- Trending Categories
- 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

# Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.

**To do:**

We have to prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.

**Solution:**

Let in a circle with centre $O, CD$ is the diameter and $AB$ is the chord

which is bisected by diameter at $E$.

Join $OA$ and $OB$.

In $\triangle OAE$ and $\triangle OBE$,

$OA = OB$ (Radii of the circle)

$OE = OE$ (Common)

$AE = EB$ (Given)

Therefore, by SSS axiom,

$\triangle OAE \cong \triangle OBE$

This implies,

$\angle AOE = \angle BOE$ (CPCT)

Therefore, the diameter bisects the angle subtended by the chord.

Hence proved.

Advertisements