![Trending Articles on Technical and Non Technical topics](/images/trending_categories.jpeg)
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Given :
Diagonals of the quadrilateral bisect each other at right angles.
To do :
We have to show that it is a rhombus.
Solution:
Let $ABCD$ be a quadrilateral in which diagonals bisect each other at right angles.
So,
$OA=OC, OB=OD$
$\angle AOB = \angle BOC =\angle COD =\angle AOD = 90^o$
To prove that it is a rhombus, we need to prove that $ABCD$ is a parallelogram and $AB = BC = CD = AD$
In $\triangle AOB$ and $\triangle BOC$,
$OA=OC$ (Given)
$OB=OB$ (Common)
$\angle AOB= \angle BOC$ ($90^o$)
Therefore, by SAS congruency, we get,
$\triangle AOB \cong \triangle BOC$
So, $AB=BC$ (CPCT)
Similarly,
$\triangle AOB \cong \triangle AOD$
So, $AB=AD$
$\triangle COD \cong \triangle BOC$
So, $CD=BC$
Therefore,
$AB=BC=CD=AD$
We can say that,
$AB=CD$ and $BC=AD$
As the opposite sides are equal, $ABCD$ is a parallelogram.
We know that a parallelogram whose diagonals intersect at right angles is a rhombus.
Therefore, $ABCD$ is a parallelogram with all sides equal and diagonals bisect each other at right angles.
So, $ABCD$ is a rhombus.
Hence proved.