In parallelogram $ABCD$, $AB>AD$. Point $E$ lies on side $AB$ with $AE=AD$. Let $F=DE\cap AC$. Take $K$ on segment $EF$, and let $M=AK\cap BC$, $N=CK\cap AB$. Prove that $AN=CM$.
In quadrilateral $ABCD$, let $M,N$ be the midpoints of $BC,AD$, respectively. Let $P=AM\cap BN$ and $Q=DM\cap CN$. Prove that $S_{PAB}+S_{QCD}=S_{MPNQ}$, where $S$ denotes area.