Diagonals AC and BD of a trapezium ABCD intersect at O , where $\displaystyle \mathrm{AB} \| \mathrm{DC}$. If $\displaystyle \frac{\mathrm{DO}}{\mathrm{OB}}=\frac{1}{2}$, then show that $\displaystyle \mathrm{AB}=2 \mathrm{CD}$

\(\displaystyle \Delta \mathrm{OAB} \sim \Delta \mathrm{OCD}\) \[\begin{aligned} & \Rightarrow \frac{O D}{O B}=\frac{C D}{A B} \\ & \because \frac{O D}{O B}=\frac{1}{2} \end{aligned} \] Therefore \(\displaystyle \frac{\mathrm{CD}}{\mathrm{AB}}=\frac{1}{2} \Rightarrow \mathrm{AB}=2 \mathrm{CD}\)