The segment AB and CD intersect at point O so that AO = BO and CO = DO. Prove that ABCD is a parallelogram.

You can always draw a plane through intersecting line segments. Consequently, the intersecting segments AB and CD lie in the same plane and the figure formed with their help also lies in the plane. The point of intersection is the point O, while O divides the segments in half. Therefore, with respect to point O, point A is symmetric to point B, and point C is symmetric to point D, since each pair of points lies on a straight line (a straight line can be drawn through any segment) and AO = OB, and CO = OD. Accordingly, the segments formed by these points will be symmetrical about the point O, AC is symmetrical to BD, AD is symmetric to CB. From the properties of symmetry, it is known that the lengths of symmetric segments are equal. In addition, the straight lines drawn through these segments will be symmetric about the point and will be parallel. This means that the segments that are symmetrical about the point O are parallel.

Thus, we have a quadrangle ACBD, in which the sides are pairwise equal and parallel, and the diagonals, when they intersect, divide each other in half. This corresponds to a single figure – a parallelogram.

The proof could also be based on the equality of the triangle segments formed at the intersection.

Finally, let us note that ACBD is a parallelogram, while the figure ABCD resembles an hourglass and is not a parallelogram.