# Prove that $AR$ passes through midpoint of $BC$

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 $$BC$$ . 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 ?