Question

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

Answer 1

Step 1. Explaining the diagram.

Let $ABCD$ be quadrilateral where $AC$ and $BD$ intersects each other at $O$ such that, $\frac{AO}{BO}=\frac{CO}{DO}$

Step 2. Showing $ABCD$ is trapezium

Construction-From the point $O$, draw a line $EO$ touching $AD$ at $E$ in such a way that, $EO\parallel DC\parallel AB$

$In\Delta DAB,EO\left|\right|AB$

By using Basic Proportionality Theorem

$\frac{DE}{EA}=\frac{DO}{OB}........................\left(i\right)$

Also, given,

$\frac{AO}{BO}=\frac{CO}{DO}$

$$\begin{gathered} \Rightarrow \frac{AO}{CO}=\frac{BO}{DO}[applyingalternendo] \\ \Rightarrow \frac{CO}{AO}=\frac{DO}{OB}[[applyinginvertendo] \\ \Rightarrow \frac{DO}{OB}=\frac{CO}{AO}..........................(ii) \end{gathered}$$

From equation $\left(i\right)and\left(ii\right),$

We have

$\frac{DE}{EA}=\frac{CO}{AO}$

Therefore, By applying converse of Basic Proportionality Theorem,

$EO\left|\right|DC$

Also $EO\left|\right|AB\Rightarrow AB\left|\right|DC.$

Hence, quadrilateral $ABCD$ is a trapezium with $AB\left|\right|CD.$