![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
$ABCD$ is a rhombus, $EABF$ is a straight line such that $EA = AB = BF$. Prove that $ED$ and $FC$ when produced meet at right angles.
Given:
$ABCD$ is a rhombus, $EABF$ is a straight line such that $EA = AB = BF$.
To do:
We have to prove that $ED$ and $FC$ when produced meet at right angles.
Solution:
Let $ED$ and $FC$ when joined meet at $G$ on producing.
We know that,
Diagonals of a rhombus bisect each other at right angles.
This implies,
$\angle AOD = \angle COD =\angle AOB =\angle BOC = 90^o$
$AO = OC, BO = OD$
In $\triangle BDE$,
$A$ and $O$ are the mid-points of $BE$ and $BD$ respectively.
This implies,
$AO \parallel ED$
Similarly,
$OC \parallel DG$
In $\triangle CFA, B$ and $O$ are the mid-points of $AF$ and $AC$ respectively.
Therefore,
$OB \parallel CF$ and $OD \parallel GC$
In quadrilateral $DOCG$,
$OC \parallel DG$ and $OD \parallel CG$
This implies,
$DOCG$ is a parallelogram.
Therefore,
$\angle DGC = \angle DOC$ (Opposite angles of a parallelogram are equal)
This implies,
$\angle DGC = 90^o$
Hence proved.