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.
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.
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.