Let $O$ be the origin of coordinate system. Because in parallelogram diagonals concurr in their midpoints and $AG$ is a median in triangle $ACD$ we know that $L$ is centroid of the triangle. Hence $3\vec{OL}=\vec{OA}+\vec{OC}+\vec{OD}$. In similar manner $3\vec{OK}=\vec{OB}+\vec{OC}+\vec{OD}$. Hence $3\vec{KL}=3\vec{OL}-3\vec{OK}=\vec{OA}-\vec{OB}=\vec{BA}$. Analogously $3\vec{IJ}=\vec{DC}$. As $ABCD$ is a parallelogram we have $\vec{BA}=\vec{CD}\implies\vec{KL}=\vec{JI}$ which means that $KLIJ$ is a parallelogram.

Let $\lbrace S\rbrace=AC\cap BD$. In parallelogram $ABCD$ we know that segments $DS$ and $AG$ are medians in triangle $ACD$ thus $L$ is centroid of the triangle. Hence homothety centered at $S$ and scale $3$ transforms $L$ onto $D$. The same we can say about points in pairs $(K,C),(J,B),(I,A)$. But quadrilateral $ABCD$ is a parallelogram so this homothety proves that quadrilateral $IJKL$ is also a parallelogram.