ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$.

Given,

ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect each other at O.

We know that,

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

Through O, draw EO || DC || AB.

In △ ADC,

OE || DC

$\therefore \dfrac{AE}{ED} = \dfrac{AO}{CO}$ ……(1)

In △ ABD,

OE || AB

$\therefore \dfrac{AE}{ED} = \dfrac{BO}{DO}$ ……(2)

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

$\Rightarrow \dfrac{AO}{CO} = \dfrac{BO}{DO} \\[1em] \Rightarrow \dfrac{AO}{BO} = \dfrac{CO}{DO}.$

Hence, proved that $\dfrac{AO}{BO} = \dfrac{CO}{DO}.$