SELECTION EXAMINATION

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: $KL\parallel K_1{L}_1$, $LM\parallel L_1{M}_1$, $MN\parallel M_1{N}_1$ and $NK\parallel N_1{K}_1$ (because the line segments $K_1{L}_1$, $L_1{M}_1$, $M_1{N}_1$ and $N_1{K}_1$ connect the middles of the sides of a triangle). Hence the quadrilaterals $KLMN$ and $K_1{L}_1{M}_1{N}_1$ 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.

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Alternative solution.

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

$$AK=OB.\text{\hspace{1em}(1)}$$

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