Question

Prove: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.

Short answer

A parallelogram is a rhombus if the diagonals of the parallelogram are perpendicular.

Step-by-step solution

Step 1. Consider the diagram.

The figure is made having diagonals AC and BD .

Here, ABCD is a parallelogram.

Step 2. State the proof.

In $\Delta AOD$and $\Delta AOB$.

$OA=OA$ (Common)

$\angle AOD\cong \angle AOB={90}^{\circ }$ (Diagonals are perpendicular)

$OD=OB$ (Diagonals of parallelogram bisects)

So, $\Delta AOD\cong \Delta AOB$ by SAS postulate.

Therefore, by CPCT $\overrightarrow{AD}\cong \overrightarrow{AB}$.

Opposite sides are congruent in parallelogram.

$\overrightarrow{AB}\cong \overrightarrow{DC}$and $\overrightarrow{AD}\cong \overrightarrow{BC}$. (As $\overrightarrow{AB}\cong \overrightarrow{AD}$)

That implies, $\overrightarrow{AD}\cong \overrightarrow{DC}\cong \overrightarrow{AB}\cong \overrightarrow{BC}$.

Therefore, parallelogram $ABCD$ is a rhombus.

Step 3. State the conclusion.

Therefore, parallelogram ABCD is a rhombus if the diagonals of the parallelogram are perpendicular (proved).