The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\displaystyle \frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{OD}}$. Show that quadrilateral ABCD is a trapezium.

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