Question

Show that if the diagonals of a quadrilateral bisect each other at right angle is a rhombus.

Answer 1

In $△AOB$ and $△AOD$

$AO=AO$(common)

$OB=OD$(Given)

$\angle AOB=\angle AOD={90}^{\circ }$

$\therefore △AOB\cong △AOD$ by SSS congruence rule

$\therefore AB=AD$ .....$\left(1\right)$ by CPCT

In $△AOB$ and $△BOC$

$BO=BO$(common)

$AO=OC$(Given)

$\angle AOB=\angle BOC={90}^{\circ }$

$\therefore △AOB\cong △BOC$ by SSS congruence rule

$\therefore AB=BC$ .....$\left(2\right)$ by CPCT

In $△BOC$ and $△COD$

$CO=CO$(common)

$BO=OD$(Given)

$\angle BOC=\angle COD={90}^{\circ }$

$\therefore △BOC\cong △COD$ by SSS congruence rule

$\therefore BC=CD$ .....$\left(3\right)$ by CPCT

In $△COD$ and $△AOD$

$DO=DO$(common)

$AO=OC$(Given)

$\angle COD=\angle DOA={90}^{\circ }$

$\therefore △COD\cong △AOD$ by SSS congruence rule

$\therefore CD=AD$ .....$\left(4\right)$ by CPCT

From $\left(1\right)$,$\left(2\right)$,$\left(3\right)$ and $\left(4\right)$ we get

$AB=BC=CD=DA$

$\therefore ABCD$ is a rhombus.