All chords have the same length, so their midpoints are at the same distance from the centre of the circle. Therefore, the midpoints of all such chords form a circle concentric with the given circle.
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In a rectangle, the diagonals are equal and bisect each other. Since the rectangle is cyclic, each diagonal is a diameter of the circle. Therefore, their intersection is the midpoint of a diameter, which is the centre.
Let ABCD be a parallelogram inscribed in a circle. Opposite angles of a cyclic quadrilateral are supplementary, while opposite angles of a parallelogram are equal. Therefore, each angle is 90°. Hence, ABCD is a rectangle.
Take a right triangle with one side 3 cm and another side 3 cm. Its hypotenuse is the required 6 cm chord. Draw its circumcircle. The distance from the centre to the chord is 3 cm.
Let equal chords AB and CD intersect at P. Since the chords are equal, their corresponding parts have equal sums. Using the intersecting-chords theorem and equality of the chords, we get AP = CP and BP = DP. Hence proved.