The diagonals of a quadrilateral ABCD intersect each other at the poin

1.	The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{AO}{BO} = \frac{CO}{DO}\). Show that ABCD is a trapezium.
Answer» Data: In the quadrilateral ABCD. the diagonals intersect at 0’ such that \(\frac{AO}{BO} = \frac{CO}{DO}\) To Prove: ABCD is a trapezium. In the qudrilateral ABCD, \(\frac{AO}{BO} = \frac{CO}{DO}\) ∴ \(\frac{AO}{Co} = \frac{OB}{OD}\) It means sides of ∆AOB and ∆DOC divides proportionately. ∴ ∆AOB \|\|\| ∆DOC. Similarly. ∆AOD \|\|\| ∆BOC. Now, ∆AOB + ∆AOD = ∆BOC + ∆DOC ∆ABD = ∆ABC. Both triangles are on the same base AB and between two pair of lines and equal in area. ∴ AB \|\| DC.

The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{AO}{BO} = \frac{CO}{DO}\). Show that ABCD is a trapezium.

Answer»

Data: In the quadrilateral ABCD. the diagonals intersect at 0’ such that \(\frac{AO}{BO} = \frac{CO}{DO}\)

To Prove: ABCD is a trapezium.

In the qudrilateral ABCD, \(\frac{AO}{BO} = \frac{CO}{DO}\)

∴ \(\frac{AO}{Co} = \frac{OB}{OD}\)

It means sides of ∆AOB and ∆DOC divides proportionately.

∴ ∆AOB ||| ∆DOC.

Similarly. ∆AOD ||| ∆BOC.

Now, ∆AOB + ∆AOD = ∆BOC + ∆DOC

∆ABD = ∆ABC.

Both triangles are on the same base AB and between two pair of lines and equal in area.

∴ AB || DC.