Consider $\Delta ABC$ is acute triangle, $O$ is circumcircle of $\Delta ABC$. $AD$ is angle bisector of $\angle BAC(\text {D} \in \text {BC})$, $E;F$ are respective in $CA,AB$ such that $CE=CD$ and $BF=BD$. $M,N$ are respective the midpoints of $DE,DF$. $EF\cap (AEM),(AFN)={Q,P}( (AEM) \text{is circumcircle of }\Delta AEM)$. And $(O)\cap (AEM),(AFN)= {L,K} $. Prove that $QL,PK$ intersect at $R$ and $AR$ passes through the midpoint of $BC$ .

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I prove that $QL,KP$ and $AR$ are concurrent, then i only need to prove $AR$ passes through $BC$

Let the intersection $AR$ and $(AEM)$ is $S$ and $T$ is mid point of $EF$ and easy to see that $EF//BC$ so we will prove $S,T,A$ are colinear.

My idea is prove $MSNT$ is cyclic quadrilateral.

$\angle MSN=\angle MSA+ \angle ASN=\angle MEC +\angle NFB=\angle DMC+\angle NDB=180-\angle MDN=180-\angle MTN $

But I wonder that i used $\angle MSN=\angle MSA+\angle ASN$ when $S,T,A$ is not colinear. Is that true ?


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