# The diagonals of a quadrilateral ABCD intersect

Question.

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

Solution:

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

$\Rightarrow \frac{A O}{O C}=\frac{B O}{O D} \quad \ldots$ (1) (given)

Through $\mathrm{O}$, we draw

$\mathrm{OE} \| \mathrm{BA}$

OE meets AD at E.

From $\Delta \mathrm{DAB}$,

$\mathrm{EO} \| \mathrm{AB}$

$\Rightarrow \frac{D E}{E A}=\frac{D O}{O B}$ (by Basic Proportionality Theorem)

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

From (1) and (2),

$\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{AE}}{\mathrm{ED}} \Rightarrow \mathrm{OE} \| \mathrm{CD}$

(by converse of basic proportionality theorem)

Now, we have BA $\| \mathrm{OE}$

and $\quad \mathrm{OE} \| \mathrm{CD}$

$\Rightarrow \quad \mathrm{AB} \| \mathrm{CD}$

$\Rightarrow$ Quadrilateral $\mathrm{ABCD}$ is a trapezium.