Question

Prove that if the diagonals of parallelograms are perpendicular, then the parallelogram is a rhombus.

Short answer

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

Step-by-step solution

Step 1. Check the figure.

Consider the figure.

Step 2. Step description.

The CPCTC theorem states that when two triangles are congruent, then every corresponding part of one triangle is congruent to the other.

Step 3. Step description.

From the figure, $\Delta AOD$and $\Delta AOB$.

$OA=OA$ (Sides are common)

As the diagonals are perpendicular such that $\angle AOB\cong \angle AOD={90}^{\circ }$and the diagonals of a parallelogram bisect such that $OB=OD$.

Therefore, by the side-angle-side postulate $\Delta BOA\cong \Delta BOA$.

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

As the opposite sides are congruent in parallelogram thus

$\begin{array}{l}\overrightarrow{AB}\cong \overrightarrow{DC}\text{\&}\overrightarrow{AD}\cong \overrightarrow{BC}\\ \therefore \left(\overrightarrow{AB}\cong \overrightarrow{AD}\right)\\ \Rightarrow \overrightarrow{AD}\cong \overrightarrow{DC}\cong \overrightarrow{AB}\cong \overrightarrow{BC}\end{array}$

Therefore, the parallelogram is a rhombus.