ABCD is a trapezium in which AB


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


We draw EOF $\| \mathrm{AB}($ also $\| \mathrm{CD})$ (see figure)

In $\Delta \mathrm{ACD}, \quad \mathrm{OE} \| \mathrm{CD}$

$\Rightarrow \frac{A E}{E D}=\frac{A O}{O C} \ldots(1)$

In $\triangle \mathrm{ABD}, \mathrm{OE} \| \mathrm{BA}$

$\Rightarrow \frac{D E}{F A}=\frac{D O}{O B}$

$\Rightarrow \frac{A E}{E D}=\frac{O B}{O D} \ldots(2)$

From (1) and (2)

$\frac{A O}{O C}=\frac{O B}{O D}$

i.e., $\frac{A O}{B O}=\frac{C O}{D O}$.

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