The line segment $AB$ connects the centers of the circles $C_A$ and $C_B$, and therefore it is the perpendicular bisector of the common chord $OK$. Since the circles $C_A$ and $C_B$ have the same radius, the quadrilateral $AOBK$ is rhombus. Thus point $K_1$ is the middle of $AB$.

Similarly, we can show that $L_1$ is the middle of $BC$, $M_1$ is the middle of $CD$ and $N_1$ is the middle of $AD$.
From the triangles $OKL$, $OLM$, $OMN$ and $ONK$ we conclude that: $KL \parallel K_1L_1$, $LM \parallel L_1M_1$, $MN \parallel M_1N_1$ and $NK \parallel N_1K_1$ (because the line segments $K_1L_1$, $L_1M_1$, $M_1N_1$ and $N_1K_1$ connect the middles of the sides of a triangle).
Hence the quadrilaterals $KLMN$ and $K_1L_1M_1N_1$ have their sides parallel. But we know that the middles of the sides of a quadrilateral define a parallelogram. So the quadrilateral $KLMN$ is parallelogram.

Since the line segment $KO$ is the common chord of the equal circles $C_A$ and $C_B$, it is the perpendicular bisector of the line of the centers $AB$ and vice versa.
Hence the quadrilateral $KAOB$ is rhombus, and so

$$AK = OB. \tag{1}$$

Similarly the quadrilateral $LCOB$ is rhombus and
$$CL = OB. \tag{2}$$
From (1) and (2) it follows that the quadrilateral $KACL$ is parallelogram, so
$$KL = AC. \tag{3}$$
Working similarly we can prove that the quadrilateral $MNAC$ is parallelogram and that
$$NM = AC. \tag{4}$$
From (3) and (4) it follows that $KLMN$ is parallelogram.