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

Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.

We have to prove, \(\frac{AO}{BO} = \frac{CO}{DO}\)

From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB

In \(\triangle ADC\), we have OE || DC

Therefore, By using Basic Proportionality Theorem
\(\frac{AE}{ED} = \frac{AO}{CO}\) ……………..(i)

Now, In \(\triangle ABD\), OE || AB

Therefore, By using Basic Proportionality Theorem
\(\frac{DE}{EA} = \frac{DO}{BO}\)…………….(ii)

From equation (i) and (ii), we get,
\(\frac{AO}{CO} = \frac{BO}{DO}\)
=>\(\frac{AO}{BO} = \frac{CO}{DO}\)

Hence, proved.