Question

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is rhombus.

Answer 1

We have a quadrilateral $ABCD$ such that the diagonal's $AC$ and $BD$ bisect each other at right angles at $O$.

In $\Delta AOB$ & $\Delta AOD$ we have

$AO=OA$ (common)

$OB=OD$ (given $AC$ & $BD$ bisect at $O$)

$\angle AOB=\angle AOD$ (each is ${90}^0$)

$\therefore$ $\Delta AOB\cong \Delta AOD$ by the $SAS$ congruence criterion.

$\Rightarrow$ Their corresponding parts are equal $AB=AD⟶\left(1\right)$

Similarly $\Delta AOD$ and $\Delta OCD$ are congruent by a $SAS$ congruence

criterion $\Rightarrow AD=CD$ (corresponding parts) $⟶\left(2\right)$

Again, $\Delta COD$ and $\Delta COB$ are congruent by a similar $SAS$ congruence

criterion $\Rightarrow CD=CB$ (corresponding parts) $⟶\left(3\right)$

Also, $\Delta COB$ and $\Delta AOB$ are congruent by $SAS$ congruence criterion $\Rightarrow BC=AB$ (corresponding parts) $⟶\left(4\right)$

Combining $\left(1\right),\left(2\right),\left(3\right),\left(4\right)$ we have

$AB=BC=CD=DA$.

Thus quadrilateral $ABCD$ is a rhombus.