Solution:

Let $M, N, P, Q$ be the midpoints of $AB, BC, CD$, and $DA$, respectively. Then $MN$ is the midline of $\triangle ABC$ opposite $AC$, so it is parallel to $AC$ and of length $\frac{1}{2} AC$. Similarly, $PQ$ is the midline of $\triangle ACD$, so $PQ \parallel AC$ and $PQ = \frac{1}{2} AC$. Thus, the opposite sides $MN$ and $PQ$ are equal and parallel, and similarly, $NP$ and $QM$ are equal and parallel. Thus, $MNPQ$ is a parallelogram.