Austria 2014

Let $\alpha =\angle NMP$ and $\beta =\angle MNP$. Because the sum of angles in $MNP$ is $180^{\circ }$, we obtain $\alpha +\beta =90^{\circ }$. Since $BP$ is a chord of the circle through $P$ and with mid-point $M$, we have $$\angle BAP=\frac{1}{2}\angle BMP=\frac{1}{2}\alpha .$$ Similarly, for the chord $CP$ in the circle $k_N$, we obtain $$\angle CDP=\frac{1}{2}\angle CNP=\frac{1}{2}\beta .$$

Since $NP$ and $MP$ are perpendicular, $NP$ is a tangent of the circle $k_M$ and $MP$ is a tangent of the circle $k_N$. Considering the chord $BP$ in the circle $k_M$, we therefore have $$\angle BPN=\angle BAP=\frac{1}{2}\alpha$$ and with the chord $CP$ in $k_N$ we have $$\angle MPC=\angle CDP=\frac{1}{2}\beta$$ It therefore follows that $$\angle CPB=\angle MPN-\angle BPN-\angle MPC=90^{\circ }-\frac{1}{2}\alpha -\frac{1}{2}\beta =45^{\circ },$$ and since $\angle APB=90^{\circ }$ we also have $$\angle APC=\angle APB-\angle CPB=45^{\circ }$$ It therefore follows that $PC$ bisects the angle $APB$ as claimed. $\square$