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
In figure, $XY$ and $X’Y’$ are two parallel tangents to a circle, $x$ with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X’Y’$ at $B$. Prove that $∠AOB = 90^o$.
"
Given:
$XY$ and $X’Y’$ are two parallel tangents to a circle, $x$ with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X’Y’$ at $B$.
To do:
We have to prove that $\angle AOB=90^{o}$.
Solution:
$XY$ and $X'Y'$ are two parallel tangents to the circle with centre $O$ touching the circle at $P$ and $Q$ respectively.
$AB$ is a tangent at the point $C$, which intersects $XY$ at $A$ and $X'Y'$ at $B$.
Follow the steps:
Join $OC$.
In $\vartriangle OAP$ and $\vartriangle OAC$,
$OP=OC\ \ \ \ \ \ \ \ \ ( Radii\ of\ the\ same\ circle)$
$AP =AC\ \ \ \ \ \ \ \ ( Length\ of\ tangents\ drawn\ from\ an\ external\ point\ to\ a\ circle\ are\ equal)$
$OA=OA\ \ \ \ \ \ \ \ \ \ ( Common\ side)$
$\vartriangle OAP\cong \vartriangle OAC\ \ \ \ \ \ \ \ \ \ \ \ ( SSS\ congruence\ criterion)$
$\angle AOP=\angle AOC\ \ .................( 1)$
Similarly, $\vartriangle OBC\cong \vartriangle OBQ$
$\angle BOQ=\angle BOC\ \ \ \ .....................( 2)$
Now, AOB is a diameter of the circle.
Hence, it is a straight line.
$\angle AOP\ +\angle AOC+\angle BOQ+\angle BOC\ =\ 180^{o}$
From $( 1)$ and $( 2)$,
We have: $2\angle AOC+2\angle BOC\ =\ 180^{o}$
$\angle AOC+\angle BOC\ =\frac{180^{o} }{2}$
$=90^{o}$
We know,
$\angle AOC+\angle BOC=\angle AOB=90^{o}$
$\Rightarrow \angle AOB=90^{o}$
Hence proved.
23 Views
