Step 1: Analyse the given question and note down the given values

            NOTE:  A quadrilateral ABCD

                        The diagonals of the quadrilateral intersecting

                          Condition \frac{AO}{BO} = \frac{CO}{DO}

Step 2:  Construct a quadrilateral ABCD

Step 3: Through intersection point(‘O’) draw a line parallel to AB which meets DA at X.    


Step 4: Apply the basic proportional theorem to the triangle ABD in the quadrilateral.

      THEOREM: If a line is drawn parallel to one side of a triangle to intersect

        the other two sides in distinct points, then the other two sides are divided

        in the same ratio.

        Given : In ∆ABC, DE || BC which intersects sides AB and

            AC at D and E respectively.

                So, \frac{AD}{DB} = = \frac{AE}{EC}

    EXAMPLE:  From the figure

                           In ∆DAB, XO || AB (by construction)

                             \frac{DX}{XA} = = \frac{DO}{OB}

                            Rewrite this

                             \frac{AX}{XD} = = \frac{BO}{OD}...................(1)

Step 5: Use the condition given in the question

                   \frac{AO}{BO} = \frac{CO}{DO}

                   \frac{AO}{BO} = \frac{CO}{DO}  

                  Rewrite the equation

                \frac{AO}{CO} = \frac{BO}{OD} .......................(2)

Step 6: Substitute the equation (2) in  (1)

              EXAMPLE:   \frac{AX}{XD} = \frac{AO}{CO} ...............(3)

Step 7: Verify the converse of the basic the proportionality theorem to the triangle ADC in the quadrilateral.

            NOTE: If a line divides two sides of a triangle in the same ratio, then the

                        line is parallel to the third side.

      EXAMPLE: From the equation (3) \frac{AX}{XD} = \frac{AO}{CO}

                          So In ∆ADC, XO is a line such that  ⇒ XO || DC

                                (by converse of the basic the proportionality theorem)


Step 8:  Prove that the quadrilateral ABCD is a trapezium

                 In ∆DAB, XO || AB

                In ∆ADC, XO || DC

          So, we can write ⇒ AB || DC In quadrilateral ABCD,

              In quadrilateral ABCD, AB || DC ⇒ ABCD is a trapezium (by definition)  

              Hence proved.