Baltic Way 2023 Shortlist

$$\angle BGF=\angle BGK=\angle BCK=\angle PCB=\angle PDB=\angle BDM=\angle BDF.$$ 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.

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Alternative solution.

Construct point $I$ as the second point of intersection of $MD$ with the circumcircle of $△BPC$. Then $IKPM$ is cyclic with $B$ being its circumcentre (using directed angles): $$\begin{array}{c}\angle BIM=\angle BID=\angle BPD=\angle DMB=\angle IMB\\ ⟹|BM|=|BP|=|BK|=|BI|.\end{array}$$ 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.

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Alternative solution.

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

Claim. $△DEF\sim △CBK$ (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 $△DEF\sim △CBK$. $\square$