- 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
Diagonals $ \mathrm{AC} $ and $ \mathrm{BD} $ of a quadrilateral $ A B C D $ intersect at $ O $ in such a way that ar $ (\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC}) $. Prove that $ \mathrm{ABCD} $ is a trapezium.
Given:
Diagonals \( \mathrm{AC} \) and \( \mathrm{BD} \) of a quadrilateral \( A B C D \) intersect at \( O \) in such a way that ar \( (\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC}) \).
To do:
We have to prove that \( \mathrm{ABCD} \) is a trapezium.
Solution:
$ar(\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC})$
Adding $ar(\triangle AOB)$ on both sides, we get,
$ar(\triangle AOD)+ar(\triangle AOB)=ar(\triangle BOC)+ar(\triangle AOB)$
$ar(\triangle ADB)=ar(\triangle ACB)$
This implies,
$ar(\triangle ADB)$ and $ar(\triangle ACB)$ lie between the same parallel lines.
Therefore,
$AB \| DC$
This implies,
$ABCD$ is a trapezium.
Hence proved.
Advertisements