Lemma:Denote the orthocenter of $\vartriangle ABC$ by $H$,$M$ is the midpoint of $BC$,$F$ is the foot of $H$-altitude on $AM$.Proof that $B,C,F,H$ are concyclic. Let $A,A'$ be a couple of antipodal points on the $\odot(ABC)$,$AM$ meets $\odot(ABC)$ at $A,K$. It is well known that $BMCH$ is a parallelogram,and we also have $\angle A'KA=\frac{\pi}{2}$.So $M$ is the midpoint of $A'H$ and the midpoint of $KF$.Due to $B,A',K,C$ are concyclic,we can know that $B,C,F,H$ are concyclic. Let $G$ be the foot $A$-altitude of $\vartriangle ABC$ and $H$ is the orthocenter of $\vartriangle ABC$.$N$ is the midpoint of $AH$.$AM$ meets $\odot MDE$ at $L,A$. Let $F'$ lies on the $AM$ such that $AF'\cdot AM=AH\cdot AG=AD\cdot AC=AB\cdot AE$.Easy to know that $B,E,M,F'$ are concyclic and $C,D,M,F'$ are concyclic.So $F=F'$. Use the lemma,we can know that $B,C,F,H$ are concyclic, so the line $XY$ is the Simson line of $F$ to $\vartriangle BCH$.Due to $A$ is the orthocenter of the $\vartriangle BCH$, so line $XY$ passes the midpoint of $AK$. Recall $N,L$ lies on the $\odot MDE$(the Euler circle of $\vartriangle ABC$).So $NL\perp AM$,so $NL\parallel HF$,so $L$ is the midpoint of $AF$. In the end,lines $AF$,$XY$ both pass the point $L$ which lies on the $\odot MDE$.

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