![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
In the figure, $AB \parallel CD$ and $P$ is any point shown in the figure. Prove that:
$\angle ABP + \angle BPD + \angle CDP = 360^o$
Given:
In the figure, $AB \parallel CD$ and $P$ is any point shown in the figure.
To do:
We have to prove that $\angle ABP + \angle BPD + \angle CDP = 360^o$.
Solution:
Through $P$, draw $PQ \parallel AB$ and $CD$
$AB \parallel PQ$
This implies,
$\angle ABP + \angle BPQ= 180^o$....……(i) (Sum of co-interior angles is $180^o$)
Similarly,
$CD \parallel PQ$
This implies,
$\angle QPD + \angle CDP = 180^o$......…(ii)
Adding equations (i) and (ii), we get,
$\angle ABP + \angle BPQ + \angle QPD + \angle CDP = 180^o+ 180^o$
$ = 360^o$
$\Rightarrow \angle ABP + \angle BPD + \angle CDP = 360^o$
Hence proved.