The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/OC = BO/OD . Show that…

Sol. Correct figure (1/2 Mark)
$\triangle AOB \sim \triangle COD$ (1 Mark)
$\therefore \angle OAB = \angle OCD$
As alternate angles are equal, so DC $\|\|$ AB (1/2 Mark)
Therefore, ABCD is a trapezium.
Alternate solution:
Correct figure. (1/2 Mark)
Draw EO $\|\|$ AB to intersect AD at E.
In $\triangle ABD$, EO $\|\|$ AB
$\therefore \frac{AE}{ED} = \frac{BO}{OD}$ (1/2 Mark)
Given, $\frac{AO}{OC} = \frac{BO}{OD}$
So, $\frac{AE}{ED} = \frac{AO}{OC}$ (1/2 Mark)
$\therefore EO\|\| DC$
So, DC $\|\|$ AB (1/2 Mark)
Therefore, ABCD is a trapezium.