Parallelogram properties

Theorem. (The property of the parallelogram diagonals)

The diagonal parallelograms intersect and the intersection point is divided in half.



Evidence.

Let ABCD be a given parallelogram. Let's draw a diagonal AC. We note on it the middle O. On the extension of the segment DO, we postpone the segment OB1 equal to DO.
By the previous theorem AB1CD - parallelogram. Therefore, straight AB1 is parallel to DC. But through point A it is possible to draw only one straight line parallel to DC. Hence, the line AB1 coincides with the line AB.
It is also proved that BC1 coincides with BC. Hence, point C coincides with C1. parallelogram ABCD coincides with parallelogram AB1CD. Consequently, the diagonals of the parallelogram intersect and the intersection point is divided in half. The theorem is proved.

Theorem. (The property of the opposite sides of the parallelogram).

The parallelogram has opposite sides.



Evidence.

Let ABCD be a given parallelogram. And let its diagonals intersect at point O.
Since Δ AOB = Δ COD by the first sign of equality of triangles (∠ AOB = ∠ COD, as vertical, AO = OC, DO = OB, by the diagonal property of the parallelogram), then AB = CD. Similarly, from the equality of the BOC and DOA triangles, it follows that BC = DA. The theorem is proved.

Theorem. (Property of the opposite angles of the parallelogram).

The parallelogram has opposite angles.



Evidence.

Let ABCD be a given parallelogram. And let its diagonals intersect at point O.
From the parallelogram Δ ABC = Δ CDA on three sides proved in the theorem on the properties of the opposite sides (AB = CD, BC = DA of the proven, AC is common). It follows from the equality of triangles that ABC = CDA.
It is also proved that ∠ DAB = ∠ BCD, which follows from ∠ ABD = CDB. The theorem is proved.