Let the quadrilateral be $ABCD$ and suppose $X$ is the point where its diagonals meet. First of all, observe that the centroids of the four triangles form a parallelogram, by vectors or complex numbers say. Now, because for any triangle, the ninepoint center is a linear combination of the centroid and orthocentre with fixed ratio, it suffices to show that the orthocentres of the four triangles form a parallelogram. Suppose these orthocentres are $H_A$ for $XAB$, $H_B$ for $XBC$ and define $H_C, H_D$ analogously. Now just observe that $H_A, H_B$ both lie on the $B$-altitude in $\triangle BAC$. In particular, $H_AH_B \perp AC$ and similarily $H_CH_D \perp AC$. This yields that $H_A H_B \parallel H_C H_D$ and we also similarily have $H_B H_C \parallel H_A H_D$ which shows that $H_A H_B H_C H_D$ is a parallelogram, as desired. This concludes the proof.