, 2022

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
$$\begin{aligned} \angle XYM &= \angle XPM = 180^\circ - \angle XDM = 180^\circ - (\angle BDM - \angle BDX) \\ &= 180^\circ - \frac{1}{2}\angle BDC + \angle BAX = 180^\circ - \frac{1}{2}(180^\circ - \angle BAC) + \angle EFX \\ &= 90^\circ + \angle PAF + \angle EFX = 180^\circ - \angle AFX = \angle XFC. \end{aligned}$$

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.