Let $O,H$ be the circumcenter, orthocenter of $\triangle ABC$ and $E$ be the midpoint of $AH$. It suffices to show $KO\perp AN$. Well-known that $AEMO$ is a parallelogram, thus $KO\parallel DE$. Since $ME\parallel AO\perp AD$ and $AE\perp MD$, $E$ is the orthocenter of $\triangle ADM$ $\implies DE\perp AM\equiv AN$ and we are done.

Let $O$ be the circumcentre of $\triangle ABC$. Note that $AD \parallel KM$ and $AK \parallel DM$ or $BC$, due to $ADMK$ being a parallelogram. This gives, $OA \perp AD \parallel KM$ and $OM \perp BC \parallel AK$, we get that $O$ is the orthocentre of $\triangle AMK$. This gives that $KO \perp AM$ or $AN$. Now, as $AN$ is a chord of the circumcircle of $\triangle ABC$, we get that $K$ lies on the perpendicular bisector of $AN$, finishing the problem.

Let $AK$ intersects $(ABC)$ again at $E$. Claim 01. Tangent to $E$ and $N$ concur at $BC$. Proof. This is equivalent to proving $CEBN$ is harmonic. But \[ A(C,B;M,AE \cap BC) = (C,B,N,E) = -1 \]Suppose they concur on $F$. Claim 02. $EKFNM$ cyclic. Proof. Let $O$ be the circumcenter of $ABC$. Now, notice that $\measuredangle OMF = \measuredangle ONF = 90^{\circ}$ easily gives us $OMNFE$ cyclic. Now, we will prove $OMKE$ is cyclic. Since $AOMD$ is cyclic, \[ \measuredangle EKM = \measuredangle AKM = \measuredangle ADM = \measuredangle AOM = \measuredangle EOM \] Therefore, \[ \measuredangle KAN = \measuredangle EAM = \measuredangle MEA = \measuredangle MEK = \measuredangle MNK = \measuredangle ANK \]which gives us $KA = KN$.

Let $P$ be reflection of $D$ over $M$. $O$ - circumcenter of $ABC$. $Q=AK \cap (ABC)$. $MP=AK \implies AMPK$ - parallelogram, $MK=DA=PQ \implies MQKP$ - isoscales trapezoid, $PQ \perp QO$ (by symmetry with $DA$) $MOQKP$ - cyclic with diameter $OP$, so $ AN \parallel KP \perp KO$ from which result is follows. Edit : I see that it's just the worse version of proof in #3. So you better read that instead.

Another proof : Reflect $K \mapsto K' , N \mapsto N'$ about perpendicular bisector of $BC$. Now we need to prove that $K'N' = DM$. Notice that $AK'DM$ - isoscales trapezoid. $AN'$ - symmedian in $\triangle ABC$, $\implies ABN'C$ - harmonic and $DN'$ is tangent to $(ABC)$ $\implies DN=DA=K'M$. With $MN' \parallel DK'$ it makes $DK'MN'$ - isoscales trapezoid and $K'N' = DM$ like its diagonals.

Let $A' \equiv AK \cap \odot (ABC)$ $\implies$ $\angle NA'K$ $=$ $\angle ABN$ $=$ $180^{\circ}-\angle ACN=\angle NMK$ $\implies$ $NMA'K$ is cyclic $\implies$ $\Delta AMA'$ $\sim$ $\Delta AKN$ $\implies$ $KA=KN$

Solution motiveted by @above solution : Let $Q = AK \cap (ABC)$ With inversion at $A$ that swaps $M,N$ , $(ABC)$ maps to $MK$ (line through $M$ parallel $DA$) so $Q$ maps to $K$ $\implies$ $\Delta AMQ$ $\sim$ $\Delta AKN$ $\implies$ $KA=KN$

Probably a different solution $OA\perp AD$ and $OM\perp DM$.Therefore $OA^2+AD^2=OM^2+MD^2$ Hence,$KA^2-KM^2=OA^2-OM^2$.So by Carnot's theorem $KO\perp AM$.Consequently $KA=KN$

IMEO 2019 P1 wrote: Let $ABC$ be a scalene triangle with circumcircle $\omega$. The tangent to $\omega$ at $A$ meets $BC$ at $D$. The $A$-median of triangle $ABC$ intersects $BC$ and $\omega$ at $M$ and $N$, respectively. Suppose that $K$ is a point such that $ADMK$ is a parallelogram. Prove that $KA = KN$. Proposed by Alexandru Lopotenco (Moldova) Let $AK\cap\omega=T$ and $TN\cap BC=R$. Claim 1:- $DARN$ is a cyclic quadrilateral. $$\angle ADM =\angle B-\angle C=\angle ACT=\angle ANR\implies DARN\text{ is a cyclic quadrlateral.}$$ Claim 2:- $NMTK$ is a cyclic quadrilateral. $$\angle MNT=\angle ADM=\angle TKM\implies NMTK\text { is a cyclic quadrilateral.}$$

Hence, $$\angle ANK=\angle MTA=\angle NAK\implies KA=KN\blacksquare$$

isn't this trivial just add the circumcenter $O$ note that $OM \perp AK$ and $OA \perp KM $ so $O$ is the orthocenter of $\triangle{AMK}$ so $KO \perp AM$ so $KA=KN$

Let tangent at $N$ intersect $BC$ at $P$. By Butterfly $DM=MP$. It's easy to see now that $MNPK$ - isoscales trapezoid and $KN=PM=KA$.

