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, Quadrilateral ABCD where AC and BD intersects each other at O such that,
\(\frac{AO}{BO} = \frac{CO}{DO}\).

We have to prove here, ABCD is a trapezium

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

In \(\triangle DAB\), EO || AB

Therefore, By using Basic Proportionality Theorem
\(\frac{DE}{EA} = \frac{DO}{OB}\) ……………………(i)

Also, given,
\(\frac{AO}{BO} = \frac{CO}{DO}\)
=> \(\frac{AO}{CO} = \frac{BO}{DO}\)
=> \(\frac{CO}{AO} = \frac{DO}{BO}\)
=> \(\frac{DO}{OB} = \frac{CO}{AO}\) …………………………..(ii)

From equation (i) and (ii), we get
\(\frac{DE}{EA} = \frac{CO}{AO}\)

Therefore, By using converse of Basic Proportionality Theorem,
EO || DC also EO || AB
=> AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.