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

Let ABCD be the quadrilateral where AC and BD intersect each other at O such that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$.

From point O,

Draw a line EO touching AD at E in such a way that, EO || AB.

We know that,

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

In △ DAB,

EO || AB

$\therefore \dfrac{DE}{EA} = \dfrac{DO}{OB}$ ……….(1)

Given,

$\Rightarrow \dfrac{AO}{BO} = \dfrac{CO}{DO} \\[1em] \Rightarrow \dfrac{AO}{CO} = \dfrac{BO}{DO} \\[1em] \Rightarrow \dfrac{CO}{AO} = \dfrac{DO}{BO} \text{ ………(2)}$

From (1) and (2), we get :

$\Rightarrow \dfrac{DE}{EA} = \dfrac{CO}{AO}$

We know that,

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

∴ EO || DC.

Since, EO || AB and EO || DC.

∴ AB || DC.

Hence, proved that ABCD is a trapezium.