Question

Prove that a quadrilateral is rhombus if and only if diagonals bisect each other at right angle.

Answer 1

Let $ABCD$ be a quadrilateral, where diagonals bisect each other

$\therefore$ $OA=OC$ ---- ( 1 )

And $OB=OD$ ---- ( 2 )

And they bisect at right angles

$\therefore$ $\angle AOB=\angle BOC=\angle COD=\angle AOD={90}^o$ ------ ( 3 )

In $△AOD$ and $△COD,$

$OA=OC$ [ From ( 1 ) ]

$\angle AOD=\angle COD$ [ From ( 3 ) ]

$OD=OD$ [ Common side ]

$\therefore$ $△AOD\cong △COD$ [ $SAS$ congruence rule ]

$\Rightarrow$ $AD=CD$ ---- ( 4 ) [ CPCT ]

Similarly, we can prove that

$AD=AB$ and $AB=BC$ ----- ( 5 )

From ( 4 ) and ( 5 )

$\Rightarrow$ $AD=CD=AB=BC$

In quadrilateral $ABCD$,

$AB=CD$ and $AD=BC$

Both pairs of opposite sides are equal

$\therefore$ $ABCD$ is a parallelogram.

Also, $AB=CD=AD=BC$

All sides of parallelogram $ABCD$ are equal

$\therefore$ $ABCD$ is a rhombus.