Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Vibuhi Khajuria 7 years, 9 months ago
- 1 answers
Posted by Jatin Gupta 7 years, 9 months ago
- 1 answers
Posted by Jatin Gupta 7 years, 9 months ago
- 0 answers
Posted by Abhishek Kumar 7 years, 9 months ago
- 2 answers
Posted by Manshu Singh 7 years, 9 months ago
- 1 answers
Posted by Vivek Singh Jadon 7 years, 9 months ago
- 0 answers
Posted by Devender Kumar 7 years, 9 months ago
- 2 answers
Posted by Shakir Chaudhary 7 years, 9 months ago
- 1 answers
Posted by Ankit Thakur 7 years, 9 months ago
- 5 answers
Posted by Ankit Sahi 7 years, 9 months ago
- 3 answers
Nidhi Patel 7 years, 9 months ago
Shivam Sharma 7 years, 9 months ago
Posted by Nidhi Patel 7 years, 9 months ago
- 3 answers
Posted by Naman Ratnani 7 years, 9 months ago
- 0 answers
Posted by H T 7 years, 9 months ago
- 5 answers
Posted by Harshita Vatyani 5 years, 8 months ago
- 5 answers
Posted by Aniket Verma 7 years, 9 months ago
- 2 answers
Posted by Yuvraj Singh 7 years, 9 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Given : In {tex}\triangle A B C{/tex}, DE || BC and intersects AB in D and AC in E.
Prove that : {tex}\frac{AD}{DB} = \frac{AE}{EC}{/tex}
Construction: Join BC, CD and draw EF {tex}\perp{/tex} BA and DG {tex}\perp{/tex} CA.
Now from the given figure we have,
EF {tex}\perp{/tex} BA (Construction)
EF is the height of ∆ADE and ∆DBE (Definition of perpendicular)
Area({tex}\triangle{/tex}ADE) ={tex}\frac{AD.EF}{2}{/tex} .....(1)
Area({tex}\triangle{/tex}DBE) = {tex}\frac{DB.EF}{2}{/tex} ....(2)
Divide the two equations we have
{tex}\frac{Area \triangle ADE}{Area \triangle DBE} = \frac{AD}{DB}{/tex} .....(3)
{tex}\frac{Area \triangle ADE}{Area \triangle DEC} = \frac{AE}{EC}{/tex} .....(4)
Therefore, {tex}\triangle \mathrm{DBE} \sim \triangle \mathrm{DEC}{/tex} (Both the ∆s are on the same base and between the same || lines).....(5)
Area({tex}\triangle{/tex}DBE) = Area({tex}\triangle{/tex}DEC) (If the two triangles are similar their areas are equal)
{tex}\frac{AD}{DB} = \frac{AE}{EC}{/tex} [from equation 3,4 and 5]
Hence proved.
Posted by Pawan Kumar Nanda 7 years, 9 months ago
- 0 answers
Posted by Dhanuja Shri 7 years, 9 months ago
- 5 answers
Posted by Lucky Charan 7 years, 9 months ago
- 7 answers
Posted by Shivansh Srivastav 7 years, 9 months ago
- 3 answers
Posted by Shivansh Srivastav 7 years, 9 months ago
- 0 answers
Posted by Shivansh Srivastav 7 years, 9 months ago
- 1 answers
Posted by Muskan Khan 7 years, 9 months ago
- 1 answers
Posted by Lucky Charan 7 years, 9 months ago
- 2 answers
Posted by Ankit Kumar 7 years, 9 months ago
- 1 answers
Posted by Satyam Yadav 7 years, 9 months ago
- 1 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Zehra Mustafa 7 years, 9 months ago
1Thank You