On the sides $AD$ and $BC$ of a rectangle $ABCD$ select points $M, N$ and $P, Q$ respectively such that $AM = MN = ND = BP = PQ = QC$. On segment $QC$ selected point $X$, different from the ends of the segment. Prove that the perimeter of $\vartriangle ANX$ is more than the perimeter of $\vartriangle MDX$.