In the exterior of the triangle $ABC$ with $m(\angle B) > 45^o$, $m(\angle C) >45°^o$ one constructs the right isosceles triangles $ACM$ and $ABN$ such that $m(\angle CAM) = m(\angle BAN) = 90^o$ and, in the interior of $ABC$, the right isosceles triangle $BCP$, with $m(\angle P) = 90^o$. Show that the triangle $MNP$ is a right isosceles triangle.