Let $O$ be the circumcenter of $\triangle{ABC}$ It is clear that $OA \perp AD$ and $OM \perp DM$ But because of the parallelogram, we have $OA \perp KM$ and $OM \perp AK$ Since $O$ is the orthocentre of $\triangle{AKM}$, $OK \perp AM \implies OK \perp AN \implies OK$ is the perpendicular bisector of $AN \implies KA=KN$

Sketch: Power of the point D with respect to w then stewart theorem in triangle AKN.

Quite surprised that this solution was not posted but I guess I might be overcomplicating things. Very nice problem Let the tangent from $D$ to $circ(ABC)$ be $T$ ($T \neq A$). We get that $TBAC$ is harmonic and therefore $AT$ is the symmedian. By the Symmedian Lemma, $TN \parallel BC \parallel AK$. Let $\angle DTA = \angle DAT = \angle TNA = \alpha$. We get that $\angle DMA = \angle TNA = \alpha$ because $NT \parallel BC$. Then $\angle ATD = \angle AMD = \alpha$ meaning that $AMTD$ is cyclic. Then $\angle TMN = \angle DMT = \angle DAT$ using the aformentioned parallelity and that $AMTD$ is cyclic. Then we get that $\angle MTN = \angle MNT = \angle DAT = \angle DTA$ therefore $$\triangle DTA \sim \triangle MTN$$Notice also that $DT = DA = MK$ because $\triangle DAT$ is isosceles and $DAKM$ is a parallelogram. Then $$\frac{MN}{TN} = \frac{DA}{AT} = \frac{MK}{AT}$$Also, $$\angle ATN = 180^{\circ} - \angle DAT = 180^{\circ} - \angle AMK = \angle KMN$$Using that $DA$ is a tangent and that $DA \parallel MK$. Thus, $$\triangle ATN \sim \triangle KMN$$Then there is a spiral similarity centered at $N$ sending $KM$ to $AT$ and therefore there exists a spiral similarity at $N$ sending $KA$ to $MN$. Then $$\triangle MTN \sim \triangle KAN$$Then $$ \frac{KA}{KN} = \frac{MT}{MN} = 1$$

WLOG, let the diagram be like in the figure. Let $O$ be the circumcenter of $\triangle ABC$. Note that $90 = \angle OMD = \angle OAD$ and so by a well known lemma, we have that $\angle OKM = \angle ODM = \angle OAM = \angle ONM \implies OKNM$ is cyclic. Now simply note that $\angle OKN = \angle OMA = \angle ODA = 90 - \angle AOD = \angle KOM - 90 = 90 - \angle KNM$. And so $KO\perp MN$, as desired. $\square$

Headsolved in $5$ minutes . Let $X$ be the $A$-Humpty point, $P$ be the midpoint of $AX$, and $F$ be the projection of $K$ onto $AM$. Because the $A$-Apollonian Circle wrt $BC$ is centered at $D$ and passes through $A$ and $X$, we know $DP \perp AX$. In addition, it's well-known that $MN = MX$. Now, parallelogram symmetry implies $AP = MF$, so $$NF = MN + MF = MN + AP = \frac{NX}{2} + \frac{AX}{2} = \frac{AN}{2}$$which finishes. $\blacksquare$

Bary bashed in 5 minutes Point $N$ is irrelevant. All we need to do is show $AM\perp OK$ where $O$ is the circumcenter as this implies that $K$ is on the perpendicular bisector of $AN$. Employ barycentric coordinates on reference triangle $\triangle ABC$. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ and let $BC=a,CA=b,AB=c$. Clearly $M=(0,\tfrac12,\tfrac12)$. Let $D=(0,t,1-t)$ since it is on $BC$. Because it also lies on the tangent at $A$, it must obey $b^2z+c^2y=0$. Plugging in $D$ gives $$b^2(1-t)+c^2(t)=0 \Rightarrow t=\frac{b^2}{b^2-c^2}$$so $D=(0,\tfrac{b^2}{b^2-c^2},\tfrac{-c^2}{b^2-c^2})$. By the parallelogram condition, $K=A+M-D=(1,\tfrac12-\tfrac{b^2}{b^2-c^2},\tfrac12-\tfrac{-c^2}{b^2-c^2})=(2b^2-2c^2-(b^2+c^2):b^2+c^2).$ To finish, we use Strong EFFT on $\overrightarrow{AM}=(-1,\tfrac12,\tfrac12)$ and $OK=(2b^2-2c^2-(b^2+c^2):b^2+c^2)$ where we set $O=\vec{0}$ and use scaled $K$. Observe, $$a^2\left(\frac{b^2+c^2}{2}-\frac{b^2+c^2}{2}\right)+b^2\left(\frac{2b^2-2c^2}{2}-b^2-c^2\right)+c^2\left(\frac{2b^2-2c^2}{2}+b^2+c^2\right)=0-b^2c^2+c^2b^2=0$$so by Strong EFFT we can conclude $AM\perp OK$ as desired.