Let $K,L,M,N$ are vertices of parallelogram, which lie on $AB,BC,CD,DA$ respectively and $LM\parallel PQ.$ We claim that $AC,KM,NL$ concur. Indeed, by the Converse of Desargues theorem it's suffice to note $$KN\cap LM=\infty_{PQ}\in PQ.$$

Moving points also kill it. Let $X,Y,Z,W$ be the vertices of the parallelogram on $AB,BC,CD,DA$. Animate $Y$ with degree 1 along $BC$. Observe that as $AX/CY$ is constant (sine theorem, angles are constant), $X$ also has degree 1. Finally $W$ also has degree 1. So if $K = XZ\cap YW$ then $K$ is just $(Y+W)/2$ which has degree 1. Thus we are left to check that $K\in AC$ in 2 cases for $Y$. $Y=C$ is trivial. If $Y=BC\cap AD$, then $K = A$ and done. Remark: an homography can also nuke this problem. Just move $ABCD$ to a square.

Let $S,K,N,M$ be points on $AD,DC,CB,BA$ and $MN || SK || PQ$. Let $SN,KM$ meet at $T$. Note that by Desargues for $PMK$ and $QNS$ for proving that $D,T,B$ are collinear we need to prove $PQ,SK,MN$ are concurrent or parallel which is true.