Question

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

• A. True
• B. False

Answer 1

The correct option is A True

$ABCD$ is a quadrilateral in which $AC$ and $BD$ are diagonals, which intersect each other at $O$.

Draw a line $OM\parallel AB.$

$\therefore$ By using basic proportionality theorem, we have

$\Rightarrow$ $\frac{DM}{MA}=\frac{DO}{OB}$

Taking reciprocals on both sides,

$\Rightarrow$ $\frac{AM}{DM}=\frac{OB}{OD}$ ----- ( 1 )

It is given that,

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

$\Rightarrow$ $\frac{AO}{CO}=\frac{OB}{OD}$ ---- ( 2 )

Comparing ( 1 ) and ( 2 ), we get

$\Rightarrow$ $\frac{AM}{DM}=\frac{OA}{OC}$

$\therefore$ By using converse of basic proportionality theorem, we have

$OM\parallel DC$

But $OM\parallel AB$ [ by construction ]

$\Rightarrow$ $AB\parallel DC$

$\Rightarrow$ One pair of opposite sides of quadrilateral is parallel.

$\therefore$ $ABCD$ is a trapezium.