![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 a $\triangle ABC$, if $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$.
Prove that $ar(\triangle ABM) = ar(\triangle ACL)$.
Given:
In a $\triangle ABC$, $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$.
To do:
We have to prove that $ar(\triangle ABM) = ar(\triangle ACL)$.
Solution:
Join $LM, LC$ and $MB$.
$L$ and $M$ are the mid points of $AB$ and $AC$.
This implies,
$LM \| BC$
$\triangle LBM$ and $\triangle LCM$ are on the same base $LM$ and between the same parallels.
Therefore,
$ar(\triangle LBM) = ar(\triangle LCM)$......…(i)
$ar(\triangle LCM) = ar(\triangle LBM)$
$\triangle LBC$ and $\triangle MBC$ are on the same base $BC$ and between the same parallels.
Therefore,
$ar(\triangle LBC) = ar(\triangle MBC)$......…(ii)
$ar(\triangle LMB) = ar(\triangle LMC)$ [From (i)]
$ar(\triangle ALM) + ar(\triangle LMB) = ar(\triangle ALM) + ar(\triangle LMC)$ [Adding $ar(\triangle ALM)$ on both sides]
$ar(\triangle ABM) = ar(\triangle ACL)$
Hence proved.