Question

State true or false:

In quadrilateral ABCD, $AD=BC$ and the diagonals $AC$ and $BD$ intersect at point $O$, such that

$\frac{OC}{OA}=\frac{OD}{OB}=\frac{1}{3}$, then the quadrilateral $ABCD$ is a trapezium.

• A. True
• B. False

Answer 1

The correct option is A True

Through $O$, draw line $EO$, where $EO||AB$, which meets $AD$ at $E$.

In $△DAB$, we have

$EO||AB$

$\frac{DE}{EA}=\frac{DO}{OB}$ ...(i) [By using Basic Proportionality Theorem]

Also, $\frac{AO}{BO}=\frac{CO}{DO}$ (Given)

$⟹\frac{AO}{CO}=\frac{BO}{DO}$

$⟹\frac{DO}{BO}=\frac{CO}{AO}$...(ii)

From equation (i) and (ii), we get

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

Therefore, By using converse of Basic Proportionality Theorem, $EO||DC$, also, $EO||AB$

$⟹AB||DC$

Hence, quadrilateral $ABCD$ is a trapezium with $AB||CD$