Question 41
In an isosceles right triangle $ABC$ the right angle is at vertex $C$. On the side $AC$ points $K$, $L$ and on the side $BC$ points $M$, $N$ are chosen so that they divide the corresponding side into three equal segments. Prove that there is exactly one point $P$ inside the triangle $ABC$ such that $\angle KPL = \angle MPN = 45^\circ$.