# A Parallelogram whose Diagonals Intersect at Right Angles is a Rhombus Theorem & Proof

Are you looking for the theorems of properties of parallelograms on various websites? If our guess is correct then you are on the right page here you can find the step-by-step explanation for all the theorems on parallelogram here. A quadrilateral in which all the four sides are equal is a rhombus. If the diagonals of a parallelogram are equal and intersect at the right angle then it is known as a rhombus. Let us prove that A Parallelogram whose Diagonals Intersect at Right Angles is a Rhombus in this article.

## A Parallelogram whose Diagonals Intersect at Right Angles is a Rhombus

Theorem: A Parallelogram whose Diagonals Intersect at Right Angles is a Rhombus.

Given:
The parallelogram ABCD with diagonals AC and BD as shown in the above figure.
The diagonals AC and BD meet at the point “O” at right angles.
To Prove that:
A parallelogram is a rhombus
Proof:
Consider ΔAOD and ΔCOD
AO = OC [diagonals AC and BD bisect each other at O]
∠AOD = ∠COD [both 90° because given that diagonals intersect at right angles]
ΔAOD ≅ ΔCOD [SAS theorem]
Therefore, we can say that the parallelogram ABCD is a rhombus.
Hence Proved

### FAQs on Parallelogram whose Diagonals Intersect at Right Angles is a Rhombus

1. Do diagonals intersect in a rhombus?

The intersection of the diagonals of a rhombus form 90 degrees. This means that they are perpendicular. The diagonals of a rhombus bisect each other.

2. Which quadrilateral has diagonals that intersect at right angles?

A quadrilateral whose diagonals bisect each other at right angles is a rhombus. each other at right angles at M. So AB = AD and by the first test above ABCD is a rhombus. If the diagonals of a parallelogram are perpendicular, then it is a rhombus.

3. Is a parallelogram a rhombus?

Not every parallelogram is a rhombus, though any parallelogram with perpendicular diagonals (the second property) is a rhombus.