- 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 is a point on the side BC of a triangle ABC, such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB.CD$.
Given:
D is a point on the side BC of a triangle ABC, such that $\angle ADC = \angle BAC$.
To do:
We have to show that $CA^2 = CB.CD$.
Solution:
In $\triangle ABC$ and $\triangle DAC$,
$\angle C=\angle C$ (common)
$\angle BAC=\angle ADC$
Therefore, by AA criterion,
$\triangle \mathrm{ABC} \sim \triangle \mathrm{DAC}$
This implies,
$\frac{\mathrm{CA}}{\mathrm{CD}}=\frac{\mathrm{CB}}{\mathrm{CA}}$
This implies,
$CA^2=CB \times CD$
Hence proved.
Advertisements
To Continue Learning Please Login
Login with Google