Junior Balkan MO

Solution:

If $\text{= ACB}$ then $\text{CAP= CBP= ACB=}$. Let $E=KN \cap AP$ and $F=KM \cap BP$. We show that points $E$ and $F$ are midpoints of $AP$ and $BP$, respectively.

Indeed, consider the triangle $AEN$. Since $KN \parallel BC$, we have $\text{ENA= BCA=}$. Moreover $\text{EAN=}$ giving that triangle $AEN$ is isosceles, i.e. $AE=EN$. Next, consider the triangle $ENP$. Since $\text{ENA=}$ we find that
$$\text{PNE=90 - ENA=90 -}$$
Now $\text{EPN=90 -}$ implies that the triangle $ENP$ is isosceles triangle, i.e. $EN=EP$.
Since $AE=EN=EP$ point $E$ is the midpoint of $AP$ and analogously, $F$ is the midpoint of $BP$. Moreover, $D$ is also midpoint of $AB$ and we conclude that $DFPE$ is parallelogram.
It follows from $DE \parallel AP$ and $KE \parallel BC$ that $\text{DEK= CBP=}$ and analogously $\text{DFK=}$.
We conclude that $\triangle EDN \cong \triangle FMD$ ($ED=FP=FM$, $EN=EP=FD$ and $\text{DEN= MFD=180 -}$) and thus $ND=MD$. Therefore $D$ is a point on the perpendicular bisector of $MN$. Further,
$$\text{FDE = FPE=360 - BPM- MPN- NPA= =360 - (90 - )- (180 - )- (90 - )=3}$$
It follows that
$$\text{MDN = FDE- FDM- EDN= FDE- END- EDN= = FDE-( END+ EDN)=3 - =2 .}$$
Finally, $KMCN$ is parallelogram, i.e. $\text{MKN= MCN=}$. Therefore $D$ is a point on the perpendicular bisector of $MN$ and $\text{MDN=2 MKN}$, so $D$ is the circumcenter of $\triangle MNK$.