49th Mathematical Olympiad in Ukraine

It is easy to see that $\angle APC=180^{\circ }-(\angle PCA+\angle PAC)=180^{\circ }-\angle BAC=\angle ANM$, since $MN\parallel AB$ as the centerline of the triangle (fig.14).

We know that $\angle MAC=\angle PCA$, and so $△MNA\sim △APC$. $N$ and $K$ are the midpoints of the respective sides in similar triangles, thus $\angle NKM=\angle PNA$.

Now since $KN\parallel BC$, we have $\angle NKM=\angle BMA=\angle PNA$, and we are done.