Theorem 4: The diagonals of a parallelogram bisect each other.

GIVEN   A parallelogram ABCD such that its disgonal AC and BD intersect at O.

To Prove OA =  OC and OB = OD

Proof  Since ABCD is a parallelogram. Therefore,

      AB DC    and   AD BC

Now, AB DC and transversal AC intersects them at A and C respectively.

    <BAC = DCA                      [   Alternate interior angles are equal ]

   <BAO = <DCO                                  .....(i)

Again, AB DC and BD intersectsthem at B and D respectively.

    <ABD = <CDB                   [ Alternate interior angles are equal ]

   <ABO = <CDO                                .....(ii)

Now, in AOB and COD, we have

      <BAO = DCO                      [From (i)]

          AB = CD                        [ Opposite sides of a gm are equal]

and, <ABO = <CDO                   [From (ii)]

So, by ASA congruence criterion

     AOB COD

  OA = OC and OB = OD

Hence, OA = OC and OB = OD

To Prove: The Quadrilateral ABCD is a Parallelogram

Proof: In and

        OA = OC        [Given ]

=       [Vertically Opposite Angles]

       OB = OD       [Given]

Therfore      [ SAS Criteria of Congruence]

              [C.P.C.T.]

But they are alternate interior angles when AB and CD are straight lines and AC is the transversal

As they are equal AB || CD

Similarly we can prove that AD || BC

As both the opposite pair of sides are parallel to each other.

Hence ABCD is a parallelogram

Solution: ABCD is a parallelogram and diagonals of the parallelogram bisect each other.

O is the midpoint of AC

OC = OA = 6.3 cm

Similarly O is the mid point of BD

OD = OB = 2.9 cm

Hence OA = 6.3 cm, OB = 2.9cm, OC = 6.3 cm and OD = 2.9cm