Cute. Let $R=(BOC) \cap (AMN)$ and by midlines it suffices to show that $R \in (PAQ)$. By twice Menelaus combined with the well-known property of $R$ (the intersection point of the circles through $A, C$ tangent to $AB$ and through $A, B$ tangent to $AC$), we have $\triangle RAC \cup \{P\} \sim \triangle RBA \cup \{Q\}$, the end.

The title comes from the fact that $R$ is the Dumpty-point in $\Delta ABC$.

Let $Q_A$ be the dumpty point, then it suffices to prove $Q_A \in (APQ)$, but this follows as $AP/PC = AN/NB$ and spiral at dumpty finishes.

Use barycentric coordinates wrt $ABC$. Let $D=(0:q:p)$. Then if $P=(x:0:z)$ from$$\begin{vmatrix}x&0&z\\1&1&0\\0&q&p\end{vmatrix}=0$$we get that $P=(-q:0:p)$. Analogously we get that $Q=(-p:q:0)$. Then the equation of $(APQ)$ has the form$$-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0$$for some $u,v,w\in \mathbb{R}$. Evaluating this at $A$ we get $u=0$, evaluating this at $P$ we get $w=\frac{qb^2}{q-p}$. Evaluating this at $Q$ we get $v=\frac{pc^2}{p-q}$. The magic point of the $A$-symmedian of $ABC$ (the so-called $A$-DM point) has coordinates $\bigg (b^2+c^2-a^2:b^2:c^2\bigg )$. This point clearly lies on $(APQ)$, and since it lies on $(AMN)$ and is not $A$, it follows that it is $R$. Hence $AR$ is the $A$-symmedian of $ABC$ and therefore of $AMN$, which means that $K$ is the magic point of the $A$-symmedian of $AMN$, so if $O$ is the circumcentre of $AMN$ then $\angle MKN=\angle MON=2\angle MAN=2\angle BAC$, as desired.

i promise i will refine and clean this up this later: Dilate at $A$ with factor $2$. Relabelling a bit, we have the following problem: Let $ABC$ be a triangle with $A$-Dumpty Point $D$, and $M,N$ midpoint of $AB,AC$. Let $R$ be ANY point on $BC$ (we generalize), and $RM\cap AC=Q$, $RN\cap AB=P$. Then $(APQ)$ passes through $D$. Dilate at $A$ AGAIN, as $D$ is midpoint of symmedian chord (well known). ignore all previous point labellings and work with below: use my really bad diagram here, with geogebra points. https://www.geogebra.org/calculator/fhxvce74 Now to show the cyclic, we will prove that $D$, the intersection of tangents at $B,C$ to $(ABC)$ has equal power to $(ABC)$ and $(AFG)$. THis is equivalent to $DB^2=DK\cdot DL$, where $K,L$ are $DB\cap(AFG)$. If $BE\cap (AFG)=H,F$, and $HG\cap BD=I$ , we can easily angle chase that $IGB\sim BFC$. DIT on $GHAF$ means that if $BF\cap AC = I'$, only want $DI\times DI'=DB^2$. $DI$ is fixed, so we just want $DI'$ fixed, as then we can make $E$ just say, $B$. We want by tangent circles, $(I'IC)$ tnagent to $CD$. This is same as $\angle CIB=\angle CBF$, easy to see this is same as $\triangle CIB\sim\triangle CBA$. Already we have $\angle CBI=\angle CAB$, so we just need say another side ratio. Now recall $A'B=BA$, and consider similar triangles $IA'B\sim BAC$. Thus, by our similar triangles we note $BI/BC=BG/CF$, thus RTP $BG/CF$ is fixed. This is a length bash, i'll add later, seems distinct enough that i posted seperately too.

I defined the point $Q_A$ and $Q_A \in \odot (AMN)$ just like above. I also got $\frac{CP}{PA} = \frac{AQ}{QB}$ but didn't know how to finish it. Can someone explain why are done by spiral? Edit: Nvm I learned Spiral sim