Question

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

Answer 1

A rhombus is a parallelogram whose all sides are equal and diagonals bisect each other at ${90}^0$.

Let $\mathrm{ABCD}$ is a $\left|\right|\mathrm{gm}$ whose diagonals bisect each other at right angle.

To prove :

$\mathrm{ABCD}$ is rhombus.

Proof:

Step 1: Observe the equal sides of the parallelogram $\mathrm{ABCD}$.

$\mathrm{ABCD}$ is a $\left|\right|\mathrm{gm}$ so it has a pair of opposite sides that are equal i.e. $\mathrm{AB}=\mathrm{CD}$ and $\mathrm{AD}=\mathrm{BC}$

Step 2: Observe the congruency in $∆\mathrm{AOD}$ and $∆\mathrm{COD}$

In $∆\mathrm{AOD}$ and $∆\mathrm{COD}$, we have

$\mathrm{AO}=\mathrm{CO}$ (diagonals of $\left|\right|\mathrm{gm}$ bisect each other)

$\angle \mathrm{AOD}=\angle \mathrm{COD}={90}^0$ (given diagonals bisect each other at right angle)

$\mathrm{OD}=\mathrm{OD}$ (common sides)

⇒$∆\mathrm{AOD}$$\cong$$∆\mathrm{COD}$ by $\mathrm{SAS}$ (Side - Angle - Side) criteria

So,$\mathrm{AD}=\mathrm{CD}$ by $\mathrm{CPCT}$ (corresponding parts of a congruent triangle)

Step 3: Use the result of the step $2$ to relate together all the sides of the $\left|\right|\mathrm{gm}$

In $\left|\right|\mathrm{gm}$$\mathrm{ABCD}$, $\mathrm{AB}=\mathrm{CD}$ , $\mathrm{AD}=\mathrm{BC}$ and from step 2, $\mathrm{AD}=\mathrm{CD}$

Based on the above result, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ i.e. all the sides of the $\left|\right|\mathrm{gm}$$\mathrm{ABCD}$ are equal and its diagonals bisect each other at right angle.

Thus, $\mathrm{ABCD}$ is a rhombus.

Hence, proved.