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Let $X$ be the other point of intersection of the circumcircles of the triangles $AEF$ and $ABC$. We have $$\angle EXF=\angle EAF=\angle BAC=\angle BXC$$ and $$\angle XFE=\angle XAB=\angle XCB,$$ so the triangles $BXC$ and $EXF$ are similar. Since $P$ is the midpoint of the segment $EF$, and $M$ is the midpoint of the segment $BC$, we conclude that the triangles $EXP$ and $BXM$ are also similar. Therefore, $\angle XPE=\angle XMB$ and so $$\angle XPD=\angle XPE+90^{\circ }=\angle XMB+90^{\circ }=\angle XMD.$$ Thus the points $X,P,M,D$ are concyclic.



Let $Y$ be the second intersection point of the circumcircle of the triangle $AEF$ and the circle passing through the points $X,P,M,D$. We will prove that $AM$ passes through $Y$. Since $\angle AYX=\angle AFX$, it is enough to prove that $\angle XYM=\angle XFC$. We have $$\begin{array}{rl}\angle XYM& =\angle XPM=180^{\circ }-\angle XDM=180^{\circ }-(\angle BDM-\angle BDX)\\ & =180^{\circ }-\frac{1}{2}\angle BDC+\angle BAX=180^{\circ }-\frac{1}{2}(180^{\circ }-\angle BAC)+\angle EFX\\ & =90^{\circ }+\angle PAF+\angle EFX=180^{\circ }-\angle AFX=\angle XFC.\end{array}$$

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Alternative solution.

Let $Q$ be the intersection of $EF$ and $BC$. We have $$\angle DMQ=\angle DMB=90^{\circ }=\angle DPE=\angle DPQ$$ so the quadrilateral $DMPQ$ is cyclic. Let $X$ be the other point of intersection of the circumcircles of the triangles $AEF$ and $ABC$. Consider the spiral similarity $f_1$ which maps $BC$ to $EF$. Since $A=BE\cap CF$, then the center of $f_1$ is the second intersection point of the circumcircles of triangles $ABC$ and $AEF$, i.e. it is the point $X$. Since $M$ and $P$ are the midpoints of $BC$ and $EF$, then $f_1$ maps $BM$ to $EP$. Since $Q=BM\cap EP$, then the center $X$ of $f_1$ is the second intersection point of the circumcircles of triangles $QBE$ and $QMP$. Therefore, we conclude that $X,P,M,D,Q$ are concyclic.



Let $Y$ be the second intersection point of the circumcircles of the quadrilaterals ($AEFX$) and ($PMDX$). We will prove that $AM$ passes through $Y$. Since spiral similarities come in pairs and $f_1$ maps $MC$ to $PF$, there exists another spiral similarity $f_2$, with the same center $X$, mapping $MP$ to $CF$. Therefore, the triangles $XMP$ and $XCF$ are similar and so $\angle XPM=\angle XFC$. We now have $$\angle AYM=\angle AYX+\angle XYM=\angle AFX+\angle XPM=\angle AFX+\angle FXC=180^{\circ }.$$ So $Y\in AM$ as required.