Let $ABCD$ be a convex quadrilateral and let $P$ be the intersection of the diagonals $AC$ and $BD$. The radii of the circles inscribed in the triangles $\vartriangle ABP$, $\vartriangle BCP$, $\vartriangle CDP$ and $\vartriangle DAP$ are the same. Prove that $ABCD$ is a rhombus,