Three equal circles $k_1, k_2, k_3$ intersect non-tangentially at a point $P$. Let $A$ and $B$ be the centers of circles $k_1$ and $k_2$. Let $D$ and $C$ be the intersection of $k_3$ with $k_1$ and $k_2$ respectively, which is different from $P$. Show that $ABCD$ is a parallelogram.