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.

Given:

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

To do:

We have to show that $ABCD$ is a trapezium.

Solution:

Draw $EF \| AB$

$\frac{AO}{BO}=\frac{OC}{OD}$

This implies,

$\frac{AO}{CO}=\frac{BO}{DO}$........(i)

In $\triangle BAD$, $EO \| AB$

This implies, by B.P.T.,

$\frac{DE}{EA}=\frac{DO}{BO}$

$\frac{AE}{ED}=\frac{BO}{DO}$...........(ii)

From (i) and (ii), we get,

$\frac{AO}{CO}=\frac{AE}{ED}$

This implies, by converse of B.P.T.,

$OE \| CD$

$AB \| OE$

This implies,

$AB \| CD$

Therefore,

$ABCD$ is a trapezium

Hence proved.