Inside $\angle BAC = 45 {} ^ \circ$ the point $P$ is selected that the conditions $\angle APB = \angle APC = 45 {} ^ \circ $ are fulfilled. Let the points $M$ and $N$ be the projections of the point $P$ on the lines $AB$ and $AC$, respectively. Prove that $BC\parallel MN $.
(Serdyuk Nazar)
Note that $ABPC$ is a concave quadrilateral. This can be found by a simple angle chase. Since $\angle BAC+\angle PBA +\angle PCA = 45^{\circ}+45^{\circ}+45^{\circ}=135^{\circ}$, reflex angle, $\angle BPC= 225^{\circ}$. Because this figure is symmetric, we know: $$\angle MNA=\angle NAM=\frac{135}{2}^{\circ}.$$Next, because $N$ and $M$ are projections, $$\angle BNM=\angle CMN=\frac{45}{2}^{\circ}.$$We can easily tell that $\angle CPN=\angle BPM$, thus $$180^{\circ}-2\cdot \frac{45}{2}^{\circ}=\angle MPN=\angle CPB=135^{\circ}.$$Using this, we can find that $\angle BCA=\angle CBA=\frac{45}{2}^{\circ}$. This means that $\angle BCN=\angle MNA=\frac{135}{2}^{\circ}$. Because they are equal and they are corresponding angles, this means that $BC \parallel MN$. $\blacksquare$