![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
Two tangent segments \( P A \) and \( P B \) are drawn to a circle with centre \( O \) such that \( \angle A P B=120^{\circ} . \) Prove that \( O P=2 A P \).
Given:
Two tangent segments \( P A \) and \( P B \) are drawn to a circle with centre \( O \) such that \( \angle A P B=120^{\circ} . \)
To do:
We have to prove that \( O P=2 A P \).
Solution:
Join $OP$.
Take mid point of $OP$ as $M$ and join $AM$. Join $OA$ and $OB$.
In right angled triangle $OAP$,
$\angle OPA = \frac{1}{2} \angle APB = \frac{1}{2}(120^o) = 60^o$
$\angle AOP = 90^o - 60^o = 30^o$
$M$ is the mid point of hypotenuse $OP$ of $\triangle OAP$
This implies,
$MO = MA = MP$
$\angle OAM = \angle AOM = 30^o$
$\angle PAM = 90^o – 30^o = 60^o$
$\triangle AMP$ is an equilateral triangle.
$MA = MP = AP$
Also, $M$ is the mid point of $OP$.
$OP = 2 MP = 2 AP$
Hence proved.