- 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
$D$ and $E$ are the points on the sides $AB$ and $AC$ respectively of a $\triangle ABC$ such that: $AD = 8\ cm, DB = 12\ cm, AE = 6\ cm$ and $CE = 9\ cm$. Prove that $BC = \frac{5}{2}DE$.
Given:
$D$ and $E$ are the points on the sides $AB$ and $AC$ respectively of a $\triangle ABC$ such that: $AD = 8\ cm, DB = 12\ cm, AE = 6\ cm$ and $CE = 9\ cm$.
To do:
We have to prove that $BC = \frac{5}{2}DE$.
Solution:
In $\triangle ADE$ and $\triangle ABC$,
$\frac{AD}{DB}=\frac{8}{12}=\frac{2}{3}$
$\frac{AE}{EC}=\frac{6}{9}=\frac{2}{3}$
$\frac{AD}{DB}=\frac{AE}{EC}$
By converse of basic proportionality theorem,
$DE \parallel BC$
In $\triangle ADE$ and $\triangle ABC$,
$\angle A=\angle A$
$\angle ADE=\angle ABC$
Therefore,
$\triangle ADE \sim\ \triangle ABC$ (By AA similarity)
$\frac{AD}{AB}=\frac{DE}{BC}$ (CPCT)
$\frac{8}{8+12}=\frac{DE}{BC}$
$\frac{8}{20}=\frac{DE}{BC}$
$BC=\frac{5}{2}DE$
Hence proved.
Advertisements