Let $ ABCD$ be a convex quadrilateral and let $ N$ on side $ AD$ and $ M$ on side $ BC$ be points such that $ \dfrac{AN}{ND}=\dfrac{BM}{MC}$. The lines $ AM$ and $ BN$ intersect at $ P$, while the lines $ CN$ and $ DM$ intersect at $ Q$. Prove that if $ S_{ABP}+S_{CDQ}=S_{MNPQ}$, then either $ AD\parallel BC$ or $ N$ is the midpoint of $ DA$.