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
If two chords of a circle are equally inclined to the diameter through their point of intersection, prove the chords are equal.
Given:
Two chords of a circle are equally inclined to the diameter through their point of intersection.
To do:
We have to prove that the chords are equal.
Solution:
$AB$ and $AC$ are two chords which are equally inclined to the diameter $AD$ of the circle with centre $O$.
Draw $OL \perp AB$ and $OM \perp AC$.
In $\triangle OLA$ and $\triangle OMA$,
$\angle OLA=\angle OMA=90^o$
$OA=OA$ (Common side)
$\angle LAO=\angle MAO$ (Given)
Therefore,
\( \triangle \mathrm{OLA} \cong \triangle \mathrm{OMA} \) (By AAS congruence)
\( \Rightarrow \mathrm{OL}=\mathrm{OM} \) (CPCT)
This implies, the chords are equidstant from the centre $O$.
\( \Rightarrow A B=A C \) (Equal chords of a circle are equidistant from the centre of the circle)
Hence proved.
95 Views
