.$ Simple angle chasing yields $KX||AY$ and $KY||AX.$ Therefore $AXKY$ is a parallelogram $\Rightarrow AK$ bisects $XY.$ $\Rightarrow K\in A-symmedian \Leftrightarrow BXYC \Leftrightarrow AX.AB=AY.AC \Leftrightarrow $ A lies on the radical axis of $(BKE)$ and $(CKD) \Leftrightarrow A,K,N $ are collinear.

First sol $\angle BNK=\angle BEK,\angle KNC=\angle KDC \implies \angle BNC=\angle BOC $ and $\angle BNK =\tfrac{1}{2}\angle BOC $ thus $N$ belongs to $(OBC)$ and $KN$ pass through the intersection of the tangents from $B,C$ which is the antipode of $O$ in $(OBC)$ so $A,K,N \iff$ $AK$ is a symmedian. second sol just by pascal with $BBECCD$... RH HAS

It is easy to see that on AL exists point N such that NBL is similiar to NLC.Now just see that N is on (BKE) and (CKD) and it is trivial now.(really simple approach).

Let $L$ be the pole of $BC$ (which is on the $A$-symmedian). Let $P = BE \cap CD$, $Q = BC \cap DE$. Since $PB \cdot PE = PC \cdot PD$, $P$ lies on $KN$, the radical axis of $(BKE)$ and $(CKD)$. From Brocard's theorem, $PK$ is the polar of $Q \in BC$, so $PK$ passes through $L$, from which the result follows.

Similar to @above's solution First the "if" direction, suppose $A,N,K$ are collinear. Let $AK \cap (ABC) = Z$ Then, $\angle AZB = \angle ACB$ and $\angle BNZ = \angle BNK = \angle BEK = \angle BEC = \angle BAC$ and similarly, we get that $\angle ZNC = \angle BAC, \angle NZC = \angle ABC$. So, $N$ is the center of spiral similarity taking $BZ$ to $ZC$. So, this means that $N$ is the midpoint of the symmedian chord of $\triangle BZC$ and so lies on the $A$ symmedian. Now the "only if" direction. Suppose $AK$ is a symmedian and let $AK \cap (ABC) = Z$. Let $N'$ be the $A$ dumpty point, which obviously lies on $AKZ$. Since $\angle BNK = \angle BNZ = \angle BAN + \angle NBA = \angle BAN + \angle NAC = \angle BAC = \angle BAC = \angle BEK$, $N$ lies on $(BEK)$, by symmetry it lies on the other circle as well and so $N'=N$ and so $A,N,K$ are collinear