, 2023

$$\angle BGF = \angle BGK = \angle BCK = \angle PCB = \angle PDB = \angle BDM = \angle BDF.$$
Similarly, $GFCE$ is cyclic.

Lastly, notice that $A$ is the radical centre of circumcircles of $GFBD$, $GFCE$ and $BDCE$. Thus, $F$, $A$ and $K$ are collinear.

Construct point $I$ as the second point of intersection of $MD$ with the circumcircle of $\triangle BPC$. Then $IKPM$ is cyclic with $B$ being its circumcentre (using directed angles):
$$\begin{gather*} \angle BIM = \angle BID = \angle BPD = \angle DMB = \angle IMB \\ \implies |BM| = |BP| = |BK| = |BI|. \end{gather*}$$
We now show that $C$, $K$ and $I$ are collinear. $\angle KCB = \angle BCP$ by reflection and $\angle BCP = \angle ICB$, since they subtend equal chords $BI$ and $BP$. Thus, $\angle KCB = \angle ICB$ and thus $C$, $K$ and $I$ are collinear.

Thus, $K$, $F$ and $A$ are collinear.

Let $\bar{F}$ denote the isogonal conjugate of $F$ with respect to $\triangle ADE$.

Claim. $\triangle DEF \sim \triangle CBK$ (with opposite orientation).

Proof. Using directed angles mod $180^\circ$, we have
$$\angle \bar{F}DE = \angle ADF = \angle ADM = \angle PDA = \angle PDB = \angle PCB = \angle BCK,$$
where we used that $BCPD$ is cyclic. Similarly, $\angle DEF = \angle KBC$. Hence $\triangle DEF \sim \triangle CBK$. $\Box$